At such low concentrations, can’t you just assume an ideal solution? Then the  change in entropy from mixing would be $\Delta S = -R(x_A\ln x_A + x_B \ln x_B)$. But because the solution is so dilute, you can see that both terms are going to be essentially zero. (For a 10 nM solution, the contribution from entropy of mixing is on the order of $10^{-8}$ J K$^{-1}$ mol$^{-1}$.)

The dilution process would thus lower the total entropy, since there’s less uncertainty in the system. (You’d get the value you compute at 10 nM, the final state, minus the value at 10 µM, the initial state.)


Generally speaking, there is no protection unless the sacrificial anode is immersed in the electrolyte. It might be better to anodize and seal the aluminum, which should protect it against most acids. There are many aluminum anti-corrosion treatments, but in any case the area under attack would have to be covered. See http://en.wikipedia.org/wiki/Anodizing.


You've described the classic preparation of sodium hypochlorite, better known as "bleach".

If you electrolyze sodium chloride solution, you form hydrogen at one electrode and chlorine at the other, and sodium hydroxide in solution. But chlorine tends to dissolve in alkaline solutions to form chloride and hypochlorite ions. So you get "sodium hypochlorite" solution, which is actually a complex equilibrium between sodium chloride, sodium hydroxide, sodium hypochlorite, free chlorine, and (of course) water.


Number of mercury atoms in a drop of mercury

You can use dimensional analysis. It is a very useful tool to solve equations such as these.

What you have is the volume of one drop, $V(ml)$, the density of mercury, $\rho(kg\cdot{}m^{-3})$, the molar mass of mercury, $Mr(g\cdot{}mol^{-1})$, and the Avogadro's constant, $N_A(mol^{-1}$).

To find out the number of atoms in a drop of mercury, all you have to do is to manipulate those variables via dimensional analysis.

You have: $\frac{Mr(g)}{(mol)},\frac{N_A(atoms)}{(mol^{})},\frac{\rho{(kg)}}{(m^3)},\frac{V(ml)}{drop}$ and you know that $\frac{1(kg)}{1000(g)},\frac{10^6(ml)}{(m^3)}$

Now, try arranging them to form an equation such that the units cancel off,in order to get the answer you need.

A: Mass of mercury:

$$\frac{kg}{atom}=\frac{1(kg)}{1000(g)}\cdot{}
\frac{Mr(g)}{(mol)}\cdot{}\frac{(mol^{})}{N_A(atom)}
$$

Notice that this is dimensionally equivalent. Therefore,

$$\frac{kg}{atom}=\frac{Mr}{1000\cdot{}N_A}
$$

B: Atoms of mercury per drop:

$$\frac{atoms}{drop}=
\frac{N_A(atoms)}{(mol^{})}\cdot{}
\frac{(mol)}{Mr(g)}\cdot{}
\frac{1000(g)}{1(kg)}\cdot{}
\frac{\rho{(kg)}}{(m^3)}\cdot{}
\frac{(m^3)}{10^6(ml)}\cdot{}
\frac{V(ml)}{drop}$$

Notice that this is dimensionally equivalent. Therefore,
$$\frac{atoms}{drop}=
\frac{N_A\cdot{\rho}\cdot{V}}{Mr\cdot{10^3}}$$


Numbering in phenolphthalein versus ascorbic acid

The subsituents do not take the highest priority here, rather, it is the parent molecule that determines the numbering. According IUPAC's organic chemistry book for nomenclature, the Blue Book, for heteromonocyclic components fused to a benzene ring:

  P-25.2.2.4 (...) The presence of a single heteroatom determines the numbering in a monocyclic compound; the heteroatom has the locant ‘1’.

This gives the following numbering for the isomers of benzofuran:

The parent molecule of ascorbic acid is a furan ring, for which the numbering is dictated by skeletal replacement ('a') nomenclature:

  P-15.4.1.2 For naming and numbering purposes, the following decreasing order of seniority is followed: $\ce{F}$ > $\ce{Cl}$ > $\ce{Br}$ > $\ce{I}$ > $\ce{At}$ > $\ce{O}$ > $\ce{S}$ > $\ce{Se}$ > $\ce{Te}$ > $\ce{Po}$ > $\ce{N}$ > $\ce{P}$ > $\ce{As}$ > $\ce{Sb}$ > $\ce{Bi}$ > $\ce{C}$ > $\ce{Si}$ > $\ce{Ge}$ > $\ce{Sn}$ > $\ce{Pb}$ > $\ce{B}$ > $\ce{Al}$ > $\ce{Ga}$ > $\ce{In}$ > $\ce{Tl}$.


Are there any substituents other than halogens which are ortho-para directors but ring deactivators?

The nitroso $\ce{-NO}$ group is deactivating as well as ortho para directing just like the halogens in electrophilic aromatic substitution.

For reasons of this behaviour, refer to this question on the website.

Here are more examples from Peter Skyes' Guidebook to Mechanism in Organic Chemistry: (least deactivating to most deactivating)

  $\ce{-F}<\ce{-Cl,-Br, -I}<\ce{CH2Cl < -CH=CH-COOH < -CH=CH-NO2}$

For all these groups $\ce{-I}$ effect is more dominating than $\ce{+M}$ effect. Hence they deactivate the ring. However, when the arenium ion is formed they stabilise the ring through the $\ce{+M}$ effect (hyperconjugation in case of $\ce{CH2-Cl}$)

The book doesn't mention the nitroso group in the order but since it has very weak +M we should expect it to be after $\ce{CH2Cl}$

$\ce{Cl-, Br- , I-}$ are almost equally deactivating therefore they have been separated by a comma.


Can I make my own graphene at home in my kitchen?

I Googled the phrase "How to make graphene" and got this result, which contains a section on how to make it at home.

The answer is, unequivocally, yes you can. Below are excerpts from the linked page.

  (1) The first DIY method is to use a lead pencil to deposit a thick layer of graphite onto a paper. Then use ordinary sticky tape to peel off a layer of graphite from the paper. Use another piece of sticky tape to remove a layer of graphite from the first sticky tape. Then, use a third piece of unused sticky tape to remove a layer from the second piece of sticky tape, an so on. Eventually, the graphite layers will get thinner and thinner, and you will end up with graphene, which is single-layer graphite in the strict sense, or bi-layer or few-layer graphite (which acts almost like graphene in certain uses). Even though this way of making graphene is only a proof-of-concept, the sticky tape method works. It takes patience and time, but it's the DIY method which the Manchester group used in 2004. And remember, they actually won a Nobel prize for their work, so there's no messing with sticky tape!

  (2) Shear exfoliation in liquids starts by pouring powdered crystal in a liquid, and then using a shear mixer to separate (exfoliate) layers of material from the crystal. The liquid that is used in such processes is chosen so that the small graphene particles don't clump back together, and the result is a liquid suspension of graphene. The suspension can then be dried to obtain graphene nanoflakes, or it could be directly used in later technological processes, such as making of graphene coatings and others.
  So, how do you make graphene at home using this method? Well, you can replace the shear mixer with a kitchen blender (since shear mixing is a fancy word for high-tech blending), and instead of special liquids you can use water and dish detergent. The detergent is added so that the particles don't clump, and acts as a surfactant. A great source of graphite powder are graphite pencils. Simply powder a few pencil leads and add the powder to the detergent solution, then blend for a while. Unless you have access to special microscopes and other equipment, you will probably be unable to confirm the existence of graphene in your detergent solution, but there's a quick rule of thumb which you can use to estimate the particle size. Generally speaking, if the graphene particles are too big, they will sink to the bottom of the vessel, while if they are smaller, they will float to the top. Nanoparticles are so small that they will be suspended in mid-water, and this is what you're aiming for if you want to make graphene. After you make enough graphene particles, you might want to filter the suspension and leave it to dry.

  (3) If you pour a non-water-soluble liquid into water it will either float on top of the water, or the water will float on top of it, depending on whether the added liquid is heavier (more dense) or lighter (less dense) than water. A common example is oil floating on top of water. One group of scientists discovered that you can make graphene by pouring two solvents which do not mix together, such as heptane and water, into a glass and adding finely ground graphite powder and putting it in a sonication bath. You can buy heptane in most chemical supply stores, water is readily available (it's a good idea to use distilled water), and you can easily obtain graphite sticks in art shops. An ultrasound bath is needed, but these are not uncommon on ebay and the like. The resulting graphene is one to four layers thick, and is chemically stable. It can be extracted and deposited on a clean glass plate, or any other substrate. The idea is to mix the finely ground graphite with water and heptane (1:1 ratio of water and heptane) and sonicate it for a while. What happens next is that graphite flakes exfoliate at the interface between water and heptane and graphene actually starts "climbing" up the glass walls of the vial. If a hydrophilic substrate, such as a glass slide, is introduced through the interface, a thin film of graphene will climb the walls of the slide as well. The glass slide can then be extracted and after drying, you'll be left with graphene-coated glass. If you use a polyethylene vial, the resulting graphene will not climb the walls of the vial, but you can still extract it with a slide made of glass or other hydrophilic material. Graphene is deposited as a thin film on both sides of the glass plate, and its slight light-absorbing properties can be seen by looking through it (Graphene absorbs about 2.3% of visible light). The explanation of the process is that graphene self-reassembles from the small exfoliated flakes on the interface between water and heptane because the surface tension of graphene (54.8 mN/m) is nearly right between the surface tension of water (72.9 mN/m) and heptane (20.1mN/m). The thickness of the graphene film is limited by capillary forces on the interface and the excess graphene will simply fall to the bottom of the vial. The formation of thick aggregates is suppressed by the diffusion and the energy requirement needed to form a new layer. This is possibly the easiest way to make DIY graphene at home as an arbitrarily large sheet. Other methods only create graphene flakes, but these flakes could possibly be used instead of graphite flakes in this method as a refinement stage.


Why is the endo product the major product in a Diels-Alder reaction?

What is the endo-selectivity of Diels-Alder reactions?
If the diene used in the Diels-Alder reaction has asymmetric substituents at the end carbons, and if the dienophile is unsymmetric, then two different isomers of the final adduct can form. The isomer, where the functional group(s) (usually carbonyl) on the alkene end up on the same side as the newly formed double bond, is called the endo-isomer [1]. The other one is the exo isomer. The naming of the isomers can be a little ambiguous in cases where there are multiple functional groups of different types on the starting alkene. Usually the electron-withdrawing group is considered when identifying which isomer is endo and which is exo.
 [2]
The experiments tell us that, in general, the endo-product is the one which is favoured kinetically i.e. it forms faster. This is not a rule as such, because there is no way to predict to what extent the endo product will be favoured. (In many cases, the exo product is the thermodynamic product i.e. more stable, so allowing the reaction to equilibriate will result in higher yield of exo product)
Why does this happen?
It's difficult to answer this question definitively, because there is still a lot of active research going on in the area. I will try to summarise the main effects which have been proposed as possible causes of the endo-selectivity.

Secondary orbital interaction

The interaction between the frontier molecular orbitals at the points where the new $\sigma$-bonds form are the first-order interactions. However, there are other interactions possible between the diene and dienophile, there are called second-order interactions, or secondary orbital interactions (SOI). This explanation was proposed by Woodward and Hoffmann, in the book The conservation of orbital symmetry[3].
The following diagram will be useful:
 [4]
As you can see, in the transition state leading to the endo-product, the carbonyl part of the alkene can interact with the back of the diene, and the interaction is favourable, as the symmetry matches. In the transition state leading to the exo-product, this interaction would not be available, as the carbonyl group is away from the diene. For a long time, SOI was considered as the right answer, and was extensively invoked in explaining not just Diels-Alder reaction, but other types of cycloaddition reactions as well.
However, there are a lot of evidences, both experimental and computational, which calls into question the validity of SOI. There is a good paper from Salvatella et.al. summarising them: "Do  Secondary  Orbital Interactions  Really  Exist?" Acc. Chem. Res., 2000, 33, 658.
For example, the dimerization of butadiene was shown to have slight endo preference.
 [5]
This was frequently used as an example of the SOI effect. However, computational studies at RHF and CASSCF level have found that the reaction proceeds via a stepwise mechanism. [6] If the mechanism is stepwise, not concerted, then it does not go through the constrained transition state which is required for the secondary orbital interactions to come into existence. So, we cannot invoke SOI for this endo-preference.

Steric effects

In certain Diels-Alder reactions, especially those involving the cyclopentadiene, steric effects have been proposed.
 [5]
Experimental data shows that when $\ce{R}$ = $\ce{CH3}$ and $\ce{X}$ = $\ce{CHO}$ the endo product is formed in $17.0%$ yield, i.e. the exo product now dominates. [7] This indicates that the steric repulsion of cyclopentadiene ring with $\ce{CH3}$ and the  is greater than that with $\ce{CHO}$, which is consistent with $\ce{CH3}$ being bulkier.
These observations can also be explained by SOI, but the trends in the endo-selectivity for a range of $\ce{R}$ and $\ce{X}$ groups seem to fit better to the steric argument.[5]
In a more recent study by Fernandez and Bickelhaupt, the selectivity was found to arise from activation strain, i.e. steric strain as the reactants approach the transition state.[12] They used $\mathrm{M05{\text -}2X/def2{\text -}TZVPP}$ to model the reaction of cyclopentadiene and maleic anhydride.

The above image, taken from their paper, shows the full reaction coordinate, with ZPE-corrected energies in kcal/mol and distances in Å. As you can see, the exo-reactant complex (exo-RC) is higher in energy than the endo one, suggesting that the strain comes into existence even before the TS. The extra strain in the exo-pathway is thought to arise from the steric repulsion between the methylene of cyclopentadiene and the oxygen of the anhydride.

Hydrogen bonds

Hydrogen bond-like interactions between C-H in alkene and the double bond has also been proposed as a cause of endo-selectivity in particular cases, such as the reaction of cyclopropene and butadiene.[8] Here, both steric effects and H-bond like interactions are thought to be present.
 [5]
This is somewhat different than the SOI interaction, because in SOI, only orbital overlap (delocalization) is considered. H-bond interactions are a sum of orbital overlap and electrostatic attraction.

Electrostatic forces

Electrostatic forces can be used to explain the unusual exo-selectivity of the furan+cyclopropenone addition.

In the exo-TS, there is a favourable electrostatic interaction between the electronegative O of furan and the electropositive carbonyl carbon of cyclopropenone. This makes the exo-isomer the preferred one. The exo-TS is about 1.81 kcal/mol lower in energy than the endo-TS, according to computational studies. [9]
The usual endo-selectivity of other Diels-Alder reactions can also be explained by the electrostatic repulsion between slightly electropositive terminal C-H of the diene, and the electropositive carbon of the electron-withdrawing group on dienophile:
 [5]

Solvent effects

Sustmann and others carried out computational studies on Diels-Alder reactions in both gas phase and with dielectric-continuum solvation.[10] They found that in the gas phase, the exo-TS is lower in energy i.e. kinetically favoured. Using a solvation model resulted in the endo-TS being the favoured one.
They suggested that solvent effect might be the cause of endo-selectivity. The endo-TS is generally more polar than the exo-TS, so it is stabilised to a greater extent in solvent. It is also observed that the preference for endo-TS increases the more polar the solvent is, which lends support to this argument. The gas phase exo-preference would also disprove the role of SOI effects, because they should be present regardless of solvation.
However, computational studies like these have to be considered carefully, because they were mostly carried out using HF or B3LYP methods, both of which severely underestimate dispersion forces (which are necessary to describe SOI). So, the results might be artificially skewed in the direction of exo-TS. [11]
Conclusion
All of this means that we are not really sure if there is a single cause that can explain the endo-selectivity in Diels-Alder reactions.
In an extensive experimental-computational study , Paddon-Row et. al.(2020) studied reactions of isotope labelled butadienes with mono- or di-substituted alkenes.[11] Surprisingly, it was found that mono-substituted alkenes (such as methyl acrylate) did not have a strong endo-preference when the reaction was done in benzene solution. Methyl acrylate and butadiene, for example gave 50:50 exo-endo product yield. Di-substituted alkenes showed higher endo-selectivity. In their analysis they suggested that SOI effects are important in some cases, whereas steric effects are forcing the selectivity in some other cases.
So, it might just be that different effects are working in different reactions, and somehow the end result is that endo-product is favoured in almost all cases. Then it might not be useful to look for one particular blanket cause of the selectivity, rather, each reaction has to be explained uniquely.

References
(1) Organic Chemistry, Clayden, Greeves, Warren, 2nd ed., Oxford University Press, p.884
(2) Image source: https://employees.csbsju.edu/cschaller/Reactivity/pericyclic/peri%20endo.htm
(3) The conservation of orbital symmetry, R B Woodward, R Hoffmann,   Weinheim : Verlag Chemie GmbH, 1971
(4) Image source: https://www.cpp.edu/~psbeauchamp/pdf/316_MO_DA_rxn.pdf
(5) Image source: Acc. Chem. Res., 2000, 33, 658
(6) Y. Li and K. N. Houk, J. Am. Chem. Soc., 1996, 118(4), 880–885
(7) Y. Kobuke, T. Fueno, J. Furukawa, J. Am. Chem. Soc., 1970, 92, 6548-6553
(8) M. Sodupe,R. Rios,V. Branchadell, J. Am. Chem. Soc., 1997, 119, 4232-4238
(9) S. M. Bachrach, J. Org. Chem., 1995, 60, 4395-4398
(10) T.  Karcher,  W.  Sicking, J. Sauer and R. Sustmann, Tetrahedron Lett., 1992, 33, 8027–8030
(11) W. J. Lording, T. Fallon, M. S. Sherburn, M. N. Paddon-Row, Chem. Sci., 2020, 11, 11915–11926
(12) I. Fernández, F. M. Bickelhaupt, J. Comput. Chem., 2014, 35, 371– 376. DOI: 10.1002/jcc.23500


What you describe is unfortunately a very common misconception. Defining orbitals does not break spatial symmetry. It is still completely arbitrary which directions you define as $x$, $y$ and $z$, and thus it is entirely arbitrary how you orient the $2p$ orbitals in space. Therefore rotational invariance is still preserved.

The other thing to remember is that all the $p$ orbitals are (essentially) degenerate and so you can take any linear superposition of them you want. For example you can write down an orbital that looks like

$$\frac{1}{\sqrt{3}}\left(\phi_{2p_x}+\phi_{2p_y}+\phi_{2p_z}\right) $$

and placing an electron in this orbital does not break rotational symmetry.


We need a better reducing agent (that itself oxidises) than $\ce {Cu^+}$.
So choose the ones which have greater reduction potential than $\ce {Cu^+}$.
It isn't $\ce {Mg^2+}$ as it cannot oxidise any further hence it has a very low(negative) reduction potential, so it cannot stop $\ce {Cu^+}$ from oxidising.


Chiral products from achiral reactants and 'principle of racemic mixture formation'

Because the starting material is achiral and the products are enantiomers, the reaction, by symmetry, must proceed via two enantiomeric transition states. Enantiomers have the same energy, so these two transition states have the same energy, so the rate of product formation is the same for both products. Therefore, they will be produced in equal amounts.

However, there is a complication that was not covered in your statement. There are examples of autocatalytic reactions like the Soai reaction where the product aids in catalysis and a tiny random preference for one enantiomer over the other leads to amplification of that preference. This is akin to how life uses preferentially one enantiomer instead of another, since presumably, we started with achiral materials in the primordial soup.


What is this red substance that has accumulated at the bottom of a plastic bottle full of petrol (see picture)?

Maybe could be some iron oxide or hydroxide from previous storage or ( if you have a iron tap) from it ore maybe directly from the plastic.

Some fuel has red dye in it but I don't think this is the case, because I don't see how it can precipitate.

Maybe it could be a residue of petroleum resin.


Unfortunately calcium sulphate, gypsum, $\ce{CaSO4}$, is depressingly stable and nothing non-deadly will dissolve it.

The other posts you have seen are about calcium carbonate, chalk/limestone/marble, $\ce{CaCO3}$, which acids will dissolve, although people do not in practice use that fact to make holes in it. One of the first difficulties would be to get the application of a liquid focused enough to make a hole-shaped hole.

If you ever did want to dissolve marble, I'd use hydrochloric acid, $\ce{HCl}$, because calcium chloride is nice and soluble. But check beforehand that it doesn't dissolve the gloves you intend to use.


How to calculate the concentration of a monoprotic organic acid that was extracted into an ethyl ether layer?

You arrive at the correct answer, but the way you get there is a bit messy. (And it only works because both phases have the same volume.) You should be a lot clearer in defining what properties you have been given in the question and which ones you need to calculate. Setting up the equation is a lot more important then calculating it in the end, because this does not only show that you know what you are doing, but it also is a way to condense a difficult sentence into a calculator understandable way.
\begin{align}
V(\mathrm{water})       &= \pu{25.0 mL}\\
c_\mathrm{tot}(\ce{AH}) &= \pu{0.456 mmol/mL}\\
V(\mathrm{ether})       &= \pu{25.0 mL}\\
V(\ce{NaOH})            &= \pu{5.45 mL}\\
c(\ce{NaOH})            &= \pu{0.523 mmol/mL}\\
\color{red}{c_\mathrm{ether}(\ce{AH})} &= ?
\end{align}
Then you must always carry the units with the quantity; they belong together. This way you will always have a way to tell if you have made a mistake setting up the equation. If the units don't work out, it cannot be correct.
Since it is a monoprotic acid the following will hold true:
\begin{align}
\ce{AH &<=>[H2O] A- + H+} &
&& n_\mathrm{tot}(\ce{AH}) &= c_\mathrm{tot}(\ce{AH}) \cdot V(\mathrm{water})
\end{align}
Now you extracted your solution and split the amount of substance between the water and the ether phase. You can calculate the remaining substance in the water phase from the neutralisation experiment.
\begin{align}
\ce{AH + NaOH &-> A- + Na+ +H2O} & n_\mathrm{water}(\ce{AH}) &= n(\ce{NaOH})\\
&& n_\mathrm{water}(\ce{AH}) &= c(\ce{NaOH}) \cdot V(\ce{NaOH})
\end{align}
The extracted amount of substance of the acid is simply the difference, from that you can easily calculate the concentration in the ether phase.
\begin{align}
c_\mathrm{ether}(\ce{AH}) &=
  \frac{n_\mathrm{ether}(\ce{AH})}{V(\mathrm{ether})}\\
c_\mathrm{ether}(\ce{AH}) &=
  \frac{n_\mathrm{tot}(\ce{AH}) - n_\mathrm{water}(\ce{AH})}
       {V(\mathrm{ether})}\\
c_\mathrm{ether}(\ce{AH}) &=
  \frac{c_\mathrm{tot}(\ce{AH}) \cdot V(\mathrm{water})
        - c(\ce{NaOH}) \cdot V(\ce{NaOH})}
       {V(\mathrm{ether})}\\
c_\mathrm{ether}(\ce{AH}) &=
  \frac{\pu{0.456 mmol/mL} \cdot \pu{25.0 mL}
        -  \pu{0.523 mmol/mL} \cdot \pu{5.45 mL}}
       {\pu{25.0 mL}}\\
c_\mathrm{ether}(\ce{AH}) &\approx \pu{0.342 mmol/mL}
\end{align}


Why must polar molecules be asymetrical?

If you start with the wikipedia definition of chemical polarity, which is standard (if not IUPAC) usage:

  In chemistry, polarity is a separation of electric charge leading to a molecule or its chemical groups having an electric dipole moment, with a negatively charged end and a positively charged end.

then it's not hard to understand why a molecule such as $\ce{CO2}$ containing a center of inversion lacks a dipole moment and is therefore "nonpolar", as are, by this definition, other molecules with high symmetry properties and lacking a net dipole moment:

  The only groups compatible with a dipole moment are Cn, Cnv and Cs. In molecules belonging to Cn or Cnv the dipole must lie along the axis of rotation.


Calculating pH of an unknown base, and the pKb

In such an solution, since the autoionization of water may not be neglected, you should expect $x=9.4 \times 10^{-8}$ do not mean $[\ce{OH-}]=9.4 \times 10^{-8}$. So you can only leave it unknown.

Then, $K_\text{b} = \frac {[\ce{XH+}][\ce{OH-}]}{[\ce{X}]} \approx \frac {9.4 \times 10^{-8} [\ce{OH-}]}{0.002} = 4.7 \times 10^{-5}[\ce{OH-}]$.

However, we still do not know what $[\ce{OH-}]$ is.

If only to consider the autoionization of water balanced, you will get

$[\ce{OH-}][\ce{H+}] = K_\text{w} = 1 \times 10^{-14}$

and this is still not enough.

The answer can only consider by the fact that the solution will be electrically neutral,

$[\ce{OH-}]=[\ce{H+}] + [\ce{XH+}] = \frac {1 \times 10^{-14}}{[\ce{OH-}]} + 9.4 \times 10^{-8}$

$[\ce{OH-}]^2 - 9.4 \times 10^{-8} [\ce{OH-}] - 1 \times 10^{-14}=0$

Since $\Delta = (9.4 \times 10^{-8})^2 + 4 \times 10^{-14} \approx 4 \times 10^{-14}$,

$[\ce{OH-}] \approx \frac {9.4 \times 10^{-8} + 2 \times 10^{-7}}{2} = 2.47 \times 10^{-7}$.

Thus, $\mathrm{pH} = -\log \left(\frac {1 \times 10^{-14}}{2.47 \times 10^{-7}}\right) = 7.39$

$\mathrm{p}K_\text{b} = -\log (4.7 \times 10^{-5} \times 2.47 \times 10^{-7}) = 10.94$.

Sometimes the “electrically neutral” trait will be very useful, this is a good example.


Does an electron's spin change when being excited?

The picture shows a triplet excited state returning to the ground state by emitting a photon, i.e. phosphorescence. This can only happen if there is also an interaction that couples angular momentum change with the transition, such as spin orbit coupling.
The electron has two quantum numbers; the spin $S=1/2$ but is not the spin of the electron that is changing but the other quantum number 'z' , 'azimuthal' or 'magnetic' spin quantum number $m_z=\pm 1/2$.


I agree absolutely with Max, and while I was still running the calculations, he already provided the answer.
The goldbook provides a nice definition for aromatic compounds:

  In the traditional sense, 'having a chemistry typified by benzene'.
  A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.
  The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

Therefore not all aromatic compounds have to strictly full-fill the Hückel (4n + 2) rule, but all that obey that rule are aromatic. Phenols certainly behave like benzene and are therefore aromatic, please note that aromaticity is an experimentally observed feature.

You can see the aromaticity from the $\pi$ type orbitals, that are in both cases quite similar and extensively delocalised. The calculations were done at the DF-BP86/def2-SVP level of theory.

Benzene exhibits a very high symmetry $D_\mathrm{6h}$, therefore the last two of the shown orbitals are degenerate.

The restriction is not present in the phenolate, as the symmetry is only $C_\mathrm{2v}$.

Also due to the missing symmetry constraint the neutral phenolic radical, $\ce{H5C6O.}$, and the phenolic cation, $\ce{[H5C6O]^{\oplus}}$ would be aromatic.


Separation of metal particles from fluid in ferrofluid

Assuming the ferrofluid is the one described at product description, then it is "an extremely fine powder, coated with a soap-like material called a surfactant, suspended in a mineral oil liquid base." From that description, it might be possible to dilute or extend the ferrofluid with mineral (or even vegetable) oil, but one would not expect ferrofluid to be compatible with water or with alcohol. It might be possible to dilute it with other aliphatic solvents.

There are directions for making your own water-based ferrofluid at Synthesis of an Aqueous Ferrofluid.


Which ideal gas law does a constant volume thermometer work on?

Gay-Lussac's Pressure Temperature Law.

The law simply states

  The pressure of a gas of fixed mass and fixed volume, is directly proportional to the gas's absolute temperature.

As telling temperature rises, the gas in the thermometer has its pressure increased in accordance with the law. Measuring this pressure gives us the temperature as per the ideal gas equation, or by using Gay-Lussac's law directly, taking the pressure ratio against a set base temperature and pressure.


Photoelectric effect (AP Chemistry question)

I think the teacher pointed out that (c) is the correct answer because the postulate is generally regarded as incorrect according to the normal convention of photons behaving as individual packets of energy.

However what I think is interesting is that more than one electron can be ejected provided the photon is sufficiently energetic (this information will not be useful during an examination, so I suggest forgeting about this when you take an exam or do your homework, but if interested read on). For more information you can search for "multielectron ejection in auger spectroscopy" which provided references [1] and [2] below. Reference [2] opens up with the following statement:

  Absorption of an energetic photon by a target should in first approximation lead to the ejection of one electron only, as this interaction is purely mono-electronic. However it is well established that sometimes several electrons can be ejected. This can happen in valence multiple ionization or after innershell ionization when secondary electrons are emitted sequentially (cascade Auger decays) or simultaneously (direct double Auger decays). Such processes and especially the ‘one-step’ or direct ones (direct double ionization or direct double Auger decay) are extremely interesting to investigate as they revealthe strongelectron correlation phenomena which are at their origin.

Not sure what Einstein would have thought of this in any case.

An Auger process does not violate the principle that intramolecular electronic transitions are quantized and therefore that the energy required to eject electrons is quantized, irrespective of the number ejected. So the first two postulates are still correct.

References

P. Lablanquie, M.A. Khalal, L. Andric, J. Palaudoux, F. Penent, J-M Bizau, D. Cubaynes, K. Jänkälä, Y. Hikosaka, K. Ito, K. Bučar, M. Žitnik, Multi-electron coincidence spectroscopy: Triple Auger decay of Ar 2p and 2s holes,
Journal of Electron Spectroscopy and Related Phenomena, Volume 220, 2017, Pages 125-132, https://doi.org/10.1016/j.elspec.2017.04.003.
https://pure.mpg.de/rest/items/item_737492/component/file_737491/content


Not sure why, but I can't modify/balance this equation

For starters, You have a problem because of a lack of charge balancing. You should also consider that at neutral pH $\ce{H2S}$ is present as $\ce{HS-}$ and $\ce{CH3SH}$ is present as $\ce{CH3S-}$.  Modifying the reaction to account for charge you get:
$$\ce{HCO3- + HS- + CH4 <=> CH3S-}$$

This will give you a start.  The other piece that will help is what the $\ce{HCO3-}$ molecule becomes. For the sake of being able to provide an answer I will assume formate ($\ce{HCO2-})$ is formed. Thus the reaction is now:
$$\ce{HCO3- + HS- + CH4 <=> CH3S- + HCO2-}$$

We know that the molecule of $\ce{CH4}$ becomes $\ce{CH3S-}$ and the molecule of $\ce{HCO3-}$ becomes $\ce{HCO2-}$ thus to balance the lost oxygen of $\ce{HCO3-}$ and the lost two hydrogens of $\ce{CH4}$ we add water to the products.
$$\ce{HCO3- + HS- + CH4 <=> CH3S- + HCO2- + H2O}$$

This equation serendipitously requires no coefficients to balance.


How to make silicic acid, potassium silicate or calcium silicate?

You can make the silicates by fusing the hydroxides with silica. Potassium silicate fertilizer grade are successfully produced by direct fusion of silica ($\ce{SiO2}$) and potasium compounds($\ce{KOH}$ and $\ce{K2CO3}$) in furnaces at temperatures up to melting point of mixture. The fusion temperature reaches around 1350 °C.
$$\ce{SiO2 + 2 KOH ->[\Delta] K2SiO3 + H2O}$$
$$\ce{SiO2 + K2CO3 ->[\Delta] K2SiO3 + CO2}$$
Calcium silicate is prepared by reacting calcium hydroxide and silica. This paper1 discussed the reaction at 30-85 °C.
References

https://www.matec-conferences.org/articles/matecconf/pdf/2016/21/matecconf_bisstech2016_01021.pdf
REACTION BETWEEN SILICA AND CALCIUM HYDROXIDE SOLUTIONS. I. KINETICS IN THE TEMPERATURE RANGE 30 TO 85°, S. A. Greenberg, The Journal of Physical Chemistry 1961 65 (1), 12-16, DOI: 10.1021/j100819a005


Does silicon dioxide react with alkalis?

It really does react:
\begin{align}
\ce{4HF + SiO2 &-> SiF4 + 2 H2O}\\
\ce{SiF4 + 2 HF &-> H2[SiF6]}\\
\ce{SiO2 + 2 NaOH &-> Na2SiO3 + H2O}\\
\ce{SiO2 + 2 F2 &-> SiF4 + O2}\\
\ce{SiO2 + 2 C &-> Si + 2 CO}\\
\ce{(2,1)Na2O + (1,1)SiO2 &-> Na4SiO4/Na2SiO3}\\
\ce{SiO2 + Si &-> 2 SiO}
\end{align}


Bond cleavage of 3 membered cyclic ring on heating

Possible mechanism for cyclopropyl ring opening of given compound in acid medium:


Bard and Faulkner are outstanding electrochemists, but that potentiostat figure always annoys me because it makes it seem as though current flows through the reference electrode and that the working electrode is literally grounded. In reality, the reference electrode should conduct extremely low currents (pA, really) and the current through the working electrode is what you care about, as a function of the electroactive species in th potentiostat's solution.

Therefore, consider Fig. 15.24 and surrounding text from from Horowitz and Hill 1:

In this three wire potentiostat, the bottom operational amplifier (op amp) simply works as a transimpedance amplifier, converting current at the working electrode to voltage. In this regard, it is basically a very low input impedance ammeter, which approximates an ideal ammeter. So the working electrode is held at virtual ground, which typically means less than a millivolt from true ground.

The potential at the inverting input of the upper op amp must equal the input reference voltage, $\mathrm V_{ref}$, to within less than a millivolt. Note that $\mathrm V_{ref}$ can be from a ramp generator, a function generator or from another op amp, as in the inverting summing op amp in the figure from Bard and Faulkner. Importantly, note that virually no current flows through the reference electrode, because of the extremely high input impedance of the op amp inputs, so it will function properly and not become polarized or damaged.

The potential between the working and reference electrodes is $\mathrm V_{ref}$ because the working electrode is clamped to virtual ground and the potential of the reference electrode is clamped at $\mathrm V_{ref}$. The solution in the cell has an impedance, which is generally complex, and the solution spans the volume between the counter electrode (also called the auxilliary electrode) and the working electrode. The reference electrode is physically placed as close to the working electrode as is feasible.

Now let $\mathrm Z_{c}$ denote the solution impedance between the counter and reference electrodes and let $\mathrm Z_{u}$ denote the solution impedance between the reference and working electrodes. Then we have a non-inverting voltage follower with gain, as shown is this figure:

The output voltage, $\mathrm V_{o}$, is the voltage at the counter electrode. Therefore, $$\mathrm V_{o} = [(Z_{c} + Z_{u})/Z_{u}] \times V_{ref} \tag{1}$$

But the voltage at the junction of $\mathrm Z_{c}$ and $\mathrm Z_{u}$ , i.e., in the solution where the reference electrode is located, is $\mathrm V_{ref}$. Therefore $$\mathrm V_{ref} = [Z_{u}/(Z_{c} + Z_{u})] \times V_{o} \tag{2}$$

If solution impedances change, the upper op amp will source or sink current, as necessary, by changing $\mathrm V_{o}$, thereby maintaining the solution potential, at the reference electrode, equal to $\mathrm V_{ref}$.

1 P. Horowitz, W. Hill, The Art of Electronics, 2nd Ed., Cambridge University Press, Cambridge, ©1989, pp. 1015-1016.


How to dissolve composite In2O3 with SnO2 without affecting stainless steel?

In case your question is about the recovery of indium from ITO, the following might be helpful:

In Recovery of indium from indium tin oxide by solvent extraction, published in Hydrometallurgy, 2011, 107, 56-61 (DOI), the abstract seems to suggest that $\ce{H2SO4}$ is more suitable than $\ce{HNO3}$. Furthermore, a solvent extraction scheme for indium, using di-(2-ethylhexyl)phosphoric acid (DEHPA) is proposed.

In Recovery of indium from used indium–tin oxide (ITO) targets, published in Hydrometallurgy, 2011, 105, 207-212 (DOI), the authors agree on using $\ce{H2SO4}$ (at 90 °C) and decrease the ymount of tin in solution by sulfide precipitation.


Is there anything we lose by using Google Colab for DeepChem?

DeepChem's project web page indicates further documentation and access to the source files used.  Just click the «fork me on github» sign in the top right hand corner leads you to this page.  From the notes there, it becomes obvious that many tasks the program suite handles are managed with Python which you should install in first place (e.g., Python's Python, or Anaconda's Python) then offering a platform independent use of deepchem in Linux, Mac, or Windows (see this report).

Once you have this installed, you just have to solve the «must have» dependencies», in Python's Python with python package installer pip3 on the CLI

pip3 install joblib pandas sklearn tensorflow pillow deepchem

and maybe the ones listed as «soft requirements» such as RDKit.

A nice thing is that a package manager like pip will check if additional requirements have to be adjusted, and if so, will offer to adjust your system accordingly «on the fly».  Just be aware that you might need a considerable chunk free memory on permanent memory for the modules deepchem relies.  Don't forget the inofficial package (wheel) index for Python modules in Windows by Christopf Gohlke which may assist you in solving the dependencies, too.


While usually short-lived, isolated hydrogen atoms (also called hydrogen radicals) are well studied and chemically important species. Bond formation in simple species is always an exothermic reaction (usually highly so) with zero activation energy, which means bond formation is thermodynamically very favourable and has no kinetic barrier stopping it from happening. Simply put, the isolated atoms really want to react with each other, and they'll attract each other at a distance, so if at some moment you create a container with pure monoatomic hydrogen, the hydrogen atoms will react as soon as possible to regenerate dihydrogen molecules (hydrogen gas).

The best you can do is expose the dihydrogen molecules to energetic conditions, so that a fair amount of bonds are being constantly broken at a rate comparable to which they're formed. A simple way to do this is to strongly heat the hydrogen gas; enough thermal energy will cause the bonds to break, though the dihydrogen bond is unusually strong ($\rm{436\ kJ\ mol^{-1}}$) so even at several thousand degrees kelvin there will be only a small portion of free hydrogen atoms in the gas (see Mithoron's comment). A more interesting option is to shine a strong ultraviolet light into the gas, as the dihydrogen molecules can separate into their constituent atoms upon absorption of photons with energy around $\rm{4.52\ eV}$, which corresponds to a photon wavelength of $\rm{274\ nm}$.

Any substance can have its bonds broken through the above two methods (among others), so the same applies to all other diatomic molecules. The stronger the bond between the atoms, the more uncommon is it to find the isolated atoms. For substances with very weak bonds, such as molecular dibromine or diiodine, mere exposure to visible light is enough to trigger formation of a fair amount of isolated atoms, something which is exploited in several reactions.

Keeping in mind that isolated atoms are very reactive, they are rather abundant in space. The average density of matter in the Universe is very, very low (about 1 atom per 5 cubic metres of space!), so isolated atoms can go a long time before finding anything to react with. This makes them extremely important for understanding astrochemistry.


Clausius–Clapeyron equation: shape of phase diagrams makes no sense

This is a summary of the equations to use to calculate phase transitions.

The Clapeyron equation $\displaystyle p_2-p_1=\frac{\Delta H}{\Delta V}\ln\left( \frac{T_2}{T_1} \right)$ is used for a solid-liquid transition. The changes in enthalpy and volume relate therefore to changes occurring in fusion.

The Clausius-Clapeyron equation describes solid-vapour and liquid-vapour changes because the final volume is far greater than the initial one, and is $\displaystyle \frac{dp}{dT}=p\frac{\Delta H}{RT^2}$ where $\Delta H$ the enthalpy change at the liquid–vapour or sublimation transition. Integrating this last equation from pressure $p_1 \to p_2$ and temperature $T_1 \to T_2$ gives $\displaystyle \ln\left(\frac{p_2}{p_1} \right) = -\frac{\Delta_{vap}H}{R}\left( \frac{1}{T_2}-\frac{1}{T_1} \right) $.

The change in the volume during fusion is $\displaystyle \Delta_{fus}V = m\left(\frac{1}{d_l}-\frac{1}{d_s} \right)$ where $m$ is the molar mass and $d_l$ and $d_s$  the densities of the liquid and solid. The pressure variation for the solid to liquid (melting or fusion) change is

$$\displaystyle p_2=p_1+\frac{\Delta_{fus}H}{\Delta_{fus}V}\ln\left(\frac{T_2}{T_1} \right)$$

and for evaporation and sublimation

$$\displaystyle p_2=p_1\exp\left( -\frac{\Delta_{vap}H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1} \right) \right)$$

with the appropriate $\Delta H$. This is $\Delta_{vap}H$ for evaporation and $\Delta_{fus}H + \Delta_{vap}H$ for sublimation. Sublimation is treated as two steps merged into one; melting and instantaneous evaporation.

The $p$ vs $T$ plot for benzene is shown in the figure. Notice how this is different to how these are generally shown. This is sometimes due to the fact that log pressure is sometimes plotted but this is not always shown on the figure. Notice also how the solid liquid line is effectively vertical.

The data used is

R = 8.314  #( J/mol/K)
dens_sol = 981.0  #( kg/m^3)
dens_liq = 879.0  #( kg/m^3)
mol_mass = 78.0/1000.0 #( kg/mol)
DH_vap   = 30.8*1000   #( J/mol)
DH_fus   = 10.6*1000   #( J/mol)
p3 = 36.0/760*101325 #( triple point pressure Pa)
T3 = 5.5 + 273.16    #( triple point temperature K)
DV_fus = mol_mass*(1/dens_liq-1/dens_sol)                          # delta volume fusion

Python/numpy  functions for the pressure are:
p_liq_vap= lambda T: p3*np.exp( (DH_vap/R)*(1/T3-1/T))
p_sol_vap= lambda T: p3*np.exp( ((DH_fus + DH_vap)/R ) * (1/T3-1/T) )
p_sol_liq= lambda T: p3 + DH_fus/DV_fus*(np.log(T) - np.log(T3))


Why does ionization violate the stable or lowest energy rule in atoms?

Textbooks won't bother to dig into details. Most of them are satisfied with saying that such thing exists. They're right I guess. I'll do a chemguide approach to the answer since I find it most understandable.

We draw this diagram, and then assume that it works for all atoms. In other words, we assume that the energies of the various levels are always going to be those we draw in this diagram.
If you stop and think about it, that has got to be wrong. As you move from element to element across the Periodic Table, you are adding extra protons to the nucleus, and extra electrons around the nucleus.
The various attractions and repulsions in the atoms are bound to change as you do this - and it is those attractions and repulsions which govern the energies of the various orbitals.

Going through the Aufbau principle, when you pass Ca (Z = 20) you will face Sc, the first element in which a d orbital is going to be filled. But, you must make a really important change in your train of thoughts.
As chemguide explained, with change in the number of protons and electrons due to the change of Z, the way their orbitals' energies are defined changes. Till Ca, the behavior insisted that 4s has a lower energy state than 3d and thus, fills sooner. But, starting from Sc, the pattern changes: In Scandium, 3d has a lower energy than 4s. The behavior changed.
This change in behavior results in the current confusion of yours. This much answers the problem you have, and you can visit the page at chemguide to learn more. If you want more info, I can freely add. Hope I've helped!


Can chiral atoms have same substituents but with different orientations?

No, it is not an asymmetric centre. To test this, you can draw the two potential stereoisomers:

Note that isomers A and B are identical (by C2 rotation), so the nitrogen cannot be an asymmetric centre. If it were one, then these isomers would by definition be distinct.

The attached groups are identical, from the point of view of the nitrogen - both have (R) stereochemistry. If one were (R) and the other (S), the nitrogen would constitute a pseudoasymmetric centre - https://goldbook.iupac.org/html/P/P04921.html

Hope this helps.

Also, don't get confused about the difference between the nitrogen being a sterogenic centre, and the molecule as a whole being chiral. The molecule is chiral, since it is non-superimposable on its mirror image. Chirality is a molecular property, and although frequently people will refer to a particular atom in a molecule as "chiral", this is a misnomer. A better description of an atom with 4 different groups attached is "sterogenic". It is entirely possible to have a molecule containing stereogenic centres but the molecule is achiral (e.g. meso compounds).

The logic behind checking the plane and centre of symmetry is that planes and centres of symmetry are conserved on reflection in a mirror, so checking for these symmetry elements is a shortcut to checking whether a molecule satisfies the definition of chirality.


Cyanide hydrolysis; could it yield an oxime?

Yes, since nitrogen is more electronegative than carbon, and also since it has the lone pair available for taking, the proton is likelier to go to it.


Why or why not can hydroquinone and 2H-pyrrole act as ligands?

Zaworotko, M. J.; Stamps, R. J.; Ledet, M. T.; Zhang, H.; Atwood, J. L. Heterocyclophane complexes of transition metals. 1. Synthesis and crystal structure of both the (.eta.5-2.2pyrroloparacyclophane)tricarbonylchromium and the (.eta.6-2.2pyrroloparacyclophane)tricarbonylchromium. Organometallics 1985, 4 (10), 1697-1700. DOI: 10.1021/om00129a001

A chromium tricarbonyl complex of a pyrrole is reported, here the pyrrole coordinates with all four carbon atoms and the nitrogen atom. I think that the pyrrole is a six electron donor to the chromium.

Bentivegna, B.; Mariani, C. I.; Smith, J. R.; Ma, S.; Rheingold, A. L.; Brunker, T. J. Formation, Stability, and Structures of Borenium and Boronium Cations Derived from Pentamethylazaferrocene–Boranes by Hydride or Chloride Abstraction Reactions. Organometallics 2014, 33 (11), 2820-2830. DOI: 10.1021/om500348u

A pyrrole binds with all five non hydrogen atoms to an iron which is also bonding to a pentamethyl cyclopentadienyl anion. This is forming a protonated version of CpFe(C4N) which looks very much like ferrocene. In this case the bonding is the same as it was in the chromium compound.

Yamamoto, K.; Kimura, S.; Murahashi, T. σ-π Continuum in Indole-Palladium(II) Complexes. Angew. Chem. 2016, 128 (17), 5408-5412. DOI: 10.1002/ange.201601992

A pyrrole binds with two carbons as a pi ligand to palladium. These selected results suggest to me that pyrrole can bind as an alkene or as if it was a Cp ligand to transition metals.

Son, S. U.; Reingold, J. A.; Carpenter, G. B.; Czech, P. T.; Sweigart, D. A. Charge-Assisted Hydrogen Bonding and Other Noncovalent Interactions in the Self-Assembly of the Organometallic Building Block [(η6-hydroquinone)Rh(P(OPh)3)2]+with a Range of Counteranions. Organometallics 2006, 25 (22), 5276-5285. DOI: 10.1021/om0604425

It reports the binding of a rhodium to all six carbon atoms of hydroquinone. Here it acts as a six electron donor.

So there is nothing to stop the molecules in the question acting as pi ligands to transition metals.


If you give enough oxygen to let it react to mercury, it still won't work just like copper and sliver. Because they are unreactive metals.

However, it will form mercury oxide when mercury and oxygen react at 350°C as the chemical equation below.

$\ce{2Hg + O_2 -> 2HgO}$

P.S: The mercury oxide will decompose back to mercury and oxygen at 400°C.

Lead is a rather more reactive subatances. So, try burning it. If it don't work, means you need a higher temperature.


The question is a bit poorly written. But it also gives a sense to the procedure of dissolution of chalk upon treatment with $\ce{HCl}$ and dilution of the solution.

The reaction that takes place after the treatment with $\ce{HCl}$ is;

$\ce{CaCO3}$+ 2$\ce{HCl}$ → $\ce{CaCl2}$ + $\ce{H2CO3}$

$\ce{H2CO3}$ decomposes into $\ce{H2O}$ and $\ce{CO2}$.
$\ce{CO2}$  is a gas. It bubbles up out of the solution.

$\ce{CaCl2}$ is soluble in water. So it should be dissolved by adding water.Further, it should then be diluted to the required volume.

"The chalk was dissolved" might signify the addition of water to dissolve the water soluble $\ce{CaCl2}$ produced during chemical reation with $\ce{HCl}$.


Why is the pKa of a ketone higher than of water?

As I'm sure you know, (p)Ka is a measure of the equilibrium between an acid and its conjugate base (what you're calling NBE?). Both the conjugate bases of water (hydroxide) and a ketone (enolate) have a negative charge on the oxygen. I'm summarizing your question as: If both conjugate bases have a negative charge on the same atom type, but one base (the enolate) is stabilized by resonance, shouldn't that corresponding acid (ketone) be more acidic? (Correct me if I'm misreading you.)

I think the key here is what (p)Ka really represents. Fot water, with room temp pKa 15.7, the equilibrium favors water but some dissociation is present. For acetone, pKa 19.2, the equilibrium lies further on the non-dissociated side than for water. We take this to mean that acetone is a weaker acid that water. If we only looked at the stability of the conjugate bases, we might conclude the opposite (previous paragraph). But as pKa measures a balance between a conjugate pair, we must consider the acid forms as well.

The ketone contains the very "stable" (energetically favorable) C=O carbonyl bond. When a ketone donates an alpha proton to become the enolate conjugate base, the carbonyl bond is disrupted. This is an extra energetic cost that is not present in the water equilibrium, and I propose it is the reason why the ketones tend to dissociate less than water.

By this logic, water is a better acid than a ketone because acting as an acid requires a ketone to disrupt an energetically preferable C=O bond. Thus the balance lies closer to the undissociated side for a ketone (lower Ka, higher pKa) than for water (lower pKa), so we say that water is a stronger acid and a ketone is weaker.

(One might argue that the carbonyl need not be disrupted, the lone pair could localize on the alpha carbon, but I hope we can agree such a carbanion is quite unstable, and a carbanion not preferable to breaking the carbonyl and localizing on the more electronegative oxygen.)


Hot water, the "universal solvent", should work nicely.


Sodium is included in the name because it was essential in the chain polymerization that forms the rubber.
Sodium served as the initiator for the anionic polymerization.


Since you're concerned about mercury from fillings, you may have already taken a look at the wikipedia article about dental amalgam toxicity. It would be worth reading through some of the sources listed, too.

I assume you're an adult, I don't know if you're male or female. Children are more sensitive to mercury than adults because their brains are actively developing. If you're female, you may become pregnant at some point and there would be some concern about fetal mercury exposure.

However, according to a 2003 WHO report, the amount of mercury ingested from dental amalgam is low, with exposures estimated from 1ug per day to 12.5ug per day, but the vast majority of people are exposed to less than 5ug per day. The mercury from fillings is elemental mercury, which is poorly absorbed, less than 0.01% is taken up, with smaller amounts reaching the brain.

Removing your existing fillings probably isn't worth it, as the removal process would expose you to more mercury than just letting the filling sit there.

I don't know if this will make you feel better or worse, but coal fired power plants are probably a larger source of mercury exposure than your fillings. The mercury from the coal plants is methylmercury, which is much more efficiently absorbed than elemental mercury. This bioaccumulates in the environment, so that fish have relatively high levels of mercury in them.

As far as your perceived concentration problems, we really can't say if it's caused by the mercury or not. If it concerns you, you should see a doctor. And it may be more efficient to treat the symptoms through ADD medication than trying to remove mercury.


How to calculate concentrations in a diluted mixture of multiple solutions?

Am I somehow supposed to use $M_1V_1=M_2V_2$?

Yep!
The problem has given you three different solutions:

Iron(III) nitrate
Potassium thiocyanate
Nitric acid

You know the concentrations of each, and you know (or can calculate) the volumes of each that are mixed to yield the final solution.  What you'll actually be using in the calculation are three different equations:

$M_{1i}V_1 = M_{1f}V_f$
$M_{2i}V_2 = M_{2f}V_f$
$M_{3i}V_3 = M_{3f}V_f$

$M_{\#i}$ is the initial concentration of each of the three chemicals in their separate starting solutions, and $M_{\#f}$ is the final concentration of each after mixing, that the problem is asking for (except for the $\ce{HNO3}$).


Interaction between a pencil and rubber (eraser)

There was much discussion in the comments to the OP regarding whether erasing a pencil mark is a chemical process or merely a mechanical process, akin to removing an oil spill with high pressure water (my analogy). But I think the answer to "Is it a Van der Waals interaction with intermolecular forces" is clearly yes.

Although the classic "pencil head" eraser generally requires a vigorous rubbing action to rapidly remove a pencil mark, even this can be done in a manner that shows pencil marks that had been bound to the paper via intermolecular forces are transferred to a material having a stronger affinity for the pencil mark than does the paper.

This Wikipedia article discusses several types of erasers that clearly illustrate a chemical process rather than a mechanical one is dominating the removal of the pencil mark. For example, regarding the "poster putty" type of eraser the article states:

  "...poster putty works much the same as traditional kneaded erasers,
  but with a greater tack or lifting strength...Repeatedly touching the
  putty to a drawing pulls ever more medium free..."

A "gum" eraser also clearly works by adsorbing graphite from paper. As the gum is dabbed or lightly rubbed on the pencil mark, the intermolecular forces holding the graphite to the paper are weaker than the intermolecular forces between the graphite and the erasure. The following photo from the above Wikipedia article shows that the graphite in the used eraser has accumulated over time. There is no evidence of paper or other material having accumulated along with the material from the pencil mark.

Furthermore, one can imagine using different materials that should make for good erasers if the pencil marks are simply being removed by the mechanical action of an eraser. For example, if that's the case, shouldn't you be able to remove the pencil mark by rubbing it off with a piece of wood, one's finger, another piece of paper, a fingernail, etc.?


"A gets eliminated" or "the elimination of A" is just plain wrong, typical sloppy lab slang by many (too often native?) speakers. Very common, but still terrible.

As you say, the small molecule X gets eliminated from substrate A. Chemical English is still just English, and that is the correct way to put the case in question into words in this language.

"A undergoes elimination" (short for "A undergoes an elimination reaction) is a widely used, and imo also correct, variant.


I think the right word for it is curdle. Have you read, for example, the Wikipedia article on curd? It says:

  “Milk that has been left to sour (raw milk alone or pasteurized milk with added lactic acid bacteria or yeast) will also naturally produce curds, and sour milk cheese is produced this way”

or this excerpt of the “Milk” page:

  When raw milk is left standing for a while, it turns "sour". This is the result of fermentation, where lactic acid bacteria ferment the lactose in the milk into lactic acid.

So, milk naturally contains bacteria which turn it acid, and hence it curdles spontaneously. You see the beginning of this process.

An invaluable source of information on this topic is McGee’s On food and cooking. Great reading if you like chemistry and food.


The following is an image of the hexagonal crystaline form of ordinary ice (Ice I$_h$) taken from S.S. Zumdahl, Chemistry, 3rd ed., copyright © 1993 by D.C. Heath and Company:

Note that the dashed lines represent hydrogen bonds. Liquid water actually has a similar "open" structure also due to hydrogen bonding. But in the case of liquid water, the hydrogen bonds are not rigid and semi-permanent as in ice. So imagine that in the image above, the hydrogen bonding network collapses. This is what happens when enough thermal energy is present to break the rigid hydrogen bonds resulting in melting. Clearly, once this crystaline structure is no longer forced into place by the rigid hydrogen bonding in ice, it can collapse into itself, resulting a greater density of water molecules.

Thus the liquid form of water, although engaged in transient hydrogen bonding, is not as open and expanded as when held into it's solid form by the rigid, semi-permanent hydrogen bonding.


One mole is a number, not a mass.

See the following figurative example:

Let's say we are dealing with a sample of pure chemistry books.

A book's weight is 1 kg.
One mol of books would be 6,022 × 10²³ books.
The mass of one mol of books (the molar mass of books) would be as follows:

1 kg (mass of one book) × 6,022 × 10²³ (# of books)

Now let's say we're dealing with a mixture (alloy) of 60% chemistry and 40% history books. What is the mass of 1 mol of this mixture?

A history book's weight is 2 kg

So, our entire collection of books, sum up to Avogadro's number = 6,022 × 10²³ units. And we know that it is 60% Chem. and 40% Hist. So the mass of the entire collection would be:

m = N × 0,6(1kg) × 0,4(2kg)

where N is the Avogadro's number.

Enough of analogies. You have the mass of the gold and palladium atoms from the periodic table. I am assuming you also have the proportion Au/Pd.

You also know that there's 1 mol of atoms. The exact same equation would apply:

m = N × P.M(Au) × P.M(Pd)

where N is the Avogadro's number, P and M are the percentages and atomic mass, respectively, of gold and palladium.


Why does oxygen not like to be a double anion?

The addition of an electron to any negatively charged ion will need to overcome the repulsive force of the negative ion. This is generally less favorable than adding an electron to a neutral atom, and the electron affinity will be positive. Even though the additional electron will complete the octet, as in the case of an  ion where the addition of electron will complete the  subshell, the electron affinity is a positive value.


This answer has technically been given in the comments, but it bears repeating.

Yes, you can stop the reaction at the epoxide stage through careful choice of conditions. If you prevent both acidic and basic aquaeous environments, then epoxide rings are stable. They are present in a wide range of molecules and even in natural products such as epothilone.

You could also open epoxides with other nucleophiles to create different 1,2-substitution patterns — with predefined stereochemistry, if the epoxide was stereoselective.

And finally, there are even methods out there to turn vicinal diols into epoxides (usually via sulphonylation, e.g. to a tosylate followed by intramolecular $\mathrm{S_N}2$ reaction).


Limescale (calcium and magnesium carbonate) precipitates in the heater, the brown colour is most likely some iron compound that co-precipitates. A fresh surface of limescale likes to adsorb other ions, which then get trapped in it as the scale grows. If you had a water kettle, you'd see the same stuff in there after using it a few times. (Not exactly, because carbonate precipitation depends on the pressure, but iron (hydroxide?) not. I wouldn't want to predict.)

The limescale first appears on the surface of the heater cell, and then breaks off during temperature changes, i.e. when you open the faucet. As you have no tank, the flow rate of the water is high in the heater and carries the scale out of there.

Very convenient, that. In a tank, the scale builds up and can fill the whole thing after a few years, depending on your water source.


When using LPCVD, is it possible to pattern a silicon wafer with poly-silicon using a mask?

In my fab experience, the answer is no, you are not going to inhibit deposition of silicon with some other material. You might well change the final grain size, but it will deposit. Furthermore, LPCVD generally has good sidewall coverage, so you can't use a lift-off mask either.

The OP asked for references, so lets dig in to those. First, it is somewhat difficult to find specific references to deposition rates on various substrates that can be broadly compared - reactor specifics, gas flow, temperatures, etc., all make it hard to compare experiment-to-experiment. But, we can pin down a few things.

First, going back to Andy Grove's (yes, Intel's Andy Grove) book "Physics and Technology of Semiconductor Devices" (John Wiley and Sons, New York NY, 1967) one finds no mention of polysilicon deposition. At that early stage in technology it was not used - metal gates were the way to go.

Polysilicon became a thing with poly-Si gates and the LOCOS (local oxidation of silicon) process. Here one might end up depositing poly over a wafer that included single crystal silicon (to make source-drain contacts), oxide (gate and field isolation), and silicon nitride (for the LOCOS process). Looking at cross sections from LOCOS devices (in various more modern text books - one I have is Stephen A. Campbell, "The Science and Engineering of Microelectronic Fabrication" (Oxford University Press, New York NY, 1996)), you see no variation in poly thickness across those different materials.

There is much more literature available around questions like polysilicon grain size vs initial surface roughness, annealing temperature, etc., such as Voutsas and Hatalis, "Surface Treatment Effect on the Grain Size and Surface Roughness of as-Deposited LPCVD Polysilicon Films", J. Electrochem. Soc. 140(1) 282-288 (1993).

Finally, a number of people have done work on the kinetics of LPCVD poly-Si growth, with one early reference being M.L. Hitchman et al., "Polysilicon growth kinetics in a low pressure chemical vapour deposition reactor", Thin Solid Films 59 231-247 (1979).

There are two takeaways from papers like this. First, under LPCVD reactor conditions, the mean free path of a silane molecule (or any other reactant) is quite long. This is, of course, highly desired to have uniform deposition across many wafers in, e.g., a tube furnace. The flip side is that the probability of a silane molecule reacting upon hitting a surface is quite low - they bounce around a lot. And, once a surface has any poly-Si on it, it all looks the same.

Second, the only way to not have poly-Si deposition is to never allow nucleation in the first place. Yet, point one above implies that lots of silane is hitting the surface, and if any one molecule manages to stick you now have a place to grow more. Further, all that seems to be needed to get some silane reactivity is the presence of hydrogen on the surface, and there will be lots of hydrogen around as silane decomposes elsewhere (plus hydrogen on surfaces is pretty much a given - it is by far the easiest surface termination on just about anything, particularly after a chemical process in an aqueous solution, like an acid clean).


Why is tetramethylsilane (TMS) used as an internal standard in NMR spectroscopy?

TMS was first proposed as a reliable internal chemical shift reference in 1958 by Tiers. Back in them good ol' days, 1H NMR was called proton nuclear spin resonance, or nsr, and the tau scale was  used for reporting chemical shifts (10ppm in the delta scale was set to 0 and positive values were read to the right. So TMS came at t10.0), and CCl4 was the one of the most common solvents used for running dilute solution samples.

TMS was proposed as a reference at the time becasue it offered the following advantages:

the chemical shifts are largely temperature and concentration independent
it offered a single phase method for referencing (That is, it was an internal standard. Most common methods at the time used an external reference, such as a coaxial insert; a method which is still acceptably used today)
The TMS peak lies outside the 'usual spectral region and is readily identifiable'
It is largely inert, and is not likely to react with most samples
It can be used for almost amy solvent (except H2O, D2O)

Today, TMS is still accepted as the single universal reference standard for NMR. For over a decade, the IUPAC recommendations have stated that, other than for 13C, all nuclei are referenced to the frequency of the 1H signal of 1% TMS solution in CDCl3. Yes, your 31P and 77Se peaks should actually be referenced to 1H of TMS. This unified reference scale is very easy to implement on modern instruments that incorporate a digital 2H lock (ask your friendly NMR custodian if they use it!). It is in fact arguably more accurate that referencing to internal solvent residues which are much more likely susceptible to concentration and temperature factors. Internal solvent residues are called secondary references, and are less accurate than the primary reference. However, they are accurate enough for the very vast majority of work done in the laboratory.

It has long been shown that the chemical shift of TMS is, in fact, both temperature and concentration dependent. But not much. There are actual corrections that should be applied to measured chemical shifts at extreme temperatures, but these are very small and rarely seen in the literature.

TMS is not water soluble, and in aqueous samples DSS is the recommended reference standard. It contains signals in the 1H spectrum at 3.1, 1.9 and 0.8ppm, all at a fraction of the intensity of the main reference signal, and these can interfere with analysis of other spectral regions. A deuterated version is widely used in the literature. An alternative, TSP, is sometimes used, although is more pH dependent. Other reference standards can also be used, however, and as long as the reference method and sample conditions are reported and verified, then all is good. For 1H NMR, compounds such as acetonitrile, methylsulfone, benzene, 1,4-dioxane and dichloromethane are all examples of internal standards that are sometimes used, both for chemical shift referencing and concentration measurements, as they all largely satisfy that important list of reference characteristics above. They all have very simple 1H spectra, soluble in common deuterated solvents and fairly unreactive. Plenty of other compounds with multiple peaks can be used as well, such as toluene and dmf.

So, to specifically answer your question, TMS is the adopted primary standard for 1H chemical shift referencing for the reasons stated above. Any other material can be used as a secondary reference, either internal or external, provided its chemical shift can be accurately determined against the chemical shift for TMS. Results reported should include a clear description of how chemical shifts were determined. For example, 1H chemical shifts of all reported compounds were measured relative to the methyl peak of n-dodecane, determined as 0.88ppm. The disadvantages of using other materials is that it needs to be shown that the chemical shift of these signals is independent of sample conditions (concentration, temperature etc), and that they will often have peaks overlapping regions of interest.


Milk contains surfactants (primarily lipids), which allow stable bubbles to form. In intense boiling, these bubbles can become mixed with liquid, supporting the liquid above them and forming what I'll call for want of a better term a "wet froth." Imagine a bunch of sponges made out of liquid. They take up more space than they would if you compressed them down and forced out the air. The inclusion of the water vapor in the bubbles lowers the density of the mixture, causing it to expand and overflow the pot.

The state is only stable because the added energy keeps creating new bubbles to replace any that have popped and keeps forcing liquid upward, while it would normally fall.

Water does not experience this mostly because its bubbles are far less stable. In fact, pure water does not form bubbles at all. The bubbles formed in water is typically due to impurities.  If you were to add soap to a boiling pot of water, you would see the exact same effect, if not worse.


I know PV=nRT but that's total what about partial?

To find partial pressures of a mixture of non-reactive gases, you can use Dalton's Law of partial pressures.

If you know the total pressure of the mixture and the number of moles of each gas, you can find the pressure of each gas by multiplying its mole fraction with the total pressure.


I think the key part of the answer is

  Assume $\ce{Fe^{2+}}$ is not present in the solution $\color{red}{\underline{\text{initially}}}$.

Since iron reacts with $\ce{Fe^3+}$ to form $\ce{Fe^2+}$ via:

$$\ce{Fe^0 + 2 Fe^3+ -> 3 Fe^2+}\tag 1$$

You have to account for this additional reaction. If the reaction did not produce any $\ce{Fe^2+}$ as you assumed then of the $\pu{0.100 mol}$ of $\ce{Fe^3+}$ in the solution then $\pu{0.051 mol}$ would remain after and you would be correct, but it does react. Now, my math says that the answer is more like

$n \times M = \dfrac{M}{3}\times \dfrac{\pu{1 mol}}{\pu{96500 C}}\times \pu{4 \frac{C}{s}} \times \pu{60 \frac{s}{min}} \times \pu{60 \frac{min}{hr} \times \pu{1 hr}}\tag 2$

$\implies n = \pu{0.0497 mol} = \pu{0.050 mol}$

Which means $\pu{0.050 mol}$ of $\ce{Fe^3+}$ remains which must be neutralized by the $\pu{0.050 mol }\ce{Fe^0}$. From equation 1 we can see that it takes one $\ce{Fe^0}$ for every 2 $\ce{Fe^3+}$ which menas $\pu{0.025 mol}$ of $\ce{Fe^0}$ is needed to react with $\pu{0.050 mol } \ce{Fe^3+}$ and your remaining $\ce{Fe^0}$ is $\pu{0.025 mol}$ or $\pu{0.0245 mol}$ in the case of $n = \pu{0.49 mol}$.


How do we identify the molecular formula from the empirical formula of a completely unknown substance?

If the empirical formula is known, then any method that will determine the molecular weight will do. Of merely historical interest is the use of freezing-point depression, nowadays, one would use mass spectrometry (electron-spray ionisation will leave most molecules intact).

This assumes that it is a reasonably small molecule, but only for those will the empirical formula be known before the molecular one.


Why is Specific heat of Ammonia higher than water?

The link in the question states that ammonia has a higher specific heat than water. This is true, but it is important to compare the specific heat values under identical (or very similar) conditions.

The isobaric specific heat of ammonia (gas) is 2.1636 kJ/kgK at 1 atmosphere and temperature = 26.9 C. https://www.engineeringtoolbox.com/ammonia-heat-capacity-specific-temperature-pressure-Cp-Cv-d_2016.html At -33.6 C (boiling point of $NH_3$ at 1 atm), the specific heat of $NH_3$ gas is 2.2946 kJ/kgK, not much different from the gas at 26.9 C; however, for liquid $NH_3$ at 1 atm and -33.6 C, the specific heat is 4.4464 kJ/kgK, a large increase from the gas at that temperature.

Liquid water has one of the highest specific heats among common substances, about 4182 J/(K kg) at 20 °C https://en.wikipedia.org/wiki/Specific_heat_capacity At 100 C, liquid water has a specific heat of 4.2157 kJ/kgK (implied pressure = 1 atm). Again, not much difference because of temperature. https://www.engineeringtoolbox.com/specific-heat-capacity-water-d_660.html Now we find water vapor data in a different place: https://www.engineeringtoolbox.com/water-vapor-d_979.html where gaseous $H_2O$ at 216 F (102.2 C) has a specific heat of 1.890 kJ/kgK, a large drop, even bigger than in the ammonia system.

So although it is often stated that water has one of the highest specific heats among common substances, it seems that either ammonia is not considered common for this issue, or yes, it's true that water has one of the highest specific heats...And that is true of the gas and the liquid states!
But then among all the data, which are truly interesting, there is one other thing that stands out: the heat of vaporization of water is 40.8 kJ/mol, whereas for ammonia, it is only 23.5 kJ/mol. So water can still be considered an extreme molecule.


Does magnesium sulfate in aqueous solution form magnesium (hydrogen-)carbonate when hydrogen carbonat is present?

The white precipitate is more likely insoluble white calcium sulfate.
Per a reference on the making of fertilizer, to quote:

Calcium, magnesium, and sulfur are also important materials in plant growth. They are only included in fertilizers in small amounts...

Net ionic reaction:
$\ce{ Ca^2+ (aq) + SO4^2- (aq) -> CaSO4 (s) }$


Yes, there are various water clusters leading to local order on the scale of a few molecules holding stabilized orientation with one another for measurable timescales.  This article(1) is a pretty good introduction to the major varieties in liquid water.  This article(2) talks more about various structures available in any phase, including the ices, which gives some pretty good context.

(1) Keutsch, Frank N., Saykally, Richard J., Water clusters: Untangling the mysteries of the liquid, one molecule at a time, 2001, PNAS, 98, 19, 10533–10540, doi:10.1073/pnas.191266498

(2) Ludwig, Ralf, Water: From Clusters to the Bulk, 2001, Angewandte Chemie International Edition, 40, 10, 1808-1827, doi:10.1002/1521-3773(20010518)40:10<1808::AID-ANIE1808>3.0.CO;2-1


Acidic KMnO4 oxidizes H2S to S. Why doesn't the oxidation proceed further to sulfur oxide, sulfur dioxide or sulfur trioxide?

Yes, the reaction can further proceed to sulfur dioxide and sulfur trioxide, but only appears for certain, per my research, to occur in alkaline conditions.
Here is a supporting opinion to this effect, which cites the corresponding reaction in alkaline conditions as:

$\ce{3 S(2-) + 8 KMnO4 → 3 SO4(2-) + 8 MnO2 }$

that implies the further conversion of $\ce{S}$ to $\ce{SO2}$ to $\ce{SO3}$, which in alkaline conditions is represented by sulfate formation.
Note, the cited source notes only in acidic conditions, to quote:

Neutral-Acid pH:
$\ce{3 H2S + 2 KMnO4 → 3S + 2 MnO2 }$

Wikipedia on KMnO4 supports this pH effect, to quote:

The Mn-containing products from redox reactions depend on the pH. Acidic solutions of permanganate are reduced to the faintly pink manganese(II) ion (Mn2+) and water. In neutral solution, permanganate is only reduced by three electrons to give manganese dioxide (MnO2), wherein manganese is in a +4 oxidation state.

Interestingly, the corresponding reaction replacing KMnO4 with either H2O2 or HOCl, can results in $\ce{SO2}$, ..., in acidic conditions.


Exo- or endothermic? Clarification required regarding a few cases

I think you are confusing terms:

exothermic means that it releases heat, that is energy, because the energy of the products is lower than the one of the reagents.

endothermic means that it absorbs heat (energy) because the products are at a higher level of energy than the reagents.

When a solution becomes hot, it's because it's releasing heat (energy), and likewise if it gets cold it's because the reagents inside are absorbing energy from the surronding environment.

The case of boling water: does it need heat to become steam? Or does it release heat? In other words, do you have to provide heat to the teakettle to make the water boil or it provides heat by itself?

I hope you can get to your answers by yourself now.


Why don't electrons pair up in an oxygen radical?

Pairing up of electrons in this case is unfavourable due to increased inter-electron repulsion; the average spatial separation of the electrons is higher in the triplet state where the two electrons are unpaired. Regular oxygen molecules also exist in a triplet state and have unpaired valence electrons; they are said to be biradicals.


Mechanism for reaction of ketene with water

The silver salt, which can also be $\ce{AgCF3COO}$, is added to the mixture to cleave off $\ce{N2}$, creating a singlet carbene which induces the 1,2-shift to generate the ketene. From then on, it has no further purpose.

Our solvent is water, which could technically act as an acid, a base or a nucleophile. Reacting as a base does not forward our mechanism, so we need to consider whether water is stronger as an acid (left-hand mechanism in scheme 1) or as a nucleophile (right-hand mechanism).

Scheme 1: Two possible mechanisms for ketene hydrolysis.

A few thoughts on the two mechanisms, first. In whichever way water reacts, it will react with the most reactive site, first. In case of a nucleophilic attack, the most electrophilic carbon is clearly the ketene carbon. Water has the possibility of cleaving either the $\ce{C=C}$ double bond or the $\ce{C=O}$ double bond. The $\ce{C=O}$ bond is preferred because it usually represents the molecule’s LUMO and the resulting carbanion would be rather unfavourable and probably immediately flip to an oxyanion. The final [1,3]-proton shift cannot occur in a concerted manner for electronic reasons. It is also highly unlikely that a single water molecule is both acceptor and donor as I drew it for brevity. I would assume a pair of molecules acting from both sides in a stepwise manner.

In case of the water-acidic mechanism, the most basic position in the molecule would be the oxygen atom so one would expect protonation there first. Again, I would assume the [1,3]-proton shift to be non-concerted which could provide major flaws to the mechanism. I also do not see the carbon position having sufficient basicity to be protonated in any noteworthy manner in rather neutral water. But let’s just assume it could happen for the sake of the following argument.

When it comes to rating water’s properties, we should note that it’s a mediocre base, a rather bad nucleophile but an even worse acid. So while one can save the nucleophilic mechanism by saying ‘yeah, water’s not a good nucleophile, but it’s the best we have floating around and the ketene position is rather reactive’ we cannot save the acidic mechanism in any similar way. Ketones are, in good approximation, unprotonated in aquaeous solution, so protonating an even worse position is all the more unlikely.

Of course, it is conceivable that both mechanisms happen to some extent. By playing around with the solution’s pH, you could probably favour either: an acidic solution will possibly facilitate the acidic mechanism while a basic solution will definitely facilitate the nucleophilic one — hydroxide is a so much better base than water.


Looking for a dye which emits around 680 nm

There are two dyes commonly used in biochemistry research with $\lambda_\mathrm{Ex}$ of around $\pu{650 nm}$. They are Alexa Fluor 647 from ThermoFisher and Cyanine5 (Cy5) from Lumiprobe. Two example of these two categories are illustrated in the following figure. Either probe can be ordered according to the way you want to conjugate to your substrate (azide and alkyne groups are shown here). For example, IgG conjugate of Alexa Fluor 647 with amide linkage is shown in the insert.


Formation of O₂, Water, and Electricity from Hydroxide?

Yes, the overall reaction can be reversed.  The reverse reaction is called the oxygen evolution reaction or simply water oxidation.  Depending on the pH, you could write it as either

$$\ce{2 H2O -> O2 + 4 H+ + 4e-}$$ or
$$\ce{4 OH- -> O2 + 2 H2O + 4e-}$$

The reverse reaction happens when water is electrolyzed to make hydrogen at one electrode and oxygen at the other.  You may have seen that demo at one time or another -- here's a Youtube version.

Oxygen is a thermodynamically powerful oxidant, which is another way of saying that it is easily reduced.  This means that the water oxidation reaction requires energy input, about 1.23 V at the chemical reference state of pH 0.  But it usually takes even more energy to effectively reverse the reaction -- notice they used a 9 V battery in the Youtube demo, so about 8 of those volts are wasted in the demo.  Developing new electrode materials that can carry out the water oxidation reaction with less energy loss is a major area of fundamental research in chemistry today.

You also said the Clark electrode works with $\ce{H2O2}$ as an intermediate.  Depending on the details of the particular materials used, in the water oxidation reaction hydrogen peroxide may or may not be an intermediate.  Working out the fundamental reaction steps that happen at electrode surfaces can be very difficult.


Is there any formula that can be used to find loss of mass due to binding forces in atomic and sub atomic particles?

Yes, there is the Semi-empirical mass formula which will give you the binding energy/mass of the nucleus.

Once you subtract the binding energy/mass from that of the nucleons, you have the mass of the nucleus.  Then you should add the mass of the electrons.  Finally, in principle at least, you would subtract the binding energy/mass of the electrons (in other words the ionization energies of the electrons), but this is a very minor correction.


Quantum mechanical model of atom and the quantum numbers

So how come we are talking about fixed shells again and returning to
  what Bohr said?

Nope, we are not returning to the old quantum theory. We do not talk about orbits anymore, we rather have orbitals. Note that it is important to bear in mind that orbitals also aren't real, i.e. this description is still approximate, but it is a much better way of thinking about atoms (and molecules) than Bohr model.

Now, to understand the concept of orbitals you have to learn the basics of quantum chemistry first, namely, the Hartree-Fock method. Until then just realize that:

We do use hydrogen-like one-electron wave functions (orbitals, or, spin-orbitals, to be more precise) to describe the state of each and every electron in any atom.
These hydrogen-like orbitals are described by a set of quantum numbers: the principal quantum number, the azimuthal quantum number, the magnetic quantum number, and the spin quantum number.
The set of orbitals belonging to a given principal quantum number $n$  constitutes what we call a shell in the orbital model.


The difference between peptide bonds and the bonds between polypeptides?

The term peptide bond refers to amide bonds. The 21 amino acids in insulin's A-chain are covalently linked by 20 amide bonds. The 30 amino acids in insulin's B-chain are linked by 29 amide bonds. The two polypeptides are covalently linked by disulfide bonds, which are not amide bonds.

The 3D structure of proteins is a complex topic and a field of intense research. It's worth pointing out though that even polypeptides (with no disulfides or other covalent linkages) have a more complex 3D structure than you might expect. I suggest you look into alpha-helices and beta-sheets as an entry into the topic.

Took some info from the Beta Cell Consortium.


What is the Chemical Compound that has a NFPA 704 (Fire Diamond) of 2-2-1-ox?

From what I found NIOSH lists 6 compounds with 2-2-1 pattern, but none of them are labeled with oxidizing properties:

ethylhexaldehyde: NFPA 704 2-2-1;
benzyl chloride: NFPA 704 2-2-1;
2-methyl-1-butanol: NFPA 704 2-2-1;
n-octyl mercaptan: NFPA 704 2-2-1;
vinyl toluene: NFPA 704 2-2-1;
methyl chloroacetate: NFPA 704 2-2-1.

On the other hand, there are oxidizers with slightly different NFPA diamond which much as well be the candidates:

ammonium dichromate NFPA 704 2-1-1-OX;
amyl nitrate NFPA 704 2-2-0-OX.

The values might deviate a little in different sources and they also depend on conditioning.


Why are hollow cathode lamps used for atomic absorption spectroscopy?

Atomic absorption lines are very narrow.
For the Beer-Lambert law to be applicable, the bandwidth of the source should be narrow in comparison with the width of the absorption peak. Otherwise, the signal-to-noise ratio and the slope of the calibration curve would be low; the resulting sensitivity would be poor.
Continuous radiation sources use monochromators; however, even good monochromators have effective bandwidths that are significantly greater than the width of atomic absorption lines.
Therefore, the use of line sources with bandwidths even narrower than the width of absorption peaks is preferred.


Precision Balance (1mg) vs. Analytical Balance (0.1mg)

Either of these balances will require some care to set up. It will need to be on a solid surface away from any drafts, etc. However, neither would have the more expensive setup and maintenance requirements of say a $\pu{0.001mg}$ balance. And in either case you can take care of the cleaning and calibration yourself, particularly as it's just a hobby with no certification requirements.

I would seriously consider looking at used equipment from a place like LabX (I have no affiliation whatsoever). I would hesitate to purchase a used $\pu{0.001mg}$ balance because that is a very sensitive piece of equipment, but a $\pu{0.1mg}$ analytical balance; my lab always bought that sort of thing used.

You can probably get a good used $\pu{0.1mg}$ Mettler for example for the same or less than a new $\pu{1mg}$ balance and there is really no downside. It's easy to imagine you wanting to weigh out a few mg of some reagent and having the ability to do that with the extra accuracy of at least a $\pu{0.1mg}$ balance would probably be desirable.


Can chemicals change crystal structure by only touching?

No, there is no "exact mimic" crystal, but the crystal habit can alter the crystal form. Consider that the environment in the uniform inside of a substance is different from that along its boundaries; growing on a different surface can alter the shape of the crystal.

The semiconductor industry, in particular, is often faced with difficulties in growing crystalline semiconductors with layers of varying composition (epitaxy). There is often strain, leading to defects, unless an intermediate layer can smooth the transition.


However my question is which values need to be used for the $A_\mathrm x$ and $A_\mathrm{IS}$ - do they need to be both morphine or just the two values from the sample peak areas?

To answer just this part of the question,
\begin{align}
A_\mathrm{x} &= A_\mathrm{unknown} \mathrm{(Morphine)} \\
A_\mathrm{IS} &= A_\mathrm{standard} \mathrm{(Codeine)} \\
C_\mathrm{x} &= C_\mathrm{unknown} \mathrm{(Morphine)} \\
C_\mathrm{IS} &= C_\mathrm{standard} \mathrm{(Codeine)} \\
\end{align}
Looking at your picture, it looks like you successfully calculated your $\mathrm {D_{rf}({morphine})}$ (though it's next to a misleading picture, as the peak area of codeine is larger than that of morphine when their concentrations are equal).
Moving down, you've successfully solved your $\mathrm{D_{rf}}$ equation for unknown concentration, but the substitution is a little off. When you're dealing with the dissolved tablet solution, $A_\mathrm{unknown}$ is indeed $17.6$ but $A_\mathrm{standard}$ would be the codeine area of $212.5$, not the earlier morphine value.
If I had to guess, the confusion is between the terms "internal standard," a known spike of (usually) a single standard (here codeine) and "standard solution," which is an reference standard of known morphine and internal standard concentrations. Two common uses of reference standards are as initial calibrants and a good way to check for interference between analytes.
Getting back to the solution, once you have your unknown concentration, in units of $\pu{g/L} = \pu{mg/mL}$, you will simply multiply by the $\pu{50 mL} = \pu{0.05 L}$ volume to have the morphine concentration of the solution and thus of the whole $\pu{200mg}$ tablet. %w/w is also easy at that point.


Time dependence of integrated rate equations

From a mathematical standpoint: If the integrated rate law is independent of time, then the rate of change of the concentration must equal zero (by the fundamental theorem of calculus). So you're looking for a reaction in which the concentration of the species you want in the rate law (presumably the reactant) does not change appreciably.

As others pointed out, this is a common case for intermediates under the steady-state approximation (rate of production = rate of consumption, net rate zero). But that's just for intermediates. For a substrate, you would need the substrate concentration to be essentially constant. That's probably not a "reaction" that will be interesting to study. A trivial solution, if you will.


What is the significance of the "Standard Temperature" in Standard Enthalpies of Formation tables?

Most enthalpies (or themodynamic observables whatever their type) are not constant across the temperature range — not even across the range of one phase. So the enthalpy of vapourisation of water is different a $273\,\mathrm{K}$ from what it is at $295\,\mathrm{K}$ or $350\,\mathrm{K}$.

Of course, standard conditions have been defined. However, the definition of standard conditions does not explicitly include temperature! Likely because some wanted $273\,\mathrm{K}$, others wanted $295\,\mathrm{K}$. So the temperature at which a certain entropy is valid must always be specified.

Some books put a ‘standard temperature’ at the top and call that their specification.


They do indeed. It could be for different reasons:

A chemical might undergo chemical changes due to different storage conditions: it could react with other chemicals present in the air or in the environment. This could change its nature (reaction to create other compounds), properties and/or purity.
A chemical might undergo physical changes as well: the temperature and pressure vary, as well as other conditions like humidity, light, electromagnetic fields. This could change it as stated above.
Certain chemicals are simply unstable and decay or degradate over time to something else, without the need of externals input / changes. This could affect the chemical as stated above.
Certain chemical suffers transpiration losses.
Human factors, like contaminations, spillings, loose caps, etc.

However, it might be noticed that these reasons could be divided to produce two different dates: shelf life vs expiration date.


Ozonolysis is a synthetic procedure deploying diluted ozone as a reagent.  This reaction may be relevant in the context of your question, because i) during the reaction, the gas typically is bubbled across a solution (and you have some exhaust to deal with), and ii) even more importantly, after the reaction, the organic products are to be isolated safely, too.

Here, I would like to suggest to use what is otherwise described as reductive workup.  Ozone, which may dissolve in solvents, hereby is quenched (deactivated) by a material which in turn is oxidized by ozone.  Because ozone is a powerful oxidizer, the deactivation is performed while keeping the concentration of ozone entering the quench solution low, and simultaneously keeping the temperature of this solution low, too.

The provision of the flux of 400 mL/min only partly describes your problem.  It neither states the concentration of ozone in this input, nor the regime of this input -- it may be in batches, it could be a contiously working line.  With these uncertainties in hand, bubbling the ozone-containing gas input into an aqueous solution of sodium hydrogensulfite (with "i", $\ce{NaHSO3}$) seems both simple to setup, manage, and scale.  The reaction basically will yield sodium sulfate ($\ce{Na2SO4}$).

As an example, Ragan et al. published about an Ozonolysis at large industrial scale I hereby refer to.  For about 19 h, their recation was running at low temperature with a constant flow of 15-20 L/min, 4-5 wt% $\ce{O3}$ in air bubbled across their reaction vessel.  After completion of reaction, their workup reported was pouring the reaction solution slowly (over 30 min) to a prechilled quench solution of 3.0 kg $\ce{NaHSO3}$ in 6 L water at 0 degC.  Only then  they allowed the mixture to warm slowly to room temperature.

Depending on the conditions in your example, maybe you do not need to cool your sulfite solution as much, yet monitoring the temperature of the stirred quench solution is recommended.  With the parameters missing in your question, the minimal quantity of bisulfite you need might be estimated.  Equally, there are KI/starch indicator stripes, which may be used to check if the concentration of oxidizing ozone still is significant; if present, they will change from colourless to dark blue (ref, example).

Literature reference: Ragan et al., Safe Execution of a Large-Scale Ozonolysis: Preparation of the Bisulfite Adduct of 2-Hydroxyindan-2-carboxaldehyde and Its Utility in a Reductive
Amination, Organic Process Research & Development 2003, 7, 155−160 (DOI: 10.1021/op0202235).


I think you could use some types of Butylrubber. If I'm not wrong from Isobutylene is derived the Post-it adhesive. However you can find Butyl tape easily in every big hardware store (especially nautical) or online.
EDIT
At this link you can find some more information this is an excerpt:

Tapes and flooring adhesives  Exxon butyl rubber can be used in
adhesive formulations for transparent tape, hot-melt pressure
sensitive adhesives, mastic for pipe wrap tape and vinyl floor tile
adhesives. It can also be used for semiconducting electrical splicing
tape. A rubber splicing tape must be self-fusing; i.e., when wrapped
upon itself, it must fuse to the underlying layer, forming a permanent
bond. The self-fusing is a result of compounding and eliminates the
need for application of a separate adhesive coat. These compounds have
very high tensile strengths and exhibit some degree of recovery or
elasticity. Electrical splicing tapes based on Exxon butyl rubber have
moisture resistance, outstanding aging resistance and good electrical
properties. Polyisobutylene elastomers have long been successfully
used in pressure sensitive adhesives both as primary elastomers and as
tackifiers and modifiers. Exxon chlorobutyl rubber can be used alone
or in blends with other polymers to achieve special properties in
pressure sensitive adhesives.


Origin of electrode potential and its connection with electric double layer

Let's see if we can tease out the main points here.  The images below are taken from Lance and Cole's Analytical Chemistry POGIL workbook.
Microscopic processes at a metal surface exposed to an aqueous solution containing the corresponding metal ion(1) will result in a slight negative charge on the metal surface due to oxidation of the metal and a slight positive charge in the region of the solution close to the metal surface owing to the sudden generation (and dissolution) of positively charged metal ions.

The amount of negative charge generated is metal specific and as of now, unknowable in an absolute sense.  Therefore, we are required to compare the extent of this process at one metal electrode to that of another.  If we compare the Zn electrode above to a Pb electrode, we would find that the Pb electrode has less negative charge on its surface relative to Zn.  Both of the electrodes, however, have a slightly negative charge at their surfaces.  Because the charge is greater at the Zn electrode, we say Zn is easier to oxidize than Pb.
To answer your question which of the above explanations is correct, the answer is both (although the clarity of the explanations is up for debate).  A negative charge is generated at both electrodes, but the the amount relative to one another differs, and that gives rise to the notions of easier/harder to oxidize/reduce.
(1) In my experience, microscopic theory of electrochemical processes typically assume the presence of the metal and metal ion in part to avoid issues which identifying what in the system is being oxidized and reduced.