How to find the change in internal energy (q+w) for an expanding monoatomc ideal gas? If a monoatomic ideal gas simultaneously expands against a constant external pressure and drops in temperature, how do you find the internal energy change?
Known values are: $T_1$, $T_2$, $p_1$, $p_2$, $p_\text{external}$ and $T_\text{external}$.

Finding the work done by the expansion is easy, but I don't understand how to change in heat.
It is not a reversible process, so it can't be found from the work, and it's not isothermic, so it isn't zero. It can be found if I have the heat capacity but that is not given.
Edit:
The pressure changes from $p_1$ to $p_2 = p_\text{external}$

The internal energy $U$ is a state function that can be expressed as a function of temperature $T$, pressure $p$, and amount of substance $n$:

$$U=U(T,p,n)$$

The change in internal energy $\Delta U$ is independent of the path by which the new state was reached. Hence, you do not necessarily have to calculate the exchanged heat $Q$ and the work $W$ done during a particular process.

The ideal gas gives simpler equations. According to Joule’s second law, the internal energy $U$ of an ideal gas depends only on the amount of gas and the temperature, and is independent of the volume or the pressure:

$$U=U(T,n)$$

For a monoatomic ideal gas (i.e. a model system of point-mass molecules) we find that the internal energy is proportional to the temperature $T$ and number of molecules $N$ or amount of substance $n$:

$$U=\tfrac32NkT=\tfrac32nRT$$

where $k$ is the Boltzmann constant and $R$ is the molar gas constant.

Since you know the temperatures $T_1$ and $T_2$, and since the amount of substance is constant $(n_1=n_2)$, you should be able to calculate the change in internal energy $\Delta U=U_2-U_1$.

Note, if you don’t know the number of molecules $N$ or amount of substance $n$, you may express the result as molar internal energy $U_\mathrm m$ (i.e. in $\mathrm{J/mol}$):

$$U_\mathrm m=\frac Un$$


Lewis Structure of OF+ Would the Lewis Structure for OF+ be drawn so that the more electronegative atom has the more negative formal charge, or would it be drawn to satisfy the octet rule?

Here are my thoughts thus far:

For one, we could make the formal charge on the oxygen +1 and the formal charge on the fluorine 0, and give them a single bond, which would break the octet rule, but make sure that the more electronegative atom (fluorine) had a less positive formal charge. Oxygen would have 4 free electrons, and fluorine would have 6 free electrons.

However, we could also make the formal charge on the oxygen 0, and the formal charge on fluorine +1, and give them a double bond, which would be wrong in terms of electronegativity, but would mean full octets. Oxygen would have 4 free electrons, and fluorine would also have 4 free electrons

How do I decide?

A strategy different from the one described by Yusuf Hasan would be to look at the neutral isoelectronic molecule. The oxygen atom is isoelectronic to the fluorine cation, so $\ce{OF+}$ is isoelectronic to $\ce{O2}$.

It is difficult to write a nice Lewis structure for $\ce{O2}$, see https://chemistry.stackexchange.com/a/15061. Experiments show that dioxygen is paramagnetic, i.e. it has unpaired electrons. If you try to write this as Lewis structure, you end up with 7 electrons on each oxygen (counting electrons in bonds twice), i.e. no octets either.

For $\ce{OF+}$, you should be able to write the same Lewis structure (and the same MO diagram). Because fluorine has a different number of protons than oxygen, however, the energy levels will be different and "the best" Lewis structure might be different as well.

As Yusuf Hasan mentioned, the struggles to write a "nice" Lewis structure for this species hints at high reactivity, so I think this is the main lesson of trying to write a Lewis structure. Which of the Lewis structure is best at approximating the real situation becomes secondary, and it makes sense to move on from a Lewis structure model to more comprehensive models of bonding for species such as this one.


Direction of mass transfer in two film theory In an extraction process compound A is transferred from one solvent to another solvent. The two solvent may be, for example, water and dichloromethane. These two solvents are immiscible and when mixing them together, one of the solvents, say solvent 1, forms droplets that are dispersed in solvent 2, as depicted [...].

Fig. 4 shows the concentration profile across the interface between the two solvents. For this concentration profile, answer the following questions:
(a) In which direction is the mass transfer, from solvent 1 to solvent 2 or vice versa? Motivate your answer by a short explanation (2 pt).

How can you tell which direction the mass transfer will go?
I would assume that because solvent 1 forms droplets in solvent 2, then solvent 1 should be the organic phase. Would the mass transfer be from solvent 2 to solvent 1? Because the aqueous phase would contain compound A.
Edit
Just to see if I have understood it correctly, if I am given the concentration profile below. Then mass transfer would occur from the aqueous phase to the organic phase because the aqueous phase has a higher concentration.

Just keep in mind: the driving force of mass transfer is a concentration gradient. Nature tries to 'equilibriate' a difference in concentration between two phases.

As such, the mass transfer in that sense will be from the phase of high concentration (S2) to the phase of lower concentration (S1).

Two film theory is only a simplification of the mass transport directly at the interface, in which we assume there are two boundary layers on each side of the interface in which the concentration gradient is constant, giving us linear concentration profiles, and the concentration directly at the interface is determined by Henrys Law. This makes dealing with such problems much easier.

Mass and Heat Transfer are often taught in one course as they have many similiarities, but Heat Transfer offers a more intuitive introduction to the principles of transport phenomena in my view:

In heat transfer the driving force is the temperature gradient. We intuitively know that heat transfer will be from the hotter system to the colder system, and formally know that from the 2nd law of thermodynamics. Can you see the similarity to mass transport?


Why does a chiral carbon affect some hydrogen's hydrogen environment? In the screenshot attached taken from Khan Academy's video (https://www.youtube.com/watch?v=p9B4s0N5yk8&list=PL0pJUwkI0YCZRYrgKqHXZ0XCIu11afbJ8&index=3) the thick pink dot refers to a chiral carbon atom. Why is it that the two hydrogens circled red and pink that are bonded to a carbon which is then bonded to a chiral carbon atom are in different hydrogen environments? However, the three green circled hydrogens that are bonded to a carbon which are then bonded to a chiral carbon atom are in the same hydrogen environment?

From my understanding of hydrogen environment, a hydrogen environment is characterised by the functional group(s) that is attached to that hydrogen. Therefore, I think that the two hydrogens circled red and pink are in the same hydrogen environment, just like the three green circled hydrogens are in the same chemical environment.

Therefore, my question is why does a chiral carbon atom affect the hydrogen environment of a hydrogen? And especially from Khan Academy's example, only some hydrogens are affected by the chiral carbon and some aren't affected by it. Is there some loop hole in my understanding of hydrogen environment?

There is a big difference between the two hydrogens and the hydrogens in the two methyl groups.

The issue is that the methyl hydrogens are part of a group that can rotate quickly and so they see an averaged environment which is the same for all the hydrogens. If the bonds were "stuck" and the methyl groups could not rotate your argument might be correct, but the is not the case.

The two hydrogens on the carbon next to the chiral centre are also connected to bonds that can rotate. But they never see the same environment even on average because they will always have a different relationship to the hydrogen and hydroxyl on the next carbon. Whatever way you rotate the bond connecting their carbon, the two hydrogens will be in a different environment, so they appear with distinct shifts.


Why are the masses of atoms less than the sum of their subatomic particles? The mass of carbon-12 is $\pu{12 u}$ by definition. However, one carbon-12 atom comprises 6 neutrons (each weighing $\pu{1.0087 u}$), 6 protons (each weighing $\pu{1.0072 u}$), and 6 electrons (each weighing $\pu{0.0005 u}$), which all add up to $\pu{12.0894 u}$.

Where does the $0.7\%$ difference in mass come from?

This is due to the mass-energy equivalence and a phenomenon called binding energy.
Forming a nucleus releases energy because the nucleons are falling into a potential energy well. Due to Einstein's mass energy equivalence this results in the mass of the new nucleus being less than that of the particles that formed it.
The binding energy of carbon-12 is quoted on Wikipedia as $\pu{92.162 MeV}$. Therefore we can estimate the mass defect of a carbon-12 atom, $\Delta m$, using $ E = (\Delta m)c^2$:
$$\Delta m = \frac{(\pu{92.162 \times 10^6 eV})(\pu{1.6022 \times 10^-19 J eV-1})}{(\pu{2.9979 \times 10^8 m s-1})^2} = \pu{1.6430 \times 10^-28 kg} = \pu{0.098943 u}$$
The difference in the mass of carbon-12 to the mass of its constituent particles is $\pu{0.08940 u}$, so we can see that our calculation is a reasonable estimate of the mass defect. The slight difference is due to other more complicated factors that I have not taken into account, but it still illustrates the main reason for the mass defect.


Looking for a dye which emits around 680 nm For a specific experiment, I'm looking for a dye which would emit around 680 nm. Preferably one I could buy easily commercially. Are there any databases which might give the emission spectra of specific dyes? Or any other methods to find dyes at around $\ce{680 nm}$.

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.


Alkoxide as a leaving group In the reduction of esters to alcohols, the RO group is pushed off. However, the negative charge on the O atom is destabilized by the electron donating R group. Why is the RO group able to function as a leaving group given that the resultant alkoxide is unstable?

Link to reaction mechanism.

The intermediate before forming the alkoxide is a tetrahedral intermediate which is also unstable. You're going from a high energy species to another high energy species.


Experiment to prove that water has air In elementary school, there was an experiment  to show that water has dissolved air, which involved boiling some water and observing the drops at the surface of the container,  and this proved that water has air. Now, in senior high school, I was thinking about the same,  and I think this conclusion is wrong. Aren't the drops formed by the conversion of liquid water to steam due to evaporation? Or is it really the dissolved air that is escaping? What are your thoughts?

There are two ways that I interpret the question you're asking, and I'll try to address each in turn.

If your teacher boiled a container of water and held a piece of glass above it to collect drops of liquid water, he or she may have been trying to show that water doesn't disappear when it boils, that instead it turns to a gas that can move through the air and can be condensed back into a liquid.  Essentially, $\ce{H2O}$ is $\ce{H2O}$, regardless of the state of matter it occupies.
If instead the teacher wanted you to observe the bubbles of gas forming in a container of hot or boiling liquid, he or she might have been on the right track, but either failed to convey the correct message to you or had the wrong idea in the first place.
If I take water from the tap and heat it strongly, bubbles will form on the inside of the vessel I'm heating it in prior to it boiling.  These are bubbles of gas that are no longer soluble because the temperature of the liquid is too high (and its pressure has decreased as compared to the tap.)  The gas is dissolved air (mostly carbon dioxide, I suspect) with a large portion of water vapor.  In any case, they are bubbles of gas that were dissolved but are no longer.  You can see this happen any time you boil a pot of water on the stove.
Once the pot reaches boiling, any bubbles are almost exclusively gaseous water formed via vaporization.
An easier, if slower, way to see this is to run a cold tap until the water is as cold as it will get, then pour a glass.  As it sits (perhaps for an hour or two), bubbles will eventually form on the walls of the glass.  These eventually will dissolve again and the gas will escape through the top of the glass, so it's not something you can set up at night and come back to the next morning.  The bubbles are air that was able to dissolve in the high pressure/low temperature water in the pipes but not in the low pressure/high temperature water in the glass.


Could you estimate the rate of reaction from the chemical structure of the reactants? If so, how? Say you have two compounds with their own unique chemical structures and you know the environment (e.g. temperature, pressure, presence of enzymes) could you then, using this information, approximate the rate of reaction? And if so, how would you? Please provide a mathematical answer, not just use some software.

As Paul J. Gans correctly pointed out in his comment, in general it is impossible.

In practice you can calculate reaction rates only for quite restrictive class of reactions - the unimolecular gas-phase reactions - using RRKM theory.
The whole potential energy surface is actually not needed, as far as I know, you just need to identify reactant, product, and transition state.
RRKM is based on few assumptions, with the first one being quite obvious: since we are talking about potential energy surfaces, the electronic motion and the nuclear motion is separated by the Born-Oppenheimer approximation.
The nuclear motion is also assumed to be well separated into rotational, vibrational, and translational one. The internal degrees of freedom are treated classically with quantum mechanical corrections.

RRKM is quite useful for predicting gas phase dissociation and isomerization rates  under low pressure (low enough so that reaction can be safely assumed to be unimolecular). For that reason, this theory is ofthen used in mass-spectrometry, since the above mentioned conditions are quite often met here, but it is of no use for reactions in the condensed phase.


What is the difference between carbon and charcoal filters? I'm confused about the difference between carbon and charcoal filters.
Ultimately, charcoal contains carbon but I have come across an exam question that asks for their individual roles in a water filter.

I have done some research and found this page that states that 'activated carbon' and 'activated charcoal' can be used interchangeably.

Would any of you be so kind as to help me decipher the individual role of carbon filter and a charcoal filter in an ideal water filter?

First of all and I want to clarify something that it often misunderstood, which is that charcoal has carbon but it is NOT carbon.  Charcoal actually have a bit of residual Hydrogen and oxygen in it and is again NOT carbon.

Carbon is a pure element that occurs naturally in a variety of allotropes.  Charcoal is the charred product from pyrolyzing wood.  Carbon is found as graphite and diamond and has no intrinsic filtering properties.  Charcoal is a porous material that can burn or adsorb pollutants. Activated charcoal is charcoal which has been specially processed to have a very high surface area.  This surface is what pollutants adsorb (not a(b)sorb) to to be remove from air and water.  If you said activated charcoal or activated carbon anyone would know what you were talking about, but activated charcoal is more proper as the material is in fact charcoal, NOT carbon.


Which will be the major monochlorination product? Which of these compounds represents the major monochlorination isomer formed in the following reaction?

According to selectivity theory, the answer should be (c). However, the given answer is (b).

c is the least likely.

Alkane reactivity order is tertiary > secondary > primary > methyl reference here

I would be interested to see an explanation of why b predominates over a


Is polylactic acid (PLA) soluble in ethylacetate? I'm searching for polylactic acid (that used in 3D printing) solvent, that is less toxic. It's used to postprocessing to polish object.

Dichloroethane is toxic, acetone and dichloromethane less, but enough. Tetrahydrofurane and chloroform usage is limited in Russia.

As I know there is D- and L- polylactic acid, I don't know what type is it, but can test using acetone.

Can I use ethylacetate for this purpose? Or there is another solvents?

Sorry about my english

Useful links:
https://www.researchgate.net/post/What_is_the_best_solvent_for_polylactic_acid2/amp
https://www.researchgate.net/post/Solubility_of_PLA/amp
https://groups.google.com/forum/m/#!topic/tvreprapug/1wTmQZJMc9g

D- and L- polylactic acids come from both enantiomers of the monomeric lactic acids.
There will be no difference in solubility between both polymers.
Besides, lactic acid is normally produced by fermentation and the L- form is obtained, so the standard PLA used in 3D printing will be L- PLA.

I have tested the solubility of a piece of PLA and it is not soluble in ethylacetate in one hour.
It does even not seem to be attacked by it.

However, when you rub a 3D printed piece with a paper soaked in ethylacetate, some polishing is attained.

I make basically structural pieces, so I normally do not polish them and therefore I do not know if this is enough for your purpose.


How to determine an oxidation number of an atom in the compound, that has at least two elements, which are not hydrogen or oxygen? Wiki says:

  [Oxidation state] is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules:

  The oxidation state of a free element (uncombined element) is zero
    for a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion.

  Hydrogen has an oxidation state of 1 and oxygen has an oxidation state of −2 when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of −1 in hydrides of active metals, e.g. LiH, and oxygen has an oxidation state of −1 in peroxides, e.g. H2O2
          the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion.

The same is written in my textbook. But how am I supposed to find the ox. number of an atom, which is in compound like $\ce{K2UO4}$?

In this case, the potassium ions are +1 each, and the oxygens are -2 as normal.  That leaves uranium in a highly oxidized +6 state.  The piece of information I think is missing is that Group 1 ions are almost always in the +1 state, because they so readily lose their outer valence electron.  A molecule that is not an ion, like your example, must have an overall charge of zero.


3,4-Methylenedioxyamphetamine (MDA) basic extraction At my lab we perform a basic extraction of seized MDA with $\ce{NaOH}$ $\pu{2M}$ prior to GC/MS analysis.  $\ce{MDA + NaOH -> MDA- + Na+ + H2O}$. This reaction should favor the products since $\mathrm{p}K_\mathrm{a}$ of MDA is about 10 and $\mathrm{p}K_\mathrm{a}$ of $\ce{H2O}$ is about 16. All of the MDA should be converted to its ionic form which is soluble in the aqueous layer. Why is there still enough of MDA in the organic layer ($\ce{CHCl3}$) to be detected by GC/MS?

MDA is an alkaloid, i.e., basic. When you talk about its $\mathrm{p}K_\mathrm{a}$ value, you're really talking about the $\mathrm{p}K_\mathrm{a}$ of the protonated ammonium ion.

The ammonium ion is water soluble, allowing us to potentially separate it from the other organic components of the reaction. After making the aqueous solution basic, the ammonium ion is deprotonated to form the free amine, which is much more soluble in the organic phase than the aqueous phase. Then, it is back extracted into clean organic solvent. This isolates (hopefully) just the free amine in the organic layer during the extraction.

So in answer to your question, what you have in the organic layer is the free amine form of MDA, not the ion.


Parallel first order reaction I am not able to solve this question. Here's my working

Have I done this in the right way? Because my textbook gives answer as 4 moles

Ok, there are two reactions for the 2 moles of A

$$\ce{A ->[rate\space =\space k_1] 2B}$$

$$\ce{A ->[rate\space =\space k_2] 2C}$$

$$\dfrac{k_1}{k_2} = \dfrac{1}{2}$$

The question asks for the total number of moles of A+B+C when the reaction is 75% complete.

At that point 25% of A remains or 0.5 moles of A
It does matter what the ratio of B and C produced is since either reaction of A creates 2 moles of product.  So there is 2*1.5 = 3 moles of product (B+C).

Thus there is 0.5 + 3 = 3.5 moles of A+B+C in the solution when the reaction is 75% complete.

There are 4 moles when the reaction is 100% complete.

So the book answer is wrong.


At what acid concentration does potassium metabisulfite react? I'm in the process of developing a well preserved food-stuff containing potassium metabisulfite and a number of acids (acetic, ascorbic, malic, tartaric, citric, lactic). Wine-making literature suggests that K-meta is added for its antibacterial and antioxidative properties.

However, other literature claims that K-meta reacts with acids. In what concentration does this acid reaction occur? How violent is the reaction, and are any very harmful (if ingested) substances formed?

A simple test shows that the pH of my food-stuff is lower than 4.0, but greater than 3.0 before the addition of K-meta.

Potassium metabisulfite ($\ce{K2S2O5}$) is a white crystalline powder, which also known as potassium pyrosulfite. It is mainly used as an antiseptic for winemaking, process engraving, and as a source for sulfurous acid. It also acts as a potent antioxidant, and hence use as a food preservative (known as E224). For example, potassium metabisulfite is sometimes added to lemon juice as a preservative (Wikipedia). Lemon juice contains citric acid ($\mathrm{p}K_{\mathrm{a1}}= 3.13$) and ascorbic acid (Vitamin C; $\mathrm{p}K_{\mathrm{a1}}= 4.17$) as ingredients, and hence, has $\mathrm{pH}$ ranging from 2 to 3. For that reason alone, your concern about reaction of $\ce{K2S2O5}$ with citric acid or a mixture of acids within $\mathrm{pH}$ 3-4 has no base (a good news).

However, a major reason for addition of potassium metabisulfite ($\ce{K2S2O5}$) as a food preservative is to have sulfur dioxide ($\ce{SO2}$) as an active ingredient. Its melting point is listed as $\pu{190 ^{\circ}C}$, but it decomposes to two compounds:
$$\ce{K2S2O5(s)  -> K2SO3(s) + SO2(g)}$$
It is also in equilibrium with $\ce{K2O}$ and $\ce{SO2}$ when contacted with water (Ref.1):
$$\ce{K2S2O5(aq)  -> K2O(aq) + 2SO2(aq)}$$
In the aqueous media, there is an equilibrium between molecular sulfur dioxide ($\ce{SO2}$; E220) and the anions, bisulfite ($\ce{HSO3-}$) and sulfite ($\ce{SO3^2-}$), which are collectively and conventionally defined as ‘free $\ce{SO2}$’ (Ref.1). The reference also states that:

  The active substance is sulfur dioxide (($\ce{SO2}$, E 220) which in the present application is generated in situ from potassium metabisulfite ($\ce{K2S2O5}$, E 224).

During their investigation:

  The ANS Panel noted that sulfur dioxide, bisulfite and sulfite ions coexist in solution in a series of equilibria and that a read across between the different sulfites used as food additives is possible.

In general, following equilibria exist in aqueous solution of $\ce{S2O5^2-}$ (potassium or sodium):
$$\ce{S2O5^2- + H2O  <=> 2 HSO3-}$$
$$\ce{HSO3- + H2O  <=> 2 SO3^2- + H3O+}$$
$$\ce{HSO3- + H3O+  <=> SO2(H2O) + H2O}$$
Where $\ce{SO2(H2O)}$ is called hydrated $\ce{SO2}$.

A final note: Iowa State University has listed following values as desired 'free $\ce{SO2}$’ by $\mathrm{pH}$ of wine:

$$
\begin{array}{ccc}
\hline\
\text{pH of wine} & '\text{free }\ce{SO2}', \pu{ppm} & & \text{pH of wine} & '\text{free }\ce{SO2}', \pu{ppm} \\\hline
2.9 & 11  &   & 3.5 & 40  \\
3.0 & 13  &   & 3.6 & 50  \\
3.1 & 16  &   & 3.7 & 63  \\
3.2 & 21 &  & 3.8 & 79  \\
3.3 & 26 & & 3.9 & 99 \\
3.4 & 32 & & 4.0 & 125  \\\hline
\end{array}
$$

References:

Claudia Bolognesi, Laurence Castle, et al. (Panel members), EFSA CEF Panel (EFSA Panel on Food Contact Materials, Enzymes, Flavourings and Processing Aids), “Safety assessment of the active substance potassium metabisulfite, for use in active food contact materials,” EFSA Journal 2016, 14(5), 4465, 8 pp. ( doi: 10.2903/j.efsa.2016.4465).


Formation of magnesium bismuthide Magnesium shows an oxidation state of $+2$ and bismuth shows an oxidation state of $+3$. Both do not show a negative oxidation state. Then how is magnesium bismuthide ($\ce{Mg_{3}Bi_{2}}$) formed?

Bismuth can have any oxidation state from +1 to +5 . As it is in the same group as nitrogen and phosphorous, its most commom oxidation states with strongly oxidizing substances such as halogens and oxygen are +3 and +5.

However with a strongly electopositive (reducing) element, bismuth can have an oxidation state of -3, just as nitrogen can. While it is true that the oxidizing strength reduces as we go down the group, the "softness" (tendency to of an element to adapt its oxidation state to the preferred oxidation states of other elements) goes up.

Without definite information I can't discount the possibility that $\ce{Mg3Bi2}$ is an alloy, but the fact that it has an even stoichiometric ratio of elements suggests that it is in fact a compound in which magnesium has an oxidation state of +2 and bismuth has an oxidation state of -3.


Is there a notion of generalized acidity for lithium ions? From my understanding, the basic notions of acidity can be explained by considering a cup of pure water, and then adding a substance to it which creates either free $\ce{H+}$, or free $\ce{H3O+}$ and $\ce{H5O2+}$ etc...
And these ions are free to react and give rise to what we consider acidic behavior. The classic example is hydrogen chloride characterized with the following reaction
$$\ce{HCl +H2O = H3O+ + Cl-} $$
Whereas $\ce{H3O+}$ is our famous hydrochloric acid.
I wonder if it is possible for an analogous reaction to happen with other polar molecules. For example liquid $\ce{Li2O}$ (I believe it melts around 1438 Celsius) should be a polar liquid for the same reasons that $\ce{H2O}$ is (though maybe not as strongly polar), and therefore: $\ce{LiCl}$ should dissolve in molten $\ce{Li2O}$ to create positive charged lithium ion species such as $\ce{Li3O+}$. Which ought to function as an "acid" of sorts, in that they react with most thing which are dissolved/reacted by hydrochloric acid by acting as a positive charge donor (clearly not a proton donor since we are using lithium and not hydrogen to do the deed here)
Is my line of reasoning sound and has this been experimentally detected before?

A complex of lithium ion with lithium hudroxide is known as the salt $\ce{[Li2OH]^+[ClO4]^-}$ [1].  As with the hydronium salt, this requires an essentially nonbasic counterion; with a chloride counterion a double salt is obtained with no evidence of a complex cation.  From the abstract (some evident faults in translation from Russian have been corrected):

An effort is made to obtain salts of Li2OH+ and Li3O+ cations, belonging to a series of simple oxonium compound cations H3O+, LiOH2+, Li2OH+, Li3O+, where proton and lithium replace each other. Binary compounds of lithium hydroxide with lithium chloride and perchlorate are obtained: LiOH·LiClO4, LiOD·LiClO4, LiOH·LiCl, LiOD·LiCl and LiClO4·2LiOH. A conclusion is made by IR absorption spectra, that an adduct with perchlorate can be built of ClO4 anions and complex [Li2OH]+ cations. For adducts with chloride, equivalence of hydroxide and chloride anions with regular alternation of the anions in the crystal lattice is most probable.

Reference
1.
Nikitina, Z. K., Nikitina, N. I., Rosolovskii, V. Y., "Complexing Lithium Salts and Lithium Oxide or Hydroxide: Synthesis and Characterization of a New Cation-Lithiumhydroxonium Li~2OH^+", Russian Journal of Inorganic Chemistry 42(9), 1295-1298  (1997).


Risks from overheated phenolphthalein in green/orange silica gel drying Recently I tried to dry out some indicating silica gel. When I googled, I found advice on line saying that it should be heated to 150 C and that such was a standard procedure, so I popped it into the toaster oven set it to 300 F which is just over 150 C and it started to make a smell and turn black. I stopped of course, cooled it down, and aired out my kitchen. Futher research indicates that the temperature I used is applicable to blue indicating or non-incdicating gel, but the newer Orange/Green gels contains an organic indicator which turns out to be phenolphthalein, rather than Cobalt Chloride and as such should not be heated above 120 C. Basically I followed outdated advice for the older type of product, a classic internet fail.
So now the question is, what have I done? What did I breath? I was cautious, watched the process stopped it early and don't have any outward symptoms, but now I'm curious how dangerous this was.
I've done a lot of searching and found nothing beyond "Toxic oxides of carbon, acrid and irritating fumes" in MSDS sheets regarding what happens when phenolphthalein combusts let alone what might happen as it degrades inside a silca gel bead. MSDS for the beads are just as vague. The lack of info is slightly alarming considering the ease with which a mistake such as mine could be made. Does anyone know what happens when heating breaks down phenolphthalein? There was a smell so something's definitely been produced other than water vapor, and the blackening of the beads might indicate that some carbon probably was left behind.
I understand that a variety of chemicals might be released, but I wonder if anyone is aware of a characterization of the types and relative quantities, or if there are any specific concerns.

I believe you are trying to reactivate moisture absorbing silica gel with a indicator dye. However, you are blaming the internet for the mishap you have faced. Yet, the website you have directed is ScinceDirect, which provides subscription-based access to a large database of scientific and medical research (Producer: Elsevier). Therefore, I find they give wrongful information on scientific research (considering almost all of these papers are peer-reviewed). I think you made a mistake where considering the indicator dye in green/orange silica gel desiccants is phenolphthalein. My conclusion is based on the inventors' remark in their patent application (Ref.1), which states that color change is from orange (when the gel is dry) to creamy white (when the gel is saturated with moisture). My best guess is it contains something different. Even with something else as an indicator dye, it should be able to stand at $\pu{150 ^\circ C}$ temperature in order to be able to reactivated. What that suggest is, your silica gel might be absorbed unknown chemical without your knowledge, which must be thermally unstable.
According to the patent, there are three kind of materials they are suggesting as indicator dyes: (1) phenolphthalein (color change Orange-red to creamy white); (2) thymolphthalein (color change dark-brown to creamy white); and (3) mixture of phenolphthalein/thymolphthalein (color change Orange-is not provided). All three types are able to get reactivated under $\pu{150-250 ^\circ C}$ temperature range (Ref.1).
References:

Waclaw Lanowy, Krystian Lukaszczyk, Przemyslaw Kasprzycki, “Silica gel with moisture indicator, a method of obtaining them, and use of phthalein dyes,” International Patent 2006, WO2006/130029 A2 (PDF).


How can the free amino acid B be enantioslectively synthesitzed? I would like to know how this reaction actually works. I think its an ring metathesis, espececially with the Grubbs catalysator II. But I dont know how to this reaction examole really works especially with respect to enantioselectivity.
May anyone help?

I'm guessing this is a homework type question. Yet, I have some concern of this being a exam question, probably one already taken. So I give a step by step answer depicted in following diagram (courtesy of https://web.chemdoodle.com/demos/sketcher/):


How do you find the average rate and instantaneous rate given a kinetics plot? What is average rate? What is instantaneous rate? How do you find these? My teacher described finding the tangent to a graph. I am not sure what is meant by this.

This is more of a question of calculus than chemistry, but in a first order reaction
$\ce{A -> B}$, the average rate is:

$$\frac{\Delta A}{\Delta t} = \frac{A_2 - A_1}{t_2 - t_1}$$

 Graciously borrowed from here

So, the average rate is the slope of a secant line between the two points,

$$ \frac {\Delta A}{\Delta t} = \frac{1 - 0.5}{400 - 200}$$

The instantaneous rate is the derivative of the plot:

$$\frac{dA(t)}{dt}$$ or $$\lim_{t\rightarrow 0} \frac{\Delta A}{\Delta t}$$

For the above first order rate law, the equation is in the form:

$$ A(t) = e^{-kt + ln(A_0)} $$ so the derivative is simply

$$ \frac{dA(t)}{dt} = -ke^{-kt + ln(A_0)} $$


Ring expansion in cyclic compounds My attempt

In the first case:
$\ce{H+}$ adds to the $\ce{OH}$ group, giving us a carbocation. The carbocation thus formed is exceptionally stable due to back bonding. I wonder why would it go under ring expansion even though the strain is not a factor here as the ring strain in a cyclobutane ring is ~$26.3\ \mathrm{kcal/mol}$, and that in a cyclopropane ring is ~$27.5\ \mathrm{kcal/mol}$.

In the second case:
Again the $\ce{H+}$ adds to the $\ce{OH}$ group, giving us a tertiary carbocation with seven hyper-conjugating structures. Why would it go under ring expansion to give secondary carbocation with just two hyper-conjugating structures? I believe is based on ring strain in this case, as the ring strain in a five-membered ring is ~$6.2\ \mathrm{kcal/mol}$, while the ring strain in a six-membered ring is ~$0.1\ \mathrm{kcal/mol}$.

Source: Advanced Problems In Organic Chemistry, MS Chouhan, 11th edition; Chapter - Hydrocarbons (Alkenes); Question 180 in latest edition

You overlooked the strain from the sp2 carbon in a small ring. The ring strain you have listed is based on sp3 carbon. sp2 supposed to have a planar triangular geometry (120°, compared to 109° for sp3) and will be more strained in a 3 or 4 membered ring. 6 membered ring will be best followed by 5 membered ring.

Also, you have discussed a lot about the stability of carboncations. Acid catalyzed elimination/rearrangement reaction is a thermodynamic condition. The stability of product is more important than the stability of intermediate.


Calculating the solubility of cuprous iodide I am clearly missing something obvious here. I get the $\mathrm{K_{sp}}$ from the CRC for $\ce{CuI}$ as $1.27\cdot10^{-12}$.

So, $\ce{CuI-> Cu+  +  I-}$

So the formula ought to be

$$\mathrm{K_{sp}} = \ce{[Cu+][I-]} =[s][s]= s^2$$

Doing the square root and all gives me a solubility of 2.1 mg/L. Hooray, except that everywhere I look, if the solubility is listed at all, it lists 4.2, exactly double my answer.

I have been wracking my brain on this one. It has to be a scalar but I don’t see how you would introduce it without throwing the entire calculation off by turning the square into a cube or something similar. What am I missing?

Edit: updated to reflect the full Ksp value in the CRC which I was using for my calculations (I had accidentally truncated it to 1 rather than 1.27 when posting this). I am using the online CRC from hbcponline.com, so this is current, not an old edition.

First of all, as MaxW pointed out in the above comment, your math seems to be wrong. From the the $K_\mathrm{sp}$ value given in your question:
$$s = \sqrt{K_\mathrm{sp}}= \sqrt{1.0 \times 10^{-12}}= \pu{1.0 \times 10^{-6} mol/L}= \pu{1.0 \times 10^{-6} mol/L} \times \frac{\pu{190.45 g}}{\pu{1 mol}}\times \frac{\pu{1000 mg}}{\pu{1 g}} \\ = \pu{0.190 mg/L}$$

Other than that, there seems to be a problem with online value of $K_\mathrm{sp}$ of $\ce{CuI}$. Some sites listed it as $1.1 \times 10^{-12}$ while other sites listed as $5.3 \times 10^{-12}$. However, I'd like the value $5.3 \times 10^{-12}$, because at least it's given in one of reliable sources (Ref.1).

If you use this value (instead of $1.1 \times 10^{-12}$), you'd get:

$$s = \sqrt{K_\mathrm{sp}}= \sqrt{5.3 \times 10^{-12}}= \pu{2.30 \times 10^{-6} mol/L}= \pu{2.30 \times 10^{-6} mol/L} \times \frac{\pu{190.45 g}}{\pu{1 mol}}\times \frac{\pu{1000 mg}}{\pu{1 g}} \\ = \pu{0.438 mg/L}$$

This value is in a good agreement with the literature value ($\pu{0.00042 g/L}$ at $\pu{25 ^\circ C}$; Ref.2), which listed the properties of $\ce{CuI}$ as:

  Copper (I) iodide is a dense, pure white solid, crystallizing with a zinc-blende structure below $300^\circ$. It is less sensitive to light than either the chloride or bromide, although passage of air over the solid at room temperature in daylight for $\pu{3 hr}$ results in the liberation of a small amount of iodine. It melts at $588^\circ$, boils at $1293^\circ$, and, unlike the other copper halides, is not associated in the vapor state. Being extremely insoluble ($\pu{0.00042 g/L}$ at $25^\circ$), it is not perceptibly decomposed by water. It is insoluble in dilute acids but dissolves in aqueous solutions of ammonia, potassium iodide, potassium cyanide, and sodium thiosulfate. It is decomposed by concentrated sulfuric and nitric acids.

References:

Steven S. Zumdahl, Donald J. DeCoste, In Chemical Principles; 8th Edition; Cengage Learning: Boston, MA, 2017, “Chapter 8: Applications of Aqueous Equilibria,” pp. 241–298 ($K_\mathrm{sp}$ of value $5.3 \times 10^{-12}$ is given for $\ce{CuI}$ in example 8.13, page 288).
George B. Kauffman, Lawrence Y. Fang (Checked by N. Viswanathan, G. Townsend), “Chapter 2: Transition metal Complexes and Compounds – 20. Purification of Copper (I) Iodide,” In Inorganic Syntheses: Volume 22; Smith L. Holt Jr., Ed.; John Wiley & Sons: New York, NY, 1983, pp. 103–107 (https://doi.org/10.1002/9780470132531.ch20).


Why can't C-C double bonds rotate? The alkyl chains of saturated fat molecules can readily rotate about their carbon-carbon single bonds, giving them flexibility and viscosity.

What is different about the carbon-carbon double bonds of unsaturated fat molecules, preventing inter-conversion of cis and trans forms?

A double bond between 2 carbon atoms means there is a $\pi$ bond with the side orbital overlaps in 1 plane, in parallel to the axial $\sigma$  bond.

The consequence is, free rotation  would need to break this $\pi$ bond, making the double bond just a single bond, what would require a strong torque and a lot of energy.

By other words, particular rotation positions around a single bond have ( almost ) the same energies. But for double  bonds, the molecule energy strongly depends on on this orientation. The $\pi$ bond can exist, only if the bystander groups on both carbon atoms   are about in the same plane. If they wete on about perpendicular planes, the  orbitals of both atoms cannot overlap to form the bond.


Boiling and melting point of trans- and cis-but-2-ene The boiling point of trans-but-2-ene is lower than that of its cis isomer but the melting point of the former is higher than the later. Why is it not following the same order? Is there any factor of structural symmetry coming playing a role here?

Yes . You are right that structural symmetry comes into play .

Boiling point depends upon intermolecular interactions which over here is more in cis due to its net dipole moment .  The dipole moment enables electronic interactions which hold molecules together .  This shows some general factors of boiling point . Also the below link to the google book covers the structural aspect of a molecule towards b.p on pg47.

For melting points , let me quote Joseph Hornback from Chapter 2.6 Organic Chemistry pg46 (google books) :

  The shape of a compound is very important and has a dramatic effect on its melting point . More symmetrical molecules pack into the crystal lattice better , allowing closer approach and larger attractive forces , resulting in higher melting points .

Melting points of alkenes depends on the packaging of the molecules. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers.(source)

Trans 2-Butene has more symmetry than its cis isomer which results in better packing and hence higher melting point .

For further reading I suggest you visit 'Molecular Shape' over here.


Effect of substituents on Ullmann reaction This question that I recently encountered asked what the most suitable substrate for the Ullmann reaction would be. The answer given is (b) but I think it has to be (d) as the nitro group acts as EWG and thus facilitates the reaction.

However I do not know the mechanism and thus am not sure if (d) is indeed the correct answer.

Any reason why the answer might be (b) and not (d)?

I think that the answer given in your textbook is wrong as they are trying to conduct ULLMANN reaction on a chloride a strong EWD needs to be present at the ortho/para positions for the reaction to be feasible.

There Are generally 2 Mechanisms Proposed for the reaction :

Free Radical Mechanism
Ionic Mechanism

  FREE RADICAL MECHANISM
  The Copper Metal (used as catalyst) Generate a Aryl Radical By transferring a electron, and then 2 Aryl Radical combine to form the final product.

  IONIC MECHANISM
  Acording to this a Organometalic intermideate of Aryl copper halide is fromed. The copper is added to the Aryl halide by Oxidative addition. then a electron is transfered to the intermidiate to for a organocuppric radical which again perform the oxidative addition on another aryl halide which result in formation of a biaryl copper complex which in turn undergo reductive elimination and make the final product.

You can easily see that we need a EWG to make the Reagent more reactive thus making the reaction feasible thus your answer must be either wrong or you might have checked it wrong. nevethless I also find this paper affirming mine and your belief.

Source:

https://link.springer.com/article/10.1007/s40828-016-0036-2
www.name-reaction.com/ullmann-reaction


What is the role of various additives in a titration of vitamin C with N-bromosuccinimide In a titration to determine the amount of vitamin C in an orange juice sample, the following were mixed with water:

oxalic acid
potassium iodide
starch
acetic acid

Then, the mixture was titrated with a solution of N-bromosuccinimide until the solution turned dark as the starch/iodine/iodide inclusion complex formed.

$$\ce{2I- + NBSucc + H+ -> I2 + Succ + Br-}$$

where NBSucc is N-bromosuccinimide and Succ is succinimide.

Source: https://media.cheggcdn.com/media%2Fd2a%2Fd2a428e0-d1b7-4cf1-bcf9-fb81813660b4%2Fphply3xn7.png

What is the role of the two different acids (acetic acid and oxalic acid) in the titration? How does the outcome of the titration depend on the concentration of potassium iodide?

Thanks, Martin, for pointing out the relevant paper with all the answers!
Basic procedure
This titration was first described by Mohamed Zakibarakat, Mohamed Fathy Abd El-Wahab, and Mohamed Mahmoudel-Sadrin in 1954 (doi.org/10.1021/ac60100a013) as an alternative to using 2,6-dichlorophenol-indophenol as titrant. Here is the procedure for a solution of pure acsorbic acid:

Into a 50-m1. conical Erlenmeyer flask, a known volume of the unknown ascorbic acid solution is introduced — i.e., 1 ml. contains 1 mg. of ascorbic acid. Then 5 ml. of 4% potassium iodide solution, 2 ml. of 3% acetic acid, and 2 drops of starch solution as an indicator are added. The aqueous N-bromosuccinimide solution (0.1% weight per volume) is introduced into the microburet and is allowed to run drop by drop, into the ascorbic acid solution, with continuous shaking. The end point is reached when the last drop of the N-bromosuccinimide solution added produces a permanent blue color in the ascorbic acid solution.

For fresh fruit, there is an additional step to prepare the analyte:

Edible Fruits. The new method has been applied to the estimation of ascorbic acid in certain edible Egyptian fruits known to contain ascorbic acid—e.g., oranges, lemons, tomatoes, grapes, watermelons, and water cress. The fruits used in the estimation of ascorbic acid were almost ripe. The sampling and extraction of the material under examination must be carried out with the minimum delay, so that no significant change in ascorbic acid content takes place prior to analysis. Each sample is squeezed or bruised and the juice is directly received in a known volume of 20% trichloroacetic acid, so that the juice is diluted with the stabilizing acid at least twice, to avoid the oxidation of ascorbic acid and also to precipitate interfering substances such as protein and thereby facilitate subsequent clarification of the extract. The acidulated juice is fil-tered and the determination is carried out on a known volume of the clear filtrate as described previously. The N-bromo-succinimide solution used in the titration is 0.01%—i.e., 1 ml. contains 0.1 mg. of N-bromosuccinimide (see Tables XI and XII).

Oxalic acid
Note that they did not mention oxalic acid, but they did use fairly acidic conditions to inhibit redox enzymes and precipitate proteins.
Oxalic acid does not interfere in the determination:

Interfering Substances. The following substances which might interfere have no influence on the titration process: carbohydrates such as glucose, lactose, sucrose, and starch; diketogulonic acid; reductones and reductic acid; urea; uric acid and creatinine; alcohols; formaldehyde; acetone; ethyl acetate and acetoacetic ester; thiamine hydrochloride and ribo-flavine; oxalates, tartrates, and citrates; amino acids such as glycine, alanine, valine, and isoleucine; and ferrous and ferric salts. The only interfering substances that are also oxidized by N-bromosuccinimide before iodine is liberated from potassium iodide in acetic acid medium include sodium sulfite, sulfide, thiosulf ate, and thiourea.

Role of iodide
They nicely describe the two relevant redox reactions used in the procedure:

The fact that ascorbic acid reacts very rapidly with N-bromo-succinimide, whereas many of the interfering substances react more slowly or even do not react at all, provides a reliable titrimetric method for the determination of ascorbic acid. N-Bromosuccinimide is an oxidizing agent and thus can liberate iodine from potassium iodide in aqueous acetic acid medium, but it oxidizes ascorbic acid to dehydroascorbic acid preferentially. Until all the ascorbic acid present in the solution is oxidized, no iodine is liberated from potassium iodide.
The slightest excess of N-bromosuccinimide added, after all the ascorbic acid content has been oxidized, will liberate iodine from potassium iodide, which is easily detected by the blue color developed with a few drops of starch solution added at the beginning of the titration process. The end point is thus easily observed by the appearance of a blue color in the ascorbic acid solution or extract. The presence of other reducing substances does not interfere with the titration with N-bromosuccinimide, since iodine is selec-tively liberated from potassium iodide before reducing substances present, other than ascorbic acid, are oxidized by N-bromosuccinimide, and consequently the end point is easily determined in the presence of starch. The end point is definitely blue in the case of pure solutions and pharmaceuticals, but violet in the case of biologicals and fruits.

So it would seem that as long as there is sufficient iodide to give the color change for the endpoint, the exact amount does not change when the endpoint is observed.


Do IUPAC names for haloalkanes require numerals? I'm new one of chemistry. I want to know the IUPAC nomenclature rules. That is if we write alkane with halogen must be indicate the numerals? Are the following names right or wrong?

chloropentane
trichloropentane

You must use the numbers unless your substance in a mixture of isomers; otherwise your molecules structure is ambiguous.  For chloropentane this name alone does not specify weather the chlorine is on the 1,2,or 3 position which is important as the position of a chlorine can dictated physical and chemical properties as well as reactivity in various organic reactions.  Same is true for trichloropentane.  there are there chlorine but without numbers I dont know if the chlorines are all on one carbon, if they are in the 2,3,4; 1,3,5; 1,2,5; 1,2,3; 2,3,5; 1,1,2; 1,3,3 1,1,4 positions or any of the many more possibilities.  all of these molecules give different products in various organic reactions.

Again though if is a mixture of isomers (as is typical of hexane solvents) it is not important to give number


Reaction of styrene oxide with hydrogen iodide The question is to find the product of the reaction of Styrene Oxide with Hydrogen Iodide.
According to me, this should be the mechanism-

As the benzylic carbocation will be more stable, therefore the product made from A should be major and the other one should be minor. But, in my book the answer to this question is the opposite of mine. Where did I go wrong?

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

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

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


Why does calcium carbonate decompose into calcium oxide? Why does calcium carbonate, when heated strongly, decompose into calcium oxide and carbon dioxide?

What is the mechanism? Can someone illustrate the means by which this happens? What bonds/associations are broken?

I'm guessing that one of the calcium ion has a strong intermolecular association with one of the partially negative oxygens, and this changes the bond character of the carbonate ion. Upon applying heat, this bond remains while carbon dioxide is off-gassed.

In other words, the positive charge on the calcium ion isolates the electron density to the nearest oxygen atom on the carbonate ion. This isolation of electron density away from the carbon and toward the calcium weakens the oxygen-carbon bond while strengthening the calcium ion-oxygen intermolecular association force.

Some quick googling did not turn up a definitive reaction mechanism, and most of the papers I did find seemed to instead be focused on the overall reaction rate and controlling factors - probably because this is such an industrially important reaction and (at least according to this article) gas transport is the rate-limiting factor - this is common for heterogeneous reactions.

However, it is not too hard to propose a plausible mechanism. I would suggest the same mechanism you did - what is likely happening is that the oxygen near one calcium ion can pull some electron density away from the carbon, leaving $\ce{CO2}$ which diffuses away as a gas. The calcium ion stabilizes the new oxygen anion and forms a new, different lattice structure.

As @Martin points out, the reason that this process is thermodynamically favorable is that the products have a higher entropy. The standard Gibbs free energy of reaction (using data from Tro's Chemistry) is:

$$
\begin{align}
\Delta G &= \Delta H - T\Delta S \\
&= 179.2\space kJ/mol - T(0.1602\space J/mol*K)
\end{align}
$$
Note that $\Delta H$ is positive, but so is $\Delta S$.
At 298 K (room temp) $\Delta G = 131.3 \space kJ/mol$ - the reaction is not favorable.

The equilibrium temperature (ignoring temperature corrections) should be around 1120 K.


Will an acyl chloride be attacked by water molecule to form carboxylic acid? $\ce{H2O}$ molecules can act as a nucleophile. So it should attack the carbonyl carbon. However, I think that that when the $\ce{O-}$ "kicks back" it will once again remove the water molecule keeping the $\ce{Cl}$ intact as water is definitely a better leaving group. But in my book they showed a reaction intermediate, which involved an acyl chloride and they used water molecule as a nucleophile to convert acyl chloride to carboxylic acid. I'm confused as to how that is possible? Isn't water a better leaving group than $\ce{Cl-}$ ?

You need to lose a proton, then chloride becomes a better leaving group than hydroxide. In general, proton transfers are incredibly fast; in this scenario there is tons of water around to accept the proton, and the tetrahedral intermediate with a positively charged oxygen is fairly acidic anyway.

Mechanism from Clayden 2nd ed. (p 199).

In the absence of an external base (e.g. pyridine), water will act as the base itself here. Another way of approaching it is to note that hydrolysis of an acyl chloride will release one equivalent of HCl:

$$\ce{RCO\color{red}{Cl} + \color{red}{H}OH -> RCOOH + \color{red}{HCl}}$$

so in the absence of a second base, you will necessarily have to produce $\ce{H3O+}$.


Why does delocalization (only) occur in molecules represented by resonance hybrids? Before I start this question, I am aware tha

Electrons in all molecules are delocalized to some extent
Delocalization is a the phenomenon and resonance is an attempt to explain it.

When I asked this question from my professor, this is the explanation he gave me:

In VBT, it is difficult to approximate the molecular orbital with
localized bonding orbitals, so the electronic configuration is a
weighted average of the infinite possible electronic configurations
i.e Ψ= Aψ1 +Bψ2+Cψ3......
For most molecules, the most stable lewis structure has the majority
of the contribution and can sufficiently approximate the actual
electronic by itself. However, for some molecules, there are multiple
equally-stable lewis structures with an equal and significant
contribution and each of them has to be taken into account. For e.g,
for nitrate ion, Ψ=1/3 (ψ1 +ψ2 +ψ3)  where ψ1,2,3  represent the 3
equal lewis structures. As a result, the electrons MUST be delocalised
between the appropriate bonds

This explanation, in simple words, argues that  since the 3 lewis strucutes are identical/indistinguishable, the electron density must be equal and thus delocalised between the bonds by symmetry.
However, this doesn't really explain the physical reasoning behind delocalization. Why is (significant) delocalization more energetically stable for some molecules but not for others? And why dpes delocalization always occur in molecules with some kind of symmetry between the possible atoms for a double/triple bond?

A short answer is that we mostly use resonance structures to represent delocalization when the bond orders or formal charges predicted by VBT are not approximately integers, simply because non-integer bond orders and charges are more cumbersome to draw. A typical example is a carboxylate, which we could draw with 1.5 bonds between both C-O pairs and -0.5 charge on each O, but it is simpler to just draw a resonance structure with one C=O double bond and one C-O single bond and a full -1 charge on one oxygen.
As others have noted, electrons are still delocalized across the entire molecule in molecules for which we do not use resonance structures. The difference is that we can approximate the bond orders in those cases as integers, so only one structure is needed to represent that.


Why is the letter J omitted in the spdf... sequence? Under spherical symmetry, the irreducible representations corresponding to $L = 0, 1, 2, \cdots$ are assigned the letters $\mathrm{S}, \mathrm{P}, \mathrm{D}, \mathrm{F}, \cdots$ after which the letters progress in alphabetical order. (The familiar names of the atomic orbitals are also labelled with this sequence of letters.) Therefore, we have:

$$\mathrm{S}, \mathrm{P}, \mathrm{D}, \mathrm{F}, \mathrm{G}, \mathrm{H}, \mathrm{I}, \mathrm{K}, \cdots$$

but as far as I can tell, $\mathrm{J}$ is conspicuously omitted. I am guessing there is a reason behind this - what is it? Is it to avoid confusion with some other $J$, like the total angular momentum quantum number $J$, or is it perhaps based on some typographical argument (J being easily confused with I)?

Omitting j when alphabetically enumerating things has a long tradition.

First of all, the alphabet did not always exist in the form we know it today.
Quoting Wikipedia:

After [...] the 1st century BC, Latin adopted the Greek letters ⟨Y⟩ and ⟨Z⟩ [...] Thus it was during the classical Latin period that the Latin alphabet contained 23 letters:
[no J, V, W]
[...]
It was not until the Middle Ages that the letter ⟨W⟩ [...] was added
[...]
only after the Renaissance did the convention of treating ⟨I⟩ and ⟨U⟩ as vowels, and ⟨J⟩ and ⟨V⟩ as consonants, become established. Prior to that, the former had been merely allographs of the latter.

In some books, this has consequences till today.

The footnotes to the Confession and Catechisms [of the Presbyterian Church], containing the proof texts, are enumerated in the traditional manner, that is, by letters of the alphabet (omitting j and v, as alternative forms for i and u in the Latin alphabet). Source

But even later, long after the letters I and J were considered distinct in terms of proper spelling, their alphabetical order (I preceding J) was not firmly established.
The New General English Dictionary of 1768 had a combined section for I and J, treating both equal with respect to alphabetical order (but not regarding spelling).
Same with Handwörterbuch der allgemeinen Chemie, a German chemistry book printed in 1818.
The latest book, I could find, is the Handwörterbuch der reinen und angewandten Chemie of 1850. This is pretty close to when the letters spdefghik.. must have been defined! (Does anyone know when exactly this was?)
And now, it perfectly makes sense to me. Even if i and j were distinct letters at the time and their order should have been commonly established in the early 20th century, the possibility that some readers could still be confused about what comes first, must have lead to the decision to leave j out.

Update (2017-12)
Concerning the spelling and order of names, I was able to find evidence that is even 100 years younger.
The Berlin telephone book contained a spelling table. The first book of 1890 contained a spelling table that assigned numbers to each letter, omitting the letter J, i.e. I=9 and K=10. That is pretty similar to our orbital labels, isn't it? In the 1903 printing, words were assigned to the letters, but J was still left out. The 1905 printing was the first to include J into the spelling table. (source)
Even the Berlin address book of 1943 did not distinguish between I and J. For instance Jutta is listed before Iwanski. (Interestingly, this book doesn't even use different glyphs for I and J in the Fraktur font.)
Of course, there are many books around that time, and earlier, that sort I before J and that use different glyphs, even if printed in Fraktur (example). Nevertheless, this shows that the convention "I before J", as we know it today, was not firmly established in early 20th century Germany.


Esterification of a carboxylic acid and the production of water In an esterification reaction between a carboxylic acid and an alcohol, a carboxylate ester and water is produced.

I found the following diagram online, and my textbook has a similar one as well

Question: In forming the water in an esterification of a carboxylic acid, why does the hydrogen ion come from the alcohol and the hydroxide come from the carboxylic acid and not the other way around?

I think it might be because carbon's bond to the OH- is weaker than the R bond to the H+ ion and requires less energy to remove those two ions from their respective compounds.

The mechanism of the reaction doesn't involve the direct exchange of ions in the way that, e.g., a metathesis/double replacement reaction might. Take a careful look at the reaction mechanism from the Wikipedia page. Notice that one of the intermediates (the product after step three) is a tetrahedral complex with two hydroxyl ($\ce{-OH}$) groups. One of those hydroxyls is indeed the original hydroxyl of the carboxylic acid moiety, but the other one is actually formed from the protonated oxygen of the carbonyl. Either one of them could, with equal likelihood, be protonated in a subsequent step to provide the water molecule that serves as a leaving group. In the image that you linked, it's convenient to circle those specific atoms, but that shouldn't lead you to believe that the mechanism of the reaction is in fact a simple exchange of ions.

The question that you didn't ask directly, but perhaps implied, is: why is protonation of the alcohol by the carboxylic acid, followed by nucleophilic attack by the carboxylate on the carbon of the protonated alcohol, not a viable mechanism? First, a protonated alcohol is more acidic than a carboxylic acid by six or seven orders of magnitude, which means that the initial proton transfer would be slow and unfavorable. Second, the resulting ester is apt to be hydrolyzed eventually (albeit slowly without catalysis, but then the whole reaction proceeds very slowly), yielding the original carboxylic acid and alcohol.


Why my boiling pot turned black? Recently I ran out of dish washing detergent, so I decided to pick up one of the recipes online, to make my own "ecological" dish washing detergent alternative.

The recipe was:

Mix ingredients:

400ml of water
200mg of kitchen salt
100ml of vinegar
juice squeezed from 3 lemons

boil for 10 minutes

After I stopped boiling and left the mixture to cool down, the fluid turned dark gray. Bottom of my stainless steel pot turned black. Here is a photo of it (after I did some scrubbing, but you can still see black on the side):

What happened? (What chemical reaction?) Additionally, are any of the formed substances known to be toxic?

You used a salt/acid mixture and, upon bringing it to a boil to clean a stainless-steel pot, parts turned black.
All stainless steel is not alike, and all are not “stainless” or non-reactive towards oxidation – given the circumstances.
Getting to your specific question:
The “black” you are seeing upon boiling is a Fe(II)/Fe(III) oxide, the same or similar to the black oxidative protection [the seasoning of a skillet] so valued by the fastidious owners of iron skillets.
One of many procedures for “seasoning” an iron skillet is similar to what you did.  I have done the following A) cleaning the skillet down to shinny metal, particularly important if you do not know the history of the skillet. B) rub a mixture of table salt dampened with an acid [vinegar] over the pan until you get the classic yellow-rust of iron we are so familiar with C) now add water to the skillet and bring it to a boil.  This step is critical.  The yellow rust will turn black upon boiling.  This is a protective Fe(II)/Fe(III) oxide coating.
Now repeat with the salt/vinegar rub and boil until you get a uniform blackish coating on the iron skillet.    Finally, fry up some bacon and eggs to give a carbon-based coating to the skillet.  The latter is more of a “paint” but with the black metal oxide coating as a base, your skillet is protected against “rust” and mild oxidation.
The “original Teflon ® coating” if you will.


Thermochemistry, calculating T given Q I have the reaction $\ce{2COCl2(g) -> CCl4(l) + CO2(g)}$

And I want to calculate $T$ (temperature) when $Q = 10^{-4}$ at equilibrium.

I tried simply putting $$T = -\frac{\Delta G^{0}}{\Delta S^{0}-R \ln Q}$$

Anyway, since $\Delta G^{0}$ for the reaction ($\pu{42000J/mol}$) is the value at standard condition (25 degrees celsius) I use that $\Delta G = \Delta H^{0} - T\times\Delta S^{0} + RT\ln Q$ where $\Delta G = 0$ since we're at equilibrium.

From this I obtain that $$T = \frac{\Delta H^{0}}{\Delta S^{0}-R\ln Q}$$

But which $\Delta H^{0}$ and $\Delta S^{0}$ values do I use here? I can calculate them from the standard conditions at 25 degrees celsius using SI, but that wouldnt be correct, would it? I'm rather confused by this task, which seems simple at first sight. Hope someone can clarify.

(The answer should be $\pu{476K}$)

We know that, at temperature T, $$\Delta G^0(T)=\Delta H^0(T)-T\Delta S^0(T)$$
From Hess's Law, we also know that
$$\Delta H^0(T)=\Delta H^0(298)+???$$
and
$$\Delta S^0(T)=\Delta S^0(298)+???$$

That should be a good enough hint.


Reacting carbon dioxide with magnesium in ribbon form My experiment involves a reaction of magnesium in ribbon form with that of carbon dioxide. I know that the product shall be magnesium oxide (white powder) and carbon (black powder) but I do not know what kind of carbon it actually is (allotrope, purity, etc.)

It would also help if you can suggest a method for separating the carbon from the magnesium oxide after the reaction.

The reaction you are working with is:

$$\ce{2 Mg(s) + CO2(g) -> 2 MgO(s) + C(s)}$$

where the carbon should be primarily graphitic.

Because you are producing two solids in this reaction, they will be physically intermixed and thus the purity of either solid will be poor, being contaminated with the other solid.

The separation of carbon from $\ce{MgO}$ is a fairly straight forward process, however. Upon contact with water, $\ce{MgO}$ becomes the basic $\ce{Mg(OH)2}$, which is readily soluble in dilute hydrochloric acid. Once the $\ce{MgO}$ has been dissolved in acid, you can filter off the acid-insoluble carbon. After the hydrochloric acid has been evaporated off, $\ce{MgO}$ will re-form from the $\ce{Mg(OH)2}$.


What is the difference in (alc) and (aq) states in reaction mechanism? In these reactions between alkyl halide and KOH in different states the product is different only because of the change in state of KOH from aqueous and alcohol .

Can someone explain the mechanism which causes this drastic change?

Nucleophilic substitution always competes with Elimination reaction in such cases.

What is the difference between them?

As put there by philipp:

  In alcoholic solution the $\ce{KOH}$ is basic enough ($\mathrm{pK}_{\mathrm{a}}$ 15.74) to deprotonate a small amount of alcohol molecules ($\mathrm{pK}_{\mathrm{a}}$ 16–17)), thus forming alcohlate salts $\ce{ROK}$. The alcoholate anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the group $R$). The higher bulkiness makes $\ce{RO-}$ a worse nucleophile than $\ce{OH-}$ and the higher basicity makes it better at $\mathrm{E}2$ eliminations.

Mechanisms below:


Use a solid or liquid stationary phase in gas chromatography In my textbook, they mention that the stationary phase used in gas chromatography is a solid, such as silica or an adsorbed high-boiling-point liquid.

My question is that what is the advantage of each one, in what situation one will be better than the other, or if it matters at all.

Also, how does the stationary phase not get carried away with the mobile phase? since a high-pressure gas is used doesn't the stationary phase just get carried along with the mobile phase, especially if it is liquid?

Solid adsorbents in gas chromatography were used about 30-40 years ago. They are obsolete now. The efficiency of solids (or packed column GC) was much lower than coated columns. Most modern GC stationary phases are coated or sometimes chemically bonded  to the silica surface inside the capillary.

  how does the stationary phase not get carried away with the mobile
  phase? since a high-pressure gas is used doesn't the stationary phase
  just get carried along with the mobile phase, especially if it is
  liquid?

The viscosity of the coating is so high that it does not get carried away. Imagine a very viscous wax like substance stuck on a glassy surface. Smear it very thinly. The coating is now a few microns thick. Try blowing away thick waxy substance from a glass surface, can you ever get rid of it?

You may think that the pressure is high, but recall that Pressure= Force/Area. If the capillary area (i.e. cross section) is so small, the force is very small too. Capillaries diameters are on the order of microns. That is why you can use a silica capillary in GC. If we had used a wide bore capillary say 4 mm i.d., and used the same pressure with the same wall thickness, it would have shattered.


Conceptualizing Multiple Equilibria Take the reaction:
$$ (1) \; \ce{A + B <=>X + Y}, \; K_1=\frac{[X][Y]}{[A][B]} $$
Now assume that X isomerizes to Z according to:
$$ (2) \; \ce{X <=> Z}, \; K_2=\frac{[Z]}{[X]} $$
We can then add the equations to obtain:
$$ (3) \; \ce{A + B <=> Z + Y}, \; K_3=K_1K_2=\frac{[Z][Y]}{[A][B]} $$
It's pretty easy to show that $ K_3=K_1K_2 $ when considering the ratio of products and reactants, but this makes less sense when considering the actual numeric value of $K_3$. The fact that $ K_3=K_1K_2 $ can be explained by the statement of multiple equilibria: all equilibria must be satisfied simultaneously. This is what I'm having trouble conceptualizing. Does the equilibrium constant in $(1)$ already take into account $(2)$, and vice versa? Equation $(3)$ will reach a state of maximum entropy in which $K_3$ applies. This must also occur when both $K_2$ and $K_3$ are satisfied? I can do the math all day, but I'm not quite getting the link between the three equilibrium constants.

When there is a situation of multiple equilibria, all of them must be satisfied. If this is the case, the whole system will be at equilibrium.

For your example and as you report, the maths let you write that

$
K_{3}=K_{1}\times K_{2} \quad (1)
$

This final equation, beyond being the mere mathematical combination of two initial equations, is obtained by considering that Gibbs free energy is a function of state: iIt describes "quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state" (Wikipedia). One may get to equilibrium via an unique straight step, or through several steps. But the final destination will be the same.

This definition allows us to write:

$
\Delta G_{3}^{o}=\Delta G_{1}^{o} + \Delta G_{2}^{o}  \quad (2)
$

Equation (2) can be converted into equation (1), by remembering that $
\Delta G^{o}=-RT\text{ln} K$.

We can first of all say that $K_{1}$ and $K_{2}$ relate to the two "pieces" of the reaction path one has to cover to reach the overall equilibrium. Knowing them is essential, if one wants to obtain the quantities for all the chemical species, at equilibrium.

$K_{3}$, on the other hand, relates to the "shortcut" (the unique "ideal" step connecting the initial and final states of a system) and is linked to $K_{1}$ and $K_{2}$, through equation (1).

Using the examples you provide and assuming initially 1 mole for $A$ and 1 mole for $B$, we write:

$
K_{1}=\frac{(y-z)y}{(1-y)^2}
$

$
K_{2}=\frac{z}{y-z}
$

We could more generally write:

$
K_{1}=f(y,z)
$

$
K_{2}=g(y,z)
$

The two constants are different functions: but they both depend on the same parameters, $y$ and $z$, which are temperature and pressure (for gases) dependent. If temperature changes, both constants will accordingly change.

To determine the equilibrium quantities, at given temperature and pressure, we'll need the values for $K_{1}$ and $K_{2}$, since knowing only $K_{3}$ won't be enough.

Beyond all this, kinetics will affect the time needed to reach system equilibrium: but that's another story.


Find amount of benzoic acid in dissolution Solubility of benzoic acid in water at $\pu{0^\circ C}$ is $\pu{1.7 g/L}$ and given the fact that $\pu{37 mL}$ was used to dissolve $\pu{1 g}$ of benzoic acid. How much of the benzoic acid would be left over given these conditions?

I have tried using $c = m/vM$ to convert $\pu{1.7 g/L}$ to a form given the $\pu{1 g}$ and the $\pu{37 mL}$ actually used. How would I find out how much is left over?

$\pu{0.037L}$ of water dissolves $0.037\times \pu{1.7g}$ of benzoic acid at $\pu{0^\circ C}$ = $\pu{0.0629g}$.

Therefore $1 - 0.0629 = \pu{0.9371g}$ benzoic acid remains undissolved


Halogenation of alcohols in the presence of ZnCl2 I know that $\ce{OH-}$ is usually a bad leaving group, but why does the following reaction occur via SN1 in presence of anhydrous $\ce{ZnCl2}$?

You are right, $\ce{OH-}$ is a bad leaving group. But the $\ce{HCl}$ in the reaction mixture protonates it. Thus, the bad leaving group $\ce{OH-}$ is converted to the good leaving group $\ce{H2O}$. Anhydrous $\ce{ZnCl2}$ might additionally facilitate the reaction by acting as a Lewis acid and it binds the expelled water thus preventing the back reaction.


How to report result to the nearest 0.01 mg/L? what is meant by reporting result to the nearest 0.01 mg/L?
The test result in certificate show 7 mg/L showing no decimal point.

Reporting to the nearest quantity means that you report as many decimal places of the quantity as suggested. This is a common analytical chemistry expression given in procedures or analytical protocols. For example, if we say weigh  $\pu{5 g} \; \ce{KOH}$ to the nearest $\pu{0.1 g}$, one would record the weight to one decimal place, e.g., $\pu{5.1 g}$ or $\pu{5.3 g}$, to one decimal place but you will not report $\pu{5.1176 g} \; \ce{KOH}$ even if your balance is capable of doing so.

Your test report should include at least two decimal places, the report should appear as $\pu{7.24 mg/L}$ or something like that number.


Polarity and acid strength in alcohols I'm totally confused about the relationship between the polarity and acid strength of a given compoud.

Take the case of alcohols itself, it's said that primary alcohols have greater acid strength compared to secondary alcohols as the electron-releasing group in secondary alcohol decreases the polarity of $\ce{O-H}$ bond.

But why is this so?
I mean, less polarity will help in liberating $\ce{H+}$ ions more easily so acidity is meant to be increased.

That means secondary alcohols are meant to have more acid strength.

Some help to solve this dilemme is appreciated.

In analysing acid strength, bond polarity of the bond to the ionisable hydrogen atom is what should be considered, not the polarity of the molecule (i.e. the dipole moment). In the case of alcohols, the polarity of the $\ce {O-H}$ bond should be considered when assessing strength as an acid.

Why does the bond dipole matter? In the process of ionisation $\ce {ROH + H2O-> RO^{-} + H3O^{+}}$, the $\ce {O-H}$ bond is heterolytically cleaved. It is easy to observe that the greater the partial positive charge on the $\ce {H}$ atom, the more easy it is for it to be abstracted by the water molecule, due to stronger electrostatic attractions. Also, the more polar the bond, the lower the covalent character of the bond as the electron density shifts more towards the oxygen. Such a bond is also more easily cleaved. Recall that the covalent character of a covalent bond is highest when the atomic orbitals are of similar energies and that electron density is shared equally between the two atoms.

In the primary alcohol, there is one electron-donating alkyl group attached to the alpha carbon while in the secondary alcohol, there are two such groups attached to the alpha carbon. Due to the greater electron donation to the oxygen atom in the secondary alcohol, where $\ce {R = CHR'R''}$ , the oxygen atom actually withdraws less electron density from the $\ce {O-H}$ bond. We can think of it this way: The oxygen atom has a fixed demand for electron density. When this demand is met by an increased supply from the alkyl chain, the demand for electron density from the hydrogen atom would decrease. Thus, the $\ce {O-H}$ bond is less polar and less easily heterolytically cleaved.

Simarly, for the primary alcohol, where $\ce {R = CH2R'}$, there is also an increased supply of electron density from the alkyl chain R, as compared to $\ce {R = CH3}$. However, the increased supply from $\ce {R}$ is less so compared to that in the secondary alcohol. Thus, the primary alcohol would be a weaker acid than the methyl alcohol but a stronger acid than the secondary alcohol.

Conventionally, acid strength is assessed using conjugate base stability, rather than bond polarity. A brief overview of assessing conjugate base stability can be found here.


Gallium makes aluminum soft like a wet tissue. Any other metals or alloys that are vulnerable to similar damage? Gallium can make aluminum soft and brittle as long as it bypasses the aluminum oxide normally formed on the surface of the aluminum (for example by scratching the surface).

Mercury can weaken aluminum as well.

Are there any other metals or alloys that can be "weakened" by coming in contact with gallium, mercury or any other substance?

Gallium and Mercury, the latter in particular are absolute pests and poisons to most metals. There are too many to mention that will suffer the same consequences. I will mention the few that don't. Iron, Platinum, Tungsten and Tantalum all tolerate them fairly well. There are certainly more examples, but the common rule is; most don't.


How does composition of air affect vapor saturation limit? Suppose we hold temperature, pressure, and gas composition (say, a standard atmosphere) constant.  Then we know that if we allow the water content of the air to vary it can range between 0 and some saturation limit S g/L.

How does that saturation limit vary with the composition of the air?  For example, does it depend on the molecular weight and proportions of the other constituents of the air?

Assuming the water (or whatever substance is in question) does not chemically react with any other constituents at that temperature and pressure are there other constituent chemical relationships that can affect the saturation limit?

I agree with "repurposer"'s comment that the saturation limit would be independent of the composition of the air.

It would just be a function of the partial vapour pressure of water.

$K_{p}=p_{H_2O}$, where $K_p$ is the equilibrium constant in pressure terms.

Consider this case similar to that of aqueous tension - it depends only on the temperature of the system, and not the other (non-reacting) components.
Whenever we want to find the pressure of the dry gas, i.e. pressure of the mixture of gases $-$ the pressure due to the water vapour.

Hope this helps!


Why does water become less dense when it cools Most materials become more dense as they cool, but ice floats in water, meaning that it's less dense. Why is this?

Under normal conditions (meaning your regular everyday freezer, or a cold day) water forms a crystal lattice below its melting point (as shown below). This structure is held together by hydrogen bonds, which are no longer overcome by the internal energy of the molecules. The hexagonal arrangement of the water molecules in this structure leads to a lower density due to the spaces left in the center of each hexagon (you can now fit fewer molecules in the same volume).

From Wikipedia

There are many other solid phases of water (at least 16 on wikipedia), some which have densities higher than normal liquid water, though most are not naturally formed (at least on Earth).


Cu2+ octahedral complexes $\ce{Cu}^{2+}$ has nine d-electrons, regardless of the ligand field strength it will have one free electron (so it is paramagnetic. Are $\ce{Cu}^{2+}$ octahedral complexes high spin or low spin?

There are not two possibilities for low spin and high spin complex for a d9 metal ion.
The only cases where you can have both is for d4, d5, d6 and d7.

Remember that both configurations arise from the possibility of organizing the d electrons in a more stable way, this is, with the lowest possible energy.
You need extra energy for doing two things:

Promoting one electron to a higher energy level (crystal field stabilization energy, in case of an octahedral compound, $\Delta_O$
Puting another electron on a semioccupied orbital (pairing energy, due to repulsions)

In the complexes where $\Delta_O$ is higher than the pairing energy, the result will be a high spin configuration (it's easier for the electrons to become promoted to a higher level than pairing them with another electron), and if it is smaller, the result will be the opposite.
But this only applies for d4, d5, d7 and d8.

Now think about d1 d2 and d3:
It's nonsense to put the electrons on the upper level, when they can be all by their own in lower energy orbitals, so there are no high spin/low spin possibilities, but only one possible case, the three electrons one on each $T_2g$ orbital.

Now d8 d9 and d10.
There is once again, only one possibility to put the electrons. For your case, d9, you have nine electrons, so 6 of then will be filling the lower energy orbitals for sure, and the other 3, on the upper orbitals.
You may think, well, there is another possibility, promote one of the lower energy electrons to the hole on the higher energy orbital. But this is again nonsense: you are looking for the most energy favorable situation, so, as long as you only have to add to the system the pairing extra energy, why would you also add the promoting energy instead of filling a semioccupied lower energy orbital?
This would result in a excited state of the complex, but this is not what you're looking for when you talk about high/low spin.
The same applies to d8 and d10, there is only one possible situation, and no high/low spin possibilities.


Is there a mnemonic to study the spectrochemical series? Mnemonics help to by-heart things which are difficult to do so and which are important. I found mnemonics for studying the periodic table, electromagnetic series etc which were of great help. So, is there a mnemonic to study the spectrochemical series since knowing it will be of great help to find the geometry of a given co-ordination compound?

The spectrochemical series is a list of common ligands for transition metals in order of increasing ligand-field splitting parameter $\Delta$, which is the energy difference between the sets of $\ce{d}$ orbitals:

$$\ce{I-} < \ce{Br-} < \ce{S^2-} < \ce{SCN-} < \ce{Cl-} < \ce{NO3-} < \ce{N3-} < \ce{F-} < \ce{OH-} < \ce{H2O} \approx \ce{C2O4^2-} \text{ (oxalate)} < \ce{NCS-} < \ce{CH3CN} < \ce{py}\text{ (pyridine)} < \ce{NH3} < \ce{en}\text{ (ethylenediamine)} < \ce{bpy}\text{ (2,2'-bipydiridyl)} < \ce{phen}\text{ (1,10-phenanthroline)} < \ce{NO2-} < \ce{PPh3} < \ce{CN-} \approx \ce{CO}$$

  Is there a mnemonic device for remembering this series?

I would not recommend a mnemonic for the spectrochemical series for a number of reasons:

Often, it depends on both metal and ligand and coordination geometry how big a certain splitting is.
The splitting of some ligands ($\ce{NH3, H2O}$) is even dependent on the pH — even though that could sort-of be overcome by including their corresponding anions in the list.
Any mnemonic is bound to be incomplete. N-heterocyclic carbenes (NHCs) are very important ligands especially in organometallic chemistry, but they aren’t on your list.
Some ligands (e.g. $\ce{NO}$) can display very complex redox chemistry that defies them being ordered in a simple list.
Rather than learning some list off by heart and rote memorising it later, it is much better to know why a ligand is strong- or weak-field and to deduce that property from the ligand’s structure.

The last point is most important. Rather than learning or attempting to remember that chloride is a weak-field ligand, ask yourself why. It is a π donor ligand that will split the $\mathrm{t_{2g}}$ orbitals of the central metal in a bonding–antibonding way, lifting the metal’s $\mathrm{t_{2g}}$ orbitals and lowering the energy difference. That way, I immediately know what to expect of (say) selenide $\ce{Se^2-}$ even if it is not present in the list. The strength of a certain ligand in this discussion is equivalent to the usual trends in the periodic table. Quickly memorised, no long mnemonic and applicable even to new ligands.


Measuring distance in a conductive liquid What is needed to  measure distance between two electrodes imersed in seawater ?
I have  tried measuring the voltage across the path (a suposed resistance) with a constant current source (about 10uA) passing the water to GND of circuit. (the resistance is suposed proportional to distance of electrodes) The probes are 100Hz reversing switched to avoid polarization. And I am using AG/ AGCl pellets as electrodoes. But until now I could not have stable results.
What may be wrong ?  Could someone help ?
Thanks

The AgAgCl sinterized disks was choose because it is very used as reference electrode for seawater potential measurements (cathodic protection) with very good results and low maintenance, but I really dont know if it is the the best for my application. The suface area is about 1 cm2 then we are expecting about 20 Ohms per cm of distance in seawater. With a 10uA as (AC square wave) current source  between electrodes voltage will range from about few milivolts up to 1 volt at 50m.  As you see the potential is bellow of water electrolisys. What you mean "with a lot of corrosion protection" for Stainless Steel plates ?
I understand that salinity, temperature etc could vary in the media but these factors are not expected  to change so quickly and I also intend to use a known distance sensors for reference as well averaging the measures.
I intend use this sistem in a dynamic environment where the reletive speed between sensors or the flow may be about 10 cm / sec.
What you think about the method ? There are a better one ?

(Initial, preliminary answer, here.)
There are several possible reasons for your unstable results:

Possibility 1:  Wrong electrodes

Ag|AgCl pellets seem like an odd choice of electrodes for this application.  What are the reasons for using them as electrodes?  What is their surface area?  Their complex shape means the electrical current maybe isn't uniform between the two electrodes.  Also, the fact that you're drawing current through an electrode with an AgCl coating may mean the coating is building-up/breaking-down (which could mess up your measurements) ...  In your current application, it seems like you'd want to use two, big, parallel stainless steel plates (with a lot of corrosion protection!) as your electrodes.

Possibility 2:  Changing salinity

Depending on the local conditions, the conductivity of the seawater may be changing.  For example, if you are by a river delta, the tides may be washing fresh water between your plates.  Or, if the electrodes are at the water surface in really sunny location, day-time evaporation may increase salinity ...  If either of those are the case, install a second set of plates with a known, fixed electrode spacing to measure salinity, and use that to modify the results from the original set of electrodes.

Possibility 3:  Convection Currents

Is the seawater between the two plates moving?  If so, how fast?  Does the speed change over time?  All of these factors could contribute to unstable electrochemical readings, too ... If so, you may add mesh or baffles to create more-consistent flow conditions.

(Follow-up answer, here)

As far as I can tell, you're method sounds alright from an electrochemical perspective.  Cathodic protection is fine for minimizing corrosion.  Having "known distance sensors" will avoid problems with changing salinity, etc (as you mentioned).


Why isn’t there LDPP, HDPP? Unlike polyethylene that there are LDPE, HDPE, or LLDPE, when it comes to polypropylene the issue becomes about copolymer (homo/ copo/ block copolymer) or location of methyl group (isotactic/ atactic/ syndiotactic)? Why does polypropylene not have designations based on density?

First off, lets understand how densities of poly(ethylene) are differentiated. The prototypical polyethylene is often represented as a simple repeating ethylene monomer units as shown below:

This is what high-density polyethlyene (HDPE) is constructed of; simple linear monomer units that form closely packed crystal structures. This polymer is synthesized by  free-radical polymerization with catalyses that encourages radical at the end of the chain, keeping the polymer linear. low density polyethylene (LDPE) is free-radically polymerized without this control, which causes branching at random points in the polymer chain. This branching prevents the crystallization that made polyethylene more dense, thus it is more amorphous and lower density. To make a linear low density polyethylene (LLDPE), ethylene is co-polymerized with longer olefins such as 1-butene, 1-hexene or other olefins but with catalysts controlling polymerization similar to polymerization of HDPE. A diagram demonstrating molecular structure differences is shown below:

Source: Plastics Technologies

Bringing this back to about polypropylene, the co-polymers are a different monkey that are used for precisely tuning properties and that is not useful for this discussion. But as you mention the tactility is important. There are two reasons for this. 1.) there is no reason to produce low-density polypropylene as LDPE and LLDPE are cheaper materials that are more mature in their development with plenty of material data available and developed recycling streams, thus there is little incentive to investigate LDPP. 2.) polypropylene is already controlled by its tacticity. Isotactic and syndiotactic polymers crystallize fairly well while atactic polymers do not. Since polypropylenes rigidity and density can be controlled by using certain polymerization catalysts there is again little incentive to produce branched polypropylene


Why can NCl3 be hydrolyzed but CCl4 cannot? While explaining why $\ce{CCl4}$ cannot be hydrolyzed, we say carbon atom doesn't have any d-orbitals and therefore a water molecule (lone pair electrons of $\ce{O}$ atom) cannot form coordinate bond with carbon. So, $\ce{CCl4}$ cannot be hydrolyzed.
In $\ce{NCl3}$, nitrogen also does not have any d-orbital.

Why it can be hydrolyzed?

Is hydrolysis mechanism for $\ce{CCl4}$ and $\ce{NCl3}$ different? If so how they are different?

Again, in some sources I have found, in the case of $\ce{NCl3}$ water forms coordinate bonds with the $\ce{Cl}$ atom and hence the molecule gets hydrolyzed. Why does the same thing thing not happen in $\ce{CCl4}$ which also has a $\ce{Cl}$ atom?

There are a few (mis-) conceptions in the question that require addressing other than giving the simple (yet correct) answer that permeakra gave.

$\ce{NCl3}$

The hydrolysis of this compound is different from the hydrolysis of the other two. This is because the $\ce{N-Cl}$ bonds are polarised towards nitrogen, not towards chlorine as they would be for $\ce{SiCl4}$ and $\ce{CCl4}$. Since the oxygen in water usually attacks nucleophilicly, the attack happens on the chlorine; it can proceed nicely with an $\mathrm{S_N2}$ mechanism (the rear side of all three chlorines are pointing into the solution). Note that another water molecule or the same will likely protonate the nitrogen either before or concertedly, so the leaving group won’t be $\ce{NCl2-}$ but rather $\ce{NHCl2}$. The other reaction product is $\ce{HOCl}$.

Other concerted mechanisms including only one water molecule might be possible, but sound less likely.

$\ce{SiCl4}$

Silicon atoms are rather large compared to both carbon and nitrogen: They have an additional electron shell. (Layman’s simplification, but it works.) As such, even in crowded cases like tri-iso-propylsilyl chloride $\ce{((CH3)2CH)3-SiCl}$ with three bulky iso-propyl groups, the silicon atom can be rear-side attacked by water (or catalysts such as 4-(N,N-dimethylamino)pyridine) liberating a chloride atom in a traditional $\mathrm{S_N2}$ mechanism.

No d-orbital contribution is required to explain it. In fact, silicon does not have ‘low-lying’ or ‘accessable’ d-orbitals; that is an old, superceded explanation.

The hydrolysis products of $\ce{SiCl4}$ are $\ce{Si(OH)4}$ (which will likely condensate to some polysilicate or $\ce{SiO2}$) and $\ce{HCl}$.

$\ce{CCl4}$

Technically, this could also proceed via an $\mathrm{S_N2}$ mechanism. However, carbon is much smaller and the chlorine atoms much larger in relation, so the rear-side attack is disfavoured sterically.

Instead, this molecule has access to the $\mathrm{S_N1}$ pathway, which can be described as follows:

$$\ce{CCl4 -> CCl3+ + Cl-} \\
\ce{CCl3+ + H2O -> Cl3C-OH2+} \\
\ce{Cl3C-OH2+ + Cl- -> CCl3-OH + HCl}$$

And so on, displacing the remaining chlorine atoms, ultimately liberating $\ce{H2CO3}$. (After the second chlorine displacement, phosgene will form which can be attacked directly by water with the $\ce{C=O}$ bond acting as an electrophile.)

Note that here again we gain $\ce{HCl}$, an indirect proof of chlorine being the more electronegative element.

Conclusion

The three molecules we considered are so different in their steric and electronic properties, that we need to consider three different mechanisms that lead us to two different sets of products (depending on which side gains the hydrogens and which formally gains hydroxides). Nowhere do we need d-orbitals.


Mathematical properties of free Gibbs energy I'm wandering, if some mathematical properties of the free Gibbs energy (also called thermodynamic potential) are well known.
For instance in chemistry, if $n_i^\alpha$ denotes the number of moles of species $i$ in phase $\alpha$ and G is the free Gibbs energy.
I know that G must be an homogeneous function of degree one in $n_i^\alpha$.

Do we have more information like continuity or coercivity properties of G ?

n is discontinuous. The least amount you can change it by is ~1 /6.022E23 of 1 mole. You'll have to be more specific about what type of (mathematical) continuity you are referring to, but it doesn't seem useful to ask about its fine grained continuity since it is a macroscopic concept.  Statistical (quantum) Thermodynamics is used both for microscopic systems and as the basis for macroscopic quantities such as Gibbs Free Energy. You need to be careful: GFE is one meaning of Thermodynamic Potential, but not the only one. I don't have any idea what you mean by G's coercivity.


Ortho-effect in substituted aromatic acids and bases When comparing o,m,p-toluidine basicities, the ortho effect is believed to explain why o-toluidine is weaker. But when comparing o,m,p-toluic acid basicities, the ortho effect is stated as a reason why o-toluic acid is stronger acid. I was told that the ortho effect is a phenomenon in which an ortho- group causes steric hindrance, forcing the $\ce{-COOH}$, $\ce{-NH2}$ or some other bulky group to move out of the plane, inhibiting resonance. Then, if the ortho effect inhibits resonance, why is o-toluic acid the strongest and o-toluidine the weakest?

Where am I going wrong in my understanding of the ortho effect?

I'd like to throw a tentative explanation for the ortho effect into the ring:

In the molecules in question, an interaction between the protons of the methyl group and the lone pair of the amine nitrogen and the negative charge on the carboxylate, respectively,  can be assumed.

In the first case, the electron density on the N atom is (slightly) reduced and thus the basicity of o-toluidine.

In the latter case, a similar interaction provides additional stabilisation of the carboxylate. As a result, o-toluic acid is more acidic than the isomers.