Can we extract KI from Lugol solution? I want to extract the $\ce{KI}$ component of the Lugol solution, is there a method to do it? If not, how can I make a $\ce{KI}$ solution?

As you know you have to remove the others species like $\ce{I2}$ and $\ce{I3-}$. $\ce{I3-}$ I'm quite sure it exists only in solution so simply evaporating the water from Lugol 's solution you should eliminate it. To remove completely $\ce{I2}$ you have to do some others steps.

$\ce{KI}$ is produced industrially by treating iodine with $\ce{KOH}$.(see Wikipedia for reference).

$$ \ce{3I2 + 6KOH -> 5KI + KIO3 + 3H2O}$$

So I think the simple way is to treat Lugol's solution with $\ce{KOH}$ make an ice bath to precipitate $\ce{KIO3}$, filter the solution to remove it and than evaporate water.
You can also try to remove iodine by means of active carbon I know this method is used to purify water see ref and ref.
You can try evaporating the Lugol's solution and re-crystalize it and than eventually  try different solvents to solubilize one of the compounds and leave the other as precipitate but if think is quite difficult to find a better solvent than  water (as you know iodine is very insoluble in water instead $\ce{KI}$ is very soluble).

If you want to make the $\ce{KI}$ from your self here there are two Home experience:

extract potassium iodide from salt
make $\ce{KI}$ from $\ce{I2}$ and $\ce{KOH}$


Yttrium -- rare earth or transition metal? The element yttrium is called a rare earth element, yet periodic tables label it as a transition metal.

Why is that?
Thanks

Comparing "transition elements" to "rare earth elements" (REE) is a bit like comparing apples and oranges. It all boils down to the different ways in which the elements can be classified.

Let's take a look at the periodic table:

(modified from this)

The rare earth elements are marked with a red outline.

Why are all rare earth elements are technically transition elements?

Strictly speaking, transition elements are the elements in groups 3 to 12 of the periodic table. The rare earth elements, in their widest sense, are Sc, Y and the lanthanides, La to Lu. You can see that if you group the lanthanides as one element, they fit nicely below Y in group 3 of the periodic table. The lanthanides are occasionally called the "inner transition elements".

Why is yttrium a rare earth element?

There are chemical and historical to the inclusion of yttrium in the rare earth elements.

Historically, yttrium was among the first elements to be isolated from the REE-bearing minerals of the Ytterby mine in Sweden. What was though as first to be "yttria" was later shown to contain also the rest of the heavy REE. It just took a longer time to purify them.

Chemically, yttrium commonly occurs with the other REE in the same mineral deposits around the world. You probably know the the chemical properties of the lanthanide series vary continuously from La to Lu. Yttrium also follows that pattern, and in terms of chemical properties it sits very nicely somewhere between Dy and Ho (that's why it's associated in mineral deposits with the heavy REE).

Yttrium is commonly used in the same high-tech applications as the other REE: it is commonly alloyed with other REE for speciality materials. It is also commonly used in combination with the other REE in lasers and various detectors in electron microscopy and analytical chemistry.

A different look

If you take a look at this figure, taken from here: https://earthscience.stackexchange.com/a/2951/725

You can see that most transition elements, as they appear in nature, usually have a cationic charge between 60 and 80 pm. The REE (including Y) are somewhat larger than that, resulting in slightly different properties.

You may also be interested in reading the answers to the following questions, also here on Stack Exchange:

What are rare earths and why do they cluster near alkaline magmatism?

Why are the rare earths erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb) named like that?


REDOX spontaneity using electrode potentials For this question, I thought the correct answer is A, because if we look at the electrode potential, that of iron(III) is higher than that of iodine, therefore the equilibrium for $\ce{Fe^3+}$ will favor the RHS, and shifts to the right, and therefore iodine will go in the opposite direction. Therefore A would be correct.

But the correct answer is D. Using the same logic, I can see it is also correct, but why isn't A correct then?

Thank you for any ideas in advance and sorry for the trivial question.

The reason is simply semantics. Iodi(n)e which is the neutral valence state of the element will not donate an electron to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$.   Iodi(d)e which is the anionic form of iodi(n)e has a negative charge and can donate an electron to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$.


Using Le Chatelier's Principle to explain why percent dissociation of acids/bases increase with dilution $$\ce{HA(aq) +H2O(l) <=>H3O+(aq) +A- (aq)}$$

Say the acid solution is currently in equilibrium. How would diluting the acid by adding water change the equilibrium position? Le Chatelier's Principle says pure liquids like water do not affect the equilibrium position. Using the equilibrium constant, it is easy to show the reaction should shift towards the right, but how can we show that with Le Chatelier's Principle?

Le Chatelier's Prinicple also says that the system will adjust itself based on a change in concentration.

Concentration of an acid is (normally) expressed in moles/liter. If you were to increase the amount of solvent (water) which is on the reactant side of the equation, you would be diluting the acid solution that is in equilibrium and causing the system to adjust accordingly.

As far as the last portion of your request is concerned I am slightly confused. Le Chatelier's Prinicple is built into the equilibrium expression. If I wanted to show how the system were to react based on the addition of more reactants/products I would just use the equilibrium expression for the system.
Hope this helps, D


Nernst equation and electrolysis Let's say we have a stable solution of
$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}$
and $\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}$
, with their respective counter-ions.

The associated redox reactions are:

$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{e}^{-}\rightleftharpoons\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}\qquad\mathcal{E}_{\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$

$\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{e}^{-}\rightleftharpoons\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}\qquad\mathcal{E}_{\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$

where $\mathcal{E}_{\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}<\mathcal{E}_{\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$
, since no reaction between $\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}$
and $\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}$
is occurring.

The questions are:

Provided that neither $\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}$
nor $\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$
are present in the solution, how can we determine, form the Nernst equation, the $\mathcal{E}$ needed to start the electrolysis process
$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}\rightleftharpoons\mathrm{A}_{\left(\mathrm{aq}\right)}^{ +}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$?
In other words, how can the Nernst equation deal with activities that are equal to zero?

Thanks in advance!

The Nernst equation only technically applies when the system is in electrochemical equilibrium—when there's no net current flow, but there's actually an exchange current. A system in which the reaction can only happen in one direction because there are no products wouldn't apply. Source:    Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008.

What I'm not sure about is the validity of ∆G values calculated from $∆G=∆G°+RT\ln Q$ at the  limits where the system is all products or all reactants. ±∞ don't seem to be meaningful values, but the system should quickly move away from these boundary values, even if only by a small amount. These chemical thermodynamics methods are well-behaved when near equilibrium, but well away from equilibrium, they can produce invalid results.


IUPAC nomenclature query The given compound is: $\ce{CH3-CH2-CH2-CH(CH3)-CH(CH2CH3)-CH2-CH2-CH3}$

The IUPAC name I came up with was 4-ethyl-5-methyloctane but the given answer states 5-ethyl-4-methyloctane.

Are both acceptable? Is there any rule I'm missing out on? I went through the rules but couldn't find the appropriate rule.

Your name is correct, while the given one is not. When the set of locants for subsituents would be the same no matter what, you would assign the lowest locant to the substituent that appears first in the name, which, by alphabetical order, it would be the ethyl group in this case. Hence 4-ethyl-5-methyloctane is the right name.


What is the structure of iodide of Millon's base? In qualitative analysis, $\ce{NH4+}$ ions are confirmed using Nessler's         reagent:-

$$\ce{NH4+ + 2[HgI4]^{2−} + 4OH− → HgO·Hg(NH2)I ↓ + 7I^{−} + 3H2O}$$

The brown precipitate formed is called iodide of Millon's base.  It is also written as  $\ce{3HgO.Hg(NH3)2I2}$ and as $\ce{NH2.Hg2I3}$. (wikipedia).

But its structure is present differently in different sites:-

Source 1 - ionic form

Source 2  - Here the reaction is given differently:-

$$\ce{([K2]HgI4 <=> 2KI + HgI2) * 2}$$
$$\ce{HgI2 + NH3 -> Hg[NH2]I + NH4I}$$
$$\ce{Hg[NH2]I + HgI2 -> NH2Hg2I3 }$$
$$\ce{[K2]HgI4 + 2NH3 -> NH2Hg2I3 + 4KI + KH4I }$$

structure -

Source 3- non ionic form

Source 4 - Hydrate form; Here also the reaction is different:-
$$\ce{[K2]HgI4 -> 2KI + HgI2}$$
$$\ce{HgI2 + NH3 -> Hg[NH2]I + HI}$$

This site also says that its structure is still uncertain. Various other structure has been proposed:-

So, what is the correct reaction, correct formula and correct structure of iodide of Millon's base?

First, ignore any monomeric formula. The compound is definitely polymeric.

The literature contains hints for existence of two separate lines of compounds. The first one is salts of polyamidomercury: $\ce{[Hg(NH2)]_{n}A_{n}}${1} with mercury-amide chains in the structure, and $\ce{(Hg2N)_{n}A_{n}}$ with structure based on crystoballite {2}{3}

It seems thatn the lines are not distinguished in earlier literature, but recent researchers use the words "Millon's base salts" for $\ce{(Hg2N)_{n}A_{n}}$ only.


What causes the iridescent colour in laboratory grown bismuth? Artificially grown bismuth crystals are arguably magnificent, as seen in the image below:

Image source: Minerals.net

The website states that:

  They have hopper-like growths in pseudocubic crystals and are usually coated with chemicals to prevent tarnish, thus maintaining the silver-white color. Sometimes the coating gives a colorful effect on the bismuth.

What coating gives the colourful iridescence? and what chemical processes cause the colourful iridescence to occur?

The colour is from a thin film of bismuth(III) oxide that forms on the surface if the crystals are formed in air. At the elevated temperatures used to melt bismuth, the oxide forms quite quickly. The iridescence is a result of thin film interference—light waves constructively or destructively interfere as they bounce off the bismuth-oxide and oxide-air interfaces. The interference is dependant on the wavelength and incidence angle of light as well as the film thickness, which results in different wavelengths of light being attenuated by different amounts, making the surface appear coloured and the colour changing with viewing angle.


How does the Rf value change when changing the eluent composition? A non-polar molecule X is analyzed by TLC chromatography on silica to determine its $R_f$ value.
The first eluent was hexane:EtOAc = $9:1$ and the $R_f$ value was $0.5$.
In the second run, the eluent was changed to hexane:EtOAc = $7:3$. Will the $R_f$ value be smaller or greater than $0.5$?

I am confused because there are two important interactions: because the 2nd system is more polar, it will form weaker interactions with substance X, and the $R_f$ value will decrease; on the other hand, the molar polar eluent will form stronger interactions with stationary phase, so the $R_f$ value will increase because the substance X is more easily eluted.

The most important interaction is that of the mobile phase with the stationary phase. If you increase the mobile phase’s polarity, that interaction will increase, decreasing the interaction of the solute with the stationary phase. Therefore, a higher ethyl acetate content in the mobile phase will increase $R_f$ values.


How do sunscreens protect the skin from UV rays? I was unsure to the largest extent about whether I should post this question in chemistry Q & A or bio Q & A until I just read that sunscreens "absorb" UV rays, not allowing the most of those rays from penetrating into the skin. So when this is about the combination, it must be chemistry. Anyhow, is there any special name for these "absorbents"? And how exactly, do they "absorb" UVA & UVB? (I'm not really definite if this absorbing is really a scientific expression or not, and that's why I use quotations when using the word) I, myself, reckon that the sunscreen products form a special compound on the skin or they make the body produce something as a shield against UV rays, but I have got no clues.

Apart from titanium dioxide, the active species in sunscreens are organic compounds from different substance classes, such as cinnamates, benzophenones, benzimidazoles, and triazines.

The do absorb in the 280-315 (UVB) and/or 315-380 nm (UVA) range. This "uptake" of energy results in an electronic excitation, the compounds undergo a transition from the $S_0$ (ground) state to an excited $S_1$ state.

The excited singlet state ($S_1)$ may

deactivate radiationless via internal conversion (IC), the energy of the photon is released as heat or
undergo intersystem crossing (ISC) to a triplet state (usually $T_1)$ with different spin multiplicity

The latter state then might again loose its energy via IC or undergo a chemical transformation, through which the suncreen might be turned to a species that not longer is able to absorb UV radiation.


Reaction between chlorobenzene and methyl chloride What will be the product of the reaction between chlorobenzene and methyl chloride in the presence of anhydrous $\ce{AlCl3}$?

I think we should get 1-chloro-4-methylbenzene as the major product, as we'll have Friedel–Crafts alkylation. Is this correct? Or will we get toluene?

You will actually get a mixture of different products, with the product with the highest yield being p-chlorotoluene, or 1-chloro-4-methylbenzene. The other products resulting from the reaction include o-chlorotoluene and m-chlorotoluene, with a significantly higher amount of the former being produced*. In fact, there would also be products with more than one methyl substituent being produced in various proportions depending on the precise conditions being used.

Yes, the reaction will proceed by Friedel-Crafts alkylation: Firstly, the Lewis acid catalyst abstracts the chloride ion from methyl chloride, giving a reactive, electrophilic methenium ion. Then, electrons from the $\pi$ system of the benzene ring of chlorobenzene then attack the electrophilic ion. Finally, a proton is lost to restore aromaticity of the ring, giving the product.

*Refer to the following post to understand why the chlorine atom as a substituent would be ortho/para-directing: Why are halogens ortho para directing even though deactivating?


NMR analysis regarding cis-1-isopropyl-3-phenylcyclohexane I have come across a question in my textbook:

The textbook gives this answer:

I understand the explanation for the first compound. However, for the second structure, I am not understanding why the 2 methyl groups are diastereotopic. Is there not free rotation about the $\ce{C-C}$ bond, connecting the $\ce{CH(CH3)2}$ group to the ring such that the 2 methyl groups become equivalent? The answer key makes it seem as though there is no rotation and the methyl groups are fixed in place. I feel as if I am misunderstanding the question. Please help me clear up the confusion.

The two methyl groups do not see the same average environment

The issue here is not whether the two methyl groups are freely rotating about the bond to the CH group: the issue is whether they see the same environment on average when the bond between the CH and the cyclohexyl ring rotates.

The easiest way to see that they don't is to build a model and play with it.

If you do you will see that their average environment can never be the same no matter how much rotation you do. Imagine rotating them so that the CH is facing the same way as the Ph with respect to the ring. The two methyls are different: one will be closer to the Ph than the other. Now rotate the CH-ring bond 180 degrees. The CH now points away to the opposite side of the ring to the Ph. But the environments of the methyl groups are not reversed: the other methyl is now closer to the Ph but is now on the opposite side of the ring so sees a different environment. So that rotation does not equalise the average environment for each methyl.

This is generally true in chiral compounds.


Movement of solid particles in colloids A liquid is made up of tiny vibrating particles of matter, such as atoms, held together by intermolecular forces. Where a solid is the same, but the atoms have much less freedom of movement.

Even if the water is still, its molecules are still moving against each other.

My question is: If you take a solid metal and create a fine dust and mix it with water, will the water molecules move the solid powder all the time?

The short answer to this question is: yes.

The movement of particles in suspension like the powder in water would be a perfect example of Brownian motion as Mithoron already said. The particles are being kicked around by the water molecules, which are in constant motion as you already said.

This is also true for small airborne particles that collide with fast moving gas molecules.


Inferring bond breakage in a reaction from InChI strings of reactans? I'm currently confronting a problem:

I would need to enumerate all chemical bond breakages that occur during a given chemical reaction, including the type of bond and the atoms participating in the bond. This is straightforward to do by visual inspection if one has, say, the 2-D structures of the reactants and products of a given reaction.

My problem is that I have a list of several thousand reactions, and I need a computer to do the job for me. To this end, I got the InChI (and SMILES) representations of all molecules participating in each reaction. At first, I thought that it would be rather easy to write a program that identifies and enumerates the broken bonds in the reactants (given the InChI strings), but by now I'm starting to become desperate.

Perhaps someone knows any software able to do this?

There is the Reaction Decoder Tool (RDT on github) that can - from the README - do:

  Reaction Annotator (Extract Bond Changes, Identify & Mark Reaction Centres)

EDIT: It potentially could have InChI support, but a quick glance at this file suggests that it currently doesn't...

In fact, that would require the CDK/RDT to implement the rInChI format

The input formats includes reaction smiles which should be possible to generate from your smiles strings. Looks like you just jam a ">" inbetween reactants and products with a "." separating the molecules in each set.

which sounds like what you want, I think. Also it should be able to handle those volumes as it has been developed to work with reaction databases at the EBI.

(Full disclosure : I worked on the project for a little while :) but it does have InChI support  - via the CDK).


Reacting calcium with warm water and magnesium oxide with carbon dioxide, what are the products? So I have several descriptions of reactions and I have to give the net ionic equation. Are these just the equations or are they the proper net ionic ones?

Solid calcium is added to warm water.
$$\ce{Ca_{(s)} + 2 H_{2}O_{(l)} -> Ca^{2+}_{(aq)} + 2 OH_{(aq)}^- + H2_{(g)}}$$

Powdered magnesium oxide is added to a container of carbon dioxide gas.
$$\ce{MgO_{(s)} + CO2_{(g)} -> MgCO3_{(s)}}$$

Did I do those right, or am I on the wrong track?

Purpose of ionic equation is to describe ions in (aqueous) solution, without the need to bother about details.

The first equation looks reasonable, you could possibly end up with $\ce{Ca(OH)_2}$, which is not very water soluble, but it's IMHO the type of detail you should not bother with.

The second one is more difficult, though. At first, there is no solution to speak about, therefore the request for ionic equation itself is invalid. The reaction written is correct, as long as it describes possibility. If it would really happen is question of reaction conditions, therefore totally outside the scope of homework. As such, I'd say, correct.


What is fused calcium chloride? As far as I have researched, fused calcium chloride is the hydrated form of calcium chloride. If that is true, then why is it called 'fused', why not simply hydrated?

The term "fused" in chemistry generally refers

to become physically joined together

as by melting, adherence or other means of consolidation [1].
At an industrial scale solutions containing $\ce{CaCl2}$ are obtained either as by-products in the production of soda (by the ammonia method) and Berthelot's salt, or by the adding $\ce{HCl}$ to limewater. Solutions are then evaporated and $\ce{CaCl2}$ is isolated as a white or greyish porous mass, which is called fused calcium chloride and is generally water-free or contains trace amounts of water.
To make it clear, in this case "fused" ≠ "melted", as it's not about melting $\ce{CaCl2}$ (m.p. = $\pu{772^\circ C}$), but rather a physical adhesion as a result of a set of steps applied to calcium chloride solution (or its deliquescent form) of gradually increasing the temperature from $\pu{70 .. 80^\circ C}$ to approx. $\pu{300^\circ C}$. This results in formation of hydrated forms ($\ce{CaCl2 * x H2O}$) with subsequent complete dehydration and solidifying above $\pu{250^\circ C}$ alongside with minor partial hydrolysis to form $\ce{CaO}$ and $\ce{HCl}$.
Fused refers to the fact that this porous mass is granulated, but not dispersed, in order to increase the surface area and improve its dehumidifying ability and at the same time to prevent the salt from deliquescing swiftly.
Reference

fused. (n.d.) American Heritage® Dictionary of the English Language, Fifth Edition. 2011.


How can I find the symmetry labels of atomic orbitals in a molecule? What are the symmetry labels for the p and d orbitals of platinum in $\ce{[PtCl4]^2-}$?

I understand how to find the point group of a molecule, but am not sure how to use the character table to find the irreducible representations corresponding to the orbitals.

First, you have to know the geometry of your compound. The complex $\ce{[PtCl4]^2-}$, for example, is square planar.
The next step is to determine the point group of the compound from this geometry by identifying the symmetry elements that it posesses (a procedure on how to do this can be found here). The point group of the complex $\ce{[PtCl4]^2-}$, for example, is $D_\mathrm{4h}$.
Now you consult the character table of the point group you determined in the previous step. The character tables for many common point groups can be found here. The character table of the point group $D_\mathrm{4h}$ is shown below and can also be found here.

$$\begin{array}{c|cccccccccc|cc} \hline
D_\mathrm{4h} & E & 2C_4 & C_2 & 2C_2' & 2C_2'' & i & 2S_4 & \sigma_\mathrm{h} & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_{2g}} & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_{1g}} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & & xy \\
\mathrm{E_g} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & (R_x,R_y) & (xz,yz) \\
\mathrm{A_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\
\mathrm{A_{2u}} & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{1u}} & 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 & & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & \\
\mathrm{E_u} & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & (x,y) & \\ \hline
\end{array}$$

Finally you need to determine the symmetry labels of the p- and d-orbitals within the point group by identifying the irreducible representations (the labels on the far left of the character table) which match the transformational properties of the basis vectors ($\color{red}{x}$, $\color{red}{y}$, $\color{red}{z}$) and some of their quadratic combinations ($\color{blue}{xy}$, $\color{blue}{xz}$, $\color{blue}{yz}$, $\color{blue}{z^2}$, $\color{blue}{x^2 - y^2}$). These can be found in the two columns on the right columns. Again this will be shown for the example compound $\ce{[PtCl4]^2-}$ where the relevant functions are highlighted.

$$\require{bbox} \begin{array}{c|cccccccccc|cc} \hline
D_\mathrm{4h} & E & 2C_4 & C_2 & 2C_2' & 2C_2'' & i & 2S_4 & \sigma_\mathrm{h} & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,\bbox[cyan]{z^2} \\
\mathrm{A_{2g}} & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_{1g}} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & & \bbox[cyan]{x^2-y^2} \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & & \bbox[cyan]{xy} \\
\mathrm{E_g} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & (R_x,R_y) & \bbox[cyan]{(xz,yz)} \\
\mathrm{A_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\
\mathrm{A_{2u}} & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & \bbox[pink]{z} & \\
\mathrm{B_{1u}} & 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 & & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & \\
\mathrm{E_u} & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & \bbox[pink]{(x,y)} & \\ \hline
\end{array}$$

Since the three p-orbitals point directly along the $x$, $y$, and $z$ axes, they show the same transformational properties as the basis vectors, and consequently get the same symmetry labels, i.e.

$$\begin{align}
\left.\begin{array}{c}
\mathrm{p}_x \\ \mathrm{p}_y \end{array}\right\} &\rightarrow \mathrm{E_u} \\
\mathrm{p}_z &\rightarrow \mathrm{A_{2u}} \ .
\end{align}$$

The five d-orbitals show the same transformational properties as their quadratic function counterparts and thus they get the symmetry labels, i.e.

$$\begin{align}
\mathrm{d}_{xy} &\rightarrow \mathrm{B_{2g}} \\
\left.\begin{array}{c}
\mathrm{d}_{xz} \\ \mathrm{d}_{yz} \end{array}\right\} &\rightarrow \mathrm{E_g} \\
\mathrm{d}_{z^2} &\rightarrow \mathrm{A_{1g}} \\
\mathrm{d}_{x^2 - y^2} &\rightarrow \mathrm{B_{1g}} \ .
\end{align}$$

Another worked example can be found in this answer of mine.


What is the preferred IUPAC name of hexamethyldisilazane? Hexamethyldisilazane (HMDS, $\ce{(SiMe3)2NH}$) and its conjugate base hexamethyldisilazide (as the Li/Na/K salt) find wide use in organic chemistry.

In my quest to only use preferred IUPAC names in my thesis1 I have come across a problem with this compound. According to P-21.2.3.1 of the Blue Book the name disilazane for the parent hydride $\ce{H3Si-NH-SiH3}$ is no longer preferred because

  When nitrogen atoms are present, amine names (see P-62) are preferred because of the higher functionality of amines. This concept does not extend to other elements.

and the name given is N-silylsilanamine.2 Just going off this, I would guess that the PIN of the hexamethyl compound is something along the lines of "trimethyl-N-(trimethylsilyl)silanamine".

However, I'm also not sure whether the addition of the methyl groups makes the choice of parent hydride different, as per P-21.2.3.2 (for example). The frequently used name bis(trimethylsilyl)amine is also a possibility, although I've personally not seen a rule suggesting that it could be a PIN.

What is the PIN of this amine?

I'm just kidding; writing "4-methylbenzene-1-sulfonyl" multiple times is quite silly. But I’m still curious about this PIN.
The original publication has this written as N-silylsilan-1-amine but this has since been corrected in the errata, which are published online.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), as already mentioned in the question, the name ‘disilazane’ is no longer recommended for $\ce{H3Si-NH-SiH3}$. Such structures are now named as amines. (Also note that the name ‘hexamethyldisilazane’ is ambiguous and therefore never was in accordance with IUPAC nomenclature.)

P-21.2.3.1 Heterogeneous parent hydrides composed of alternating heteroatoms, i.e., [a(ba)n hydrides], excluding carbon atoms and the halogen atoms.
Compounds containing an unbranched chain of alternating atoms terminated by two identical atoms of the element coming later in the seniority order O > S > Se > Te > N > P > As > Sb > Bi > Si > Ge > Sn > Pb > B > Al > Ga > In > Tl may be named by citing successively a multiplying prefix denoting the number of atoms of the terminal element followed by the ‘a’ term for that element, then the ‘a’ term of the other element in the chain and the ending ‘ane’. The terminal letter ‘a’ of an ‘a’ term is elided when followed by a vowel; the terminal vowel of a numerical prefix is not elided even when the ‘a’ term begins with the same vowel. When nitrogen atoms are present, amine names (see P-62) are preferred because of the higher functionality of amines. This concept does not extend to other elements. These parent hydrides are preselected parent hydrides (see P-12.2) and have priority to receive preferred IUPAC names, as long as they are used to name compounds containing carbon.

Symmetrical secondary amines are named according to Subsection P-62.2.2.1, which describes three different methods.

P-62.2.2.1 Symmetrical and unsymmetrical secondary and tertiary amines are named only by the same methods described in P-62.2.1.2.
(1) substitutively using the retained name ‘aniline’ or the suffix ‘amine’ and the name of a parent hydride with further N-substitution;
(2) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘azane’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.
(3) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘amine’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.
Method (1) generates preferred IUPAC names. Functional parent names like diethylamine and triethylamine are deprecated. The prefixes in names of such secondary and tertiary amines formed by method (2)[sic: read (3)] are set off by parentheses to distinguish them from these deprecated names.

The corresponding names for $\ce{H3Si-NH-SiH3}$ are actually given as examples in the Blue Book:

(1) N-silylsilanamine
(2) disilylazane
(3) disilylamine
(not disilazane; see P-21.2.3.1)

Note that, since $\ce{H3Si-NH-SiH3}$ does not contain any carbon, no preferred IUPAC name (PIN) is indicated in the Blue Book for this compound.
The corresponding names for the complete compound that is given in the question are:
(1) 1,1,1-trimethyl-N-(trimethylsilyl)silanamine
(2) bis(trimethylsilyl)azane
(3) bis(trimethylsilyl)amine
Since P-62.2.2.1 explicitly states that Method (1) generates preferred IUPAC names even for symmetrical amines, the preferred name for the complete compound that is given in the question is 1,1,1-trimethyl-N-(trimethylsilyl)silanamine.


Which is easier, the nitration of toluene or the nitration of mesitylene (1,3,5-trimethylbenzene)? Does the toluene can undergo nitration easier than mesitylene (1,3,5-trimethylbenzene), due to the high repulsion that will occur between $\ce{NO2}$ and the two adjacent methyl groups (in mesitylene nitration case)?

The kinetics of aromatic nitration depend on the nitrating method, which also bear directly or indirectly on the subject of nitration. For example, aromatic nitration displays second-order kinetics in sulphuric acid (when used in nitrating mixture), and first-order kinetics in nitric acid. Aromatic nitration also displays first-order kinetics in either nitromethane or acetic acid (when used as a solvent) with nitric acid in constant excess. It exhibits zeroth-order kinetics for sufficiently reactive aromatic compounds, and first-order kinetics for sufficiently unreactive compounds used for nitration (Ref.1). The reference predicts that:

  The numerical results vary with the conditions, but are not very sensitive to them; so that the following statements may be taken as approximately true over a considerable range of temperatures and nitrating media. Toluene is attacked 24 times faster than benzene, fluoro- and iodo-benzene 6 times more slowly than benzene, chloro- and bromo-benzene 30 times more slowly than benzene, and ethyl benzoate 300 times more slowly than benzene. Qualitatively we know that benzenesulphonic acid is attacked still more slowly, and nitrobenzene even more slowly. Thus the relative effects of substituents in activating or deactivating the benzene nucleus towards this type of attack is expressed by the following sequence:
  $$\ce{Me > (H) > F, I > Cl, Br > CO2Et > SO3H > NO2}$$

Thus, it is fair to conclude that the nitration of mesitylene is faster than that of toluene regardless of the steric effects enforced by neighboring 1,3-dimethyl functions. That conclusion is confirmed by the relative rates listed in Ref.1 & 2 where it is the case in all conditions considered (obs. rates are at $\pu{25 ^{\circ}C}$):

$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\textrm{Compound} & 7.5\% \: \textrm{aq.} & 15\% \: \textrm{aq.}  & 61.01\% \: \textrm{aq.}  & 68.3\% \: \textrm{aq.}  & \textrm{Estimated}  \\
\textrm{nitrated} & \textrm{Sulfolane} & \ce{CH3NO2} & \ce{HClO4} & \ce{H2SO4} & \textrm{Rel. Rate}  \\ \hline
 \textrm{Benzene} & 1 & 1 & 1  & 1 & 1 \\ \hline
\textrm{Toluene} & 20 & 25 & 19  & 17 & 23 \\ \hline
\textrm{p-Xylene} & 114 & 130 & 84 & 38 & 50 \\ \hline
\textrm{m-Xylene} & 100 & 146 & - & 38 & 400 \\ \hline
\textrm{Mesitylene} & 350 & 400 & 78  & 36 & 16000 \\ \hline
\end{array}
$$
Estimated relative rates are from reference 3 (the authors have calculated from the observed value for toluene and the known isomer ratios by assuming the additivity principle).

References:

E. D. Hughes, C. K. Ingold, R. I. Reed, “493. Kinetics and mechanism of aromatic nitration. Part II. Nitration by the nitronium ion, $\ce{NO2^+}$, derived from nitric acid,” J. Chem. Soc. 1950, 2400–2440 (DOI: 10.1039/JR9500002400).
J. G. Hoggett, R. B. Moodie, J. R. Penton, K. chofield, In Nitration and aromatic reactivity; Cambridge University Press: London, United Kingdom, 1971, “Chapter 3: Nitrating Systems: B. Inert organic solvents,” pp. 32–49.
R. G. Coombes, R. B. Moodie, K. chofield, “Electrophilic aromatic substitution. Part I. The nitration of some reactive aromatic compounds in concentrated sulphuric and perchloric acids,” Journal of the Chemical Society B: Physical Organic 1968, 800–804 (DOI: 10.1039/J29680000800).


Rule of thumb of how much vacuum does lab glass stand? Sometimes I have to apply vacuum to all kind of different borosilicate glass equipment like:

round bottom flasks
liebig condensers
fractional distillation apparatus
spiral condensers
chromatography column
and so on...

While I know that a thick walled round bottom flask will pretty much stand any vacuum my vacuum pump can produce, I am not so sure for something as fragile as a spiral condenser.

Is there a kind of rule of thumb how much mBar thin walled glass equipment like spiral condensers or chromatography columns can stand without risk of implosion?

The basic rule is that the stress on an evacuated glass vessel is pretty much the same at 0.01 mbar (oil pump), 1 mbar (membrane) or 25 mbar (water jet) absolute internal pressure. 0.99999, 0.999 or 0.975 bar $\Delta p$ to the outside make very little difference if the thing goes pop.

And basically everything made out of borosilicate and with no flat surface is safe. Condensors definitely, they're not delicate at all. It's basic procedure to make them oxygen-free by evacuating.

Be careful with chromatography columns and flasks in the > 1 liter range. They're principally safe to be evacuated, too, but then are also a bomb, armed and ready to drop. If it goes off, you better not stand next to it. The danger from evacuated vessels is more or less proportional to their volume.


Where does the potential energy of capillary action come from? Capillary action is when water pulls itself up a paper towel, glass tube or a tree. It is going up, which results in an increase in potential energy. According to the law of conversation of energy, energy cannot be created or destroyed. Where does this potential energy of capillary action come from, or is this an exception or something? My question is different from How does water defy gravity in paper towels?, because the other question is asking why it moves up, but mine is asking where the energy comes from.

Where does this potential energy of capillary action come from, or is this an exception or something?

First, the laws of thermodynamics still apply. Second, you have to account for the energy balance when you go from dry to wet on the one hand, and continuous flow on the other hand.

Initial wetting of the paper towel or glass tube: Quoting from one of the comments, "the increase in gravitational potential energy is countered by a reduction in the chemical potential of the substances due to a change in the areas of mutual contact". That explains why the capillary fills up in the first place, and in this case it is fine to say "water pulls itself up".
Continuous flow. If you are asking about a tree, where there is continual flow, you are not increasing the water-capillary contact area anymore, it remains constant. The energy has to come from somewhere else.
It takes some energy to remove a water molecule from the top of the
capillary because as you do, you are pulling all the water molecules
up a tiny bit. As explained in the excellent video by Veritasium, water can have negative pressure in the capillary, so trees can grow higher than atmospheric pressure might suggest (thanks, A.K., for the pointer to the video). This is not a problem for the tree because the top of
the tree is the site of photosynthesis, so there is free energy
available to do the "pulling". So you can think of the water-filled
capillary serving both in the role of the rope and the bucket, if
you compare it to fetching water from a deep well. In cases where the water simply evaporates, the energy comes from cooling down the surrounding as the water molecule pulls up another water molecule and evaporates. In the case of continuous flow, water does not "pull itself up", something on the top (or bottom) has to provide the energy. For example, if you put a towel in a bucket of water with the towel reaching over the rim a bit, the towel will get wet. However, if the end of the towel outside of the bucket reaches below the water level, it will continuously siphon water out of the bucket, with water dripping from the towel.

As an aside, it does not take much work to lift a water molecule up a tree (1.8 J/mol for a 10 m tree). Hydrolysis of a single ATP affords about 50 kJ/mol under physiological conditions, so it is sufficient to get 25,000 water molecules up the tree.Gravity is weak compared to electromagnetic interactions, which dominate in chemistry.


How does increasing the Hydrogen Concentration of CuSO4 (electrolyte) affect electroplating of copper I'm trying to find an answer as to how lowering the pH of the CuSO4 solution during electroplating of copper affects the amount of copper deposited onto the cathode.

My very basic understanding is that if the concentration of hydrogen is too high, then there is greater competition and hydrogen will be reduced instead of the copper. (But why?)

Also the reason why CuSO4 is acidified is to prevent copper from hydrolysing from the solution as it prevents the oxidised Cu2+ from reacting with the OH- ions to form Cu(OH)2 precipitate.

However, I was wondering if this could be related to electrode potentials. That the increase of H+ concentration in the electrolyte affects the electrode potential, which is why increasing H+ increases the competition and makes it more likely for H+ to be reduced.

Im trying to explain why there is an optimal pH for the electroplating of copper and I need background information as to why too low pH or too high pH decreases the amount of copper deposition on the cathode.

In electrolysis, $Cu^{2+}$ is reduced in two steps, first to $Cu^+$ and later on to $ Cu$. Of course $Cu^+$ is automatically disproportioned into $Cu$ and $Cu^{2+}$ in acidic conditions. But in neutral solutions, $Cu^+$ may react with $H_2O$, forming an unwanted red precipitate of $Cu_2O$. This will affect electroplating of copper.


Why CO2 is not reducing the pH of NaOH as the way it is expected I am using $\ce{CO2}$ for the reduction of pH of $\ce{NaOH}$ solution ($\pu{2.5g}\; \ce{NaOH} + \pu{250ml}\; \ce{H2O}$, with $\mathrm{pH}\; 13+$).
I am introducing $\ce{CO2}$ with different dosages ($\ce{50cc, 100cc, 150cc, 200cc, 250cc, 300cc, 600cc, 900cc, 1200cc}$, and $\pu{1500cc}$) using syringe into $\pu{35 ml}$ solution of $\ce{NaOH}$ prepared as mentioned above. The $\mathrm{pH}$ first increases and then starts decreasing but stop at the same level of initial $\mathrm{pH}$ value of the solution (The amount of $\ce{CO2}$: $\pu{50cc}=\pu{0.03g}$).
Trials run for $\pu{60 min}$ after that, the samples are preserved at $\pu{4 ^\circ C}$ and $\mathrm{pH}$ measured again after $\pu{24 h}$, but the value does not reduce.

Ideally, if everything works properly, the $\mathrm{pH}$ of the solution should decrease gradually. However, instead of $\mathrm{pH}$ reduction first it increases and then decrease at the same level, when we mix $\ce{CO2}$.

I will appreciate if someone share his or her experience or comments as to what is the possible cause of this unexpected trend and how to fix it.

There are a so-called acid error and an alkaline error with glass electrodes. Which means that the pH electrode shows a systematic bias when your solution is too acidic or too basic and if it contains lot of sodium ions. The value which you quote for +13 are unreliable. So if you are bubbling carbon dioxide into NaOH solution, you are indeed consuming NaOH and converting it into sodium carbonate. The Wikipedia article on glass electrode is poorly written however if you search the term "alkaline error of glass electrodes" you will hundreds of better results.

May be you should follow the pH changes spectrophotometrically using a proper indicator (dyes) in NaOH.

Also keep in mind that you shouldn't store NaOH in glass containers because glass starts to dissolve in NaOH (very slow process though)


How does methyl vinyl ketone dimerise with heating? (A) to (B) is simply reduction, tosylation followed by intramolecular reaction and removal of tosylate. However, I am not able to guess how dimerization will take place, seems similar to Rauhut–Currier reaction, but can't guess. Any hint?

The dimer of methyl vinyl ketone (1) is not shown in the choices in the question. The dimer is A shown below. It is a "tail-to-tail" dimer. Product B is an intramolecular ketal as a probable mixture of epimers. Compare with this paper.

B. P. Mundy, et al., J. Org. Chem., 1978, 43, 2347.


Is flour or baking soda really suitable for extinguishing grease fires? I think most have heard the advise that you should put flour (or baking soda) on a grease fire to absorb the oil before. While this is certainly better than using water, I'm skeptical.

For starters it is know that flour is excellent at demolishing flour mills,  but in these instances it is a dust explosion (actually a deflagration) in a confined space versus a splash of mostly unaerated powder onto a fire. Even the fine particles that fall off should only cause mild flames compared to the on going grease fire. The main cause of skepticism is the potential charring of the flour releasing water and causing splattering (i.e. bigger fire).

Alternatively baking soda has been suggested, but this also seems dodgy as it starts to decompose at 50 °C releasing carbon dioxide and water, and how many people have enough baking soda to fill a pan in their kitchen ready to go? It seems that a lid would be the most appropriate response and of course would not recommend someone to do their own fire fighting if not prepared to do so, but when seconds count help is only minutes away.

I suppose that decomposition would be less of an issue if sufficiently slow kinetically. But debating physical sciences is a fools errand so I ask: Is flour or baking soda really suitable for extinguishing grease fires?

Do not use flour, it may explode!

Use the pot lid to smother the fire.
Use a fire blanket - read the instructions, watch a you tube demo. People who think they know what they're doing often spread it or end up burned.
Salt, will put out small fires, such as a pan fire, you won't have enough salt to smother a pot or absorb the heat.
Good for putting out small fat flare ups on a barbie plate.


Wedge Dash Diagram of 1,2-Dichloroethane I'm in class right now and my teacher put this up on the board as a representation of 1,2-dichloroethane. He also projected a model. I'm thinking ... his board drawing is wrong. Shouldn't the chlorines also be on dashes too? There's no way the chlorines can be in the same plane as the carbon atoms; the model makes it too obvious.

The teacher's drawing is correct.  Of course there are other possible conformers that could be drawn, but there is nothing wrong with the one that is pictured.   Basic rule:  a dashed wedge indicates a bond projecting behind the plane of the screen, a bold wedge indicates a bond projecting out in front of the plane of the screen, and a simple solid line indicates a bond existing in the plane of the screen.  Also note that this drawing notation can be used for any molecule, not just carbon systems.


How do you experimentally calculate the ionization constant of the carbonate ion? Given sodium hydrogen carbonate, $\ce{NaHCO3}$, as well as hydrochloric acid, $\ce{HCl}$, how would you experimentally determine the ionization constant for the carbonate ion, $\ce{HCO3-}$? By experimentally, I mean by actually doing the reaction and measuring the pH and such, and ignoring the already known theoretical values.

Here's what I've got so far.

The sodium hydrogen carbonate dissociates in water.
$$\ce{NaHCO3->Na+_{(aq)} + HCO3-_{(aq)}}$$

Then, titrate the ion with the hydrochloric acid, a strong acid.
$$\ce{HCO3-_{(aq)} + HCl_{(aq)} <=> H2O_{(l)} + H2CO3_{(aq)} + Cl-_{(aq)}}$$

It's at this point that I would measure the $\mathrm{pH}$ and determine the necessary concentrations, and then plug it into the formula

$$K_\mathrm{b} = \frac{\ce{[H2CO3]}}{\ce{[HCO3-][HCl]}}$$

But I'm wondering if I need to go further with the $\ce{H2CO3}$ since it would hydrolize in water:
$$\ce{H2CO3_{(aq)} + H2O_{(l)} <=> H3O+_{(aq)} + HCO3-_{(aq)}}$$

So far, so good. You'll have to graph the $\mathrm{pH}$ of your system with the volume of titrant you've used to construct what's called a titration curve.

When you do the titration curve for this experiment, depending on the relative strengths of the bases involved, you will see one or two points of inflection points that will signal that the equilibrium conditions have changed. This will tell you if you need to take into account the second ionization. If there is a second inflection point, that's the second ionization.

Without actually going ahead and looking at the $\mathrm{p}K_\mathrm{a}$ of $\ce{H2CO3}$, I think for most computations, it could be omitted, since it is a second ionization which has a much lower constant than the first. Not unless you're looking for zwitterionic states or something like that.


Help with determining concentration of magnesium hydroxide for a specific pH level This seems like such an obvious and elementary-level question for someone who has taken high school chemistry, but yet I'm having a difficult time solving it. I need to find the right amount of $\ce{Mg(OH)2}$ to have a $\pu{250 mL}$ solution with a $\mathrm{pH}$ of $8$ and $9.$
I first started by trying to determine the concentration of $\ce{Mg(OH)2}$ needed to obtain a solution with a certain pH. Since $\ce{Mg(OH)2}$ is a weak base, we must use a RICE table to determine its concentration at equilibrium.
$$
\begin{array}{lcccc}
  & \ce{Mg(OH)2 &<=> & Mg^2+ &+ & 2OH-} \\
\text{I} & x & & 0 && 0 & \\
\text{C} & x & & +x && +2x \\
\text{E} & x & & x && 2x
\end{array}
$$
However, the equation involving $K_\mathrm{sp}$ only involves the $x$ values in the RICE table above, not the actual concentration. How would I calculate the $\mathrm{pH}$ level for $\ce{Mg(OH)2}$ with varying molarity? I thought it was a weak base, so I'd have to use a RICE table.

If $c$ is the concentration of $\ce{Mg^2+}$ in the $\ce{Mg(OH)2}$ solution, the concentration $[\ce{OH-}] = 2c.$
At $\mathrm{pH}~9,$ $[\ce{OH-}] = \pu{1E-5 M},$ then $c = \pu{5E-6 M}.$ So you have to dissolve $\pu{1.25 μmol}$ $\ce{Mg(OH)2}$ in $\pu{250 mL}$ water. This is $\pu{71.3 μg}$ of $\ce{Mg(OH)2}.$ This is difficult to do in practice, as such a small amount is difficult to weigh. It may be done in two steps: first prepare a moderately concentrated solution, then dilute it to $\mathrm{pH}~9.$


Effect of temperature on equilibrium Why is exothermic reactions are favoured by low temperature while endothermic reactions are favoured by high temperature?

I'll try and simplify the answer.

Looking at the image above, imagine the products sitting above the letter D and the reactants sitting below the letter B.  Now, imagine the particles involved in the reaction moving since they have kinetic energy.  It is much easier for the reactants to climb up the top of the hill labeled C then it is for the products to climb up the hill.  Once a reactant climbs the hill and slides down to products it will not be able to go back to reactants by climbing up the hill unless we give them more energy by raising temperature.  Raising temperature allows the products to climb up the hill and go to reactants.

The forward reaction in the diagram is exothermic and the reverse is endothermic.  So, the exo is favored at low T and the endo at high T.

You can also look at the equation delta G = delta H - Tdelta S.

Make up some numbers for delta H and delta S.  You will see that endo reactions becomes more spontaneous as you raise T and exo reactions become more spontaneous as you lower T regardless of delta S.


A useful way to reuse non-recyclable plastics: Turn them to petrochemicals, shall we? I read here (https://chemistry.stackexchange.com/a/34016/101483) that strongly acidic catalysts can facilitate dimerization of alkene. With catalysts on zeolite to allow only the isooctane (octane rating 100) to come out as the only product, can non-recyclable plastics be turned into useful petrochemicals.
In this particular case of isooctane, polystyrene (C8H8)n is broken down into styrene (C8H8) the monomer. Styrene is hydrogenated into octenes (C8H16). The hydrogenation on octenes at the crucial third stage with strongly acidic catalysts on zeolite only yields isooctane.
Is the whole process feasible?

The broad question here is whether "non-recyclable" plastics can be turned into fuel. The short answer to that is yes. More on that below. However, the specific proposal to convert polystyrene to isooctane is not particularly feasible. I'll address that first.
The proposed process begins with depolymerization of polystyrene back to styrene. Although lab-scale processes have been reported that give as high as 80% yield of styrene monomer, there is not yet an economically feasible process for scale-up. In fact, if styrene could be depolymerized practically, it would not be considered "non-recyclable". Furthermore, if the styrene could be recovered, it would be much more valuable as styrene than as gasoline, so there would be no incentive to spend the money on the subsequent steps described by OP.
Nonetheless, if one wanted to make liquid fuel from styrene, there is no need to produce isooctane specifically. Liquid fuels are generally mixtures of a variety of hydrocarbons, and the exact chemical composition does not especially matter as long as the autocombustion temp, vapor pressure and other relevant properties fall within an acceptable range.
So rather than make isooctane, one could simply hydrogenate styrene to ethylbenzene, an already common anti-knock agent or all the way to ethylcyclohexane (a more likely final product than octene).
Going back to the original premise of getting liquid fuel from plastics, just as there is no need to produce a specific compound from the styrene, there is no need to recover styrene specifically from the polymer, nor even restrict one's self to a specific polymer type such as polystyrene. Instead, bulk plastic mixtures can be pyrolyzed to produce a mixture not all that different from some crude oils and gases. These liquids and gases can be used as fuels themselves or mixed with existing petroleum products and refined.
Many challenges remain to make such a process economically viable, however, not least of which is the cost of collecting and transporting plastic waste (generally quite low density) compared to recovering oil from the ground.


What type(s) of gloves are effective against DCM and acetone? I know that standard latex and nitrile gloves don't stop common organic solvents such as DCM and acetone. However, I was reading a Reddit thread and there were comments that if you are coming into contact with solvents you are just wearing the wrong type of gloves. In a non-manufacturer specific fashion, what types of gloves will stop DCM and acetone? What will it slow down long enough for me to have a chance to finish what I'm doing and change my gloves in a safe manner? I am looking for a guide on what gloves to use. I just had DCM on my mind since nitrile doesn't slow it down, and it burns. And acetone, since it is hard to stop.

I just did a bit of an experiment. My lab uses Kimberly Clark nitrile gloves, and we have a huge jug of acetone. First, I put a glove on my hand, and poured acetone all over it. My hand felt cold, but not nearly as cold as it feels when I spill acetone on my bare hand. I allowed the acetone to evaporate from the glove (approximately 30 s), and removed the glove. There was absolutely no acetone on my hand or the inside of the glove. I then took a fresh glove, and poured enough acetone into it to fill the fingers with acetone. I held the opening shut for about 5 minutes, and poured out the acetone. During that time, the glove slowly felt thinner and easier to stretch. At the end, it was so thin that it tore with almost no effort. So for the particular pair of nitrile gloves and acetone, it looks like somewhere between 1 and 5 minutes' exposure will render the glove useless.


At what pressure will hydrogen start to liquefy at room temperature? I want to increase a fixed-size object's internal gas pressure by generating hydrogen in it, but I could not find the proper phase diagram for it. So I am wondering how high pressures I can get.

$\ce{H2}$ cannot be liquified at room temperature, whatever the pressure. Generally speaking, all gases can only be liquified when the temperature is under its critical value.


Partial pressure of water in a submerged balloon A balloon is half filled with water at sea level ($P = \pu{1 atm}$, $T = \pu{298K}$) such that the partial pressure of water is $\pu{24 mmHg}$.  It is then submerged to a point where $P = \pu{2 atm}, T = \pu{298K}$.  What will be the vapor pressure of water at that point?

My first instinct is to find the mole fraction of water in the balloon.  Using Dalton's law,

$P_{total} = P_{\ce{H2O}} + P_{\mathrm{atm}} = 24 + 760 = \pu{784 mmHg}$

$P_{\ce{H2O}} = X_{\ce{H2O}}\,P_{total}$

Solving for $X_{\ce{H2O}}$, I get $24/784$.  Then I use that to find the new partial pressure at the new depth, $(24/784)\cdot 1520$.  But the answer seems to be that the partial pressure of water remains constant. Can someone explain why the partial pressure will be constant?

Excellent question. On the one hand, the vapor pressure of water (or any other volatile liquid) is a function of the temperature alone, but on the other, if you assume that the mole fraction is constant you find that the vapor pressure of the water changes even though the temperature is constant.

The short answer

The water vapor pressure stays constant because as the pressure in the balloon increases, the mole fraction of the water reduces proportionally to maintain the equilibrium it has established.

The longer answer

Let's dig into the former approach I mentioned in the introduction first. We can begin with the Clausius-Clapeyron relationship.

The Clausius-Clapeyron relationship (often is simplified to):

$$\ln \left(P_{vapor}\right) = -\frac{L}{R}\left(\frac{1}{T}\right) + C$$

Where $L$ and $C$ are constants and $R$ is the ideal gas constant.

Using this model, we would expect the vapor pressure of the water to remain constant and therefore the mole fraction of water in the gas phase to decrease.

Dalton's law tells us: $$X_{water}=\frac{n_{water}}{n_{tot}} = \frac{P_{water}}{P_{tot}}$$

Using this expression we can see that, as the total pressure of the gas increases and the vapor pressure of the water stays constant, the mole fraction of water in the gas phase must go down. If we assume, as a first approximation, that the total number of gas molecules in the gas phase is constant, then a reduction in mole fraction can only be accomplished by some of the molecules of water in the gas phase being reabsorbed into the liquid water phase. This consequence is in agreement with the Le Chatelier principle prediction. A decrease in volume available to the water molecules will shift the equilibrium below to the left:

$$H_2O\,(l) \rightleftharpoons H_2O\,(g)$$

Thank you for such a rich question! How interesting that there are so many topics it relates too! Excellent!

Other thoughts

When we calculate the initial pressure of the other gases in the volume, we should use:

$$P_{tot} = P_{water} + P_{all\,other\,gas}$$

$$P_{tot} = P_{external} = 760\,mm\,Hg$$

$$P_{all\,other\,gas} = P_{tot} - P_{water} = 760\,mm\,Hg - 24\,mm\,Hg = 736\,mm \,Hg$$

The dissolved gas in the water in the balloon will also increase at higher pressures. How will this affect our calculations?
We have not considered energy concepts explicitly in this argument about the pressures being equal (in the first bullet above), but we could have considered the elastic energy stored in the bonds of the balloon. How could we include this to get a better estimate of the pressure of the other gases in the balloon?


How does mass affect bond stretching in UV spectroscopy? If, for example, you have a bond on the form:

H - X, where H is hydrogen and X is some other element.

How does the mass of X affect the stretching of the bond during UV spectroscopy? Why do heavier atoms require a less 'energetic' wave to cause them to stretch?

In a very crude approximation, every bond can be considered to be a quantum harmonic oscillator. As you might know, the energies of such an oscillator are given by

$$ E_\nu = hc\omega_e \left ( \nu+\tfrac{1}{2}\right )$$

where $h$ is Planck's constant, $c$ the speed of light and $\omega_e=\sqrt{k/\mu}$. The constant $k$ is the force constant of the bond and $\mu$ is the reduced mass
$$ \mu=\frac{m_1m_2}{m_1+m_2}$$
So if you increase the mass of one of the particles, while $k$ stays roughly the same, $\omega_e$ becomes smaller and the energy spacing between the vibrational levels decreases.


Use of repeated numerical prefixes for substituents on methane Take trichloromethane for example. Is its preferred IUPAC name:

  1,1,1-trichloromethane

or simply

  trichloromethane

And which is that rule in IUPAC that governs this naming? Thank you!

According to Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013,

  P-14.3.4.2 The locant '1' is omitted in substituted mononuclear parent hydrides.

For the compound $\ce{CHCl3}$ (chloroform), the parent hydride is methane, which is mononuclear. Hence the locant '1' should be omitted in the preferred IUPAC name (PIN); it is simply trichloromethane.


Nomenclature priority of prefixes consisting of identical letters but containing different locants I have no idea how to name this compound. Can someone please help give the IUPAC name for this? Also mention the IUPAC nomenclature rules followed while naming.

The compound that is given in the question consists of a cyclohexane ring with two substituent groups: 1-chloroethyl and 2-chloroethyl.

The correct numbering of locants is stipulated by Rule 14.4 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

  P-14.4 NUMBERING
  When several structural features appear in cyclic and acyclic compounds, low locants
  are assigned to them in the following decreasing order of seniority:
  (…)

However, the given case is not explicitly mentioned in this rule.

The correct numbering is actually indirectly given by Rule 14.4 (g):

  (g) lowest locants for the substituent cited first as a prefix in the name;

Since the correct order of citation as a prefix in the name is given by Rule P-14.5.4:

  P-14.5.4 When two or more prefixes consist of identical Roman letters, priority for order of citation is given to the group that contains the lowest locant(s) at the first point of difference.

Thus, ‘1-chloroethyl’ is cited before ‘2-chloroethyl’ in the name, and thus a low locant is assigned first to the 1-chloroethyl group.

Therefore, the complete name for the compound that is given in the question is 1-(1-chloroethyl)-4-(2-chloroethyl)cyclohexane.


Derivation of formula for Hydrolysis constant While studying acid-base equilibrium I thought I would derive the formula for hydrolysis constant myself. But I ended up getting a weird result and I am not able to understand where I went wrong.

The equilibrium constants for reactions that tend to favour products are large. As the concentration of the reactants decrease the equilibrium constant slowly should move toward infinity. For example the ionisation of strong acids is taken to be a reaction that goes to completion. According to my understanding the equilibrium constant $K_\mathrm a=\infty$. The equilibrium constant for its reverse reaction would be zero. Similarly the equilibrium constant for the ionisation of salts should be $K_\mathrm s=\infty$.
Let us consider the reaction $\ce{HA + BOH <=> AB + H2O}$. This can be carried out in a number of steps:
$\ce{HA <=> A- + H+}$  with $K_\mathrm a$
$\ce{BOH <=> AB + H2}$ with $K_\mathrm b$
$\ce{H+ +OH- <=>H2O}$ with  $1/K_\mathrm w$
$\ce{A- +OH1+<=> AB}$ with $1/K_\mathrm s=0$
So the hydrolysis constant should be
$1/K_\mathrm h=K_\mathrm aK_\mathrm b/(K_\mathrm wK_\mathrm s)=0\cdot K_\mathrm aK_\mathrm b/K_\mathrm w=0$
hence $K_\mathrm h=\infty$
I am sure there is something I havent understood well and hence I've ended up on this result. Can somebody explain where I am wrong?
Note: In most proofs the equilibrium constant of ionisation of salts is not considered but I see no reason to not consider it.

Sorry, I didn't get that. first of all, which hydrolysis are you talking about? I will talk about weak acid and strong base equilibrium. You have three concentrations [HA], [H+], [A-]. For instance, if they took 1L of 1 mM of HOAc (acetic acid) and added 0.3 mmoles of NaOH (concentrated, so no dilution) you get [A-]+[HA] = 1mM Then you say HA <-> H+ + A- has Ka. Then

[H+][A-]/[HA] = Ka. Then you say  [HA] = x, so [A-] = concentration_total- x (in this case 1mM -x). Then you get an equation
[H+]*[concentration_total-x]/x=Ka. Then you solve this square equation.

Regarding your question. I think you are mixing up to things: protonation of weak acid (stuff, relevant to analytical chemistry with a lot of drama, pH swings etc) and the behavior of other salts (they just dissociate completely and we forget about them). In stories like "to a solution of NaCl and K-acetat a drop of acid was added. What happen?" you should first forget about NaCl. It has no impact on the question. You should also forget about K+. Now you have "to a solution of Acetate- a drop of acid was added. What happen to pH?" Now you can easily calculate it using a square equation.

Run away from chemistry. Carcinogens, stinky chemicals, no  jobs... It teaches you how to make poisons and drugs, but how useful is that?


Axial or planar chirality in spiro[3.3]hepta-1,5-diene The two spirocyclic compounds drawn below are enantiomers. My question is, do they exhibit axial or planar chirality?

3D representations:

Useful Definitions (from the IUPAC Gold Book)

chirality plane - "A planar unit connected to an adjacent part of the structure by a bond which results in restricted torsion so that the plane cannot lie in a symmetry plane. For example with (E)-cyclooctene the chiral plane includes the double bond carbon atoms and all four atoms attached to the double bond; with a monosubstituted paracyclophane the chiral plane includes the monosubstituted benzene ring with its three hydrogen atoms and the three other atoms linked to the ring (i.e. from the substituent and the two chains linking the two benzene rings)."
planar chirality - chirality resulting from a chirality plane.
chirality axis - "An axis about which a set of ligands is held so that it results in a spatial arrangement which is not superposable on its mirror image. For example with an allene abC=C=Ccd the chiral axis is defined by the C=C=C bonds; and with an ortho-substituted biphenyl the atoms C-1, C-1', C-4 and C-4' lie on the chiral axis."
axial chirality - chirality resulting from a chirality axis

Interpretation

According to the above definition, a chirality plane must involve restricted torsion (restricted rotation) about a bond.
A molecule can contain both a chirality plane and a chirality axis; an example would be an ortho-substituted biphenyl where the restricted rotation is responsible for chirality (atropisomers).

Conclusion

  Axial or planar chirality in spiro[3.3]hepta-1,5-diene

There is no restricted rotation that brings about chirality in this molecule.  Unlike trans-cyclooctene or a biphenyl, this molecule cannot be made to lie in a plane by torsional motion (at least not as long as we maintain the tetrahedral nature about the spiro carbon). Therefore, this molecule does not possess a chirality plane.

Rather, this molecule is analogous to the allenes. It does have a chirality axis passing through the $\ce{CH}$ group in each ring furthest from the spiro carbon along with the spiro carbon. The molecule can be said to possess axial chirality.


What is octanamide used for? I read a paper about enzyme-catalysed synthesis of octanamide.

What is octanamide used for?

Well, one thing is as plastics. Nylon is a linear polyamide; it's traditionally produced from alkyl-diamines and a dicarboxylic acid (to donate the acyl group of the amides, as well as a second alkyl chain joined directly to the carboxyl), and the typical product is a poly-hexanamide, but nylon-8,x variants are also possible, as are reactions involving a more direct route of welding the monomers together and arriving at the same place.

Some Google-fu turns up a number of possible end products for which octanamide aka caprylamide would be a useful intermediate. By itself, it's an emulsifier and common solvent in organic synthesis. Other alkyl amides including derivatives of octanamide, often with a few additional alkyl or amino groups welded on, are used in drugs, inks/dyes, fragrances, and skin care items. N,N-dimethyloctanamide and N,N-diethyloctanamide appear to be common ingredients in deodorants and sunblocks, though I'm not sure if they're an actual fragrance or a carrier/stabilizer (Edit: they seem to be emulsifiers and stabilizers for essential oils of the fragrances in water-based products).


Why are the nitrate salts hygroscopic? Any metal is never extracted from its nitrate salt because nitrate salts are highly soluble in water, so flow away along with rainwater, or any other source.
But why are the nitrate salts so hygroscopic in nature? Is there any relationship with hydrogen bonding? Or something else is the cause?

Anions like nitrate forms strong hydrogen bond with water. Here water bridges the nitrate oxygen with hydrogen bonds giving it a very stable arrangement.


If all molecules have basically the same KE at a given temperature, why do lighter isotopes evaporate more readily? One of the proxies used for paleoclimate is the ratio of oxygen-18 to oxygen-16 in ice cores and in sedimentary rocks.  The idea is that water molecules with oxygen-18 generally evaporate less readily than oxygen-16, and condense more readily;  and this proclivity varies with temperature.  One can then use the deviations of oxygen-18 concentration from a standard ratio to say something about Earth's temperature in the distant past.
Why do light water molecules evaporate more readily?  It makes some intuitive sense that lighter molecules would evaporate more easily.  But how does this square with the equipartition theorem and the idea that all molecules have the same amount of kinetic energy at a given temperature?
It would seem to me that because the intermolecular forces between two water molecules are independent of their mass, it should take a certain amount of energy to remove a water molecule from its neighbors in the liquid phase and move it far away.  This energy requirement doesn't depend on the mass of the molecules, only on how the force on it varies with distance ($W = \int \vec{F} \cdot d\vec{r}$.)  So by this argument, molecules with the same amount of kinetic energy would have the same likelihood of evaporation.  And by the equipartition theorem, the translational kinetic energy of a water molecule is only dependent on the temperature, not on the mass of the molecule.
EDIT:  Just to pre-empt the "it's velocity that matters, not energy" argument:  I agree that lighter molecules would, on average, have a higher velocity.  But from classical physics, we know that applying a fixed force to a smaller mass leads to a larger change in velocity, while a larger mass gets a smaller change in velocity.  That would imply that the "escape velocity" required for a heavier molecule would be smaller; since it's heavier, its motion would not be affected as much by the force it feels from the other molecules. This would imply that while heavier molecules are moving more slowly on average, they also don't need to be moving as quickly to escape.  In other words, the velocity threshold for evaporation should be lower for heavier molecules.  These two effects (lower speeds and lower threshold for evaporation) should cancel each other out.
Where does my logic fail?

Evaporation depends on molecular velocity not on the energy
The simple mistake in your assumption is that evaporation is driven by molecules having enough energy to escape the liquid phase. This is wrong. Evaporation depends on whether the molecule is travelling fast enough to escape the liquid.
The molecules in the liquid will have, on average, the same kinetic energy. But this means that, in practice, the molecules will have a range of different kinetic energies across a wide distribution. But, since kinetic energy is a function of both molecular mass and molecular velocity, there will also be a distribution of velocities. And, crucially, the more massive molecules (in this case those with 18O) will have slightly lower velocities at the same average kinetic energy. So the lighter isotopes will be slightly more likely to have a high enough velocity to escape the liquid phase.
You can work through the mathematical details (taking into account the statistical distribution of kinetic energy and converting that to the distribution of velocities and then calculating the proportion of molecules with a velocity above the threshold for evaporation) and check that the proportion of light molecules is higher than the proportion of heavy molecules.

Clarification: why is the force-based view wrong
The argument that applying a fixed force leads to a smaller effect on a heavier molecule so heavier molecules don't need to moves as quickly to escape the liquid is flawed for several reasons.
It sounds relevant but isn't because the vast majority of all collisions don't lead to escape from the liquid. There are so many momentum (and KE) redistributing collisions that all we have to consider is the distribution of energy/momentum across the ensemble of molecules in the liquid. And what we need to know is the average proportion of molecules having enough velocity to escape after their last collision.
Yes, the force argument matters for individual collisions (if they are not purely elastic), but, if you simulated a drop of liquid down to the molecule level, you would find that the distribution of energy/momentum would match the statistical one expected from theory and you would have gained no new knowledge. Follow an individual molecule and, yes, the thinking about force will apply. But follow the distribution of velocities across the ensemble of molecules and what matters is the proportion of molecules having a high enough escape velocity and that is governed by the statistical distribution. Think about the properties of the ensemble of molecules not the individual molecule.
And the force argument ("the escape velocity will also be lower") doesn't work at the ensemble level. All that matters is the velocity after the last collision and the only way to know that is to look at the statistics of the whole system not the individual forces on individual molecules.


Equilibrium between R and S configuration? We have learnt that the energy evel of both R and S optical isomer of a compoud is same, just as well the entropy of each isomer is also same. If we consider the conversion of R form of a compound to it's S isomer we observe that the change in gibbs free energy is "0". This implies that the two isomers must exist in equilibrium. But we also know that any compound does not change from its R configuration to its S configuration for example inversion of configuration in Sn1 mechanisms lead to only one configuration and not both are obtained after the reaction.

So the question is that why don't any compounds R isomer convert into its S isomer on its own so as to maintain equilibrium ?

Because even though neither the R nor the S isomer is thermodynamically favored (in an achiral environment), the free energy barrier (or activation energy) between those two states is huge. This is the most common case, when R/S interconversion requires breaking and reforming a chemical bond. Then, even though the thermodynamic equilibrium would be to have a racemic mixture, kinetics prevent the solution from reaching this equilibrium state.


The effects of adding neutrons to an atom So neutrons are neutral in terms of charge, and adding neutrons to an atom affects its atomic mass. But when neutrons are added to the nucleus, the nuclear radius would be affected. Couldn't that affect how the charge of the nucleus is distributed within the nucleus?

With an atom large enough that, say, 40 neutrons could be added (and still be stable), I would imagine that the nucleus would be slightly rearranged to accommodate this (so that neutrons and protons are distributed throughout the nucleus and there isn't a shell of 40 neutrons surrounding what is the pre-existing nucleus).

I imagine that this redistribution (if it does occur) would dilute the charge coming off of the nucleus (as the protons are farther apart and the nucleus is less dense in terms of charge), which would lead the electron shells to be farther from the nucleus. This would increase atomic radius, affecting a wide-range of chemical characteristics. I could equally see an opposite effect, where the nuclear charge is strengthened (which would decrease atomic radius).

Is this an actual reaction to the adding of nuclear charges? If so, does it occur in large enough quantities to play a significant role in chemistry? Or have I just made a false assumption?

Even if such a phenomenon were to occur, and the resulting nuclei were sufficiently stable to participate in chemistry (they would not be sufficiently stable, in reality), the "inflation" of the charge of the nucleus would almost certainly have a negligible effect on the chemistry.  The nucleus is really really really tiny with respect to the length scales of bonds and the electron cloud.

To give an idea of the scales involved: the largest stable element, uranium, has a nuclear radius of about fifteen femtometers $\left(15~\mathrm{fm}\right)$, per Wikipedia, or $1.5\times 10^{-14}~\mathrm m$.  Bonds and electron clouds are measured in Angstroms $\left(10^{-10}~\mathrm m\right)$.  Thus, the nuclear radius would have to expand by well over $\sim\!\!100$-fold in order for the assumption of a point-charge nucleus to begin to fail.

It is well accepted that the effect of isotopic substitutions on chemistry primarily is due to the change in mass of the nucleus, leading to the kinetic isotope effect.  Another potential isotope effect on reactivity involves the change in nuclear spin, leading to a "magnetic isotope effect" only relevant for certain classes of reactions between species with unpaired electrons (e.g., organic radicals and/or various inorganic species).

Isotopic substitution does affect the behavior of nuclei in certain spectroscopies, however, leading to isotopic shifts.  The isotopic mass change can have noticeable effects on the vibrational spectrum, due to the effect on the local distribution of mass. As described in the link to ADF's documentation provided by Aesin in a comment to the question, these effects can also be seen in the hyperfine structure of atomic spectra.  As well, the NMR isotopic shift of light elements, especially hydrogen, can sometimes be observed.


Confusion about the ICE table Calculate the pH of a $\pu{2M}$ $\ce{H_3PO_4}$.Calculate the equilibrium concentrations of $\ce{H_3PO_4 , H_2PO_4^{-} , HPO_4^{2-}}$.

First equation:

$\ce{H_3PO_4  <=> H^{+} + H_2PO_4^{-}}\ \ \ K_\mathrm{a1} =7.5\times 10^{-3}\\$

$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{H_3PO_4}] & [\ce{H^{+}}] & [\ce{H_2PO_4^{-}}]\\ \hline
\text{I} & 2 & 0 & 0 \\
\text{C} & -x & +x & +x \\
\text{E} & 2-x & +x & x\\ \hline \end{array}\\$

$$K_\mathrm{a1}=\frac{[\ce{H_2PO_4^{-}}]*[\ce{H^{+}}]}{[\ce{H_3PO_4}]}=\frac{x^2}{2-x}=7.5\times 10^{-3}$$

Solving for $x$ we get $x=\pu{0.118M}$.

$[\ce{H_3PO_4}]=2-x=2-0.118=\pu{1.882M}$

$[\ce{H_2PO_4^{-}}]=x=\pu{0.118M}$

Second equation:

$\ce{H_2PO_4^{-}  <=> H^{+} + HPO_4^{2-}}\ \ \ K_\mathrm{a2} =6.2\times 10^{-8}\\$

The solution manual says:$$K_\mathrm{a2}=\frac{[\ce{HPO_4^{2-}}]*[\ce{H^{+}}]}{[\ce{H_2PO_4^{-}}]}=\frac{0.118\times[\ce{HPO_4^{2-}}]}{0.118}$$

$[\ce{HPO_4^{2-}}]=K_\mathrm{a2}$

However I'm not sure why didn't we use an ICE table for the second equation as follows:

$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{H_2PO_4^{-}}] & [\ce{H^{+}}] & [\ce{HPO_4^{2-}}]\\ \hline
\text{I}& 0.118 & 0.118 & 0 \\
\text{C}& -x & +x & +x \\
\text{E}& 0.118-x & 0.118+x & x\\ \hline \end{array}\\$

Is it because $x$ is negligible with respect to 0.118? Or should the concentrations precalculated above be in the equilibrium row of the ICE table?

Yes, x << $\mathrm{0.118 M}$ that it is not gonna hurt even if it is ignored. But for a rigid perfectionist,
$$K_\mathrm{a2}=\frac{[\ce{HPO_4^{2-}}][\ce{H^{+}}]}{[\ce{H_2PO_4^{-}}]}=\frac{x\times (0.118+x)}{0.118-x}$$
Hence we get a quadratic equation,
$$x^2 + 0.118000062 - 7.316\times 10^{-9}=0$$
Solving for $x$, we get $x = 6.199993484584665\times 10^{-8}$ which is approximately equal to the value of $K_\mathrm{a2}$.


Why there is a non-linear trend on the melting points for period 2 elements, especially Na to Si? Using the knowledge of periodicity from high school cannot explain this, what it tells me is just metallic bonding.

The model of delocalised electron from a metallic lattice can explain the increase of melting point from Na to Si, but it fails to explain why the relationship is non-linear?

If your textbook says that the melting point increases linearly, that is incorrect, as you point out. Generally high school textbooks can only make broad, qualitative statements about characteristics like melting points since the physics involved can depend to a larg degree on very complicated (quantum mechanical) interactions between the atoms (and their electrons) in a solid.

See for instance http://www.ptable.com/, and try playing with the slider on the top right to see melting points for different elements. There is a general trend of heavier metals in the same period having higher melting points, but there are some exceptions, such as manganese (Mn) around 1570 K.

As for why specifically Si's melting point is so much higher than Al's, the only truly correct answer is something along the lines of "that's just because of the differences in the way Si and Al atoms interact with each other". Because that's not a very satisfying explanation, I can try to paint a more intuitive picture: (UPDATE: it has been pointed out in the comments that even this simplified picture is wrong, see the linked wikipedia article for more details.) Aluminium has three valence electrons, which means that it would need to form five covalent bonds with its neighbours in order to reach the favourable octet of electrons. Unfortunately there is no regular crystal structure where every atom has exactly five nearest neighbours, so what ends up happening is that one electron from each atom delocalises, and the resulting postively charged ions are weakly (i.e. noncovalently) bound to each other, resulting in a low melting point (in other words, the atoms can be shaken apart easily).

Silicon on the other hand, has four valence electrons, only one more than Al, but this means that it only needs to bind to four neighbours to get an octet, and it turns out that there is a regular arrangement of atoms such that each atom has exactly four nearest neighbours (called the diamond structure). So, in solid Silicon, each atom is covalently (and thus strongly) bound to four neighbours, (and no electrons delocalised) which means that the atoms are hard to shake apart, which leads to a much higher melting point.

Now, this might be a more pleasing explanation (at least I hope so), but note that it is only meant for illustrative purposes: the same logic, dealing with 3 vs. 4 valence atoms can also be applied to for example Thallium and Lead, but these two metals have virtually the same melting point. Why is that? Well, there are other effects at play for these heavier atoms. In other words, it really does come down to how groups of the same atom interact with each other, and much of that can only be understood through the framework of quantum mechanics, and a rigorous treatment of solid state physics is probably too much to expect from a high school textbook :-)


Karl Fischer Titration simple explanation I'm looking for a simple explanation how the Karl Fischer method to determine the content of water works. I'm not a chemist but an engineer from an electronic-physical field and my chemistry courses are a while ago, meanwhile. That's why I can't understand what's going on but I don't really need a detail explanation, just low-level enough that I can say "Yep, makes sense" and high-level enough, that I can understand it.
Hence, could someone please describe roughly what's going on?

Attempting an almost non-chemical description, you need to know that the concentration of a sample, $c$, equates the number of moles $n$ per unit of volume $V$; expressed as $c = n/V$.
For once, a known volume $V$ of your liquid to be analyzed for its concentration $c$ of water into is injected into a chamber equipped with a stirrer and electrodes.  There is a stock of reagent solution for which you know ahead of the analysis its concentration. You equally know ahead how many molecules of the reagent will react with how many water molecules to yield a balanced reaction equation (stoichiometry).  The reagent solution shall react with water only.
The addition of the reagent solution to the analyte is monitored; for one, you know the volume of the reagent solution added.  Because of this known volume added, and the known concentration of the stock of reagent solution, you know how many molecules of the reagent were added.  The mixing and stirring ensures that each molecule of the reagent quickly encounters a molecule of water to react with.
For two, during the addition of the reagent solution to the analyte, the electric potential between the electrodes dipped into the analyte is monitored. Initially, the addition of a little volume of reagent solution changes this potential only a little.  However, continuing the addition of small volumes of reagent solution, there is one point where this change of potential suddenly is large.  This is the point when 1) basically all the water molecules reacted with the reagent, 2) the electrodes now monitor an excess of the reagent in respect of the water molecules, and 3) the titration may be stopped.
Because you know the volume of the reagent (of known concentration) added and the balanced reaction equations used in this analysis, you may calculate the number $n$ of water molecules consumed to reach this point of sudden change of the potential.  However, because you know both the number of water molecules consumed as well as the volume $V$ of analyte initially injected, you may compute the concentration of water molecules in the analyte, too ($c = n/V$).

The Karl-Fischer analysis offers a rapid characterization of organic liquids (minute scale), provided they do not react differently with the reagent.  Samples containing alcohol, for example, may differ in the underlying reaction equations and stoichiometries from samples containing water.  Generally speaking, to work well, typical samples submitted do not contain more than 1 mol/L of water (about 18 g/L).


Do the salt bridge ions have to match the ions in the half-cells solutions? Does the salt bridge in a galvanic cell have to have the same ions as the ones in the solutions? E.g. does it need to have nitrate ions if the half-cells contain copper nitrate and zinc nitrate, or could nitrate ions still be used if the solutions had sulfate instead?

In short: a priori no, it need not.

The purpose of the salt brigde is to close the electric circuit (transport of charges) between the two half-cells.  To work as intended, all you need to do is to fill the bridge with any salt whose ions, both cation and anion do not react with either half cell components, nor under the external potential applied, and ideally have the same mobility in the electrolyte of the bridge, expressed by the transfer number.  This is why aqueous solutions of KCl are used so often for this purpose, but you are free to choose differently.

And not to forget, the tube normally ends by a diaphragm at each side, to slow down considerbly the mechanical exchange of matter (ions) between the bridge's content and the half cells.


Why we can't see 3-d structures of certain compounds on PubChem? It's common to read statements like one given below on PubChem:

  Conformer generation is disallowed since MMFF94s unsupported element.

What's the meaning of the statement, and why are certain compounds unsupported? And what is MMFF94s?
This might be something really basic, but since I have never come across I asked it.

Since it's been a couple of days, I guess I convert my comment into an answer.

MMFF94s is Merck Molecular Force Field.
From what I understand, the issue is that this FF has certain limitations when it comes to electrostatic interaction, estimating polarizability and intramolecular interactions, which confuses the method trying to predict 3D conformers for ionic compounds, metal complexes or molecules with twisted geometry.

From the Concluding Discussion section of the original publication [1, p. 515]:

  Despite encouraging success, certain limitations are evident.
  One of particular importance arises from the fact that MMFF94 uses static atom-centered charges. As such, it neglects both higher order multipoles and electrostatic effects that arise from molecular polarizability.
  Because of these simplifications, MMFF94, like a number of other force fields, employs “enhanced” charge distributions that emulate the effect of polarizability in amplifying electrostatic interactions for favorable contacts in a high-dielectric medium. Unfortunately, these enhanced charge distributions also amplify electrostatically unfavorable iteractions, whereas proper account of polarizability would
  diminish them.
  They also improperly enhance electrostatic interactions in gas-phase or low-dielectric environments. Furthermore, they may not be optimal for describing intramolecular interactions, and may thereby limit the ability of the force field to account for differences in conformational energies.
  Indeed, compounds containing two or more strongly polar functional groups in close
  proximity have proven to be the most problematic in this respect, though good results have been obtained in most cases to date. …  Other significant limitations include: the overly simplistic nature of the bond-charge-increment scheme used to assemble the partial atomic charges …; the lack of conformational dependence of the resultant charges …;
  and the omission of bond-torsion (and certain other) cross terms needed to account for significant geometrical changes that can occur when a torsion angle varies,… an example being the elongation of an amide partial C−N double bond by up to 0.1 Å when conjugation is broken.
  A further significant limitation is that no account is taken of metal-ligand interactions beyond that afforded by a relatively simplistic model that includes only electrostatic and van der Waals nonbonded interactions.

References

Halgren, T. A. Merck Molecular Force Field. I. Basis, Form, Scope, Parameterization, and Performance of MMFF94. Journal of Computational Chemistry 1996, 17 (5–6), 490–519. DOI: 10.1002/(SICI)1096-987X(199604)17:5/6<490::AID-JCC1>3.0.CO;2-P.


Why does Sulfur dioxide incompletely freeze when entrained in a helium stream inside tubing dipped in liquid nitrogen? Background - In a laboratory setting, I use typical elemental analysis (EA) to generate about 50 µg of sulfur in the form of sulfur dioxide (SO2). The EA uses a helium carrier. As an example, a sample of silver sulfide wrapped in tin falls into the reactor set to 1000 °C at the same time as an oxygen pulse, hopefully leading to flash combustion and creation of SO2. The SO2 is carried through a desiccant (Magnesium perchlorate) followed by a quartz chip reactor set to 800 °C, and a GC column to separate out impurities by polarity. The helium flow rate through the EA is typically 80 mL / min. The pressure of helium at the top of the EA is set to approximately 135 kPa. The end detector is a stable isotope ratio mass spectrometer where we measure mass / charge (m/z) 64 and 66 (the hot quartz chip reactor exchanges oxygen atoms so as to force all SO2 to have the same oxygen isotope composition so that our only variability is from sulfur isotopes 34 and 32. The above scenario is routine and can be found in Fry B, Silva SR, Kendall C, Anderson RK. (2002) Oxygen isotope corrections for online δ34S analysis. Rapid Communications in Mass Spectrometry 16, 854-858. doi: 10.1002/rcm.651.
Observation - We were experimenting with analyzing smaller quantities of sulfur by tightening up the SO2 peak using a stainless steal loop dipped in liquid nitrogen for a standard amount of time followed by warming it to room temperature. The tubing is 1/16” OD and allowed to completely cool to liquid nitrogen temperatures before the sample is dropped into the reactor. I observed SO2 at the detector downstream of the loop still residing in liquid nitrogen at helium carrier flow rates above 30 mL / min. I had to reduce the flow rate to about 10 mL / min to completely trap all SO2 in the liquid nitrogen immersed tubing. Based on the phase diagram of SO2, even at this higher helium pressure, it seems like I am still well within the solid phase space.
Question - Are these observations expected and if so please help me understand why?
UPDATE - To add information that addresses the existing answers and comments, I have indeed tried larger diameter tubing, longer tubing, and tubing that loops into the liquid nitrogen back out and back in again which should handle the snow flake example. Also, this approach of freezing out of a helium stream works perfectly at high flow rates with N2O and CO2. My question to this Chemistry Stack then, is what about SO2 prevents easy freezing as I can accomplish with CO2 or N2O?

Phase diagrams are about thermodynamics, while you have to consider kinetics as well.
The stream may have too high linear speed at 30+ mL/min to reach equilibrium while still in the loop.
Another  option is dragging  particles of/containing SO2 by the gas stream by mechanical way, not allowing it to settle. Like wind taking snowflakes, not allowing them to land and stay.


How to correctly write the condensed formula for salicylic acid I am struggling to find the right way to write the condensed formula for salicylic acid

Some people say CHCHCHCHC(OH)C(COOH)
Or (CH)4C(OH)C(COOH)

Others say C6H4(OH)COOH (which doesn't seem to make sense since it looks to me like it could represent many isomers and not just one molecule like a condensed formula is supposed to)

Which (if any) is correct?

Some people say $\ce{CHCHCHCHC(OH)C(COOH)}$ Or $\ce{(CH)4C(OH)C(COOH)}$

Both of these are definitely wrong since it seems to indicate some sort of alkene.

  Others say $\ce{C6H4(OH)COOH}$

This is much better since it "hints" that $\ce{C6H4}$ is a benzene ring. With only 4 hydrogens and what seems to be two substitution groups that fits. Note there is no indication of the relative position of the groups. So this would represent either 3-Hydroxybenzoic acid or 4-Hydroxybenzoic acid as well.

The basic empirical formula would be $\ce{C7H6O3}$ which would represent probably dozens of molecules.

So the best way to denote the particular molecule specifically is to name it, or to draw out the whole 2D structure.


Delta G in an equilbrium reaction(forward and backward) Consider a reaction say

$$\ce{A + B <=> C + D}$$

Now $\Delta G$ for the forward reaction is:
$$\Delta H - T\Delta S$$
For the backward reaction $\Delta H$ and $\Delta S$ are clearly the same in magnitude but opposite in sign. Therefore,

$$ \Delta G_\text{forward} =-\Delta G_\text{backward}$$
Which implies that if the forward reaction is feasible then the backward isn’t or vice versa.

But both the reactions occur simultaneously in an equilibrium reaction.
Can anyone please help me sort this discrepancy?

I feel Koushal has not seen the difference between $\Delta G$ and $\Delta G°$, because $G$ of all reactants and products change during the reaction. $G$ and $G°$ are very different concepts.

$G°$ is the Gibbs energy of $1$ mole of any reactant and of any product in the pure state at 25°C and 1 atm. $G°$ is a constant of the substance, independent of the equation chosen, and independent of the concentrations.  $\Delta G°$ is obtained by calculating the sum of the Gibbs energies of the products in the pure state, and substracting the sum of the Gibbs energies of the reactants in the pure state, before the initial mixture is done.

During the reaction, the $G$ values are changing all the time, because they are proportional to the concentration of the reactant (or of the products) . The $G°$ values are never changing during the course of a reaction.

As soon as you mix the initial reactants, the Gibbs energy of each reactant decreases from $G°$ to a lower value $G$ < $G°$ even if no chemical reaction occurs. And the Gibbs energy of each reactant continues to decrease due to the chemical reaction. Simultaneously the Gibbs energy of the products are moving in the opposite direction. But they are is very low at the beginning of the chemical reaction. The reaction is finished when the sum of the Gibbs energies of the reactants is equal to the sum of the Gibbs energies of the products.