IUPAC naming - choosing the main branch What are the rules for choosing main branch?
Wikipedia states:

Identification of the parent hydrocarbon chain. This chain must obey the following rules, in order of precedence:

It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.
It should have the maximum number of multiple bonds
It should have the maximum number of single bonds.
It should have the maximum length.

I don't understand how 3 and 4 differ. Which single bonds are we counting? Only carbon-carbon single bonds in the main chain, right?
Also, I heard there is a rule that says if you've two equal length chains, then choose the more substituted one. Why is it not given here?
And now now imagine you've two choices for a chain, like this:

The chemdoodle program gives the name as 2-(1-Hydroxybutyl)-3-methylhexanoic acid .
I simply do not understand why the methyl substituted chain is chosen instead of hydroxy substituted one and how that fits with IUPAC rules given in Wikipedia.

For the given structure, the principal chain could be a 3-hydroxyhexanoic acid or a 3-methylhexanoic acid; i.e. the name of the structure could be 3-hydroxy-2-(pentan-2-yl)hexanoic acid or 2-(1-hydroxybutyl)-3-methylhexanoic acid.

The IUPAC recommendations on this matter have been changed.
According to the old IUPAC recommendations (1979, retained in 1993), the principal chain was the 3-hydroxyhexanoic acid. Thus, the preferred name was 3-hydroxy-2-(pentan-2-yl)hexanoic acid.
However, the relevant rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-45 SELECTION OF PREFERRED IUPAC NAMES
(…)
P-45.5 CRITERIA RELATED TO ALPHANUMERICAL ORDER OF NAMES
P-45.5.1 The preferred IUPAC name is the name that is earlier in alphanumerical order (…). Alphabetic letters are considered first in the order that they appear in the name; (…)

Therefore, the preferred name is 2-(1-hydroxybutyl)-3-methylhexanoic acid (not 3-hydroxy-2-(pentan-2-yl)hexanoic acid) since ‘hydroxybutyl…’ is earlier alphabetically than ‘hydroxy-pentan-yl…’.


How exactly do we superimpose molecules while checking optical activity? I am struggling with these type of questions where in some cases we rotate a molecule's mirror image by 180 degrees to check superimposability.

Like when checking any tetrahedral chiral compound, we draw its mirror image and rotate it by 180 degrees and say that it's not superimposable.
Suppose the mirror image of the left-handed version is the right-handed version, hence both are not superimposable. If we rotate the right-handed version by 180 degrees and then try to check, will it be superimposable now? Similarly, trans 1,2 dimethyl cyclopentane is chiral but if we rotate its mirror image by 180, is it superimposable?

You're being confused by the rotation. You take the mirror image. Any rotation or sequence of rotations that superimpose the mirror image over the original means that the molecule is achiral. It's not just a 180-degree rotation.


How to calculate the energy of a single proton? Using e.g. the VWN5 functional, what is a simple way to calculate the energy of a single proton? Of course, it is not possible to perform a calculation of a proton in isolation, as it bears no electrons. So I first attempted to use the experimentally determined value (see e.g. M. Dewar and K. Dieter, J. Am. Chem. Soc. 1986, 108, 8075.), but this leads to highly erratic proton affinities.

Could it be that I need to determine the proton affinity for any given functional + basis set (e.g. VWN5, 6-31+G** in my case)? What would be an effective way to get a good rough estimate of the energy of a proton?

An isolated proton will have energy given by the Schrodinger equation (just like any molecular system):

$$
-\frac{\hbar^2}{2\mu}\nabla^2\psi + V\psi = E\psi
$$

($\mu$ is reduced mass; $\hbar$ is $h/2\pi$; $V$ is potential energy; $\psi$ is the wavefunction)

The potential of this single-proton system is zero; the kinetic energy will depend on the momentum of the particle.

If you're using a generic quantum-chemistry software, the Born-Oppenheimer approximation is probably being applied, i.e. the proton ($\ce{H+}$) will be fixed in space with zero momentum. Thus its total energy will be zero, unless you deactivate BO-approximation or introduce another particle.

It's not typical to consider the energy of an isolated proton for obvious reasons (try finding one in nature!). Nonetheless it is calculable if the wavefunction is known.

Another way to consider its kinetic energy is by the classical equation $K = \frac{1}{2} mv^2$; if you consider an approximation of the lonely proton's speed as roughly that of atoms in liquid water, 1 angstrom per picosecond, you obtain ~ $5.03$ kJ/mol. This is similar to the thermal energy available at room temperature, $k_B T$, ~ 2.48 kJ/mol.

You're better off calculating energy of $\ce{H+}$ by getting energy for a molecule, and subtracting the energy of its deprotonated form. In that way you can evaluate proton affinity for a given molecule.


Has this way of comparing internal energy got any sense? Can internal energy between molecules be compared?
For example, $\ce{H2}$, $\ce{O2}$ and $\ce{N2}$ have some internal energy ($U$) but taking infinite distance between atoms as zero. But each reference point is different. So, can $U$ for those molecules be compared?

Yes. This is the binding energy.

In general, the binding energy is defined as the amount of energy released upon bringing infinitely separated particles in to contact. This would then be a negative number. Another sign convention is the binding energy can be viewed as the amount of work required to separate two atoms or molecules (or whatever) which are in contact. This binding energy is the negative of the first one (is a positive number).

For molecules, you might be more familiar with the term bond energy instead of binding energy. For a diatomic, these two things are the same. Notice that when we talk about the bond energy in a molecule, we treat that bond as being a diatomic molecule.

For the diatomics you mentioned, the electronic energy for the heavier diatomics will be artificially larger than the lighter ones simply because there are more electrons. For this reason, when comparisons between systems like this are made, people usually divide by the total number of electrons. So, in this case you can visualize putting in positive work to bring the protons near to each other and then adding the electrons one at a time to the system until it has no net charge. The binding energy per electron is then a measure of the average amount of stabilization the system gained for each new electron. The same thing can be visualized for the nucleus of an atom, but in that context you will read of the binding energy per nucleon.

This same concept is also quite useful for molecular clusters and other non-covalently bound systems as a measure of the strength of intermolecular interactions. In that case it is infinitely separated molecules which define the zero of energy.

This concept of energy relative to infinitely separated components of a system is a very useful one as it gives an unambiguous definition of the strength of interactions. As a fun thing to ponder, consider how to determine the strength of an internal hydrogen bond? As it happens, to the best of my knowledge, there is no general method for determining the strength of internal hydrogen bonds as there is no natural zero of energy to choose. Some people have tried using the red shift in vibrational frequency, but this too is still a bit ambiguous and hard to determine.


In the absence of a physical model, how can I use free software to visualize stereochemistry? In a different question, the task was to turn a wedge-and-dash representation into a Fischer projection. This involved rotation of a fairly complex molecule - either as a physical model or a mental model - to orient each chiral carbon in preparation for drawing the Fischer projection. How can you use free software to visualize this process?

I used Jmol to make the following animated GIF:

First I used NCBI's Pubchem Sketcher to get the SMILES code for the molecule: CC(C(C(C)F)(Br)F)O
Then, I loaded it into JSmol running here using the load molecule by name button.
Internally, this triggers the command
load /*file*/"https://cactus.nci.nih.gov/chemical/structure/CC(C(C(C)F)(Br)F)O/file?format=sdf&get3d=true
I fixed the stereochemistry by clicking on the carbons with incorrect chirality after issuing the command "set picking INVERTSTEREO". To rotate around the bonds, I used "rotate branch @3 @2 180 180" and "rotate branch @3 @4 180 180". Finally, I chose the viewing orientations, displayed the orientationwith "show moveto" and issued the resulting moveto commands before and after rotating around the single bonds.
A more elaborate example with glucose (4 chiral centers) can be found on Proteopedia.


How does oil on the surface of water prevent rust? I distinctly remember a side-by-side comparison from a book where there are two nails submerged in water, in two beakers: one nail had a layer of oil on top of the water, and that nail didn't rust; the other did.

It seems that oiling iron/steel products is supposed to prevent them from rusting:
http://www.justanswer.com/home-improvement/0n1r2-does-olive-oil-prevent-rust.html
https://outdoors.stackexchange.com/questions/1404/can-oiling-keep-tools-from-rusting

The explanation given, I think, is that the oil somehow prevents oxygen from reacting with the metal.

Can't help but wonder why. Oxygen is a non-polar molecule, so it should actually dissolve better in oils (which are also non-polar) than in water (which is highly polar) and therefore allow more oxygen molecules to contact the iron (metal) surface in some sort of equilibrium.

Let's say that the conductivity of water and its ability to form electrochemical chains between metals is irrelevant in this thought experiment (i.e., a nail submerged in water in a glass). From what I understand, the other key item that has an effect on the formation of rust in water is that the oxygen molecules have better contact with the iron surface.

This is supported by data: the concentration of oxygen in air is far higher than that in water. We know from experience, though, that wet iron rusts way quicker than dry iron does.

If I am reading it correctly, the same data shows higher oxygen concentrations in non-polar solvents than in water.

Here is the data: http://www.nist.gov/data/PDFfiles/jpcrd219.pdf

Please correct me if I am reading it wrong.

My question is:

How does oiling prevent rust?

Edit: For the sake of clarity, let's say we use deionized water in a vacuum, drop in the nail, then put some oil (let's say it's caprylic acid) on it, and expose it to standard conditions.

Well, I went and searched for the solubility of oxygen in water and oil, and found this summary paper on the NIST web site: "The Solubility of Oxygen and Ozone in Liquids" by Battino, Rettich and Tominaga, J. Phys. Chem.Ref. Data., vol 12, no. 2, 1983.

Conveniently, the paper gives solubility data for oxygen in both water and olive oil.  The solubility is given both as molar fractions, and also as the Ostwald coefficient $L$, defined as the volume of gas absorbed per volume of absorbing liquid.

Given the very different molar volumes of water and olive oil, I believe the latter provides a much more useful comparison here.  Comparing tables 1 (water; note the $10^2$ factor in the header) and 25 (olive oil) in the paper, we see that the Ostwald solubility coefficient for oxygen in water is about 2,000 times higher than in olive oil:

 Temp. (K) | Temp. (°C) | L (water) | L (olive oil)
-----------+------------+-----------+---------------
    298.15 |         25 |     310.4 |        0.1269
    308.15 |         35 |     276.6 |        0.1326
    318.15 |         45 |     254.5 |        0.1383
    328.15 |         55 |     240.9 |        0.1441

So, yes, oxygen is a lot more soluble in water than in oil.  But why?

Well, if this web site is to be believed, it's because of the strong hydrogen bonding between water molecules, which causes even liquid water to maintain some degree of an ice-like lattice structure, with gaps that small gas molecules like $\ce{O2}$ can easily slip into.  Once there, the gas becomes solvated, with the strongly polar water molecules inducing a temporary dipole in the $\ce{O2}$ molecule and thus creating an attractive Debye interaction.  The overall solvation process is exothermic (i.e. releases heat), explaining why the solubility of oxygen in water decreases with temperature.


Why do chalcogens (Group VI) stink so badly? For more fun with Dr. Derek Lowe, see this for a primer: Things I Won't Work With - Carbon Diselenide. The short of it is that we carbon-based life forms generally like oxygen. However, move just one row down in the group, to sulfur, and things start going horribly wrong. Sulfur, and the thiol/mercaptan analogs of otherwise relatively benign hydrocarbons and alcohols, gives us the thrilling eaudeur of rotten eggs, burning tires, skunks, farts, etc.

Move one row further, to selenium, and abandon all hope ye who enter here; the tamest common example of selenium in our daily lives is garlic, an instantly recognizable scent, especially when your body starts excreting the selenium-containing essential oil back out of your pores. It's supposed to be the last word in mosquito repellent; down here in Texas the little biters apparently have a taste for Italian. Selenium analogs of thiols often defy chemists' descriptions of their olfactory power; one of Dr. Lowe's other selenium-based posts, on selenophenol (the selenium analog of thiophenol, which is bad enough), puts it thusly: "Imagine 6 skunks wrapped in rubber innertubes and the whole thing is set ablaze; that might approach the metaphysical stench of this material".

This poses the rather interesting question; why do things change so suddenly with this group? I'm willing to believe that all of them have similar chemical properties, and we as humans and other oxygen-respiring life forms differentiate oxygen from the rest of the group because oxygen's good for us while the others are bad. But, that borders on the simplistic, and poses more questions, like why we work so much better on oxygen than sulfur (there are a few bacterial species hanging out near deep-sea vents that just can't get enough of those sulfur gases, so obviously there's sulfur-based "organic" metabolism just as there is oxygen-based). I guess what I'm looking for is an explanation of why two chalcogens, both obeying the octet rule as Periods 2 and 3 tend to, are treated so differently by our sensibilities.

I think that there are two reasons: the sulfur is quite common in biochemistry and at the same time a lot of sulfur compounds are toxic. The bio-degradation of sulfur compounds mostly leads to some sulfides which very strongly bind transition metals, i.e. they can destroy/block Metalloenzymes. If you add some millions years of evolution it's a v.good feature to smell them (that would also work for small phosphines and amines). Your chances for food-poisoning are lower, chance for survival higher (Long Live "The selfish gene").


Combustion of NaCl(aq) when testing conductivity In a experiment that dealt with electric conduction of many concentrated solutions, NaCl(s) created an unknown, to me, substance after adding a drop or two of water. Furthermore, it isn't the reaction of water and NaCl(s) that I don't understand, but the reaction between the anode and cathode when they came in contact with NaCl(aq), sparks were present and it seemed to have "burned" NaCl(s). The question is: what is the correct name for the part that "burnt"?

There is not "burning", one might see a spark. It is a well known phenomenon. Do you see lightening in the sky? Air is an insulator, yet when very high voltage is present it becomes a conductor. It is called dielectric barrier discharge. In the same way, discharge between two electrodes can occur in pure water (almost an insulator). Adding NaCl to water makes it a good conductor but still it is mostly water, and if the electrodes are close enough, electrical breakdown can occur. Instead of ions carrying the current as it should happen in an aqueous conducting solution, the current becomes electronic even in the presence of water. Apparently, the local heating between the electrodes is so high, that a gaseous plasma is formed between the electrodes and plasma are electrically conducting.
As suggested in the comment, always exercise cause caution and all experiments should be done under proper supervision by a trained person/teacher.


Thought Experiment: Is it possible to preserve the charge imbalance present during molten lead(II) bromide electrolysis? Suppose I were to electrolyse some molten lead(II) bromide. According to what I have been taught, the lead ions travel towards the cathode and are reduced at it, while the bromide ions are oxidised at the anode. The movement of the ions creates a charge imbalance (there will be more lead ions near the cathode and vice versa) in the molten fluid.

Suppose I were to let this run for a little while in a vacuum (and siphoning off the $\ce{Br2}$ as soon as it is produced) and non-conducting container. Now suppose I was to quickly put an impermeable non-conducting plate between the cathode and the anode and at the same time remove both the cathode and anode, so separating the molten $\ce{PbBr2}$ into two halves. This will stop the electrolysis but there is already a charge imbalance between the two halves (e.g. the half which originally had the cathode has more $\ce{Pb^2+}$ than $\ce{Br-}$) which has no way to correct itself.

If I cooled down both halves of the mixture now so that they both solidified there would be excess ions on both side which wouldn't be able to recombine. It might even be possible to remove the re-formed $\ce{PbBr2}$ by some method and have just the uncombined ions left on each side. My question is whether this is even possible, and if so what would happen next.

An electret is a dielectric (non-conducting) material that has a semi-permanent electric charge or polarization, the electrostatic equivalent of a permanent magnet. Electrets were made by first melting a polymer or wax containing polar molecules, and then allowing it to re-solidify in a powerful electrostatic field (several kilovolts/cm). The polar molecules in the dielectric align to the direction of the electrostatic field, producing a dipole with a permanent electrostatic bias when the melt solidifies.
Although electrets are only in a metastable state, those fashioned from very low leakage materials can retain excess charge or polarization for many years. The Japanese used microphones polarized by electrets in WWII.
https://en.wikipedia.org/wiki/Electret
So, using the electret as an example, the likely charge imbalance in the solidified lead bromide melt would be very small because the electric field across the molten salt is nowhere near kilovolts per centimeter. It might be measurable, but very small, and probably not very long-lived.


Rate Expression for Reactions Involving 2° Benzylic Alkyl Halides My teacher took off marks because he said the reaction was SN1. Before I go confront him, please tell me if my reasoning is correct.

So, basically, I thought: we have two possibilities:
 1. SN1: a 2°benzylic has the same stability of carbocation as a tertiary carbocation so, SN1 could work.
 2. SN2: we have a secondary carbocation and a weak base cyanide: SN2 could work.

So I just randomly put the rate law for SN2 since the problem did not specify SN1 or SN2 and there is no way I can know which mechanism is prominent.

My inclination is that you are correct and the reaction proceeds by an SN2 reaction. Secondary alkyl halides can react by either an SN1 or SN2 mechanism, so the other reaction conditions are important to make a decision of the predominant pathway. Since cyanide is a strong nucleophile (and often utilized as the prototypical SN2 nucleophile), that suggests that the reaction should go through an SN2 pathway. One other piece of information that we don't have is the solvent. If this is run in a protic solvent, that could tip the pathway toward SN1, while polar aprotic will favor SN2. I also would not be surprised if both pathways are operating.

It is true that the aromatic ring will stabilize a carbocation at the benzylic position, but it will also stabilize the transition state of the SN2 reaction.

A brief literature search has revealed less insight than I expected. The closest result I have been able to find is that an optically active benzylic bromide treated with trimethylsilyl cyanide in acetonitrile (polar aprotic solvent) does give an optically active benzylic cyanide, suggesting SN2.

DOI: 10.1055/s-0030-1260187


Can carbon have quadruple bonds? Carbon has 4 valence electrons, so it means that it can have a total of four (covalent) bonds, correct?

If that's the case, why can't carbon share four bonds, such as with another carbon atom (to form $\ce{C2}$, for example)?

s orbitals have spherical symmetry, and p orbitals have a dumb-bell shape.  When these orbitals overlap, then they form the bonding orbitals (and corresponding antibonding orbitals).  The p-orbitals are aligned along Cartesian (xyz) axes.  If all three p-orbitals overlap, then the s-orbitals can't get 'close enough' to overlap and form a fourth bond.  If we start mixing s and p orbitals, the geometry always puts electron density away from the other bonding atom.

Concisely: there are no arrangements which allow overlap of atomic orbitals that create four bonding molecular orbitals with the proper orientation.

C2 as a molecule can exist, it will simply only have 3 bonds plus two unpaired electrons confined to each carbon.

If you want a more elaborate answer we have to unpack the heavy quantum mechanics, and you won't be seeing that until your junior year.


Interpreting from phase diagram whether O2 will float or sink The normal melting and boiling points of $\ce{O2}$ are $\pu{-218 ^\circ C}$ and $\pu{-183 ^\circ C}$, respectively. Its triple point is at $\pu{-219 ^\circ C}$ and $\pu{1.14 Torr}$, and it's critical point is $\pu{-119 ^\circ C}$ and $\pu{49.8 atm}$. Will $\ce{O2 (s)}$  float or sink on $\ce{O2 (l)}$?

I quickly sketched the phase diagram of oxygen:

Pressure of the molecules will affect how compact the molecules are in solid and liquid, so I guess we can explain whether it will float or sink using the slope?

How does the slope tell us about the densities of $\ce{O2}$ solid and liquid ?

And in addition, how do I use intermolecular forces to explain this?

Yes, we can do that.

In fact, it is only the sign of the slope that matters. Say, you stand right below the gas/liquid curve; you apply a little more pressure, and you step from gas to liquid. How does that correspond to Le Chatelier's principle? Well, just like that: the system responds to increased pressure by decreasing its own volume, hence the liquid must be denser that the gas (which we of course already knew from independent sources).

Guess what?

The same reasoning applies verbatim to the liquid/solid curve, since it is sloped in the same direction.

As to the intermolecular forces, I think the usual hand-waving along the lines of "less freedom $\to$ less motion $\to$ lower volume" will do, and even that is redundant. Having solid denser than liquid is normal. If anything, it is the opposite situation that would require an explanation.


Is the inductive effect always measured relative to hydrogen? Wikipedia has defined the Inductive effect thus:

  "In Chemistry and Physics, the inductive effect is an experimentally observable effect of the transmission of charge through a chain of atoms in a molecule, resulting in a permanent dipole in a bond."

Recently, I learned from a teacher that the inductive effect of any atom/group is always measured with respect to hydrogen.

Under this definition, we would ideally attach the atom/group with an atom of hydrogen and observe the direction of the dipole moment caused by the difference in electronegativity between hydrogen and the atom/group in question. If the electrons-initially positioned at the middle of the bond-are pushed toward hydrogen, the atom/group is termed as a "$+I$", electron-donating group. On the other hand, if the electrons are pulled toward the atom/group, it is termed as a "$-I$", electron-withdrawing group. Thereafter, this atom/group will always exhibit this type of inductive effect. Even if a situation does arise, in which it appears as though an atom/group that exhibits the $-I$ effect with hydrogen is exhibiting the $+I$ effect with another atom/group, this must be the cause of other dominating effects such as the mesomeric effect and in those cases also, it is said that the atom exhibits the $-I$ and not the $+I$ effect.

To illustrate the clear distinction between the definition put forward by the teacher and the one on Wikipedia, let us consider a molecule of $\ce{Cl2}$.

Under the definition on Wikipedia, it is clear that $\ce{Cl2}$ is a non-polar molecule (i.e. No permanent dipole moment has been experimentally observed in this molecule). Hence, no inductive effect is in play and neither of the two $\ce{Cl}$ atoms show either the $+I$ or $-I$ effect.

On the other hand, if we work with the definition put forward by the teacher, we observe that an atom of $\ce{Cl}$ exhibits the $-I$ effect when attached to an atom of hydrogen, being more electronegative than it. Therefore, in the $\ce{Cl2}$ molecule, no permanent dipole exists as a result of the $-I$ effect of both chlorine atoms cancelling away from opposite directions.

It appears that only one of these two definitions is definitely correct and I am confused as to which one that is.

Your definition is falling apart because you are over-applying it. In the definition, you can see that it says:

  ...transmission of charge through a chain of atoms in a molecule...

It doesn't make sense to apply this in the case of a two-atom molecule, because the inductive effect is used to measure how one bond effects the electron density of other bonds in the molecule. You can assign any bond a dipole moment, but inductive effects are used specifically to explain phenomena observed in organic chemistry, where chains of molecules are (exceedingly) common.

To answer your other question: yes, inductive effects are always measured relative to hydrogen. Linear free-energy relationships are a quantifiable way of measuring a reaction's sensitivity with respect to a certain parameter. The Hammett equation is used to study how sensitive a molecule is in a given reaction to a change in its subsituent's inductive and resonance effects, for which the reference subsituent in all cases is a hydrogen atom. This is perhaps because the $\ce{C-H}$ is the most common bond in all of organic chemistry.

Linear free-energy relationships are particularly useful because similar as to what you said, substituent groups will always have the same relative inductive effects no matter the molecule. A quaternary ammonium group will always have a larger $-I$ effect (and larger $\sigma$ value) than a nitro group than a trifluoromethyl group. A molecule's response to these changes can give great mechanistic insight into organic reactions and the pathways through which they proceed.

A table of the different types of Hammett relations$^{[1]}$:

Their reference reactions:

$\hspace{3.25cm}$

$[1]$ Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science: Sausalito, CA, 2006.


Are there any sources for tabulated Rf-values, especially 1-phenylazo-2-naphthol? I  have synthesized 1-phenylazo-2-naphthol in my laboratory tutorial and have to analyze it. I've done a thin layer chromatography (TLC) of it:

solvent: ethanol
eluent: toluene
stationary phase: silica gel plate

My only problem is, I need an theoretical value for comparison but haven't found any source.

Did you react aniline (or its hydrochloride) with sodium nitrite in the presence of 2-naphthol?
I'm afraid there will not be any table with $R_\mathrm{f}$ values.
These values are not only dependent on the compounds and the mobile phases, but also on the stationary phase: e.g. silica, aluminium oxide, etc.
Your best option is to use reference samples and measure the $R_\mathrm{f}$ value yourself for the conditions you want to apply or transfer from TLC to column chromatography.
My suggestion is to collect the crude reaction product (Büchner funnel, etc.) dry it, note the yield and measure the melting point.
For purification, recrystallize from ethanol. You should receive red needles with a melting point of $\pu{134 ^\circ C}$.
NOTE: Wear proper protection when handling the azo dye, which is a cancer suspect.

References:

IOC-Praktikum (PDF in German)
GESTIS-database on hazardous substances


What is torsional strain? I am having difficulty in understanding the concept of torsional strain. Can someone explain it to me in a more elaborate manner?

I mean why should bond restrict movement in eclipsed conformation? Doesn't that make it unstable! I tried this Wikipedia article but I'm still having difficulty in understanding the concept.

For the case of a single bond, consider a simple case like ethane, which can take an eclipsed conformation, a staggered conformation, or anything in between depending on the torsion angle of the C-C bond.  Each carbon has a region of especially high electron density where it's bonded to a hydrogen.  When the torsion angle is changed from staggered to eclipsed, these regions approach eachother and repel, and thus the eclipsed conformation is higher in energy than the staggered.

In the case of a double bond, as in ethylene, the energy associated with torsional strain is much higher because the P orbitals involved in the pi bond have to be oriented the same way.  Thus, applying torsional strain involves breaking this pi bond, and has a high energy.


What are the differences between D and L glucopyranose? I have just studied the concept of cyclotron in carbohydrates. My reference book used Glucose as an example and began by saying we can find D-glucose or L-glucose:

Depending on what enantiomer we have, we will get opposite chair conformations at cyclisation:

After saying this, my book states that significant differences appear at cyclisation of these enantiomers, and there I just do not follow. The only difference I see is that one is the mirror image of the other. But visually, I see both structures identically equal.

Could someone explain to me what differences I should be observing in both chair conformations, and also, what consequences these have on the molecules' behaviors?

They are mirror images of one another and therefore their chemical and physical properties are the same, except for the deflection of polarized light, which will be in opposite directions.

Mirror image compounds that are not the same are called enantiomers and it does not matter if they are in the open or cyclic form.
They will, however, have different properties if they are in a chiral medium.
For instance, if placed in a chiral solvent, they will have different solubilities.
Chiral means that the compound has a mirror image that can not be superimposed on the original structure.


Does solubility matter when measuring density with displacement method? Can the volume of a solid (specifically a granular crystal or powder) be accurately determined using the displacement method with a liquid in which the solid is soluble?

Originally I thought by dissolving in the liquid there would (or could) somehow be a change to the density of the mixture which would affect the displacement. As such, it would be a good idea to ensure the substance is not soluble in the liquid being used to determine the solid's volume. But if the dissolution of the solid is only a physical process (i.e., dissolving sugar or salt in water), would this at all affect one's ability to measure the volume of the amount of substance added?

I guess when dealing with an unknown substance, we might not be able to easily determine if the dissolution took place as a physical or chemical process without further testing, so it would make more sense to conduct the measurement with a liquid in which the substance was insoluble.

Volume is not a conserved extensive property of mixtures in the same way that other quantities like mass, net electric charge, number of particles, etc. are (within closed systems). That is to say, in reality, the volumes of components physically mixed together are not simply additive, with the total volume not necessarily equal to the sum of the individual volumes. I.e., $\sum{V_i} \not= V_{mix}$ generally. This occurs chiefly because intermolecular forces of attraction between particles will vary with the composition of the mixture. Mixtures of ethanol and water, for example, are always reduced in volume by comparison to the summed volumes of ethanol and water, respectively, in isolation.

In fact, the total volume can be calculated given the partial molar volumes of the individual components for a specific composition under given conditions. Specifically, $V_{mix} = \sum{n_i \bar{V_i}}$, where $\bar{V_i}$ and $n_i$ are the partial molar volumes and moles, respectively, of the separate constituents of the mixture.

A few good references I found for additional study: 1, 2. Any decent physical chemistry textbook should also cover partial molar properties.


At the freezing point, why is the vapor pressure of the liquid phase equal to the vapor pressure of the solid phase? At the freezing point of a substance, the solid phase is in dynamic equilibrium with the liquid phase. My textbook defines the freezing point of a substance as: "the temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase". Why is this so? How does the vapor phase even come into the picture during freezing? Shouldn't it be all about the solid and the liquid?

My book also states that when we add a non volatile solute to a liquid solvent, "the solution will freeze at the temperature when its vapor pressure is equal to the vapor pressure of the pure solid solvent". This doesn't make sense to me as well. Assuming the first statement is true, shouldn't the second statement read, "...is equal to the vapor pressure of the solid solution"? Are the two statements somehow related to each other?And why are they true?

If the vapor pressure of the substance in its liquid phase would be different from the vapor pressure of its solid phase then there would be a transfer of mass from the phase with the higher vapor pressure to the phase with the lower vapor pressure.

This would involve releasing or consuming energy (heat of fusion) even in a closed system in thermal equilibrium, which is obviously a contradiction.


Why is the cyanide ion toxic? As the title implies, what is the molecular basis of cyanide toxicity? I did some searching around at the CDC and it only states that it prevents cells from using oxygen. I also read how it could take as little as $131~\mathrm{ppm}$ to kill. How can a simple $\ce{CN-}$ ion do so much damage?

Cyanide is a pretty good ligand for coordination compounds. The electron pair on carbon (which, incidentally, also carries the Lewis structure’s formal charge) is located in the HOMO — much like as in $\ce{CO}$, whose molecular orbitals can be found in this answer by Martin (replace oxygen with nitrogen to arrive at $\ce{CN-}$) — making it a good Lewis base and a good σ donor.

It is remarkable for being a strong-field ligand, i.e. one that often creates low-spin complexes even for 3d-metals (which generally prefer high-spin if there are no competing reasons for low-spin). While one could draw analogies to $\ce{CO}$ or even $\ce{NO+}$ whose π-antibonding orbitals are strongly Lewis acidic and participate in metal to ligand backbonding, it turns out that for cyanide the main reason is a high covalency of the metal–ligand-σ-bond due to the relevant orbitals being very similar in energy. Thus, cyanide is unique in that it both stabilises higher oxidation states by being an anion and low-spin complexes due to its binding modalities.

One might think, in analogy to $\ce{K4[Fe(CN)6]}$, that cyanide preferentially binds to haemoglobin’s iron(II) centre and thereby inhibits oxygen transport, but that is in fact not the case as this centre requires a less-oxidised iron(II). Oxygen (due to the resulting redox-mechanism) and carbon monoxide are much better ligands for haemoglobin.

What cyanide does strongly inhibit is cytochrome c oxidase, the last enzyme in the respiratory cycle which collects electrons from four cytochrome c’s and transfers them to an oxygen molecule to create two water molecules:

$$\ce{4 Fe^2+-cytochrome~$c$ + 4 H+ + O2 -> 4 Fe^3+-cytochrome~$c$ + 2 H2O}$$

This is done by a plethora of metal centres consisting of two haems, to cytochromes and two copper centres. I can’t tell you exactly which one, but there will be at least one which needs accessable iron(III) at some point in the catalytic cycle to which $\ce{CN-}$ can bind well and strongly inhibiting the catalysis of the enzyme.

Since the inhibited process is directly responsible for cellular respiration by transferring electrons to molecular oxygen (and being the only enzyme dealing with molecular oxygen in the respiratory chain), inhibition of this protein will cause the breakdown of the entire respiratory chain and lead to cellular suffocation.

Considering that an amount of $200~\mathrm{mg}$ can be toxic according to Wikipedia, and also considering that a human being contains some 10 trillion cells also according to Wikipedia, that’s still some 400 million cyanide ions per cell demonstrating the efficiency of binding and inhibition.

Note that popular antidotes also directly address the fact that cyanide is a strong ligand:

One antidote, hydroxycobalamin, presents the cyanide ion with a $\ce{Co^{III}}$ centre and a weakly bound, easily displaced hydroxido ligand. Hydroxide is displaced by cyanide creating cyanidocobalamin, a variant of vitamin B12 that the body has no problems dealing with. It is ejected via the kidneys. Injected into the circulatory system, it scavenges cyanides before they can enter cells.
Another antidote, also injected into the circulatory system, is 4-dimethylaminophenol (not to be abbreviated as DMAP to prevent confusion with 4-dimethylaminopyridine). It is an oxidising agent designed to oxidise haemoglobin’s iron(II) to methaemoglobin (containing iron(III)) thereby creating a much better acceptor for cyanide. Since there is a lot of iron(II) (a quick estimate with $8~\mathrm{mmol/l}$ gives $40~\mathrm{mmol}$ haemoglobin in human blood) in the blood for oxygen transport, small amounts of this can be sacrificed to capture cyanide before that reaches the respiratory cycle.


Why does Stokes raman not relax to the ground state while anti-Stokes does? I'm just wondering why stoke-Raman doesn't go to the ground state. Where does the rest of energy from exciting laser leave?
On the other hand, why does anti-stoke go back to ground state? I know that the overall energy of anti-stoke is higher but why it has to go back to the ground state?
My question is based on this diagram.

As I understand, the anti-stoke has higher energy as it is excited from v1 due to an enhancement. I'm just confused with the last process of anti-stoke when it comes down to the ground state. Why doesn't it come down to the v2 or higher vibrational state like stoke?

There is no law that a transition must go to the ground state. Now, each molecule has a multitude of electronic states, and within each of them a multitude of vibrational states. Thus each transition between two pure electronic states is accompanied by a variety of related transitions between higher-to-lower and lower-to-higher vibrational states. Which of these are the Stokes lines and which are the anti-Stokes ones, you'd better figure out for yourself.


Phases of Atomic or Molecular Orbitals What does phase mean in orbitals?

I know that phases are separated by nodes. They are in some way related to wavefunctions, I can't understand, how? How can wavefunctions be negative, since they are complex numbers (as I know)?

Complex Numbers

As you point out, general a wavefunction if complex-valued.  A complex number and be described by two real numbers and hence are of depicted on the 2D complex plane (sometimes called the argand plane).

Numbers in the complex plane can be expressed in multiple forms.  The one usually taught first is as the sum of a real part and imaginary part:

$$\begin{align} z = x + iy\;{} & ; x,y \in \mathbb{R}\\
\Re(z) = x\; {} & ; \Im(z) = y \\
|z| = {} & \sqrt{x^2+y^2}
\end{align}$$

Alternatively it can be described by a magnitude ($r$) and a phase ($\phi$):

$$\begin{align}
z = r e^{i\phi}\; {} & ;r\in\mathbb{R}^{+0}, \phi\in[0:2\pi)\\
\Re(z) = r \cos(\phi) \; {} &; \Im(z) = r \sin(\phi) \\
|z|{} & = r
\end{align}$$

Two signals or functions with the same magnitude are said to be perfectly/exactly 'out of phase' if there phases differ by $\pi$ as $e^{i \pi}=-1, and perfectly 'in phase' if they have the same phase.

Orbitals

Now we can apply this to some orbitals.  Consider a cross-section of the 2p$_{\text{x}}$ orbital in the $x$-$y$ plane through $z=0$:

$$\psi_{2px}(x,y,z) = r(x,y,z) e^{i \phi(x,y,z)} \propto x \exp\left(-\sqrt{x^2+y^2+z^2}\right)$$

The intensity of the colour represents magnitude, $r(x,y,z)$, of the wavefunction and the colour itself represents the phase - red for $\phi=0$, blue for $\phi=\pi$, with a colourless node separating the two lobes.

As the probability density is equal to the square modulus of the wavefunction: $$\rho(x,y,z) = |\psi(x,y,z)|^2$$
we can multiply the wavefunction by a constant phase factor $e^{i\theta}$ and still represent the same particle:

Consider:
$$\begin{align}
\tilde{\psi} {} & = \psi e^{i\theta} \\
\to \tilde{\rho} {} & = |\tilde{\psi}|^2 \\
& = |\psi|^2 e^{i\theta} e^{-i\theta} \\
& = |\psi|^2 \\
& = \rho
\end{align}$$
So both $\psi$ and $\tilde{\psi}$ are valid wavefunctions with either one is, and the wavefunction can only be defined up to multiplicative phase factor, so we can talk about absolute phase, but we can talk about relative phase and phase differences.

Consider the out of phase 2p$_{\text{x}}$ orbital:
$$\tilde{\psi}_{2px}(x,y,z) = \psi_{2px}(x,y,z)e^{i\pi} = - \psi_{2px}(x,y,z) $$

Molecular Orbitals

As we just showed they both have the same probability density, but when we begin to combine them, the phases dictate how much the wavefunctions constructively/destructively interfere.  We can combine linear combinations of atomic orbitals to make molecular orbitals.

Consider adding together two adjacent wavefunctions with the same phase on different atoms:

The adjacent regions of different phase cancel one another out.  When the probability density is examined:

we can see than the chance of the particle being between the two atoms is very low and it is localised around each nuclei.  This would be an anti-bonding molecular orbital.

On the other hand, adding together two adjacent wavefunctions with the opposite phase on different atoms:

The adjacent regions of similar phase add together.  When the probability density is examined:

we can see than the chance of the particle being between the two atoms is very much higher and it is localised in a bond.  This would be a bonding molecular orbital.

This is why we talk about in phase and out of phase orbital when talking about bonding - the regions of in phase overlap of wavefunctions increases the probability that a particle will be found in there, and out of phase overlap decreases the probability.

The physical result is the wide variety of bonding phenomena we see in nature

covalent bonds: local in phase overlap
anti-bonding orbitals: local out of phase overlap
conjugation: extended in phase overlap etc...


Tin in silver nitrate, any soln color change? I have lab question in which a $\ce{AgNO3}$ solution with a tin strip placed in it is mentioned. This isn't a reaction we actually did in the lab, but due to a color change in the solution of one we did do, I was wondering if there would be any for this reaction. Googling I found that the $\ce{AgNO3}$ solution is initially colorless, but that's all I could find. So would there be for this one, or would the solution remain colorless?

The solution will remain colourless until silver particles will participated from the solution.

Silver nitrate solution is unstable, its decomposition is accelerated by light:

$\ce{2 AgNO3(l) -> 2 Ag(s) + O2(g) + 2 NO2(g)}$

That will change the color of the solution to brown. After silver participation, the solution will be colorless again.

A tin strip added to the nitrate solution will dissolve into tin(II) nitrate and will consequently react with silver nitrate accordingly:

$\ce{Sn(NO3)2 + 2AgNO3 + H2O -> Sn(OH)4 + 4HNO3 + 2Ag}$

Analogous silver reactions are described by Tollens' reagent, mirror preparation and gelatin silver process.


Is the mole fraction used in Raoult's law calculated initially or at equilibrium? Raoult’s law states that:

The partial pressure of any volatile constituent of a solution at a constant temperature is equal to the vapour pressure of pure constituent multiplied by the mole fraction of that constituent in the solution

My question:
Is the mole fraction of the constituent of the solution to be considered at the time when equilibrium between vapours dissolving into the solution and rate of evaporation has been achieved or at the beginning when the solution is just formed and evaporation has just begun change the mole fraction of the constituent which we have just calculated?
Also, experimentally, is there much difference in the mole fractions at equilibrium and at the beginning when there is apparently no vapour pressure (for zero vapour pressure in the beginning, I think  we can just close the lid of the container)?

Assume the system to be in equilibrium, otherwise you can't use the standard equations of thermodynamics. As for your second question, it's impossible to answer because it depends on the solvent and details of the experimental setup (size of container relative to volume of solvent, temperature, etc)


How to Dissolve Glyphosate? I am a microbiologist forming a growth media which has glyphosate in it. Unfortunately, I am having troubles dissolving it in water. At a .2M solution in 200ml of water, it is not completely dissolving. Does anyone know why this is chemically happening? Even upon adjusting pH and raising temperatures, I am unable to dissolve it.
https://pubchem.ncbi.nlm.nih.gov/compound/glyphosate
If you have any ideas please let me know.

This structure has a phosphonic and a carboxylic acid moieties with a minimum number of carbon atoms. As such, it should be soluble in basic water to deprotonate at least one site


How do I turn room temperature water into boiling water for drinking using an chemical/additive? I am trying to create instant boiling water using a chemical of sorts, so that it can be consumed by a human. For example; someone could make tea without a kettle or any available water heating systems.

It must be completely safe and if possible, leave no bad tasting remains. Preferably, the chemicals used would have to be easily accessible and stable in its bought state.

I am in high school and just want to do a cool science experiment. I guess this could have endless potential applications. I am also open to other ideas, like making a disposable bottle, that when you twisted the lid, the water would heat up as the chemicals dropped down or the bottle itself heated up. Just any way of doing it would be awesome.

I saw this page but is wasn't consumable ideas:

Can you heat water with additives?

Thanks,
Charles

Adding something to water that heats it to 100°C and leaves only water is of course nonsense.

A self-heating cup is easy: You take a double-walled beaker, fill the space between with two compounds that react with great heat but no gas formation, separated by an extra wall with a valve.

Google the self-cooling beer keg. It works very similar, only it gets hot outside, not inside.

Or just google "self-heating coffe cup".


How to prove that reaction is possible I have been given task to write possible reaction equations, determine oxidizer, reducer, write corresponding electron-equations and calculate electromotive force (I do not know if that is what you call it in English) of reactions:

Iron reaction with diluted sulfuric acid
Aluminium reaction with solution of sodium hydroxide

Could you be so kind and help me with this task?

The ionic equation for this reaction is:

$\ce{Fe + 2H+ ~(+\;SO4^2- ) -> Fe^2+ + H2 ~(+\;SO4^2- )}$

As two half-reactions, it is seen that the iron is oxidized:

$\ce{Fe -> Fe^2+ + 2e- }$

The reductant (iron) transfers electrons to hydrogen, and is thus itself oxidized.

And the hydrogen is reduced:

$\ce{2H+ + 2e- -> H2}$

The oxidant (hydrogen) removes electrons from iron, and is thus itself reduced.

To calculate the reduction potential you have to solve the following equation:

$\Delta E = E_\text{acceptor} - E_\text{donor} = E_{\ce{2H+/H2}}- E_{\ce{Fe/Fe^2+}}$

$\Delta E = 0\ \mathrm{V} - (-0.41\ \mathrm{V}) = 0.41\ \mathrm{V}$

The reactions runs voluntarily if $\Delta E > 0$.

You can do the same for b).


Double bond formation in alkenes during cracking In thermal or catalytic cracking, energy is used to break a long chain alkane into smaller alkanes/alkenes. Since the energy goes into breaking the bonds, how is the C=C bond in alkenes formed?

I don't know if I have understood well your question.
During the process of cracking, homolysis of long chains occurs. Radical intermediates result from breaking the bonds. These intermediates are highly unstable and undergo rearrangements. As a consequence of the rearrangements,  smaller alkanes and alkenes are formed. We have to point out that, due the the high reactivity of radicals, the control of radical reactions is very difficult and we can not readily predict what happens inside the chemical reactor.
I hope I've answered your question.


What is the difference between BET and Langmuir model for calculating the specific surface area? Brunauer–Emmett–Teller (BET) surface area is measured by specific surface area of the adsorbent. BET surface area is measured by using non-corrosive gases like $\ce{N2}$, $\ce{CO2}$, $\ce{Ar}$ etc. BET surface area depends on size and number of gas molecules adsorbed. Langmuir surface area depends on adsorption capacity of the adsorbent.

Why is Langmuir surface area always higher than BET surface area?

Langmuir adsorption model considers that only a monolayer of adsorbate can be formed on top the surface.
In contrast, BET model considers multilayer formation.

Then, if (as generally occurs) multilayers are formed, BET is able to consider that they are not being formed by directly covering the surface, but by covering extra layers. Langmuir will consider that multilayers can only form as the first adsorbed layer, thus resulting in an exaggerate (unrealistic) surface area.

You can clearly see the differences in the Wikipedia page for Langmuir adsorption.


Birch Reduction of Benzene According to my Organic Chemistry Textbook:

Benzene can be reduced to 1,4-cyclohexadiene by treating it with an alkali metal (sodium, lithium, or potassium) in a mixture of liquid ammonia and an alcohol. This reaction is called the Birch reduction.

Why the conjugated 1,3-cyclohexadiene is not formed in this reaction (as conjugated molecules are more stable)?

According to the article on Birch Reduction on the Organic Chemistry portal:

The question of why the 1,3-diene is not formed, even though it would be more stable through conjugation, can be rationalized with a simple mnemonic. When viewed in valence bond terms, electron-electron repulsions in the radical anion will preferentially have the nonbonding electrons separated as much as possible, in a 1,4-relationship.

http://www.organic-chemistry.org/namedreactions/birch-reduction.shtm


Acid solution — lower pH when diluted Today I came across a "thought experiment" regarding $\mathrm{pH}$ of a diluted acid solution. This idea looks somewhat logical but obviously contradicts the reality. Can you show me where is the error?

Let's suppose we have $\pu{100 g}$ of concentrated 95% (mass) sulfuric acid. That means we have a mixture of $\pu{95 g}$ of H2SO4 and $\pu{5 g}$ of $\ce{H2O}.$ Let's calculate the amount of both compounds.

$$n = \frac{m}{M}$$

$$n(\ce{H2SO4}) = \frac{\pu{95 g}}{\pu{98.08 g mol-1}} = \pu{0.9686 mol}$$

$$n(\ce{H2O}) = \frac{\pu{5 g}}{\pu{18.02 g mol-1}} = \pu{0.2778 mol}$$

Since the amount of the solvent molecules is much much lower than the amount of the $\ce{H2SO4}$ molecules, it is not possible for the acid to be completely dissolved, which leads to the relatively small amount of $\ce{H3O+}$ cations, and $\mathrm{pH}$ of the solution is relatively high $(\mathrm{pH} = -\log [\ce{H3O+}]).$

$$\ce{H2SO4 + H2O -> HSO4^- +  H3O+}$$

If we increase the amount of the $\ce{H2O}$ molecules (dilute solution), more $\ce{H3O+}$ cations can be formed, and the $\mathrm{pH}$ will decrease.

Conclusion: dilution of acid-water solutions decreases $\mathrm{pH}.$

In reality it is completely opposite. So what is the catch?

You run into all sorts of problems when the pH is outside of the range of 1 to 13. For example, in Environ. Sci. Technol. (2000) 342, p. 254-258, they say:

  [...] the former National Bureau of Standards (NBS) established a set of conventions that limits measurements to 1 < pH < 13 and to ionic strength, I < 0.1. The main limitations are the activity coefficient expression, the range of defined standard pH buffers, and interferences with the reversible response of the glass H+-sensitive membrane electrode.

They go on to use standard of concentrated sulfuric acid to define the pH, using a Pitzer model to predict activity coefficients (figure from researchgate) :

  In reality it is completely opposite. So what is the catch?

If you do the math, the extreme pH of -5 shown in the figure corresponds to a hydrogen ion activity of hundred-thousand. The concentration of water for pure water is "only" ~55 mol/L, so this must reflect an activity coefficient far from one.


Options for long term storage of helium? Specifically, if I wanted to purchase a large amounts of helium, what are my options for purchasing helium in a state that is suitable for long term storage?

It's my understanding that if I were to buy helium in bulk it would be shipped cold, as a liquid.  Keeping helium in liquid form for 10-20+ years would involve a lot of serious refrigeration.

Are there other reasonable options for bulk, long term storage?

This may not seem like much of a "chemistry" question but since this is used in chemical sciences, I assume this is a need that some may be aware and may have a good answer since this is an elemental material that some may use on a very common basis.

In my experience, for small quantities in laboratories helium usually comes compressed in cylinders. Except possibly in the case that you require extremely high purity, I don't see why helium couldn't be stored for long periods in this form, it would be relatively safe and would avoid the need for refrigeration.

I don't know what sort of quantities you are thinking about, but it seems that you can get big cylinders. For example the largest cyclinder size available from BOC gases, a British company, is 194.7 $m^3$ filled to a pressure of 300 $bar$ (source: BOC website).


State functions and their arguments $U$ and $H$ are defined as
$$U=\sum E_{pot} +\sum E_{kin} ; dU=dq+dw $$
$$H=U+pV$$

However, it was stated in a lecture note that for a fixed molar number, $U$ and $H$ can be defined using any two properties, example: $U(Temperature,Volume)$ and $H(Temperature, Pressure)$.

However, I cant figure out why only two quantities are enough, even if we consider the above example.

Can somebody show how by using just these two quantities is enough to figure out $U$ and $H$? (Just proving for the above example would be fine)

This follows from the first law of thermodynamics; i.e., $$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V,$$ which identifies $U$ as a function of $S$ and $V$. If in addition we have other forms of work, such as chemical work, then we add a chemical work term $\mu\mathrm{d}N$ to the first law, and introduce an additional dependence of $U$ on $N$.

In addition, we can also replace a given independent variable by its conjugate variable, since, by thermodynamic stability, the independent variable is a monotonic function of its conjugate and vice versa. That is, the following sets of independent variables can all be used to characterize $U$: $$(S, V, N), (T, V, N), (S, P, N), (S, V, \mu), (T, P, N), (T, V, \mu), (S, P, N).$$ Note in particular that $(T, P, \mu)$ cannot be used, because these three variables, all being intensive, are not independent and are related by the Gibbs-Duhem equation.


Heating of Ethers with HI and Cleavage of Ethers Question:

Give the major product formed by the heating of the following ether with HI.
        CH3
        |
CH3-CH2-CH-CH2-O-CH2-CH3

My Attempt:
According to me, the product should be this due to delocalization of positive charge among the neighboring carbon atoms.

        CH3
        |
CH3-CH2-CH-CH2-I + CH3-CH2-OH

Answer:
I checked the answer in my book and it is completely different:

        CH3
        |
CH3-CH2-CH-CH2-OH + CH3-CH2-I

Now, I think they may have considered steric hindrance to be the deciding factor of site of attack of OH but will not delocalization be dominant over steric effect?
Also, what will be the effect of hyperconjugation on the reaction?

Your ether has $\ce{CH2}$ groups at both sides of the oxygen atom.

I would exclude a $S_N1$ reaction here! Instead, cleavage of the ether proceeds via $S_N2$. The oxonium ion resulting from addition of a proton is attacked by iodide at the least hindered carbon atom. Iodide is quite big ;-)


Can you experimentally determine activation energy if the rate expression and rate constant are unknown? When drawing a graph to find the activation energy of a reaction, is it possible to use ln(1/time taken to reach certain point) instead of ln(k), as k is proportional to 1/time?

For example, if an experiment I'm doing has a specific end point, can I use ln(1/time taken to reach that endpoint) instead of ln(k), and do this for a range of temperatures in order to work out the activation energy? Or would I have to use the rate equation and determine the rate constant for each temperature, as I am not sure of the rate equation for the reaction.

Thank you:)

This is valid only provided that the reaction is strictly first-order and provided that the “specific end-point” is a fixed concentration ratio. Note that the slope of $\ln(t_\text{end})$ vs. $\frac{1}{T}$ is $\frac{E_a}{R}$ but the intercept is not $\ln(A)$.

If the rection is first order ( $X {\xrightarrow{k}} Y$) then the product concentration at time t is:

$[Y]_t = [X]_0 .e^{k. t}$

ie:

$\ln\left( \frac{[Y]_t}{[X]_0}\right) = k. t $

If the end point is some fixed concentration ratio ($\frac{[Y]_{end}}{[X]_0}$), the time to reach the endpoint is:

$ t_{end} = \frac{\ln\left( \frac{[Y]_{end}}{[X]_0}\right)}{k}$

If we substitute this into the Arrhenius equation ($\ln(k) = \ln(A) - \frac{E_{a}}{R.T}$), we get:

$\ln(t_{end}) = \ln\left( \ln\left( \frac{[Y]_{end}}{[X]_0}\right)\right) - \ln A + \frac{E_{a}}{R.T} $

so a plot of $ln(t_{end})$ vs. $\frac{1}{T}$ will have a slope of $\frac{E_{a}}{R}$


Why are metalloids found in a zig-zag line on the periodic table I am confused about the zig-zag pattern of metalloids in the periodic table. Why are metalloids arranged in a zig-zag line? Can anyone answer my question?

In addition to Pritt Balagopal's answer, in which he has clarified the trends in metallic character across the Periodic Table, I would also like to explain why the metalloids occur in a zig-zag fashion.

Firstly, I would like to again clarify that it is possibly a coincidence that the metalloids start from boron, not carbon or any other non-metal in that period. My guess could be the start of electron occupation of the p subshells. Note that boron has one electron in its p subshell. Thus, its chemical properties start to change to become more non-metallic in nature.

Diagonal relationships is a common occurrence in the Periodic Table. However, it is often used to describe the relationship between two elements which are diagonal to each other, not a whole line of them. Thus, I would like to clarify again that this is just my conjecture, not what is the true reason. But I believe this should be part of the correct explanation as it is logically sound.

This site offers a more in-depth explanation as to why diagonal relationships occur:
https://chem.libretexts.org/LibreTexts/University_of_Denver/Chem_2132%3A_Chemistry_of_the_Elements/Unit_3%3A_Descriptive_Chemistry/09%3A_Building_a_Network_of_Ideas_to_Make_Sense_of_the_Periodic_Table/9.3%3A_The_Diagonal_Effect

In short, the diagonal relationships are based on two effects: the increasing effective nuclear charge across a period and the decreasing effective nuclear charge down a group. Effective nuclear charge increases across a period because the number of shielding electrons in the inner shells is constant while there are additional protons in the nucleus of the atoms of the elements across the period. Effective nuclear charge decreases down the group as there is an addition shell of shielding electrons from one element to the next in the group. (These are similar to the trends in metallic character pointed out in Pritt Balagopal's answer.)

These two competing effects cancel each other out and thus, chemists believe that the chemical properties of elements in a diagonal should be similar. The image shows some examples.


Preparing an inert atmosphere reaction vessel with only an input? One time, the graduate student that I work with was showing me how to prepare an inert atmosphere in a double necked round bottom flask. After drying it in the oven, she used a torch to drive out all the water vapor. Then she put septums over both openings, and inserted a needle connected to the nitrogen input through one of them, but with no outlet! I asked her to explain how any gas could escape in this set-up, and she tried but I was still very confused.

Her setup was wrong. There needs to be an outlet, the simplest solution in this case being a second needle with nothing connected to it, puncturing either septum to act as a vent to the atmosphere. After a while, depending on the volume of the flask and amount of nitrogen flow, most of the atmospheric air is removed.

This procedure is, however, quite inefficient, and likely unsuitable for particularly sensitive reactions. While most oxygen and moisture is quickly removed, it becomes exponentially slower to remove the last remnants. A much quicker and more reliable method is to perform vacuum/nitrogen backfill cycles. Three or four cycles will fully replace the interior atmosphere with nitrogen gas of the same purity as the input. This can be performed using a single needle and a Schlenk line (or any system connected to nitrogen and a vacuum pump with a intermediating stopcock), which essentially switches the same needle between a gas inlet and a gas outlet.


Is this a feasible synthetic route for 1-(phenylethynyl)cyclopentan-1-ol? From my notes $\ce{NaNH_2}$ is a good base for the deprotonation of alkynes. After the deprotonation I thought the alkyne would attack the carbon on the cyclopentanone because of the $\delta^+$ charge on the carbonyl carbon due to the EWG effects of the oxygen. Should the protonation of the $\ce{O^-}$ be carried out with conditions specified or can I just leave it as the addition of $\ce{H^+}$?

The route definitely looks feasible and it is more or less what I would have done to arrive at your desired product. A few notes:

Make sure you actually have sodium amide $\ce{NaNH2}$ and not sodium in ammonia $\ce{Na/NH3}$. As pointed out in the question comments, they react very differently; the latter generates a trans-alkene.
I’m not sure I would have chosen the same base. But the choice of base is likely arbitrary anyway. n-Butyllithium would definitely also do the trick but might be overkill.
Typically in a lab setting, reactions with strong bases (as is the case here) are carefully quenched with something aquaeous. So you can be sure that there will be an aquaeous proton source ($\ce{H2O}$ or $\ce{H+/H2O}$, depending on your markers’ preferences) in the end. There should be no need to additionally specify conditions. It would make no sense to attempt to isolate the alcoholate anyway.


What reagents change the time of a chemical clock? I am doing a research task involving an iodine chemical clock reaction. I need to figure out which reagents in the experiment are responsible for changing the time before the colour change in the clock. The current method (see image attached) shows the quantity of materials required for a ~2 minute iodine clock. For this research task, we are required to find the quantity of materials necessary for a ~10 minute iodine clock. Can you please clarify which reagents cause the increase/decrease in time of the chemical clock reaction?

First, a review of the underlying chemistry per an educational source:

A clock can be formed from molecules that react at a rate from the time the chemicals are mixed to the time the reaction occurs.
Explanation: There are two forms of iodine:
Element form (Iodine): I2 + starch -------> blue
Ion form (Iodide): I- + starch --------> colorless
There are two reactions taking place simultaneously in the solution:
#1. 2I- + H2O2 ---(slow)-------> I2 + H2O
#2. I2 + Vitamin C ----(fast)----> I (colorless)
In Reaction # 1 iodide ions react with hydrogen peroxide to produce iodine element which is blue in the presence of starch. BUT, BUT......
In Reaction # 2 The Vitamin C is immediately reacting with any iodine formed in reaction # 1. The net result, at least for part of the time is that the solution remains colorless with excess of iodide ions being present.
Now after a short time as the reactions keep proceeding in this fashion, the Vitamin C gets gradually used up. The Vitamin C creates a clock reaction (1 - 2 minutes) and once it is used up, the solution turns blue, because now the iodine element and starch are present.

So, to answer the question: "which reagents in the experiment are responsible for changing the time before the colour change in the clock", the answer is H2O2 and Vitamin C. Now, while the rate-controlling step, associated with the slowest reaction is H2O2 (creating the colourful I2 highlighted by starch), some Ascorbic acid must be present to turn the solution clear again.
On the question:"Can you please clarify which reagents cause the increase/decrease in time of the chemical clock reaction?", the strength of the H2O2 (try changing from 3% to 10%) and the pH of the hydrogen peroxide (try adding alkaline Washing Soda Na2CO3 to the H2O2), both I would likely expect to have impacts on the Vitamin Clock timing.


Why do we use external pressure to find force exerted by gas on a piston in a closed vessel during its expansion? I was reading a book in which i got confused due to this. A gas is in a vessel covered by a friction-less piston. Gas pressure is more than outer atmospheric pressure (which is constant) due to which it expands. Since it expands and move the piston outward, it does work on it which is calculated by force times displacement. In finding force we use Force = Pressure times Cross section area of piston. In finding this force to calculate work why are we using Pressure on piston by atmosphere and its weight and not pressure exerted by gas. Why? Can you also tell how the correct work done by gas can be calculated.
Thanks.

For a frictionless, massless piston, the force per unit area that the gas exerts on the inner face of the piston is always equal to the force per unit area applied on the outer face of the piston (since the piston is massless, ma = 0).  However, for an irreversible process, the force per unit area exerted by the gas at the interface is composed both of the "thermodynamic pressure"  that one could determine from the equation of state for the gas (e.g., the ideal gas law) applied locally at the interface,  plus "viscous stresses" in the gas that are determined by the rate at which the gas is deforming.  Furthermore, if the expansion or compression is rapid enough, the conditions within the cylinder are not even uniform spatially.  So figuring out the force per unit area that the gas exerts at the interface for irreversible expansion is problematic.  However, fortunately, if the force per unit area on the outside is specified by manually imposing a force per unit area (e.g., if the atmosphere is imposing the pressure outside), the work done by the gas on the surroundings can still be determined.  This is because the force per unit area exerted by the gas at the interface will adjust to always match the force per unit are on the outside (again, since the piston is massless and ma = 0).

In the case of a reversible process, the situation is much simpler, since here, viscous stresses are negligible, and the force that the gas exerts per unit area on the inner face of the piston can be determined by applying the ideal gas law (or real gas equation of state) to the spatially homogeneous gas in the cylinder.  In this case, however, the force imposed on the outside must be controlled very carefully, so that it differs only slightly from the average inside gas pressure during the entire expansion or compression (and the gas pressure within the cylinder remains essentially uniform spatially).


Why does $\text{Cr}^{3+}$ not have the same electron configuration as $\text{Sc}$? To be explicit, $\text{Cr}^{3+}$ and $\text{Sc}$ have the same number of electrons, the only difference is the nuclear charge.

This is not an isolated anomaly, it seems: the book I'm using claim that with all fourth period transition metal cations, the $4 \text{s}$ electrons are removed first. This indicates that the $4 \text{s}$ electrons are at a higher energy than $3\text{d}$.

My question: why is this the case, given that the $4\text{s}$ electrons are 'added' before the $3\text{d}$ electrons as one moves rightwards across the fourth period, indicating that they are of a  lower energy? I cannot reconcile this apparent contradiction.

The electron configuration of chromium is : $\ce{[Ar] 4s^1 3d^5 }$ (chromium is one of those examples of special electron configurations : the electron configuration is not $\ce{[Ar] 4s^2 3d^4 }$ as many people might suspect, an electron is moved to the 3d orbital because this configuration is more stable  --> that is the reason why chromium has this special electron configuration)

So the $\ce{Cr^3+}$ ion loses an electron from $\ce{4s}$ and loses 2 electrons from $\ce{3d}$ to give it an electron configuration of $\ce{[Ar] 3d^3 }$

Scandium has an electron configuration of $\ce{[Ar] 4s^2 3d^1  }$ and isn't one of those exceptions.

These are the reasons why the chromium(+III) ion doesn't have the same electron configuration as scandium, despite the fact that they have equal amounts of electrons.

As for why the 4s orbital is emptied first:

  In all the chemistry of the transition elements, the 4s orbital
  behaves as the outermost, highest energy orbital. The reversed order
  of the 3d and 4s orbitals only seems to apply to building the atom up
  in the first place. (Using the Aufbau principle) In all other respects, the 4s electrons are always
  the electrons you need to think about first.

(this link is interesting for you in case you wanted a more detailed explanation about the 4s / 3d problem :http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html )

EDIT : Chromium, just like all other elements, tries to obtain more stable configurations by losing certain amounts of electrons. Now, in the case of $\ce{Cr^3+}$ half/fully filled subshells can't be achieved ( you get $\ce{[Ar] 3d^3}$ ) but that doesn't necessarily mean that it is not stable. But chromium does have other ions in which it does have more stable electron configurations. Take a look at the $\ce{Cr^6+}$ ion : here the ion has an electron configuration that is equal to the one of argon, which does have completely filled valence subshells. ($\ce{[Ne] 3s^2 3p^6}$)

So ions can't always achieve these more stable half/fully filled subshells, but that doesn't necessarily mean that they are not stable.


Calculating Ratio of Stability Constants Solution I contains a mixture of $\ce{FeCl2}$ and $\ce{FeCl3}$, and solution II contains a mixture of $\ce{K4Fe(CN)6}$ and $\ce{K3Fe(CN)6}$. The concentrations of iron-containing species satisfy the relations $[\ce{Fe^2+}] = [\ce{Fe(CN)_6^4-}]$ and $[\ce{Fe^3+}] = [\ce{Fe(CN)_6^3-}]$. The potential of platinum electrode immersed into the solution I is $0.652\ \mathrm{V}$, while the potential of platinum electrode immersed into solution II is $0.242\ \mathrm{V}$.
Standard redox potential for $\ce{Fe^3+ + e- -> Fe^2+}$ is $0.771\ \mathrm{V}$
Calculate the ratio of the stability constants $\beta [\ce{Fe(CN)_6^3-}] / \beta [\ce{Fe(CN)_6^4-}$]

My attempt
First I wrote out the Nernst Equation for both reactions:
$$0.652 = 0.771 - 0.059 \times\mathrm{log}\ce{ \frac{[Fe^2+]}{[Fe^3+]}}$$
$$0.242 = \mathrm{E_{cell}^\circ} - 0.059 \times\mathrm{log}\ce{ \frac{[Fe(CN)_6^{3-}]}{[Fe(CN)_6^{4-}]}}$$
Also I wrote out the expression for the ratio of stability constants"
$$\frac{\beta _1}{\beta _2} = \ce{ \frac{[Fe(CN)_6^{3-}][Fe^2+]}{[Fe(CN)_6^{4-}][Fe^3+]}}$$
Now I know that the ratio is just the sum of the 2 logs in the Nernst equations that I have wrote above. However I am not sure how to calculate the $\mathrm{E_{cell}^\circ}$ for the second reaction. Can we assume that $\mathrm{E_{cell}^\circ}$ for the second reaction also equals $0.771\ \mathrm{V}$?
The answer is $8.90 \times 10^6$
Any help is greatly appreciated.

You used all the information from the question except the line that would have helped you solve it

  The concentrations of iron-containing species satisfy the relations $[\ce{Fe^2+}] = [\ce{Fe(CN)_6^4-}]$ and $[\ce{Fe^3+}] = [\ce{Fe(CN)_6^3-}]$

From your first cell you find that

$$\frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} = 103.97$$

which means that in your second cell

$$\frac{[\ce{Fe(CN)6^4-}]}{[\ce{Fe(CN)6^3-}]} = 103.97$$

and $E^\circ$ for the cell with cyanide is $+0.361~\mathrm{V}$. (You have a typo in your equation, the 4- should be in the numerator and 3- should be in the denominator. Also your equation is missing units but I won't bother with that.) This is quite far off from the accepted literature value (around $+0.430~\mathrm{V}$) but let's move on.

  Now I know that the ratio is just the sum of the 2 logs in the Nernst equations that I have wrote above

No, that is wrong. The ratio of the stability constants refers to that expression you have written, when all of the compounds are placed into the same solution. You cannot simply use the two equations earlier because those describe two separate solutions.

You have to use the difference in $E^\circ$ to work out the ratio of $\beta$.

The way to do this would be to realise that if you take the difference between the $E^\circ$ you get

$$\begin{align}
\ce{Fe^3+ + Fe(CN)6^4-} &\longrightarrow \ce{Fe^2+ + Fe(CN)6^3-} & E^\circ &= +0.410~\mathrm{V}
\end{align}$$

The equilibrium constant for this reaction is the quantity that you are looking for

$$K = \frac{[\ce{Fe^2+}][\ce{Fe(CN)6^3-}]}{[\ce{Fe^3+}][\ce{Fe(CN)6^4-}]}$$

and so

$$K = \exp\left(\frac{-\Delta_\mathrm{r} G^\circ}{RT}\right) = \exp\left(\frac{-nFE^\circ}{RT}\right) = 8.589 \times 10^6$$


Piece of equipment similar to syringe for gas collection I'm looking for the name of a piece of equipment similar to a syringe. The only difference is that this item is made entirely from glass. The purpose of this equipment is to collect gases.

Here is a photo of two of these pieces of equipment connected to a gas-washing bottle:

This piece of laboratory glassware is a gas syringe. Apparently, gas syringes are not standardized; however, typical models have a capacity of 100 ml. Some models include a GL 14 screwthread tube, a stopcock, or a three-way stopcock.


What is the order of stability of the resonance structures of furan? I want to know which are the most stable furan resonance structures. I know that structure number 1 is the most stable, but which one comes after?

A quick post so you can get started on thinking about this.

Remember, resonance is just a model. The electrons are delocalized all the time; the furan does not morph back and forth among the various structures.

You can do a "formal" evaluation of which resonance forms have the most contribution to the structure by looking at formal charges, and there are rules to help you decide, from the formal charges, which resonance forms are most important.

If you're looking at reactivity, the property you're interested in may not be that the one that is "more stable is less reactive". What you might look at instead, is the consequences of attack at the 2 and 3 position, look at the stability of the intermediates formed...a more stable intermediate would tend to drive the reaction along that path.

After looking at these ideas, feel free to come back with specific questions.


How to dissolve candle wax (paraffin)? Wikipedia states that waxes can be dissolved by non-polar organic solvents.

I failed to dissolve candle wax in both acetone and ethanol, which are the only organic solvents I have at home.

I have some clothes with wax stuck on them. How can I dissolve the wax and drive it away without damaging my clothing?

Ethanol and acetone are not non-polar organic solvents. Each one has  a slight dipole moment; due to the difference of electronegativity between $\ce{H}$ and  $\ce{O}$ in ethanol and between $\ce{C}$ and  $\ce{O}$ in acetone.  Wax is composed of heavy, long-chain alkanes. And as "Like dissolves like" try to dissolve your wax in toluene or in xylene.


What is the balanced equation for the reaction of potassium permanganate, iron sulfate, and sulfuric acid? I'm trying to balance the following redox equation. I think this is happening in acidic solution, the textbook doesn't specify anything more. Can you please help me understand if I got the half reactions correct?
$$\ce{KMnO4 + FeSO4 + H2SO4 -> K2SO4 + MnSO4 + Fe2(SO4)3 + H2O}$$

I found out that manganese is reduced as the oxidation number goes from $+7$ to $+2$ and iron gets oxidized as the oxidation number goes from $+2$ to $+3$.
I wrote the two half reactions, \eqref{Q:red} for the reduction and \eqref{Q:ox} for the oxidation. Did I get them right?

\begin{align}
\tag{A}\label{Q:red}\ce{KMnO4 + H2SO4 &-> K2SO4 + MnSO4}\\
\tag{B}\label{Q:ox}\ce{FeSO4 &-> Fe2(SO4)3}\\
\end{align}

I started working on the reduction half-equation and this is what I came up with after balancing atoms and charge:

$$\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 -> K2SO4 + 2 MnSO4 + 8 H2O}$$

Now comes the trouble. How do I balance iron and sulfur in the oxidation half-reaction \eqref{Q:ox}?

Your first half reaction, for the reduction, is correct:
$$\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 -> K2SO4 + 2 MnSO4 + 8 H2O}\tag1\label{red}$$

For the second half reaction, the oxidation, start by balancing iron:
\begin{align}
\ce{FeSO4 &-> Fe2(SO4)3 }\tag{2a}\\
\ce{2 FeSO4 &-> Fe2(SO4)3 }\tag{2b}
\end{align}

Add $\ce{H2SO4}$ on the left so you can balance sulfur:
\begin{align}
\ce{2 FeSO4 &-> Fe2(SO4)3 }\tag{2c}\\
\ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 }\tag{2d}
\end{align}

Now balance the protons and electrons:
\begin{align}
\ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 }\tag{2d}\\
\ce{H2SO4 + 2 FeSO4 &-> Fe2(SO4)3 + 2 H+ + 2 e-}\tag{2e}\label{ox}\\
\end{align}

Now add \eqref{red} and five times \eqref{ox} so that the electrons are equal on every side and coincidentally the protons also balance:
\begin{align}
\ce{10 e- + 10 H+ + 2 KMnO4 + 3 H2SO4 &-> K2SO4 + 2 MnSO4 + 8 H2O}\tag1\\
\ce{5 H2SO4 + 10 FeSO4 &-> 5 Fe2(SO4)3 + 10 H+ + 10 e-}\tag{$5\times$2e}
\end{align}

And the final result is:
\begin{align}
\ce{2 KMnO4 + 8 H2SO4 + 10 FeSO4 &-> K2SO4 + 2 MnSO4 + 8 H2O + 5 Fe2(SO4)3}\tag3
\end{align}


Carbon and dilute sulfuric acid I have to distinguish between copper oxide and carbon. I thought of dilute sulfuric acid because it forms copper sulfate with copper oxide. But I don't know how carbon reacts with dilute sulfuric acid. I know how carbon reacts with concentrated sulfuric acid but dilute?

Sulfuric acid is  unreactive with carbon- indeed only hot concentrated sulfuric acid reacts with carbon to produce carbon dioxide.


Relative humidity and vapor-pressure in equilibrium I have quite an unclear understanding of relative humidity. I am writing my understanding on the topic and please correct me wherever I'm wrong:
What I understood:
Let's say $\ce{H2O}$ has a (saturated) vapour pressure of $x~\pu{atm}$ at a certain temperature. Now, relative humidity (RH) is the ratio of the partial pressure of water vapour to the equilibrium vapour pressure of water at a given temperature.
So what this means is that if RH = 100% then the maximum amount of water vapours is present in the air. Now if RH = 50% means that the amount of water vapours present in the air is only half of the maximum possible.

So, does this mean the water has not been fully evaporated up to the
limit it can?
Does this mean that the equilibrium has not yet been established, and
the reaction tends to shift in the forward direction? If yes, then
what is stopping the reaction to move forward?
Does 100% RH mean the water and it's vapour are in equilibrium?

Can relative humidity be compared with an example of a sponge? There's a limit of the amount of water a sponge can absorb water. If it has only absorbed 50% means there is still room for more absorption until 100% capacity is reached.

What I understood:
Let's say $\ce{H2O}$ has a (saturated) vapour pressure of $x$ atm at a certain temperature. Now Relative humidity (RH) is the ratio of the partial pressure of water vapour to the equilibrium vapour pressure of water at a given temperature.
So what this means is that if RH = 100% then the maximum amount of water vapours is present in the air. Now if RH = 50% means that the amount of water vapours present in the air is only half of the maximum possible.

So, does this mean the water has not been fully evaporated up to the limit it can?

Less than 100% humidity would mean exactly that. Assuming that there is still liquid water left, then an equilibrium has not been reached.

Does this mean that the equilibrium has not yet been established, and the reaction tends to shift in the forward direction? If yes, then what is stopping the reaction to move forward?

Exactly. Assuming an isothermal system with more liquid water available, then yes, more water would evaporate. Remember when water evaporates it cools the remaining water.

Does 100% RH mean the water and it's vapour are in equilibrium?

Yes, assuming both phases exists, and that the gas and liquid phase are at the same temperature.
Some  extra points:

The saturated vapor pressure of water is a complex function of temperature. So it is $x_1$ at $T_1$ but $x_2$ at $T_2$, and so on.

Ice sublimes and thus will also have an equilibrium water vapor pressure. The equilibrium water vapor pressure again depends on the temperature of the ice.


Na(s) vs NaCl(s) Entropy I would think the crystalline, ionic solid would have the lower entropy, however, the tablature indicates otherwise. Why does $\ce{NaCl(s)}$ have a greater entropy than $\ce{Na(s)}$?

If you don't mind me asking, why would you expect the $\ce{Na}$ to have a higher entropy than the $\ce{NaCl}$? The entropy of a state is defined as,$$S=k_b\ln W$$where $W$ is the number of possible microstates which can make up that macrostate.

In this case, we are interested in the number of possible ways that we can arrange $\ce{Na}$ into a lattice or the number of ways we can arrange $\ce{Na}$ and $\ce{Cl}$ ions into a lattice. The latter has many, many more microstates for a given macrostate, and hence a larger entropy.

The standard entropies can be found in this table.

It is worth noting that what I have written above is not exactly the same as the standard entropy $S^\circ$ which is defined as,
$$S^\circ=\sum_{i=1}^N\int\frac{dq_i}{T}dT$$which just says that the standard entropy going from zero kelvin to 298 kelvin depends on the heat put into all $N$ atoms of a system. Thus, there is a dependence on the energy of any state the system passes through.

For the solids you're talking about, if the energetics of the systems were exactly identical, then the ratio of standard molar entropies would be the same as the ratio $\frac{\ln W_1}{\ln W_2}$. But, because sodium has much weaker bonds than sodium chloride, it has a higher entropy change per unit change in temperature than sodium chloride. This is because for weaker bonds, the atoms have greater mobility in space which increases entropy, and it is more likely for there to be lattice defects when the bonds are weak which also increases entropy.


How is chlorocarbene synthetised? The wikipedia page on cyclononatetraene says that the aromatic cyclononatetraenyl anion can be synthetised by a rearangement of 9-chlorobicyclo[6.1.0]nona-2,4,6-triene.

My first idea on how to synthetise this compound is to react cyclooctatetraene with chlorocarbene $(\ce{:\!CHCl})$. However, in contrast to dichlorocarbene $(\ce{:\!CCl2})$, chlorocarbene $(\ce{:\!CHCl})$ is much less stable. Can it be synthetised by an action of a strong base on dichloromethane?

It seems, from this pdf, that $t$-$\ce{BuOK}$ will make $\ce{:\!CPhCl}$ from $\ce{CHPhCl2}$. Is it, however, a strong enough base to deprotonate $\ce{CH2Cl2}$? Or must some stronger one, e.g. LDA, be used?

The formation of dihalocarbenes (at least cloro- and bromo-counterparts) in reactions of haloforms with bases has been well established (e.g., Ref.1,2). In 1959, Closs and Closs introduced a novel reaction to prepare chlorocarbene as a reaction intermediate in situ in order to synthesize chlorocyclopropanes derivatives (Ref.3). They have reported that when n-butyllithium is added to methylene chloride in the presence of olefins at $\pu{—25 °C}$, chlorocyclopropanes are formed in fair to good yields depending on the structure of the olefin. They further claimed that the novel procedure strongly suggests the existence of chlorocarbene as a reaction intermediate.

They further developed the method and published a full paper as a general method (Ref.4) where abstract states:

  Alkyllithium compounds react with methylene chloride in the presence of olefins to form chlorocyclopropanes. This reaction is believed to proceed through the intermediate chlorocarbene. Seven alkyl-substituted chlorocyclopropanes have been prepared by this method. Configurations are assigned to the products by considerations of predictible non-bonded interactions in the transition states of the addition of chlorocarbene to the olefins. The configurations are confirmed by n.m.r. spectra.

Later, Billups and Haley have used the same method to synthesize spiropentadiene (Ref.5).

References:

W. von E. Doering, A. Kentaro Hoffmann, “The Addition of Dichlorocarbene to Olefins,” J. Am. Chem. Soc. 1954, 76(23), 6162–6165 (https://doi.org/10.1021/ja01652a087).
W. von E. Doering, Wm. A. Henderson, Jr., “The Electron-seeking Demands of Dichlorocarbene in its Addition to Olefins,” J. Am. Chem. Soc. 1958, 80(19), 5274–5277 (https://doi.org/10.1021/ja01552a065).
Gerhard L. Closs, Lisellote E. Closs, “Synthesis of Chlorocyclopropanes from Methylene Chloride and Olefins,” J. Am. Chem. Soc. 1959, 81(18), 4996–4997 (https://doi.org/10.1021/ja01527a057).
Gerhard L. Closs, Lisellote E. Closs, “Carbenes from Alkyl Halides and Organolithium Compounds. I. Synthesis of Chlorocyclopropanes,” J. Am. Chem. Soc. 1960, 82(21), 5723–5728 (https://doi.org/10.1021/ja01506a041).
W. E. Billups, Michael M. Haley, “Spiropentadiene,” J. Am. Chem. Soc. 1991, 113(13), 5084–5085 (https://doi.org/10.1021/ja00013a067).


How do cells form a solid structure? How do cells unite to form a solid structure. For example a plant body is made up of a number of cells. The cells are not solid structures. Their consistency is in between solid and plasma. So how come they form solids. In solids i know that the constituent particles which may be ions or atoms or molecules are present very close to each other forming an amorphous or crystalline solid structure. But in plant body there are cells. So how are the cells are arranged to form a solid structure a solid body. why not a liquid or a gaseous. I have looked up in various sources but i have not found any mention of this topic. I am reallt desperate to know more on this. So any help will be appreciated. This is not a homework question and if anybody requires me to edit the question then please say so. Thanks

So how are the cells are arranged to form a solid structure a solid body?

Plant cells have cell walls, are filled with liquid, and have strong cell-cell adhesion. This gives a fairly rigid structure that may seem like a solid. Take a carrot for example, it seems solid. But if you put the carrot in a blender, you get carrot juice - not a solid. Also, when the carrot dries out, it loses its rigidity.

Trees

Woody plants contain a lot of cellulose, a water-insoluble polymer, outside of cells. This gives them even more of a solid (and dry) appearance, but the insides of their cells are aqueous as well.

Cells other than plant cells

There are cells that lack a cell wall, they are just surrounded by a lipid bilayer. Tissues made from these cells (muscle, brain, liver, etc.) don't seem solid at all. Animals do have connective tissue (mostly protein-based) and a skeleton that holds everything together unless there is trauma (predators, accidents, etc).

  How do cells form a solid structure?

Let me un-ask the question a bit. Sometimes they do form a solid-like structure, and sometimes they don't. Fibrous biological macromolecules play a large role in making tissues rigid, either directly or in combination with having cells filled with aqueous solution.


Titration of aspirin I have performed an experiment to test how temperature affects how fast aspirin dissolves by titrating solutions at various time intervals. According to this website, aspirin reacts with water as such:

$$\ce{C9H8O4(aq) + H2O(aq) -> C9H7O4-(aq) + H3O+(l)}$$

However, I've also read that aspirin undergoes hydrolysis with water, according to this site, where Aspirin (acetylsalicylic acid) hydrolyses to produce 2-hydroxybenzoic
acid and ethanoic acid.

So, I'm confused — when I'm dissolving the aspirin tablets in water, am I just making acetylsalicylic ions and hydronium ions which I can then titrate to find the concentration, or am I making two other acids as per the second site suggests? I have all my titre values, I just need to calculate the concentration of the aspirin the solutions, and I'm now confused about what chemical I'm actually finding the concentration of, e.g.

$$\ce{NaOH + HCl -> NaCl + H2O}$$

Here, sodium hydroxide reacts with hydrochloric acid. In my experiment, does the $\ce{NaOH}$ react with acetylsalicylic acid, or with 2-hydroxybenzoic acid and ethanoic acid?

I am afraid, direct acid-base titration is not the right way to analyze aspirin exactly because of the you stated - hydrolysis. The rule number no of any titration is that there should be no side reaction and it should go to completion almost instantly. The direct titration of aspirin is problematic because hydrolyzes pretty fast to salicylic acid- an unwanted side reaction which may or may not go to completion.

In order to circumvent this, analytical chemists do a back-titration. You add an excess of base, heat the mixture so that neutralization as well as hydrolysis are complete. The remaining base is titrated. The rest of the calculations are normally done.


How will my studies in mathematical statistics be applied to and assist me with problems in chemistry? I'm currently working towards a major in both chemistry and mathematical statistics as a part of my science degree, and I got to wondering:

How will these two go hand in hand and help each other?

Now, I know there's the obvious applications of statistical inference to any experimentation I do, but I'm talking about less obvious relationships. Could my knowledge of solving SDEs ever help me with my experiments? What about continuous Markov processes? Will I ever even think about a discrete random variable again if I go on to get a PhD in chemistry? Also not looking for the standard "well we use statistical physics to model electrons" example, either (welp, unless you can offer some nice specifics). Surely there's other ways that these skills I've found can interact with each other?

Most definitely yes.

I took multiple statistics classes as an undergraduate and they have helped me considerably. (I can argue fairly that it helped me get tenure.)

A few suggestions from my own work:

Statistical design of multivariate experiments: Many scientists think you should test one variable at a time. No. I taught a grad student (in another group) about optimization theory and proper design and three months later, a great paper came out.
Data mining and exploration: Many chemistry data sets are extremely complicated, with cross-correlated factors. Understanding how to draw empirical inferences, find the most important factors, etc. is enormously helpful across a wide range of chemistry.
Principal component analysis: An emerging area in spectroscopy and imaging is taking complicated spectra and performing PCA to help with quantifying chemical components in mixtures and following everything from cancer growth to materials design.
Multivariate regression and ANOVA: My group has created numerous quick statistical screens using multivariate surrogates for complex simulations or experiments.
Cross-validated analysis: Again, the understanding of training and test sets, bootstrapping, etc. is very useful.

Now, you also mention SDEs and Markov processes. These are crucial in chemical kinetics, including areas like reaction-diffusion systems. Monte Carlo methods are also highly used in studying dynamical processes.

The list would be enormous. Basically, I think it's a great skill to have and you'll use it throughout your career.


Is the redox method applicable to all chemical reactions? Is it possible that some chemical reactions can't be balanced through the redox method? Because it seems to me that the elements on the reactants side have the same oxidation states with the product side.

Sometimes you aren't able to use the redox method to balance a chemical equation. This is because the redox method is only used to balanced an equation where a element gets oxidised or reduced. However if there is a chemical equation where no element is getting oxidised or reduced, then you won't be able to use the redox method to balance the equation. However, fortunately most of the time these type of equations can be easily balanced. Consider the following unbalanced equation: $$\ce{Ba(OH)2.8H2O (s) + NH4SCN (s) -> Ba(SCN)2 (s) +  H2O (l) + NH3 (g)}$$

Here none of the elements are reduced or oxidised, hence you would not be able to balance this equation using the redox method.


How many tetrahedral and octahedral holes are there in a bcc unit cell? When I tried to count, I found 24 tetrahedral holes (each formed by 2 body centers of 2 unit cells sharing a face and 2 vertices of the same edge of that face). As each hole is being shared by 2 unit cells, there are 12 holes per unit cell.

Also,I found 6 octahedral holes (6 at face centers being shard by 2 unit cells each and 12 at every edge center being shared by 4 unit cells each).

Generally I was taught that for $n$ packed spheres, there should be $2n$ tetrahedral holes. Does the analysis above mean that this rule doesn't apply here?

BCC doesn't have any tetrahedral voids, however we have octahedral holes at face centres. Four adjacent corners and two closest body centres. In a homogeneous lattice the octahedral is compressed along the body-centre-atom axis. However I believe we can form a hetrogenous lattice and form a close approximation to octahedral.

We have thus $6\times1/2 = 3$ distorted octahedral voids.