Finding work done for a chemical reaction at non-constant pressure and temperature I have studied the expression for work done for a chemical reaction which is basically $W=-\Delta n_{g}RT$. Along with this, the assumption that has been made is that chemical reactions take place at constant pressure and temperature.
Is this always the case? What would the general expression for work done look like if we were to consider temperature and pressure changes during the chemical reaction? Thanks.

There are a series of five assumptions you need to make to get from the general expression for work, to the expression you wrote in your OP.  With each assumption, you are applying successively more constraints to your system.  In the post below I give the assumptions, and the equation that results from each, so that you can see how the form of the equation evolves as successively more constraints are applied.
We start with the general expression for work:
$$\text{đ}w = \vec{F} \cdot d \vec{x}$$
[We use the form $\text{đ}w$ rather than $dw$ because work is path-dependent, and its differential is thus inexact; the dyet glyph, $\text{đ}$, is used to indicate this.]
Now we integrate and express the dot product in terms of scalar quantities:
$$w = \int_{path} \vec{F} \cdot d \vec{x} = \int_{path} |F| \,d |x| \cos(\theta)$$
Because work is path-dependent, it is necessary to integrate over the actual path, as indicated in the above integrals, rather than merely from the initial state to the final state.
Dropping the explict scalar notation, we can write it as:
$$w =  \int_{path} F d x \cos(\theta)$$
Now let's add some assumptions:
Assumption 1: The force is applied in the direction of motion.
As the angle between the direction of motion and the direction of force is $0$, and $\cos(0) = 1$, we have:
$$w =  \int_{path}F dx \cos(\theta) = \int_{path}F dx$$
Assumption 2:  The work is $\bf{pV\text{– work.}}$
For $pV\text{– work}$, we can take work done on the system to be either positive or negative.  I'll use the former convention. And we can only reliably use the external pressure to determine the work; the internal pressure is not always defined, particularly for rapid expansions or compressions.  To indicate this explicitly, we write "$p$" as $p_{ext}$. Then (using "$A$" for area):
$$p_{ext} = F/A;\,\,dV = -A dx\,\, \text{(negative sign is because as $x$ increases, $V$ decreases)}$$
$$\Rightarrow  \int_{path} F dx =  -\int_{path}(p_{ext}A) (dV/A) = -\int_{path}p_{ext} dV$$
Thus:
$$w = -\int_{path}p_{ext}dV$$
This is the general expression for $pV\text{– work}$.  It applies regardless of the conditions.
Assumption 3: The system is in mechanical equilibrium with its surroundings.
In this case, $p_{ext} = p_{sys}$.  By convention, we designate $p_{sys}$ simply as "$p$".
Note that this assumption restricts us to paths in which the system is continously in mechanical equilibrium with its surroundings.  This limits us to paths whose work depends only on the difference between the initial and the final state.  Thus we do not need to concern ourselves with doing an explicit path integral. I.e., because the path restriction is now implicit, we can simply integrate between intitial and final states (which I've designated as $i$ and $f$):
$$w =  -\int_{i}^{f}pdV$$
Assumption 3a: The system is at constant pressure.
Since $p$ is a constant, we can take it out of the integral:
$$w =  -\int_{i}^{f}pdV=  -p\int_{i}^{f}dV = -p\Delta V$$
I labeled this assumption as "3a" rather than 4 because we didn't need to both specify the system is at constant pressure and in mechanical equilibrium with its surroundings, since if it is at constant pressure it is also in mechanical equilibrium with its surroundings.
But I broke the steps up, since I wanted to give the expression following assumption 3 for completeness.
Assumption 4: The system is an ideal gas.
Substituting $V=nRT/p$, and remembering that the pressure is constant, gives:
$$w =  -p\int_{i}^{f}dV =  -p\int_{i}^{f}d(n R T/p)=-R\int_{i}^{f}d(n T)=-R\int_{i}^{f}ndT-R\int_{i}^{f}Tdn$$
Assumption 5: The temperature is constant.
Since $T$ is a constant, we can take it out of the integral:
$$w =-R\int_{i}^{f}d(n T)=-RT\int_{i}^{f}dn = -RT \Delta n$$
Or, equivalently:
$$w =-R\int_{i}^{f}ndT-R\int_{i}^{f}Tdn = -R\int_{i}^{f}Tdn = -RT\int_{i}^{f}dn = -RT \Delta n$$


How are the parameters of a force field obtained for the bond angle HCH? In my head, I can't see molecules that I could vary the HCH bond angle without varying other bond lengths, angles and dihedrals to build an energy vs. HCH bond angle curve.

If I vary a single HCH angle of a simple methane molecule, for example, I would automatically be varying the other HCH angles of that same molecule, which would affect the construction of the actual energy vs. HCH bond angle curve.

I may be talking nonsense.

You don't necessarily need a potential energy curve to fit a force field. I answered a related question about fitting a force field from quantum calculations.

Let's assume you have a harmonic angle term. You could either do this in terms of the angle bending or the distance between the end atoms, but let's take the angle for our example:

$E_{angle} = k (\theta - \theta_0)^2$

The first parameter you need is the optimal angle $\theta_0$ which you can get from a geometry optimization, experiment, etc.

Your question stems from the second parameter, $k$ - the force constant for whatever this angle type happens to be (e.g. H-C-H in methane for your example).

There are a few ways to get the force constant, but it comes from the second derivative, right? So you can get that from a Hessian or vibrational calculation. My favorite discussions of this center around the "Badger's rule for angle force constants, e.g."

"Maximally diagonal force constants in dependent angle-bending coordinates. II. Implications for the design of empirical force fields" J. Am. Chem. Soc. 1990, 112, 12, 4710-4723

  Unfortunately, most such force fields are defined in well-determined
  sets of internal coordinates, whereas empirical potentials use larger
  sets of dependent coordinates. This paper illustrates a unique
  “localized” representation of the angle-deformation potential in
  dependent coordinates which is exactly diagonal for in-plane bending
  at trigonal-planar centers and is nearly diagonal for angle bending at
  tetra coordinate centers.

Modern force field fitting methods typically use scripts that minimize the differences between an in-development parameter set and a set of Hessians and/or experimental data.

As I mentioned in the other question, there's a huge pile of such papers, many with code to derive force fields from quantum chemical data.

QubeKit J. Chem. Inf. Model. 2019, 59, 4, 1366-1381 - code at GitHub
QuickFF J. Comput. Chem. 2015, 36, 1015– 1027 - code at GitHub
ForceBalance J. Chem. Theory Compute. 2013, 9, 1, 452-460 and J. Phys. Chem. Lett. 2014, 5, 11, 1885-1891 - code at GitHub
ForceFit J. Comput. Chem., 2010 31: 2307-2316
Parfit J. Chem. Inf. Model. 2017, 57, 3, 391-396 - code at GitHub


What is the correct way to pronounce iodine? I have always said it like 'io-deen' because whenever 'ine' is used (eg. amine, or fluorine) it is said that way. Sometimes I'm corrected by people who think it should be pronounced 'io-dyne', which is correct?

Manshu is correct,in British English we pronounce iodine as in seen while in North American English we pronounce it as in mine.


Does the octet rule only apply to the second and third shell or does it apply further? In my textbook, examples for the octet rule are only given for up to the third shell.
What about the fourth and fifth shell?
I still get confused over the fact that the $n=3$ shell has 18 electrons yet for chemical purposes atoms seek out 8 electrons. My understanding is that the energy difference between the 3p and 3d orbitals are just so large that it creates a certain stability when the 3p orbital is full. If that is true then things should get messier for the fourth and fifth shells due to more overlapping.
The first shell contains the 1s orbitals. The second shell the 2s and 2p. The third shell the 3s and 3p. What orbitals do the fourth and fifth shells include and how many electrons fill these shells?
Another way of viewing my issue. Take the electronic configuration of Calcium which is 2,8,8,2. There are 2 electrons in the fourth shell. What is the maximum number that can go in that fourth shell before a fifth shell is introduced?

The shell concept is useful rather for 1-electron hydrogen-like atoms, where energies of ns, np, nd, nf orbitals are all the same ( well, not fully true due the very tine QED effect of interaction of electrons with quantum vacuum ).
For multielectron atoms, energy wise, $\mathrm{nd}$ orbitals from originally nth shell belong now rather to the (n+1)th shell. Similarly, $\mathrm{nf}$ orbitals from originally nth shell belong now rather to the (n+2)th shell.
So the octet rule for the 4th and higher periods have limited applicability for the $\mathrm{s}$ and $\mathrm{p}$ orbitals only for groups 1-2 and 13-18. It is not applicable for transition metals involving partially occupied $\mathrm{d}$ or $\mathrm{f}$ orbitals.


Understanding lattice matching of graphene on silver (111) crystal surface and meaning of the √3 x √3 lattice? Figure 1 in Ivor Lončarić and Vito Despoja, Phys. Rev. B 90, 075414 Benchmarking van der Waals functionals with noncontact RPA calculations on graphene-Ag(111) (available in Researchgate) shows a schematic of the graphene structure on what I believe is a silver (111) surface.

The caption reads:

  FIG. 1. (Color online) Geometry of the supercell used for our vdW DFT calculations. Upper part: 2 × 2 graphene lattice matched to √3 × √3 Ag(111) lattice. Lower part: Geometry of the supercell in perpendicular direction.

What exactly is a "√3 × √3 Ag(111) lattice" and how exactly does the square root of three come in to this?

"Root 3" or "$\sqrt{3}$" or "$\sqrt{3} \times \sqrt{3}$" are slang for the Wood's notation (1, 2) $ \left( \sqrt{3} \times \sqrt{3}\right)R30$, a commonly occurring configuration of one hexagonal 2D lattice with respect another.

In this example the substrate is the (111) face of silver. Since it is face-centered cubic with lattice constant of about 4.08 Angstroms the (111) surface is hexagonal with a lattice constant 4.08 / $\sqrt{2}$ = 2.88 Angstroms.

The figure caption says "2 × 2 graphene lattice matched to √3 × √3 Ag(111) lattice."

The annotation in the OP's drawing copied below shows solid arrows for the Ag(111) lattice vectors the dashed arrows are rotated by 30 degrees and $\sqrt{3}$ longer, and they match the periodicity of a $2 \times 2$ graphene cell.

The lattice constant of free graphene is about 2.46 Angstroms and double that is 4.92. Compared to $\sqrt{3}$ times 2.88 = 4.99 Angstroms it's a fairly good lattice match.


Concentrations at equililibrium At $25 ^\circ C$, $K=0.090$ for the reaction
  $$\ce{H2O (g) + Cl2O (g) -> 2HOCl (g)}$$
  calculate the concentrations of all species at equilibrium for the case: $\pu{1.0 g} \ce{H2O}$ and $\pu{2.0 g} \ce{Cl2O}$ are mixed in a $\pu{1.0 L}$ flask.

I don't know where I go wrong in my calculations:

We have $\pu{0.0555 M}$ $\ce{H2O}$ and $\pu{0.023 M}$ $\ce{Cl2O}$. So at equilibrium,
\begin{align}
  \ce{[H2O]} &= 0.0555 - x,\\
  \ce{[Cl2O]} &= 0.023 - x\\
  \ce{[HOCl]} &= 2x\\
\therefore K &= \frac{[\ce{HOCl}]^2}{[\ce{H2O}][\ce{Cl2O}]}
\end{align}

After some moving about I get to
$$-3.91x^2 - 7.1\times10^{-3}x + 1.15\times10^{-4} = 0$$
Where I end up getting $x = -8.74\times10^{-3}$ or $x=8.4\times10^{-3}$.

And this is where it all goes wrong, doing some quick calculations, I see that it does not match what the book says the answer is. So I'm doing something wrong, but I don't know what it is or where it's at.


Why do nitro groups confer explosive tendencies? When suggesting nitration of an aromatic compound in the synthesis of some organic molecule, it was raised that this route should be avoided as to prevent things from going "ka boom."

An explanation was not forthcoming.

So, why do nitro groups tend to make organic molecules explosive? Is it because the $\ce{NO2}$ group really "wants" to be $\ce{N2}$ since a) diatomic nitrogen has a super high bond strength and b) diatomic nitrogen is a gas usually and therefore conversion to diatomic nitrogen would be entropically favorable? Is it because carbon's most oxidized state is also a gas and the $\ce{C=O}$ bond isn't a wimp either - hence the reason many organic compounds are flammable?

Also, what happens first in an explosion? Does carbon become oxidized first, which then provides the activation energy necessary to decompose the nitro group? I ask this because organic carbon compounds by themselves which are flammable - i.e. toluene - aren't considered explosive, but trinitrotoluene is an explosive.

Explosives chemistry is a rather complex topic. I've heard that this book is a good source of information about it (I haven't read it).

In a nutshell, your intuition about the nitro group is accurate. Formation of $\ce{N2}$ is highly energetically favourable. To get an explosive, what we need is a rapid reaction that produces a lot of heat and gas to both cause the obvious effects of an explosion and to propagate the reaction to other molecules of explosive. The rapid part is what separates something like TNT from toluene. The combustion of toluene is also energetically favourable, but in TNT, to create $\ce{N2}$ is not dependant on mass transport of any other species and can thus happen very rapidly, whereas the combustion of toluene is limited by how quickly oxygen is transported to it. If one were to vaporize toluene in the correct concentration in air, this transport problem goes away and an explosion can occur. TNT also has an advantage in this regard because the nitro groups provide a source of oxygen to react with the carbon and nitrogen remaining (not enough for all of it, but it helps). Explosives are often mixed with fuels or oxidizing agents to produce a more oxygen-balanced mixture for a more efficient explosion.

As for what happens first in an explosion, many different reactions can occur during an explosion, but it may be helpful to consider what it takes to actually detonate TNT and think about the timescale of the reactions. TNT is a solid at room temperature and has a flash point of 163 °C making it difficult to even ignite, and while it will burn in a fire, there is no risk of explosion. For an explosion to occur, enough gas and heat has to be produced to propagate the reaction through the bulk of the material. In practice this is done using a much more sensitive explosive (other explosives like lead azide or nitroglycerine are unstable enough to be set off by heat or pressure) to produce a small shockwave that provides the activation energy to initiate a reaction as it travels through the explosive which then sustains the shockwave through the rest of the material. In a normal explosive (not a fuel-air explosive or the like), the reactions that contribute to the bulk of the explosion are limited to what the explosive is made of because the speed of the explosion is too fast for air to play much of a role initially.

In the case of TNT, the experimentally-measured time it takes for a shockwave to pass through is 100–200 fs (no idea how one measures that), so any oxidation of the carbons seem unlikely to contribute much to the the initial explosion, given the only readily available source of oxygen is from the nitro groups which must presumably decompose first. This group proposed a few decomposition pathways for TNT, including homolytic cleavage of the $\ce{C-NO2}$ bond, rearrangement from $\ce{C-NO2}$ to $\ce{C-ONO}$ followed by homolytic $\ce{O-NO}$ cleavage, and $\ce{C-H}$ attack from an adjacent nitro group to the methyl ring substituent, but they found that only the first was fast enough to occur during detonation, the others possible only for lower temperature thermal decomposition. This initial decomposition step is the only thing fast enough to contribute to the shockwave that sets off the rest of the TNT, while the reactions that produce the final products occur (relatively) long after the initial blast has initiated the rest of the explosive.


Insoluble Group I/ammonium salts Most solubility references (e.g. this one) say that there are "few" insoluble salts of alkali metals.

Are there any exceptions?

Lithium carbonate has poor water solubility.  Cesium triphenylcyanoborate is insoluble in water.


Pi bond need sigma bond to exist My teacher was teaching about chemical bonding and she said that pi bond needs sigma bond to exist, but she didn't know why. It is a doubt which I get in my mind, can someone explain it?

I think it would be more correct to say that pi bonds tend to exist in conjunction with sigma bonds, rather than that they need sigma bonds to exist - as pointed out in the comments.

This can be explained rather easily by basic molecular orbital (MO) theory, which seems like an appropriate framework for the question. Let's consider the simple case of a homoatomic bond such as C=C. When you have two C atoms coming into contact so as to form MOs, you form sigma MOs (bonding and antibonding) from the set of atomic S orbitals, and pi MOs (again, bonding and antibonding) from the Px and Py atomic orbitals.

Among the (molecular) orbitals, the sigma bonding MO generally has the lowest energy, followed by the pi bonding MOs. This means that the sigma bonding orbitals will 'fill' before the pi bonding orbitals. Because electrons preferentially occupy MOs from lower to higher energy, as long as the sigma bonding MOs are lower energy than the pi bonding MOs you will see sigma bonding orbitals populated before pi bonding MOs. Thus, to have a pi bond (occupied pi bonding MOs), you generally also have a sigma bond (occupied sigma bonding MO).

As pointed out in the comments to your question, this is somewhat of a generalization - but I think it's an accurate description of the phenomenon your instructor described.


Which cyclohexane conformation is more stable? I got this question wrong on a test and I don't know why:

  Question: Which conformation is more stable?

  Answer: the chair on the left is more stable.

I thought that the ring conformation has zero ring strain so I'm confused.

By symmetry, the cyclohexane on the right requires each carbon atom to have a $\ce{C-C-C}$ angle of 120 degrees. From VSEPR theory, this is not optimal for tetrahedral carbon. A regular tetrahedron has angles between vertices of approximately 109.5 degrees and the equilibrium bond angles of a secondary carbon won't be too discrepant from this.


Does the enthalpy of formation include translational and rotational energy? I can look up the enthalpy of formation of water vapor and get -241,826 KJ/mol. I want to use this to compute the total energy of water vapor at STP from the energies of O2 and H2 gas. What I don't know is whether I should use the electronic energies of H2 gas and O2 gas or the total energies of H2 gas and O2 gas, which include the contributions from the translational and rotational partition functions.
Basically, which of these two is the correct usage of Enthalpy of Formation?

ETotal(H2O) = Eelectronic(O2) + Eelectronic(H2) -241,826 KJ/mol + ETranslational(H2O) + ERotational(H2O)
ETotal(H2O) = ETotal(O2) + ETotal(H2) -241,826 KJ/mol

To add to @Kexanone's answer above, the enthalpy of formation refers to how much heat is evolved or adsorbed from the environment in a reaction. So it's the total energy, and in an ideal gas it has the form: dft + translational + rotational + vibrational + PdV, where the PdV refers to the energy needed to make space for the gas (grand canonical ensemble; in the canonical ensemble (fixed volume) this does not show up). At room temperature the vibrational and PdV terms are typically small compared to DFT energy errors (order 1 millihartree), with the PdV term only coming into play at pressures within an order of magnitude or two of those found in the detonation of thermonuclear bombs.


Confusion in barometer experiment Now at A, there is pressure by mercury column which is nothing but it's weight. Then, at C there is pressure which is atmospheric pressure. Also, the density at A is more than at C. What they say according to the law is that the pressure at C and A are same. I am not able to understand why the pressures are same actually because of the difference I listed in my text. Also, the pressure at C is inside the barometer and at A is outside. So, how can they ever be same and if they are, what is the condition for it.
Pressure at A = Pressure at B +pgh. Now how is this possible? Pressure at A should be more than pressure at B since the height for pressure at is more. So, why is that? Acceding to me Z it should be Pressure at A + pgh= pressure at B.
Then, they also say that pgh is atmospheric pressure. Where they used denstiy of mercury. Now, This should be mercury pressure and not atmospheric right. Density for atmosphere is very low right

Said by other words, you do not believe that $3 + 2 = 3 + 2$

There are 2 $\ce{Hg}$ columns, sharing the same pressure at the bottom.
Both have the same height, so the pressure difference is the same.
Yet you have doubts the pressures at their tops are the same.

Pressure at A = Pressure at B +pgh

It should be $p_\mathrm{A} = p_\mathrm{B} + \rho gh$ where $\rho$ is the density of mercury.

The vertical gradient of the hydrostatic pressure in liquid of the density $\rho$ at gravitational accelerationg $g$ is:
$$\mathrm{d}p = - g \cdot \rho \cdot \mathrm{d}h$$
As we can consider $\rho$ and $g$ as constants, the difference of the pressures is then :
$$p_C - p_A = - g \cdot \rho \cdot \int_{h_A}^{h_C} {\mathrm{d}h}$$
$$h_C=h_A \implies p_C=p_A$$

If pressure at the same altitude within the connected liquid of constant density was different, there would be the liquid flow until it is equal.
The pressures at the same altitude would be slightly different, if e.g. mercury at both columns had different temperature and density.

Feedback to comments:
In fact, it does not matter if it is mercury or water, the important thing is the density is constant across the liquid. What other than atmospheric pressure would you expect at C ( regardless of the liquid below ) ? Your scenario description is not exactly clear. Now at A, there is pressure by mercury column which is nothing but it's weight.
If the density along the thought path from A down and then up to C is constant, then pressure at A and C are the same, if the liquid is in the rest. Any pressure difference would cause motion of the liquid.

Let assume the liquid is mercury with density $\rho$.
Let assume the atmospheric pressure at C is $p_0$.
Let assign to C the relative height $h=0$.
Then pressure anywhere below C ( with negative $h$ ) is
$$ p = p_0 - \rho \cdot g \cdot h \gt p_0$$
The same is anywhere below A. At A is the same pressure as at C, i.e. $p_0$.
Between A and B, at height $h$, is the pressure:
$$ p = p_0 - \rho \cdot g \cdot h \lt p_0,$$
but this time $h$ is positive.
B is then at the height
$$ h_\mathrm{B}= \frac {p_0}{\rho \cdot g}$$
with $p=0$
( if we neglect very low vapour pressure of mercury.)


Ksp calculation of Cr(OH)2 At $25~\mathrm{^\circ C}$, $10.24~\mathrm{mg}$ of $\ce{Cr(OH)2}$ are dissolved in enough water to make $125~\mathrm{mL}$ of solution. When equilibrium is established, the solution has a $\mathrm{pH}$ of $8.49$. Estimate $K_\text{sp}$ for $\ce{Cr(OH)2}$ (Ans : $1.47\times 10^{-17}$)

I have calculated it but not have the same answer as this.
I tried it this way:

$\mathrm{pH}$ is $8.49$, then $\mathrm{pOH}$ is $5.51$. I use this formula: $\mathrm{pOH} = -\log[\ce{OH}]$
The $[\ce{OH}]$ is $10^{-5.51}$
Find molarity $\ce{Cr}$: $(10.24\cdot(1/1000)\cdot(1/86))/(0.125)=9.525\times 10^{-4}$
Substitute into $K_\text{sp} = [\ce{Cr}][\ce{OH}]^2$

But after I substitute value of $[\ce{Cr}] = 9.525\times 10^{-4}$ and $[\ce{OH}]$ of $(2\times 10^{-5.51})^2$, I get answer of $3.6\times 10^{-14}$.

Where have I gone wrong? Any Suggestions?

Your calculation of $[\ce{OH-}] = 10^{-5.51}\ \mathrm{mol\ l^{-1}}$ is correct.

The pH is well above 7. Therefore, the present $[\ce{OH-}]$ is mainly a result of the dissociation of $\ce{Cr(OH)2}$:

$$\ce{Cr(OH)2 <=> Cr^2+ + 2 OH- }$$

Thus, you know
$[\ce{Cr^2+}] \approx \tfrac{1}{2} [\ce{OH-}] = \tfrac{1}{2}\cdot10^{-5.51}\ \mathrm{mol\ l^{-1}} $.

$K_\text{sp} = [\ce{Cr^2+}][\ce{OH-}]^2 = \tfrac{1}{2}\cdot10^{-5.51} \cdot \left(10^{-5.51}\right)^2 = 1.48\cdot 10^{-17}$


Can a single monomer be a macromolecule? If so, when? In our science class, we're learning about the four fundamental macromolecules in every living organism: carbohydrates, proteins, lipids, and nucleic acids. Our class definition of a macromolecule essential is:

  A macromolecule is a large molecule that is essential to all life, and present in all living cells. It is a polymer- a chain of monomers.

Our definition of a monomer is like so:

  A monomer is the simplest building block of a macromolecule with the properties of that macromolecule. They can be strung together to produce a macromolecule (usually by dehydration synthesis).

I would have no problem with these definitions if not for my teacher mentioning once that some monomers can also be macromolecules by themselves. Because some monomers of certain macromolecules- such as the monosaccharide glucose vs. the disaccharide sucrose or the polysaccharide amylose - can act on their own as an essential and functional carbohydrate, they are macromolecules by themselves.

Is this true? For example, could glucose be a macromolecule by itself?

Thanks.

I can't think of an example where a biological monomer would be a macromolecule.

Definitions of macromolecule vary, usually by molecular weight or number of monomers (repeat units).

Personally, I'd go with ~1000 Dalton for a minimum, but the original definition of 1000 atoms is a good start too.

In any case, no biological monomer, including glucose will function the same as a macromolecule.

Consider starch - a macromolecule of sugars. It doesn't dissolve as quickly as simple sugar and has different physical properties.

Update

To clarify my comments.. Macromolecules or polymers are made up of monomers the way words are made up of letters. So no, a glucose molecule isn't really the same as a macromolecule, just like "R" is not a word.

Yes, macromolecules can be used to make larger assemblies like microtubules, filaments, etc., much the same way that words can form sentences and paragraphs.

In the polymer literature there's even the concept of a "macromonomer" referring to a monomer that is already large in size.

In my opinion though, the basic constituents (monomers) are still amino acids, nucleic acids, sugars, etc. These are not macromolecules.


Entropy of Activation and Temperature Dependence I am trying to understand the extrapolation of enthalpy $\Delta H^{\ddagger}$ and entropy of activation $\Delta S^{\ddagger}$ from the Eyring equation. It's typically cast as:

$$\ln\left(\frac{k}{T}\right) = \frac{-\Delta H^{\ddagger}}{RT} + \frac{\Delta S^{\ddagger}}{R} + \ln\left(\frac{k_\mathrm{B}}{h}\right)$$

where $k$ is the rate, $T$ is the temperature, $R$ is the ideal gas constant, $k_\mathrm{B}$ is the Bolztmann constant, and $h$ is the Planck constant. However, the entropy itself also has a temperature dependence in principle (i.e. harmonic oscillator entropy is a function of temperature). Is there a way to determine the temperature dependence of the entropy of activation from the Eyring equation for a real experiment?

The equation
$$\ln\left(\frac{k}{T}\right) = \frac{-\Delta H^{\ddagger}}{RT} + \frac{\Delta S^{\ddagger}}{R} + \ln\left(\frac{k_\mathrm{B}}{h}\right)$$
does not assume that $\Delta S^{\ddagger}$ is temperature independent. To evaluate its T dependence you might proceed as follows:
$$\left(\frac{\partial \Delta G^{\ddagger} }{\partial T}\right)_p = -\Delta S^{\ddagger}$$
$$\rightarrow\left(\frac{\partial [RT\ln K^{\ddagger}] }{\partial T}\right)_p =  \Delta S^{\ddagger}$$
$$\rightarrow \Delta S^{\ddagger}=\left(\frac{ \partial [RT \ln (hk/k_\mathrm{B}T)] }{\partial T}\right)_p $$
It follows that you can evaluate $\Delta S^{\ddagger}(T)$, $\Delta S^{\ddagger}$  at different temperatures, from the change of $[RT \ln (hk/k_\mathrm{B}T)]$ with respect to temperature.


What is an example of a biodegradable polyethylene? I would like an example of a Biodegradable Polyethylene.
I am using polylactic acid (PLA) as an example of a biopolymer/ bioplastic but I can not find a biodegradable polyethylene, that is a substance that contains polyethylene but has additives added so it biodegrades quicker

In a short research I found little to no evidence, that a there is  polypropylene/polyethylene, that can actually be called biodegradeable. I have found a couple of articles that have this as a topic. I don't have the time to actually read all of them and since this is going to be to long for a comment, I'll share it as an answer.

S. Bonhomme, A. Cuer, A-M. Delort, J. Lemaire, M. Sancelme, G. Scott, Polymer Degradation and Stability 2003, 81, 441–452.
This is a general overview of the mechanisms of biodegradation of such materials.
Ignacy Jakubowicz, Polymer Degradation and Stability 2003, 80, 39–43.
This article investigates the degradation of polymers, which contain pro-oxidant additives. It mainly focusses on the kinetics of the thermo-oxidative degradation. I am certain you can learn from this article which kind of additives are used to make braking long polymers into oligomers easier and hence complete degradation faster.
Prasun K. Roy, Minna Hakkarainen, Indra K. Varma, Ann-Christine Albertsson, Environ. Sci. Technol. 2011, 45 (10), 4217–4227.
This article is very promising as it also deals with some long-term effects of polymers with pro-oxidative additives. The abstract already lays out the main problem, when talking about biodegradation: entering the eco-cycle harmlessly.

  However, the “degradable polyethylene” which is presently being promoted as an environmentally friendly alternative to the nondegradable counterpart, does not seem to meet this criterion. [...] It appears from the existing literature that our search for biodegradable polyethylene has not yet been realized.
  (Emphasis by me.)

All in all I think it is very important to carefully outline, what biodegradation is or should be. I hope these few articles can be a starting point for your further research.


Does the chemical in an alkaline battery make battery leaks unavoidable? Battery leaks was an issue in the 80s and 90s, and since quality and innovation is constantly improving, I thought battery leak might be slowly going away.  But I found that even nowadays, name brand alkaline batteries still leak after a while, even if originally concealed in a tight package of the metal shell.  Is it true that the chemical in an alkaline battery inherently expands in volume over time, and make the alkaline battery leak issue unavoidable?

The reason for leaks in alkaline batteries is that as batteries discharge — either through usage or gradual self-discharge — the chemistry of the cells changes and some hydrogen gas is generated. This out-gassing increases pressure in the battery. Eventually, the excess pressure either ruptures the insulating seals at the end of the battery, or the outer metal canister, or both. In addition, as the battery ages, its steel outer canister may gradually corrode (in alkaline batteries there is some potassium hydroxide, a very strong base) or rust, which can further contribute to containment failure.
(see http://en.wikipedia.org/wiki/Alkaline_battery#Leaks )

I also found this online :

Heat is the biggest culprit. Storing your flashlights under the seat of your car, or in direct sunlight will cause the batteries to leak. The reason they leak is because they are filled with a chemical paste of which the liquid portion is composed of a base called Potassium Hydroxide. When the batteries get hot, the paste inside expands, and forces chemicals out of the seals in the bottom and top of the battery.
[...]
Regardless of the brand you buy,[...] all batteries will leak eventually, so if you don't use the flashlight for extended periods of 3 months or more, remove the batteries, or change them out every couple months.
(from here)


How is the process of creating PVC fabrics harmful for the environment? I often hear that the production process for PVC fabrics (such as those used in Gucci and Versace handbags) is very harmful to the environment. What part of the process is harmful and can this be prevented?

I wouldn't think that the process of producing PVC fabrics is that much of a problem, for reasons:

PVC is produced by experts that have taken all the necessary precautions for all possible reductions of safety risks it may cause in the lab or in mass scale production.
The process of synthesizing PVC is not a secret. If it was too harmful (rather than its profitability), then companies would have been banned from producing such a thing.

Phew, I thought there was something serious about PVC fabrics.

There is! The problem is not the process, but VCM and phthalates.
Issue 1: VCM
With a chemical formula of $\ce{C2H3Cl}$, a boiling point of $-13.9°C$ and an approximate molecular weight of $62.5 \frac{g}{mol}$, VCM is the raw material for PVC. According to here, the death of four workers had been approved due to VCM in 1974. (Though wikipedia says it was three workers dead) It's reported that in 1975, an annual 5-billion-pounds rate of production of VCM was confirmed.
The industry had the worst ways of producing VCM, and has suffered for decades until better ways of producing it have come around. According to here, a traditional way of producing VCM was the reaction of acetylene with Hydrochloric acid, as following: $$\ce{C2H2} + \ce{HCl} ~->~ \ce{C2H3Cl}$$
It was in the presence of Mercury(II) Chloride as a catalyst! Nothing has been done to reverse the harmful effects done to environment because of this very reaction.
In the modern process, "polymerization" is known to be the most dangerous stage and the peak of the overexposure.
VCM is believed to be carcinogen. Severe symptoms have been reported due to the overexposure of VCM; such as: Hepatotoxicity (liver damage and tumor growth caused by chemicals and more precisely, angiosarcoma).
Issue 2: phthalates
VCM is the fundamental raw material for the synthesis of all kinds of PVC; rigid or soft. But Plasticizers are being added to rigid PVC to "soften it up"; so it would be applicable for the future use as "PVC fabrics". Phthalates are common plasticizers in the production of soft PVCs.

They are divided into two distinct groups, with very different applications, toxicological properties, and classification, based on the number of carbon atoms in their alcohol chain.

Phthalates aren't covalently bonded to the matrix of PVC. Therefore, the lighter ones can leach relatively very easy, usually due to heat or strong solvents.
The most common phthalate is diethylhexyl phthalate(DEHP). Studies show that Americans are exposed to DEHP. DEHP is believed to cause health complications in a variety of ages.
Oh, wait: Issue 3: PVC itself! But when it degrades
PVC is known for its strong linear polymer; but when it degrades (usually due to careless use or disposal) it degrades! Degradation is excessive reduction in a plastic's average molar weight. PVC will "microcrack" when it's degraded, into macroparticles that are a big threat to the environment. These particles "soak up" POPs and afterwards are usually ingested by living things, making it a big deal of our everyday plastic pollution exposure. (as wikipedia puts it)

What should I do now?! Panic?

Considering the advancement of technology and our anxiety to preserve our health we humans aren't sitting around to watch ourselves getting sick! I don't mean the hard-to-come-by sites that say overexposure to PVC isn't carcinogenic, but real efforts into solving the problem. News has been spread that REACH attempts to authorize DEHP and two other phthalates and ban their use in specific materials. DINP is another plasticizer, approved by the EU risk assesment, that will eventually replace the former plasticizers, hopefully.
Scientific researches for over ten years have proven that the amount of exposure to the chemicals mentioned is far from dangerous in current conditions and uses. So sit back relaxed; it's true that these dangers exist, but they're not serious threats to a healthy person, at least not yet.


Final pressure of an ideal gas mixture, when two separate gases are put into one vessel At constant of 200 cm³ of $\ce{N2}$ at 720 mmHg and 400 cm³ of $\ce{O2}$ at 750 mmHg are put together into one litre flask. The final pressure of the mixture is:

My question is how to use ideal gas law for mixture of two gases. For one gas, it is easy but for two it becomes difficult for me. I am assuming that these gases won't react or else it would have been given in the question. Can somebody tell me how to apply ideal gas law for two gases and what happens when they are kept together?

Note that since $1\ \mathrm{cm^3} = 1\ \mathrm{mL}$, I will talk in mL.

I doubt that you are supposed to use the ideal gas law for this problem since no temperature is specified, and moles are not talked about. Instead, let's look at a few instances of Boyle's Law ($P_1V_1=P_2V_2$ -- for comparing the pressures and volumes of gasses).

The gasses are, indeed, unreactive, and, according to Kinetic-Molecular Theory, have nearly identical properties where gas laws are concerned. Let's find the partial pressure of each gas with Boyle's Law and sum them with Dalton's Law of Partial Pressures ($P_\mathrm{tot} = P_1 + P_2 +\dots$).

Note that to make the units cancel, I will use $1\ \mathrm L = 1000\ \mathrm{mL}$.

$$720\ \mathrm{mmHg} \times 200\ \mathrm{mL} = x\ \mathrm{mmHg} \times 1000\ \mathrm{mL}$$

$$\frac{720\ \mathrm{mmHg} \times 200\ \mathrm{mL}}{1000\ \mathrm{mL}} = x\ \mathrm{mmHg} = 144\ \mathrm{mmHg}$$

$$750\ \mathrm{mmHg} \times 400\ \mathrm{mL} = x\ \mathrm{mmHg} \times 1000\ \mathrm{mL}$$

$$\frac{750\ \mathrm{mmHg} \times 400\ \mathrm{mL}}{1000\ \mathrm{mL}} = x\ \mathrm{mmHg} = 300\ \mathrm{mmHg}$$

$$144\ \mathrm{mmHg} + 300\ \mathrm{mmHg} = P_\mathrm{tot} = 444\ \mathrm{mmHg}$$

444 mmHg is the final pressure in the container.


Difference between HCD and CID collision induced dissociations? With the Thermo-Fisher Fusion Orbitrap mass spectrometer I can do both CID (collision-induced dissociation) and HCD (higher energy C trap dissociation) collision-induced dissociation studies.

When I apply HCD on inorganic molecules I observe more fragmentation products than with just CID suggesting multiple fragmentation steps. However I don't get why physically this is possible.

Is HCD design to keep the fragmentation products in the trap so that they collide many times with the inert gas?

EDIT: according to this manual found on the Termo Fisher website:

  HCD tends to produce EI-like fragmentation
  and records ions from multiple steps of collision as seen
  in Figure 3.

https://assets.thermofisher.com/TFS-Assets/CMD/Application-Notes/AN-540-LC-MS-Small-Molecule-Characterization-AN63396-EN.pdf

And according to a post found in research-gate:

  In beam type CID ("HCD") multiple collisions with a collison gas take place under an elevated pressure regime. Ions can be activated multiply, which results in richer spectra compared to resonance type CID ("CID"). In resonance type activation, only the precursor ion is activated - in beam type CID also the fragments (collision gas).

https://www.researchgate.net/post/Why_sometimes_there_are_a-ions_when_you_perform_HCD_fragmentation_for_peptides_on_Q-exactive_plus

From the answer given here the reason of a richer spectra is that:

  The reason HCD gives you more fragment ions is that ions do not have time to equilibrate.

So, do I have more fragments when I apply HCD because:
1) I have multiple collisions(multiple activation)where also the fragments collides?
2) It's a non-equilibrium process, so bonds are broken independently of the strength?
3) both?

Brief overview of CID vs. HCD

"Collision-induced dissociation" or CID is a much older and more general term in mass spectrometry than HCD.  HCD is a vendor-specific term invented to describe a new modification on how ions can be dissociated in Orbitrap mass spectrometers.  CID is a "universal" term that applies to many different types of mass spectrometers.  Unfortunately, how exactly CID works depends on which type of mass spectrometer you are talking about.

Wikipedia has a good overview of how CID implementations vary according to instrument type.  Understanding that the meaning of "CID" varies strongly by instrument type is important.  The naive assumption that CID on e.g.a triple-quadrupole or qTOF instrument is like CID on an ion trap instrument is understandable -- it is the same term after all -- but very wrong.

CID on an Orbitrap

On an ion-trap instrument like your Fusion instrument, CID is done in an ion trap.  This means that on an Orbitrap, CID is a resonant excitation technique.  Thus, to reach a certain collision energy, ions must be accelerated relatively slowly from their starting energy to the final energy.  (The final energy is a user-controllable parameter.)  They thus collide many times with the inert gas during the excitation process.  Each of these collisions may "heat" or "cool" the ion.  In an ion trap, resonant excitation will only accelerate the precursor ions, so after dissociation has taken place, fragment collisions with the gas molecules will cool the ions, eliminating or at least reducing further fragmentation into smaller pieces.

A 2014 paper in the Journal of Mass Spectrometry by Ichou et al. explains CID on an ion-trap-equipped Orbitrap fairly well:

  In ion trap devices, each collision can activate or cool the stored precursor ion. The evolution of this equilibrium depends greatly on the ion kinetic energy. The second step relies on the unimolecular dissociation pathways of the precursor ion. It has been extensively modeled using the Rice–Rampserger–Kassel–Marcus (RRKM) theory18-21 and the quasi‐equilibrium theory.22 Note that the internal energy distribution of ions greatly relies upon the rate of these two steps. So, the internal energy distribution depends on the regime of collisions under study.23-25 The ‘slow’ or ‘rapid’ heating regime is defined according to the unimolecular dissociation constant rate. If the latter is slower than the constant rate of the activation/deactivation process, the regime is defined as a slow heating technique. The slow heating regime in an ion trap instrument for the conventional CID mode may be described as a random walk process,25, 26 where a large number of ion‐neutral gas collisions are needed to ‘heat’ the ions in order to dissociate them. Some studies25-29 have demonstrated that the ion internal energy distribution for ion trap instruments is close to the temperature due to the multiple collision regime.

Thus, on an ion-trap Orbitrap like your Fusion, CID can be at least close to an equilibrium process.  Whether it is or not depends on how quickly ions are ramped in energy to the final user-decided collision energy.  If ramp times are long compared to the unimolecular dissociation rates, ions have the opportunity to distribute the excitation energy across most of the bonds in the molecule.  They bump into many gas molecules, sometimes being heated by the collision, and sometimes being cooled.  These collisions are what allows the energy to distribute itself among all the bonds of the excited ion.  As a result, in slow resonant CID, the weakest bonds break first.  If ramp times are short, then CID (on an ion-trap instrument) ions do not have time to equilibrate and dissociation spectra can be more like what is observed for HCD.

HCD on an Orbitrap

The same paper explains HCD fairly well too:

  'HCD' is an excitation mode that belongs to the category of non‐resonant activation techniques like the CID conducted in tandem‐in‐space instruments.

Thus, 'HCD' in an Orbitrap is like CID on a triple-quadrupole or qTOF instrument.  It is a non-resonant dissociation technique.  Ions do not experience a long series of "equilibrating" collisions with gas molecules.  Effectively, a single bolus of energy is imparted to them very quickly. This type of dissociation is a non-equilibrium process.  Both weak bonds and stronger bonds may have a chance to break, depending on which ones are excited by the smaller number of collisions with gas molecules.  Product ions remain excited, and do not experience enough collisions to "cool" them down, meaning that if they do further collide with gas molecules, they may be further fragmented.

Answers to your questions

  Is HCD design to keep the fragmentation products in the trap so that they collide many times with the inert gas?

No, actually the opposite, as hopefully the discussion above makes clear.  On an ion-trap instrument (but not on other instruments), it is CID that collides ions many times with the inert gas.

  When I apply HCD on inorganic molecules I observe more fragmentation products than with just CID suggesting multiple fragmentation steps. However I don't get why physically this is possible.

The reason HCD gives you more fragment ions is that ions do not have time to equilibrate, so many bonds can be broken, each with some probability.  Parent ions are energized rapidly, in a near-single-step, to highly excited states.  A variety of vibrational modes can get excited, leading to a variety of bonds breaking in the dissociation step.  In (slow) CID, ions are excited slowly; as energy accumulates during the resonant excitation, ions collide with and have time to redistribute energy over the whole molecule, meaning that only the weakest bonds tend to break, resulting in fewer fragments.


Confusion in hybridisation of coordination compound I want to know the configuration and hybridisation of the compound, hexaaquavanadium (III) ion that is $\ce{[V(H2O)6]^{+3}}$.

In this Vanadium is in $+3$ oxidation state its electronic configuration is $\mathrm{[Ar]4s^0 3d^2}$.

So there are $3$ more empty $\mathrm{d}$ orbitals.

Hence the hybridisation should be $\mathrm{d^3sp^2}$.

But it is $\mathrm{d^2sp^3}$ and paramagnetic .

I could not understand how .

Can you explain me?

Related

Basically, it isn't really wise to use hybridization on coordination complexes to begin with. However, the hybridization of $d^2sp^3$ makes sense if you think of it as just describing the shape of the molecule. This should be octahedral, due to its 6 equivalent groups around the metal center, and $d^2sp^3$ is the valence bond description of the hybridization of an octahedron.

Using Ligand Field theory and MO diagrams is probably a better way to understand how the orbitals overlap and where electrons are in the final structure. This diagram is for $\ce{[Ni(H2O)6]^2+}$, but the diagram will look very similar for the $\ce{V}$ complex, just with fewer electrons. (Image source: dx.doi.org/10.1021/jp4100813 | J. Phys. Chem. B 2013, 117, 16512−16521)


Why does the cathode ray tube only start glowing at low pressures? I am considering the cathode ray tube of the set up, when the cathode rays were discovered:

Voltage = very high
Gas = any
Atmospheric Pressure = changing

At different pressures, there are different observations:

1 atm: the tube does not glow.
0.1 atm: the whole tube lightens up
0.01 atm: the tube glows at the anode end (Glowing is shown as there is a detector coating of ZnS.)

Note: I don't have this set-up, this is just taken by me from theory.

Why does, when I am just reducing the pressure from 1 atm to 0.01 atm, the tube at first not glow, then it glows completely, and only at low pressure it glows just at one end?

At 1 atm. it is difficult create stable plasma, because breakdown voltage is very high.
Your electrodes are located at least 10-15 cm I think, and it is simply you do not have enough electrical potential between them to create initial plasma channel.

Then pressure goes down, voltage is fixed. When you reach breakdown voltage you get your glow (the Townsend discharge happens just as glow discharge becomes visible)-normal glow discharge. What happens? You ionized your gas, you have free electrons and ions, which are floating in different directions: ions are floating to negatively charged cathode and electrons flow to positively charged anode.

Because charge carriers are moving in space. They are rubbing against each other, part of the energy is dissipated and goes into heat and other is emitted and you see your glow. Ions are moving relatively slow, electrons, on contrary,  move very fast comparing to ions, because they are light weighted. So this fast free electrons, accelerated by a sufficiently strong electric field, give rise to electrical conduction through a gas by avalanche multiplication caused by the ionization of molecules by ion impact.

Then you continue to reduce your pressure more and more, less and less gas molecules can be ionized by free electron flow, less and les collisions are happening (less energy is dissipated), so the intensity of glow is going down, and finally it stops, because the breakdown voltage increased and overcome your applied voltage. (see Paschen Curve, lower is a pressure higher is the breakdown voltage.)

This is generally how it works.


Why does 2-chloropropane's mass spectrum have a peak at 27? I have noticed that all halogenoalkanes of the form 2-Xpropane seem to have a mass spec peak at 27, implying that a $\ce{C2H3+}$ ion is being formed.

Forming a $\ce{C2H3+}$ ion, however, would mean breaking three bonds and, given its abundance, that wouldn't make sense - especially in the case of 2-chloropropane where it has an abundance of 25.36 according to this
 website.

Why exactly does the $\ce{C2H3+}$ ion form?

EDIT: You can also see this peak in the mass spectrum of 2-bromopropane, 2-iodopropane and 2-fluoropropane, but it is less pronounced.

EDIT 2: as pointed out by @PLD, it appears that other molecules of this form have peaks at 27 for a different reason to that of 2-chloropropane, so my previous edit is irrelevant.

Here is a little mechanism I made up about it. Not quite sure if its correct, seems to fit the mass spectrum though.

Definitely have a look at McLafferty classical textbook. Maybe you will find the answer there.


Why does magnesium deposit during electrolysis of a molten mixture of MgI2 and AlBr3? During the electrolysis of mixed salts, I read, and was taught, that the cation with a higher ionization energy would deposit at the cathode. I was also told that you cannot use the standard electrode potential to determine which cation would be reduced.

Is this ionization energy referring to the first ionization energy? If it is, why doesn't the second and third ionization energy factor into this problem seeing that magnesium would have a 2+ charge and aluminum would have a 3+ charge?

Yes, it is true that you cannot use the standard electrode potential to determine which cation would be reduced and vise versa during an electrolysis of mixed molten salts, in your case, $\ce{AlBr3}$ and $\ce{MgI2}$. That's because most tabulated values of standard electrode potentials are determined for aqueous solutions. Therefore, they can be used only as approximate guidance. Instead, electronegativity values (EN) can be used to estimate the stronger oxidizing and reducing agents. For example, EN values of $\ce{Br}$ and $\ce{I}$ are 2.8 and 2.5, respectively. Therefore, we can conclude that $\ce{I-}$ is the stronger reducing agent between two elements (because $\ce{I}$ is less EN than $\ce{Br}$). Similarly, between $\ce{Al}$ and $\ce{Mg}$, $\ce{Al^3+}$ is the stronger oxidizing agent because $\ce{Al}$ is more EN than $\ce{Mg}$ (1.5 vs 1,2, respectively), so $\ce{Al^3+}$ gains electrons easier.

Therefore, possible cathode half-reaction (reduction): $\ce{Al^3+(l) + 3e- -> Al(l)}$
And Possible anode half-reactions (oxidation): $\ce{2I-(l) -> I2(l) + 2e-}$


Is there a least reactive chemical species? As a total chemistry layman I enjoyed reading "Why doesn't $\ce{H2O}$ burn?", but it prompted another question in my mind. One of the answers there was that $\ce{H2O}$ can burn in the presence of a stronger oxidizer like fluorine, so burning stable compounds is just a question of using a stronger oxidizers.

Or is it? Is there a chemical species, or a "least reactive compound" that, once made, is difficult or impossible to chemically transform into something else? A chemical black hole, if you will.

Does the above answer change if we allow high temperatures and pressures versus restricting ourselves to roughly standard temperature and pressure?

I think a good argument can be made for either helium or neon, the most noble of the noble gasses. Those are the two prototypical unreactive elements. They are the only two stable elements for which no more complex compounds (i.e., other than the single atoms themselves) have yet been isolated, at any temperature. The slightly more reactive element argon will admit formation of compounds such as argon hydrofluoride ($\ce{HArF}$). This compound is only stable up to $\mathrm{17\ K}$, because any hotter and the frail bonds are overcome by random thermal collisions which break the compound apart into $\ce{Ar}$ and $\ce{HF}$.

Picking which of helium or neon is less reactive is a bit more difficult. A naive analysis of periodic trends would point to helium as the most inert, but more detailed computational studies suggest that at least in some cases neon may be less reactive. For example, the extremely Lewis acidic compound beryllium monoxide ($\ce{BeO}$) may potentially form an isolable, if very weakly bound, compound with helium, $\ce{HeBeO}$, but neon is not thought to form the analogous $\ce{NeBeO}$. $\ce{HHeF}$ may also be just barely stable, whereas $\ce{HNeF}$ is not thought to form. None of these have yet been observed in the laboratory, but there is definitely ongoing research into coaxing helium and neon to make isolable compounds.

All that said, we can get helium and neon to react, if we drop the requirement that the product must be isolated (that is, "put into a bottle"). Chemical species such as $\ce{He2^+}$, $\ce{Ne2^+}$ and $\ce{HeNe^+}$ have long been known from mass spectrometry experiments, they just can't be isolated because that would require the presence of a counterbalancing negative ion, which would immediately proceed to react with the positive ion and cause decomposition with release of the noble gas. You can also bring what is arguably the most reactive species in chemistry, the hydrogen cation ($\ce{H^+}$) into the fray. Helium and neon will both easily react with $\ce{H^+}$ to form $\ce{HeH+}$ and $\ce{NeH+}$ , as shown by the exothermic proton affinities of $\ce{He}$ and $\ce{Ne}$. Again, these composite particles cannot be isolated, as they are amongst the strongest Brønsted-Lowry acids in existence, and will protonate anything they come into contact with in order to release the neutral noble gas atom. Some other possible non-isolable relevant noble gas ions are $\ce{FHeO^{-}}$, $\ce{HeCCH+}$ and $\ce{PbHe15^{2+}}$ (!), among others.

Edit: I forgot to mention that there are a few other cases of helium or neon binding to other atoms. By exciting one of the electrons in the atom, it is possible to coax it to bond with another one, forming an excimer or exciplex. See for example, the dihelium excimer $\mathrm{He_2^{*}}$. This is a short-lived species that still can't be isolated, however, because in a matter of microseconds it releases a photon and de-excites, promptly separating into the two free atoms.


Condensation of Ethylene Glycol I'm looking to reflux a 250 mL of ethylene glycol solution. Would an Allihn condenser be suitable; if so what length?

Ethylene glycol was a boiling point of 193 °C using any water cooled condenser will crack. Use a air cooled condenser. It is just a glass rod. The length of tube depends on how much you want to condense and how quickly you want it to come over. Remember visible condensation should always take place in the condenser. If it goes further down the tube then some hot vapours will be lost to atmosphere if your system is open. If under vacuume the ethylene glycol will boil at a lower temperature but you should use a cold trap at minus 40 °C or below to catch vapours which come over. Remember liquid ethylene glycol is toxic and the vapours are more dangerous check the MSDS. I suggest that you should work with a closed system in a fume cupboard for this job or you might never see your grandchildren. Good luck.


Why do noble gases like Xe and Kr form compounds, despite offending the octet rule? According to the octet rule noble gases should not form compounds but xenon and krypton form compounds like $\ce{XeF2}$, $\ce{XeF4}$, $\ce{XeOF2}$ and $\ce{KrF2}$ respectively. Why?

When we say that noble gases are stable it means that they are more stable than the rest. Yet it is to be considered that even these have ionisation potential ;though high. if we make a noble gases react with such ions that help in their bonding then we can have a compound of noble gas. Search this on Google.

  The first published report, in June 1962, of a noble gas compound was by Neil Bartlett, who noticed that the highly oxidising compound platinum hexafluoride ionised O2 to O+
  2. As the ionisation energy of O2 to O+
  2 (1165 kJ mol−1) is nearly equal to the ionisation energy of Xe to Xe+ (1170 kJ mol−1), he tried the reaction of Xe with PtF6. This yielded a crystalline product, xenon hexafluoroplatinate, whose formula was proposed to be Xe+
  [PtF
  6]−
  . It was later shown that the compound is actually more complex, containing both XeFPtF5 and XeFPt2F11. - Wikipedia, noble gas compounds

This was the first real compound of any noble gas.


How there are 7 enolizable hydrogens in 2‐methylcyclohex‐2‐en‐1‐one? 49. How many enolizable hydrogens are there in the following compound?

(a) 2
(b) 4
(c) 5
(d) 7

I know that enolate ion is the anion formed when an α-hydrogen in the molecule of an aldehyde or ketone is removed as a hydrogen ion.
In the above question I find only two α-hydrogens which can participate (denoted with red color):

But the answer is (d) 7.
Where am I getting wrong?

Actually there are exactly 7 alpha hydrogen in the given structure.  Just look that there is also a double bond and it is also considered as functional group.  So, you should count alpha hydrogen w.r.t  double bond and they are 5 .

Finally 2 + 5 you get the required answer = 7


Find increase in OH- ion concentration after base is added The pH of a solution is 7.00. To this solution sufficient base is added to increase the pH to 12.0. the increase $\ce{OH-}$ ion concentration is:

  (A) $5$ times
  (B) $1000$ times
  (C) $10^5$ times
  (D) $4$ times

The answer is (C) $10^5$ times.

Now, the temperature hasn't been mentioned ergo we can't assume that it's a neutral soultion, right? So, I tried doing it this way:

\begin{array}{llll}
&\mathrm{pH} &= 7 &\qquad &\mathrm{pH} &= 12 \\
&[\ce{H+}] &= 10^{-7} &\qquad &[\ce{H+}] &= 10^{-12} \\
& & &[\ce{OH-}]~\text{added} \\
& & &= 10^{-7} - 10^{-12} \\
& & &= 10^{-7} (1 - 10^{-5}) \\
& & &\approx 10^{-7}
\end{array}

$[\ce{OH-}]$ added is the increase in its concentration, or so I think. I don't know what I'm doing wrong. Please, help.

Also, can you please point out the loopholes I might have in my concept if it's obvious from my doubt, so I can look up and take care of it?

The problem is that you are trying to subtract the values, whereas all you need to do is find an increase in ratio $\frac{[\ce{OH-}]_2}{[\ce{OH-}]_1}$ knowing that $\mathrm{pOH} = - \log{[\ce{OH-}]}$ and $\mathrm{pOH} = \mathrm{p}K_\mathrm{w} - \mathrm{pH}$:

$$\frac{[\ce{OH-}]_2}{[\ce{OH-}]_1} = \frac{10^{-\mathrm{pOH_2}}}{10^{-\mathrm{pOH_1}}} = 10^{\mathrm{pOH_1} - \mathrm{pOH_2}} = 10^{\mathrm{p}K_\mathrm{w} - \mathrm{pH_1} - \mathrm{p}K_\mathrm{w} + \mathrm{pH_2}} = 10^{\mathrm{pH_2} - \mathrm{pH_1}} = 10^{12 - 7} = 10^5$$


Predict NMR spectra from the Mass Spectrometry Imagine, you have a tissue and the Mass Spectrometry for it.
How would you get the corresponding NMR spectra knowing only the Mass Spec?
Is there is a special formula or a well-known technique?

I have an idea to build an ML model based on http://hmdb.ca metabolites data where you have those pairs (MS and NMR). But I worry it is completelly wrong.

Do you think is it a stupid idea?
If there is an easy way of converting MS -> NMR which everybody knows?

I'm sorry for a probably silly question, I came from mathematics.

Sorry, but in my opinion this endeavour is unlikely to be successful.  The techniques are quite different by their underlying principles. This is, at least in part, why chemists working in (especially, but not limited to) organic sythesis use these and additional ones to get a sufficiently "rounded picture" about the substances analyzed, putting the information each technique provides together.

mass spectroscopy typically relies on fragmenting the molecules in question, and then separates these then charged fragments by their mass/charge ratio.  There are techniques keeping the molecules almost intact (softer ionization techniques, used for example in MS-mapping of tissues), and others (electon impact ionization) where the molecules are intentionally shattered into many differently sized fragments.  Depending on the ionization technique applied, the eventually intensity of these individual signals will vary. And if the molecules of interest are embedded in a matrix (e.g., a serum, a tissue), identification of this molecules by MS becomes even more complex.
NMR spectroscopy typically characterizes probes soluted, keeping the molecules of interest (smaller ones, or large proteins) basically intact, without fragmentation.  You basically probe your sample for specific atoms of a kind (isotopes) and their nearest chemical environment. Yet the NMR spectrum of a given sample -- say of an amino acid (building block for proteins) -- already looks differently for scrutinizing the either peripheral hydrogens, the backbone of carbon atoms, or nitrogens.

Addendum

I think a few pictures may provide a glimpse about the complexity here.  For a fairly simple molecule like acetophenone, given the molecular structure is known, a sketcher like Chemdoodle may compute within a blink of an eye an estimate how the NMR spectrum either about the hydrogens, or about the carbon atoms will look like.  This is possible because  the underlying physical principles are known.  In addition, there are very many similar compounds analyzed successfully by NMR, allowing (fine tuning) of details in the spectra simulated, e.g. where to expect the signals to appear.  Regarding MS, a conservative approach is to predict only the peak(s) that by theory should appear without change of the molecule in terms of its mass.  In other words, without fragmentation.  Hence Chemdoodle's estimation is this:

(accessed demo page)

There are equally projects to simulate MS spectra, like CFM-ID.  Under the menu Utilities -> Spectra Simulation, you have the possibilty to try out different modi of ionization (shortcut).  Acetophenone's structure, expressed in a machine-readable Smiles code is

C1=CC=C(C=C1)C(=O)C

Then, just choosing ESI mode about positive ions, without considering adducts, the three MS spectra simulated by CFM-ID of actephenone, in increasing hardness of ionization are these:

As you see, not only the presence / absence of the signals changes in function of the probability to record certain fragments of a specific mass/charge ratio (m/z), their relative intensity equally is affected by changing the parameters of ionization.

(Hovering the mouse over the streaks allows you to display the suggested fragments responsible for the signals in these simulations.  Because of the special conditions in a mass spectrometer, unstable specis you wouldn't be able to store in a shelf may be seen, too.)


Does chlorine trifluoride react with candlewax? I've run across references to chlorine trifluoride quite a lot. It pops up fairly regularly on the Worldbuilding SE, and whatever idea people have with it is usually shot down very quickly - For obvious reasons.

One of the things that comes up in almost any web search about chlorine trifluoride is that it doesn't react with "candle wax" pretty much at all. (Example here)

Is this true? If it is, why? What is so special about candle wax that makes it immune to this substance that reacts extremely well with almost anything on, in, or around Earth?

(Worth noting: I'm not a chemist, so a basic overview would be appreciated in addition to any extensive scientific answer. Said extensive scientific answer is also appreciated because I'm sure there's others that would like the answer to this as well.)

I don't think paraffin would last long, since PTFE doesn't seem to last in the presence of CTF.  From IGNITION! (John D. Clark), which you really, really need to read if you are interested in chlorine trifluoride or rocket propellents in general:

  The results were excellent, but the difficulties were infuriating.
  Ignition was beautiful —so smooth that it was like turning on a hose.
  Performance was high —very close to theoretical. And the reaction was
  so fast that you could burn it in a surprisingly small chamber. But.
  If your hardware was dirty, and there was a smear of oil or grease
  somewhere inside a feed line, said feed line would ignite and cleverly
  reduce itself to ashes. Gaskets and O-rings generally had to be of
  metal; no organic material could be restrained from ignition. Teflon
  would stand up under static conditions, but if the CTF flowed over it
  with any speed at all, it would erode away like so much sugar in hot
  water, even if it didn't ignite. So joints had to be welded whenever
  possible, and the welds had to be good. An enclosure of slag in the
  weld could react and touch off a fire without even trying. So the
  welds had to be made, and inspected and polished smooth and
  reinspected, and then all the plumbing had to be cleaned out and
  passivated before you dared put the CTF into the system. First there
  was a water flush, and the lines were blown dry with nitrogen. Then
  came one with ethylene trichloride to catch any traces of oil or
  grease, followed by another nitrogen blow-down. Then gaseous CTF was
  introduced into the system, and left there for some hours to catch
  anything the flushing might have missed, and then the liquid chlorine
  trifluoride could be let into the propellant lines.


Why does chlorine trifluoride ignites concrete and sand? Concrete and sand are highly unreactive to most of the chemical attacks. But chlorine triflouride is so dangerous that it will ignite concrete, sand and other fire retarding substances. Which property does ClF3 possess to cause this ignition and why?

Concrete, sand, and asbestos are all, in large part, silicon dioxide (SiO₂). It's energetically favorable for available fluorine to take the place of oxygen, producing SiF₄. This occurs even in weak solutions of hydrogen fluoride, and it's why HF solutions are used for etching glass.

ClF₃ has a much greater amount of potential bond energy than HF or even F₂. This means that, when it decomposes, such as upon contact with silicon dioxide, it gives off a lot of energy, which is released as fire.


Sn + HNO3: When will the metastannic acid H2SnO3 appear? I am dissolving tin in nitric acid.
What reactions will take place depending on the conentration of $\rm{HNO}_3$ and temperature?

Particularly, I am interested when the metastannic acid $\rm{H_2SnO_3}$ is likely to be produced, according to

$\rm{Sn + 4 HNO_3 → H_2O + H_2SnO_3 + 4 NO_2}$

P. S.
References to books or scientific articles are welcome.

It is a common belief that tin dissolves in dilute nitric acid to produce tin(II) nitrate. The reactions are addressed in a previous answer of mine. However, employing a more hot and concentrated nitric acid yields an unstable compound called hydrated stannic oxide also known as β-stannic acid. The compound also has a α-form which is later converted to β form. The relevant information can be found here:

  [...]tin dissolves slowly, in very dilute nitric acid to produce
  stannous nitrate (but) when hot and more concentrated acid is
  employed, is very unstable and quickly decomposes, yielding the form
  of hydrated stannic oxide known as β-stannic acid. Probably
  α-stannic acid is first produced [...] which then passes into the
  β-form. The product is, therefore, liable to contain both α- and
  β-forms, and the pure β-acid is obtained by dissolving this product
  in sodium hydroxide solution, and then adding excess of concentrated
  soda, which precipitates sodium p-stannate whilst the α-salt remains
  in solution. Pure β-stannic acid, or β-metastannic acid, is then
  obtained by decomposing the sodium salt with acid. It is also formed
  by the hydrolysis of its sodium salt at 60° C., and by boiling a
  dilute solution of stannic chloride, with or without the addition of
  nitric acid.[...]

  [...]Gelatinous, precipitated β-stannic acid has the empirical
  composition $\ce{SnO2.4H2O}$, when air-dried $\ce{SnO2.2H2O}$, and
  when dried in a vacuum, $\ce{SnO2.H2O}$; [...]


Can a battery powered by iron and air really provide a feasible power source for automobiles? I'm referring to the Shuttle Battery recently announced (read more on it here) that suggests that:

  The energy density of the device is up to 7,700 Wh/L. It is more than 5 times
  higher than state-of-the-art (SOA) Li-ion (550Wh/L), and comparable to fossil
  fuel (10,000 Wh/L).

It goes on to say that:

  It is rechargeable because it is based on a simple reversible chemical reaction, 3Fe + 4H2O = 4H2 + Fe3O4 and 2H2 + O2 = 2H2O. The proof of concept cell demonstrated zero capacity loss after 200 cycles of discharge and charge.

Now the question is: Is this even remotely possible? Or is there a red flag here somewhere that I can't seem to pick up?

This is certainly possible.  In fact, there are a large number of metal-air batteries being developed all over the world.

It is, however, strange that such high energy density is achieved, because iron ion itself would not offer very high energy density.  My guess is that likely they did it by having an extremely high matter density.  So this battery may be small in size for its capacity, but at the same time it may also be much heavier than a conventional battery.

That could be why they brand it as a shuttle power supply(well, energy density isn't really relevant for a bus anyways).  A very heavy battery on a hand held device is never good.   Cost is the more important factor there, but it is hard to tell at the moment.  0 loss over 200 cycles seems like a fairy tale, but I will give them the benefit of the doubt at the moment.

This battery will also likely to have a considerable loss of charge when not in use.   Again, this might not be a huge problem for buses, where you keep driving non-stop during days then kept it in the charging station overnight.

Anyways, don't expect this to be on your iPhone 10 or your laptop.  Even if it works as presented, it might not be that ideal for portable electronic devices, where people worry much more about weight and stand-by time than if it occasionally explodes, sadly.


How to read electron energy levels abbreviations? The electron energy transfer diagrams I've been coming across all have energy levels labeled as e.g. $^{2}F_{5/2}$ or $^{4}S_{3/2}$. I've tried to Google "quantum energy level abbreviations" but haven't come across anything too useful, just explanations of electron configurations. Can anyone explain or point to a website or textbook that goes through what these numbers/letters/fractions mean for someone who's taken general chemistry but not quantum mechanics?

Thanks!

Those are Russel Saunders atomic term symbols and represent the quantum numbers of the many-electron wave function.

The notation scheme is $^{2S+1}L_J$ where $2S+1$ is the spin multiplicity, $L$ is the angular momentum quantum number and $J$ is the total angular momentum due to the coupling between the former two.

There are also molecular term symbols using greek instead of latin letters for the angular momentum. Those are commonly used for diatomic molecules.

You should find the Russel Saunders Coupling scheme in standard physical and quantum chemistry books. Molecular term  symbols are for example explained in detail in Molecular Spectra and Molecular Structure: I. Spectra of Diatomic Molecules by Herzberg and Huber (1950).
This link may give you a preliminary idea : http://wwwchem.uwimona.edu.jm/courses/RScoupling.html


Net production of CO2 in plants I am trying to understand in a qualitative way photosynthesis. I am not a chemist so please have some patience.

I read on the Wikipedia page for photosynthesis that the way plants use $\ce{CO2}$ is through the reaction:

$\ce{energy + 6 CO2 + 6 H2O -> C6H12O6 + 6O2}$

but then for cellular aerobic respiration, I find exactly the opposite reaction:

$\ce{ C6H12O6 + 6O2 -> 6 CO2 + 6 H2O + energy}$

So, from this simplified view where only net reactions are written it looks that this mechanism can be used to transfer and store energy. But from a $\ce{CO2}$ point of view it looks that plants do not "consume" carbon dioxide since what they use via photosynthesis they give back through respiration. Clearly this cannot be true. So where is the catch ?

Plants are able to store energy as carbohydrates, i.e. they can make more carbohydrates than they need for their metabolism  when there is no sunlight, and store it for later use. So the amount of energy in your equations is not equal.

Some plants are offering this stored energy as sugars as nectar or fruits in exchange for pollination and spreading their seeds or use it as energy storage for times when they can't make enough carbohydrates for quick use, e.g. in winter or as storage for high demand, e.g. quick growth to outsmart competitors for space.
In addition, quite much of this fixed CO2 goes into structures like stems, roots and leaves and thus is sequestered for some time from the atmosphere.


Liquid-liquid caffeine extraction I am performing a lab experiment in which I am extracting crude caffeine from different samples of tea. I have read a couple of similar experiments in which methylene chloride has been used to extract caffeine, owing to its volatile nature and caffeine's affinity to methylene chloride. I know chloroform can be used as a substitute for methylene chloride. I was wondering if carbon disulphide can used as a substitute in the experiment.
Thank you very much!

I suggest to use a 1:1 mixture of dichloromethane and 0.2 M NaOH at room temperature.

The protocol was described by Tetsuo Onami and Hitoshi Kanazawa in J. Chem. Educ, 1996, 73, 556-557.

The authors mention that only trace amounts of caffeine are extracted from tea leaves with dichloromethane alone. They reason that due to the hight amount of tannines in the tea leaves, most of the caffeine exist in protonated form and is therefore bound to the tannines.

As an experimentally proven alternative, they suggest to treat the tea leaves with a mixture of dichloromethane and 0.2 M NaOH at room temperature!

According to the article, 20-30 mg of caffeine were extracted from a 2 g tea bag after shaking the above mixture for about ten minutes.

UPDATE
If you insist on the liquid-liquid extraction of previously prepared aqueous solutions and feel the urge to set up a percolator-type apparatus for continuous extraction, the Kutscher-Steudel glassware will not work with solvents that have a higher density than water.

For continuous extractions with solvents like $\ce{CH2Cl2}$, some different glassware has been designed and was described by S. Wehrli in Ein Apparat zur Extraktion von Lösungen mit schweren Lösungsmitteln and was published in Helv. Chim. Acta, 1937, 20, 927–931 (DOI)


Are van der Waals coefficients independent of temperature? In the van der Waals equation for real gases, the coefficients $a$ and $b$ are used as respective correction terms for pressure and volume in the ideal gas equation.

I want to know whether the van der Waals coefficients are independent of absolute temperature or not.

The van der waals constant can be directly obtained from the critical temperature and pressure. We have,

${a = \frac{27 R^2 T_c^2}  {64 P_c}}$
${b =  \frac{R T_c }{ 8 P_c}}$

Since, $T_c$ and $P_c$ are constants, we can conclude that the van der waals constants are really "constants."


How much can the pH change through dilution? Consider an acidic solution with Hydrogen ion concentration, $\ce{[H+]}$ of $10^{-5}\:\mathrm{M}$. Since $\:\mathrm{pH} = -\log \ce{[H+]}$ the $\:\mathrm{pH}$ of solution is $5$. Suppose we dilute solution 10 times with water. Now, $\ce{[H+]}$ is $10^{-6}\:\mathrm{M}$ and $\:\mathrm{pH}$ is $6$. Further dilution should increase $\:\mathrm{pH}$ from $6$ to $7$ and then from $7$ to $8$ and so on. Can this go in for ever? Does this not imply that an acidic solution can be made basic/alkaline simply by adding water? But that doesn't happen? What prevents it?

Is there anyone already found answer for this problem or it is just an unsolved basic problem of chemistry?

Water undergoes autoionization, i.e., it reacts as follows:

$$
\ce{H2O + H2O <=> H3O+ + OH-}
$$

The equilibrium constant for this reaction at standard conditions is $K_w = [\ce{H3O+}][\ce{OH-}] \approx 1.0 \cdot 10^{-14}$. In pure water, $[\ce{H3O+}] = [\ce{OH-}]$, hence $[\ce{H3O+}] = \sqrt{K_w} \approx 1.0 \cdot 10^{-7}\ \textrm{M}$.

Suppose we dilute a solution with some initial concentration of $[\ce{H3O+}]_i = \frac{n_i}{V_i}$, where $n_i$ is the initial number of moles of $\ce{H3O+}$ and $V_i$ is the initial volume. If I now add a volume $\Delta V$ of pure water (which would contain $(1.0 \cdot 10^{-7}) \cdot \Delta V$ moles of $\ce{H3O+}$) the resulting final concentration can be crudely approximated as:

$$
[\ce{H3O+}]_f \approx \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V}
$$

We can see what value this expression approaches as we make the solution more and more dilute by taking the limit as $\Delta V \to \infty$:

$$
\lim_{\Delta V \to \infty} \frac{n_i + (1.0 \cdot 10^{-7}) \cdot \Delta V}{V_i + \Delta V} = 1.0 \cdot 10^{-7}
$$

Therefore, $[\ce{H3O+}]$ tends towards $1.0 \cdot 10^{-7}$ with further dilution, so $pH$ will approach a value of $7.0$.


Isomer Identification Using Condensed Structural Formulae Which of the following pairs are isomers?
a) $\ce{C5H10}$ and $\ce{C10H20}$
b) $\ce{CH3(CH2)4CH3}$ and $\ce{CH3(CH2)3CH3}$
c) $\ce{CH3CH(CH3)(CH2)2CH3}$ and $\ce{CH3(CH2)2CH(CH3)2}$
d) $\ce{(CH3)3CH}$ and $\ce{CH3CH2CH2CH3}$

My textbook says the answer is C). I beg to differ. I think it is D). C) is just a different way of writing 2-methylpentane.

isomer
One of several species (or molecular entities ) that have the same atomic composition (molecular formula) but different line formulae or different stereochemical formulae and hence different physical and/or chemical properties.

Source:
Pure and Applied Chemistry, 1994, 66, 1077 (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994))
Pure and Applied Chemistry, 1996, 68, 2193 (Basic terminology of stereochemistry (IUPAC Recommendations 1996))
a)
C5H10
and
C10H20
different chemical formula → not isomeric
b)
CH3(CH2)4CH3 = hexane (C6H14)
and
CH3(CH2)3CH3 = pentane (C5H12)
different chemical formula → not isomeric
c)
CH3CH(CH3)(CH2)2CH3 = 2-methylpentane (C6H14)
and
CH3(CH2)2CH(CH3)2 = 2-methylpentane (C6H14)
same chemical formula and same structural formula → identical, not isomeric
d)
(CH3)3CH = isobutane (C4H10)
and
CH3CH2CH2CH3 = butane (C4H10)
same chemical formula and different structural formula → isomeric


Calculating pH: Weak Acid, Strong Base This is a question that had me puzzled for quite a while.  I feel that information is missing.

A weak acid, $\ce{HA}$, $K_\textrm{a} = \pu{1.0E-4 M}$, is titrated with $\ce{NaOH}$.  The concentration of $\ce{NaA}$ at the equivalence point is $\pu{0.010 M}$.  What is the $\mathrm{pH}$ at the equivalence point?
a) 4.0
b) 6.0
c) 7.0
d) 8.0
e) 11.0

The correct answer is :

 d) 8.0

The answer is fairly easy to guess, as the $\mathrm{pH}$ needs to be greater than 7, but 11 is much too basic.  However, I'm not sure how I would be able to figure that out from the information given?  I've tried using the Henderson-Hasselbalch equation, but I can't figure out the relationship between $\ce{[NaA]}$ or $\ce{[A-]}$ and $\ce{[HA]}$.

At the equivalence point, you have an aqueous solution of weak base $\ce{Na}A$ at the concentration of $c$. The pH of the solution is given by the equation: $$\mathrm{pH}=\frac{1}{2}(\mathrm{p}K_a +\mathrm{p}K_w -\mathrm{p}c)$$
In your case: $\mathrm{pH}=\frac{1}{2}(4.0 +14 -2.0)=8.0$
Now, if you are not familiar with the above equation, you can write the reaction that could happen in the solution; the reaction of  $\ce{Na}A$, completely dissociated, with water:
$$A^- +\ce{ H2O <=> OH- +H}A$$
The constant of this equilibrium is: $$K=\frac{K_w}{K_a}= 10^{-10}$$

On the other hand,  you have:

$$K=\frac{x^2}{c-x}=\frac{x^2}{1.0\times10^{-2}-x}\approx\frac{x^2}{1.0\times10^{-2}} =1.0\times10^{-10}$$
Where $x$ is the concentration of ion hydroxide and the weak acid $\ce{H}A$.

We have, $x=1.0\times10^{-6}\mathrm{M}$.

$$[\ce{H+_\mathrm{(aq)}}]=\frac{10^{-14}}{1.0\times10^{-6}}\mathrm{M}$$
 $$\mathrm{pH}=8.0$$


Are Hartree-Fock solutions eigenfunctions of the electronic hamiltonian? $\newcommand{\Ket}[1]{\left|#1\right>}$
$\newcommand{\Bra}[1]{\left<#1\right|}$

I'm currently revising for my qualifiers, and I've come across a point of confusion, one that several places give seemingly contradictory results.

To my understanding:

In Hartree-Fock, the variationally determined ground state, $\Ket\Phi$, is not an eigenfunction of the electronic hamiltonian. It's composed of eigenfunctions of the Fock operator. It's an approximation of the ground state of the Hamiltonian.

(1) Is this all correct so far?

(2) If so, is it justified to use it in numerical derivations as an eigenfunction of the Hamiltonian,
i.e. can we write $\hat{H}\Ket\Phi\overset{?}{=}E_{0}\Ket\Phi$ or are we restricted to $\Bra\Phi\hat{H}\Ket\Phi=E_{0}$

$\newcommand{\ket}[1]{\left|#1\right>}$
$\newcommand{\bra}[1]{\left<#1\right|}$

  (1) Is this all correct so far?

Yep.

  (2) If so, is it justified to use it in numerical derivations as an
  eigenfunction of the Hamiltonian, i.e. can we write
  $\hat{H}\ket\Phi\overset{?}{=}E_{0}\ket\Phi$ or are we restricted to
  $\bra\Phi\hat{H}\ket\Phi=E_{0}$?

First, you can write $\hat{H}\ket\Phi = E\ket\Phi$ where $\Phi$ is the HF wave function but then $\hat{H}$ has to interpreted as the so-called mean-field Hamiltonian. It is the approximate Hamiltonian in which the exact potential which represents the energy of the Coulomb repulsions between the
electrons ($\hat{V}_\mathrm{ee}$) is replaced with some approximate mean-field potential ($\hat{V}_\mathrm{MF}$) which describes the model system in which electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons. That is basically the definition of the HF wave function, i.e. it is an eigenfunction of the mean-field Hamiltonian:
$$
\hat{H}_\mathrm{MF} \ket\Phi = E_\mathrm{MF} \ket\Phi \, ,
\quad \text{where} \quad
\hat{H}_\mathrm{MF} = \hat{T}_\mathrm{e} + \hat{V}_\mathrm{en} + \hat{V}_\mathrm{MF} \, .
$$

Secondly, despite the fact that the HF wave function is not an eigenfunction of the exact electronic Hamiltonian
$$
\hat{H}_\mathrm{e} \ket\Phi \neq E_\mathrm{e} \ket\Phi \, ,
\quad \text{where} \quad
\hat{H}_\mathrm{e} = \hat{T}_\mathrm{e} + \hat{V}_\mathrm{en} + \hat{V}_\mathrm{ee} \, ,
$$
we could evaluate it energy $\bra\Phi\hat{H}_\mathrm{e}\ket\Phi$ using Slater rules and minimize it to find an upper bound to the ground state energy
$$
\bra\Phi\hat{H}_\mathrm{e}\ket\Phi \geq E_\mathrm{e0}
$$
But the resulting Slater determinant obtained by minimizing the energy is only an approximation to the ground state wave function.

Finally, yes, can we use $\ket\Phi$ as our trial wave function despite the fact that it is not an eigenfunction of $\hat{H}_\mathrm{e}$ since it is not required by the variation method. $\ket\Phi$ is normalizable, it (presumably) satisfies same boundary conditions as the exact wave function, and it has adjustable parameters. That is basically enough to qualify as the trial wave function. And at the end of the day, starting from a known solution for a simpler problem is a very usual practice for variation method.


Could someone please explain the difference between m/z and m/e in mass spectroscopy The definition:

  The abbreviation m/z is used to denote the dimensionless quantity formed by dividing the mass number of an ion by its charge number. It has long been called the mass-to-charge ratio although m is not the ionic mass nor is z a multiple or the elementary (electronic) charge, e. The abbreviation m/e is, therefore, not recommended. Thus, for example, for the ion $\ce{C7H7^2+}$, m/z equals 45.5.

Could someone please explain the definition? I found it a bit difficult to comprehend.

What is "ionic mass"?
What do they mean by "z a multiple or elementary charge"?
$\ce{C7H7^2+}$, m/z equals 45.5 -> what did they do here?

$m$ is the atomic mass of the ion (in units ($u$); $1u=1.665402\times10^{-27} \mathrm{kg}$), $z$ is the number of its positive/negative charges. $z$ should not be used in interchange with $e$ because it is dimensionless, while $e$ has the unit $\mathrm{C}$ (Coulomb). For $\ce{[C7H7]^2+}$, $m/z$ is given by (the unit of $m$ is omitted):

$$m(\ce{C7H7})/2=(7 \cdot 12+7 \cdot 1)/2=91/2=45.5$$


Finding the energy given out when 2 moles of reactants are mixed and ignited from a balanced equation The question I am trying to answer is this:

I only know about Hess' law, bond enthalpies and enthalpies of combustion and have tried to use all concepts in order to somehow get an answer but none of them have worked. I first tried to write a balanced equation for what the reaction of 2 moles of reactants would look like but I don't believe it could be balanced. Then, I tried using bond enthalpies but I couldn't have done that either since the equation couldn't be balanced. Next, I tried using enthalpies of combustion, and the same thing happened. So, I'm wondering if there's anyway to answer this question, preferably with the concepts mentioned above.

The N2O4 is going to define the maximum amount that will be used up in the reaction, since it has the higher stoichiometric number of the reagents. That is, the second term will always be less and the amount consumed will be defined by the N2O4 because that two moles will limit the N2O4 and in doing so reduce the amount of the second term consumed- there will be excess second compound after the N2O4 has been consumed.

For example, the ratio of first to second term is 5:4, so if we consume all of the second term (2 mols) this requires 2.5 mols of N2O4, which we don't have. So we are limited with the N2O4.

So, knowing that, with two mols of N2O4 the amount is reduced to 2/5 of the enthalpic calculation. That becomes 5116*2/5 = 2046, or answer B.

Apologies for awkwardness of this answer, on a mobile device and have to keep swapping from question to answer.


In Hückel's rule, can n be any integer? Hückel's rule says that any planar compound with a ring of conjugated p orbitals with $4n+2$ electrons is aromatic. Here, can $n$ be any integer, or does $n$ have to be related to the number of p orbitals in the conjugated π system (as shown in the Frost circle)? For example, in benzene, there are 6 electrons in the pi system ($n=1$), which fill all 3 bonding orbitals. Thus, benzene is very stable. But say we remove 4 electrons so that there's only 2 electrons in the π system. Here, $n=0$, so is this cation considered aromatic?

On the other hand, if we add 4 electrons to benzene's π system, it has 10 electrons ($n=2$). Is this anion aromatic? It doesn't seem very stable to me, since it has now filled 2 antibonding orbitals as well.

In general, it is a requirement that the $4n+2$ p electrons exactly fill all the bonding orbitals of the pi system for the molecule to be aromatic?

No, you do not have to have as many π electrons as p-orbitals or any type of direct relationship. As long as you have a continuous, circular arrangement of p-orbitals and an electron count $4n+2$ (for monocycles), you can expect your molecule to be aromatic.

Pyrrole, furan and related molecules are aromatic. They have 6 π electrons like benzene, but only five p-orbitals.
The cyclobutadienyl dianion has been synthesised as a dilithium salt. The authors claim evidence for aromaticity.[1]
The dianion of cyclooctatetraene has also been synthesised and examined by ESR as early as the 1960’s. The authors state:

  The assumption that both the dianion and the radical-anion are planar […] is consistend with [their equation] within experimental error.

Thus it can be assumed that $\ce{cot^2-}$ is also an aromatic system.[2]
Disubstituted dications of cyclooctatetraene have been synthesised a decade later by Olah et al. The published proton NMR data implies that the protons are more deshielded than a simple triene moiety would imply. The summary the authors give states the following:

  The $\ce{^{1}H}$ and $\ce{^{13}C}$ NMR spectroscopic characterization of the reported new cyclooctatetraene dications indicates these ions possess those characteristics expected of a $\ce{C8}$ 6 π aromatic system and hence further extends the evidence for the great predictive utility of the $4n+2$ Hückel rule.

$\ce{cot^2+}$ is thus also an aromatic system.[3]

These last two bullet points must be the ones you are looking for. The cyclooctatetraene dianion, a cyclic $\ce{C8}$ species with 10 π electrons ($n=2$) is aromatic, as is also the cyclooctatetraene dication, the same $\ce{C8}$ species albeit with 6 π electrons ($n=1$).

For most systems, aromaticity is either already given (as they have e.g. six π electrons in their neutral state; compare benzene etc.) or they are only one electron away (cyclopentadienyl, cycloheptatrienyl or tropylium). The way these molecules obtain aromaticity is relatively fixed since greater charges destabilise molecules.

Only for those systems starting with $4n$ π electrons such as cyclooctatetraene, which are antiaromatic by the Hückel definition, can there be a choice. Both the dication and the dianion are generally unfavourable, but they are similarly unfavourable thus the system can be pushed into either direction.

It is highly unlikely that a tetraanion or tetracation of benzene exist outside of a $4~\mathrm{K}$ matrix, much like it is very unlikely for a cyclopentadienyl-trication or a cycloheptatrienyl-trianion to exist outside of similar conditions due to their great charge density.

References:

[1]: A. Sekiguchi, T. Matsuo, H. Watanabe, J. Am. Chem. Soc. 2000, 122, 5652. DOI: 10.1021/ja0004175.

[2]: H. L. Strauss, T. J. Katz, G. K. Fraenkel, J. Am. Chem. Soc. 1963, 85, 2360. DOI: 10.1021/ja00899a004.

[3]: G. A. Olah, J. S. Staral, G. Liang, L. A. Paquette, W. P. Melega, M. J. Carmody J. Am. Chem. Soc. 1977, 99, 3349. DOI: 10.1021/ja00452a027.


What is the rationale of the IUPAC name for DDT? The structure of DDT is

One of the benzene molecules should be the parent chain as it has a greater number of carbon atoms than the straight chain (which has just two carbon atoms). Taking benzene as the parent chain, I think that the name of the compound should be 1-Chloro-4-((1'-(4"-chlorophenyl))-2',2',2'-tricholoroethyl) benzene.

But the Wikipedia article states that the IUPAC name of the compound is
1,1,1-Trichloro-2,2-bis(4-chlorophenyl)ethane.

Why is the ethyl chain taken as the parent chain when it has lesser number of carbon atoms than the benzene ring?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature. However, the IUPAC recommendations define a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations.

P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES
P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.
When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The given structure of the DDT molecule fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

P-15.3.1.3 Multiplicative name formation
Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:
(a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: 1,1'

(b) the name of the linking multiplicative substituent atom or group;

Here: 2,2,2-trichloroethane-1,1-diyl

(c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: bis

(d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: 4-chlorobenzene

The numbering of the identical parent structural unit is retained and, when there is a choice, the locants of the point of substitution by the linking multiplicative substituent groups on the identical parent structure are as low as possible.

Here: 1,1'-(…)bis(4-chlorobenzene) (not 4,4'-(…)bis(1-chlorobenzene))
Therefore, the preferred IUPAC name is 1,1'-(2,2,2-trichloroethane-1,1-diyl)bis(4-chlorobenzene)


Which base pair is more stable, U:G or I:C? As a result of damage (usually deamination), DNA will sometimes contain uracil and inosine bases. I am trying to determine which unnatural base pairing combo would be more stable: uracil paired with guanine or inosine paired with cytosine. See the images below.

One can discuss the strength of H-bonds, but that would be half of the picture. It is more important how well (or poorly) this pair fits in the whole helical structure. Stacking interactions (arene-arene interactions) are playing important role in stabilization of DNA. If the pair disturbs backbone shape the binding will be poor. Google scholar search for relevant papers might be the most productive way to go.