Can I use lye as a cooking leavener instead of baking soda? I asked this question on the cooking site, but got unsatisfactory answers: https://cooking.stackexchange.com/questions/65520/can-i-use-lye-as-a-leavener-instead-of-baking-soda

I'm interested to try to make waffles using a concentrated food-grade lye water mixture instead of baking soda as a leavening agent. I have three questions related to this subject:

Would this potentially make my waffles fluffier (due to increased CO2 producing reactions or otherwise)?
Would it impact the flavor of the waffles?
Is it safe to ingest waffles that have had a lye water reaction occur in them? (Obviously I should let the lye reaction complete before ingesting).

How long should I let a lye reaction occur before cooking?

Notably, I often use sourdough starter in my waffles, which has an acid component - I've noticed that I get a strong reaction from the baking soda as a result, and it got me thinking that I would get an even stronger result from lye, but I'm worried about taste and ingestion safety.

I'm not an expert here, but, as a chemist with a strong gastronomic interest (and a penchant for waffles), can suggest some things which may be helpful. Lye is a caustic alkaline solution, traditionally KOH, and more commonly these days NaOH.

First, my understanding of lye as an agent in cooking is limited to its use in pretzels and "laugen" bread and ramen soups. The pretzel making process involves a final dipping of the dough in a lye solution in order to promote the Maillard reaction during cooking; this is not the same as the leavening process. Ramen noodles are a type of "alkaline noodle" and I believe use the lye to break down the gluten to make the flour more pliable and less elastic. Again, not the same as the leavening process.

The use of lye as a leavening agent comes from taking lye (produced by washing fireplace ash) and boiling with lard to produce potash. The purified version, pearl ash, was one of the first chemical leaveners used in the kitchen. Pearl ash acts as a leavening agent by reacting with an acidic ingredient; fruit juice, sour milk or molasses, to produce CO2.

So to answer your questions:

I can't see how it would make your waffles fluffier than conventional modern rising agents, even if you did get your chemistry right. You still need to get your CO2 from somewhere and substituting your bicarb for lye actually removes this element. You would need to rework your entire recipe, I think.
Yes. Without doubt. More than likely it will degrade the taste of what should otherwise be a delicious dish. Alkaline foods are well known to have a bitter after taste (think kale and broccoli). Even scones made with a little too much baking soda can leave your mouth tasting and feeling like you've just had your annual clean and scale at the dentist.
While lye is used as a cooking agent in some instances, even the food grade solutions are extremely caustic, and, undiluted, are actually classed as poisons in some instances. It is corrosive to the body, especially mouth, throat and stomach, and can cause serious damage to the eyes. This may be stating the obvious, but best to assume ignorance and work up from there. Apart form mooncakes, I am not aware of a dough recipe (or anything else for that matter) that incorporates lye within the dough itself. Unless you have experience handling this ingredient in the kitchen, alternatives are recommended. I cannot tell you how you would determine whether your 'lye reaction' is complete within the dough; certainly gauging the completion of 'the rise' would not be sufficient, and the cooking process may need to be adjusted to accommodate this. Mooncake dough, for instance, is allowed to sit overnight, and even for up to 3 days before cooking. Too long to wait for my waffles!

For your interest, there is a very good comparison of various traditional and modern leavening agents here. Interestingly, the author concluded that the best agent overall was baking soda. Just like modern research chemistry in the lab - someone has done the reaction already.


Result of glucose with acetic anhydride and pyridine? What is the result when glucose is mixed with excess acetic anhydride and excess pyridine?

Most probably it forms glucose pentaacetate. The acetic anhydride esterifies with all the alcohol groups on the glucose ring.


Separating two sugars according to their heavy atoms - Steered Molecular Dynamics I am trying to separate two methyl-N,N'-diacetylchitobiose sugars using steered molecular dynamics in AMBER. I have tested out:

&rst
iat=-1,-1,
r2=2.5,
rk2=10.0,
r2a=12.5,
iresid=0,
fxyz=1,1,1,
outxyz=1,
igr1=2,5,6,8,9,11,14,16,18,20,22,24,26,27,28,32,33,35,36,38,41,43,45,47,49,51,53,54,55,59,60,
igr2=65,68,69,71,72,74,77,79,81,83,85,87,89,90,91,95,96,98,99,101,104,106,108,110,112,114,116,117,118,122,123
/

Which should apparently separate the two sugars relative to their heavy atoms from a distance (between each other) of 2.5 A to 12.5 A along the x,y and z axes and using a force constant of 10 (I think kcal/mol). The solvent I am using is water.

I ran this separation for 100 ns and the two sugars refused to separate. I am not sure if my force constant is too small, or if something else is wrong with my reference file. I have googled it and spoken to m supervisor about it, and unfortunately I have not succeeded in finding a solution.

Any help would be appreciated.

I have realized the problem. The force constant was too low. Increasing it to 25.0 from 10.0 sorted things out.


Identifying MgCl2 or CaCl2 in solution I have an aqueous solution of either $\ce{MgCl2}$ or $\ce{CaCl2}$ in solution with $\ce{K2SO4}$. There is a white precipitate that forms on the bottom that does not dissolve when $\ce{HCl}$ is added (maybe it needs to be in excess?). Is there any way to find out if it's $\ce{Mg^2+}$ or $\ce{Ca^2+}$? I have already tried adding $\ce{NaOH}$, and the result was a not very cloudy, white precipitate, which is not really jelly like, but when compared to stock samples of $\ce{MgCl2}$ and $\ce{CaCl2}$ with $\ce{NaOH}$, they don't match. I tried a flame test but only could see the $\ce{K+}$ (violet flame).

$\ce{MgSO4}$, otherwise known as epsom salt, is soluble. $\ce{CaSO4}$ ($K_\mathrm{sp}=7.1 \times 10^{-5}$), is slightly soluble. It is extremely likely that unless there are some other unknown metal ions in this solution, the white precipitate you describe in the solution is $\ce{CaSO4}$.


Effect of Base on E2 product One base is bulkier than the other, that's my observation. How does it affect the product? Could someone please give proper reasoning and the product? Thanks a lot.

P.S.
I guess both follow E2

Under $E_2$ reaction conditions, the MeO$^-$ product will be

while the Me$_3$CO$^-$ product will be

.

This is due to the bulkiness of Me$_3$CO$^-$ being unable to reach the hydrogen at the substituted carbon so instead it goes to the primary position instead.


Types of radioactive decay Besides alpha, beta, and gamma, are there any other types of radioactive decay? If so, what are they? Is there any type of radioactive decay that is more powerful than gamma?

Here's a list of types of radioactive decay. The most notable types of decay that are not among the classic three involve the direct emission of a free proton or a neutron, the emission of atomic clusters other than helium nuclei (alpha particles), the absorption of one of the innermost shell electrons into the nucleus, or spontaneous fission of the unstable nucleus.

Regarding the question about gamma ray energies, it is first important to notice that "gamma ray" is a generic term that refers to any electromagnetic energy (photons) coming from a nucleus or subatomic particle decays, and has no direct reference to the energy of the photon (though in the past it did). For example, a nuclear excited state of thorium-229, $^{229m}\ce{Th}$, decays through the emission of gamma rays with an energy of 7.6 eV, which corresponds to photons in the ultraviolet region of the electromagnetic spectrum.

Also, the answer depends on exactly what you mean by "power". Assuming you're comparing kinetic energies, Wikipedia says gamma rays from nuclear decay rarely have energy above 10 MeV, and another source says electrons from beta decay usually have energies up to 4 MeV. The energies are all of the same order of magnitude, which makes sense because all the particles are coming from the same source (an unstable nucleus). Theoretically gamma rays can have a slightly higher kinetic energy than other decay particles because photons have zero rest mass, therefore all the decay energy is converted into kinetic energy, whilst for electrons 511 keV is "wasted" on creating the rest mass (and even higher "wastes" for heavier particles such as protons/neutrons/alpha particles etc). I think one can also consider that some energy is used up in "climbing out" of the electrical potential and residual strong nuclear potential wells of the nucleus, neither of which applies to gamma ray photons (though positively charged species would in fact gain energy, being electrically accelerated away from the nucleus by repulsion, but this contribution is probably smaller than the nuclear force one). However, different types of nucleus have different allowed decay paths, and depending on complicated details the most energetic beta particles or proton/neutron emissions might have a slightly higher energy than the most energetic nuclear decay gamma ray.

As a final note, it is also possible to create even more energetic photons than any that could be produced via nuclear decay by putting electrons in a sufficiently powerful particle accelerator. Any charged particle under acceleration/deceleration loses some energy in the form of electromagnetic radiation. For ultrarelativistic charged particles, the photons emitted can be of extremely high energy. These are a type of x-ray termed "hard x-rays" (more energetic than so called "soft x-rays").

Edit: Correction regarding energy lost in rest masses of particles from decay. Energy is used up in the creation of electrons, but protons/neutrons/clusters already exist inside the nucleus, and so are not created. These particles could never be produced directly as a single one would have upwards of 930 MeV, far above typical nuclear decay energies.


How can a Hydrogen make such a difference? Looking at familiar bodily fluids for a learning session. At first glance they both look the same. Then I noticed the N=C in biliverdin and HN-C in bilirubin. How can a N=C (imine group) and HN-C (secondary amine) make such a difference in colour? They look like structural isomers of each other to me. Or is there any stereoisomeric differences as well to this colour functional group change?

(Jaundice is directly related to bilirubin, urine is a product of bilirubin, I reckon there must be a real good reason and help aid from the colour change representation.)

e.g.

image 1: bilirubin is a brown pigmented substance while biliverdin is green pigment

image2

The difference in structure of these two compounds is  perhaps the presence of a more conjugated structure in biliverdin compared to bilirubin. These products (degradation products of haeme1), just as most pigments, make use of the conjugated electron systems to absorb visible light, giving rise to strong colors. The long chain conjugated system in bilverdin leads to its strong green color, the reduced biliribin (methenyl bridge) has a less conjugated system with a different absorption consenquently different color.

This is a case of effects of conjugation in pigments than merely a difference in functional groups as I see it.

1Note The degradation of haeme occurs at  the end of the red cells’  lifetime, where they are removed from the circulation and their components degraded, thus depends of many factors most importantly regulations carried out in the body, enzymes etc.

References

Voet and Voet Biochemistry Section 26-4. Amino Acids as Biosynthetic Precursors.
Conjugated system


Which of the following atoms has the largest first ionization energy? Of the following atoms, which has the largest first ionization energy?

$\ce{Br}$
$\ce{O}$
$\ce{C}$
$\ce{P}$
$\ce{I}$

I got confused between $\ce{O}$ and $\ce{Br}$. The answer given is $\ce{O}$ although in the periodic table $\ce{Br}$ is closer to Group 18 (noble gases). Why is the answer $\ce{O}$ then?

I actually think this is an unfair question, unless you were given some additional information, or you were specifically instructed to memorize specific ionization energies, rather than learning the trends.  The trend for ionization energies is that they increase from left to right across rows, and from bottom to top in columns.  Based on only that information, I can't see how one should distinguish between $O$ and $Br$ ... yes, it happens to be true that $O$ has a higher IP than $Br$, but how are you supposed to tell? Suppose the example had used $S$ instead of $O$, and $I$ instead of $Br$?  Those elements have the same relative positions in the table, but in that case, it would be $I$ that has the larger ionization energy.

So, in this case I think your confusion was quite understandable .. I would never construct a test question like this unless the student had some additional information.  It's not even true that you can always use the rule about them increasing from left to right along rows .. in the first row, $N$ actually has a slightly higher IP than $O$, because there is a penalty for disrupting the half-filled p-shell in $N$, which is anomalously stable.  You can see the periodic trends here.


Redox reactions - alkene to alcohol How is the "correct" answer not a redox reaction?

Isn't the "correct" choice a textbook example of an organic redox reaction? After all, we are breaking a C-C bond and replacing that with a C-O bond. If that's not redox I don't know what is redox.

Or am I missing something fundamental here?

The reaction in question involves the addition of water across a double bond. The proton from the water is added to one carbon in the double bond - this carbon is reduced.  The hydroxyl group is added to the other end of the double bond - this carbon is oxidized. Even though the oxidation number of one olefinic carbon (the one that winds up with 2 hydrogens attached) changes from-1 to -2, and the other from -1 to 0, overall, for the entire molecule, there is no net oxidation or reduction.


Finding orbit radius using the Bohr model and Rydberg equation To start with its a homework problem, quite lengthy.

  A particle of mass equal to 208 times the mass of an electron moves in a circular orbit around a nucleus of charge $+3e$. Assuming the Bohr model of the atom is applicable to this system,

  Derive an expression for radius of $n$th Bohr orbit.
  Find the value of $n$ for which the radius is equal to the radii of first orbit of hydrogen.
  Find the wavelength of radiation emitted when revolving particle jumps from third orbit to the first.

Now, I did the first part and got the answer correct. Here's what I did.

Suppose the mass of the particle revolving is $M$, its speed is $v$, and $M = 208 m_{e}$. Electrostatic force is the centripetal force. Therefore

$$
\begin{align}
\frac{Mv^2}{r} &= \frac{(ke)(3e)}{r^2} \\
v^2 &= \frac{3ke^2}{208m_{e}r}
\end{align}
$$

From the Bohr model,

$$
m_{e}vr=\frac{nh}{2\pi}
$$

where $h$ is Planck's constant. Therefore,

$$
v = \frac{nh}{2\pi m_{e}r}
$$

Squaring it,

$$
v^2 = \frac{n^2h^2}{4(\pi)^2(m_e)^2r^2}
$$

Equating the two equations which have $v^2$ in them,

$$
\frac{n^2h^2}{4(\pi)^2(m_e)^2r^2} = \frac{3ke^2}{208m_{e}r}
$$

After solving for $r$, we get something like this,

$$
r = \frac{n^2h^2}{4(\pi)^{2}3ke^{2}208m_e}
$$

All of the above is correct. The problem is in the second and third parts; when I put $r =\pu{0.53*10^{-10} m}$ I do NOT get the required answer. To approach the third part, I started with the standard Rydberg equation,

$$
\frac{1}{\lambda} = \mathcal{R}Z^2 \left( \frac{1}{n_f^2}-\frac{1}{n_i^2} \right)
$$

I plugged in each value, $n_i = 3, n_f = 1, Z = 3$; but again didn't get the answer correct.

  The answer to the second part is 25 $(n=25)$; and to the third is 55.2 picometers.

To answer the second part:
We know $M=208m_e$, $Z=3$, $\hbar=\frac{h}{2\pi}$.
Part one has a mistake, as it is
$$
\begin{align}
&&\frac{Mv^2}{r}&=\mathcal{k_e}\cdot\frac{(\mathcal{e})(Z\mathcal{e})}{r^2}\\
&&Mvr&=n\hbar\\
\implies&& r &=\frac{n^2\hbar^2}{M\cdot\mathcal{k_e}\cdot Z \mathcal{e}^2}
\end{align}
$$
We also know the Bohr radius:
$$a_0= \mathcal{k_e}\cdot\frac{\hbar^2}{m_e\cdot\mathcal{e}^2} \approx 5{,}29 \cdot 10^{-11}\mathrm{m}$$
Therefore we can write and cancel:
$$
\begin{align}
&& r&=a_0\\
&& \frac{\color{\green}{\hbar^2}}{\color{\red}{\mathcal{k_e}}\cdot m_e\cdot\color{\navy}{\mathcal{e}^2}} &=\frac{n^2\color{\green}{\hbar^2}}{M\cdot\color{\red}{\mathcal{k_e}}\cdot Z \color{\navy}{\mathcal{e}^2}}\\
\therefore&& Z\frac{M}{m_e}&=n^2\\
\therefore&& n&=\sqrt{Z\cdot208}\approx25
\end{align}
$$
The third part:
The Rydberg Formula is given as
$$\frac{1}{\lambda_{\mathrm{vac}}} = \mathcal{R}Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
with the Rydberg $\mathcal{R}$ constant defined for a photon emitted by an electron. We'll assume that the mass of the nucleus is 7 atomic units (three protons + four neutrons). Taking into account that $m_p\approx 1836m_e$, we arrive at
$$\mathcal{R} =\frac{\mathcal{R}_\infty}{1+\frac{M}{T}} = \frac{\mathcal{R}_\infty}{1+\frac{208m_e}{7\cdot1836m_e}}$$
Now the Rydberg constant has to be modified to include the mass of the particle:
$$\mathcal{R}_\infty = \frac{M e^4}{8 c \varepsilon_0^2 h^3}=208\mathcal{R}_{m_e,\infty}$$
With $\mathcal{R}_{m_e,\infty} = 1.097\cdot 10^7~\mathrm{m^{-1}}$ (wikipedia), I got to $\lambda_\mathrm{vac}=55.6~\mathrm{pm}$.
Without taking the reduced mass into account, i.e. $\mathcal{R}\approx\mathcal{R}_\infty$ I arrived at $\lambda_\mathrm{vac}=54.8~\mathrm{pm}$.
Both values are reasonably close to the solution given.
(If the question was really about the muon, the more accurate weight ratio is 206.77 and the corresponding wavelengths 55.1 pm and 56.0 pm.)


Why is the molecular structure of water bent? Water has the formula $\ce{H2O}$ and we can draw a Lewis structure with two lone pairs on the central oxygen. As a physics student and not a chemist, I think to myself, "Okay, there are two lone pairs, they will repel each other" and so we arrive at the conclusion of bent shape.

The issue in my mind is that I don't see why the lone electron pairs have to exist on the same side of the atom. Also, wouldn't the Schrödinger equation provide an equally plausible structure for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or on the bottom of the oxygen in the Lewis structure)? If that were true, then there would be a resonance structure between the two states and we would get a linear geometry. Clearly I'm running around in circles here, please someone enlighten me!

I mean, there is a time and place for VSEPR, and this is probably as good a time as any, because all beginning chemistry students go through it. The actual model has already been explained multiple times, so I will only briefly say that according to this theory, there are four pairs of electrons around the central oxygen. In order to minimise electron-electron repulsions, these pairs adopt a tetrahedral arrangement around the oxygen. It does not matter which two are lone pairs and which two are connected to hydrogen atoms; the resulting shape is always bent.

What's worth bearing in mind (and hasn't been explained very carefully so far) is that VSEPR is a model that chemists use to predict the shape of a molecule. The truth is that there is no real way to predict the shape of a molecule, apart from solving the Schrodinger equation, which is not analytically possible for water. Everything else is an approximation to the truth. Some of these approximations are pretty accurate, such as the use of density functional theory. Some of them are extremely crude, and VSEPR falls into this category: it essentially treats electrons as classical point charges, and seeks to minimise the electrostatic repulsion between these point charges. As a physics student you should know better than to do this. Thus, while it predicts the correct result in this case, it is more in spite of the model rather than because of the model. And you should not be surprised to hear that in some slightly more complicated cases, VSEPR can predict entirely wrong outcomes.

As you learn more chemistry you will find that there are increasingly sophisticated ways of explaining molecular geometry. Most revolve around molecular orbital theory. For a qualitative method, you have Walsh diagrams which have been explained at Why does bond angle decrease in the order H2O, H2S, H2Se?. For a more rigorous method you would likely have to run some quantum chemical computations, e.g. Are the lone pairs in water equivalent?. Of course, the drawback of this is that it becomes more and more difficult to extract true chemical understanding from the numbers. Although it should also be said that you cannot extract any true chemical understanding from the VSEPR model.

What interests me more is the followup question:

  Also, wouldn't the Schrödinger equation provide an equally plausible structure for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or on the bottom of the oxygen in the Lewis structure)?

Because the Hamiltonian of the water molecule is invariant upon rotation, this means that indeed, any orientation of the water molecule is equally likely. However, this only refers to the orientation of the water molecule as a whole. It does not say anything about the internal degrees of freedom, such as the bond angle.

In the absence of any external force, the molecule is free to bend in whichever direction it likes, and most water molecules indeed do do this as they float through space or swim in a lake. But it will always be bent.

  If that were true, then there would be a resonance structure between the two states and we would get a linear geometry.

If you were to think of a single particle in a double-well potential, say something with

$$V = \begin{cases}
\infty & x < -b \\
0      & -b \le x \le -a \\
\infty & -a < x < a \\
0      & a \le x \le b \\
\infty & x > b
\end{cases}$$

then because of the symmetry of your system, in every eigenstate of your system, the expectation value of $x$ would be $\langle x \rangle = 0$. This is quite similar to your argument. In the case of water, let's set the oxygen nucleus to be at the origin. Because it can point either up or down, the expectation value of the hydrogen nucleus position along the up-down axis would be exactly level with the oxygen atom, i.e. 0. In fact, don't stop there: it can point to the left or the right, and to the front or the back. So the hydrogen nucleus has a position expectation value of exactly $(0, 0, 0)$, i.e. right inside the oxygen nucleus.

Does that mean it's actually there, though? In our contrived double-well system, it's patently impossible for the particle to be at $x = 0$, because $V = \infty$ there. If you were to measure its position, you would never find it at $x = 0$; you would only find it in the left-hand side $[-b, -a]$, or the right-hand side $[a,b]$. Just because the particle has an expectation value of $\langle x \rangle = 0$ does not mean that it is physically there, or that $x = 0$ is somehow its equilibrium state. You're confusing an expectation value with a genuine eigenstate (which is what a resonance structure is).

In exactly the same way, if you ever were to measure the properties of water (and bear in mind that practically every interaction with a water molecule is, in effect, a measurement), we would find that it is indeed always bent.


Which is more stable? CF2 (singlet) or CF2 (triplet) Generally, we know that triplet carbene is more stable than singlet. But in my book, it is written that singlet difluorocarbene is more stable than triplet difluorocarbene.
What's the reason for it?

So generally singlet is less stable as triplet has half-filled orbitals which leads to stability.

However whenever a group that can give  electrons by back donation (examples being lone pair carriers like $\ce{-OH},$ $\ce{-NH2},$ etc. are attached, they stabilise the singlet by back donation of their lone pair into the vacant orbital of the singlet carbene.

This is similar to back bonding in $\ce{BF3}.$


Sum of a rate constant gives the total decay rate I want to ask a question about the sum of a rate constant to yield the total decay rate.

From previous studies, the rate of a reaction can be expressed evidently as

$$\ce{v_{rate} = k[A]}$$

where $\ce{A}$ is a molecule in a reaction.

I studied today about unimolecular photochemistry and that a molecule in its excited state $\ce{S_1}$ can do a number of processes:

Fluoresence (emit a photon and return to its ground state), $\ce{F}$
Intersystem crossing (convert to a triplet state and then relax), $\ce{ISC}$
Internal conversion (release heat to the environment and relax back to the ground state), IC
Photochemistry (be involved in a reaction), PC

and according to the lecture material, the total decay rate can be expressed as:

$$\ce{k_F + k_{ISC} + k_{IC} + K_{PC}}$$

i.e. the sum of the rate constants.

But I thought that from my studies, the total decay rate is expressed as:

$$\ce{v_{total \ decay \ rate} = (k_F + k_{ISC} + k_{IC} + K_{PC})[A]}$$

so why do I not include the concentration term as part of the total decay rate?

My thought was that $\frac{v_{\ total \ decay \ rate}}{[A]}$ is expressed as the total decay rate and hence it can be expressed as the sum of the rate constants, but I want to check that my logic is correct.

Why is the total decay rate expressed as a sum of all the rate constants of the expressions above?

I am simply confusing two different variables here, rate and rate constants.

In this situation, the sum of the rate constants is the sum of the individual rate constants of the processes occuring in the reaction i.e.

$$\ce{k_F + k_{ISC} + k_{IC} + K_{PC}}$$

This is consistent with my current knowledge.

It should be referred to as "total decay rate constant"

Of course, the rate equation can still be written here as:

$$\ce{v_{rate} = k[A]}$$


Why does Avogadro's Law hold true for ideal gases? Avogadro's Law states that the volumes of ideal gases are proportional to the number of molecules that are present in the gas (or the number of moles in the gas). I've been trying to find out more about it, but I can't seem to find out why this holds true, at least for ideal gases.

Avogadro's Law actually states that the volumes of ideal gases at a fixed temperature and pressure are proportional to the number of molecules that are present. This is actually a sub-case of the more general ideal gas law, which states $PV = nRT$ (or $PV = nkT$, if you're working on a per-particle basis) .

The ideal gas law actually falls out of a basic statistical mechanics analysis of the behavior of a collection of molecules which interact solely by elastic collision. (Here's a 25 minute video of one such derivation.)

Basically, the more particles you have in a given region with a certain fixed speed (i.e. a fixed temperature), the more frequently they collide - not only with each other, but with the walls of the container. This means that putting more particles in the (same size) container (at the same temperature) increases the outward pressure on the container walls. In fact, the rate of collisions (and thus the pressure) is directly proportional to the number of particles (doubling the number of particles at a fixed temperature doubles the rate of collisions).

But we want to hold the pressure constant! If we can't change the temperature, one way to reduce the pressure is to increase the volume (e.g. let the container expand like a balloon.) As the volume grows, each particle has to travel farther before it can collide with the walls of the container, meaning that the rate of collisions and thus the pressure on the walls gets less and less.

It just so happens that this relationship between pressure and volume is inversely proportional (that is, Boyle's Law). So to halve the pressure, we need to double the volume. So if we double the number of particles, to keep the same pressure we need to also double the volume. Or more generally, the volume is increased in direct proportion to the number of particles - that is, each mole of gas takes up a certain volume (at a fixed temperature and pressure). If we try and put it into a different volume, the pressure (or the temperature) will change.

Okay, but why doesn't a fixed number of different particles take up different volumes? For example, why doesn't doubling the mass of the particles make it take up half the space (or double, or a quarter ...)? - Heavier particles might hit the walls of the container harder, but at any given temperature heavier particles are moving slower then lighter ones, so the collision rate is less. When you run the numbers and calculate the velocity of the particles, the rates of collision, and then how hard they hit the container walls, the mass cancels out. At a fixed temperature, the contribution to pressure from a single particle colliding with the walls of the container is independent of its mass.

You can do the same sort of analysis with other properties. They are either 1) are explicitly discounted by the assumption of "ideal gas" (e.g. no particle interactions) 2) don't factor into the calculation at all (e.g the absorption spectra of the atoms) or 3) precisely balance out - leaving you with $PV = nRT$ being the relation for an ideal gas.


How does solubility of corrosion product affects rate of corrosion? Why is it that if the corrosion product are soluble, then the rate of corrosion increases while insoluble corrosion products cause rate of corrosion to decrease?(How exactly would insoluble corrosion product protect the material?)

The process in general is called passivation. It means, that the surface of a metal is corroded and the corrosion product forms a protective layer. If the corrosion products are hardly soluble, it therefore blocks the metal beneath it from further corrosion. If the corrosion products are soluble, this protective layer can be washed away and the material is exposed to further oxidation.

In iron, for example, the initially formed iron hydroxides are somewhat soluble in water. The formed rust is flaky and friable, which basically means it is removable by little or no stress, i.e. rain can break the surface and expose more metal to the oxidising agent. It therefore usually always decomposes completely.

In another example, copper does form such a protective layer of copper oxides. This is also called patina. These layers are hard(er) to remove and protect the underlying metal from further oxidation. This is also the reason, why for example many church roofs and the Statue of Liberty are green.

This method is often used to protect metals from corrosion, i.e. the formation of the protective layer is induced chemically.


Are the following two compounds isomers or not? Are the following two compounds isomers or not?
I feel they are enantiomers(non-superimposable mirror images) however my book states that they are not isomers?
I cannot understand why?

You are right, these two molecules are enantiomers. If you rotate the bottom molecule 180 degrees, you can see that it is the mirror image of the top molecule. Enantiomers are stereoisomers, so the only reason your book would say they are not isomers is that it (unfortunately) contains a typo.


How to determine the temperature dependence of equilibrium constants? Given $K_c=2.8\times10^2$ at $727~^\circ\mathrm{C}$ for
  $$\ce{ 2SO2 + O2 <=> 2SO3},$$
  calculate $K_c$ and $K_p$ when 8 grams of $\ce{SO2}$ result in 0.32 grams
  of $\ce{O2}$ forming at equilibrium (temperature is not $727~^\circ\mathrm{C}$).

Now at first I thought this was straightforward :

determine $K_c$ at this temperature from the equilibrium conditions
apply Van 't Hoff equation to determine the temperature
calculate $K_p$ from $K_c$ and the temperature

However right after that part of the question the next part asks

  Is the new temperature higher or lower than $727~^\circ\mathrm{C}$?

This implies that it is possible to calculate $K_p$ without calculating the temperature? With the new $K$ values then provide a qualitative argument for the temperature increasing or decreasing without having to calculate it?

I agree that the second question seems a bit leading, but personally I would just calculate all quantities of interest as you laid on in your plan. The van't Hoff equation requires the reaction enthalpy change, so I assume using a tabulated value is fair game. If so, you will know whether the forward reaction is endo- or exothermic. If you calculate the equilibrium constant at the unknown temperature as you planned, you can compare it to the equilibrium constant given in the problem (for 727 C). Comparing these two equilibrium constants will tell you which way the reaction proceeded, and with the sign of the enthalpy, you should be able to deduce whether heat was added or lost... i.e. whether the new temperature is higher or lower than the initial state, 727 C (without having to go through the entire calculation).


Most effective means of removing coffee stains from carpet I have several old coffee stains on a carpet. Googling "coffee stain removal" reveals a variety of suggested cleaning substances-- vinegar, baking soda, laundry detergent, etc. I have used all of these on stains in the past to varying success, but always have a hard time totally eliminating the stain. Typically the instructions suggest iterative applications of the substance, and I usually end up with a huge pile of wet, wadded-up paper towels, each tinged slightly brown from the tiny amount of coffee absorbed.

What is the most effective way to remove old coffee stains from carpet using common household chemicals? Chemically, why is this the most effective means?

Oxidise it using a mild oxidizing agent like $\ce{H2O2}$ (hydrogen peroxide). When it undergoes oxidation coloured compounds get decolorized. Good luck on cleaning it.


What is the geometry of a tri-substituted methyl free radical? My teacher explained that free radicals could have a geometry that is planar or pyramidal(tending to tetrahedral) based on whether or not it is substituted by lone-pair containing atoms.
For example,
$\ce{(CH3)_3C}$ would exhibit sp2 hybrid geometry of planar.
But when tri-substituted with Flourine/Chlorine, it will exhibit a geometry that is tending towards tetrahedral(pyramidal) because the free radical electron in the P-orbital repels the lone-pairs on the Halogen.
It does not seem convincing because, the P-orbital is above and below the plane, then why does the electron in P-orbital repel the groups into a geometry tending towards tetrahedral?
That's like as if the electron was present only on the top and was repelling the groups down.
Here is a picture that might help visualise the p-orbital above and below the plane: https://www.google.com/imgres?imgurl=https://chemistryonline.guru/wp-content/uploads/2016/09/carbonium-1280x720.jpg&imgrefurl=https://chemistryonline.guru/carbonium-ion/&tbnid=1bNjv-81SP8QxM&vet=1&docid=MPke176Wg2AS2M&w=1280&h=720&source=sh/x/im

A priori, I would expect substituted methyl radicals to adopt a planar structure due to Bent’s rule.
However, I consider it possible for methyl radicals with electron-withdrawing substituents to adopt a trigonal-pyramidal structure too. In this case, Bent’s rule would imply that the electron-withdrawing substituents are ‘more electron-withdrawing’ than the radical is electron-poor. In such a case, it might be beneficial to distort the planar structure, causing rehybridisation and thus resulting in $\mathrm{sp}^n$ orbitals across the board so that all orbitals together add up to $\mathrm{sp^3}$. This structure may not be perfectly pseudotetrahedral but close enough to have a significant s-contribution to the singly-occupied orbital.
Note that $\mathrm{sp^3}$-type radicals are not inherently unstable (unlike $\mathrm{sp^3}$-type cations) as can be seen in bicyclo[2.2.2]octa-1-yl radicals (or bridgehead radicals in general). This points to $\mathrm{sp^3}$ radicals being accessible and thus the finer electronic environments mattering.

Addendum: Ron has shared links to two other questions he answered (1, 2) which argue the case that methyl radicals would be (slightly) pyramidal rather than planar in their ground state. So my a priori expectation might be wrong. I believe it might be very hard to prove the structure conclusively, as it is expected for a pyramidal radical to rapidly inverse as nitrogen does in ammonia and its derivatives so that any non-short timescale measurements would result in an averaged structure (which would be planar).


How to solve 1-deuterio-3-methylcyclohexene dehydrohalogenation? Ive been struggling to solve this dehydrohalogenataion excercise.
How does the E2 mechanism work in this case and how is it affected by the stereoselectivity and antiperiplanar position? How do I get to the transition state and how does the chair conformations for the cyclohexane ring look like?

As you noted, the chlorine and the other group being eliminated must be antiperiplanar.  In the case of the chair form of cyclohexane, this means that the chlorine and other group being eliminated must be trans-diaxial.  In the figure below I've drawn the 2 possible conformations (interconverted by a chair flip) for your deuterated cyclohexane.  Only in the top conformation is the chlorine axial, and the hydrogen on the adjacent (non-methylated) carbon is also positioned trans and axial.  Only this hydrogen can be eliminated as it is the only one that is axial and antiperiplanar to the chlorine.  Therefore, the only product produced when this material is treated with base and an E2 occurs, is the pictured 1-deuterio, 3-methylcyclohexene.  The hydrogen on the methylated carbon cannot be eliminated since it is never axial when the chlorine is axial.


How do water models fit into force field methods? What I Understand:

In molecular mechanics, it is recognised that molecules tend to be composed of units that are structurally similar in a vast array of environments so atom types are defined from which molecular models can be built.

A force field is defined that describes the energy required to distort the molecule from equilibrium.[1]

$$E_\mathrm{FF} = E_\mathrm{str} + E_\mathrm{bend} + E_\mathrm{tors} + E_\mathrm{vdw} + E_\mathrm{el} + E_\mathrm{cross}$$

Once a functional form has been decided on, the field is parameterised based on empirical results.

What I Don't:

A large array of force fields have been developed, each different in their functional form or parameterisation. Numerous "water models" exist, and these appear to be force fields in their own right, often with far fewer energy terms.

The SPC model, for example, appears to contain only Coloumbic and Lennard-Jones terms, as described in Berendsen.[2]

  We developed a three—point charge model (on hydrogen and oxygen
  positions) with a Lennard— Jones 6—12 potential on the oxygen
  positions only.

What is the purpose of these water models?

Question:

Why are standalone water models required, and how are they incorporated into more general forcefields like MM2?

References

F. Jensen, Introduction to Computational Chemistry, John Wiley &  Sons, Chichester, 2007.
Berendsen, H. J. C., Postma, J. P. M., van Gunsteren, W. F.,
and Hermans, J., Interaction Models for Water in Relation to Protein Hydration, in: Intermolecular Forces: Proceedings of the Fourteenth Jerusalem Symposium on Quantum Chemistry and Biochemistry Held in Jerusalem, Israel, April 13--16, 1981. DOI: 10.1007/978-94-015-7658-1_21; pdf mirror

I guess for starters, try finding a torsion angle on a 3 point molecule.

Is it fast enough, accurate enough?

The real issue is why do some force-fields have intra-molecular bonds and others do not.

Including bond and angle forces in water models make their computation more expensive. There are classes of flexible forcefields for water. There are even more indepth forcefields that are polarizable.

Why don't we use the most sophisticated forcefield models for water?

The more sophisticated the model, the longer our simulations take. Often, rigid water models are good enough for certain properties. So, we use the simplest model that gives us the right answer for the property we are after. This is just a general good rule.

Parameters

Force-fields use parameters, and proteins use ALOT of parameters. These parameters are often fit for a specific water model. Hence, you are kind of stuck using that water model if you are using that Force-Field for the protein/solute molecule.

Complexity

As our interests move towards more involved calculations that involve water it does become important to include more effects such as flexibility and polarizability. This is increasingly the case in calculating Free Energies, whether it is the Free Energy of hydration or the Free Energy of binding a ligand to a protein. Another example is modelling supercritical water - only polarizable models have been successful. Rigid models such as SPC/E and TIP3P simply are not good enough for many properties under many conditions. You can of course fudge parameters to make anything fit a particular problem, but we want to use generalizable force-fields as much as possible. In this case, models with more parameters (flexible, polarisable) give us additional accuracy for the difficult problems.

Bandwagon

Another reason many people use one model more than another is because their colleagues use it, so they use it because other people use it and it likely won't be questioned since other people use it. This is not a good reason for using a certain model! usually the first people to use a certain model had a reason, it solved their problem. However often the following bandwagon users don't read the strength/weaknesses of the model but instead just use it because they see it used lots for seemingly similar areas of research. Welcome to molecular modelling.

Mixing Rules

Generally atoms have dispersion and electrostatic terms. In the case of dispersion, when the force or energy between two atoms is being calculated a single parameter for each term is required for the calculation. Take the well known Lennard-Jones for instance

$$U(r_{ij}) = 4 \epsilon_{ij} \left[ \left( \frac{\sigma_{ij}}{r_{ij}}\right)^{12} - \left( \frac{\sigma_{ij}}{r_{ij}}\right)^6 \right]$$

different forcefields may have different numeric values for $\epsilon_i$, $\epsilon_j$, $\sigma_i$,$\sigma_j$ but they will all have a value. You need to use mixing rules to take the respective $i$ and $j$ parameters and create an $\epsilon_{ij}$ and $\sigma_{ij}$. We call these mixing rules a.k.a. combining rules. Wikipedia lists several of the common ones. https://en.wikipedia.org/wiki/Combining_rules

For instance for the Lorentz-Berthelot
$$\epsilon_{ij} = \sqrt{\epsilon_{i} \epsilon_{j} }$$
and
$$\sigma_{ij} = \frac{\sigma_i + \sigma_j}{2}$$

Finally, in SPC and SPC/E the hydrogens do not have Lennard-Jones parameters. It is not uncommon for hydrogens to have no dispersion parameters. This can lead to problems though, since dispersion parameters keep other atoms/particles from getting too close.

So a specific example of an SPC water molecule interacting with a Sodium ion, the electrostatics just needs to know that the Oxygen has a negative charge of $q_o =−0.82$ and the sodium has a charge of $q_{N_a} =+1$. A naive approach would be to treat this as a bare coulomb interaction which is just

$$ U(r_{ij})^{\rm electrostatic} = k \frac{q_o q_{N_a}} {r_{ij}}$$

where

$$k = \frac{e^2}{4 \pi \beta}$$

$e$ is the charge of an electron and $\beta$ is the permittivity and a global and constant value for the simulation.

A better approach is to use EWALD's summation for electrostatics. It is more complicated, but the only charge specific parameters it needs to know is still just $q_o$ and $q_{N_a}$. The other parameters it uses for damping can be considered global parameters. Note: also do this for each of the hydrogen charges interacting with the Sodium ion.

For Lennard-Jones you would need to use mixing rules for the oxygen (there is no LJ parameters for the hydrogens)

$$\sigma_{(o-N_a)} = \frac{\sigma_o + \sigma_{N_a}}{2}$$

and

$$\epsilon_{(o-N_a)} = \sqrt{\epsilon_o \epsilon_{Na}}$$

then do the standard LJ calculation for force and or potential energy.


Balancing the chemical equation using Oxidation Number Method Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.

Please balance the equation using the oxidation number method. The skeletal ionic equation is:

$$\ce{MnO4-(aq) + Br-(aq) -> MnO2(s) + BrO3-(aq)}$$

This question is asked in Problem 8.9, NCERT Textbook Class XI, Pg. no. 267. My answer to this question is different from the answer given in the book. I'd love to know the correct answer to this question.

It's hard to pinpoint what the problem here is since I don't have the mentioned book to compare an answer with.
The simplest way is write down half-reactions for reduction (red) and oxidation (ox) processes once you've assigned oxidation numbers (denoted above the symbols of the elements which are participating in a redox reaction), then balance the number of the transferred electrons and the algebraic sum of both half-reactions will yield in a complete balanced redox reaction:

$$
\begin{align}
\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- &→ \overset{+4}{Mn}O2 + 4 OH-} &\quad |\cdot 2 \tag{red}\\
\ce{\overset{-1}{Br}^- + 6 OH- &→ \overset{+5}{Br}O3- + 3 H2O + 6 e-} &\quad \tag{ox}\\
\hline
\ce{2 MnO4- + Br- + H2O &→ 2 MnO2 + BrO3- + 2 OH-} \tag{redox}
\end{align}
$$

As a sanity check, you can see that there are no protons $\ce{H+}$ participating in either of the reactions, and the resulting medium is indeed basic (slight excess of $\ce{OH-}$), which is in a good agreement with the conditions mentioned in the problem.


Electronic configuration and the periodic table I came across the following question:

  If each orbital can hold a max. of 3 e–  what is the number of elements in the 4th period of the periodic table?

I was unable to even start the thought process of answering the question because I'm rather horrible at discerning the period number using the electronic configuration, the number of elements that can be accommodated in a period etc. If someone could please explain it to me, that would be so great, especially since I've googled this too many times already and haven't been able to wrap my head around it yet!

Thanks in advance :) Cheers!

In the 4th period of the periodic table there is 1 s orbital, 3 p orbitals, and 5 d orbitals. That's a total of 9 orbitals. $9 \times 3=27$

That's 27 elements.


Reducing pressure in a binary liquid solution Consider the following diagram for a binary liquid system comprised of A and B, where A is the more volatile component.

My professor says

Consider an isopleth from $a$ to $a_4$. At the start, the solution is in pure liquid phase at state $a$.
As we reduce the pressure, the solution starts to boil at $a_1$. The composition of the liquid solution is denoted by the point $a_1$ which is the same as the overall composition of the solution since the solution has only just started to boil. The composition of the first trace of vapor formed is $a_1'$.
As the we reduce the pressure, the solution continues to boil to reach point $a_2''$. We may draw a tie line to find the composition of the liquid $a_2$ and the composition of the vapor $a_2'$. At every pressure where the solution exists as a mixture of liquid and vapor, we may use a tie line to find out its composition in the liquid and vapor phase.
Upon reaching $a_3'$, the solution has just been completely changed into vapor phase and the last trace of liquid has the composition $a_3$. After all the solution is vaporized at $a_4$, the composition of the vapor is again the overall composition of the solution.

I am slightly confused about we are able to make the pressure drop from $a_1$ to $a_3'$. My reasoning is: since the liquid and vapor are in equilibrium, shouldn't enough liquid keep on evaporating to maintain a constant vapor pressure?

Consider a system with a variable volume (e.g., a syringe, but on a larger scale). As the volume increases, liquid becomes vapor without changing the amount of material present.

In a system where there is only one component, changing the volume of the container will change the fraction of material in the vapor vs the amount of material in the liquid. But as you concluded, the vapor pressure will remain constant (assume constant temperature, since it was not mentioned), because there is an equilibrium between liquid and vapor for the one material in each phase.

However, when there are two components of different volatility, the vapor composition varies, depending on how much vapor there is, compared to the fraction left as liquid. So the equilibrium is no longer between a vapor and a liquid of the same composition.

It might be easier to visualize the change going on in the liquid/vapor if you change the dimension of the vertical axis to read 1/Volume. Then the vapor over the solution is (intuitively) richer in A, always (again, intuitively), until the liquid is all gone, when the vapor attains the original composition. Then, BTW, you realize that the pressure was changing while you changed the volume, and pressure might be a more relevant parameter.


The algebraic problem in the equilibrium of weak acids and bases I have had some trouble verifying the systematic treatment in a weak acid base weak balance. The great equation I must arrive at is this:

and I just can't see clearly how to get that equation from these:

This sort of derivation is a just painful, but straightforward. As Poutnik mentioned it is just a series of equations where you eliminate variables by substitution.
So start with the basics.

$\ce{HA <=> H+ + A-}\tag{A}$
$\ce{BOH <=> B+ + OH-}\tag{B}$
$\ce{H2O <=> H+ + OH-}\tag{C}$
Given:

$\mathrm{C_A}$ is the concentration of the HA reagent
$\mathrm{V_A}$ is the volume of the HA reagent added to the solution
$\mathrm{C_B}$ is the concentration of the BOH reagent
$\mathrm{V_A}$ is the volume of the BOH reagent added to the solution

So there are six unknowns: $\ce{[HA], [A-], [BOH], [B+], [H+]}$ and $\ce{[OH-]}$
Thus we need 6 independent equations to solve the problem.
Step 1
$\mathrm{K_a} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}\tag{1-1}$
$\mathrm{K_b} = \dfrac{\ce{[B+][OH-]}}{\ce{[BOH]}}\tag{2-1}$
$\mathrm{K_w} = \ce{[H+][OH-]}\tag{3-1}$
$\ce{[H+] + [B+] = [OH-] + [A-]}\tag{4-1}$
$\ce{([HA] + [A-])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_A\cdot C_A}}\tag{5-1}$
$\ce{([BOH] + [B+])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_B\cdot C_B}}\tag{6-1}$
So now for the 6 equations just eliminate equations one at a time.
Step 2
So for example we want to know the final equation to only have the unknown $\ce{[H+]}$.  Rearrange equation (3-1) to:
$\ce{[OH-]} = \dfrac{\mathrm{K_w}}{\ce{[H+]}}\tag{3-2}$
Now in equations (1-1), (2-1), (4-1), (5-1), and (6-1) substitute for $\ce{[OH-]}$ and equation (3-1) is eliminated.  So now you're left with 5 equations and 5 unknowns.
$\mathrm{K_a} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}\tag{1-2}$
$\mathrm{K_b} = \dfrac{\ce{[B+]\cdot \mathrm{K_w}}}{\ce{[BOH][H+]}}\tag{2-2}$
$\ce{[H+] + [B+] = \dfrac{\mathrm{K_w}}{\ce{[H+]}} + [A-]}\tag{4-2}$
$\ce{([HA] + [A-])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_A\cdot C_A}}\tag{5-2}$
$\ce{([BOH] + [B+])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_B\cdot C_B}}\tag{6-2}$
Obviously as you continue it gets more and more tedious, but the process is straightforward.

The point is that you have 6 unknowns and 7 equations below. So all of the equations can't be independent. With a problem that is going to be this tedious I like to write out everything explicitly so that:
(1) I don't get confused
(2) So that I can check my work step by step.


In a condensation reaction, which of the reactants loses their OH group? You have Propanoic Acid ($\ce{CH3CH2CO-O-H}$) so an $\ce{OH}$ on the end
You have Ethanol ($\ce{CH3CH2-OH}$) so an $\ce{OH}$ on the end

When you combine the two for a condensation reaction you have the $\ce{OH}$ group from 1 reactant and the $\ce H$ group from the other that joins to release a water molecule correct?

So which of these groups loses its $\ce{OH}$ group and which group loses its $\ce H$ group?

How can i determine this is im guessing it may have something to do with strength of bonding and shielding but can somebody please explain i cant find the answer

Acid loses its OH and alcohol loses its H, that's the way it is.

Long story short, oxygen from alcohol's $\ce{-OH}$ attacks the carbon from the acid's $\ce{-COOH}$. That carbon, being linked to two oxygens, carries significant positive charge, so this is much more likely than the other way around.

Also, it has been proven rigorously with the use of isotope-labeled compounds. Say, you take some acid with "normal" oxygen ($^{16}\ce O$) and alcohol with $^{18}\ce O$, and behold, the ester comes out with one $^{16}\ce O$ and one $^{18}\ce O$, so the latter must have come from alcohol, so it is the oxygen from acid that was lost in the form of water.


How to tell the difference between octane-1-ol and oct-1-ene? How to tell the difference between octane-1-ol and oct-1-ene? The things I can do are:

check their solubility in water
use bromine water
use $\ce{KMnO4}$
use sulfuric acid

I assumed that checking the solubility would be a good idea because of the presence of an alcohol but will I notice a difference since the alcohol has 8 carbon atoms. And they all react with bromine water, $\ce{KMnO4}$ and $\ce{H2SO4}$. I think I'm missing something.

Alkene is best characterized, then, by its property
of decolorizing both a solution of bromine in carbon tetrachloride  and
a cold, dilute, neutral permanganate solution (the Baeyer test). Both
tests are easily carried out; in one, a red color disappears, and in the other, a purple color disappears and is replaced by brown manganese dioxide.


Enthalpy of neutralization of strong acid and strong base differing from enthalpy of formation of water from ions If the neutralization between strong acid and bases has enthalpy of neutralization of around $\pu{-57.1 kJ mol^-1}$, why is it that when I try to calculate the enthalpy of formation of water from $\ce{H+}$ and $\ce{OH-}$ ions I get approximately $\pu{-55.8 kJ mol^-1}$ using the reference values?

Why is there this difference in enthalpy if the reaction between say $\ce{NaOH}$ and $\ce{HCl}$ is only between $\ce{H+}$ and $\ce{OH-}$ ions, while $\ce{Na}$ and $\ce{Cl}$ remain dissociated?

Calculations: From the Wikipedia page, the standard enthalpies of formation are:

$$\begin{align}
\ce{H2(g) + 1/2 O2(g) &-> H2O(l)} &\quad \Delta_\mathrm{f,1}H &= \pu{-285.8 kJ mol-1}\tag{1}\\
\ce{1/2 H2(g) &-> H+(aq) + e-} &\quad \Delta_\mathrm{f,2}H &= 0 \tag{2}\\
\ce{1/2 H2(g) + 1/2 O2(g) + e- &-> OH-(aq)}&\quad \Delta_\mathrm{f,3}H &= \pu{-230 kJ mol-1} \tag{3}
\end{align}$$

Computing $\Delta_\mathrm{f,1}H - \Delta_\mathrm{f,2}H - \Delta_\mathrm{f,3}H$ gives us the reaction $\ce{H+ (aq) + OH- (aq) -> H2O (l)}$ with the associated enthalpy change $\Delta_\mathrm{f}H(\ce{H2O(l)}) = \pu{-55.8 kJ mol-1}.$

-55.8 is pretty close -57.1 kJ/mol. There could be other reasons, but remember that:

1) The enthalpy of neutralization was measured colorimetrically. Different authors reporting different results. I personally only trust articles from the late 80's and onwards since accurate solution calorimeters were not available prior to that era. It is typical to see +- 2kJ/mol difference.

2) many use a single concentration of acid and bases to be reacted and repeat the experiment multiple times to get and average value. What I do instead, I do multiple experiments in incremental amounts of acid to be titrated on excess of base. This way, you're changing one variable and probing against the change in that variable only (the amount). You will get a slope of J/mol in the best statistical way: linear regression.

2) Many authors are not careful about using only dilute solutions and correcting for the enthalpy of dilution when titrating in a calorimeter

3) Hess's law method combines all the errors in the heats of formation


What would chemistry seniors like to hear about industrial hygiene? I have a master degree in physical chemistry with close to 10 years experience as a chromatographer in pharma.  About two years ago I made a career change into industrial hygiene.

Right now I am prepping to give a senior seminar talk about my professional experiences as an industrial hygienist at the college where I earned my BS in chemistry.  The talk will be half lecture and half lab - I'll be bringing in some equipment to demonstrate.

What would chemistry students be particularly interested in hearing about from someone in industry?

To keep this answer as far from opinionated as possible, I'd suggest some of the W's:

What is industrial hygiene? (I've never heard of the term before this post and actually had to look it up myself.)
How do you do industrial hygiene? Procedures, risk assessment...
Why is industrial hygiene important? You might give examples where disaster was averted thanks to IH, and an example where it would have helped if done right.
Where is industrial hygiene applied? (I guess the office building will have less rigorous guidelines than the production facility.)
When did industrial hygiene start to be a thing? A small historical overview could be given, you could even give a who to this one if it was some distinct person who came up with the idea.

Always good is a personal anecdote, something you witnessed or someone (whom you trust as a source) told you, highlighting the bullet points above.


Why doesn't water burn? Hydrogen is flammable, and for any fire to burn it needs oxygen. Why does a compound made of hydrogen and oxygen put out fires instead of catalyzing them? I understand that hydrogen and water are chemically different compounds, but what causes water to be non-flammable?

You can think of water as the ash from burning hydrogen: it's already given off as much energy as possible from reacting hydrogen with oxygen.

You can, however, still burn it.  You just need an even stronger oxidizer than oxygen.  There aren't many of them, but fluorine will work,

$$
\ce{2F2 + 2H2O -> 4HF + O2}
$$

as will chlorine trifluoride:

$$
\ce{ClF3 + 2H2O -> 3HF + HCl + O2}
$$


Are matrices a good way to do stoichiometry? I was never taught in school to use matrices to balance my equation in stoichiometry. But I recently started to review my chemistry and, as I have taken linear algebra, I was suddenly struck by the idea this would be a really efficient way to balance equations. It might save me from making a lot of errors.

Are matrices a common tool for chemists when balancing equations? Does anyone have a good reference for learning about this?

Sure, most of the equation-balancing Javascript calculators on the web use that method. It's overkill to balance a single equation that way, but if you're automating the process or if you have a lot of equations to balance and check it can be convenient.

If you search on "balancing equations matrix" you'll find many simple matrix inversion  approaches for balancing equations that fail if you don't have a single unique solution.  Here's one that's more general:

L. R. Thorne, An Innovative Approach to Balancing Chemical-Reaction Equations: A
Simplified Matrix-Inversion Technique for Determining The Matrix Null
Space, Chem. Educator 2010, 15, 304–308

The basic idea is that coefficients in balanced equations are found in the nullspace of the formula composition matrix (that is, the coefficients are solutions $\bf x$ of the matrix equation $\bf C x = 0$, if $\bf C$ is the formula composition matrix). For example, if I want to find a balanced equation that relates $\rm HCl$, $\rm K_2Cr_2O_7$, $\rm FeCl_2$, $\rm CrCl_3$, $\rm H_2O$, $\rm FeCl_3$, and $\rm KCl$, I can build a formula composition table

$$\begin{matrix} & \ce{HCl} & \ce{K2Cr2O7} & \ce{FeCl2} & \ce{CrCl3} & \ce{H2O} & \ce{FeCl3} & \ce{KCl}\cr H & 1 & 0 & 0 & 0 & 2 & 0 & 0\cr Cl & 1 & 0 & 2 & 3 & 0 & 3 & 1\cr K & 0 & 2 & 0 & 0 & 0 & 0 & 1\cr Cr & 0 & 2 & 0 & 1 & 0 & 0 & 0\cr O & 0 & 7 & 0 & 0 & 1 & 0 & 0\cr Fe & 0 & 0 & 1 & 0 & 0 & 1 & 0\end{matrix}$$

I can find the nullspace vectors using the multistep procedure outlined in that paper, or I can do it very simply in Maxima:

C : matrix([1,0,0,0,2,0,0],
           [1,0,2,3,0,3,1],
           [0,2,0,0,0,0,1],
           [0,2,0,1,0,0,0],
           [0,7,0,0,1,0,0],
           [0,0,1,0,0,1,0])$
nullspace(C);

$$\mathrm{span}\left( \begin{bmatrix}14\cr 1\cr 6\cr -2\cr -7\cr -6\cr -2\end{bmatrix}\right)$$

which corresponds to the equation

$$\ce{ 2CrCl3 + 7H2O + 6FeCl3 + 2KCl\ =\ 14HCl + K2Cr2O7 + 6FeCl2}$$

As chemists, we'd write that equation reversed,

$$\ce{ 14HCl + K2Cr2O7 + 6FeCl2\ =\ 2CrCl3 + 7H2O + 6FeCl3 + 2KCl }$$

Here are a few more nails you can pound with your linear algebra hammer: you can determine the minimum number of chemical equations required in a reaction mechanism, or discover how many equations in a reaction mechanism are stoichiometrically independent. For example, consider the following halogen/hydrogen reactions:

$$\begin{array}{rcl}
\rm H_2 + X_2&\xrightarrow{k_1}&\rm 2HX\\
\rm H_2 + X_2&\xrightarrow{k_2}&\rm 2X + H_2\\
\rm H_2 + X_2&\xrightarrow{k_3}&\rm 2H + X_2\\
\rm H + X_2&\xrightarrow{k_4}&\rm X + HX\\
\rm H_2 + X&\xrightarrow{k_5}&\rm H + HX\\
\rm HX + H&\xrightarrow{k_6}&\rm X + H_2\\
\rm H_2 + HX&\xrightarrow{k_7}&\rm X_2 + H_2\\
\rm 2HX &\xrightarrow{k_8}&\rm X_2 + H_2\\
\end{array}$$

Let's make a formula matrix for the five species in the reactions above:

$$\begin{pmatrix}{H}_{2}\cr H\cr HX\cr X\cr {X}_{2}\end{pmatrix}=\begin{pmatrix}2 & 0\cr 1 & 0\cr 1 & 1\cr 0 & 1\cr 0 & 2\end{pmatrix}\begin{pmatrix}H\cr X\end{pmatrix}$$

Any set of $S-C$ stoichiometrically independent equations is a "complete set" of reactions, where $S$ is the number of species and $C$ is the number of independent equations for the conservation of elements (components). There are five species ($S=5$) and the rank of the formula matrix is 2, so the system has two components ($C=2$). There must be $S - C = 5 - 2 = 3$ stoichiometrically independent reactions for the mechanism to be complete.

Do the reactions above constitute a complete set? We can write a stoichiometric matrix like so:

$$\begin{pmatrix}-1 & 0 & 2 & 0 & -1\cr 0 & 0 & 0 & 2 & -1\cr -1 & 2 & 0 & 0 & 0\cr 0 & -1 & 1 & 1 & -1\cr -1 & 1 & 1 & -1 & 0\cr 1 & -1 & -1 & 1 & 0\cr 0 & 0 & 0 & -2 & 1\cr 1 & 0 & -2 & 0 & 1\end{pmatrix} \begin{pmatrix}{H}_{2}\cr H\cr HX\cr X\cr {X}_{2}\end{pmatrix} = \begin{pmatrix}0\cr 0\cr 0\cr 0\cr 0\cr 0 \cr 0 \cr 0\end{pmatrix}$$

The rank of the stoichiometric matrix is 3, so the eight equations above are equivalent to three independent equations. The mechanism is complete. Whether it's correct is of course a different matter. Stoichiometric constraints don't determine a mechanism; they limit the number of reactions in a particular reaction channel, but you can have more than one channel (more than one mechanism running side-by-side).

Why is this useful---why would you want to know what the number of independent stoichiometric equations ($R$) are? Well, according to Gibb's rule of stoichiometry, $R \le S-C$. If you have $R < S-C$, that tells you that you have a restricted equilibrium: the components must be constrained by conditions other than stoichiometric equations. The restricted equilibrium composition will depend on which species are present initially in that case. There are also experimental design implications, since in a single experiment you can get only $S-C$ reaction rates. You may have to figure out how to favor one channel over another in separate experiments to get all of the rates.

TL;DR: If you're looking for applications of linear algebra in chemistry, chemical kinetics is one playground; there are many others.


Unit of time used in rate law equations If doing a problem where you use the integrated rate law for a second order reaction

($\frac{1}{[\mathrm{A}]}-\frac{1}{[\mathrm{A_0}]}=kt$) to first find $k$, then use that to find the half-life. Does it matter seconds are used, or minutes, or whatever?

The unit of $k$ is $\frac{\mathrm{concentration}}{\mathrm{unit~time}}$. It is not necessary that you use any specific unit for concentration or time, but most often $t$ is measured in seconds, and the unit of $k$ is $\mathrm{mol~L^{-1}~s^{-1}}$.


If hydrogen bonding in water was weaker, what happens to H+ ion concentration? Water ionization becomes much less evident if the hydrogen bonds are just a few percent stronger but pure water contains considerably more $\ce{H+}$ ions if they are few percent weaker.

I found this line in some article. You can even copy/paste this line on search bar of Google. I didn't understood the latter part of this statement which is "but pure water contains considerably more $\ce{H+}$ ions if they are few percent weaker". The author is indicating to which of the following three situations?

only $\ce{H+}$ ion concentration will increase
both $\ce{H+}$ and $\ce{OH-}$ will increase
$\ce{H+}$ will increase and $\ce{OH-}$ will decrease that is the equilibrium constant remains same.

The article by Martin Chaplin 'Water's Hydrogen Bond Strength' (arXiv:0706.1355 [cond-mat.soft], no evidence of peer review) is discussing how changes to the properties of water would alter life.

If the hydrogen bonding in water was stronger, there would be less ionization (less $\ce{H+}$ and less $\ce{OH-}$).

If the hydrogen bonding in water was weaker, there would be more ionization (more $\ce{H+}$ and more $\ce{OH-}$).


Why does silver (I) have a larger lattice enthalpy and hydration enthalpy than sodium? The ionic radius of the $\ce{Ag^+}$ ion is $129$ pm, and that of the $\ce{Na^+}$ ion is $116$ pm.

Since the sodium ion is smaller than the silver ion, it makes sense that it has a stronger polarizing power than the silver ion. Due to the higher charge density, the lattice energy of $\ce{NaX}$ compounds is predicted to be larger than those of $\ce{AgX}$ compounds. However, that is not the case, as shown here:

Furthermore, the enthalpy of hydration of $\ce{Na^+}$ is expected to be more exothermic than that of $\ce{Ag^+}$, because $\ce{Na^+}$ is more polarizing than $\ce{Ag^+}$. However, that is not the case:

Why are the lattice enthalpies of silver compounds larger than those of sodium compounds, and why are hydration enthalpies of silver compounds more exothermic than those of sodium compounds, despite the silver ion being larger than the sodium ion?

Sources:

R. D. Shannon (1976). "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides". Acta Crystallogr A. 32: 751–767. Bibcode:1976AcCrA..32..751S. doi:10.1107/S0567739476001551.

Atkin's Physical Chemistry

The electron configuration of sodium ion is that of the inert gas neon (1s^2 2s^2 2p^6), a spherical shell with a positive charge at the nucleus. There are 10 electrons inside this shell.

The electron configuration of silver ion is that of krypton plus 4d^10, a spherical shell with a positive charge at the center. There are 46 electrons inside this shell.

While the slightly smaller sodium ion might have slightly more polarizing power than silver ion, silver ion has far more polarizability than sodium ion. Polarizability is the distortion of the electron cloud as a result of a point charge (or dipole). Dispersion forces are a form of polarizability where two polarizable entities polarize each other. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Specific_Interactions/Polarizability

The effects of dispersion forces due to the polarizability of the anion can be seen by comparing the enthalpies. The crystal structures for the fluorides, chlorides and bromides and NaI are FCC, rock salt. AgI is wurtzite. The difference in enthalpies for the fluorides is 43, but as the polarizability of the chloride anion increases, the attraction of dispersion forces increases the difference to 125, and for bromide, 148. Even tho the crystal structure of AgI is different, it still seems to have a polarizablity-polarizability effect (difference = 181).

This is not what you would call a predictable effect, but does a good job of explaining the trend. It is not a first-order attraction, nor a second-order polarizability, but a third-order polarizability-induced polarization effect.

And very perspicacious of you to have observed this trend.


Copper metal with silver ion My book states that $\ce{2AgNO3(aq) +Cu ->Cu(NO3^-)2(aq) +2Ag(s)}$. In addition to the formula, it also shows the following image:

As in the image it shows the salmon colored copper atoms are donating an electron to the $\ce{Ag+}$ ions. In the 2nd picture on the page it shows the $\ce{Ag}$ atoms on top of the salmon colored copper atoms. I would assume that the $\ce{Cu}$ and $\ce{Ag}$ form somewhat of an ionic bond, is this correct to say? If that is the case, why does the formula not indicate a bond between the copper and silver instead the overall product is just copper nitrate plus silver. I would think it would be something like $\ce{CuAg}$?

Actually the copper and silver aren't bonded. The silver that precipitates is pure. Also, a bond between two metals could not be ionic. Ionic bonds are always been metals and non-metals. Looking at the picture again, I think what may have thrown you off is that it looks like the silver at the end is Ag+. However, the silver is actually Ag+ in the beginning of the reaction when it is in the silver nitrate. That's how it is bonded with the nitrate. When the copper donates an electron to the silver, the silver goes from having a +1 charge to having no charge. It becomes neutral, and hence it doesn't bond with the copper.


Removing sulfur in Barton-Kellogg reaction without triphenylphosphine I'm trying to understand how the sulfur was removed in the following reaction, taken from a paper by Ben Feringa:1

After some research I found out that this is a Barton-Kellogg reaction. Based on this I went through the mechanism as follows:

I'm stuck on the final step of how to remove the sulfur. I looked over multiple papers and all of them used $\ce{PPh3}$ to open the episulfide but the reaction from the paper I'm considering does not have $\ce{PPh3}$. I looked at all the sources and supporting information the paper had linked but couldn't find anything to explain how the double bond forms. My only guess is that something happens when they purify it using $\ce{SiO2}$ flash column chromatography. Here is the excerpt from the SI (L2 is the alkene product):

  Freshly prepared diazafluorene 1 ($\pu{350 mg}$, $\pu{1.80 mmol}$) and thioketone (±)-2 ($\pu{383 mg}$, $\pu{1.80 mmol}$) in THF ($\pu{10 mL}$) were heated at reflux for 14 h under a $\ce{N2}$ atmosphere. The solution was cooled to rt, concentrated and purified by FC ($\ce{SiO2; EtOAc}$) to afford (±)-L2 ($\pu{327 mg}$, $52\%$) as yellow solid [...]

I would like any suggestions on what people think may be happening.

Reference

Wezenberg, S. J.; Chen, K.-Y.; Feringa, B. L. Visible-Light-Driven Photoisomerization and Increased Rotation Speed of a Molecular Motor Acting as a Ligand in a Ruthenium(II) Complex. Angew. Chem. Int. Ed. 2015, 54 (39), 11457–11461. DOI: 10.1002/anie.201505781.

After contacting the authors of the paper, one author, Prof. Wezenberg said that a conclusive mechanism is not yet available for this case but he believes that the unstable episulfide undergoes desulfurization by itself under reflux.


Action of KMno4/H+ and heat in this reaction The first step uses LAH which is a hydride donor. So H- attacks any one carbonyl carbon and forms an aldehyde and acid attached to the ring.

Then Conc. KMno4 is used. As far as i know, it cleaves c=c and oxidises them into acids.
 The third step then says heat, so it leads to the formation of the anhydride on both sides.

However, this product has nothing to do with the correct answer which is option (A) . Somehow the product formed consists of 2 seprate 5 membered rings. Any idea on where i went wrong?

edit::

The $\ce{LiAlH4}$ will reduce the cyclic anhydride to the diol.

$\ce{KMnO4}$ re-oxidises both the alcohol groups (diol formed) to carboxylic acids and also oxidises the double bond to the diacid so the product you have drawn is correct, but it is an intermediate. You then have to consider the effect of heating this (to a pretty high temperature, I would assume).

Product A Heating causes a double dehydration to produce the di-anhydride shown but the right hand pair of acids do not recombine, instead the top acids combine and the bottom acids combine.

Product B Heating dehydrates the right hand diacid to the anhydride, the left hand diacid does a Claisen-ester-type thermal condensation followed by decarboxylation to give a ketone in  the left hand ring.


What happens when heating water in a pressure vessel? Suppose that I take a pressure vessel that can withstand up to 2000 PSI. Now I fill it with water (not completely, but around 90%) and start to heat it with a direct flame. What I want to know is, what will happen?

Although I don't know anything about it, here are some guesses that I think might happen:

The water will begin to convert into gas. But since the volume is limited, therefore the pressure would increase (and keep on increasing) and the vessel may burst after the internal pressure crosses 2000PSI.
The water won't convert into gas and would just take all the heat energy into itself. In other words, nothing would happen.

Those are my guesses and it would be great if you can help me out.

Look at the phase diagram of water to see what happens. Josiah Willard Gibbs invented these diagrams o help understand what happens as a substance is heated under pressure, and transitions from solid to liquid to gas and supercritical fluid (neither quite gas nor liquid).

Given sufficient heat, the vessel will burst. Above ~650 K, the water can no longer be liquefied by pressure, and behaves more like an ideal gas, so pressure continues to increase.


Why are noble gases used in "neon" lamps Neon lamps are lamps that contain noble gases... They light due to the presence of energetic levels for electrons (according to the definition of it in books). But I don't understand yet why noble gases are used in such lamps. Any help please?

Strictly speaking "neon" lamps contain neon but the term is often used colloquially for a whole range of coloured lighting probably because the red neon tubes are one of the commonest.

But it isn't just noble gases that are used, though they are the most common. Carbon dioxide is sometimes used. And there are many discharge lamp that add metals or metal salts, though these are more common in other uses such as high intensity street lighting.

The noble gases and mixtures of them are commonly used in coloured advertising lighting because they offer a range of colours in a simple low-pressure discharge lamp without the added complexity of fluorescent coatings (as in common fluorescent tubes) or high-pressure systems (as is required in high intensity systems used in street or industrial lighting). Advertising signage often consists of long glass tubes in custom shapes and the simplicity of simple gas discharge makes the results cheaper than more sophisticated alternatives.

The colours are derived from the energy levels of the electrons in the gases and it just so happens that the range of those energy levels in noble gases correspond to a useful variety of colours.

There is a good summary of the colours (and other types of discharge lamps) in this Wiki article.


Why should alkylation of enols/carbonyl compounds be done at low temperature with a strong base? Alkylation of acetone:

This reaction is done with a strong base (LDA in this case). Is this because the protons aren't particularly acidic? Secondly it should be done at low temperatures; why is this so? -78 degrees Celsius is typical.

Why should alkylation of enols/carbonyl compounds be done at low
  temperature with a strong base?

If you start with an unsymmetrical ketone, then there are two different sides of the molecule you could remove a proton from. Let's use methyl ethyl ketone as an example.  Removing a proton from the methyl group leads to what is termed the kinetically favored anion (e.g. this hydrogen is the most acidic and the corresponding enolate anion is the fastest formed anion).  Removing a proton from the methylene group on the ethyl side leads to the thermodynamically favored anion (e.g the enolate anion with the most highly substituted [most stable] double bond).

Alkylation of carbonyls/enols is not always done using strong base and low temperature.  Strong base and low temperature generally leads to the kinetically favored enolate anion, while weaker bases and higher temperatures generally lead to the thermodynamically favored enolate anion.

In your case, acetone is symmetric so it doesn't really matter which side you alkylate.  But, once you form your desired product, methyl ethyl ketone, it can react further to give polyalkylated products.  These secondary products are more stable and thermodynamically preferred.  To reduce the likelihood of thermodynamic control leading to these undesired products, the reaction is run under kinetic control - strong base and low temperature - to maximize formation of the monoalkylated product.  Here is an earlier answer that further highlights kinetic vs. thermodynamic control in alkylation reactions.


Are crown ethers "recycled", typically? I know that crown ethers have the interesting property of strongly complexing alkali metals, which allows them to be used in organic solvents (eg. as "dissolving-activity enhancing agents" for potassium permanganate or potassium tert-butylate).

Having never used such molecules myself, I was wondering: is there a procedure employed in order to remove the complexed metal, hence allowing a recycling of that ether?

I am sorry if the question might be trivial, but I couldn't find an aswer online.

Crown ethers are typically not recycled because they remain in the organic phase where normally also all the byproducts are (so the purification is very complicated) and they are still complexing the alkali metal after the reaction.
You want to recycle them to again complex some alkali metal, but then you have first to break down the previous complex, which is not straightforward.

To recycle crown ethers you must use a different approach.
The trick is to immobilize them on some solid support.
Typically, you have some crown ether functionalized monomer and polymerize it, or you have some reactive polymer and introduce the crown ether at the end.
A review can be found here.

These solid phase crown ethers are used in the same way as the better known ion exchange resins.
The problem is that you will have to synthesize them, because they are not readily available for purchase.


Is adding the word "solution" necessary when listing the materials used in a synthesis? I am editing a chemistry-related paper and the synthesis procedure described is as follows:

  Then, 10 mL of the intermediate were mixed with 10 mL of 1% (w/v)
  VCPL and 2 mL of 0.1% (w/v) KPS.

I was wondering if one should add the word "solution" after the name of the ingredients as follows:

  Then, 10 mL of the intermediate were mixed with 15 mL of a 2% (w/v)
  VCPL solution and 3 mL of a 0.2% (w/v) KPT solution.

I think that indicating a volume and a % (w/v) gives it away that you are referring to a solution containing the active ingredient, but should the word "solution" still be added?

Thanks.

No, the repeated addtion of the word "solution" is not needed.

All you want here is to provide a measured quantity (e.g., volume), concentration of your reagent in the solvent, and inferred quantity (e.g., mmol).  This may be reported in the following pattern, too:

  "1-Cinnamyl-4-methylbenzene (4a): The compound 4a was synthesized from
  1-methoxy-4-{[(2 E)-3-phenylprop-2-en-1-yl]oxy}benzene 2b ($\pu{240.9 mg}$,
  $\pu{1.00 mmol}$) with p-tolylmagnesium bromide 3a ($\pu{1.4 mL}$, $\pu{1.1 M}$ in THF, $\pu{1.5 mmol}$) in the presence of 1b ($\pu{4.1 mg}$, $\pu{1 mol\%}$) in $\ce{Et2O}$ ($\pu{5 mL}$) at room temperature for $\pu{24 h}$ ($\pu{198.0 mg}$, $\pu{95\%}$ yield, colorless liquid)."

(source: Hashimoto et al. in Molecules 2019, 24, 2296 (doi.org/10.3390/molecules24122296), section 3.2; open access publication.)

This pattern is seen for inorganic reagents like $\ce{HCl}$ dissolved in water, yet equally available in methanol, 1,4-dioxane, etc. (example); and organometallic reagents, e.g. butyllithium, methylmagnesium bromide.


Preparation of a fluoride solution using CaF2 I need to prepare several solutions with different concentrations of $\ce{F-}$, in order to carry out some investigations on fluoridated water. Would this be achievable by dissolving calcium fluoride, $\ce{CaF2}$, in water? Wikipedia writes that $\ce{CaF2}$ is insoluble in water, so I am not sure.

Calcium fluoride has a $K_\mathrm{sp}=3.9\times 10^{-11}$. That means a saturated solution of calcium fluoride will have a fluoride concentration of $4.28\times 10^{-4}\mathrm{\ M}$, which, assuming the density of this dilute solution is approximately that of water, is $\mathrm{8.13\ mg/L}$ fluoride.

Fluoridated water has fluoride concentration ranging from $\mathrm{0.5\ mg/L}$ to $\mathrm{1.5\ mg/L}$, so your saturated calcium fluoride solution is right in that range. If you need to test lower concentrations, dilute your solution. If you need to test higher concentrations, you are out of luck. You cannot make more concentrated fluoride solutions with calcium fluoride.


Why do distilled alcoholic beverages taste different from each other? I was wondering about vodka vs. whiskey mostly. The way I understand it, a distilled beverage is created by heating a non-distilled beverage, say, beer, and then collecting and cooling the vapors.

Why would these vapors contain anything but alcohol and some water? Certainly the impurities that color, say, whiskey do not vaporize at the low temperatures involved? Why don't all distilled beverages taste and look like vodka (i.e. water and alcohol)?

Two different aspects come into play here. For once, most of the whiskey aroma stems from aging the fresh destillate in oak barrels previously used to store wines, e.g. sherry. Colour an aroma thus, at least in part, stem from a  slow extraction process using the fresh destillate as a solvent.

But thinking in other beverages, particular the clear fruit spirits, there's more in it. The key to a beverage that doesn't smell and taste just like ethanol but has a cherry aroma is the sloppyness of the destillation :) In the lab, a destillation or rectification is performed to isolate a particular compound from a mixture at a sharp boiling point.

In the production of alcoholic beverages, the destillation typically aims to obtain three fractions over different boiling point ranges.

At lower boiling points, methanol, acetone and ethylacetate are removed.
The upper range mostly furnishes long chain alcohols.

These two fractions aren't meant to be digested ;)

The central fraction, however, will contains all the fragrances, ethanol and water.


Why does increasing the pressure cause ice to melt more? This question was given under chemical equilibrium.

  Why does increasing the pressure on this system cause ice to melt more?
  $$\ce{H2O(s) <=> H2O(l)}$$

I'm pretty sure this has to be solved using Le Chatelier's principle; the hint given is that The molar volume of water is more that of ice.

My understanding of Le Chatelier's principle is that a system tries to oppose any change.

So the equilibrium must shift towards the side where the pressure decreases.

Given this hint, I fail to see the correlation between molar volumes and pressure.

At equilibrium, the change in Gibbs free energy is zero.  So, to maintain equilibrium, if the pressure increases, the temperature must change in tandem in such a way that the change in free energy of the solid is the same as that of the liquid.  We have, for either phase, $$\mathrm{d}G=-S\mathrm{d}T+V\mathrm{d}P$$So, $$\mathrm{d}(\Delta G)=-\Delta S \mathrm{d}T+\Delta V\mathrm{d}P=0$$where the deltas refer to the changes between liquid and solid.  We also have that, at equilibrium, $$\Delta S=\frac{\Delta H}{T}$$So, combining these equations gives $$\mathrm{d}T=\frac{T(V_\mathrm{l}-V_\mathrm{s})}{(H_\mathrm{l}-H_\mathrm{s})}\mathrm{d}P$$Since $V_\mathrm{l}<V_\mathrm{s}$, this equation indicates that the equilibrium temperature for the solid-liquid transition decreases as the pressure increases.

This shows that the melting temperature decreases. So if the pressure is increased, and the actual temperature is not changed, the temperature of the ice will be higher than the melting temperature, and the ice will melt.


Chemical reactions in plasma state I am wondering about chemical reactions when the reactants are in a plasma state.  Consider hydrogen and oxygen.  Heated, these would react to form water.

If hydrogen and oxygen plasma were combined, would the separation of electrons from nuclei prevent normal chemical reactions?  Or would the reaction take place in the plasma state?

Yes, chemical reactions do take place in high-temperature plasmas. In fact, these reactions present some problems in the analytical technique known as Inductively Coupled Plasma - Mass Spectroscopy (ICP-MS), which typically uses an argon plasma.

As noted in the comments, the conditions within a plasma are very different from those in a laboratory at STP, for example. But that just means that the chemical reactions that are observed are different, not that they don't exist. In ICP-MS for example, one is trying to detect the masses of very low concentrations of various elements in the presence of relatively huge quantities of argon, oxygen and hydrogen (the latter assuming that an aqueous sample is being analyzed). Species that are formed within the plasma include $\ce{Ar2+}$, $\ce{ArO+}$ and more. When one of these species has the same mass as the trace isotope of interest, it only takes a very small percentage of these products to interfere with the analysis.

This example of argon plasma can largely be extended to your specific question "If hydrogen and oxygen plasma were combined, would the separation of electrons from nuclei prevent normal chemical reactions?". Depending on the conditions of the plasma (temperature, concentrations, degree of ionization, etc.) the answer is that the reaction would only form some small degree of transient water molecules, as compared to say molecular hydrogen and oxygen gas at high temperatures or in the presence of a catalyst, which would overwhelmingly form water. In other words, the former (plasma state oxygen and hydrogen) would form trace amounts of water, and probably hydrogen peroxide and other oxygen-hydrogen species, where under "normal" conditions the formation of water would be strongly favored (thermodynamically anyway). It's just the nature and definition of a plasma that whatever species you have in your plasma will want to be broken into it's elemental constituents and the formation of molecules will be strongly unfavored.

Summary:
The conditions within a plasma are very different from conditions typically encountered in a non-plasma environment, and thus the chemical reactions are different. The formation of otherwise exotic species like $\ce{Ar2+}$ are relatively commonplace within an argon plasma for example ("common" being a relative term of course, as concentrations of $\ce{Ar2+}$ may still be relatively low, just much higher than would be seen outside of the conditions present in the plasma). Plasma conditions themselves can also vary widely, from conditions where only a small portion of the gas molecules are charged to conditions where the majority are charged. Regardless, the high energy state of a plasma tends to result in the destruction of molecules to their atomic precursors, but occasionally there are reactions to form products that would not be seen outside of a plasma.


Velocity of electrons ejected from Cl atom When neutral $\ce{Cl}$ atoms are bombarded by high energy photons, causing ejection of electrons, which subshells' ejected electrons will have highest velocity?

I chose $1s$ because I thought more energy is needed to ejected those electrons so they might have higher velocity. My second choice $3s$ is right, but I can't explain why.

It would be the $3s$ shell because less energy is required to bump it off course. Due to electron repulsion and the shielding effect, the $e-$ in the $3s$ is more prone to get dislodged. And of course because of this, it will have a higher velocity when it gets bumped off. Another concept to note: $e-$ will have more energy at higher energy levels.

A reason why it wouldn't be the $1s$ shell is because due to the strong pull on the $1s$ electron. Since it is on the first energy level, it is strongly held by the nucleus. Even it were to get dislodged it would move very slow because of the strong pull by the nucleus.


What is the trend in hydration entropy of alkali metal ions? Is there a trend in alkali metal ion hydration entropy?
There's much talk of hydration enthalpy, but I have a hard time finding information about the change in entropy.
Am I right in thinking that Li$^+$ would have a more negative $\Delta S^o_{hyd}$ than Cs$^+$ because more water ligands will be ordered around it, due to it's higher charge density?

Yes, there is. The following is a plot of the hydration enthalpy versus row in the Periodic Table. Data was extracted from the Wikipedia, not the primary source (Ref 1).
Your thinking is right. The hydration entropy is evidently negative due to structuring of water (including loss of rotational degrees of freedom), becoming less negative for the larger ions (smallest loss of entropy for the cesium cation).

References

Burgess, John (1978). Metal Ions in Solution. Chichester: Ellis Horwood. ISBN 978-0-85312-027-8.


Question about reactions that are always spontaneous and non spontaneous I wanted to ask for examples of spontaneous chemical reactions meaning that their enthalpy would be negative (exothermic) and their variation of entropy would increase, therefore the Gibbs energy would be also negative and the reactions would be spontaneous at all temperatures (?).

I was thinking about the boiling of water as the reaction feels hot and the entropy change is positive as it goes from liquid to gas (entropy increases) but then I thought that to do that, we need an external energy and so the reaction would be non-spontaneous?

Another one could be dropping mentos into a carbonated drink. I can be sure that the reaction doesn't requiere an external force to happen BUT the reaction itself doesn't feel hot (I think I saw it once when they did it in primary school but I'm assuming it isn't releasing any heat?)

I know that when enthalpy is negative and entropy is negative, it would only be spontaneous if the temperature was low enough.

For the non spontaneous reactions, I can't really think about one that has positive enthalpy and negative entropy. I CAN think of a reaction that needs a push to "occur" like taking off a stain of a shirt or even a reaction of decomposition of water into h2 and o2.

I think I'm getting something wrong.

I was thinking about the boiling of water

That is a good example of a process that goes forward at higher temperature (water boils when heated above the normal boiling point) and goes in the opposite direction at lower temperature (steam condenses below the normal boiling point). This process also depends on the pressure, but my statements are valid at atmospheric pressure.

For this process, the entropy change is positive (water has more freedom to move in the gas phase compared to the liquid phase) and the enthalpy change is positive (endothermic, it takes energy to break the hydrogen bonds between water molecules).

  Reactions that are always spontaneous

How about combustion of cyclohexane? The entropy change is positive (lots of gas molecules are formed - six carbon dioxide molecules per hexane) and the enthalpy change is negative (exothermic).

  Reactions that are always non-spontaneous

Once you are certain that a reaction is always spontaneous, it is easy to find one that is always non-spontaneous - just swap reactants and products. So according to this argument, cyclohexane and oxygen will never form from carbon dioxide and water spontaneously. That sounds reasonable.


General methods of Analysis for FTIR Results What can we say about a molecule from the position and number of bands in FTIR spectrum. I know that the more symmetric a molecule is, the less number of bands it has. What else can we say, for example about the position of the band? Does it show the frequency of vibrations and therefore the strength of the bonds?
For example, the spectrum of MoF4 has one band at 674 cm -1. XeF4 has 3 bands at 586, 291, and 161. How can we compare the structure of these materials from the above information?

Let me start by saying that the XeF$_4$ is a standard example of group theory applied to IR, which is generally treated in Uni courses, for which you can find plenty of notes on the web. For these reasons, I will not explain it, but you can rather put a bit of effort and read it yourself.

In general, the level of information you can extract depends on the nature of the molecules you analyse. For example, for organic molecules, the analysis is qualitative and limited to functional group identification. This is because usually they have a high number of atoms, thus there will be many vibrations. Since the chemical composition is quite homogeneous, all these vibrations have similar frequencies, and they overlap. Thus a quantitative analysis is difficult, and the only information you extract concerns some particular bond whose signal is isolated from the mess (usually carbonylic and carboxylic bonds, triple bonds, X-H bonds). An additional problem lies in conformational flexibility.

Now, for molecules like yours, which are small, have few vibrations, are symmetric etc, you can achieve a very detailed level of analysis. But, as I said, I will not discuss it, since you can read it yourself with a quick google search.
The basic thing to keep in mind is the frequency of oscillation of a classical harmonic oscillator, given by

$$ \omega = \sqrt{\frac{k}{\mu}} \\
\mu = \frac{m_1 m_2}{m_1 + m_2}$$

where $k$ is the force constant of the bond, and $\mu$ is its reduced mass.
So, what this tells you is that stronger bonds vibrate at a higher frequency. It also tells you that bonds involving lighter atoms vibrate at a higher frequency. Another thing that you can deduce is that bending frequencies are lower than stretching frequencies.
So, to give you a hint, what you can say immediately by looking at those frequencies you reported is that the 586 cm$^{-1}$ vibration is a stretching.


What other intermoelcular forces can exist between neutral molecules? What kind of intermolecular forces can exist between neutral molecules?

My answer was London dispersion.

My logic was this: London dispersion is the only thing weaker than dipole-dipole, and as far as I am concerned two neutral molecules can't be attraced by dipole-dipole forces because you need a positive end on one and a negative one on the other: but there cannot be because, well, both are neutral, no?

  What are the intermolecular forces present in $\ce{CH3CH2NH2}$?

My answer was also London dispersion because it appears like this molecule is neutral.

Both answers are wrong. It appears I don't properly understand intermolecular forces. Why are my answers wrong?

This can be a bit tricky when you are first starting your journey into the world of intermolecular forces. What I like to look for first are what (if any) heteroatoms are present when the backbone of the molecule is composed of carbon atoms.

In your case there is a nitrogen present and it is attached to 2 hydrogen atoms. This is a perfect example of a molecule that will exhibit hydrogen bonding as an intermolecular force. Hydrogen bonding, at least in my opinion, is the easiest to pick out. N, O, and F atoms bonded to Hydrogen are the only species in which this attractive force between molecules is observed. Now if you were to remove that $\ce{NH2}$ group and just had $\ce{CH3CH3}$ you would be correct. The only intermolecular force between the molecules would be London forces.