Why is the enthalpy of vaporization greater than the enthalpy of fusion? Why is the molar enthalpy of vaporization of a substance larger than its molar enthalpy of fusion (at constant pressure); for example, in the case of ice and water.

Enthalpies of phase changes are fundamentally connected to the electrostatic potential energies between molecules. The first thing you need to know is:

  There is an attractive force between all molecules at long(ish) distances, and a repelling force at short distances.

If you make a graph of potential energy vs. distance between two molecules, it will look something like this:

Here the y-axis represents electrostatic potential energy, the x-axis is radial separation (distance between the centers), and the spheres are "molecules."

Since this is a potential energy curve, you can imagine the system as if it were the surface of the earth, and gravity was the potential. In other words, the white molecule "wants" to roll down the valley until it sits next to the gray molecule. If it were any closer than just touching, it would have to climb up another very steep hill. If you try to pull them away, again you have to climb a hill (although it isn't as tall or steep). The result is that unless there is enough kinetic energy for the molecules to move apart, they tend to stick together.

Now, the potential energy function between any two types of molecules will be different, but it will always have the same basic shape. What will change is the "steepness," width, and depth of the valley (or "potential energy well"), and the slope of the infinitely long "hill" to the right of the well.

Since we are talking about relative enthalpies of fusion and vaporization for a given system, we don't have to worry about how this changes for different molecules. We just have to think about what it means to vaporize or melt something, in the context of the spatial separation or relativity of molecules, and how that relates to the shape of this surface.

First let's think about what happens when you add heat to a system of molecules (positive enthalpy change). Heat is a transfer of thermal energy between a hot substance and a cold one. It is defined by a change in temperature, which means that when you add heat to something, its temperature increases (this might be common sense, but in thermodynamics it is important to be very specific). The main thing we need to know about this is:

  Temperature is a measure of the average kinetic energy of all molecules in a system

In other words, as the temperature increases, the average kinetic energy (the speed) of the molecules increases.

Let's go back to the potential energy diagram between two molecules. You know that energy is conserved, and so ignoring losses due to friction (there won't be any for molecules) the potential energy that can be gained by a particle is equal to the kinetic energy it started with. In other words, if the particle is at the bottom of the well and has no kinetic energy, it is not going anywhere:

If it literally has no kinetic energy, we are at absolute zero, and this is an ideal crystal (a solid). Real substances in the real world always have some thermal energy, so the molecules are always sort of "wiggling" around at the bottom of their potential energy wells, even in a solid material.

The question is, how much kinetic energy do you need to melt the material?

  In a liquid, molecules are free to move but stay close together

This means you need enough energy to let the molecules climb up the well at least a little bit, so that they can slide around each other.

If we draw a "liquid" line approximating how much energy that would take, it might look something like this:

The red line shows the average kinetic energy needed for the particles to pull apart just a little - enough that they can "slide" around each other - but not so much that there is any significant space between them. The height of this line compared to the bottom of the well (times Avogadro's number) is the enthalpy of fusion.

What if we want to vaporize the substance?

  In a gas, the molecules are free to move and are very far apart

As the kinetic energy increases, eventually there is enough that the molecules can actually fly apart (their radial separation can approach infinity). That line might look something like this:

I have drawn the line a little bit shy of the "zero" point - where the average molecule would get to infinite distance - because kinetic energies follow a statistical distribution, which means that some are higher than average, some are lower, and right around this point is where enough molecules would be able to vaporize that we would call it a phase transition. Depending on the particular substance, the line might be higher or lower.

In any case, the height of this line compared to the bottom of the well (times Avogadro's number) is the enthalpy of vaporization.

As you can see, it's a lot higher up. The reason is that for melting, the molecules just need enough energy to "slide" around each other, while for vaporization, they need enough energy to completely escape the well. This means that the enthalpy of vaporization is always going to be higher than the enthalpy of fusion.


Why is crystallography still used in some cases for small molecule structure determination? I was wondering for some time why in some cases x-ray crystallography and not NMR is used for structure determination of small molecules? Does crystallization maybe require a smaller amount of the compound than NMR?

Our group uses both methods a lot so here are some examples why you would use X-ray, in addition to NMR, in organic synthesis.

Your compound isn't soluble enough:

Colleagues produce very large aromatic systems for organic electronics which are basically insoluble in all solvents. With a good NMR machine (600MHz and up) it might still be enough to get spectra, but full structure elucidation might be difficult. Since the compounds crystallize well it's quite easy to get a X-ray structure.

Low number of carbon and hydrogen:

I've produced compounds with a C:H:N ratio of 2:3:5. The 1H is two singlets, the 13C is two signals. COSY, HSQC, HMBC,... won't tell you much either and it's much easier to use the nice crystals and get a structure that way than trying some correlated NMR experiments including nitrogen.

3-D orientation/stereochemistry:

Imagine this compound which is quite similar to something we produced:

Since you got two quaternary centers here it's really hard to tell the configuration on those centers relative to each other using NMR. There's no coupling to determine the stereochemistry and a NOE experiment isn't very conclusive here. With X-Ray it's no problem to find out.

Actually, these problems are definitely the ones we use X-ray the most.

Absolute Configuration:

Theoretically it's also possible to determine absolute configuration with single crystal X-ray diffraction, while it's extremely hard to do, if not impossible, with NMR.

Summary:

Single crystal X-ray diffraction gives you the configuration of your molecule, including relative configuration of stereocenters and possibly even absolute configuration, right away. If you only care about configuration (and not behaviour in solution) X-ray is definitely superior to NMR, there's basically no question on how your molecule looks.

On the other hand you need crystals and it's much more complicated to perform than NMR, so this definitely stays the number 1 method for synthetic chemists.


What is the liquid coming out of these Duracell batteries? I picked up a remote control to a DVD player that I had not used in a couple months.  The remote control was stored in a clean plastic bag.

The bottom of the remote control is covered with a liquid.  Upon further inspection, I discovered that the liquid is coming out of the Duracell Coppertop (non-rechargeable) batteries in the remote.

The amount of liquid is substantial.  It almost completely covers the bottom of the remote control.

Typically, when batteries have problems, you see crystals forming.  But not so much in this case.  In this case, it is almost all liquid that has been discharged.

I did not initially notice the liquid, and so I picked up the remote, and then set it down on the furniture.

What is this liquid?  What chemical process took place that resulted in the liquid being discharged from the batteries?

(And, as a bonus, how to I get the liquid off the furniture?)

Well, it depends on which type of battery you have. Alkaline batteries such as Duracell Coppertop usually leaks KOH (potassium hydroxide) under conditions of misuse as a white foam. However, KOH is hygroscopic and will easily form a oily like liquid which is what you might be seeing in your batteries. Handle it with proper care since KOH is caustic and can cause tissue damage.


What is the chemistry of this household cleaning recipe? What would be the product if baking soda and common salt and vinegar are added together?

Baking soda is sodium bicarbonate, and vinegar is dilute acetic acid. Common salt (table salt) is sodium chloride. Sodium bicarbonate will react with acetic acid to form sodium acetate (in solution), carbon dioxide gas, and water. Sodium chloride will not participate in the reaction. The reaction equation would be:

$$
\ce{NaHCO3(s) + HCH3COO(aq) -> NaCH3COO(aq) + CO2(g) + H2O(l)}
$$


If a chemical reaction is repeated multiple times, will it produce the same result? If a chemical reaction is repeated multiple(infinity) times, will it produce the same result?
For example, I put two $\ce{H}$ and one $\ce{O}$ together, and repeat it many (infinity) times within an exact same environment, will it produce $\ce{H2O}$ every time?
My chemistry teacher tells me it will. But I still wonder if it's true. Hope you will help me. Thanks.

The answer depends on whether you are looking at the level of a single atom or a large population of atoms. Because of quantum effects, there is a lot of randomness in the behavior of single atoms. If there are multiple possible outcomes, we can fairly accurately determine the probability of each outcome, but which outcome occurs is not predetermined.
With a large population of atoms (such as we have in any normal laboratory experiment like the ones your teacher is talking about), the large sample size means that we get the same result every time if we control every relevant variable (which is actually quite hard to do).


Define, (actually define) the "stability" and "energy" of a compound When we say a conformer or compound is of "higher energy" than another, are we quantifying energy in terms of the bond strength? Or are we going off of how much energy it takes to break the bond?  I don't think we can just assign energies to each compound... so what is this based off of?

One might answer that its based off of stability. Well, what is "Stability?" Is it how rigid/strong the bonds are? Is it how reactive the compound is?

Often times i find myself reading explanations with circular language on this topic. A higher energy conformer means lower stability or A stable compound is one that has low energy. I'm finding these definitions to be crazy ambiguous, and don't thoroughly define what these are.

If someone can clearly explain this in terms that a first year undergraduate student can understand, you have earned my gratitude.

As a chemist, I would agree that textbooks are not clear about the term stability and energy. It is not your fault and undergraduate organic chemistry books make the situation worse. I cannot recall the text, it was an old book, however it clearly said that stability of a compound does not mean anything. We should always ask, stability with respect to what? Are we talking about thermodynamic stability or kinetic stability? For example H2+O2 mixture is not stable. This statement is meaningless because we are not saying stability with respect to what? This mixture should form water right away but people actually waited for 30 years and did not find much water in a mixture of H2 and O2. Thermodynamically, this mixture is not stable and it will react violently once you provide the right activation energy. When you are reading about stability, ask this question yourself.

In the same vein, the term energy should also be clarified by authors. As a crude analogy is that you can think of energy as the potential energy of the whole molecule when the authors talk about conformers. Have a overall picture of the molecule. Imagine two springs, A) one which is less stretched and the other one (B) is more stretched. You can say spring B has more (potential) energy than (A). An eclipsed conformer has higher potential energy (overall, no need to think locally) than a staggered one.


Why does zinc sulfide glow when hit by electrons? In cathode ray tube experiment in order to check the direction of flow of electrons a hole was made in a note and behind it phosphorescent material zinc sulfide it was coated.

Why does zinc sulfide glow when hit by electrons?
What causes this glow?

This is an example of phosphorescence. The Wikipedia page of zinc sulfide and phosphorescence can explain it better than I do, but in short, when zinc sulfide get hit by electrons, electrons transfer some of its energy to zinc sulfide and excites its electron. Zinc sulfide stores the energy for a while, then emit light when the electron goes back to its ground level.


Structure of Buckminsterfullerene In several pictures that I have seen of buckminsterfullerene and graphite (such as below), each carbon is bonded to only 3 other carbons, so it only forms 3 covalent bonds.

I asked my teacher about this and he says that each carbon only make 3 bonds and has a lone pair. These lone pairs are de-localised hence, that is why buckminsterfullerene is able to conduct electricity.

However, I am doubtful about this as how can a lone pair on an atom be de-localised. Would the electrons be in their orbitals, and not be able to freely move around the compound. I try searching the internet for any information, but there is no mention on how many bonds each carbon atoms makes.

I am guessing that there is alternating double bonds between each carbon so that each carbon has 4 bonds. It is these pi bonds which along the compound that makes it a  hyper-conjugated compound. This would explain why buckminsterfullerene and graphite are able to conduct electricity. Am I correct?

Have you ever seen a picture of graphite, to begin with? It shares the same feature, i.e., that each carbon is bonded to exactly 3 other carbons. You may call it "alternating double bonds", but this is not quite correct either, because in fact all bonds are equal. Think of it as a resonance structure, much like benzene. This is the reason for graphite conductivity, too.

The idea about that lone pair is plain wrong. Carbon has but 4 valence electrons; once 3 of them are spent for sigma bonds, how could it have a lone pair?

Pure buckminsterfullerene does not conduct electricity (strictly speaking, it is a semiconductor, not a conductor like any metal). Some of its derivatives do, but that's another story.


how is the esterification of amino acid carried on in this reaction i was asked to synthesis this compound from this amino acid, i first alkylated the alpha carbon using enolate alkylating method then i esterficated the amino acid but i'm doubtful that my steps are right, do i need to protect the amino group in the esterification step? or even before ?

I think you need a protection strategy and there are several choices. This one uses O'Donnell's benzophenone imine strategy details here.
First you prepare glycine ethyl ester hydrochloride from glycine with EtOH/HCl, a method is here patent.
From the ethyl ester hydrochloride prepare the benzophenone imine by O'Donnell's method as described in ref 1. This gives a protected species that can be easily alkylated under phase-transfer conditions with allyl bromide as described in ref 1
The allyl product is then epoxidised (MCPBA) and the epoxide opened with iodide procedure here
Deprotection with acid hydrolysis gives your target molecule.


What fraction of the vapour in electronic cigarettes is water? Not a chemist but looking for an answer I can understand.

Electronic cigarettes use propylene glycol and/or vegetable glycerin as a base liquid. This is heated and vaporized at a low temperature (my e-cig is 7 watt). When exhaled the VG/PG absorbs water from the air producing the visible vapour cloud.

Assuming standard temperature ($\pu{20^\circ C}$) and humidity ($\pu{50\%}$), what ratio (or percentage) of the visible cloud is water?

It's not that I can't find an answer to this, just can't find an answer I believe. Most based on rumor and warm fuzzy feelings of what people want to believe.

The short answer to this is that there is no single answer to this problem as the vapor composition that is produced by any given atomizer is dependant upon a large number of components. To name a few:

The initial composition of the liquid, which will vary flavor to flavor and manufacturer to manufacturer. You can find out a bit more about this with the following articles

Characterisation of mainstream and passive vapours emitted by selected electronic cigarettes DOI 10.1016/j.ijheh.2014.10.001
Electronic cigarettes: overview of chemical composition and exposure estimation PMCID: PMC4304610

     A note to consider with the above is that many toxic organic solvents such as Formaldehyde, Acetaldehyde, Acetone, Acrolein, Propanal, Ethylene glycol, and    Diethylene glycol are found in eLiquids, in various concentrations.

The temperature that the eLiquid is heated to internally (this affects the vapor pressures of the various components in the liquid)
The concentrations of all of the, solvents, flavorings and such which are adsorbed by the person who is using the atomizer

Given all of these factors, and the large number that I am sure that I am missing, there is simply too many confounding variables to be able to give a straight answer to your question without directly measuring the concentration - not that I would know how you would make such a measurement.


How to explain the discrepancy between the calculated absorption wavelength within the particle in the box theory and the observed wavelength? The experimental observed wavelength for this molecule is 425 nm.

We are supposed to estimate this wavelength theoretically using this equation (energy change in a 1D box), where $N$ is the number of pi electrons.
$$\Delta E=\frac{h^2(N+1)}{8m_\mathrm{e}L^2}$$

$L$ is given as 11.84 angstroms.

I'm getting a wavelength of 200 nm. It seems to be too far off.

Molecule:

I'm counting 22 pi electrons and using that as N

This question is different from How to identify the number of pi electrons in a conjugated system to calculate the HOMO-LUMO gap with the particle in the box approach? in that this question is specifically addressing the discrepancy between the theoretical and experimental wavelengths, whereas the previous question was specific to the number to use for $N$.

Also how does this even make sense? As conjugation increases the energy gap between HOMO and LUMO should decrease. With this relationship as $N$ increases as does the energy gap. To me it makes no sense, how can it be explained?

Luckily I found a source containing your molecule as an example.

You need your equation for the energy difference between the HOMO and the LUMO, which is (nearly) defined as you wrote:
$$\Delta E = \frac{(N+1)h^2}{8 m_e L^2}$$

As you want a wavelength, you need to convert the energy difference, which yields then:
$$\lambda = \frac{8m_ecL^2}{h(N+1)}$$

Now they state, that the length of the box can be guessed by the amount of carbons (p) between both nitrogens by using this formula:
$$L = (p+3+\alpha)l$$
... with $l$ being the length of typical benzene-C-C-bonds, thus being 0.139 nm and $\alpha$ is an empirical correction factor for polarizable end groups.

The amount of pi-electrons (N) that you need to inlude can be calculated by a similar equation:
$$N = p+3$$
... where the additional "+3" come from both nitrogens (would be actually +4, but it's +3 because one nitrogen atom had to form an ion).

This leads to their equation 4:
$$\lambda = 63.7\frac{(p+3+\alpha)^2}{p+4}$$
... where all the physical constants have been combined to give 63.7.

As you have three carbon atoms between both nitrogens, you come up with 328 nm and when you include their "correction factor for polarizable end groups" of $\alpha = 0.675$, you come up with 405 nm, which is more than close to the experimental result.

When I use your box length and use 12 electrons (3 from the "left" benzene ring, 4 from the two double bonds to the "right" nitrogen" and 1 from the lone pair of this nitrogen), I come up with about 356 nm, which is not that far away as it sounds. In eV this is an error of 20% (the 405 nm are wrong by about 5%), whereas your 200 nm would have been an error of 120%!


The pH of a neutralized solution If pH is defined as the concentration of hydrogen ions in solution, then how can a ‘neutralized’ solution (defined as having an equal amount of hydrogen and hydroxide ions) have a pH other than 7?

Wikipedia writes: “In a reaction in water, neutralization results in there being no excess of hydrogen or hydroxide ions present in solution.” In my mind, an equivalent amount of hydrogen ions and hydroxide ions means a pH of 7 (in water), but Wikipedia carries on to write: “The pH of the neutralized solution depends on the acid strength of the reactants.”

This question doesn’t seem to answer my question.

TL;DR
The neutral point is not the same as the equivalence point.
A neutral aqueous solution at room temperature, $25~^\circ\mathrm{C}$ and standard pressure $1~\mathrm{atm}$, has always $\ce{pH}=7$.

If you look for credible sources of definitions of neutral solutions on the internet, most likely you will find something along the lines of this:

  Neutral solution
  (Science) Has a $\ce{pH}$ level of $7$: a solution in which the concentration of hydrogen ions and hydroxide ions are equal (biology-online.org)

Or even worse:

  Neutral Solution Definition:
  An aqueous solution with a $\ce{pH}$ of $7.0$ $(\ce{[H^{+}]} = 1.0 \times 10^{-7}~\mathrm{M})$. (chemistry.about.com)

The second one especially is negligent of a lot of features.
Unfortunately the IUPAC does not provide an actual definition of a neutral solution. There are two possible definitions, which come to the same conclusion (compare the above), given the same external conditions.

The lazy definition:
The concentrations of hydronium and hydroxide ions are identical. $$c(\ce{OH-})= c(\ce{H3O+})$$
In terms of $\ce{pH}, \ce{pOH}, \mathrm{p}K_w$ one arrives at a more complete side.
Pure water is a neutral solution. It's autoprotolysis provides
\begin{align}
\ce{2H2O &~<=>~ OH- + H3O+}.\tag{1}\\
\end{align}
Now we can formulate the equilibrium constant and furthermore assume that autoprotolysis is small compared to the overall concentration of water.
\begin{align}
  K_c &= \frac{c(\ce{OH-})\cdot c(\ce{H3O+})}{c^2(\ce{H2O})}\\
  K_w &= c(\ce{OH-})\cdot c(\ce{H3O+})\tag{2}\\
\end{align}
Equation $(1)$ also provides the lazy definition. $$c(\ce{OH-})= c(\ce{H3O+})\tag3$$
We plug that into $(2)$ and we arrive at
$$c(\ce{H3O+})=\sqrt{K_w}\tag{4}$$
Consider the actual definition of
$$\ce{pH} = −\lg a(\ce{H+}) = −\lg\left( m(\ce{H+}) \gamma_m(\ce{H+}) / m^\circ\right)\tag{5}$$ rewritten using concentrations
$$\ce{pH} = −\lg a(\ce{H+}) = −\lg\left( c(\ce{H+}) \gamma_c(\ce{H+}) / c^\circ\right)\tag{5a}$$
The identities used here are $a(\ce{H+})$ for the activity of a proton in aqueous solution, $\ce{H+ (aq)}$ and we will consider $\ce{H+ (aq) = H3O+}$ to be the same. Now we are going to assume, that the activity coefficient is $\gamma_c(\ce{H+})\approx 1$ for very diluted systems. Using the standard concentration $c^\circ$ we do not have to care about units.
$$\ce{pH} = −\lg\left( \frac{c(\ce{H+})}{c^\circ}\right)\tag{5b}$$
Now we can substitute in $(4)$ and we will have a nice definition of the $\ce{pH}$ of a neutral solution.
$$\ce{pH} = −\lg\left( \frac{\sqrt{K_w}}{c^\circ}\right)\tag{6}$$

The second formulation includes a very crucial point. The autoprotolysis $(1)$ is temperature, $T$, dependent, which is obvious following the definition of the standard equilibrium constant
$$ K^\circ = \exp\left\{\frac{-\Delta G^\circ}{\mathcal{R}T}\right\}\approx K_c.$$
Therefore also the ion product of water is temperature dependent, $K_w(T)$.
Wikipedia actually has a couple of values for the $\mathrm{p}K_w$. Reaction $(1)$ is endothermic. A higher temperature means providing energy, means a higher $\mathrm{p}K_w$. For example, in the human body, $37~^\circ\mathrm{C}$, this value is slightly higher than for the one we usually assume as the "neutral point" $\mathrm{p}K_w(25~^\circ\mathrm{C})=14$.

Since $(6)$ is more correct in the following form,
$$\ce{pH}(T) = −\lg\left( \frac{\sqrt{K_w(T)}}{c^\circ}\right)\tag{6a},$$
the neutral $\ce{pH}(T)$ therefore changes with temperature.

Another note is that it is also pressure dependent, but that might take it a little too far.

The equivalence point is usually defined for an arbitrary chemical reaction between an acid and a base, and it refers to the situation where there are stoichiometric quantities of acid and base are present.
$$\ce{AH + B <=>[\ce{H2O}] A^- + HB+}$$
This means, that at the equivalence point the following statement holds:
$$c_\text{initial}(\ce{HA}) = c_\text{final}(\ce{HB+}).$$
Therefore the $\ce{pH}$ at the equivalence point will only be governed by the reactions of $\ce{HB+}$ and might not be neutral.

The issue of solvent dependency is addressed by Fred Senese quite concise on antoine.frostburg.edu:

  pH is often used to compare solution acidities. For example, a solution of pH 1 is said to be 10 times as acidic as a solution of pH 2, because the hydrogen ion concentration at pH 1 is ten times the hydrogen ion concentration at pH 2. This is correct as long as the solutions being compared both use the same solvent. You can't use pH to compare the acidities in different solvents because the neutral pH is different for each solvent. For example, the concentration of hydrogen ions in pure ethanol is about 1.58 × 10-10 M, so ethanol is neutral at pH 9.8. A solution with a pH of 8 would be considered acidic in ethanol, but basic in water!


Why are all 3 of propane, butane and isobutane used in deodorant sprays? Here are the back ingredient labels of a few popular commercial alcohol-based deodorant sprays.
As one can clearly see, they all have many things in common, but the 3 which are of importance in lieu of this question are-

Butane
Isobutane
Propane

After a quick search on the internet, one can find all 3 have the same common function i.e. acting as a propellant to give a perfect aerosol spray.
My questions lies, why are all 3 of them used together when they have the same function? There are also many deodorant sprays that use only butane as the propellant. Is it something related to pressure factors or its ability to be used in different conditions(hot/cold weather)?
If there's no such strong reason behind it, and if cost effectiveness isn't harmed either, then why not just use one of the 3? And if so, what would be the best of the 3 alternatives?
Image source: Google Images 1
2
3
4

Bu/iBu or Bu/iBu/Pr mixtures are easier/cheaper to get instead of pure components as Ivan says, as there is extra effort for their separation, not the extra effort for their mixing.
Butane(Bu)/isobutane(iBu)/propane(Pr) boiling points are $\pu{0 ^\circ C}$, $\pu{-11  ^\circ C}$ , $\pu{-33  ^\circ C}$  respectively. Real boiling points would be higher than boiling points of pure hydrocarbon mixture, as deodorants are solution of active components in volatile liquid hydrocarbons, what increases the boiling point.
Propane (and isobutane too to a lesser extent) shifts usability toward lower temperatures, similar as Pr/Bu mixture of liquefied gas fuel cartridges for outdoor usage. Butane itself does not work well around $\pu{0 ^\circ C}$ either as propelent either as fuel, as it has not enough pressure to overcome the external one.
The goal is to provide the vapour pressure within the given pressure range for the given operational temperature range. Pure compounds have usually too low vapour pressure at the minimal temperature or too high vapour pressure at the maximal temperature.
Another reason is the control of evaporation rate of the solvent when it is being sprayed out. Different compositions of deodorant matrix may prefer different rates of hydrocarbon evaporation and different related cooling effect.


Catenane optical activity Would a [2]catenate whose each ring consist of a different number of methylene groups only, be optically active if one hydrogen atom of a random methylene group is replaced with e.g. chlorine? I think yes, but I need a second opinion.

The compounds attached seem to be enantiomers. I don't see how, by rotation, they could be super-imposable.

If a molecule, or any conformation of a molecule, contains a plane of symmetry, then the molecule is not chiral (achiral).

The all hydrocarbon [2]catenane will be achiral. The molecule contains a plane of symmetry; picture a plane that contains one of the rings and bisects the other ring. In the following image, you can see that the plane of the screen contains the ring on the right and bisects the ring on the left.

You might say that if we replaced the circles with cyclohexane rings, because of their chair-like shape, the plane of symmetry is lost.  That is true, but you can flatten one of the cyclohexane rings and the plane of symmetry is restored.  Remember, if any thermally accessible conformation of a molecule is achiral, then the molecule is achiral.

Sometimes it helps to build a model or use a Newman projection to help with the stereochemical analysis.  Here is such a projection for your chloro-[2]catenane.  Note that it still possesses a plane of symmetry (a plane perpendicular to the screen).

If we go further and add another chlorine atom, but now to the back ring, then the molecule becomes chiral — just like 1,3-dichloroallene.

Regarding:

  The compounds attached seem to be enantiomers. I don't see how by rotation they could be super-imposable.

Look at the bottom figure in my second drawing.  I've taken your interlocking 5- and 6-membered catenane and rotated the 6-membered ring a bit further along (a conformational isomer; remember, as I mentioned above, if a molecule, or any conformation of a molecule, contains a plane of symmetry, then the molecule is not chiral [achiral]).  The plane of the screen now contains the 5-membered ring and it bisects the cyclohexane ring.  The plane of the screen also contains the methyl group.  The plane of the screen is a symmetry element of this molecule; therefore this molecule is achiral.


Formation of Tropylium carbocation I've read that Tropylium carbocation is formed by the ring expansion of benzyl carbocation. But in benzyl carbocation, the $\ce{C}$ atoms in the benzene ring are $sp^2$ hybridized i.e. $120^\circ$  is the required angle for stability. And in the benzene ring of the benzyl carbocation, the angle is already 120 degrees. Therefore benzyl ring need not expand for stability. Therefore why does benzyl rearrange to give tropylium?

More stable products are lower in energy.  At first glance the bonding arrangements might seem equivalent in both cases in terms of number of $\ce{C-C}$ and $\ce{C-H}$ bonds from $\ce{sp^2}$ carbons, which would result in no net change.  However, the change in molecular structure leads to a change in the molecular orbitals and the resulting energies of the molecule for the aromatic/resonance component to the energy. Using the Frost diagram technique we want to compare the orbital energies, nicely pictured here:

The key point to notice is the filled orbitals for seven member ring are slightly lower in energy than the filled orbitals for the six-member ring.  This means there is an increase in resonance stabilization energy for making a fully aromatic system rather than just a cation in resonance with benzene ring.


Should the 'a' in the acid dissociation constant (Ka or pKa) be capitalised? In science, the negative logarithm of the acid dissociation constant is denoted $\mathrm{p}K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{A}$ depending on the source (lowercase "a" or uppercase "A"). Since it is related to the acid dissociation constant, defined as

$$K_\mathrm{A} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{AH}]}$$

I would write $\mathrm{p}K_\mathrm{A}$ (the LaTeX package chemmacros renders it so, but doesn't give any source it in its documentation). It seems to me that $\mathrm{p}K_\mathrm{a}$ is the "modern" version of pKa, which was written like this in the last century because of the lack of subscript character on typesetting machines.

I cannot find any source or styling guide saying explicitly which one should be used today (like the IUPAC Green Book), and why.

TL;DR: for the documents written in English use $\mathrm{p}K_\mathrm{a}$ (upright lowercase "a").

Current (2017Q3) chemmacros' documentation (p. 9) includes a reference for how acid-base module handles \pKa[...]:

  \pKa: $\mathrm{p}K_\mathrm{a}$, \pKa[1]: $\mathrm{p}K_\mathrm{a1}$, depends on language settings, see section 6.5 starting on page 30. The translations can be adapted. [...]

  As you can see the default subscripts of \Kw, \Ka and \Kb are lowercase letters. The literature is inconclusive about if this is the right way or if uppercase letters should be preferred. In textbooks the uppercase variant usually seems to be used while journals seem to prefer the lowercase variant. chemmacros’ default follows the usage in The IUPAC Green Book [Coh+08].

From IUPAC Green Book [1, p. 59]:

  The equilibrium constant of dissolution of an electrolyte (describing the equilibrium between excess solid phase and solvated ions) is often called a solubility product, denoted $K_\mathrm{sol}$ or $K_\mathrm{s}$ (or $K_\mathrm{sol}^⦵$ or $K_\mathrm{s}^⦵$ as appropriate). In a similar way the equilibrium constant for an acid dissociation is often written $K_\mathrm{a}$, for base hydrolysis $K_\mathrm{b}$, and for water dissociation $K_\mathrm{w}$.

Additionally, "a" should be typed as an upright letter [1, p. 5]:

  Subscripts and superscripts that are themselves symbols for physical quantities or for numbers should be printed in italic type; other subscripts and
  superscripts should be printed in Roman (upright) type.

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen R. E., Ed.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.


Rotation energy barrier for carboxypyrazolate In this molecule, which is 3,5-dimethyl-4-carboxypyrazolate (unless I got it wrong):

I do not think there is free rotation around the bond indicated in red, because one can write a resonance structure with significant reasonable weight where a double bond would be located between the two groups.

However, how can I go further than that and qualitatively describe the energy barrier for the rotation. I have found (here and there) that in the case of benzoates, the barrier is in the range 3–6 kcal/mol. Should I expect it to be higher or lower in the above molecule?

One of way to answer this question is to put it in some QM software. The system is small enough and has higher symmetry so computations shouldn't take to long, even on an older computer.

For co-planar geometry the energy calculated at B3LYP/6-31G(d,p) is –492.132206748 a.u.
The geometry with dihedral angle of 90° between $\ce{CO2}$ and $\ce{C3N2}$ is –492.116719625 a.u. That makes roughly 41 kJ/mol (10 kcal/mol).

Now if you really want to be sure if that's the right energy you should make frequency calculations on both structures to find out if they are minima or transition states. This can take a bit longer (I haven't done it) and I would expect that the better structure (with lower energy) should have no imaginary frequencies and the higher one should have one (TS). If by some accident both of them are minima than TS has probably a dihedral angel somewhere around 45°.

Of course this is only computation. It gives as clues but not the exact answer. You can go for higher level of theory, add solvent modelling and... or you can take your compound and try to make temperature depending NMR spectra. With a bit of luck you can freeze your TS and find out how high is your barrier.

p.s.

It would be good if you mention in your post what tools (in this case computational) are available to you. I really would prefer to explain how to get the number instead of giving just dry answer.


Calculate the heat of a reaction Say I would like to calculate the heat produced by the following reaction:

$$\ce{2SO2 + O2\longrightarrow2SO3}$$

I have calculated the moles of the reagents. I can't seem to figure out how to calculate the heat of the reaction.

How can I calculate the heat of a reaction with moles of reagents?

in order to calculate the heat of reaction, you need the bond energies (which are measured in $\mathrm{kJ/mol}$ and usually provided in the question or in an appendix in the back of the chemistry book). The heat of reaction is equal to the energy of bonds broken minus the energy of the bonds created. It would be equal:

$[4*(\ce{S-O} $ bond energy$) ~+ ~1*(\ce{O=O} $ bond energy$)] - [6*(\ce{S-O} $ bond energy$)]$.

This will give a negative answer if the reaction is exothermic and vice versa. Since it is measured in $\mathrm{kJ/mol}$, multiply by moles to get the total energy released.


How is the yield of a complex organic synthesis defined? I just watched the Periodic Table of Videos episode Deadly Strychnine that describes the history of this particularly challenging synthesis. University of Nottingham chemist Rob Stockman calls the problem "an "Everest" of synthetic chemistry."

What is the definition of yield in this case? How is it calculated? Is it as simple as mass of product divided by total mass of all reactants, or something more specific and meaningful?

In theory

For simplicity's sake, let's consider a two-step sequence:

The usual terms for these are as follows: A is the "starting material", D is an "intermediate", F is the "final product", and B, C, and E are "reagents".

Why are B and C listed over the same arrow? This means that they have been added together to the starting material A, in the same flask, and together they effect the transformation of A to D.†

Now, depending on the stoichiometry of the reactions, as well as the quantities of A, B, C, and E which are added while performing the synthesis, it will turn out that one of A, B, C, or E will be a limiting reagent for the overall synthesis. Note that D cannot be a limiting reagent, as it is not an "input".

For example, let's say that the exact stoichiometry of the two reactions is as follows:

$$\begin{align}
\ce{A + 2B + 2C &-> D} \\
\ce{D + E &-> F}
\end{align}$$

Now, it is clear that if you add 1 mol of A, and if everything else is added in excess, then the theoretical maximum you can get is 1 mol of D and hence 1 mol of F. In this scenario, A is therefore the limiting reagent, because the quantity of A limits how much of F you can get.

Sadly, reactions don't ever work perfectly. You perform the synthesis, and at the end of the day you get 0.05 mol of F instead of the theoretical maximum of 1 mol. Your yield is therefore:

$$\frac{\pu{0.05 mol}}{\pu{1 mol}} = 0.05 = 5\%.$$

Identifying the limiting reagent

Usually the limiting reagent is chosen to be the most expensive, or most precious reagent you have (or else part of it goes to waste). This is most often an organic compound, and is nearly always the given starting material. So, one usually does not have to calculate the theoretical yield from every single input, as the limiting reagent can usually be identified at a glance.

MacMillan's strychnine synthesis can be used as an example here. The sequence of synthetic steps is given on SynArchive. In this case, it is reasonable to assume that the organic starting material is the limiting reagent, and that all the other "inputs" are added in excess.

In practice

In the lab, organic chemistry is done one step at a time; therefore the intermediate D is usually isolated and stored for a period of time.

If the above theoretical analysis is to hold true, then all of the D that you formed in the first step must be carried forward to the second step. i.e. if you form 0.8 mol of D in the first step, the second step must start with all of the 0.8 mol.

In practice this is never the case, due to some reasons:

Some sample must be sacrificed to obtain spectroscopic data on them; this helps the chemist know that they have made the right thing. In other words, we need to make sure that what we think is D is really D.
Some sample must also be sacrificed to small-scale experiments. Reactions often need to be tested before being performed on a full scale. If a chemist made 0.8 mol of D from the first step, and happily carried all of it forward to the second step, they would be left fuming should the second step fail, because they would have to perform the first step again. The way around this is to take a tiny amount: say 0.01 mol of D, and carry out the reaction on it. That way, if the test run works, you can then take the remaining 0.79 mol and do the reaction again; and if not, you still have plenty of D left to experiment on. Ultimately, this leads to losses in the amount of D which is carried forward.

So, the better way to do it is to measure the yield of each individual step. That is to say, measure the yield of D that you made from A; and then measure the yield of F that you made from D. If you multiply those two numbers together, you will obtain the overall yield.

For example, if you put in 1 mol of A and got 0.8 mol of D, the yield of the first step is 80%. Let's say you then took 0.75 mol of D, treated it with E, and got 0.075 mol of F. The yield of the second step is then 10%.

It looks like the overall yield is 0.075/1 = 7.5%, but in truth the yield is (80%)(10%) = 8%. Why is the true yield larger than expected? It's because you didn't carry forward all the D that you could have. In theory, if you repeated the synthesis from start to end and carried forward the entire 0.8 mol of D, you would then have gotten 0.08 mol of F, which corresponds to the correct yield of 8%.

In a paper, yields for every step are reported. This helps people identify which steps work well, and is also useful for other chemists who wish to only replicate a certain step or steps from the paper (for other purposes). The quoted yield of the overall synthesis is usually simply the product of the yields for the individual steps.

† The exact definition of a "step" has been subject to some recent controversy (1, 2), but it's not worthwhile getting into it here.


IUPAC nomenclature of chelates I could finally get a copy of the Red Book 2005 from a library. In the chapter dedicated to nomenclature of coordinate complexes, while explaining the κ-system of specifying donor atoms, the following two examples were given (among many others):

Can anyone explain how the naming is done for the two compounds (labelled 2 and 3 in the picture)? The problem that I am facing is with the closed rings. From where is the naming started?

The image gives a thousand words, it is difficult to say how many words the colorized image gives, but might help.


Why is monohydrogen phosphate an incorrect name? In my Ap chem class I got this sheet with the names of a bunch of common polyatomic ions. One of them is named monohydrogen phosphate ($\ce{HPO4^2-}$). In naming, I learned not to use ‘mono-’ if it’s at the beginning of a name.

Is this simply a mistake? Or is there a reason?

The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) reads as follows:

(…)
(…) The prefix ‘mono’ is, strictly speaking, superfluous and is only needed for emphasizing stoichiometry when discussing compositionally related substances (…).
(…)

Therefore, hydrogenphosphate ($\ce{HPO4^2-}$) may be called monohydrogenphosphate in order to emphasize the difference to dihydrogenphosphate $\ce{H2PO4-}$.
Note that, according to IUPAC Recommendations, the systematic hydrogen name of $\ce{HPO4^2-}$ actually is ‘hydrogen(tetraoxidophosphate)(2−)’; however, ‘hydrogenphosphate’ is an accepted simplified hydrogen name.


What kind of chemical used in a chromatography run could be denoted with the letters D.H.? A fellow translator is having trouble translating (into Russian) the mysterious D.H. abbreviation in the following passage:

  Slight variations of the ratio of the mobile phase constituents or adjustments of the mobile phase flow-rate can be made occasionally, in order to provide a suitable elution times for Methyl p-hydroxybenzoate and D.H., and to meet the requirements of the system suitability tests.

If by some chance it's a widely known compound, or at least there are several chemicals that could be abbreviated thus, I'd be grateful for answers. A crapshoot, I know, but I became curious.

It could be some rare chemical; in that case it would be futile to try figuring it out.

Recently I had to read up on HPLC-MS technique and I encountered D.H./DH several times as an acronym for dehydrogenase, also, like in the quoted section, in the context of parabens analysis with HPLC.

For example, DH is explicitly defined as an acronym for dehydrogenase in Kastner's Protein liquid chromatography [1, p. 5] or in the paper by Zimmerling et al. [2].

Other examples for the exact dehydrogenase composition include acronyms such as ADH (alcohol dehydrogenase), FDH (formate dehydrogenase), $\small\text{D}$-LDH ($\small\text{D}$-lactate dehydrogenase), 11βDH (11β-hydroxysteroid dehydrogenase) and many more.

References

Protein Liquid Chromatography; Kastner, M., Ed.; Journal of chromatography library; Elsevier: Amsterdam; New York, 2000; Vol. 61. ISBN 978-0-444-50210-0.
Zimmerling, J.; Tischler, D.; Großmann, C.; Schlömann, M.; Oelschlägel, M. Characterization of Aldehyde Dehydrogenases Applying an Enzyme Assay with In Situ Formation of Phenylacetaldehydes. Appl Biochem Biotechnol 2017, 182 (3), 1095–1107. DOI: 10.1007/s12010-016-2384-1.


Will these K-alum crystals combine into a "cohesive/well-glued" mass? The attempt to make a single solid mass of K-alum, using over 125 grams of McCormick K-alum powder, has resulted, in less than 2 hours, in:

So besides the growth around the loop of the seed "crystal", there are 3 other regions where crystals are forming.
Context: I am seeking a solid mass of K-alum to use as an astringent and antiseptic after shave (starting at 30 seconds into this video).
Questions: (1) Assuming that the over 125 grams of powder is large enough, will the bottom two or three masses combine into a big "cohesive/well-glued" mass? (2) Or should I dismantle the setup -- keep the seed-block, but heat the water to dissolve the other three masses to become very slightly under-saturated and put the seed back in?
Update: It has been under 3 hours, and the bottom three masses have combined into a single mass. I do not know how "cohesive/well-glued" that single mass will be when it is eventually removed and dried with a paper-towel.
The comment by @tstone-1 suggests this simpler approach for creating a single mass of K-alum: Heat about 200 ml of water to about 88C; saturate this with K-alum powder (about 100 grams); filter this solution into a paper coffee cup; cover cup and wait till most of the water evaporates; tear off paper cup to get solid mass. Question: how cohesive/well-glued will this mass be?
Update 2: Did end up with two consolidated masses, one from the starter near the rim and the other from the bottom three chunks;  both are suitable for use after-shave as an astringent and antiseptic:

According to the Wikipedia article Potassium alum, the dodecahydrate melts at 92 to 95 °C. So if I wanted a large lump, I'd put the salt in a test tube (or some other container), dunk the container in boiling water to melt the salt, then pour the melted salt into some type of a mold to cool and solidify.


Chemical formula of fluoroapatite My textbook says the formula for fluoroapatite is $\ce{CaF2 * 3Ca3(PO4)2}$, whereas according to Wikipedia it's $\ce{Ca5(PO4)3F}$.

What is the correct chemical formula of fluoroapatite and why there are different representation of the same compound?

As it's been stated in the comments, both formulas are equivalent, however, I'd like to clarify the terminology and use cases for both.

$\ce{CaF2 * 3Ca3(PO4)2}$ is a formula of a formal addition compound (often used for clathrates, crystallohydrates and multiple salts). Widely used several decades ago when crystal structure of a given compound hasn't been elucidated enough to speak of exact atomic positions, hence it's often been attempted to present it as a ratio of simpler compounds packed in a crystal lattice.

$\ce{Ca5(PO4)3F}$ is a formula unit. Constituents are ordered from the most electropositive to the most electronegative groups; the ratio between them is denoted with numerical indices. By multiplying the formula unit $\ce{Ca5(PO4)3F}$ by $Z = 2$ ($Z$ is the number of chemical formula units per unit cell), you get $\ce{Ca10(PO4)6F2}$ — the content of the entire unit cell. $\ce{Ca10(PO4)6F2}$ is equal to $\ce{CaF2 * 3Ca3(PO4)2}$ in terms of stoichiometry, but the use of latter middle-dot-notation is deprecated for multiple salts as it doesn't carry additional structural information and often leads to confusion.

I think it's important to mention that not only $\ce{Ca5(PO4)3F}$, but $\ce{CaF2}$ and $\ce{Ca3(PO4)2}$ are also formula units, and not molecules!


Why is 295 nm a good wavelength to check absorbance when testing for aspirin? Take aspirin (acetylsalicylic acid) and hydrolyze it with NaOH. This results in sodium acetylsalicylate and water. I think the sodium is held on with an ionic bond, so it is really salicylate dianions that are the important thing in the solution.
This is then diluted to several different concentrations. Using UV-VIS, the absorbance of these solutions are measured at 295 nm. Then a calibration line is plotted, that is, absorbance verse concentration. It will be a (pretty much) straight line according to the Beer–Lambert law. You can use this calibration line to calculate the concentration of a similar, but unknown, solution from its absorbance.
Why was 295 nm a good choice of wavelength to measure at?
The only idea I currently have is that at 295 nm there is a distinct difference between the absorbance of acetylsalicylic acid and salicylic acid. Apparently, the salicylic acid has a strong response at 295 nm, whereas the acetylsalicylic acid does not. However, I am not sure this is exactly relevant, since I'm not sure I have salicylic acid in my solution.
If I am on the right track here, then I have a follow-up question:
Apparently the lambda-max for salicylic acid is 302 nm. So, why would 295 nm be used in this experiment instead of 302 nm? Or, would anything between, say, 295 nm and 310 nm be just as good?

I think this was basically an educational experiment. Your analytical method assumes that the hydrolysis was complete (this is the key requirement of a successful calibration curve). Your experiment did not include a test to assess the completion of hydrolysis. Acetylsalicylic acid begins to hydrolyze in alkaline solution to salicylic acid and basically it is an indirect method of detection.
You want to analyze acetylsalicylic acid but you are detecting salicylic acid. Guess where does this method fail? It will fail if you had a mixture of acetylsalicylic acid and salicylic acid in the beginning and you were asked to quantify both.
As EdV explained to you, being 1 nm off from the lambda max is not a big deal in UV-Vis spectra of solutions. Even the monochromator may be slightly off!


Dependence of carbocation stability with respect to the length of alkyl group Which carbocation is more stable, the ethyl carbocation or n-propyl carbocation?
What is the effect of alkyl group length on the stability of a carbocation?

Carbocation stability is widely known to increase going from primary -> secondary -> tertiary.

In the absence of other factors (such as lone pair donation etc.), the major contribution to the greater stability of tertiary carbocations vs primary is due to hyperconjugation between the vacant p-orbital (carbocation) and the C-H sigma bonding MO's.

Based on hyper conjugation alone, you can predict that length of the alkyl chain will have no effect on the stability by itself, only the number of neighbouring C-H bonds with which hyper conjugation can take place.
In specific regard to ethyl vs n-propyl cation, neither are considered stable.


Is there a Material that is High Heat Resistent and soluble? What i‘m looking for is any sort of material that dissolves, while being able to resist temperatures of up to 1000°C.
And it doesn‘t have to resist during dissolution, only while it‘s solid.
And the solvent should not be corrosive, or at least only be corrosive for the soluble Material, but not for other (or most) materials (mainly metals).

Also it would be good if said material can be reused afterwards.
(Like you can reuse salt if you dissolved it in water and then dried it again.)
And the solvent should be environmentaly friendly.

Now, the first Part is basically what i‘m looking for and the Second is just some bonus, so it‘s good even if it‘s not consistent with the 2nd Part.

If anyone knows such a Material, it‘d be nice if you could tell me.

Thanks in advance

I cannot guarantee the materials in this answer will meet all your requirements, so it would still be very helpful if anyone else could pitch in (especially an inorganic chemist or someone with hands-on knowledge of refractory materials). I will describe my thought process, which may ultimately be more useful than the suggestions themselves.

Immediately, the requirement to resist temperatures up to 1000 °C implies organic materials are very unlikely to be suitable. Almost all reasonable organic compounds would boil, decompose or burn (in air) well before then.

So now we focus on inorganic materials. Simple examples of very heat-resistant compounds are quartz ($\ce{SiO2}$) and alumina ($\ce{Al2O3}$), both melting only past 1500 °C and with no decomposition. However, these materials also resist dissolution; quartz dissolves only in concentrated hot aqueous base, whereas alumina dissolves in both of concentrated hot aqueous base or aqueous acid.

That said, it is possible to add other compounds to these materials and change their properties. For example, mixing $\ce{SiO2}$ with a number of other inorganic substances causes its melting point to lower to more manageable temperatures. This is how we make most kinds of glass. Particularly interesting is that the addition of sufficient alkali metal ions can turn the glass water-soluble. For example, there are a variety of sodium silicates, whose exact composition and properties will vary with the sodium content, and at high sodium content they become very water-soluble. However, it appears these silicates melt close to 1000 °C, which may preclude their use.

Fortunately the same approach works with alumina; increasing its alkali metal ion content eventually renders the material highly water-soluble. Sodium aluminate with a 1:1 mol ratio of sodium to aluminium forms $\ce{NaAlO2}$, which appears to be very water soluble and still melts at >1600 °C. In principle, this material could be made into a concentrated, thick aqueous solution or slurry, poured into a mould, dried at low heat to solidify then calcined to obtain a sodium aluminate object of the desired shape. This could then be used for your applications, and redissolved in water when necessary.

In both of the previous examples, it's certainly possible to add further inorganic compounds (e.g. borax, lime, etc.) to tweak properties further. However, much of this kind of research has already been performed, so I would recommend digging into glass fabrication/cement research rather than rediscovering the wheel.

I should also mention there are ways in which my above suggestions may be inadequate. Alkali aluminates/silicates appear to be strongly hygroscopic, which may make their use more cumbersome. More worryingly, at high alkali metal content they become strongly basic, which could corrode some of the metals you plan to use, both during dissolution in water, or by direct reaction with the aluminate/silicate surface at high temperatures. You'll have to see whether this is a potential deal-breaker.


Chemical Equilibrium - Le Chatelier's principle If the concentration of reactants is increased such that one is a limiting agent, wouldn't the concentration of products also increase (because excess of a reactant will get carried to the other(product) side  of the equation considering that the limiting agent is in aqueous medium?

e.g.
  (Consider 1 mole of each reactant)
   $$\ce{H2O + NH3 <=> NH4+ + OH-}$$

Now, let us take another scenario in which  there are two moles of NH3 and 1 mole of H20. Then concentration of one reactant has increased (from the previous scenario). Wouldn't the concentration of the products also increase because one mole of NH3 does not react and gets carried to the products' side of the equation?

I am confused with how  Le Chatelier's principle is applied in the 2nd scenario. Why does the equilibrium shift to the right if the concentration of the reactants and products increase (if my assumption is right)

If my assumption is wrong (concentration of product increases to balance excess reactant), my concern is such an equation is not balanced. I am unable to reconcile how the Le Chatelier's principle is applied in the 2nd scenario.

(Based on the answers below, I would like to restate my question)
I understand that

$$\ce{2H2O <=> H3O+ + OH-}$$

Now if I add HCl, I have

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

The way I interpret this is : the concentration of reactants  is 1 + 1 = 2 moles and the concentration of products is also  1 + 1 = 2 moles. Since the concentration on both sides of the equation is same, why do we say equilibrium shifts left. What am I missing?

For the case you're describing, having left-over product species will likely not cause the reverse reaction to increase.

When you define a reaction equation like: $\ce{aA + bB <=> cC + dD}$, you are explicitly defining A and B as 'reactants' and C and D as 'products'. In the scenario you've described, just because you have excess reactants because you have a limiting concentration of either A or B, doesn't mean the concentration of products will increase. This is because we can think of concentration of products as $[Products](t) = [C](t) + [D](t)$ and not $[Products](t) = [C](t) + [D](t) + [A](t) + [B](t)$.

--

A nice definition of Le Chatlier's principle is the following: "A change in one of the variables that describe a system at equilibrium produces a shift in the position of the equilibrium that counteracts the effect of this change."

If one is to get a little more technical with this, equilibrium is a function of many factors, primarily Temperature (T), Pressure (P), and mixture species (n). This is typically described by thermodynamics by the state property, Gibbs Free Energy, or $\Delta G$.

For the purposes of an example to quantify Le Chatlier's principle, let's look at the differential form of Gibbs Free Energy: $dG = -SdT +VdP + \sum^{n}_{i=1}\mu_idn_i$. If we consider a case where temperature isn't changing (isothermal), and where pressure isn't changing (isobaric), then the Gibbs Free Energy equation is affected by changes in system composition: $dG = \sum^{n}_{i=1}\mu_idn_i$. For an equilibrium case, which means that $\Delta G = 0$, it can be shown that if you expand this expression out for a reaction mechanism (like the one mentioned above), you end up with an expression like: $\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$. This particular equation is defined for standard conditions (0 degC, 1 atm).

--

We're almost to an answer, let's look a little closer at $K_{eq}$. Let's use a gas phase system $\ce{aA + bB <=> cC + dD}$ as an example. If you look up the derivation of the $\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$ in any standard thermodynamics textbook, you will see that $K_{eq}$ is actually equal to the ratio of the mol fractions of each species multiplied by the system pressure: $K_{eq} = \frac{y_C^c y_D^d}{y_A^a y_B^b} P^{c + d - a - b}$. If one assumes that the pressure is fixed and that $a = b = c = d = 1$, then this expression would reduce down to: $K_{eq} = \frac{y_C y_D}{y_A y_B}$.

You can further add detail to this expression by defining the mol fraction in terms of the extent of reaction $\xi$ for a reaction system.
Where, $n_i = n_{io} + v_in_{io}\xi$--which essentially says that the mols of a given species is equal to inital mols +/- the extent to which that chemical is generated or reacted away (based on the specified chemical of reference $n_{io}$). If you plug this into the expression for $K_{eq}$ you can see that now the equilibrium conversion of a reaction system is dependent on the amount of mols of each species. For the system described above, the total mols $N_{total}$ will cancel out of the expression leaving:

$K_{eq} = \frac{(n_{Ao} + v_An_{io})(n_{Bo} + v_Bn_{io})}{(n_{Co} + v_CN_{io})(n_{Do} + v_Dn_{io})} = exp(-\frac{\Delta G^o_{T_o}}{RT_o})$

For a given temperature and pressure, the Gibbs Free Energy ($\Delta G$) will be a fixed value. Therefore the conditions in the equation (composition of the mixture of reaction system) must balance in such a way that the equilibrium conversion $\xi$ will satisfy the condition. You can see here that as one either varies products (C,D) or reactants (A,B) that the solution for $\xi$ will change. You can generally show that an excess in either products or reactants will shift the system to maximum equilibrium conversion $\xi$. And as such, if one suddenly added or removed products or reactants you would cause a shift in the equilibrium conversion $\xi$.


Solvent for copper(I) iodide, copper(I) thiocyanate, silver iodide for HTM deposition; safety for methylammonium lead iodide I'm a student working in a high school laboratory that is looking to pursue research into perovskite cells for next year, but I have some solubility and safety questions.

What would be a good solvent (please be specific) of copper(I) iodide?, copper(I) thiocyanate?  Silver iodide?

I want to deposit these materials out of solution onto the substrate.  Christians et al. in 2014 deposited copper(I) iodide from a 1:39 chlorobenzene/di-n-propyl sulfide solution, but my mentors have rejected chlorobenzene as a substance with high environmental toxicity and difficult disposal procedures.

Ideally, the solvent for any of the three compounds would dissolve the compound well, and then evaporate off with heating, while having no to low toxicity.

Additionally, I have received approval for purchasing methylammonium iodide and lead(II) iodide which I will dissolve in gamma-butyrolactone in 1:1 molar.  There are no safety sheets for the product, methylammonium lead iodide, so my mentors wanted me to see if there were any harmful properties that would come about from making this compound, ie: is there anything more dangerous or something more to watch for from methylammonium lead iodide than methylammonium iodide or lead(II) iodide?

Article: An Inorganic Hole Conductor for Organo-Lead Halide Perovskite Solar Cells. Improved Hole Conductivity with Copper Iodide

Tell your mentor that to be concerned about chlorobenzene toxicity on perovskites is riddiculous. The most toxic thing in perovskites is the lead iodide...chlorobenzene´s toxicity is much lower than that. If he is so worried about toxicity, he should not work on perovskites.


Buffer Solution Problem (Acid/Conjugate Base) To create a buffer mixture, I understand that mixing an equal concentration of a weak acid and it's salt (conjugate base) will form an acidic buffer.

However, why is it that when $\pu{25 cm^3, 9.100 M}$ $\ce{CH3COOH}$ and $\pu{50 cm^3, 0.100 M}$ $\ce{CH3COONa}$ can also form a buffer solution?

Is there any specific rule for the creation of buffer solution from weak acid and it's salt?

There is no such rule that you need equal concentrations of weak acid and salt of conjugate base to make a buffer. (In this case acidic buffer)

A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.

In this case, if the solution contained equal molar concentrations of both the acid and the salt, it would have a $\mathrm{pH}$ of $\mathrm{4.76}$ because $\mathrm{pK_a}$ of acetic acid is $\mathrm{4.76}$. It doesn’t matter what the concentrations were, as long as they were the same, the $\mathrm{pH}$ remains at $\mathrm{4.76}$.

You can change the pH of the buffer solution by changing the ratio of salt to acid, or by taking a different acid and one of its salts.

Note: The $\mathrm{pH}$ of a buffer is calculated using Henderson-Haselbalch equation, which is,
$$\mathrm{pH = pK_a + log\frac{[Conjugate Base]}{[Acid]} }$$

Also, as indicated by @VicNeal in the comments, the buffer solution has highest buffer capacity at equal concentrations of the constituents, meaning its more effective at resisting the change in $\mathrm{pH}$.


What is the melting point of diamond? Textbooks and the online reference differ about this and there are more than two answers.

Diamond (carbon) does not melt at 1 atm. It sublimes to vapor.

Using carbon's theoretical phase diagram below (from Wikimedia), "liquid diamond" could be achieved at about 10 GPa (99 thousand atmospheres) and 5000 K (4700 °C).

Edit:

In fact, heating up diamonds at 1 atm turns them into graphite first. Only further heating would cause sublimation, as expected by observing the phase diagram. Here's a video:

http://www.youtube.com/watch?v=7L7BV3IBfFA


How do quaternary ammonium salts catalyse SN2 reactions? I am asked to identify the "best catalyst" for the reaction:

$$\ce{CH3(CH2)8CH2Br->[\ce{KCN}][\text{benzene}]CH3(CH2)8CH2CN}$$

The options are (A) benzyl chloride (B) aniline (C) N,N,N-trimethylbenzylamine chloride, and (D) N-phenylethanamide.

Now, at first, I immediately thought that the catalyst chosen may have to prevent a possible ring formation, given there are already ten carbon atoms. However, I am not sure how any of those catalysts help prevent that (I doubt if a ring would even form though, because there isn't any nucleophile, but those ten carbons look suspicious!)

Anyway, I realized that this catalyst may have to be polar aprotic for the given SN2 to occur. But, all of the given catalysts are pretty protic (resonance delocalisation of the benzyl carbanion with the ring) I would rank them on proticity as (D) > (C) > (B) > (A).

Hence, although all are protic, I may make a vague guess that (A) is the least protic and hence the answer. However, the answer given is (C) and the solution instead states that:

  Solvation of nucleophile is very less in presence of quaternary ammonium salt.

So, how? Where did the part about acidity go? And why worry about solvation? There doesn't appear to be any factor related to it at all (I read from here). Even so, how exactly would quaternary ammonium salts be poorer "solvators" than the others, say option (A)? Can I even predict this theoretically?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q11 in Amines

Trimethylbenzylammonium chloride is in this reaction a good example of a phase transfer catalyst. It first forms the trimethylbenzylammonium cyanide in aqueous phase (or on the surface of solid phase) which is soluble in the organic phase. Because of the polar nature of the cyanide ion it is not going to be heavily solvated by organic solvent (in this case benzene) molecules so the result is 'naked' cyanide which is highly reactive. This reacts with the substrate forming, in this case, the cyanodecane and trimethylbenzylammonium bromide. The trimethylbenzylammonium bromide is then free to migrate back to the aqueous phase and exchange anion to reform trimethylbenzylammonium cyanide. Most phase transfer reactions benefit from vigorous stirring to accelerate the exchange between the phases.

Further reading in this PDF.


Is hydrogen (on its own) considered an acid? An acid is:

  A molecule or other species which can donate a proton

So when hydrogen appears by itself on one side of a dissociation reaction e.g. $\ce{HOCl  ->   H+ + OCl-}$ is the hydrogen an acid? As the hydrogen can donate itself?

Well, you can never have H+ by itself. In aqueous media it's always associated with the lone pairs of water molecules and exists as H3O+. But yes, theoretically speaking H+ is a Brønsted–Lowry acid as the definition of such is to donate a proton and it donates itself to water in the case. It's also a Lewis acid as by donating itself it accepts electrons in its 1s emptry orbital.

But be aware that in water, you never have H+, it's always H3O+ which is the strongest acid that can exist in water solutions, all acids are limited by the strength of H3O+ when dissolved in water. Therefore, a species is an acid in water if it generates H3O+. In other solvents, for example liquid NH3, the corresponding species is NH4+ and an acid in ammonia is defined as species that generates NH4+. This whole paragraph relates to the solvent definition of acids and bases.

I hope all that answers your question, please comment if you want anything clarified.


What is the proper description/name of this amide/ester structure? What is the proper description/name of this amide/ester structure?

In IUPAC speak it is a carbamic acid methyl ester.

The IUPAC name for the compound drawn according to Chemspider structure search is Carbamic acid, N-[5-(propylthio)-1H-benzimidazol-2-yl]-, methyl ester  1

Most working chemists would call it a methyl carbamate


How do I make a ZMAT or XYZ file from scratch? From a structure representation like the one below for Bacteriochlorophyll A, how do I make the ZMAT or XYZ file?
The closest similar question asked here was about converting a PDB entry into XYZ, but what I have here is not in PDB format, it is just the image of the molecular structure, reported in a paper.

I know of two three ways to do this, but in either case it's quite a bit of work since I don't know any software that can read molecular structures from pictures. The second and third way use freeware only.

Draw the structure in Chem3D. For whatever stupid reason Chem3D doesn't support xyz files but you can save as *.pdb. If you need *.xyz then:

save as Gaussian input
delete the first 3 lines
Put the total number of atoms in the resulting first line
remove any "atoms" labeled as LP (lone pair, Chem3D uses them for force field optimization)
remove the second column next to the elemental symbol (if you sue Notepad++ you can select that by using ALT)
change file extension

Use openbabel 2.2.0 or higher to convert SMILES into *.xyz or *.pdb using the "gen3d" keyword. Of course for this you need the SMILES so you either have to produce that yourself or you need to draw it in a software that is able to produce SMILES like, again, Chem3D or this free website
It turns out this NIH website is also able to produce 3D structures, also in pdb file format, directly from SMILES or drawing the structure in their editor.


Is Gibbs free energy the sole determiner for spontaneity? If a reaction has negative $\Delta G$, can we then say with certainty that the reaction will be spontaneous?

Is negative $\Delta G$ the sole determinant for spontaneity?

For a spontaneous process at constant temperature and pressure, $\Delta G$ must (emphasis on must) be negative.

Spontaneity of a process , explicitly, depends on the enthalpy change and the entropy change. The Gibbs free energy change includes both these factors to predict the spontaneity of a process.

So, yes it can be said that Gibbs free energy change is the main factor that decides the spontaneity of a process.

Putting into better words, $\Delta G$ being negative is a consequence of a process being spontaneous.


Angular Momentum Of S-subshell of an atom? The angular momentum of every S-subshell of an atom is 0 by Azimuthal Quantum No. Relation. But if angular momentum of S-subshell is zero. Then by,
Angular Momentum =Mass×Velocity×Radius;
Radius of S-subshell shoul be 0 which is not the the case. Then why do we consider Angular Momentum of S-subshell zero.

Your confusion arises from the wrong choice of equation for the angular momentum. Your equation

\begin{align}
\lVert \vec{l} \rVert = m \lVert \vec{v} \rVert \lVert \vec{r} \rVert \ ,
\end{align}

where $\vec{l}$ denotes the angular momentum vector ($\lVert \vec{l} \rVert$ is its length, i.e. the angular momentum), $\vec{v}$ the velocity vector, $\vec{r}$ the radius vector  and $m$ the mass, does already entail some assumptions that are not necessarily valid. It derives from the more general equation

\begin{align}
\vec{l} = \vec{r} \times  \vec{p} ,
\end{align}

where $\vec{p}$ is the momentum vector which under certain circumstances can indeed be written as $\vec{p} = m \vec{v}$ (there are a few subtleties that I don't want to touch here, keyword: canonical momentum, since they don't relate to the problem at hand) and $\times$ denotes the cross product between two vectors. If you now take the absolute value of the angular momentum vector and use the properties of the cross product you arrive at

\begin{align}
\lVert \vec{l} \rVert &= \lVert \vec{r} \times  \vec{p} \rVert \\
&= \lVert \vec{r} \rVert \lVert \vec{p} \rVert \sin \theta \ ,
\end{align}

where $\theta$ is the angle between the vectors $\vec{r}$ and $\vec{p}$. So, the factor that is missing in your equation is the $\sin \theta$ term, whose absence means that your equation implicitly assumes that $\theta = 90°$ since $\sin 90° = 1$. And indeed if this would be the case, then for $ \lVert \vec{l} \rVert = 0$ to be true either $ \lVert \vec{r} \rVert$ or $ \lVert \vec{p} \rVert$ had to be equal to zero. However, since this assumption isn't made anywhere in the quantum mechanical description of an atom the $\sin \theta$ term has to be kept and there is another way to achieve $\lVert \vec{l} \rVert = 0$, namely $\vec{r}$ and $\vec{p}$ being either parallel ($\theta = 0°$) or antiparallel ($\theta = 180°$) to each other because then the $\sin \theta$ term is equal to zero.

Of course, for the quantum mechanical description you have to replace the vectors by the quantum mechanical operators and the absolute values of the vectors by the expectation values of the operators.


Clemmensen Reduction If 4-hydroxy-3-methoxybenzaldehyde undergoes Clemmensen reduction, will the hydroxyl group (OH) be converted to a chloro group (Cl)?

I know that it happens in cyclohexanes, and I am not sure if this is true for aromatic compounds as well.

You're right about the side reaction (hydroxy-halogen exchange) in the case of cyclohexanols. It is nucleophilic substitution on a (secondary) alkanol under acidic conditions.

In the case of vanillin, your starting material, the situation is different. Under the conditions of the Clemmensen reaction, that is in the presence of (amalgamated) zinc and hydrochloric acid, a nucleophilic aromatic substitution can be ruled out.


Lennard-Jones diagram - what is on the x-axis? I have a Lennard-Jones diagram here where I can't figure out, what is on the x-axis.

G. Ertl, Angew. Chem. Int. Ed. 2008, 47

We are looking at a diagram for a molecule $\mathrm A_2$ composed of two $\mathrm A$ atoms that comes into close contact with a surface.

If the molecule $\mathrm A_2$ forms a bond with the surface, a shallow minimum ($\mathrm A_{2,ad}$) is reached.
$E_{diss}$ is the dissociation energy needed to seperate/dissociate the two $\mathrm A$ atoms from each other (breaking the molecule). In this form (as two individual atoms), they could both seperately bind to the surface, resulting in much stronger bond formation $2 \mathrm A_{ad}$. This would be called chemisorption, if I'm not wrong.
Now combinding these two processes, we would have dissociative chemisorption. The crossing point between the two curves markes the activation energy of this.

This was just an explanation of the graph. My question is, how I should understand the horizontal dimension here - the x-axis. On a google search many diagrams pop up, but never with the dissociation included. Many of those as well as the Wikipedia article about Lennard-Jones potential show the x-axis as a distance between atoms - the interatomic distance $r$.

I cannot make sense of that on the diagram I am showing above. The distance between atoms should be smaller for the molecule than for the seperated atoms. What is on the x-axis in this case?

Ahh, silly me. The x-coordinate is of course just the distance between the $\mathrm A$ atoms / $\mathrm A_2$ molecule and the surface, to which they bind. This diagram doesn't show the interaction between $\mathrm A$ atoms but between them and the surface. I will close this question as solved.


Difference between conformational, constitutional, and structural isomers and the same and different molecules Consider the following sets of problems:

I was hoping someone could clarify what each checkbox means exactly. All of these answers are incorrect (except the first one with the check marks signifies the correct answers).

For example, why are the first pair different molecules? What makes something a conformation/constitution/structural isomer?

What does it mean to say something is an "identical conformation of the same molecule" vs "different conformation of the same molecule?"

When can I definitiviely tell if something is an enantiomer or just a diastereomer? It seems like (for an enantiomer) you can't just do the check of "switch all dashes to wedges and all wedges to dashes" because, as in the third pair of newman projections, they are enantiomers of each other but I don't understand why that is.

I'm going to assume that you don't need an explanation for the checkboxes the same molecule and difference molecules, so that leaves the following:

The terms constitutional isomers and stereoisomers make up two broad categories of isomers (molecules with the same chemical formula).  Constitutional isomers typically have different connectivities and stereoisomers have the same connectivities but differ in spatial arrangements.

enantiomers are stereoisomers that are non-superimposeable mirror images of one another.  Your hands are (roughly) enantiomers.  both are optically active is true of enantiomers.

diastereomers also known as geometric isomers are also stereoisomers that are not mirror images of one another. cis- and trans-dichloroethene are diastereomers.

If you do a web image search of isomer flowchart you'll see a lot of informative information on isomers.  Linked below is the public domain version from Wikipedia which is an example (although not the best in my opinion.)


Reaction of phenol with concentrated sulfuric acid What is the product when phenol reacts with conc. $\ce{H2SO4}$ in excess?

I have been told that the major product will be ortho at low temperature and para at high temperature. Why cannot it form both ortho- and para-substituted product as in the case of picric acid?

It's a reversible reaction, so at higher temperature the thermodynamic para product is formed from the kinetic ortho product.

Further reading here 1


Flame test color for non-ionic compounds What color would a non-ionic compound, such as $\ce{NH4OH}$ or $\ce{HCl}$, produce in a flame test? Is no color produced, or is it different for each compound?

Flame emission (or better atomic emission) does not care whether your starting compound was ionic or not. Everything decomposes into flame either into atoms or very simple molecules. The blue color you see from the flame is also from very small molecules which exist in the flame.

https://en.wikipedia.org/wiki/File:Spectrum_of_blue_flame.png

Ordinary flame like the Bunsen burner is way too cold for non-metals. However if we were to heat HCl or ammonium hydroxide in a confined "container" at very high temperatures say, 5000 - 10,000 $^oC$, eventually you would start to see atomic emission from hydrogen, oxygen, nitrogen and chlorine. Most of the non-metal emission (p-block) lies in the UV or deep UV. You won't be able to see anything with bare eyes.


How many hydrogen bonds in these compounds? We usually get some questions about how many hydrogen bonds these compounds can create, we use it to differentiate between the boiling point of each one of them so it would be great if you could help me.

So what I know that $\ce{H2O}$ has 4 hydrogen bonds those are 2 from hydrogen atoms and 2 from other molecules that has 2 hydrogen atoms connected to the 2 lone pair of electron that the oxygen has.

In $\ce{HF}$, It has 2 hydrogen bonds. The same explanation implies here.
In $\ce{NH3}$, 2 hydrogen bonds because $\ce{N}$ doesn't have the required amount of lone pairs to make more hydrogen bonds.

But what I really want to know is about alcohols. In the way that I am explaining it, it should have 2-3 hydrogen bonds but when I check some books and articles they say that it only creates 1.

So if you can explain how many each of them creates and what it depends on that would be awesome.

Hydrogen bonds require a $\ce{\delta{+}}$ hydrogen atom capable of forming a hydrogen-bond, so $\ce{H2O}$ makes 2 H-bonds, and $\ce{NH3}$ has 3 H$\ce{\delta{+}}$ atoms capable of forming these bonds.

However hydrogen bonds also require an "active" lone pair capable of accepting a hydrogen-bond, where $\ce{H2O}$ can do 2 of these again, but $\ce{NH3}$ can only accept 1.

So, does water make 4 bonds or 2? Does ammonia make 1 bond or 2?  The answer is both, but it depends on how you count it.  If you look at an individual molecule in a cluster of water molecules it may appear to be surrounded and held in place by four H-bonds; it makes two with the hydrogen and accepts two with the lone pairs.  That molecule is involved in 4 hydrogen bonds.

But if you take 100 water molecules and count how many hydrogen bonds there are between them, the answer will be about 200 because each molecules makes 2 bonds.  If you consider each molecule making 4 bonds then you are double counting each bond being made and accepted.  The same is true for ammonia, each molecule makes one hydrogen-bond, it also accepts one.  You could say that an individual molecule is involved in two bonds, but if you count them all up each molecule only makes one each.

Because of these two different viewpoints, some texts will say water makes 4 and alcohols make 2 bonds, and some will say water makes 2 and alcohols make 1 bond.  They are both right, but right in different ways. Whichever way you count them, you must be consistent. If this is related to an exam syllabus, the correct way is defined by that.


How to calculate the heat capacity of a calorimeter based on a neutralisation reaction? I've looked at the other similarly worded questions but I can't seem to make heads or tails out of them.

So here's my case. I need to find the heat capacity of the calorimeter (in Joule/degree Celsius), and I've been provided with the following information:

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

$\Delta T = 11.5\ ^\circ \ \text{C}$
$\Delta_r H = -58.3 \text{ kJ/mol}$
$s_{p,\ce{NaCl}} = 3.91 \text{ J/(g}\cdot \ ^\circ \text{C)}$
$d_{1\ \text{M NaCl}}=1.0037\text{ g/mL}$
$V= 100 \text{ mL}$

So far I've calculated the mass of $\ce{NaCl}$:

$$m = d\cdot V = (1.037 \text{ g/mL})(100 \text{ mL}) = 103.7 \text{ g}$$

Then I calculated the heat capacity of $\ce{NaCl}$...at least I think that's what I did. I'm not even sure if I was supposed to do this:

$q = m\cdot s_{p,\ce{NaCl}}\cdot \Delta T$
$q = (103.7 \text{ g})(3.91 \text{J/(g} \cdot\  ^\circ \text{C}))(11.5 \ ^\circ \text{ C})$
$q = 4662.8 \text{ J}$

Now I'm confused as to the next step, and I don't know how to use $\Delta_r H$ in this problem.

Any help would be immensely appreciated.

Edit: Below I've added more things I've tried in my attempt to find the answer.

I saw this equation in a textbook so I gave it a shot:

$q = - \Delta H\cdot n_{\text{limiting reactant}}$
$q = (- 58,300 \text{ J/mol})(1.62 \text{ mol HCl})$
$q = 94,446 \text{ J}$

But I don't know how/if this fits anywhere in the problem.

Another equation I tried:

$q = C_P\Delta T$
$-58,300 \text{ J}= C (11.5 \ ^\circ \text{C})$
$C = 58,300 \text{ J} / 11.5 \ ^\circ \text{C}= 5069.5 \text{ J/}\ ^\circ \text{C}$

Which is at least the correct unit of the heat capacity I'm supposed to find.

Finally, I tried subtracting the heat capacity I found for $\ce{NaCl} \ (4662.8 \text{ J})$ from the supposed heat capacity of the system $(5069.5 \text{ J})$, I get $406.76 \text{ J}$

I'm kind of at a loss...

First, you have to find out the concentrations so you can calculate the amount of heat released. Since the density is given for a 1 M solution, I will assume this concentration.

Next, the mass of our solution:
$$ m = \rho \times V = 103.7~\mathrm{g}$$

Now we have to calculate the amount of energy that gets released through the reaction, but viewed from the environment and not the reaction (that's the reason for the negative sign before $q$):
$$ -q = \Delta_\text{r}H \times n = -58.3 \times 0.1 ~\mathrm{kJ\, mol^{-1}\, mol} = -5.83~\mathrm{kJ}$$

The total heat capacity of your system is given as
$$ c_\text{tot} = c_\text{s, sol}\times m + c_\text{cal}$$
where $c_\text{s, sol}$ is the specific heat capacity of your solution and $c_\text{cal}$ is the heat capacity of the calorimeter.

Now we know all the quantities needed for the following equation:
$$ q = c_\text{tot}\Delta T$$

We expand:
$$ q =( c_\text{s, sol}\times m + c_\text{cal}) \times \Delta T$$

Solve for $c_\text{cal}$:
$$ c_\text{cal} = \frac{q-c_\text{s, sol}m\Delta T}{\Delta T} = \frac{5.83~\mathrm{kJ} - 3.91~\mathrm{J\, g^{-1}\, K^{-1}} \times 103.7~\mathrm{g} \times 11.5~\mathrm{K}}{11.5~\mathrm{K}} = 101~\mathrm{J\, K^{-1}}$$

And now you know the heat capacity of the calorimeter.


Is FeS2 both an ionic and covalent compound? Iron (II) disulfide.  Iron pyrite.  Fool's gold to some.  But what bond does it fall under?  It could possibly be used for both?

This would give a deeper understanding into other like examples, where it is a fine line between the determination of an ionic compound or a covalent compound.

Why do you feel a need to classify it under one or another? Chemistry is not about having a list of ionic compounds and another list of covalent compounds.

There are "ionic" bonds (see below) between $\ce{Fe^2+}$ and $\ce{S2^2-}$, and there is a $\ce{S-S}$ covalent bond in $\ce{S2^2-}$. That's all there is to it.

If I absolutely had to pick one, I would say it is an ionic compound. $\ce{S2^2-}$ is a covalently bonded species; $\ce{FeS2}$ is an ionic crystalline solid. The fact that it melts at $1193~^\circ\mathrm{C}$ supports this "classification".

As @Mith notes in the comments, all ionic bonds are not 100% ionic, and a lot of ionic bonds are in fact covalent to a large degree. The fact that $\ce{Fe^2+}$ is a transition metal ion (meaning it has empty 3d and 4s orbitals available for overlap) and also that sulfur is a rather polarisable ion (because of its larger size) means that there is likely to be significant covalency in the $\ce{Fe-S}$ bonds.

What does this mean? It just shows that trying to classify something as "ionic" or "covalent" is futile. $\ce{FeS2}$ is likely somewhere right in the middle of the spectrum.


Carboxylic Acid to Acyl Chloride Instead of using oxalyl chloride and DMF then $\ce{AlCl3}$, could $\ce{SOCl2}$ followed by $\ce{AlCl3}$ been used to achieve the same result? If yes, is one pathway preferred over the other and why?

This reaction comes from Simmons, E. M.; Sarpong, R. Ga(III)-Catalyzed Cycloisomerization Strategy for the Synthesis of Icetexane Diterpenoids:  Total Synthesis of (±)-Salviasperanol. Org. Lett. 2006, 8 (13), 2883–2886.

They should both work. I would use oxalyl chloride with a solvent like DCM. DMF should act as a catalyst here for the acid chloride formation. (I don't have access to the paper but this is most likely what they have done) The oxalyl chloride reaction is maybe easier to work-up although I don't think there is a big difference when working in lab scale. In this case you can skip work-up (assuming you have used DCM as solvent and 1-2 equivalents of oxalyl chloride) and just add $\ce{AlCl3}$ for the cyclisation step.

EDIT: You can read the experimental in the supporting info -page S7


How many grams of sodium sulfate and water in 100ml of 0.5M solution Calculate how many grams of sodium sulfate ($\ce{Na2SO4}$) and water ($\ce{H2O}$) should be used to prepare $\pu{370 g}$ of 15% solution of this salt. How many mL of this 15% solution should be taken to prepare $\pu{100 mL}$ of $\pu{0.5 M}$ solution of $\ce{Na2SO4}$?

My calculations:

$\pu{370 g}$ = 85% of water, 15% of sodium. So we need $\pu{370 g} \times 0.85 = \pu{314.5 g}$ of $\ce{H2O}$ and $\pu{370 g} \times 0.15 = \pu{55.5 g}$ of sodium sulfate.

We need $\pu{100 mL}$ of $\pu{0.5 M}$ solution.

The amount of sodium sulfate we need:
$$
\begin{array}{cc}
\pu{1 L} & -\quad \pu{0.5 mol} \\
\pu{100 mL} & -\quad \pu{0.05 mol}
\end{array}
$$
$$m = \pu{0.05 mol} \times \pu{142 g/mol} = \pu{7.1 g}$$

To get $\pu{7.1 g}$ of sodium sulfate we need
$$
\begin{array}{ll}
\pu{100g} & -\quad \pu{15 g} \\
x & -\quad \pu{7.1 g}
\end{array}
$$

$x = \pu{47.3 g}$ of solution that's over $\pu{50 mL}$ of it.

But the answer states we need $\pu{3.75 mL}$ of solution.

Could you tell me where my reasoning is wrong or it is a problem with the answer?

As Jason indicated your calculations are corrected.

370 grams of 15% solution of sodium sulfate does have 314.5g $\ce{H2O}$ and 55.5g $\ce{Na2SO4}$.
Your calculation for grams of solution are correct. 47.3g of solution is required.

However the density would be expected to be somewhat above 1.00 g/ml not less. So a little less than 47.3 ml of solution would be required. However the density of the solution is nowhere near 10 g/ml so the book answer is horribly wrong.

To solve for ml of solution, I'd have just stated an assumption that the density of the solution was 1.00 g/ml since you weren't given that information. There is absolutely no way to calculate the density given the information in the problem.


Do organopolonium compounds exist? Analogues of alcohols exist for all the heavier Group 16 elements, namely sulfur, selenium, and tellurium. Would polonium also be able to form a "polonol" like $\ce{CH3PoH}$?

As one might expect for an element with no long-lived isotopes, little is known about organopolonium compounds. What is known may be broadly divided into two categories [1,2; cited by Wikipedia]:

Compounds where polonium takes the place of a lighter chalcogen, acting more nearly parallel to sulfur, selenium and tellurium than to oxygen. Examples include polonoethers $\ce{R2Po}$ and arylpolonium halides $\ce{Ar2PoX2,Ar3PoX}$.

Complexes with chelating agents similar to metal complexes.  WP cites complexes with 2,3-butsnediol and theoretical from Ref. [1].

All such compounds are known only in tracer amounts because of both chemical instability and rapid radioactive decay.
References
1.
Zingaro, Ralph A. (2011). "Polonium: Organometallic Chemistry". Encyclopedia of Inorganic and Bioinorganic Chemistry. John Wiley & Sons. p. 1–3. doi:10.1002/9781119951438.eibc0182.
2.
Murin, A. N.; Nefedov, V. D.; Zaitsev, V. M.; Grachev, S. A. (1960). "Production of organopolonium compounds by using chemical alterations taking place during the β-decay of RaE" (PDF). Dokl. Akad. Nauk SSSR (in Russian). 133 (1): 123–125. Retrieved 12 April 2020.


Explaining why CN- is a soft base So, I am having difficulty with the following explanation/question.

  (Based on your MO diagram and pi bonding theory) explain why $\ce{CN-}$ is a soft base and prefers to bind to soft acids.

Now, I know that the HOMO is the 5 σ MO and that the electron pair in the 5 σ MO is, to a large extent, more concentrated around the carbon.  So, is $\ce{CN-}$ a soft base because it has a low energy HOMO but large magnitude HOMO coefficient?  Does polarizability have anything to do with it?

When $\ce{CN-}$ reacts with a soft acid what physical changes in the cyanide molecule would such a reaction lead to?

Not really sure about this one.  Would it have to do with the donation of the electron pair in the HOMO and therefore change the HOMO?

In one of his great answers, Martin gave us the calculated molecular orbitals for carbon monoxide I am copying below. Note that while cyanide has a nitrogen atom in place of the oxygen and features a single negative charge, we can still qualitatively consider the MO schemes identical; thus I will base my discussion on what you see in the image.

In this scheme, the HOMO is referred to as MO number 7; since 5 and 6 are π-type MOs, this corresponds to the 5σ you mentioned.

One thing we immediately note is that the carbon-centred lobe of the HOMO is very diffuse and extends far into space; much farther than any other orbital of the molecule. It is also the orbital that an acid (either proton or Lewis acid) would bond to. In Pearson’s HSAB theory, large and diffuse orbitals — often also called polarisable — are associated with soft acids or bases, as the charge per volume value is considered comparatively low. Therefore, just by looking at the size of this orbital, we can identify it as a ‘soft’ one.

It should also be explained why soft bases tend to create more stable complexes with soft acids. This is a direct consequence of the more diffuse nature of the orbital: the overlap between two orbitals (which will later form a bond) is typically best if the orbitals have similar shapes and sizes. Thus, because the HOMO has a rather diffuse lobe on carbon’s side it will prefer binding to other compounds with rather diffuse orbitals in a Lewis acid-base reaction.

That should be all that is required to explain cyanide’s behaviour under Pearson’s theory.


What are the benefits of using zero capillary voltage for discarding fractions in LC–ESI–MS/MS? In the second episode of The Association for
Mass Spectrometry & Advances in the Clinical Lab (MSACL) podcast, namely Getting going with mass spectrometry : Josh learns chromatography (aired 2021-01-07), Dr. Russell Grant mentioned the following tip at 01:03:11 (I apologize in advance for any inaccuracies in the transcript):

The other thing that is often forgotten there is that you can drop the capillary voltage to zero. It's not as good as the diverter valve, but you are not going to be pulling the stuff towards the mass-spectrometer. If you only have a couple of windows and a bunch of junk, you can just keep capillary voltage on during these desired windows and save your diverter valves' maintenance time because they can become maintenance-limiting.

I would think it would lead to a more frequent cleaning of the ionization chamber. Is the gain in postponed maintenance of a diverter valve really worth it?
Are there any other use cases when disabling capillary voltage could be a superior alternative to using a diverter valve?

I think there is no perfect answer to this situation. In general when the flow is stopped to any detector in an HPLC experiment, there is a change in baseline, and when the flow is re-started, ghost peaks appear in chromatogram. I am talking about optical detectors such as UV, fluorescence, refractive index etc. Same thing happens in MS and perhaps LC ESI-MS is not an exception, in the sense that diverting the flow can lead to the appearance of so-called "ghost" peaks or raised baseline. I have seen that in GC-MS experiments where a change in flow causes weird baseline shifts. This is an interesting phenomenon worth investigating.
So the authors suggest that let us continue the flow, but let us stop the ionization process so that the junk does not go inside the quadrupoles and stay there forever. It is always easy the clean the external ESI system but the internal system needs professional service.
Someone talks about these problems in this forum Diverter valve issues