Heat and work of glucose combustion $$\ce{C6H12O6(s) + 6 O2(g) -> 6 CO2(g) + 6 H2O(g)}$$
When looking at glucose combustion I found that the change in Gibbs free energy is greater than the change in enthalpy. If we assume the process is reversible we can set the change in Gibbs free energy to be the work available and if we assume there is no work we can set the enthalpy to be the heat. Why then would the work be less than the heat? I suspect that $\Delta G = \Delta H-\Delta (TS)$ is at work.

Cool, so can you substitute work, W and heat, Q into the equation and write W=Q−TΔS to directly show that ΔS>0? Also would you please explain how an increase in entropy would imply the discrepancy between heat and work (not mathematically).

For the process to go forward, the entropy of the universe has to increase. For this specific case, the reaction is exothermic and the entropy of reaction (entropy change of the system) is positive. If all of the enthalpy is released as heat, the entropy of the universe will increase (the entropy of the surrounding increases because of the heat transfer, and the entropy of the system increases because molecules are more dispersed in the gas phase).

If there is a mix of work and heat exchange, you just have to make sure the entropy of the universe still increases. In this case, the work possible exceeds the enthalpy change. When you do maximal work, you have heat transfer into the system. The entropy decrease of the surroundings is offset by the entropy increase in the system.

So in this case $$|\text{work}| > |\Delta H|$$

For a picture, see https://chemistry.stackexchange.com/a/112958 scenario 2.

  Cool, so can you substitute work, W and heat, Q into the equation and write W=Q−TΔS to directly show that ΔS>0?

No, where there is work, enthalpy change is no longer equal to heat.

Also, you have to specify which ΔS you are talking about. If it is the entropy change in the system (due to the reaction) $\Delta S_\text{sys}$, the equation below is a variation of the second law (the left-hand side is equal to the entropy increase of the universe times temperature):

$$T \Delta S_\text{sys} - q_\text{rev} > 0$$

And if you want to relate work to enthalpy, you use the first law:

$$\Delta H = q + w$$

and insert it into equation above to get:

$$T \Delta S_\text{sys} + w - \Delta H > 0$$

or for maximal work with a bit of rearrangement,

$$  w_\text{max} = \Delta H - T\Delta S_\text{sys}$$

So you can say that if the work you get out of a reaction is larger than the enthalpy change, the entropy of the reaction has to be positive.


How does a Lewis acid differ from an oxidizing agent; Lewis bases from reducing agents? A Lewis acid is defined as an electron acceptor. An oxidizing agent is also defined as an electron acceptor (in the electronic theory of oxidation).

So then what is the basic difference between an Lewis acid and an oxidizing agent?

Similarly;

A Lewis base is defined as an electron donor (Lewis theory). A reducing agent is also defined as an electron donor (electronic theory). Therefore, what is the difference between Lewis bases and reducing agents? Are they exactly same thing?

There is a similar question which, however, only considers bases and nucleophiles and does not go into detail about oxidation and reduction.

An acid and an oxidising agent are very different concepts that apply to different phenomena — although a single compound can be both at the same time, e.g. $\ce{H2SO4}$. However, a compound can also be an acid and a reducing agent, e.g. $\ce{HI}$.

The Brønsted acid/base theory is clearly distinct from any type of redox reaction, so I suppose that the confusion lies with the extension of said theory that the Lewis theory provides. But first a short word on the Brønsted theory: therein, an acid is a compound that can transfer a $\ce{H+}$ ion onto another compound.

The Lewis theory realises, that the acidic transfer reagent need not be hydrogen. Instead, anything with an unoccupied low-energy orbital can function as a Lewis acid; most notably boron compounds and metal cations. Upon creation of a Lewis acid-base pair, a coordinate bond is formed. Basically, this means that the energy of the lone pair of the Lewis base is lowered because it interacts with the empty orbital of the Lewis acid whose energy is raised. This can be depicted as $\ce{LB\bond{->}LA}$. The key concept is that the interaction is fully reversible and that upon dissociation the Lewis base keeps its lone pair.

When discussing oxidising or reducing agents, however, the electrons are irreversibly transferred. They actually move from orbitals of the reducing agent (which will be oxidised) into orbitals of the oxidising agent (which will be reduced). This transfer is thermodynamically favourable and thus the electrons won’t just ‘go back’.

To exemplify this, let’s consider two reactions of $\ce{H2SO4}$; one where it is an oxidising agent and one where it is an acid (Brønsted type; the proton of $\ce{H2SO4}$ is actually the Lewis acid).

$$\text{Acid-base:}\\
\ce{H2SO4 (l) + NH3 (g) <=>> NH4+ HSO4- (s)}$$

Note that ammonia, our base, is a gaseous compound at room temperature. If you employ appropriate conditions (i.e. temperature and low pressure), the equilibrium will be drawn to the left, releasing ammonia as ammonia ($\ce{NH3}$ molecules) including its lone pair. $\ce{H2SO4}$ will stay behind.

$$\text{Oxidising agent:}\\
\ce{2H2SO4 + Cu -> CuSO4 + SO2 ^ + 2 H2O}$$

$\ce{SO2}$, being a gas, will diffuse away. However, even if you perform this in a pressurised vessel, there will be no significant back-reaction because the electrons have been fully transferred from copper to sulfur.

The argument for Lewis bases/reducing agents is the same in reverse.


Rate constants for water exchange Are rate constants for water exchange of transition metal aqua-complexes tabulated anywhere ? I haven't been able to find a resource which is anywhere near comprehensive.


NMR spectrum of cyclohexanol - hints on integration I've been given this NMR along with the IR and Mass Spec and assigned the task of figuring out the unknown compound. I know that it is cyclohexanol and by analysing the NMR I know that the peak at around 2.8 is the OH and the peak around 3.6 is the deshielded CH. However I can't pair up the other protons and the peaks between 1-2 ppm. I know that the difference between the peaks/protons is their vicinity to the OH group.

My questions is:
Is there any way to do the integral of the points lying between 1-2 ppm. This will be useful as I think that the peak at around 1.3 ppm represents more protons than the others. Also could the peaks be different due axial and equatorial protons resonating at different frequencies, or would the temperature

Or is there a better way to pair up the peaks between 1-2 ppm and the remaining protons in the cyclic.

To answer your first question about measuring the integrals for the peaks between 1 and 2 ppm: If you seriously need to do a very accurate integral of these peaks without having access to NMR processing software (ie you've just been given the spectrum as a handout), you can resort to one of the methods of choice from the good 'ol days.

Print your spectrum out on a nice big piece of paper, get a nice sharp pair of scissors and cut out your peaks and weight them (a 4 or 5 figure  balance may be required). These method gives incredibly accurate integrals
Print out your spectrum on graph/grid paper and count the grids. Again, very accurate, depending on the scale of grid you choose to print on.

As for 'pairing up' the remainder of the ring protons, this is not a trivial exercise, made even less so with the dispersion of the spectrum you are working with. You are quite right in your assessment. The protons closer to the -OH group will be at higher chemical shifts; equatorial and axial protons will be non-equivalent (equatorial at higher chemical shifts); and the whole fluxional ring dynamic will be temperature, concentration, solvent and field dependent. So, probably it is not worth trying to pair up any of these protons. You could (and hopefully would, if you are a thorough researcher) assign all of your peaks with the assistance of 2D COSY, HSQC/HMBC methods.


Does potassium nitrate solution degrade? I have created a solution of $\ce{KNO3}$ and table sugar in order to create fuse from hemp rope. I wet the rope in the solution and let it dry - I repeat this process several times.

It works and produces very slowly burning rope which is easy to set off. The problem is that after 24 hours, my solution started to stink. Something between garlic, onion and vinegar. I have a bad feeling about that.

Does my solution degrade, or is this stink caused by some impurities (the solution isn't very pure)?

As mentioned in the comments, a likely culprit is microbial contamination.  In a $\ce{KNO3}$ solution, you are providing plenty of water, a nitrogen source (nitrate), and even a high-energy, respiratory means for growth (nitrate respiration).

Living things also need (i) food, (ii) trace amounts of sulfur, (iii) and even smaller trace amounts of phosphorus.  But if you are soaking hemp in this solution, enough leachable S and P-containing impurities are probably extracted into the solution to provide these.  (Or also low levels of S and P may already be present in your $\ce{KNO3}$, or water.)  The food could be organic compounds leached from the hemp.  Or (less likely) the microbial contaminants might be capable of photosynthesis, in which case their food is atmospheric $\ce{CO2}%edit$.


What ratio of bicarb to vinegar do I need in order for the result to be pH neutral? This question stems from one I just asked over at Seasoned Advice, where my recipe tasted decidedly off because the pH was wrong.
The recipe called for $\pu{16 g}$ bicarb of soda (sodium bicarbonate, $\ce{NaHCO3}$) and $\pu{15 g}$ of vinegar (mine was 5% acidity, and I assumed a density of $\pu{1 g ml-1}$). I just googled the corresponding pHs of these two, and apparently vinegar is more acidic than bicarb is basic: vinegar has a pH of 2.4–3.4 depending on concentration, and bicarb is pH 9.5.
How can I calculate whether I put too much vinegar or too much bicarb into my recipe? The ratio in the recipe is almost $1:1$.
What ratio in grams do I need of bicarb vs vinegar in order for the result to be neutral? I know there are other contributing factors to the pH in the recipe but I just want to focus on these two first. Assuming the vinegar is 5% which mine was.
Edit:
So I found the explanation for the chemical reaction between bicarb and vinegar here. Relevant bit:

$$\ce{CH3COOH + NaHCO3 → CH3COONa + CO2 + H2O}$$
$$\frac{\pu{1 g}\;\ce{NaHCO3}}{\pu{84.0068 g mol-1}} \times \frac{1}{1} \times \pu{60.0522 g mol-1} = \pu{0.715 g}\;\ce{CH3COOH}$$
What this formula tells you is that every gram of baking soda requires \pu{0.715 g} of acetic acid. But the remaining problem is that vinegar is not pure acetic acid.  In fact, most vinegars vary from 4% to 18%.
Using those numbers, you would need somewhere between $\pu{0.715 g}/0.04 = \pu{17.9 g}$ and $\pu{0.715 g}/0.18 = \pu{3.97 g}$ of vinegar, for each gram of baking soda.
If you're lucky you can get some vinegar with the percent acetic acid listed on the label, then you can substitute that number into the equation above to determine how much vinegar you'll need.

So if I plug in my 5%, I get $\pu{0.715 g}/0.05 = \pu{14.3 g}$ vinegar for $\pu{1 g}$ bicarb.
Yikes! This is way more than $1:1$. But I'm not sure if it answers my question. Is the result, $\ce{CH3COONa + CO2 + H2O}$, pH neutral? If yes, then my vinegar does hardly anything to cancel out the base, is that right?
Edit 2:
Okay so according to Wikipedia, $\ce{CH3COONa}$ is a base. So even though my reaction formula shows what happens when all the bicarb/vinegar is used up, the result is not neutral (how much does the water and carbon dioxide contribute to bringing the pH down?). I need more acid to make my sodium acetate neutral (or at least neutral enough to be palatable?). This means that the needed ratio is higher than 14:1!? So almost certainly way too much bicarb?

Your calculation is correct (14.3 grams of 5% acetic acid completely reacts with 1 g sodium bicarbonate). But you're not interested really in "completely reacts"; you want something like pH 7.

You've got it just about right with your mix; you're at the endpoint, and this is a point where the pH will drop very rapidly with even a small addition of vinegar. So it would only take a few additional drops to make the pH go down to 7... but that may not be what you really want.

Sodium acetate has a mild salty taste by itself. So I'm not sure adding extra vinegar will be worthwhile, it'll make it a bit more vinegary.

Here's what I'd do: Mix vinegar and baking soda in the ratio you've figured out, and then see how that tastes. Make adjustments at that point according to taste, rather than by "titrating" to an arbitrary pH.


Deciding locants of double bonds vs. other substituents After having decided that the keto-group is the highest priority, this is a ketone. However, does the double bond become C2, or does the amino become C3?
In locant placement, do we prioritize double bonds over substituents, or other substituents over double bonds? (Or is it just a matter of having the smallest starting number?)

5-aminocyclohex-2-enone
3-aminocyclohex-5-enone

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) while numbering the parent chain, an unsaturation is preferred over a substitutent.

P-14.4 Numbering
When several structural features appear in cyclic and acylic compounds, low locants are assigned  to them in the following decreasing order of seniority.
.. (a) fixed numbering of a ring or ring system;
.. (b) heteroatoms in chains and in rings and ring systems;
.. (c) indicated hydrogen;
.. (d) principal characteristic groups and free valences (suffixes);
.. (e) added hydrogen;
.. (f) saturation/unsaturation (hydro/dehydro prefixes and ‘ene’ and ‘yne’ endings);
.. (g) detachable alphabetized prefixes, all considered together in a series of increasing numerical  order;
.. (h) lowest locants for the substituent cited first as a prefix in the name;
..

As per the preference order the ketone will get the priority first as it is the principal characteristic group. Thus the name will end in an -one.
Now as the unsaturation is preferred over a substituent the numbering will be done in such a way that the double bond gets the lower number.
Thus the preferred IUPAC name of this molecule would be (5S)-5-aminocyclohex-2-en-1-one.


Why do the 4f and 5d orbitals have similar energies? I've come to know this phenomena when learning about the exceptions to the aufbau principle. But there is no explanation of why 4f and 5d have similar energies (as I've read in some websites).

The energy of a hydrogenic orbital with principal quantum number $n$ is proportional to $-(Z/n)^2$, so as $n$ increases the orbitals become more and more bunched together.

In a multi-electron atom the degeneracy of orbitals with different orbital angular momentum $l$ is lifted, i.e. s/p/d/f orbitals are no longer equal in energy. Since the gap between $n = 2$ and $n = 3$ is fairly large, one would not expect the 2p and 3s orbitals to be particularly similar in energy.

But the gap between $n = 4$ and $n = 5$ is smaller, so it is inevitable and not all that surprising that some of those constituent subshells have similar energies when the degeneracy is lifted. This is how it looks like for lithium (diagram from Modern Spectroscopy 4th ed., Hollas, p 214). The diagram will not be the same for the 4f/5d elements – in particular, I would expect the 4p/4d/4f orbitals to have a more significant difference in energy – but the idea is the same.


How do I calculate the pressure of a known liquid in a sealed container heated above boiling point? If I have water in a sealed container heated to say 150 degrees, how do I determine the amount of pressure being generated in the container? What about for other liquids? I have searched extensively and cannot figure this out.

I was looking for a formula of sorts, but for example reasons let's say a 5000 ml container with 4500 ml of water in it, with the rest of the space air, heated to 150 degrees Celsius.

Note: for convenience, "gas" refers to any gas at a temperature beyond its boiling point; "vapor" refers to any gas evaporated from its liquid state, also implying that the liquid itself is below its boiling point, such as the 150 degrees water in your example.

Dalton's law of partial pressure would come in handy here. The law states that the partial pressure generated by each type of gas particles sum up to the total pressure in a sealed container. The partial pressure of "gasses" can directly be calculated using the ideal gas law. Of course the partial pressure of "gasses" is 0Pa if there is no gas but only vapor in your container.

Now the partial pressure of the vapor of liquids. In a sealed container, if left to attain dynamic equilibrium (which is the steady state in this case) the partial pressure of the vapor of any liquid would be equal to its equilibrium vapor pressure, which is a function of temperature only. Though not excessively accurate, the Antoine equation can be used to estimate this:
$\log P = A- {B \over C+T}$
where P is the equilibrium vapor pressure, A, B and C are substance-specific constants and T is the thermodynamic temperature of the substance. The three substance-specific constants can be checked up on the internet, or provided by other sources, whereas conducting experiments can be a last resort (as it is a very tedious experiment that you need the whole curve of the function to approximate the constants).

If the liquid in question is a mixture of more than one liquid substance, then you'd need a third law - Raoult's law to determine the partial vapor pressure of each component of the mixture:
$p_i = p_i^\star x_i$
where $p_i$ is the partial vapor pressure of component $i$ of the mixture, $p_i^\star$ is the equilibrium vapor pressure of a pure sample of component $i$(in itself, not in a mixture), and $x_i$ is the mole fraction of $i$ in the liquid.
At dynamic equilibrium, combining Dalton's law of partial pressure and Raoult's law yields the expression for total vapor pressure:
$p_{total} = \sum p_i^\star x_i$


Why doesn't Sodium (Na) form duplet? The atomic number of Na is 11 (2, 8, 1). Na loses an electron to achieve octet stability but why can't it just accept an electron to complete duplet stability?

Well, sodium does have non-zero electron affinity, so it surely can acquire another electron, provided it can get one for free. That's what happens when a lone sodium atom meets a lone electron somewhere in deep space. But here on Earth, with all that condensed matter around, we don't have many free electrons (*). They all belong to some other atoms, and if you want to get an electron (that is, to get reduced), you have to oxidize something else.

(*) Then again, some might recall metals in which the electrons are kinda free, but wait: they are not really free. They are bound by the collective field of metal cations, and you need to apply certain energy to take one out.

It turns out that $\ce{Na}$ and other alkali metals can actually be reduced to form alkalides, but those are exotic compounds. They are not particularly stable, and get oxidized with pretty much anything. It is much easier for $\ce{Na}$ to form a cation, thus achieving the stable octet, and call it a day.

There is a whole different dimension to the problem: why wouldn't two sodium atoms share their electrons, establish a bond, and form a diatomic molecule? Again, they would, but only in a vacuum! When you have many such molecules, the sodium atoms would share their other empty orbitals, find out they don't have enough electrons to fill them all, and just resort to being a metal.


Why does the surface of an acid appear to have a stronger "dissolving power"? In this Periodic Videos YouTube video by Brady Haran and the University of Nottingham, we see a piece of gold being dissolved in aqua regia, while Professor Poliakoff explains why aqua regia can dissolve gold even though its constituent acids can't.

The piece of gold was placed partway into the acid. I noticed that the acid appears to attack the gold the strongest at the liquid's surface, and the gold was first "dissolved through" along a line at the surface, leaving a little plate of undissolved gold dangling in the acid by one corner. You can see this best from about 4:10 to 4:30 into the video.

I was wondering if there's some effect at the surface of the liquid that causes this.

(I apologize if this is more physics than chemistry,  and also for my Google-fu not being enough to find ready answers.)

Besides mechanical aspects

bubbles might form a sort of layer at the immersed part while at the air interface they evolve and detache easily renewing the metal/acid (oxidizer) contact

a more fundamental reason can be° surface excess. Normally we treat interfaces as very thin and therefore negligible (in mass or volume) as compared to the bulk.

However at an interface the chemical potential (and all related thermodynamic properties) differs from that of the bulk owing to the fact that a layer can be assumed to have the same number of entities per unit of surface.

While such a layer at depth is surrounded by the very same, at the interface it sees two different environments.

This aspect is quite familiar in physical phenomena, such as surfactivity, the shape of droplets, and even the tip effect in electrostatic. Another example  possibly familiar to many chemists is the difference in product of solubility / solubility for solids at very different sizes (digestion of precipitates to attain bigger grains/crystals is based on this, tough intermixed with the merely geometric fact that small particles dissolve faster).

As for in water the interface has higher potential and this results in spherical droplets to minimize it, in your specific case you can see the aqua regia doing the same by being more reactive. It is like the oxidizer trying to escape the air interface (tough a new one is continually formed, of course) by reacting more energically than the layers below, if this pictorial helps you.

If you dig for "potential at acid/air interfaces" or "pH of an acid at the air interface" and so on, you can find some material dealing with example somehow related to your example. For instance I have found study of the dissociation of weak acids (ie pKa at the interface vs bulk), The vast majority is however studies in which the very interface is at focus, like in monolayer LB films, and the surface excess comes naturally at play whatever explicitly considered or not.

(For instance a title such as "On the dissociation constant of carboxylic acids in monolayer thin films" implicitly assumes that some difference exist as compared to the bulk of the solution).

° can be as for I do not know if the evident effect you have spotted is mainly due to the "stirring effect" of evolving bubbles or to the higher activity of the interface, or their mix.


Why Inductive effect operates only through single bonds? I am not able to find any examples of the inductive effect operating through multiple bonds. Why can't electron displacement take place in multiple bonds?

If I'm not mistaken a good example of the inductive effect through multiple bonds is trifluoro-acetic-acid, or $\ce{F3CCOOH}$. The proton is so electron deficient that it is easily abstracted by anything that can act as a base. The not 'fluorinated' equivalent, acetic acid, is not nearly as strong an acid.

The reason for the difference in acidity, is the difference in electro negativity between hydrogen atoms and fluorine. Due to the very large electro-negativity of fluorine atoms the electrons in the acid protons are 'pulled' towards it, leading to a clear example of the inductive effect over multiple bonds.

A comment on one of the comments below the question: Electron displacement does occur in double bonds. A good example is the carbonyl functional group (C=O) in a molecule. The inductive effect results in a slightly more positive charge on the Carbon atom and slightly higher at the O-atom. This is also due to electro-negativity and results in the reactivity of the C=O functional group.

Hope this helped!


Is there any reason why halogenation of substituted alkanes wouldn't occur through free radical substitution? Say for example, you have isopropyl alcohol. Is there any reason why if you react it with bromine that it won't form a radical, as one of the bromines removes one of the protons adjacent to the hydroxyl group on the alkane, and then propagate into 2-bromo,2-propanol?

In principle, this reaction should be possible. However, the intermediately formed 2-hydroxy-2-propyl radical is not stabilized by the OH substituent, as the alternative mesomeric structure with C=OH bond and unpaired electron at O would violate the octet rule. 2-bromo-2-propanol should still be the major product, but there are also a lot of possible side products like multiply brominated compounds. Other substituents which stabilize an adjacent radical (by conjugation, for example) will increase the selectivity of the reaction. The possibility of reaction of the bromine radicals with the functional groups and sensitivity of the substrate to the reaction conditions (UV radiation etc.) also need to be considered. For the selective halogenation of certain substrates, often more suitable reagents than $\ce{Br2}/ h\nu$ are available, for example, NBS/AIBN for allylic bromination.


Hydration vs Dissolution While revising Thermodynamics and Thermochemistry from revision material published by Arihant publications I encountered the following mentioned statement.

During dissolution, the physical state of the compound changes while during hydration, there is no change in the physical state of the compound.

How does dissolution result in a change in physical state?

Should not hydration (or solvation in general) be the same as dissolution as in both the bonds of solid salt to be dissolved or solvated weaken and a new bond is formed between constituents of salt and the solvent? Could someone please explain what I'm missing?

Before dissolution, the substance is usually a solid and forms crystals. Its physical state is solid. After dissolution, the substance is not visible any more : it is not a solid any more. Its physical state has changed.
Before hydration, the substance is usually solid. After hydration, the color and the volume may have changed, but the physical state has not changed : it is still a solid –


How do I figure out how many hydrogens my compound actually has using a mass and NMR spectrum? Question 3:

It said $m/z = 122,$ and $m/z = 124$ is in a $3:1$ ratio, so I figured that meant that chlorine is present. Then I thought $m/z$ was the actual compound's molecular mass.

So I used the rule of 13, and did:

Chlorine's molar mass = 35

122 - 35 = 87

Using the rule of 13

87/13 = 6C + 9H/13

So the molecular formula is $\ce{C6H9Cl}.$ But the integral values I rounded are: 2,2,2,2 and 3 adding up to 11. Did I use the wrong mass to charge ratio? What am I missing?

Being an NMR fan myself I would inspect that NMR spectrum:

The integrals suggest you have 11 $\ce{^1H}$ or a multiple thereof (the number under each peak is the normalized integral, which is proportional to the number of protons represented by the multiplet). That leaves you with $\pu{87 Da -11 Da}=\pu{76 Da}$ to explain. If you throw in an oxygen you get $\pu{60 Da}$, which is neatly divisible by 12, for a formula $\ce{C5H11OCl}$. Next use the NMR spectrum again to infer the connectivity. The rightmost integrated multiplets (multiplet and triplet) are typical of $\ce{-CH2CH3}$ with $\ce{CH2}$ split by methyl and something else. From the integrals we see clearly there is only one methyl group, and there are no lone H which seems to eliminate the possibility of carbonyl or hydroxyl groups, suggesting an ether.

In the end I just "cheated" (it's 2019 after all) and used the online NMR simulator predictor:

This nice piece of software lets you tinker quickly with different structures and inspect assignments interactively, helping you develop your intuition.

Next you need to verify the MS and IR spectra, but the NMR spectrum is a giveaway that this is the correct hit.


Weighing reaction mixture At school we do an experiment with baking soda and vinegar (in a cup) both sealed in a ziplock bag. The bag is masses on a triple beam balance (example 32.7 g). Bag is tilted to allow the vinegar and baking soda to mix.

The students note the signs of chemical reaction and mass the bag again. There is always a 0.3 g – 0.5 g loss of mass. How is this explained? Cheap ziplock bags? Is some of the mass from the endothermic reaction converted?

Think about the reaction:
$$\ce{NaCHO3(s) + CH3COOH(l) \to CO2(g) + H2O(g) + Na+(aq) + CH3COO- (aq)}$$
The resulting $\ce{CO2}$ will make the bag expand, therefore increasing the volume and the subsequent buoyant force. This force opposes the force of gravity, making it seem like the mass of the matter inside the bag has decreased.


What the balanced chemical equation for the conversion of starch to dextrin by dry heating? This question is related to the toasting of bread. When we toast a slice of bread, the starch in it breaks down into dextrin. I want to know the balanced chemical equation for the process.

Let's take a look at the structure of both molecules. First, starch:

Now, dextrin:

Do you see the difference? They are the same molecule (both are glucose polymers), the only difference is the molecular weight, or the length of the polymer chain.

Starch, by convention, is 300-600 monomers long. Dextrin is something less than that.

Since the number of dextrin molecules you get will depend on the polydispersity (size distribution), average chain length, and extent of the reaction, there is no way to write a balanced reaction equation for a real reaction.

You could assume that a starch molecule of a given length breaks into dextrin molecules of a given length, and write an equation. It would look like this:

$\ce{C6H11O5[C6H10O5]_{m}C6H11O6 + (\frac{$m-n$}{2})H2O->} \frac{(m+2)}{(n+2)}\ce{C6H11O5[C6H10O5]_{n}C6H11O6}$

This would only work for one length of starch and dextrin (in reality there could be 10 - 1000) and would also only be true if $m > n$, and $m$ is evenly divisible by $n$.

Also, in the question you specify dry heat, but there is no way to balance the equation without including water (unless you use hydrogen and oxygen gas). I suspect that even under dry heat conditions, there is moisture on the bread, and it is consumed in the reaction. This could be part of the reason that toasting drys bread.


What does the prime symbol mean in organic nomenclature (e.g 2')? I encountered the following compound on Wikipedia:

$\hspace{6cm}$

Its systematic IUPAC name is:

  4,4',5,5',6,6'-hexahydroxy-2,2'-biphenyldicarboxylic acid

What do the prime symbols in the locants 4', 5', 6', and 2' signify?

Primes, double primes, triple primes, etc., are added to differentiate between the same locant in different parts of the structure. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), they are used

to differentiate the nitrogen atoms of hydrazides, imidamides (amidines), amidrazones, and hydrazidines;

  N'-acetyl-N'-ethyl-N-methylpropanehydrazide

with superscripted numerical locants indicating the position of the suffix on the parent structure, to differentiate the nitrogen atoms in other cases, especially when symmetry conditions are not fulfilled;*

  N3-ethyl-N'3-methylhexane-3,3-diamine

in multiplicative nomenclature to denote multiplied units and modify locants accordingy;

  2,2',2''-nitrilotri(ethan-1-ol)

in spiro-fused compounds, to denote positions in polycyclic systems

  7,7'-spirobi[bicyclo[4.1.0]heptane]

in ring assemblies, to number identical ring components;

  1,1'-biphenyl

in fusion nomenclature, to identify first and higher attached components, identical attached components, and multiparent names;

  difuro[3,2-b:2',3'-e]pyridine

in fullerenes ortho-fused to organic ring or ring systems, to identify positions in the nonfullerene component;

  3'H-cyclopropa[1,9](C60-Ih)[5,6]fullerene

in natural product nomenclature, to identify positions in ring(s) fused to a fundamental parent hydride;

  naphtho[2',1':2,3]-5α-androstane

The example that is given in the question is a ring assembly. The numbering of locants corresponds to the numbering of 1,1'-biphenyl, where primes are used to differentiate between the same locant in different phenyl groups. Therefore, the correct systematic name is 4,4',5,5',6,6'-hexahydroxy-[1,1'-biphenyl]-2,2'-dicarboxylic acid.

* Note that, according to current IUPAC recommendations, primes are no longer used to differentiate the nitrogen atoms of symmetrical geminal diamines, diimines, diamides, di(imidamides), (diamidines), di(hydrazonamides), di(amidrazones), di(hydrazonohydrazides), imidohydrazides, and poylimidopolycarbonic and inorganic oxoacids.


Why are p-type solar cells more prone to degradation in space than the n-type ones? Why are p-type (boron doped substrate) solar cells more prone to degradation in space than n-type (phosphorus doped substrate)?

All the sources I've looked at simply state it was found that p-type degrades slower, hence they were used for space application.

What is the main damaging radiation from the sun?
Why is it more damaging to p-type than n-type?

The space radiation environment is a quite complex subject. Generally speaking one needs to be careful about just where in space you are concerned about - near earth orbit (within the Van Allen Belts), outside the belts, regularly passing through the belts, interplanetary, etc.

That being said, the three main areas of concern are protons from the solar wind, electrons from the solar wind, and cosmic rays from the rest of the universe. For testing purposes the standards for the solar wind are 10MeV protons and 1MeV electrons. I will focus on these for the purposes of this answer.

As the protons or electrons hit a solar cell, the will go in some ways and lose energy through interactions with electrons and the nuclei (electronic and nuclear stopping, respectively). Electronic excitations at best make a little more current in the depletion region. The interactions with nuclei are more problematic - either particle can transfer enough energy to a nucleus to displace it off a crystalline lattice site (protons are about 1000 times more effective at it then electrons because their mass is much higher). So, as the cell accumulates dose, more and more atoms are off their sites.

OK, what does that do? Well, point defects in semiconductors often create mid-gap electronic states that allow carriers to recombine - your cell is now less efficient at making current since some gets lost. Further, point defects (vacancies and interstitials) can combine with either other point defects or impurities to form more complex defects, and these may be even better recombination centers.

Of particular relevance for your question is the impact of boron. To quote from
T. Hisamatsu et al. Radiation degradation of large fluence irradiated space silicon solar cells (1998):

  Radiation-induced defects associated with boron atoms are said to form
  recombination centers for minority carrier electron and this phenomenon is said to
  cause the degradation in low-fluence regions. Therefore, using highly boron-doped
  substrates in order to avoid the sudden drop of cell performance should not be
  a proper means to provide cells with higher radiation tolerance.


Why is the melting point of p-dichlorobenzene higher than those of o-dichlorobenzene and m-dichlorobenzene? I was going through alkyl and aryl halides and noted the following statement in my textbook:

  p-dichlorobenzene has higher melting point and solubility than those of o-dichlorobenzene and m-dichlorobenzene.

I thought over it but could not come to a conclusion. Why is the above assertion true?

Generally, the melting point of para-isomer is quite higher than that of ortho- or meta-isomers. This is due to the fact that it has a symmetrical structure and therefore, its molecules can easily pack closely in the crystal lattice. As a result, intermolecular forces of attraction are stronger and therefore, greater energy is required to break its lattice and its melts at higher temperature. This question is most likely to be answered easily.
If you consider solubility. These are insoluble in water but soluble in organic solvents. Generally, para-isomers are highly soluble in organic solvents than ortho-isomers. This is interesting question.
Every compound (more specifically solids) does not dissolve in a given compound. In general, a liquid dissolves in a liquid if the intermolecular interactions are similar. This is in accordance with the rule "like dissolves like," ionic or polar compounds dissolve more readily in polar solvents. Non-polar compounds (covalent or organic) are soluble in non-polar solvents.
If you consider p-dichlorobenzene, it has zero dipole moment and thus is more non-polar than o-dichlorobenzene which has dipole moment of about $\pu{2.54 D}$. So, non-polar p-dichlorobenzene dissolves more readily than o-dichlorobenzene in non-polar (organic) solvents.


Racemic compound widely available Can racemic compounds be found at the local drug store non-prescription? Where might one be found most readily/cheaply? Only one chiral center is needed. More may actually be unhelpful. The purpose here is to practice separating enantiomers.

I know that ibuprofen is sold as racemic mixture, it's a drug for fever treatment. It's very cheap in Brazil, one dollar each box (400mg per tablet).


What is the typical resistance of a piece of 1 cm^2 of carbon paper? I have been using carbon paper (Carbon sheet 99.8% from STERM CHEMICALS) to load catalysts for fuel cell study. Just now, I realized that for a 1 cm^2 piece the resistance is about 3 ohms!
Is that a typical value? I can't find any info on Google or the seller's webpage.
Thanks!

If there is no published resistance for that carbon sheet, then the best choice is to do exactly what you have done, i.e., measure the resistance per square. However, that specification is in given in the Strem catalog! It might be difficult o find if you do not spell the company's name correctly.
Two notes:

Resistance of a uniform, homogeneous sheet is measured per square, not per cm2. Whether the sheet is 1 mm on a side, or 100 km on a side, the resistance from one edge to the other would be the same; in your case, 3 Ω. [Consider that making the separation wider between edges is exactly compensated by the increased length of the conducting sheet.]
"Carbon paper" referred, in English, to paper with a mix of lampblack carbon and wax, though now it also applies carbon microfiber paper. For clarity, I suggest using the name Strem uses in it's catalog.


Why won't removing the salt bridge cause the battery to recharge? If I understand correctly, the salt bridge in a battery is used to move positive ions towards the side that is becoming more negative, and negative ions towards the side that's becoming positively charged, to prevent the reaction from quickly reaching equilibrium. The ions in the salt bridge never leave the salt bridge.

But if we removed the salt bridge, wouldn't the forces between the $\ce{SO_4^{2-}}$ ions or $\ce{Zn^{2+}}$ ions cause the reaction to go in reverse and cause them to reattach to the electrodes?

The ions in the salt bridge never leave the salt bridge.

That's not true.  The ions literally have to move to balance the charge of the gain / loss of the electrons at the electrodes.  (You even allude to this in your question.)  One of the ways that a battery can run out is that the bridge is used up.  You could views this by using a deeply colored ions (e.g. MnO4-) to see the migration.

While the salt bridge is in place, there is no build up of charge.  Electrons moving through the wires are offset by the ions moving through the salt bridge and into the respective half-cell.  Thus, if you remove the salt bridge, there is no net charge that will cause anything to reverse; there is no 'force' that will cause ions to revert.

If you want to reverse the reaction (and charge the battery), you have to force electricity back in the other direction, using more voltage than what you got out of the battery.  This doesn't work for all batteries.  (I.e. this is why only some batteries are rechargeable).  As you force electricity backwards through a battery, any movement of ions from the salt bridge means that the half-cell might now be contaminated with chemicals other than the ones you want in your half-reaction (unless you designed your salt bridge just right).  Thus, the "right" ions may not go back to being uncharged (or vice versa).


Custom-defined functionals in Gaussian 09 As stated before I want to try out custom-defined functionals. Besides the question about B3LYP itself, there is another related question.

Short excerpt:

  $$\text{XC-Functional}=a E_x^\text{local}+(1-a)E_x^\text{HF}+b E_x^\text{non-local}+c E_c^\text{non-local}+(1-c) E_c^\text{local}$$

  The Gaussian documentation for DFT inputs specifies how one could change the amounts of each "sub-functional" by using the IOp-Statements through:

  IOp(3/76=mmmmmnnnnn) IOp(3/77=mmmmmnnnnn) IOp(3/78=mmmmmnnnnn)

  Where mmmmm and nnnnn will get divided by 10000 and 3/76=$P_1P_2$, 3/77=$P_3P_4$ and 3/88=$P_5P_6$. Those $P_{1,..,6}$ values are part of the following equation, which is a modified version of the equation above.

  $$\text{XC-Functional}=P_2 E_X^\text{HF} + P_1 \left(P_4 E_X^\text{Slater} + P_3\Delta E_x^\text{non-local}\right) + P_6 E_C^\text{local} + P_5 \Delta E_C^\text{non-local}$$

Now Gaussian's description site about DFT tells that one simply has to choose one exchange and one correlation functional, i.e. B and LYP. But then B would be used for $E_x^\text{non-local}$ and LYP probably for $E_c^\text{non-local}$.

Does that mean, that the $E_c^\text{local}$-part

cannot be defined,
is always the VWN functional as in B3LYP, or
is also the chosen correlation functional?

Take a look at the IOp manual. Under IOp(3/74),

  IOp(3/74) Type of exchange and correlation potentials.

  -5 Becke3 using VWN/LYP for correlation.
  02 Lee-Yang-Parr correlation.
  05 VWN 80 (LSD) correlation.
  200 Hartree-Fock-Slater exchange (Alpha = 2/3).
  400 Becke 1988 exchange.

  So 100 is Hartree-Fock, 200 is Hartree-Fock-Slater, 205 is Local Spin Density, and 402 is BLYP.

They're unclear about it, but this means you can take the "sum" of a 1- or 2-digit number (which specifies the correlation) and a 3- or 4-digit number (which specifies the exchange) to build an exchange-correlation functional. So, IOp(3/74=-5) == IOp(3/74=402), and an identical way to define B3LYP is

IOp(3/74=402) IOp(3/76=1000002000) IOp(3/77=0720008000) IOp(3/78=0810010000)

  Does that mean, that the $E_{c}^{\text{local}}$-part

  cannot be defined,
  is always the VWN functional as in B3LYP, or
  is also the chosen correlation functional?

Looking at the available correlation functionals (incomplete, these are the lowest few),

01 VWN5
02 LYP
03 P81 (PL in keyword list)
04 P81 + P86
05 VWN3
06 VWN3 + P86
07 OS1
08 PW91
09 PBE
10 VSXC
11 Bc96
18 VWN5 + P86
19 LYP + VWN5 for scaling

your choices for local correlation functionals are rather limited. Either you choose a correlation functional that's purely local (VWN3, VWN5, or P81), or a non-local/gradient-corrected functional which in Gaussian all seem to have the local component as part of their definition (as in LYP), or they're forcibly combined (as in P86).


Number of electrons transferred & Gibbs free energy To determine the number of electrons transferred for the following reaction:

$$\ce{CH3OH + \frac{3}{2} O2 -> CO2 + 2 H2O}$$

I found the half reactions to be:

$$\text{Anode: } \ce{CH3OH + H2O -> 6H+ + 6e- + CO2}$$
$$\text{Cathode: } \ce{\frac{3}{2}O2 + 6 H+ + 6 e- -> 3 H2O}$$

The number of electrons transferred is therefore 6. If I was given the following overall reaction:
$$\ce{2 CH3OH + 3 O2 -> 2 CO2 + 4 H2O}$$

Which is just multiplying everything by 2 for the original reaction, would I say 12 electrons have been transferred or still 6? Since the Gibbs free energy change = nFE where n is the number of electrons transferred should surely be the same?

You should use $12$ electrons. Another way to think of it is when you calculate the $\Delta G$ of the first reaction, that value will be units of energy per mole of reaction. So, since you doubled the moles of reaction, the $\Delta G$ would double.


How to find the units of the rate constant? Consider the reaction $$\ce{2A + B -> Products},$$ when the concentration of $\ce{B}$ alone was doubled, the half-life did not change. When the concentration of $\ce{A}$ alone was doubled, the rate increased by two times. The units of the rate constant is:
a) $\mathrm{s^{-1}}$
b) $\mathrm{L\ mol^{-1}\ s^{-1}}$
c) Unitless
d) $\mathrm{mol\ L^{-1}\ s^{-1}}$

I have tried it solving through this method:
According to me, when the concentration of $\ce{A}$ alone was doubled, reaction rate also increased by two time, implies that reaction is first order according to $\ce{A}$. In same way, the reaction should be 0th order according to $\ce{B}$. So the net order would be 1. So this gives me answer $\mathrm{s^{-1}}$ but the answer is $\mathrm{L\ mol^{-1}\ s^{-1}}$. Where am I going wrong?

You are doing it wrong in situation when concentration of B alone was doubled but half life did not change.

In this situation, the reaction order should be 1 according to B as Half life in first order reaction doesn't depend upon initial concentration of reactants.

$$t_{1/2} = \ln2 / k $$

So the final order will be $2$ and the answer of the question will be b.


Why do we need three equations to find the pH of NaCN, given Ka(HCN)? To find the pH of a 1.0 M solution of $\ce{NaCN}$, given
$\ce{Ka(HCN) =4.9E(-10)}$

The solution (refer end of question) uses three equations:

Equation 1 : $\ce{HCN + H2O <-> CN- + H+}$

Equation 2: $\ce{NaCN <-> CN- + Na+}$

Equation 3: $\ce{CN- + H2O <-> HCN + OH-}$

I am confused as to:

1) the logic behind why we need these three equations (up to this point in the course, all the similar questions have been of the form e.g. find pH of NH3 given Kb(NH3), and so we only used the NH3 acid base reaction equation.

I assume equation 1 is necessary because the Ka is given for HCN and Equation 2 is necessary because we are asked to find the pH of NaCN.  Where does Equation 3 come from? We did talk about spectator ions in class so does it have something to do with Na being a spectator ion?

2) why the ICE (intial, change, at equillibrium) table to find x is written in terms of Equation 3 and not, for example, Equation 2.

The solution from the textbook is given below:

Thank you.

So strictly speaking you only really need equation 3 to solve this problem, but the first two equations help you figure out how to solve it if you're not too familiar. What do I mean:

  Equation 1: $\ce{HCN + H2O <=> H3O+ + CN-}$

As you correctly note, this is to remind you of the definition for $K_a$

  Equation 2: $\ce{NaCN -> Na+ + CN-}$

Note that I've changed $\ce{<=>}$ to $\ce{->}$. This is because most sodium salts fully dissociate in aqueous solution and so this "equilibrium" doesn't exist and you can then consider all $\ce{NaCN}$ to be converted to $\ce{CN-}$ for your "initial" stage in the ICE table. Hence there's no need to write an ICE Table with respect to Equation 2.

Value of writing this out:
In the case that you were given something sparingly soluble or did not fully dissociate, you would have to consider another equilibrium and this equation 2 would become significant. But in this case, $\ce{Na+}$ is a spectator ion which you can ignore.

  Where does Equation 3 come from?

Equation 3 comes from the question itself (or rather what is happening in the system described). Remember, you've got a 1.0M solution of $\ce{NaCN}$. Since we've already established that it fully dissociates, at the "initial" stage of the system you have only $\ce{Na+}$ and $\ce{CN-}$ ions floating around. But from Equation 1 you know that $\ce{HCN}$ is a weak acid and so free $\ce{CN-}$ in water may act as a base and take a proton from water, and that is given by the equation:

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


How to interpret a luminescence intensity vs wavelength graph? Luminescence is defined as the amount of light emitted by a atom. But what confounds me of this graph is the fact that all the peaks are of the same height. In addition, there are no peaks in between two peaks.
 why are the peaks of the same height?
Andwhy are there only six peaks, if photons can be emitted having the wave length of say, 110nm?

I am sure those are ultraviolet rays.

This is a partial spectrum of the luminescence from a hydrogen atom, the graph presents only the Lyman series of emissions.

When a hydrogen atom is excited (absorbs energy), the electron can be promoted to the n=2, 3, 4, 5...infinite level.  This excited electron can eventually emit its excess energy and return to the n=1, 2, 3, 4...etc. state.  All returns to the n=1 state (ground state) are termed the Lyman series,  all returns to the n=2 state are termed the Balmer series and so on.

For the hydrogen atom, the wavelength of these various emmissions is given by the Rydberg formula

$${1 \over \lambda} = R \left( {1 \over (n^\prime)^2} - {1 \over n^2} \right) \qquad \left( R = 1.097373 \times 10^7 \ \mathrm{m}^{-1} \right)$$

for the Lyman series this equation becomes

$${1 \over \lambda} = R \left( {1} - {1 \over n^2} \right) \qquad$$

and six lines can be seen in the ultraviolet.  After the 5th line (n=6 -> n=1), all other transitions are so closely spaced that they appear as one line (the sixth line).

\begin{array}{|c|c|}\hline
n & λ (nm) \\ \hline
2 & 122 \\ \hline
3 & 103 \\ \hline
4 & 97.3 \\ \hline
5 & 95.0 \\ \hline
6 & 93.8 \\ \hline
\infty & 91.2 \\ \hline
\end{array}

As to the relative intensities, I'm not sure why they are all the same in your graph.  The intensities do depend upon how many electrons are excited to the various excited states.  If for some reason all excited states were created in a manner such that they were equally populated, then something like your spectrum would result.  Usually the relative intensities differ and are used much like a fingerprint.  For example, the relative intensities of the Lyman series from the hydrogen atoms in one galaxy differ from those in another galaxy due to the difference in temperature of the two galaxies.  My guess is that your data was artificially created (e.g. the title in your graph says "Model").


Can neutrons be ejected from an atom? In a neutral atom, can neutrons be ejected, and if so, by what (nuclear) process or reaction? I have read Neutron emission and Can a proton be ejected from an atom, but I don't understand how and if that would work for removing neutrons from a heavy atom such as lead-208. Can neutrons be ejected from lead-208 so that it is lead-197, and if not, why?

The typical binding energy per nucleon $(E_\mathrm B/A)$ of most nuclides is about $5{-}8\ \mathrm{MeV}$. Such values are higher than the released energy of a typical radioactive decay. Therefore, emission of neutrons doesn’t occur during most usual radioactive decay processes.

In order to make a neutron source, you have to find a suitable target (not $^{208}\mathrm{Pb}$) and a corresponding nuclear reaction that releases a neutron and that is also energetically favourable, for example
$$^9\mathrm{Be}\ (\alpha,\mathrm n)\ ^{12}\mathrm C$$
$$^9\mathrm{Be}\ (\gamma,\mathrm n)\ 2\ ^{4}\mathrm{He}$$
Note that $^{12}\mathrm C$ and $^{4}\mathrm{He}$ are very stable nuclides with a relatively high binding energy per nucleon so that these reactions can release a large amount of energy, but still you need to irradiate the beryllium targets with high-energy radiation (high-energy $\gamma$ radiation of more than $E=1.665\ \mathrm{MeV}$ e.g. of $^{124}\mathrm{Sb}$ or $\alpha$ radiation) to release neutrons.

Nevertheless, neutron emission can occur during radioactive decay if the neutron excess is large and the beta decay energy is high enough, i.e. at the edge of the valley of beta stability. At very high values, spontaneous emission of neutrons becomes possible.

More important, however, are so-called neutron precursors, e.g. $^{87}\mathrm{Br}$ or $^{137}\mathrm{I}$, which have a very high decay energy so that the product nuclides ($^{87}\mathrm{Kr}$ and $^{137}\mathrm{Xe}$, respectively) can get enough energy to make neutron emission possible in the next step.
$${}^{87}\mathrm{Br}\xrightarrow[{E_\mathrm{max}=6.83\ \mathrm{MeV}}]{\beta^-}{}^{87}\mathrm{Kr}^*\xrightarrow[\quad]{}{}^{86}\mathrm{Kr}+\mathrm n$$
$${}^{137}\mathrm{I}\xrightarrow[{E_\mathrm{max}=5.89\ \mathrm{MeV}}]{\beta^-}{}^{137}\mathrm{Xe}^*\xrightarrow[\quad]{}{}^{136}\mathrm{Xe}+\mathrm n$$
Such reactions are responsible for the so-called delayed neutrons, which are very important for most nuclear reactors – without this effect, controlling a nuclear chain reaction would be difficult.

Of course, the most important neutron sources are nuclear reactors, which generate neutrons (about 2–4 neutrons per fission) in a controlled nuclear chain reaction.


What is the solublity of water in liquid carbon dioxide at low temperature? The solubility of water in liquid carbon dioxide has been investigated by Hosmer W. Stone in 1943 from
-29 to about 22 °C.

Is there any research done on the solubility of water in liquid $\ce{CO2}$ at still lower temperatures ?

The lower the temperature and higher the pressure, the more soluble CO2 is in water.  Pepsi and Coke have staudiesd this for years.


Why is current non-zero when applied potential is still below E$^0$ in cyclic voltammetry? In this diagram (Evans et al., J. Chem. Edu.) the current is non-zero, but the potential is still below the normal potential. How can the compound become oxidized at the working electrode, when the applied potential is still below the normal potential of the compound?

I think you may be coming at this question from the wrong angle. Since I'm not entirely knowledgeable about electrochemistry, I'll give a somewhat qualitative answer, the general idea of which I hope to be understandable.

Fundamentally, what you are doing is taking a system containing a species, say $\ce{A}$, and its oxidized form $\ce{A+}$ (of course this same argument applies for the reduced form $\ce{A-}$ if you sweep in the other direction of voltage). In the absence of an applied voltage, the system is at equilibrium with a certain concentration of $\ce{A}$ and $\ce{A+}$. Let us consider that $\ce{[A]}$ $\gg $ $\ce{[A+]}$. This implies that the equilibrium constant for $\ce{A <=> A+ + e-}$ is much smaller than 1, and therefore that the oxidation reaction has a high positive standard reaction free energy change $\Delta _r G^{o} \gg 0$ (loosely speaking, the reaction is "not spontaneous" in standard conditions, though this sort of thinking is misleading, as I explain later).

As you start to sweep the voltage, you begin adding electrical free energy into the system, such that the reaction free energy (in non-standard conditions) now becomes a sum of the initial chemical free energy change and the electrical free energy $\Delta _r G = \Delta _r G^{o} + G_{electrical} = \Delta _r G^{o} + nFE$ (I think that you can argue that by altering the voltage, you are altering the free energy of the electron in the equilibrium). If you sweep in the right direction, then the electrical free energy term can contribute to decrease the reaction free energy. At some defined voltage, the two factors cancel out, giving $\Delta _r G = 0\ \ \rightarrow \ \ \ \Delta _r G^{o} = -nFE^*$, where to $E^*$ we give the special name "oxidation potential", $E^{ox}$.

The important thing to notice is that all this time, the potential sweep is altering the reaction free energy incrementally. Since the free energy is shifting, the associated equilibrium constant is also changing. At voltages much than lower $E^{ox}$,  $\Delta _r G = \Delta _r G^{o} + nFE \gg 0$, and so even though the reaction free energy is lowering, it is still quite positive and the associated equilibrium constant is still much smaller than 1. Thus the reaction is only slowly shifting towards products, and therefore there are few electrons being passed into the circuit, and the current is close to zero. However, when the voltage applied starts to get close to the oxidation potential, even if $\Delta _r G < 0$, the equilibrium constant will be smaller than but somewhat close to 1, and there is some conversion of $\ce{A}$ into $\ce{A+}$, creating a current below the formal oxidation potential.

The whole problem comes from thinking that as long as $\Delta _r G < 0$, a reaction "happens", while for $\Delta _r G > 0$, the reaction "doesn't happen". In truth, there is a gradual change of equilibrium concentrations as a reaction goes from one regime to the other. Even if $\Delta _r G > 0$, if it sufficiently close to zero then the equilibrium constant is only a bit smaller than 1, and you can expect appreciable amounts of products. Notice that in the voltammogram, the current is practically zero until about 0.1 V away from the oxidation potential, and it only really starts to pick up around a mere 0.05 V away. Even though these values mean $\Delta _r G > 0$, they are sufficiently close to the oxidation potential that the equilibrium constants aren't so small, and so you can actually measure a significant amount of species being oxidized.


Negative s - orbital chatacter I am learning about how to calculate s-orbital character. While doing so, I came across a formula
$$\cos\theta = \frac{s}{s-1}.$$
For cycloproprane, I used $\theta = 60^\circ,$ and got $s = -1.$

How can the s-orbital character be negative?
What does the negative sign mean?

Question which I want to solve
Find out the stability of the following carbanion:

I found Does negative sign of s-orbital exist?, but failed to understand it. If that answers my questions, please tell.

The formula only works when the orbital is directed along the bond line between two atoms. Since no orbitals can give you a $60^o$ bond angle, the orbitals are directed outside the triangle, forming a “banana” bond and the formula is being used inappropriately. So the negative sign in your Question #1 is not acceptable. We must do something different.
Let’s look at the H-C-H bond angle. The H-C-H bond angle in cyclopropane is $115^o$ Ref 1. Using that angle, and assuming that the orbitals are directed exactly between the atoms, s = 29.7% for the C-H bond, which is broken to form the carbanion. The acidity of cyclopropane is also indicated by its pKa: 46.
Cyclopropane is interesting because you might think, starting with atomic orbitals, that such a structure would not form, since the smallest angle you could imagine for a carbon atom would be $90^o$; how could you possibly get $60^o$?
Well, that's where your imagination comes into play. Don't go to negative s-bonds (that's too imaginative). Consider how the simple (i.e., positive) addition of individual atomic orbitals could lead to something like the bonding that you do observe:

With simple $sp^3$ orbitals, you can point one pair from each carbon just outside the ring (there will still be considerable overlap: Figure b) and each carbon will be attached to 2 hydrogens above and below the ring with the other two $sp^3$ orbitals. These ring bonds are called banana bonds - not directly in line with the atoms, but bonding nevertheless. Figure a) shows a more complicated bond scheme where the carbon is hybridized $sp^2p$ with the $p$ obital in the plane of the carbons: you get two bonds and one antibond (the whites overlap, the reds overlap, and the red-white antibonds). But then you get three $sp^2$ orbitals poking right into the ring to form a 3-atom bond - again, the bonding is not in line with the atoms. (The hydrogens are bound with $sp^2$ bonds, above and below the ring. You might use the equation to determine the s and p character of these C-H bonds if you know the H-C-H angle (Ref 1).
A very interesting (11 slides) presentation about the bonding is made at Ref 2.
Ref 1: https://cyclopropanelasalle.weebly.com/structure.html
Ref 2: https://www.slideshare.net/DanielMorton26/using-molecular-orbital-theory-to-explain-bonding-in-cyclopropane
But the upshot is that all (most?) equations have limited ranges of usefulness. Whenever some experimenter makes something that goes beyond, a new equation is needed.
Cyclohexane, at the other end of your question, has a pKa of 52 (very non-acidic!), and a probable H-C-H bond angle of $109.5^o$, and s-character in the C-H bond of 25% (note that the molecule has a chair structure and is not planar).
Other pKa’s of interest are:
acetylene, pKa =15, C-H bond is $sp$, 50% s
ethylene, pKa = 44, C-H bond is $sp^2$, 33% s
methane, pKa = 48, C-H bond is $sp^3$, 25% s
(interesting that cyclohexane is similarly hybridized but even less acidic)
Perhaps the best answer to the stability question is that the smaller rings are more acidic, i.e., have more stable anions because the acidic proton has more s-character in the more strained ring.
Question #2 discussed sign in your link. “Sign” (+/-) in the equation for s-character is a measure of amount: how much s-character, in %, is in the bond. But when describing orbitals, we use + and - as indicators of the polarity of the wavefunction, not the amount of s or p character. Plusses combine in such a way that electrons flow from one orbital to another without hindrance, in a bond. A - polarity next to a + requires a node: no electrons there at the node, therefore no bonding. So s-orbitals are designated with one sign (either + or -, depending on whether it is bonding with an adjacent atom; higher levels have internal nodes, but that is no concern for us here). p-Orbitals have a node at the nucleus which is indicated by calling one lobe + and the other lobe -. Bonding molecular orbitals have the same sign of the wavefunction adjacent; in the picture, red-red and blue-blue (in (c)) are bonding; red-blue (in (a) and (b)) is antibonding because the wavefunction splits between the atoms.


Raman transition and IR transition In a character table one can find the reducible representation of vibration of a molecule. In such a table we also see things like $xy,z,x...$. If this corresponds with an irrep we say that IR transitions can occur if $x,y,z$ and quadratic terms give Raman transitions. This is because the dipole moment and polarizability are non-zero and can change in that case. My question is: where does this come from? I prefer a physical or mathematical answer involving quantum mechanics since I am from physics.

To simplify slightly both Raman and IR spectroscopy show the vibrational modes of a molecule (though the techniques used to reveal these are very different).

IR spectroscopy relies on coupling between the electromagnetic field of light passing through a sample and the electric dipole of the molecule. But that absorption is only possible is the vibration in the molecule changes the molecules dipole. Consider, for example, carbon dioxide, a linear molecule with no net dipole moment. Symmetric stretches (ie both oxygens moving the same amount in opposite directions) don't change the dipole so can't be detected by IR. But asymmetric stretches or bending will change the dipole and can be detected by IR.

Raman spectroscopy (simplifying a little) relies on detecting changes to the molecules polarisability not the dipole. So the symmetric vibrations of carbon dioxide are detectable because they do change the polarisability of the molecule.

In both cases the symmetry tables tell you whether the symmetry of the vibrational mode changes the electric dipole or the polarisability of the molecule. In simple molecules physical intuition can usually tell the same thing but symmetry tables are more reliable.

A full understanding would involve a quantum mechanical description of the interaction of EM radiation with the possible vibrational modes of a molecule considering the allowed and forbidden transitions. But the physical intuition above captures the essentials without getting too mathematical.


Can I obtain a different density of a recycled elemental material? I am recycling Aluminum products/scrap. I melted it and then form cylindrical ingots. When I measure their density, I calculated $\pu{2.4 g/cm^3}$ but Al has density of $2.7$. I think all the materials are aluminum since they all melt in the same temperature using a home setup. Can I claim that the ingot is Al?

I think it is risky. The density observed lower than the expected density of pure Al is indicative for the presence of something of even lower density than Al itself.

Similar in appearance to aluminium (density $\pu{2.7 g/cm^3}$, melting point $\pu{660 ^\circ{}C}$, source) is magnesium (density $\pu{1.7 g/cm^3}$, melting point $\pu{650 ^\circ{}C}$, source).  Both are used, pure as well as in mixture of other metals (alloys), widely.  Even to the point that they are mixed to form alloys with each other, for example in the frames of laptops

(source)

or items that are used near the sea / in contact with sea water (source).


Role of alcoholic potassium hydroxide in an E2 reaction Which of the following compound will give least substituted alkene as major product with alcoholic $\ce{KOH}$.
The options are:
1.
2.

3.
4.

I am confused between options 1 and 3. Both seem to give more substituted alkene as the major product. Does alcoholic $\ce{KOH}$ always mean it is an E2 elimination?
What is the mechanism here?

Haloalkanes react with alc. $\ce{KOH}$ mainly via $E_2$ mechanism. But, in some cases, where $\beta-\ce{H}$ isn't present, it proceeds via $E_1$, as in case of $(4)$ one. Let's analyze each option,
Option $(1)$:
Firstly, the $\ce{-Cl}$ departs, and then the base picks up a $\beta-\ce{H}$ from the terminal methyl group (i.e., $\ce{C-5}$), resulting in mono- substituted alkene (Hofmann product as major). Steric factors are significant in selectivity of $\beta-\ce{H}$. Therefore, Zaitsev product will be formed in very minor concentrations, bcoz $\ce{-C(CH3)2Ph}$ is more bulkier substituent than $\ce{^tBu}$.
Option $(3)$:
This one is similar to $(1)$ case, except for the fact that due to less steric hindrance, it will give di- substituted alkene (Zaitsev product).
Option $(2)$: It's clear that this molecule will form a tri- substituted alkene.
Option $(4)$: As hinted above, it will proceed through $E_1$, resulting in a tri- substituted alkene.

Different levels of substitution are shown below, (source: www.chemistrysteps.com)


What is the physical quantity/magnitude of pH? I have a database to classify variables/parameters in water resources as follows:

Variable: Precipitation (P); Quantity: Length; Units of measurement: mm (millimeters).

Following the example, what would be the properly quantity of variable pH?

Variable: Potential Hydrogen (pH); Quantity: ?; Unit of measurement: pHU (pH unit, unit less, dimensionless).

Maybe concentration or activity?

[I know what pH is, It is not necessery explain, I'm looking for the most appropriate word to describe its quantity]

I disagree with the answer that says "pH measures acidity or basicity" (no matter what is in Wikipedia).

pH is the negative of the log of the hydrogen ion (H+) activity. Activity is a thermodynamic concept that is related to but not the same as concentration. Activity is similar to chemical potential. The same concentration of hydrogen ions will have different activities, depending on what else is in solution. In weak solutions, the activity coefficient that relates the concentration to the activity depends mainly on the charges of the other ions in solution and not on the particular ion species. But at higher amounts of other species, the particular chemicals and chemical interactions become increasingly important.

On the other hand, acidity is the ability to provide hydrogen ions to a reaction and considers not only the hydrogen activity but also the amount of acid-base reaction that can occur. You may have heard of pH 'buffers'. These are solutions that are designed to have a balance of chemical species so that the pH is very stable, even if small amounts of acid or base are added or if the ionic strength of the solution changes. If you add a bit of base to the buffer solution, the solution reequilibrates to provide acid to react with the base. You have to add a fair bit of base before the buffer capacity is used up and the pH changes very much. Then, with further addition, the pH will change a very quickly until it approaches the pH of the base that is being added. So a pH buffer has high acidity and alkalinity, keeping the pH stable.

In groundwater geochemistry, acidity is measured by reacting the solution with strong base until the acid is essentially used up. This is done by titration either to a fixed endpoint pH or to the inflection point in the curve of base added vs pH curve.

Bottom line, pH doesn't measure acidity. In terms of the original question, I would say:

Variable: pH, Quantity: negative log of the hydrogen ion activity, Units: dimensionless pH units.


Clarification on absorption spectra and crystal field theory My textbook, Chemistry: The Central Science, gives this figure to explain the purple color of the hexaaquatitanium(III) ion:

Such absorption spectra are common to chemistry and biochemistry. However, as the crystal field theory explains, the peak wavelength absorbed is equal to the splitting energy $\Delta$ between the two split energy levels of the d subshell. This is illustrated in the figure below.

So the splitting of the d subshell explains the peak of the absorption spectrum. However, I’m not comprehending why the curve slowly descends on either side of that peak—if only 495 nm light is absorbed, then shouldn’t the absorption spectrum show absorption only at 495 nm, with no absorption anywhere else? Why does absorption slowly decrease as you move in either direction away from 495 nm? What mechanism allows these other wavelengths to be absorbed as well?

Moreover, normally, when things absorb light, they reemit the energy in the form of infrared. That’s why that wavelength is taken out of the wavelengths that reflect back to your eyes; if the same wavelength were reemitted, you would perceive it too. I assume infrared is radiated because electrons can take alternate pathways to move back to ground state, jumping down in small steps that release less energy each time. However, with d subshell splitting, there is only one possible way for the electron to climb down to ground state. Shouldn’t that mean that the wavelength it absorbed will be reradiated out? Shouldn’t we see that wavelength, then?

You would expect a sharp peak for a well-defined structure which has bond lengths and angles that do not significantly change. In a crystal structure at low temperatures you may be able to observe sharp peaks rather than broad bands.

In solution, however, there is a number of effects that broaden the peaks, making the entire thing less sharp:

aqua ligands can dissociate and reassociate; if the photon hits the complex while the structure is different, the wavelength will be, too
typically, the ligands of a complex vibrate both symmetrically and antisymmetrically. Both vibrations lead to structure changes, therefore electronic changes and thus changes in the absorption wavelength.

Different solvents will give different peak broadening (I saw a graph once, but I can’t remember where it was).

The titanium(III) spectrum is also more complex, by the way. If you look closely, you can see a shoulder on the low-energy side of the main peak. This is because the excited state is Jahn-Teller distorted, and there is not one but two transitions possible that slightly differ in energy. (The ground state is very, very weakly Jahn-Teller distorted, because there is not much energy to be gained.)

After a molecule has absorbed the energy of a photon and went into an excited state, there is not only the simple rerelaxation that would emit a photon of the same wavelength to return to the ground state. Rather, by a set of complicated mechanisms, the energy can be transformed into vibration (more and more vivid than before) and thus slowly released in form of heat. That is why a solution that absorbs light will warm slightly.


Can you fix rubber on a handle after it has all but turned into a liquid? I picked up a splitting maul at a yard sale last year. I used it for the rest of the season with no problems. It has been in the wood shed and exposed to some sun, but mostly shade.

I picked it up the other day and it stuck to my hand. When I tried to let go, it slid from my hand, leaving behind black goo and rubber from the handle. Now it's so bad the whole rubber part is gooey and if I try to rub anything on it the rubber comes off. It's permanently soft and sticky and is coming off the handle.

Can I save it or is it done?

I've had the same problem with other types of rubber-coated tool handles. The best solution is to completely remove the rubber coating, which takes a lot of scraping and then scrubbing (outdoors) with paint thinner on paper towels. Invariably, as you have already noticed, everything touched gets coated with hard-to-remove black goop.

Another choice would be to wipe off the loose stuff and then over-wrap the handle with a strong cloth tape, checking carefully that it will not slip when you use the maul.

You might cover the tape with heat-shrink tubing, if you have the right diameter stuff, or with tubing cut from an old bicycle inner-tube (cheap or free, and makes a good grip).


Why do we need the rms, mean, and most probable velocities? In the kinetic theory of gases, we have rms (root mean square), mean, and mp (most probable) velocities. I understand the concept well. But my question is why do we have three different kinds of velocities? Why don't we stick onto any one of them? What is the significance of these velocities?

And it would be helpful if someone could explain how:

$$\bar{v} = \sqrt{\frac{8RT}{\pi M}}$$

Alright, here's my best attempt at it.

Root Mean Squared $v_\mathrm{rms}$

The root-mean-square speed (r.m.s. speed) of the molecules. This quantity is defined as the square root of the mean value of the squares of the speeds,$v$, of the molecules. That is, for a sample consisting of $N$ molecules with speeds $v_{1}, v_2,...,v_N$, we square each speed, add the squares together, divide by the total number of molecules (to get the mean, denoted by angle brackets $\langle\rangle$), and finally take the square root of the result:

$$\begin{align}
v_{rms} &= \langle v^2\rangle^{1/2}
\\&= \left(\frac{v_1^2+v_2^2+...+v_N^2}{N}\right)^{1/2}
\end{align}
$$

Mean Speed $\bar{v}$

The r.m.s speed might at first sight seem to be rather peculiar measure of the mean speeds of the molecules, but its significance becomes clear when we make use of the fact that the kinetic energy of a molecule of mass $m$ travelling at a speed $v$ is $E_\mathrm{k} = \frac{1}{2}mv^2$, which implies that the mean kinetic energy of a collection of molecules, $\langle E_\mathrm{k} \rangle$, is the average of this quantity, or $\frac{1}{2}mv_{rms}^2$. It follows from the relation $\frac{1}{2}mv_\mathrm{rms}^2= \langle E_\mathrm{k}\rangle$, that

$$v_\mathrm{rms} = \left(\frac{2\langle E_\mathrm{k}\rangle }{m}\right)^{1/2}$$

Therefore, wherever $v_\mathrm{rms}$ appears, we can think of it as a measure of the mean kinetic energy of the molecules of the gas. The r.m.s speed is quite close in value to another and more readily visualized measure of molecular speed, the mean speed, $\bar{v}$, of the molecules:

$$\bar{v} = \frac{v_1+v_2+...+v_N}{N}$$

For samples consisting of large numbers of molecules, the mean speed is slightly smaller than the r.m.s. speed. The precise relation is

$$\bar{v} = \left(\frac{8}{3 \pi}\right)^{1/2}v_\mathrm{rms} = 0.921v_\mathrm{rms}$$

For elementary purposes, and for qualitative arguements, we don't need to distinguish between the two measures of average speed, but for precise work the distinction is important.

Most Probable $v_\mathrm{mp}$

$v_\mathrm{mp}$ is the most probable speed of the molecules, and is the maximum value for the velocity distribution curve. When putting all of of these equations together we see the relation

$$v_\mathrm{rms} > \bar{v} > v_\mathrm{mp}$$

That $v_\mathrm{mp}$ is the smallest of the three speeds is due to the asymmetry of the curve.

References

Atkins, P.; de Paula, J. Elements of Physical Chemistry 6th ed. W.H. Freeman and Company. New York, NY. 2013.

Chang, R. Physical Chemistry for the Chemical and Biological Sciences. University Science Books. Sausalito, CA. 2000.


Naming compound containing a ring (cycloalkane) side chain? Why is the correct name 5-cyclopentyl-2-methylheptane rather than 2-methyl-5-cyclopentylheptane? I am using the 2nd edition Hornback Organic Chemistry textbook. I got this answer from the solutions manual.

I have searched the textbook and the internet for the explanation and am still very confused. I thought the order of the side chain in the name was alphabetical, not including the prefixes? So by this logic, "methyl" should be before "pentyl," right?

In view of the release date of your book (2005), the used names for organic compounds are probably in accordance with the 1993 Guide to IUPAC Nomenclature of Organic Compounds. However, note that the current version of the recommendations can be found in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

According to the 2013 recommendations, two methods are possible for naming systems composed of rings and chains: either the ring or the chain can have seniority. To construct the preferred IUPAC name (PIN), the ring is always selected as the parent structure. Therefore, the PIN for the compound given in the question is (6-methylheptan-3-yl)cyclopentane since the cyclopentane ring has seniority over the heptane chain.

In general nomenclature, however, the context may favour the chain, so that the parent structure has the greater number of atoms in the principal chain. The corresponding systematic name for the compound given in the question is 5-cyclopentyl-2-methylheptane.

According to the 1979 recommendations, a hydrocarbon containing a small ring attached to a long chain was generally named as a derivative of the acyclic hydrocarbon. The 1993 recommendations do not explicitly mention this rule but refer to the 1979 recommendations. Hence, it is not surprising that your book still uses such a name, although the 1998 Guide to IUPAC Recommendations already explains that the ring is always selected to be the parent structure.

Now coming to your actual question:

According to the 2013 recommendations, the prefixes ‘cyclopentyl’ and ‘methyl’ in the name 5-cyclopentyl-2-methylheptane are arranged alphabetically.

  P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

  P-14.5.2 The name of a prefix for a substituent is considered to begin with the first letter of its complete name.

Therefore, ‘cyclopentyl’ begins with ‘c’ and ‘methyl’ begins with ‘m’. Note that, for example, ‘2,3-dimethyl’ is considered to begin with ‘m’ since the multiplicative prefix ‘di’ does not alter the alphabetical order. Whereas, for example, the compound substituent name ‘(1,2-difluorobutyl)’ in 7-(1,2-difluorobutyl)-5-ethyltridecane is considered to begin with ‘d’.


Decreasing order of electron density in benzene ring among the halobenzenes What should be the order of electron density in monosubstituted halogen derivatives of benzene?
I am aware that fluorobenzene is more ring activating as compared to chlorobenzene due to the more effective +M effect of F over Cl, but I would like to know the full order among all the monohalogenated benzene compounds.
Is there any significant +M effect shown by Br, I with the benzene ring?
If yes, then how strong is it as compared to their –I effect?
Please let me know the correct final order of electron density in the benzene ring among Ph—F, Ph—Cl, Ph—Br, Ph—I, as well.

Firstly, I would like to talk about the order of withdrawing effect among halogens. Since all the four halogens are electron-rich, they tend to donate electrons (+M effect). However, the high electronegativity of these elements hinders the donation by applying an even greater electron-withdrawing effect (-I effect) that overcomes the +M (resonance) effect.
In the case of halogens, the -I effect varies as -$\ce{F}$ > -$\ce{Cl}$ > -$\ce{Br}$ > -$\ce{I}$ due to the decreasing electronegative character as we move from fluorine to iodine.
The +M effect also varies as -$\ce{F}$ > -$\ce{Cl}$ > -$\ce{Br}$ > -$\ce{I}$ due to the decreasing overlap between the 2p orbitals of carbon and the 2p, 3p, 4p and 5p orbitals of the halogens respectively. Since the -I effect is stronger than the +M effect as mentioned earlier, the final order of electron-withdrawing capability is
-$\ce{F}$ > -$\ce{Cl}$ > -$\ce{Br}$ > -$\ce{I}$
Equipped with this knowledge, we come back to the question of comparing the electron densities of mono halogen benzene derivatives. From the above-mentioned trend, we can infer that the order of electron density is
$\ce{Ph-F}$ < $\ce{Ph-Cl}$ < $\ce{Ph-Br}$ < $\ce{Ph-I}$
This trend is validated by the order of reactivity of these compounds towards electrophilic substitution reactions.
Hope it helps!


Numbering in chlordiazepoxide The IUPAC name of chlordiazepoxide is given as : 7-chloro-2-methylamino-5-phenyl-3H-1,4-benzodiazepine-4-oxide, but when I started numbering the structure, I am unable to find chlorine on 7th carbon, it is on 10th. Then why 7-chloro... is given is it wrong? Or I am wrong? Please help by numbering the structure in a diagram even on a paper an upload the photo.

The parent structure of chlordiazepoxide consist of the heteromonocyclic component 2H-1,4-diazepine that is fused to a benzene ring.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the name of the fused parent structure is generated as follows:

  P-25.2.2.4 Heteromonocyclic components fused to a benzene ring

  Unless listed as a retained name (…), a benzene ring fused to a heteromonocycle of five or more members (a benzoheterocycle) is named by placing the locant(s) indicating the position(s) of the heteroatom(s) at the front of the name consisting of the fusion prefix ‘benzo’ followed by a retained name, a Hantzsch-Widman systematic name, or a name formed by skeletal replacement (‘a’) nomenclature (…). The locants cited correspond to the full bicyclic structure. As in Hantzsch-Widman names, locants are placed in the order corresponding to the order of citation of the heteroatoms in the heterocyclic component. The locant ‘1’ is always assigned to the atom of the heterocyclic component next to a fusion atom. Heteroatoms are allocated lowest locants as a set, without regard to kind; if there is a choice, lowest locants are assigned in accordance with the seniority of the ‘a’ prefixes (…). In general nomenclature locants may be omitted when the name is unambiguous; for preferred IUPAC names locants must be cited. The letter ‘o’ of the ‘benzo’ prefix is elided when followed by a vowel. Indicated hydrogen is placed at the front of the name, when required.
  (…)

Accordingly, the name ‘benzodiazepine’ consists of the prefix ‘benzo’ followed by the name ‘diazepine’. The locant ‘1’ is assigned to the atom of the heterocyclic component next to a fusion atom and the heteroatoms are allocated lowest locants as a set (here: 1,4):

The same numbering of locants applies to the complete structure of 7-chloro-2-(methylamino)-5-phenyl-3H-1,4-benzodiazepine 4-oxide:


Software to assist drawing of complex, three-dimensional skeletal formula from scratch or from existing crystal structure Is there a tool that would help in creating rather intricate line drawings (not pre-rendered balls-and-sticks/wireframe/sticks/ORTEP diagrams) for a skeletal formula in 3D space?
To illustrate the question a bit further, here are two examples:

Organic molecules — An all-benzene catenane and a trefoil knot [1]:

Inorganic molecules — proton transfer in polyoxoniobate $\ce{[H_xNb10O28]^{(6-x)-}}$ [2, p. 111]:

I also saw a much larger line-art structure of an entire peptide drawn the similar way, but unfortunately I cannot find the reference at the moment.
Line-drawn structures like these are usually presented alongside with the determined crystal structures, so I naturally assume that there is a tool that would allow to convert a CIF file to an illustration like the ones shown above.
However, I wasn't able to find anything relevant by screening the IUCr software list, and this already concerns me for several years.
Do people really use molecular editors like ChemDraw for this, or there is a more optimal way?
Also, if ChemDraw is the tool for the job, are there any general recommendations as to how to efficiently draw structures like this in 3D space without loosing one's mind on estimating z-buffer for proper bonds' greyscaling/thickness/type based on the distance from the viewer as well as overlapping atoms and bonds?
I would really appreciate if someone who published similar work would share some tips regarding their workflow.
References

Segawa, Y.; Kuwayama, M.; Hijikata, Y.; Fushimi, M.; Nishihara, T.; Pirillo, J.; Shirasaki, J.; Kubota, N.; Itami, K. Topological Molecular Nanocarbons: All-Benzene Catenane and Trefoil Knot. Science 2019, 365 (6450), 272–276. DOI: 10.1126/science.aav5021.
Polyoxometalate Chemistry, 1st ed.; van Eldik, R., Cronin, L., Eds.; Advances in Inorganic Chemistry; AP, Academic Press, an imprint of Elsevier: Cambridge, MA San Diego, CA Oxford, 2017; Vol. 69.

Thanks to vapid and their comment

  There are editors that have 3D mode and they can display a 3D molecule in 2D manner. MarvinSketch is first that comes to my mind, but there are probably more.

I figured a crude solution, which I hope might as well serve as a starting point for other answers.

First of all, I completely forgot that MarvinSketch (free (registration required), available for Windows, Linux, macOS) has support for 3D.
Drawing in 3D from scratch with MarvinSketch is probably not the best idea as even though it has a support for optimizing (cleaning) 3D structures, it sometimes gives weird results for organics and practically not suitable for inorganic complexes and especially clusters.

Arguably, the best and the most affordable third-party tool for creating such structures is Avogadro (open source, available for Windows, Linux, macOS).

If there is a crystal structure at hand (e.g. CIF data), then one must do two things first:

crop/grow the fragment of interest;
assign proper bond orders and export the structure in one of the formats that preserve bond orders (such as MOL, MOL2, PDB or CML).

I believe Avogadro can do both, but I personally like to do part 1 in Olex2 (free (registration required), available for Windows, Linux, macOS) by force of habit.

As a proof of concept, I took the crystal structure for the trefoil knot, allowed it to grow from assymetric unit to a complete molecule as well as cut out solvent and guest molecules, so I was left with the initial shape:

which I immediately exported to the MOL file by running file trefoil-knot.mol.

Next, I opened this MOL file with Avogadro, which automatically assigned bond orders:

and the file was exported to MOL once again, and subsequently opened in MarvinSketch (I reduced fog in settings, made implicit hydrogens hidden and used F7 button to align the molecule on screen):

And the final result came out to look like this:

The image is, however, not quite as perfect as the one in the paper.
For instance, there are some artifacts about inflection points as if the program was confused how to place double bonds.
Also, if you look closer, the grayed out part which was supposed to be further from the viewer, actually appears in front, which I attribute to the fact that Marvin was designed as a 2D molecular editor first and foremost, so it doesn't always handles depth (or the aforementioned z-buffer) adequately.


How long does it take to change states of matter? I'm interested to know how long it takes to change states of matter and why it takes the time that it does.  For example, what is the very quickest that we could freeze a packet of chicken breasts?  In the food industry, flash freezing is often used to prevent damage from occurring by freezing meat very quickly.  But why does it take the time that it does?  Would it be possible to freeze my packet of chicken in say, a nano second instead of a few minutes?

Part of the answer is that the rate of freezing depends on the heat of fusion of the material being frozen (relatively large for water, the chief ingredient of uncooked chicken), but smaller for other substances).

Another piece of the answer is that an object can lose heat by convection, conduction, and/or radiation. You would increase the rate of heat loss by:

(1) increasing the surface area of the chicken as much as possible (flattening it),

(2) placing it against a surface held as cold as possible and made of a material such as metal with high heat capacity;

(3) hitting it with a continuous blast of extremely cold air or gas at the highest possible wind speed (remember wind chill factor?) and maximizing radiant heat loss by previously having painted the chicken with flat, black paint and conducting the experiment in a mirrored, insulated chamber. Some unusual materials can be induced to cool by irradiation with a laser (this is how we have obtained the lowest recorded temperatures ever) but this will probably not work with chicken.

I do not think there is an absolute limit on cooling rate, just diminishing returns for increasing energy expenditure. But I may be wrong. I hope you pursue this until you get an answer that satisfies you. That is how new discoveries are made.


Is this chemical name wrong? 2-bromobutane vs 1-bromobutane Well, technically it's already wrong because it should end in "-propane" but my textbook is so crappy let's just ignore that for a second. The naming convention goes that substituent groups should be named so that they have the lowest number possible. Then, in both of the bromopropanes below, shouldn't they both be 1-bromopropane because there are no other substituent groups besides the bromine?

The numbering of locants in the question is correct. In acyclic hydrocarbons, the longest chain is numbered from one end to the other.

The corresponding subsection of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-15.1.7.1.4 The longest chain is numbered from one end to the other by arabic numbers, the direction being chosen so as to give the lower locants to the substituent groups (side chains) [see P-14.4(f)]. (…)


Doubts on IUPAC name 1,1'-methylenebis(4-isocyanatobenzene) Wikipedia states  IUPAC name of the above compound as 1,1'-methylenebis(4-isocyanatobenzene) .

I have a confusion. Shouldn't isocyanate group be given preference over the methylene group? I assume the name as 4,4'-methylenebis(1-isocyanatobenzene) .

Similar is the case with DDT.  Wikipedia states its IUPAC  name as 1,1'-(2,2,2-Trichloroethane-1,1-diyl)bis(4-chlorobenzene) .

Why is the numbering of carbon with three chlorine atoms done $1$ ? Moreover,  can it be  $1,1'$ in the first place in place of $4,4'$ ? I guess the $\text{diyl}$ term indicates that it is connected to two different carbons(of benzene)  .

To sum up,  I want to know which chain should I select as parent chain and why .

The isocyanato group is named by prefix (like halogeno-), it's not a principal characteristic group named by suffix. It can be said that these groups have lower priority than the “joint” position.
See the different numberings (examples from the IUPAC nomenclature of organic chemistry, 2013):

P-15.3.2.4.1
(…)

1,1′-oxybis(4-bromobenzene) (PIN)

4,4’-oxybis(2-bromobenzoic acid) (PIN)

Therefore it is named 1,1′-methylenebis(4-isocyanatobenzene), not 4,4′-methylenebis(1-isocyanatobenzene).
(You will probably argue that here, the oxy might have higher priority than bromo group or methylene connector, but “connector priority” is not taken into account.)
See also more rigirous answer to Q: What is the rationale of the IUPAC name for DDT?


Retention of surface texture after combustion Just some while ago I was burning a box of cardboard in my backyard and I saw a very curious phenomenon.

I saw that as the cardboard box started burning it started to curl.  Up after a while as I came back to check if the fire was out I saw that though the cardboard box wasn't a box any longer but individual sides/faces of the box had retained their ridges and texture though they had tuned greyish white and crumbly.

My questions:

Why did the cardboard start to curl up as it began to burn?
Why did it retain its texture if not shape and size?

The curling up has a simple reason: the burning was not uniform. Paper contains water (really!), and heating it makes it lose the water.

Now, paper is a bad conductor of heat. When you burn it, inevitably the flame will be mainly on one side(the side that's pointing up). The flame will transfer heat to nearby regions of paper, and evaporate some (or possibly all) of the water. The loss of water leads to a contraction of the cellulose fibers in the paper (I'm not too sure of this, comments appreciated). But, this only happens on one side of the paper. Even though paper is razor-thin, the fact that it still has two "faces" is important here. Now, due to the contraction, the paper curls in that direction (similar to how a bimetallic strip works). That's it.

The reason for retaining texture, is something I'm not too sure about. The burning reaction throws up a lot of soot, but a lot of the products stay right where they are. This results in the charred/burnt paper occupying the previous position of the cellulose. While charred paper isn't a polymer and thus isn't too strong, it can "mimic" the initial structure just by occupying the position of the cellulose. But, unlike cellulose, if we try bending it, it will crumble since there are no long polymer "chains"/intermolecular forces holding it together. It's like the difference between a house of cards and a similar "house", made of a single piece of rubber. The structure is the same, but the former case is only staying together because it had the "luck" to exist in a stable configuration. Disturbing it will destroy it.