Why should I not use different battery types together? Everyone knows about this but I'm curious about why one should never mix batteries of different types. By type I mean the same shape of the cell like AAA but with different chemical reactions and voltages inside.

Would it just mess up the voltage and the instrument that is operated with the battery? Or would it lead to unexpected reactions in the battery itself that could lead to leakage/explosion or other catastrophies?

Serial usage

For the serial usage, one must avoid combination of cells of different kind and health. They should be as identical as possible.

The best is using

the same chemistry
the same format
the same vendor
the same brand
the same lot
the same aging

Otherwise the cells will have the different capacity and different discharge profiles under the load.

If the serial cell chain is deeply discharged, the "weakest" cell gets discharged completely.

And what is worse, other cells start to electrolyze the unlucky cell in the opposite direction that would be its charging.

This would lead almost inevitably to heavy damage or more likely destruction of the cell, what is unwanted for rechargebles.

Physical destruction with electrolyte leakage is also possible.

For primaries, it would be rather annoyance of not fully used better cells and no easy possibility to find a matching cell.

Parallel usage

It is generally adviced not to use it, unless you know exactly, what you are doing. See also notes for non rechargeable cells below.

Dangers

If cells of different chemistry and therefore voltage range are used in parallel, they overcharge or overdischarge each other when their common voltage does not fit the overlapping of their save voltage ranges.

But even within the overlapping voltage range, cells can be overheated and thermally functionally damaged/destroyed or can even explode, depending on the cell voltage difference and cell internal resistence.

The last paragraph is also valid for cells of the same chemistry. They cannot overcharge each other, but the thermal damage from above applies to them as well.

How to do it well

Only cells of the same chemistry and performance class with about the same actual voltage (check it ) should be connected in parallel.

The voltage difference must respect expected internal resistance and acceptable resulting balancing current. Note that typical internal resistance of Modern AA NiMH cells is $\pu{30-100 m\Omega}$, so $\pu{60 mV}$ difference can cause the current $\pu{1 A}$.

Once connected and balanced, they are safe, as they autobalance themselves during charging and discharging, but consider notes below.

Cell autobalancing of already connected parallel cells

This analysis is aimed for rechargeable cells., supported by Excel simulation.

Cells in an open circurt

Balanced parallel batteries have always the same external (behind their internal resistance ) and internal voltage.

Cells under load

The initial current is in reciprocal ratio of their resistances, so high performance cells would get drained initially faster despite their lower capacity.

The cells with lower $resistance * capacity$ parameter gradually get the internal voltage lover than the others and provide therefore gradually lower current than the other cells.

This leads to the cell balancing, with the current ratio possibly even getting reversed to the initial one. The current ratio slowly converges to the capacity ratio.

Important parameters are the ratio of cell capacities and the ratio of cell internal resistances.

The best results are achieved if these ratios are reciprocal. In such an ideal case, cells are discharging in ratio of their capacities and no internal autobalancing ever occurs.

The good thing is, that the cell resistance increases with the degrading the cell capacity, as cells are aging.

Best autobalancing is achieved on low currents.  High currents lead to higher internal voltage disbalance and are more demanding for good cell matching.

If the cell resistance ratio is not optimal for the capacity ratio, autobalancing kicks on, but does not fully compensate. The internal voltages get therefore little disbalanced.

After disconnection, there is ongoing small temporary compensating current.

It is not good to combine lower capacity high performance cells with low resistance with high capacity low performance cells, especially for high loads. As it significantly raises the internal voltage disbalance and after load compensation

Cells under charging

It is generally adviced against parallel charging, as it affects charge stop detection and can cause overcharging of some cells.

That affects serial charging as well.

Non rechargable cells

Parallel connection is not generally good for different . As there can occur small balancing load after disconnection of the powered device.

OTOH, if the identical cells ( like in the serial section) are used in parallel connection all their lifetime, it is not a big deal, as these currents are minimal and short time.


Mechanism for Chromyl Chloride Formation This is the very famous chromyl chloride test, used for detection of chloride ion in solution -

$$\ce{K2Cr2O7 + 4NaCl + 6H2SO4 -> 2CrO2Cl2 + 2KHSO4 + 4NaHSO4 +3H2O}$$

When a mixture containing chloride ion is heated with potassium dichromate and concentrated sulphuric acid, deep orange-red fumes of chromyl chloride are formed. My question is, how?

I want to know the proper mechanism for this reaction, particularly for chromyl chloride formation. I couldn't find it in several texts (and online), such as JD Lee and Atkins.

Also, if instead of chloride ion, bromide ion (or iodide) is used then 'chromyl bromide' is not formed. Instead, bromine gas is liberated, and chromium (III) sulphate is formed along with potassium (I) sulphate. Why? What is so special about the chloride ion?

The answer may (and does) lie in the oxidation potential of the halogens. As we move down the group 17 the tendency for oxidation increases.

Refer:
$$\ce{H2SO4 + HBr -> Br2 + SO2} \tag{unbalanced}$$
but
$$\ce{H2SO4 + HCl -> X} \tag{does not occur}$$

$\ce{K2Cr2O7}$ is not able to oxidise $\ce{Cl-}$ to $\ce{Cl2}$ but it does so with $\ce{Br2}$, thus we observe reddish brown fumes of $\ce{Br2}$ as if $\ce{Br-}$ had been present, which are then passed through $\ce{NaOH}$ to confirm the presence of chloride or bromide.

Refer:

\begin{align*}
\ce{CrO2Cl2 + NaOH ->& Na2CrO4} \tag{yellow soln,unbalanced} \\
\ce{above + H+ ->& (Cr2O7)2-} \tag{orange solution} \\
\ce{Br2 + NaOH ->& NaBr + NaBrO3} \tag{light yellow soln}
\end{align*}


Can I use harvested plants to gain electric power? I was just thinking - plants are mostly acid. Even grass is quite acid. Could you create battery cell from squashed plants (after cutting them in pieces)?

How much effective would it be? Could you compare it to solar panels? Would you need to replace electrodes often?

A company in the USA called Voltree Power has gone one step further to harvest electricity generated directly from live trees and other plants!

It seems trees are able to provide somewhere between 20 and a few hundred millivolts as well as some current, derived from a difference in pH between the trees and the soil, maintained by the trees metabolic processes.

The science behind this technology was published in 2008 by researches at MIT.

The company claims to have demonstrated the technology by developing the first wireless sensor network powered exclusively by trees to detect forest fires! They are also developing the technology to monitor humidity and temperatures in the forest using electrical power harvested from the trees themselves and transmit the data using low-power wiremess mesh technology.

Another collaborative group from University of Michigan, MIT & UC Berkeley published their research in 2009 into micro-power electricity scavenged from evaporative flows within the microchannels in trees.


Why doesn't sulfate get electrolyzed in sulfuric acid electrolysis? If I have aqueous sulfuric acid solution as electrolyte and the anode electrode is made out of copper and the cathode electrode is made out of graphite, what are the two reactions at the respective electrodes and why?

\begin{align}
\text{Anode:}  && \ce{Cu &-> Cu^2+ + 2e-}\\
\text{Cathode:}&& \ce{2e- + 2H+ &-> H2 (g)}\\\hline
\text{Total:}  && \ce{2H+ + Cu &-> H2 + Cu^2+}
\end{align}

That should be the reaction, but why? Why isn't sulfuric acid participating? Why does that specific reaction happen?

To analyse this kind of problem, the first step should be identifying all the species you have in solution that could take part in any electrolytic reaction. These are:

$$\begin{array}{}\ce{SO4^2-} & \ce{H+} & \ce{H2O} & \ce{Cu}\end{array}$$

I did not add graphite to this list since it generally never participates in electrolytic processes. The next step would be to identify what can be reduced and what can be oxidised. Thankfully, all species are either in their lowest or highest common oxidation state. See the list below with oxidation states added:

$$\begin{array}{cccc}\ce{\overset{{+VI}}{S}\overset{-II}{O4}^2-} & \overset{\mathrm{+I}}{\ce{H+}} & \ce{\overset{{+I}}{H2}\overset{-II}{O}} & \overset{\pm 0}{\ce{Cu}}\end{array}$$

Obviously, the only candidates for oxidation are oxygen or copper, and the only candidates for reduction are sulfur or hydrogen. I’ll concentrate on the oxidation first, as the equations are simpler. We need to find a redox pair for each component and then decide which is more likely. For copper, the only real possibilty is generating copper(II), while for oxygen the only real possibility is generating oxygen gas. See equations $(1)$ and $(2)$:

$$\begin{align}\ce{Cu &-> Cu^2+ + 2 e-}&&E^0 = +0.337~\mathrm{V}\tag{1}\\
\ce{2H2O &-> O2 + 4 e- + 4 H+}&&E^0 = +1.229~\mathrm{V}\tag{2}\end{align}$$

I added the standard electrode potentials $E^0$, too. We can easily compare which oxidation is more likely: the smaller potential will act first, thus, copper is oxidised. Now on to the reductions. For hydrogen, it is naturally the known equation. What about sulfur in sulfate? Well, the most obvious choice is assuming $\ce{SO3^2-}$ as a reduction target. See equations $(3)$ and $(4)$:

$$\begin{align}\ce{2H+ + 2e- &-> H2 ^}&&E^0 = \pm 0~\mathrm{V}\tag{3}\\
\ce{SO4^2- + 2 e- + 2 H+ &-> SO3^2- + H2O}&&E^0 = +0.17~\mathrm{V}\tag{4}\end{align}$$

The reduction potential is actually given for the target $\ce{SO2 (aq)}$; however, that has the same oxidation state and is in equilibrium with sulfite, so they are comparable. Here, the more positive potential should actually react first. However, using the graphite electrode typically favours hydrogen evolution via overvoltage mechanisms. This is, unfortunately, something you will have to memorise.

Summing up:

There are only four reactions that can happen
of these, two can be chosen as more favourable by their respective electrode potentials
one choice needs to be adjusted due to overvoltage.


Why aren't all bonds in cyclobutadiene not equal? Cyclobutadiene is like benzene in that it has alternating single and double bonds in a ring. However, it's bonds are not the same length, the double bonds being shorter than single bonds. The molecule is rectangular, not square.

How would this be explained?
I have one plausible explanation that I tried to give...that is as follows

Cyclobutadiene is anti aromatic with 4n = 4 pi electrons, moreover it has huge strain in the molecule. To avoid such strain the molecule goes out of plane to compensate, but thus loses its ability to resonate.

Thus resonance structures like ones in benzene cannot be drawn for cyclobutadiene.

Is this correct? Please correct me if I am wrong..and some more data about this would be appreciated

Strictly speaking cyclobutadiene is not antiaromatic but non aromatic, ie the same energy 4A + 4B is attained by applying the Hückel LCAO, as for two isolated double bonds, 2 times 2A + 2B.

Given that cyclobutadiene can only be isolated as non delocalized system when in matrixes and a very low T points to strain as a reason for instability of both forms, the delocalized one remaining hypothetical.

Edit for the readers not familiar with the terms: A system obeying all the requisites (cyclic, pi orbitals on each atoms) is antiaromatic when the hypothetical delocalization destabilizes it as compared to the localized system. If no energy difference arises, the system is called non aromatic.

Beware there are plenty of pages discussing non aromaticity in a trivial way, e.g. aliphatic or linear compounds. Cyclobutadiene is non aromatic within the Hückel picture. Also it seems worth to point out again that antiaromatic refers anyway to an hypothetical structure, not to an actual delocalization. A system which cannot avoid it by distorting would be unstable tout-court. Cyclobutadiene, albeit being non aromatic, is on the same line. Indeed you can find antithetical assertions about it, such as it being "the archetype of antiaromatics“ along with “it reacts as diradical“ as if the two things would be straight forward correlated. Normally a system escape antiaromaticity. Being cyclobutadiene non aromatic results in a distorted rectangular molecule, but only when it cannot escape this anyway unstable situation. See also Resonance structure of cyclobutadiene?. Although the situation of cyclobutadiene is again erroneously described as an antiaromatic case, it gives the answer. Also worth reading is the answer by Georg.


Interpretation of Raoult law In attachment I have tried to derive Raoult's law by using the concept of equilibrium constant. There is nothing mentioned in my text book about how this equation comes about, so I decided to do it myself. I may like to know whether there is some mistake in what I have done.

The partial pressure of a component in a mixture is always less than that of the pure component at the same temperature. (Calculating the chemical potential shows that if it were not the case the mixture would be unstable compared to the pure liquid).  The question then becomes what is the partial pressure of the component A above a mixture $p_A$ related to that above the pure liquid $p_A^*$.  Raoult's law states that the ratio is equal to the mole fraction $x_A$ of A, $p_A/p_A^*=x_A$.

Physically, this is justified by considering that the chance of finding a molecule of A in the vapour, relative to the chance for the pure liquid, is equal to its mole fraction. This is reasonable if dealing with ideal mixtures and assuming that the vapour behaves as an ideal gas. Thus there is no derivation as such, we proceed directly from the physical argument to the law.


How to obtain nitrogen using only homemade utilities? I am now constructing a RC plane. I would like to fill the tires of the landing gear with nitrogen to avoid combustion of oxygen during landing. I would therefore like to obtain a small amount of nitrogen using (only) home-made utilities. How can it be done?

Crude process, but does the trick: You can use a combustion/oxygen consuming process, like a candle, to burn the excess oxygen. Eventually the candle will be too oxygen starved to sustain a flame, so some residual oxygen would be left (although this would already mean there is no fear of combustion of said tyres). Now to remove the remaining oxygen you'd use something like iron files (since you wanted to use only "household" stuff). Together with some humidity in that air, they will rust rapidly and should drive down the oxygen contents towards negligible. So now we have:

a small trace remainder of oxygen,
the equivalent of $\ce{CO2}$ from the combustion process,
some impurities from the candle wax probably,
trace gases such as helium etc. and ... mostly
nitrogen.

This already is not sustaining any combustion. But if you're a purist or want to use $\ce{N2}$ because it is lighter than $\ce{CO2}$, to remove the $\ce{CO2}$ you could use something like "roasted limestone", i.e. the stuff from you water kettle, which you heat so that you get lime, which then binds with the $\ce{CO2}$ to revert to limestone $\ce{CaCO3}$. Now you have something, I'd assume, like 98% nitrogen and surely to few traces of $\ce{O2}$ to cause any harm to your equipment. Of course, I'd buy bottled $\ce{N2}$, but for the sake of using household implements only, this might do the trick.


Notation in atomic structure In the book, The theory of atomic structure and spectra, Robert D. Cowan says that the allowed terms of $p^2$ are $^1S$, $^1D$, $^3P$ and so the possible terms of $p^2 + p\quad$ (obtained by using the vector model) are:

$$^2P^\circ,\quad ^2P^\circ D^\circ F^\circ,\quad ^{2,4}S^\circ P^\circ D^\circ
 \tag1$$

I am having trouble in understand the notation in $(1)$. For example  the term $^2P^\circ$ means that $s=1/2,l=1$ and $^\circ$ means odd parity, but for the term $^{2,4}S^\circ P^\circ D^\circ$, I do not know what it means. Can anyone explain me it please?

Copied table below from "Quantum chemistry", Donald A. McQuarrie, University Science Books,  2008, page 472

This illustrates the notion that $p^2$ and $p^4$ are equivalent.

I think the 2,4 superscript in Cowan's book for $^{2,4}S^\circ P^\circ D^\circ$ is a peculiar notation of his. Most other books seem to presume the reader knows the equivalency.


How would you separate sugar from water? I'm a student and I've been getting interested in chemistry. I really like experiments and was brainstorming of possible things to try out, one of them was separating a homogeneous mixture. I thought of sugar water and tried looking it up online but couldn't find anything like I wanted. Everything says to use crystallization but I don't understand why you can't just boil away water and be left with sugar like you can with salt water. Anyways, what methods/ways can I separate sugar from water?

If you start with a known 'pure' mixture of sugar and water, then what you propose is fine, to an extent, if you are just after separating the two ingredients.

Crystallisation is a purification method for extracting a solid from solution; it will allow you to separate pure solute from the solvent. The remaining solution would still contain a small amount of dissolved material, and hopefully all of your impurities.

If you wanted to isolate pure solvent from the solution, you would need to use distillation. After collecting the distillate, you would be left with a gummy residue of semi-decomposed sugar (and your impurities).

Sugar is hygroscopic, and starts decomposing at just over 180°C (~365°F). The problem that you would have with your proposed method is that in order to remove the final remnants of the water, you will need to heat well above the boiling point of water. Google 'making caramel' if you can't think of what will happen. Even a very small amount of decomposition of your sugar will mean you are left with an impure solid. Back to square one. The reason that you can simply boil the water off salt is that it is far more stable and can be heated safely to much higher temperatures.

So, there are a few ways to separate these two ingredients, depending on your end goals:

To collect just solute (sugar): heat over very low heat to evaporate as much water as possible to give yourself a saturated solution. Then crystallise sugar and filter (in the lab you'd wash with a suitable solvent and dry under vacuum)
To collect just the solvent (water): distill over moderate heat to collect water. To collect as much as possible, the heat used to boil off the solvent needs to be slowly increased, especially as you approach the final portion.
To collect both solute (sugar) and solvent (water): You would remove solid first, then distil solvent. Unless you are starting with a saturated solution, you would need to do a partial distillation to remove some solvent, then crystallise your sugar, then complete the distillation. You'd still be left with a horrible brown gummy residue.

Sounds like a great home project, though. Just be sure to have done some suitable research to allow you to make use of your materials if it doesn't quite go to plan :-)


Batch conversion of Waters mass spectrometry file formats Are there any free tools for converting raw mass-spectrometry files into open formats? In particular, is there any way to convert Water's .raw MS files into a readable format in a batch-wise fashion (talking about hundreds of mass spectra, so it needs to be at the command line ...)? After a long search, I only found MassWolf which seems very outdated and which also doesn't seem to provide any command line batch processing. Isn't there something like OpenBabel for analytical chemistry file formats?

Waters Databridge actually allows the conversion of multiple files at the same time, but somehow it is not very intuitive to use (not a native UI) and also does not really work in batch mode. Still, it is somewhat possible to quickly export Waters .raw files using Databridge, and it is the best solution I found so far. From there, a simple python script allows me to average the mass spectra (that are still chromatograms), so that it at least doesn't take hours to export every single mass spectrum by hand.

For other people in my position, having to convert hundreds of Waters .raw files, I hope that helps at least somewhat.


Can nanotechnology reduce electrode losses in electrochemical cells? I'm just curious about this topic, as most portable, "clean" energy sources or carriers (in the sense of not being a primary energy source -fuel cell hydrogen, for example) are in essence electrochemical reactors, and most of the current work in the field is aimed at reducing irreversible losses at the electrodes, be it because of discharge overpotential, or polarization because of mass transfer along the membrane electrolyte; and of course, in making this structural elements cheaper.

Also, a lot of proposed and existing waste treatment and recycling processes aimed at recovering rare and/or active metals (such as lithium from Li-ion batteries) involve at least one electrochemical unit op.

I wonder if nanotechnological advances, such as increasing the surface area of electrode materials or increasing the effecive diffusivity of relevant species in solid porous electrolytes, can reduce this irreversibilities, perhaps making use of cheaper raw materials than noble metals (i.e. platinum). I would also appreciate any references to current work in this applied field.

Nanotechnology offers enormous opportunities for improving a huge range of electrochemical processes.

The use of platinum as a catalyst is been widespread throughout the electrochemical industry, helping reduce the activation energy of reactions such as in hydrogen and methanol fuel cells. However, platinum is a rare and expensive metallic element. Given that the reactions only take place on the surface of the catalyst, the use of nanoparticles of platinum reduces the amount of platinum needed to achieve the same surface area, thereby achieving the same performance at a fraction of the cost.

Another interesting application is in lowering the operating temperature in solid oxide fuel cells.

A good reference on Nanotechnology in Electrocatalysis for Energy, which covers background fundamentals and some applications, can be read on Google Books for free.


Oxidation numbers of atoms in Ti(SeO3)2? What are the oxidation numbers of atoms in $\ce{Ti(SeO3)2}$? I'm having trouble figuring this out since I don't know the charge for either $\ce{Ti}$ or $\ce{Se}$.

The selenite anion ($\ce{SeO3^{2-}}$) has a net charge of -2. Titanium's oxidation state should be obvious given this information. As for selenium's oxidation state, since we know each oxygen has a charge of -2, and since the entire polyatomic anion has a net charge of -2, the following equation must balance:

$$\ce{Se}+3 \cdot (-2)=-2$$


Calculating the heat capacity of a calorimeter I've been struggling with this problem that I can't seem to figure out. I think I know how to solve it but I think there's missing information.

  The combustion of 1 mole of glucose $\ce{C6H12O6}$ releases $\pu{2.82\times10^3 kJ}$ of heat. If $\pu{1.25 g}$ of glucose are burnt in a calorimeter containing $\pu{0.95 kg}$ of water and the temperature of the entire system raises from $\pu{20.10 ^\circ C}$ to $\pu{23.25 ^\circ C}$. What is the heat capacity of the calorimeter?

I think I need the specific heat of glucose (which I haven't found yet), but also I don't know why they give me the heat released by 1 mole of glucose.  I need to know the method of how to solve such problems.

$12.5\ \mathrm{kJ}$ of heat was absorbed by the surroundings.

I found this by using the mcat formula and the specific heat capacity of water (4.18 J/(g °C)):

$Q=m\cdot c\cdot\Delta T$

$Q = 950\ \mathrm{g} \times (4.18\ \mathrm{J\cdot g^{-1}\cdot {^\circ C^{-1}}}) \times (23.25\ \mathrm{^\circ C} - 20.10\ \mathrm{^\circ C}) = 12508.7\ \mathrm{J}$

If you wanted to use this whole formula for solving the calorimeter's specific heat capacity, you would need to know the mass of the calorimeter as well, which is not given.

What your book is probably asking is for what is called the "calorimeter constant". This is given in units of $\pu{J/^\circ C}$ notice that it does not include mass.

Note: Sometimes "the calorimeter's specific heat capcity" is used instead of referring to the calorimeter constant, but in this case we cannot find a value which will include mass in the units, so I think it is more clear to use the term "calorimeter constant."

You can determine the constant by this formula: $$Q_\text{cal} = C_\text{cal} \times \Delta T_\text{cal}$$

Where $Q_\text{cal}$ is the energy absorbed, $C$ is the constant and $\Delta T$ is the same as the change in temperature of the water.

You may calculate $Q_\text{cal}$ by using this formula: $$Q_\text{cal} = -(Q_\text{water} + Q_\text{glucose})$$

It may also help to think of $Q_\text{water}$ = $Q_\text{surroundings}$ and $Q_\text{glucose}$ = $Q_\text{system}$

To find $Q_\text{glucose}$ I did: (glucose has lost energy, it is negative value)

$-2820\ \mathrm{kJ} \times 0.007\ \mathrm{mol}$ and $Q_\text{water}$ is simply the $12508.7\ \mathrm{J}$ positive because $\Delta T$ is positive for the surroundings (the system/glucose lost energy)

$Q_\text{cal} = -(12508.7\ \mathrm{J} + (-19740\ \mathrm{J}))$

So my final answer is then: $2.3\times10^3\ \mathrm{J/{^\circ C}}$

It is important that heat capacities are positive, think about what it would mean if this were a negative value.

In the laboratory, it is necessary to do a calculation such as this one before using a calorimeter for anything. Normally it can be done by heating a piece of nickel or something, recording the temperature of the metal and the water, and then dropping the metal into the calorimeter to find the final temperatures, and then calculate the calorimeter constant. You can then proceed with further experimentation using that calorimeter, but only after this constant has been found can you find the specific heat capacity of other materials.


How to calculate saturation point of a solution? Is there a way to calculate theoretical saturation point between 2 materials?

For example, let's say I want to determine solubility of vitamin A in water, or vitamin B in olive oil, or vitamin D in milk. In these examples, if I know the molecular structure of the vitamins and the various liquids, can I figure out at what point saturation will be reached?

To be clear: If there's a way, I want to learn how to calculate these theoretical saturation values. I don't want to just get them from a table somewhere.

Yes. But it would probably be faster to simply find some empirical data.

You would need to:

approximate the hydrophobicity of your vitamins (for rather big molecules like vitamins, this would need some mathematical modeling, but there are some computer programs that can do this rather accurate.
this data could be used to approximate the solubility in water and probably also olive oil. But milk is quite complex, as there will be a solubility equilibrium in/between both the fat phase and the water phase. And all the salts and other compounds in the milk (and to a small degree in olive oil) will affect the solubility.

If you know A-level chemistry and know someone to help you irl (or some fab ppl here :-) ), I'd guess there's some days of work in it!


Does water have a chemical name? Does water have a chemical name?
If so, what is it?
P.S. I checked up the web and got all sorts of crazy answers like dihydrogenmonoxide, oxidane, hydrogendihydride etc. Please validate.

TL;DR IUPAC hasn’t made up their mind, but plain old water appears to be an appropriate name. However, chemical derivatives of water may not be named using water.

In Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 it is stated (P-21.1.1.2) that

  The common names water, ammonia, [...] are used in these recommendations but their use as preferred IUPAC names is deferred pending publication of recommendations for the selection of preferred inorganic names; thus, no PIN label will be assigned in names including them.

I’m not sure when the next edition of the Red Book (i.e. inorganic chemistry nomenclature recommendations) is coming out, but I can quote the relevant section from the most recent version.

Table IR-6.1 of Nomenclature of Inorganic Chemistry: IUPAC Recommendations 2005 lists "oxidane" as the parent hydride name for $\ce{H2O}$. However, it also adds the caveat that

  The names ‘azane’ and ‘oxidane’ are only intended for use in naming derivatives of ammonia and water, respectively, by substitutive nomenclature, and they form the basis for naming polynuclear entities (e.g. triazane, dioxidane). Examples of such use may be found in Section IR-6.4 and Table IX.

Therefore the compound $\ce{ONONO} = \ce{(ON)2O}$, a nitrosylated derivative of water, is named "dinitrosooxidane" and not "dinitrosowater". More examples may be found in Table IX of the same publication.

Water itself is still called water. For example, tritiated water $\ce{H^3HO}$ is named (3H1)water (Section IR-2.2.3.2).


Consequence of interactions between filled orbitals It is known that the interaction between filled orbitals from the perspective of molecular orbital theory is overall destabilising because antibonding MOs are more antibonding than bonding MOs are bonding. Consider a large molecule (e.g. $\ce {I2}$). Due to the presence of multiple net destabilising interactions between the filled inner shell atomic orbitals (e.g. $\ce {1s, 2s, 2p}$ ...), wouldn't its formation be almost impossible since the single stabilising interaction due to the valence atomic orbitals would be small, compared with the total destabilisation from all the interactions between the inner shell orbitals, which are in fact, very numerous?

On p. 261, Atkins & Friedman (2005) mention two criteria when considering the significance of orbital interactions. On the first criterion, they write:

  First, to participate significantly in bond formation, atomic orbitals must
  be neither too diffuse nor too compact. In either case, there would be only
  weak constructive or destructive overlap between neighbouring atoms, and
  only feeble bonds would result. It follows that in Period 2, ($\ce {1s,1s}$)-overlap can be largely neglected in comparison with ($\ce {2s,2s}$)-overlap, for $\ce {1s}$ orbitals are too compact to have significant overlap with each other. Indeed, it is generally safe, for qualitative discussions at least, to consider only overlap between orbitals of the valence shell, for only these orbitals are neither too compact nor too diffuse to have significant overlap.

This means that it is likely that for the $\ce {I2}$ molecule I mentioned in the question, the destabilisation from the interaction of filled inner shell orbitals (e.g. $\ce {1s, 2s, 2p}$ etc.) is probably very insignificant. As orthocresol writes in the comments, this can also be related to the overlap integral $\ce {S}$.

In orthocresol's answer here, we note that the reason behind antibonding MOs being more antibonding than bonding MOs are bonding is that the value of the overlap integral is not negligible (i.e. $\ce {S}$ is not much less than $\ce {1}$ such that $\ce {S}$ $\approx$ $\ce {0}$). However, in the case of the $\ce {1s-1s}$ overlap in the $\ce {I2}$ molecule, the orbitals may be too compact to interact such that $\ce {S}$ may actually become negligible. Thus, the antibonding and bonding MOS are destabilised and stabilised to the same extent respectively. Thus, these filled orbital interactions of the inner shell orbitals do not constitute much destabilisation.

Reference

Atkins, P.; Friedman, R. Molecular Quantum Mechanics (4th ed.). Oxford University Press, 2005.


Nuclear power: what material is the control rod made from? What type of material is used in the control rod of a nuclear reactor? (Nuclear power plant)

There are various designs and materials for control rods, which are more technically correctly called control assemblies. The choice depends on the type and purpose of the reactor, the size of the core, the availability of other means for power control (e.g. boric acid in the reactor coolant, burnable poison rods, $\ce{UO2-Gd2O3}$ fuels, recirculation pumps, or thermal-hydraulic feedback), patent rights, and tradition.

The control assemblies of typical boiling water reactors (BWRs) are inserted from the bottom because, in a BWR, the neutron flux is larger in the lower part of the core and the space above the core is occupied by the water separator and the steam dryer. Therefore, the design should consider saving weight. BWRs usually use cross-shaped control assemblies. The four blades of one cruciform control assembly are inserted between four fuel assemblies. The blades are made of stainless steel and contain tubes that are filled with boron carbide ($\ce{B4C}$) as neutron-absorbing material. In addition, hafnium ($\ce{Hf}$) absorber may be placed in selected locations of selected control assemblies.

The control assemblies of typical pressurized water reactors (PWRs) are inserted from above. Each control assembly has multiple rod fingers, which enter openings at the upper end of a fuel assembly. These control assemblies are very heavy because, when the reactor is tripped, they have to quickly drop into the core against the pressure of the reactor coolant. The neutron absorbing material mainly is an $\ce{Ag-In-Cd}$ alloy (usually 80 % $\ce{Ag}$, 15 % $\ce{In}$, and 5 % $\ce{Cd}$) encapsulated in an austenitic steel cladding. These three neutron absorbers together effectively block the entire energy spectrum of fast, epithermal, and thermal neutrons. In other PWR core designs, the control rods are divided into two groups: so-called black assemblies made up of rods consisting of an $\ce{Ag-In-Cd}$ alloy in the lower section and of $\ce{B4C}$ in the upper part, and so-called grey assemblies containing absorber rods made of $\ce{Ag-In-Cd}$ alloy and steel.


Are glycans, such as glucose, considered alicyclic compounds? That's my question. If true, it would make my job a lot easier. I will appreciate any feedback on the matter.

Seems I underestimated this problem.

First passage from Wikipedia is crucial:

  An alicyclic compound is an organic compound that is both aliphatic and cyclic.

And second is wrong:

  They contain one or more all-carbon rings(...)

(Source)

Glucose is carbohydrate, in its circular form which is found in glycans (also glucose isn't a glycan, but glucose residue is found in many of them) it has oxygen atom in the ring - according to second passage from Wikipedia it therefore shouldn't be alicyclic, but one can easily find mention about alicyclic ethers for ex. here, with oxygen in ring. So I'm changing my answer to:

Yes, they are.


What is an equivalent? I want to know what is an equivalent and how is it different from equivalent mass. How to calculate and why is it used in chemistry for knowing concentration?

Equivalent concept is an archaic unit of measurement that was used in chemistry and the biological sciences in the era before researchers knew how to determine the chemical formula for a compound.
An equivalent (symbol: officially equiv; unofficially but often Eq) is the amount of a substance that reacts with (or is equivalent to) an arbitrary amount of another substance in a given chemical reaction. The mass of an equivalent is called its equivalent weight.
In a more formal definition, the equivalent is the amount of a substance needed to do one of the following:

react with or supply one mole of hydrogen ions ($\ce{H+}$) in an acid-base reaction.
react with or supply one mole of electrons in a redox reaction.

I know that it is a bit difficult to understand by the means of only definitions, so let me give you some intuition about it.
The basic idea of equivalent concept is that,

$1$ eqiv of a substance reacts with exactly $1$ equiv of another substances to produce $1$ equiv of each product.

Let us start with the example of Acid-Base reaction, so for acids we define one equivalent as the amount of the acid (in moles) that supplies one mole of $\ce{H+}$ ions and for acid we define it as the amount of base (in moles) that supplies 1 mole of $\ce{OH-}$.
For Example:-

$1$ mole $\ce{HCl}$ supplies $1$ mole $\ce{H+}$ ions, so $1$ equiv of $\ce{HCl}$ is equal to $1$ mole $\ce{HCl}$.
Similarly, $1$ equiv of $\ce{H2SO4}$ = $1/2$ mole $\ce{H2SO4}$, $1$ Eq $\ce{NaOH= 1}$ mole $\ce{NaOH}$,
$1$ equiv $\ce{Al(OH)3}$ = $1/3$ mole $\ce{Al(OH)3}$.

Now if you look at the above closely, you will notice that since 1 equiv of acid supplies 1 mole $\ce{H+}$ ions and 1 equiv of base supplies 1 mole $\ce{OH-}$ ions so 1 equiv of an acid should completely react with 1 equiv of base to produce 1 equiv of each product ( eg. 1 mole $\ce{H2O}$, since its a neutralization reaction).
In a similar manner we can define equivalent for all other types of reactions, such that the main idea i.e. "1 eqiv of a substance reacts with exactly 1 equiv of another substances to produce 1 equiv of each product", remains unaltered.
Now lets define equivalent for other things,

Element
The equivalent of an element can be defined as the amount of the element that reacts with or supplies 1 mole electron in a reaction, so if you try a bit you can conclude that 1 equiv of an element is equal to 1/n mole of element where n is the valency of the element. (NOTE:- the equivalent of a substance can be different for different reactions, eg. 1 equiv of $\ce{Fe=}$ 1/2 mole $\ce{Fe}$ if it under goes to an oxidation state $\ce{Fe^2+}$ and it will be 1/3 mole if it undergoes to an oxidation state $\ce{Fe^3+}$).

Ions
Similar to element equivalent of ions can be defined as 1/n mole of ion where n is the charge on ion (try to understand it yourself using concept of electrons used earlier).

Salt
A salt have an equal number of cations and anions, so we need to define an equivalent in such a way that, 1 equiv of salt supplies 1 mole of cations (with +1 charge or 1/n moles of cation with +n charge) and anions (with -1 charge or 1/n moles of anion with -n charge) each so that these cations and anions react completely with another one mole of anions and cations (having -1 and +1 charge) that may have been supplied 1 equiv of another salt/ substance.
Now if you think a bit harder you can get to the point that one equiv of a salt is equal to 1/n mole of salt where n is the total number of charge either on cationic part or on anionic part (eg. in NaCl, total cationic/anionic charge = +1 so, 1 eq of NaCl= 1 mole NaCl, now if you pay your attention, you can notice that 1 mole NaCl will give 1 mole Na+ and 1 mole Cl- which is in accordance to what we said earlier.)
Some more examples for your help: 1 equiv of $\ce{AgCl2= 1/2 AgCl2}$ as it will supply 1/2 mole of $\ce{Ag^2+}$A and 1/2 mole of (Cl-)2, in simpler terms it will give one mole of positive charge and one mole of negative charge.
1 equiv of AlCl3= 1/3 moles of AlCl3.

Oxidizing agent/ Reducing agent
Now if you have come this far without dozing off, so you should be able to define that 1 equiv of an Oxidizing/ Reducing agent is equal to 1/n mole of that agent where n is the number of electrons gained or lost.

Now lets get to the question " Why is equivalent concept is used in chemistry for knowing concentration? ", well you might yourself have got the answer that it is because it is much simpler to apply as you do not need the Stoichiometry of the reaction going on, equivalent concept works fine without it.
And also to know the equivalent mass you simply need to find the mass of one equivalent of a substance using mole concept.
So that was all, I knew. Lets end this with a quote " Science is nothing but simple logic which is often not so simple."


Why do alkyl bromides and iodides develop colour when exposed to light? My book says alkyl halides are colourless when pure. However, bromides and iodides develop colour when exposed to light. Why do they develop colour when exposed to light? It is important to note that alkyl fluorides, and alkyl chlorides are not said to develop colour.

This is because of spontaneous breakdown in case of iodides and bromides which have a lower bond-dissociation enthalpy as compared to fluorides and chlorides. This low enthalpy means that even at normal temperature or on exposure to sunlight, the halides (iodides and bromides) decompose to a certain extent forming free iodine and bromine, both of which are coloured and hence impart colour to the original halides.


Does acrolein give Fehling's test? Aromatic aldehydes do not give Fehling's test, which makes me believe that for the same reason acrolein should not. It has a double bond conjugated with the carbon-oxygen double bond as in benzaldehyde. Is my assumption correct? Or is extended conjugation a factor?

Glyoxal(CHO-CHO) and Glyoxalic acid(CHO-COOH) also don't give Fehling's test. Any reasons for that?

It has nothing to do with aromaticity. It is related to this answer.

Why does benzaldehyde not respond to Fehling's test?

Any aldehyde without an alpha hydrogen will not give Fehling's test.
Acrolein also doesn't have an alpha hydrogen, so it will not give Fehling's test.


Is there a way to convert aniline to phenol without diazotizing it? I am a high school student and while reading about reactions on amines, I was trying to figure out a way to convert aniline $(\ce{C6H5NH2})$ to phenol $(\ce{C6H5OH})$ without diazotizing it with sodium nitrite $(\ce{NaNO2})$ and dilute hydrochloric acid $(\ce{HCl})$ in temperature conditions of $\pu{273 K}$ to $\pu{278 K}$ and then reacting it with $\ce{H2O},$ i.e. anything that does not involve the following reaction:
$$\ce{RNH_2 + NaNO_2 + HCl ->[$273–\pu{278 K}$] [RN2^+Cl^-] ->[H2O] ROH}\tag{R1}$$
I tried converting it to an isocyanide via the carbylamine reaction, but that seemed to make things more complicated. The closest I found to solving this was Tilden's reagent $(\ce{NOCl}),$ which reacts with primary amines in the following way:
$$\ce{RNH2 + NOCl -> RCl + N2 + H2O}\tag{R2}$$
However, this is limited to primary amines by definition.
Is there any way to convert aromatic amines to aromatic alcohols without diazotizing the compound? More specifically, is there a way to convert aniline to phenol without converting it to a benzene diazonium ion?

Benzenediazonium salt is so versatile that it can be used to make many compounds. That's why, diazotization reaction is so popular and useful. The reaction conditions is not that harsh and is easily achievable. Coming to the reaction, diazotization is a straight one-way procedure to achieve aniline to phenol conversion. If you talk about other ways of converting aniline to phenol, the reaction conditions for such ways are generally harsh and hence not done practically to obtain phenol. In addition, there are so many steps, that the conversion is practically not feasible.
One of the ways of converting aniline to phenol without diazotizing is to convert aniline to acetanilide and then hydrogenolyze it to make benzene which can be then used to make phenol.

Step 1

Step 2

The reaction is taken from a paper1. The paper discussed the use of $\ce{NiMo}$ based catalyst along with novel neo-microporous composite material HSM/ZSM-5 as support. The reaction result is compared with other supports like $\ce{\gamma-Al2O3}$ and lone ZSM-5 and HSM system.

Step 3

Reference

Sajedeh Jafarian, Ahmad Tavasoli, Hasan Nikkhah, "Catalytic hydrotreating of pyro-oil derived from green microalgae spirulina the (Arthrospira) plantensis over $\ce{NiMo}$ catalysts impregnated over a novel hybrid support," International Journal of Hydrogen Energy 2019, 44(36), 19855-19867 (https://doi.org/10.1016/j.ijhydene.2019.05.182) 44. 10.1016/j.ijhydene.2019.05.182(PDF).


DSC Metal hysteresis (melt/cryst) So we were analyzing Polymers using DSC and calibrated the system using Indium. I expected it to be a single transition only as Indium only has 1 known crystal structure and as I never worked with DSC nor seen a real thermogram before I was quite surprised to a relatively large hysteresis between metling and recrystallizing. I expected it to be a big gap for polymers but for something as uniform as a metal with only one crystal structure it forms I thought the point of melting and crystallizing should be quite close together.

I remember the lecture on solid state chemistry about DSC's of crystals where Ehrenfest and displacive/reconstructive changes in for example quartz-phases had hysteresis and I think displacive ones didn't have any but it was never really explained what caused this gap. I always imagined that it takes more time to form a symmetry like going against entropy and this is why more energy has to be taken out of the system.

But is there a better explanation for pure metals? Does this gap tell me anything about the structure in the end? I imagine that structures which tend to me rather amorphous might have a smaller gap because it's not so much symmetry involved in the process which would then also imply why it is big (relatively, I haven't compared it to any other compound yet) for Indium.

Disclaimer: the following is based on using DSC data for organic materials.
Is the melting temperature of Indium (156 C) just at the temperature where your sample exhibits a phase transition / decomposition?  In general, I suggest to use the flat indium pellets and to record their melting for calibration of the DSC, rather than their recrystallisation. This is likely described in the manual of your DSC.  (I does not hurt to have a look at DOI: 10.1007/s10973-015-5186-8, or similar, either.)
Because of its transient character, it is worth to keep an eye on the heating rate / cooling rate while recording DSC data.  Provided enough sample material, sometimes a quick screening with 10 K/min is followed by a second recording at a rate of 4  K/min (or slower).  While the endothermic peaks related to melting may be sharp, the ones for the recrystallisation are often shifted towards lower values because without proper seeds of crystallisation, supercooling may occur, especially organic samples and polymers.
I recommend to consider cyclic DSC, instead of a simple single-heating experiment, as this offers you additional insight about the sample in question. It is applied to characterize pure metals (like DOI: 10.1098/rspa.1996.0113), and may shed a light on the degree of crystallinity and processes of reorganisation of your polymer sample, too.  In the following, a DSC of polylactic acid:

(source)
Sometimes these differences about recording the melting point (pristine sample, after 1st / 2nd heating cycle) are spot on small molecule materials, too.  They may appear even if the sample did not incorporate solvent molecules.


Equilibrium, temperature change, and prediction in constant volume When it's constant volume, change of $T$ (Kelvin temperature) in a reaction equilibrium system won't change the volume. Let's say we change the temperature from $T$ to $2T$ of this reaction at equilibrium

$$\ce{3A + B -> C + D}$$

Because of the change of temperature, $K_\mathrm{p}$ will change, depending on whether the reaction is exo- or endothermic. At the same time, because of constant volume and change of temperature, reaction quotient $Q_\mathrm{p}$ at $2T$ is newly formed (originally $Q_\mathrm{p}(T)=K_\mathrm{p}(T)$ in equilibrium turn to $Q_\mathrm{p}(2T)$)

Here's a question. Say, this reaction is exothermic, and $T$ is changed to $2T$. So, $K_\mathrm{p}$ should be lowered, and with $K_\mathrm{p}$ lowered, reaction also proceeds to reverse. This can also be explained by Le Chatelier's principle.

But when it is at constant volume, $Q_\mathrm{p}$ could newly form simultaneously. And if $Q_\mathrm{p}(2T)<K_\mathrm{p}(2T)$, the reaction could proceed forward even though it's exothermic with the increase in temperature.

I've done some math and it seems it only depends on how big $\Delta H$ is. So can that really happen? Or am I missing something? If what I'm saying is right, is this an expception of the Le Chatelier's principle?

(I've explained everything here, but I can explain with an example if anything is still unclear)

For this particular scenario, the system in equilibrium satisfies (assuming ideal gases):

$$K_{\text{p}}^{(T)} = \frac{p_\text{C} p_\text{D}}{p_{\text{A}}^3 p_\text{B}} = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(T)\,2}}}$$

The equilibrium constant at temperature $2T$ can be calculated as:

$$K_\text{p}^{(2T)} = K_\text{p}^{(T)} \exp \left(\frac{\Delta H}{2RT}\right)$$

After increasing the temperature at constant volume, a new total pressure, $p_\text{tot}^{(2T)}$ is attained. Since the molar fractions are not changed at the beginning, the reaction quotient is:

$$Q_{\text{p}}^{(2T)} = \frac{p_\text{C} p_\text{D}}{p_{\text{A}}^3 p_\text{B}} = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(2T)\,2}}}$$

So, for the reaction to proceed forward, we want $Q_{\text{p}}^{(2T)} < K_\text{p}^{(2T)}$, that is:

$$\frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(2T)\,2}}} < K_\text{p}^{(T)} \exp \left(\frac{\Delta H}{2RT}\right) = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(T)\,2}}} \exp \left(\frac{\Delta H}{2RT}\right)$$

Simplifying:

$$\frac{1}{{p_\text{tot}^{(2T)\,2}}} < \frac{1}{{p_\text{tot}^{(T)\,2}}} \exp \left(\frac{\Delta H}{2RT}\right)$$

But $\displaystyle p_\text{tot}^{(T)} = \frac{n_\text{tot}RT}{V}$ and $\displaystyle p_\text{tot}^{(2T)} = \frac{2n_\text{tot}RT}{V}$, so:

$$\frac{1}{4} < \exp \left(\frac{\Delta H}{2RT}\right)$$
$$\Delta H > - 2RT \ln{4}$$

As long as $\Delta H$ satisfies this relationship (which will always imply a range of exothermic reactions since $T$ is always positive), this reaction will proceed when increasing the temperature at constant volume.

Notice that if this was a reaction where $\Delta n = 0$, the $p_\text{tot}$ cancels out and we get:

$$1 < \exp \left(\frac{\Delta H}{2RT}\right)$$
$$\Delta H > 0$$

Which implies that, when change in number of moles is not a factor to consider, the equilibrium will only shift to the right upon temperature increase at constant volume for reactions with $\Delta H > 0$, i.e. endothermic reactions.


What defines an element's taste? A useful post by @Martin indicated that probably the naming of Sweetwater town is because of the sweet tasting lead compounds in it's water.
Then my question arose. I know that the taste of any material is dependent on which tongue buds it provokes. Now, my question is, which chemical property of an element defines or is directly related or is causing the chemical property of taste? (I deliberately avoid asking this for all chemical species to prevent very broad answers, if you think you can give not-too-long answers, my question is about all chemical species.)
Note that by a responsible chemical property I mean things related specifically to an atom, like e.config.

TL;DR: Don't taste pure elements. They either taste of nothing, taste foul, or kill you. Or all of the above.

Edit: Also, for clarification, elements don't have a taste. Taste is a biological reaction to chemical interactions happening with an element, not a property of the element itself. So, like everything else in chemistry, the taste is decided by the electron configuration of the element.

Taste as we know it happens in chemical reactions between our taste buds and the chemical species in question. Most of the taste is actually governed by the nose, where very specific receptors interact with very specific chemicals to send very specific signals to the brain, the chemistry of which is, while not trivial, also not interesting. It's your basic, run of the mill presumptive test.

In the taste buds, however, the receptors look for certain specific characteristics of the species being tasted. As an example, acids are sour, indicating that our sour taste buds are actually pH meters (although there might be more to it than this).

Similarly, most sugars have similar characteristics, that can be detected by a relatively simple chemical detector.

If, however, you go around tasting elements, you'll find two very interesting things.

First, you'll find that you're dying from all kinds of poisonings. At a rough count, all of the periodic table is poisonous in its pure form, with only a few exceptions (carbon, oxygen, hydrogen, nitrogen, silicon, titanium, noble gases, gold and platinum, possibly a few others).

Second, you'll find that the pure elements don't taste much at all, if anything. Taste is a completely biological concept, and these elements simply aren't found in nature, meaning that the tongue and nose have no frame of reference.

Most metals will taste metallic, due to tarnishes forming on their surface, mostly oxides, which all have the familiar, metallic taste. The non-metals are mostly fairly reactive, and taste absolutely foul as they react to the air in your moist mouth.

Gases like chlorine and fluorine may react and form chloride and fluoride, which taste sour, and then dissolve your mouth and nose, and then kill you.

The more radioactive metals will seem to taste of blood, but that's actually just acute radiation poisoning setting in, actually filling your mouth with blood. If they didn't kill you, they'd probably taste metallic like the rest of the metals, but they do, so they don't.

Finally, the alkali metals like lithium and its friends will taste bitter, also kill you.

"But wait!", I hear you scream, "What of, say, lithium in pharmaceuticals?"

When elements like lithium are ingested, they are done so in a rather more controlled fashion. You don't get a brick of pure, elemental lithium and told to go crazy when you go crazy, but rather, delivered in pills, in conjunction with a delivery agent, in this case, lithium citrate.

Most elements have some biological role or other, but if they are in the kind of quantity you could taste, they are most likely poisonous, and in many cases, fatally so.


Why do methylamine and ethylamine have nearly the same basicity constant? It is known that ethyl has a stronger inductive effect than methyl.
Its positive inductive effect is greater than methyl.

Then why does ethylamine has a $\mathrm{p}K_\mathrm{b}$ of $3.35$ and methylamine has the $\mathrm{p}K_\mathrm{b}$ of $3.34$
($\mathrm{p}K_\mathrm{b}$ values in water)
Shouldn't the $\mathrm{p}K_\mathrm{b}$ of ethylamine be lower than methylamine?

Ethylamine has a pKb slightly greater than that of methylamine.

As you are saying

  It is known that ethyl is better than methyl for inductive effect. It's positive inductive effect is greater than methyl.

is absolutely true. But the basicity in aqueous medium is not only determined by +I effect but also solvation of substituted ammonium cation.

Even though +I effect in ethylamine is more than that in methylamine, ethylamine  is more bulkier and sterically hinders the formation of intermolecular H−bonds with water.


Converting pure nylon salt to polymer? During one of the methods of  polymerization:

  Nylon 6,6 is prepared from polycondensation of hexamethylene diamine
  (HMD) and adipic acid. The repeat unit of the polymer is $\ce{–[NH(CH2)6
NHCO(CH2)4CO]}$. The diamine, which melts at 40.87 °C, is normally used
  in the form of a concentrated aqueous solution. The dibasic acid is
  used in its pure solid form (m.p.= 152.1 °C). In the preparation of
  nylon 6,6 polymer, the first step is a preparation of salt formed from
  a mixture of precise stoichiometric quantities of the two monomers.
  The salt can be prepared by mixing the alcoholic solutions of the two
  components, the pure salt precipitates and is dissolved in water to
  give a solution. Alternatively, the salt can be prepared by mixing a
  dispersion of the diacid in water with a solution of the diamine. The
  aqueous salt solution is prepared so that it has a concentration of
  about 50-60%. For the preparation of nylon 6,6 polymer, the salt
  solution is subjected to evaporation at the boil, possibly at elevated
  pressures, until concentration ≥ 60% are achieved. The concentrated
  salt solution together with a small amount (0.5%) of a molecular
  weight stabilizer is then heated in a reactor under a blanket of
  nitrogen so that temperature increases gradually and pressure reaches,
  typically, 1.73 MPa (250 psi). As water evaporates and temperature
  increases from about 212°C to 275°C, the molecular weight of the
  polymer reaches about 4000. The further reaction is achieved by a
  gradual decrease in reactor pressure to atmospheric pressure and then
  holding the polymer under these conditions for about one hour. At this
  point, the polymer is not quite equilibrated but molecular weights are
  in the range of 12000 to 17000. The process is devised to remove all
  the liquid water present in the salt solution as well as almost all
  the potential water, which may get liberated on the reaction of
  carboxyl (-COOH) and amine end groups present in the polymer. High
  molecular weight nylon for industrial applications is obtained by
  conducting the final stages of melt polymerization under reduced
  pressure (to remove water) or by addition of chain coupling agents (to
  help improve stoichiometric balance). The finished polymer is then
  extruded in the form of a ribbon or strand, quenched with water and
  cut to form chips. After drying, the chips are sent for subsequent
  spinning operation. Alternatively, the polymer may be sent directly to
  a spinning machine without prior solidification. In nylon 6,6 the
  polymerization reaction goes almost to completion.

Now my question is in the above method of polymerization involves formation of aqueous nylon salt from pure salt. I want to know why the precipitated pure salt isn't converted to polymer but is made aqueous and then converted to polymer? Isn't there anyway of forming the polymer from pure nylon salt (fully crystal dehydrated salt) instead of the aqueous one from the reactants?

Note: I'm ok with reacting the pure salt with any chemical substance to form the polymer.

The problems with reacting pure nylon salt are as follows:

The molecules of hexamethylene diamine (HMD) and adipic acid will have a very hard time accessing each other's reacting groups, because of limited mobility within the solid structures.
The reaction is a condensation, so it will produce a substantial amount of water anyway, so pretty much any reason for keeping the reactants dry is irrelevant, because the final polymer will end up wet anyway.

It might be possible, but because of (1), your product might be impure, containing a lot of small unreacted adipic acid and HMD particles.

Also, If you don't want to deal with the tedious procedures above, considering that they were probably written for industrial applications, consider using this procedure. It's a lot more lab friendly.


Balmer's formula and lagurre polynomials How can you explain the Balmer's formula using the properties of Laguerre polynomials?
It's a homework question but I really don't get it.
1.What is the property of the Lagurre polynomials in the first place?
2.How is Balmer's formula related to the Lagurre polynomials??

When solving Schrodinger equation different potentials are included depending on the problem at hand, for example, particle in a box, harmonic oscillator, H atom etc. Then boundary conditions are applied; for a particle in a box, for example, the wavefunction has to be zero at either end of the box and this ensures that the solutions are quantised, i.e the wavefunction depends both on position x and an integer n and the energy also depends on an integer.

The different solutions to different types of differential equations were worked out many many years ago, and often by French mathematicians consequently the polynomial solutions obtained often carry their names, Laguerre, Hermite, Legendre and Chebychev (although he was Russian). These polynomials all depend on position x and one or more integers. The Hermite polynomials, for example, which form solutions to the harmonic oscillator, are often labelled as $H_n(x)$ where different polynomials are found with  $n=0,1,2 ...$, e.g. $H_0(x)=1, H_1(x)=1, H_2(x)=4x^2-2 \cdots$.

In solving the H atom the radial and angular solutions are separated and each solved independently. The associated Laguerre polynomials ($L_n(x)$) are used in finding the radial solutions, and the angular solutions  depend on the spherical harmonic polynomials which themselves depend on the associated Legendre polynomials $P_l^m(\cos(\theta)$.

(Notes. The polynomials are often described by recursion formulae meaning that once the first couple are known the rest are easily found. The associated polynomials are found by repeated differentiation of the polynomial, via Rodrigues' formulae. The polynomials are orthogonal.)


Phase rule for a pure substance in vapor-liquid region I'm confused with how the Gibbs Phase Rule works for pure substances in liquid-vapor mixtures. According to the phase rule, there should only be one degree of freedom for a two-phase mixture with a pure substance. For example, if we pick a temperature for two phases to exist, there will only be one pressure that satisfies this condition. But how is the amount of vapor in the mixture, defined?
Looking at the figure below, it appears as if the pressure and specific volume are both independent of the two-phase region. Does this contradict the phase rule? If I wanted to obtain a vessel containing a mixture of water with 50% vapor and 50% liquid, could it be done by changing the pressure and temperature alone?

Note that the phase rule is qualitative. It says nothing about the particular amounts. It simply enumerates the theoretical number of independent parameters needed to determine amounts and compositions for the given number of components and phases.
The definition of a system, like total amounts of components, does not count among degrees of freedom.
The number of degrees of freedom of a system is the number of independent parameters, that are needed for the given system to determine its state.
For the given container volume and the substance amount, a given temperature determines the equilibrium pressure and vice versa. So there is 1 degree of freedom.
Amounts of particular  phases are determined by the temperature as the  single degree of freedom.


Competition between substitution and elimination (uni- or bimolecular) Whats is/are the main product/s of the following reaction:

I think the main product would be the one obtained by $S_N2$, and then the one pair of enantiomers obtained by $E2$. Is $E1$ with subsequent rearrangement possible?

For an $\mathrm{E1}$ mechanism, we would need to generate a carbocation. Our halide is attached to a secondary carbon so the generated carbocation would be secondary. A quick check of neighbouring stabilising groups is negative. And as a professor of mine once put it (possible slightly oversimplified but with some general truth in it):

  Exactly one secondary carbocation exists and that is the prop-2-yl cation.

Thus, $\mathrm{E1}$ is not possible since the carbocation will not form in any measurable amounts. For the same reason, $\mathrm{S_N1}$ is not an option. For everything else, let’s consider the molecule’s 3D structure:

The tert-butyl group acts as a conformation ‘anchor’ meaning that the other chair form is not viable. Luckily, bromine is pointing upwards, which gives us access both to $\mathrm{S_N2}$ and $\mathrm{E2}$ mechanisms without having to invert or use a boat conformation.

Our base/nucleophile is methanolate which is both a good enough base and a good enough nucleophile, able to access both pathways. The centre for nucleophilic attack is also well accessable. If I had to choose, I would indeed go with your choice of saying the main products will be $\mathrm{S_N2}$ adduct and two enantiomers of elimination product will also be formed. (They stem from either abstracting the proton I drew or the enantiotopic proton on the back half of the six-membered ring.)


Comparing the reactivity towards aromatic electrophilic substitution reaction In the following pairs of molecules, which is more reactive towards electrophilic substitution reaction?

1)  1,4-dinitrobenzene or 1,3-dinitrobenzene (don't consider the ortho isomer)

2)  benzene-1,3-diol or benzene-1,4-diol (don't consider the ortho isomer)

3)  4-methylphenol or 3-methylphenol

4)  4-nitrotoluene or 3-nitrotoluene

I am having a conflict of concept.

In the first pair of molecules, according to resonance the presence of one nitro will make C3 negative and C4 positive. So if another nitro comes at C4 then the positive charge at C4 will intensify as nitro draws more electron density. If the second nitro comes at C3 then the negative charge at C3 will be absorbed by nitro. So which of these will reduce the electron density on benzene more.

In my opinion 1,3-dinitrobenzene will have less electron density on the benzene ring as (C3 is already negative due to nitro at C1) much of the electron density at C3 is absorbed. But in 1,4-dinitrobenzene the second nitro group is absorbing electron density from C4 (in which electron density is very less already due to presence of nitro at C1). So since the electron density is less at C4 then it won't be able to take much electron from ring (as its already positive and deficient of electron density).

If my concept is wrong with respect to above context, please correct me.

All of the remaining isomer pairs have same problem, but each is a bit different. I would like an explanation for each answer and especially the 1st one in detail. And try not to provide an answer by directly copying from a source (especially large) thanks.

We can answer your question by first looking at each individual, unsubstituted position in the benzene ring and determining the relative reactivity at each of these positions. Next we can compare the molecules and see which one has the most activated positions.

I've carried out the first step of this exercise and show the results in the following figure.

Looking at the top molecule, m-dinitrobenzene, we see that 3 positions are deactivated by the resonance effects of both nitro groups (that is, these positions are either ortho or para to each nitro group).  One position is not deactivated since it is meta to both nitro groups. We can analyze the para-dinitro isomer in a similar manner.  In this case we find that each position is only deactivated by one of the nitro groups.  Clearly, a position that is doubly deactivated will react much slower than a position that is singly deactivated.  Now, if we try to estimate the relative molecular reactivities by summing up and averaging the reactivities at each position, we see that the meta isomer is roughly deactivated by an average of (0+2+2+2/4) 1.5 at each position, whereas the para isomer is roughly deactivated by an average of (1+1+1+1/4) 1.0 at each position.  Therefore our analysis suggests that the para isomer will be the least deactivated and consequently react faster.

In the dihydroxybenzene case, hydroxyl groups are ortho-para activating.  Applying the same type of analysis leads us to conclude that the various positions in the meta isomer will, on average, be more reactive than the positions in the para isomer.  Hence the meta isomer will react faster in an EAS reaction.

Similarly, in the cresol case both substituents are again ortho-para activating but with the hydroxyl group being more activating than the methyl substituent. In the meta isomer, 3 positions are doubly activated and one position is not activated (or deactivated).  In the para isomer, each position is singly activated, and only 2 of these positions are activated by the strongly activating hydroxyl group.  This would lead us to suspect that the meta isomer will react the fastest.


How do I determine molality from mole fraction without knowing how much of the solution I have? What is the value of $x$ if $x$ molal solution of a compound in benzene has a mole fraction of solute equal to 0.2?

I tried calculating the moles of $x$ through the value given but there are two variables, i.e mole of $x$ and mole of benzene so I am stuck.

Solution 1

Let's assume there is 1 mole total of solvent and solute. With a mole fraction of 0.2 solute, there is 0.2 mole of solute, the rest (0.8 mole) is benzene. We can calculate the mass of 0.8 mole benzene from the molar mass, and express it in kilograms:

$$ m_\mathrm{benzene} = n_\mathrm{benzene} \cdot M_\mathrm{benzene}$$
$$= \pu{0.8 mol} \cdot \pu{78.11 g/mol} = \pu{62.49 g} = \pu{0.06249 kg}$$

The molality $b_\mathrm{solute}$ is defined as the amount of solute divided by the mass of solvent:

$$ b_\mathrm{solute} = \frac{n_\mathrm{solute}}{m_\mathrm{benzene}} = \frac{\pu{0.2 mol}}{\pu{0.06249 kg}} = \pu{3.201 mol/kg}$$

Solution 2

Let's assume there is 1 kg of benzene. That corresponds to 12.80 moles of benzene. If the mole fraction of solute is 0.2 and that of the solvent benzene is 0.8, they are present in a molar ratio of 1 to 4 (for every four molecules of benzene, there is one molecule of solute). So the amount of solute in that sample is 3.201 mole. We are ready to calculate the molality:

$$ b_\mathrm{solute} = \frac{n_\mathrm{solute}}{m_\mathrm{benzene}} = \frac{\pu{3.201 mol}}{\pu{1 kg}} = \pu{3.201 mol/kg}$$

Problems lacking extensive quantities

It is helpful to understand the difference between extensive and intensive quantities. Intensive quantities do not depend on the size of your sample (examples include density, concentration, molar mass, partial pressure, coefficients in the formula of a compound). On the other hand, extensive quantities do depend on how much you have (mass, volume, amount of substance). Often, you define intensive quantities by dividing one extensive quantity by another (example: density is mass divided by volume).

In your question, you want to determine an intensive quantity (here: molality) from intensive quantities only (here: mole fraction). There is a larger set of exercises of that type. If it helps, you can arbitrarily come up with an extensive quantity to solve the problem. You can set the mass of benzene to 1 kg or 2 kg. Then, you can solve the problem. The answer will not depend on the mass of benzene, it cancels out. Other options are setting the amount of solute or the amount of benzene.

There are common exercises asking you for intensive quantities given other intensive quantities:

What is the formula of an oxide given the mass percentages of elements?
What is the density of xenon at STP given its molar mass?
What is the concentration of chloride ions in a solution containing 0.4 mol/L $\ce{MgCl2}$?
What is the partial pressure of dioxygen at STP if an air sample contains 20% of dioxygen?

For all of these, you may assume an arbitrary amount, volume, or mass (but just one) if it helps to solve the problem. Again, the actual value does not matter because it will cancel out.

For more on intensive and extensive, see e.g. https://chemistry.stackexchange.com/a/6538


Why is the viscosity of water and ethanol lower than that of a water-ethanol mixture? We found in an experiment using Ostwalds method that
ethanol is more viscous than water, which
is already surprising as water has stronger intermolecular bonds than ethanol as @MaxW pointed out.

The mixture
of both substances furthermore is significantly more viscous than both individual substances. This
is interesting. We did not come up with a
satisfying explanation for this behaviour.

I know that one shouldn't expect that a mixture behaves the same way or proportional to the concentration of the components, but I would appreciate an explanation of this effect on a molecular level.

This answer just popped up when I searched for viscosity of water-ethanol mixtures, so I just copied the abstract:
Viscosities of Binary and Ternary Mixtures of Water, Alcohol, Acetone, and Hexane
https://www.tandfonline.com/doi/abs/10.1080/01932690802313006?journalCode=ldis20
"This articles studied and determined the viscosities of the binary mixtures of water–methanol, water–ethanol, water–propanol, water–acetone, acetone–ethanol, methanol–ethanol, and acetone–hexane and the ternary mixtures of water–methanol–ethanol and water–ethanol–acetone at 20°C. It is shown that the mixing of water with the alcohols and acetone resulted in a positive deviation of viscosity, which reached the maximum value at the water mole fraction x 1 ∼ 0.7 for water–methanol, x 1 ∼ 0.72 for water–ethanol, x 1 ∼ 0.74 for water–propanol, and x 1 ∼ 0.83 for water–acetone binary mixture. This viscosity deviation can be mainly attributed to the formation of micelles of alcohol or acetone molecules in water because of the hydrophobic attraction between the hydrocarbon chains. The micelle surfaces are surrounded by hydration layers, leading to the positive viscosity deviation in the liquid mixtures because the water in hydration layers has a much higher viscosity than bulk water. Also, the contrary observation was found in the binary mixtures of acetone–ethanol and acetone–hexane, having a negative viscosity deviation."


Methods of preparation of propane My book States part C as the answer of the following question.

But according to me, part D, that is soda lime decarboxylation can also be a perfect answer besides the Corey House synthesis( that is the part C )the answer given by my book.
Please help in clearing my misconceptions.

Simple carboxylic acid or carboxylate ions (like CH3CH2CH2COO–) do not undergo decarboxylation to yield corresponding alkanes because in this case the leaving group would be a carbanion (in this case CH3CH2CH2–) which is very strong base and hence a very poor leaving group (the reason behind this is that strong base wants to give electrons, but leaving group has to take electrons and leave the compound, strong base is poor leaving group and weak base is good leaving group)

For decarboxylation to happen a keto group must exist to stabilise the negative charge, which is not present in salt of butanoic acid CH3CH2CH2COO–Na+

Yield of propane is very less if you chose decarboxylation, that is why Corey-House synthesis is best for preparation of propane.

I want to know why do you think that Option D is correct, mention in comment.


Why enol content of ethyl acetoacetate is more in toluene than water? Ethyl Acetoacetate has enol content of 0.4% in water and 19.8% in toluene. What is the reason behind this? Does this anything have to do with the intramolecular hydrogen bonding in the enol form?

The extent of enolization is affected by the solvent, concentration and temperature. Ethyl acetoacetate has enol content of 0.4% in water and 19.8% in toluene. This is because water reduces the enol content by hydrogen bonding with the carbonyl group making this group less available for intramolecular hydrogen bonding. The effectiveness of intramolecular hydrogen bonding in stabilizing the enol, with respect to the keto form is thus seen on varying the solvent and particularly on transfer to a hydroxylic solvent.


Length of C-H bonds in hybridized bonding orbitals What is the effect of hybridization and double-bond formation on the rest of the atoms/molecules attached to the central atom?

For example, in a benzene molecule, the Carbon-Carbon bonds are intermediate between single and double bonds. But what is the effect of the presence of the double bonds between the carbons on the bonds between the carbon and hydrogen? Do the bond lengths decrease because of the sp2 hybridized bonding orbitals?

Yes, the bond length decreases.  One way to think about it is that a sp3 center with 109.5 degree angles is more crowded than an sp2 center with 120 degree angles.  More crowding = more repulsion = longer bond length.

Another way to think about it is that the e-pair in an s orbital is sitting on the nucleus.  Adding a p-orbital moves the electron pair further away from the nucleus.  More p-orbital character = longer bond lengths.

You can test it for yourself using http://molcalc.org/


How to calculate percent error for absolute zero? In chemistry lab, we did an experiment with water of various temperatures to extrapolate a value for absolute zero based on experimental pressure values for water at different temperatures. Naturally, the data is a little off -- even though absolute zero is $0 K$, my extrapolated value is about $15 K$.

One of the questions in the post-lab asks us to calculate percent error, and I'm a little confused. I thought (and this post confirms) that percent error was calculated by dividing by the expected or "actual" value. In this case, however, the actual value is $0$, and you can't divide by zero.

How do you calculate percent error for absolute zero?

Using Celsius values doesn't really make much sense to me, because the Celsius scale is not absolute; $-273.15$ is essentially an arbitrary number.

It sounds like this is one of those canned labs that ask you to do standard things, even when they do not apply.

As you noticed, the percent-error formula needs you to use a value in the denominator.  If you use 0K in the denominator, you get a divide by zero.  If you use your 15K extrapolation, the percent error will always be 100%, no matter what the error actually was!

In such cases, I would look at the percent error with respect to your data.  Why?  Well, because somebody said you need to calculate a percent error and you might as well use a real datapoint rather than divide by 0 or some other silly result.  If you choose your coldest datapoint, you can say something about the percent error of the extrapolation process at the least.  Thus, if your coldest temperature recorded was 2C, you could say "My extrapolation suggests that absolute zero is 260C below my coldest recorded value.  The actual absolute zero is 275C below my coldest recorded value.  This gives me a 5.7% error on my measure of absolute zero relative to my coldest recorded value."


Essential oil extraction using ethanol I want to make extracts of Black seeds. I do not posses any professional equipment, I just want to ask if it's possible to make extracts by soaking the seeds in ethanol for a long time and then evaporating it. Will the dissolved extracts evaporate with the solvent?

The primary compounds of potential theraputic interest are all far less volatile than ethanol.  Ethanol based tinctures are sold and even have   available MSDS, so at least some of the oils of interest are soluble.  In looking at the chemical structures of the predominant oils of nigella sativa I would guess that they should all be fairly soluble in ethanol.  In some ways it seems that pure acetone (available at pharmacy) might make a better solvent and is even much more volatile even than ethanol.  If you used acetone, be certain to dry it via heating and time such that there is no scent of acetone remaining.  This shouldn't be difficult.

Anyway, to answer your question: ethanol should be a good solvent and it should not be difficult to remove the majority of the ethanol from the far less volatile oils just by drying in air for sufficient time (until you don't smell alcohol).


Why is -OCH3 more strongly activating than -CH3 in electrophilic aromatic substitution? Facts I know:
1) More the electron density in benzene ring, the faster the reaction.
2) Lone pair on -OCH3 group undergoes resonance and makes ortho-para positions electron rich, and so does -CH3 by hyperconjugation.
3) -OCH3 group shows -I (electron withdrawing) and -CH3 shows +I (electron releasing).

Now, -OCH3 and -CH3 both undergo resonance and hyperconjugation respectively and both make ortho/para positions equally electron rich.

So the factor that should come in play is I effect. As the -CH3 group releases electron into benzene ring by +I effect, and -OCH3 withdraws electron by -I, what should be more activating is the -CH3 group.

Where am i wrong?

Since we are talking about the rates of these reactions, we really want to examine the structures of the transition states in the rate determining step in these reactions, rather than the starting materials. For electrophilic aromatic substitution, there is a high energy carbocation, the Wheland intermediate, formed when the benzene derivative attacks the electrophile. Formation of this intermediate is normally the rate determining step. Since this carbocation is high in energy, the transition state leading to it is going to be similar in structure, according to the Hammond Postulate. So, looking at the influence of varying substituents on the stability of these Wheland intermediates is a good approach to justifying differences in reaction rates (not to mention selectivities).

If we take the resonance theory approach, we can draw many resonance structures for eg. the para-substitution intermediate, putting the positive charge ortho and para to the "E" substituent we just added. Looking at the methyl-substituted benzene substrate (toluene), resonance theory would allow us to draw an additional "no bond" resonance structure, where we break the C-H bond and have a proton associated with a neutral methylidene cyclohexadiene. This is the resonance theory approach to hyperconjugation. This resonance contributor can justify why toluene is more reactive than benzene. However, how important is the contribution of a resonance structure where we have one less σ-bond and a positive charge on a tiny hydrogen atom?

Looking at anisole (methoxybenzene) the same way, resonance theory would lead us to draw a resonance structure where the non-bonding (lone) pair of electrons on oxygen is used to make an additional π-bond and the oxygen ends up with the formal positive charge. It may not feel nice to put a formal positive charge on the oxygen, but this resonance contributor has a lot going for it - we have gained an additional π-bond relative to the other resonance structures and all non-hydrogen atoms satisfy the octet rule. The "loss" of the lone pair is no problem - as a non-bonding pair of electrons they were already relatively high in energy. For these reasons, this is an excellent contributor to the overall structure of this intermediate - probably the most important.

Taken together, and using resonance theory, the above can help justify the differences in reactivity. Another approach would be to use molecular orbital theory. Hyperconjugation would be viewed as overlap of the C-H σ-orbital with the adjacent carbocation π-system. The efficiency of this overlap is poor because there is a big energy difference between a low energy C-H σ-bond and π-system of the carbocation. In contrast the energy difference between the oxygen non-bonding (lone) pair and the π-system is relatively small and overlap is therefore efficient.


What is the meaning of [2,1-c][1,4] in Pyrrolo[2,1-c][1,4]benzodiazepine? Pyrrolobenzodiazepines are a group of DNA-alkylating natural products. What is the exact meaning of the [2,1-c][1,4] part in the full descriptive name Pyrrolo[2,1-c][1,4]benzodiazepine?

General structure of pyrrolo[2,1-c][1,4]benzodiazepines. The 2-3 double bond is not present in all natural products.

Thanks to Loong I may be able to figure it out

1,4-Benzodiazepine is pretty trivial, [1,4] defines the position of the nitrogens likeso:

For clarification [1,5] could look like 1H-1,5-Benzodiazepine:

[2,1-c] is answered in What does the "h" "k" and "(5.2.2.02,6)" mean in Dibenzo(h,k)-4-oxatricyclo(5.2.2.02,6)undec-8,10- diene-3, 5-dione

  According to the current version of Nomenclature of Organic Chemistry –
  IUPAC Recommendations and Preferred Names 2013 (Blue Book), isomers of
  fused polycyclic ring systems are distinguished by lettering each
  peripheral side of the parent component using the italic letters a, b,
  c, etc., beginning with a for the side numbered ‘1,2’, b for ‘2,3’,
  etc.

So the letters would be:

So Pyrrolo[2,1-c] would be attached to the c-side.
I still need clarification in one point though:

Does the direction 2,1 (NOT 1,2) show the direction of the attachment?

Yes, the order of the numbers indicates the relative orientation of the fused ring system. The direction of the numbers of the attached component (e.g. ‘1,2’ or ‘2,1’) corresponds to the direction of the lettering of the parent component (i.e. a, b, c, etc.).

For example:

The correspondig Subsection P-25.3.1.3 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), reads as follows:

  (…)
  Isomers are distinguished by lettering, continuously, each peripheral side of the parent component (…) using the italic letters a, b, c, etc., beginning with a for the side numbered ‘1,2’, b for ‘2,3’ etc. To the letter as early in the alphabet as possible that denotes the side where the fusion occurs are prefixed, if necessary, the numbers of the positions of attachment of the other component. These numbers are chosen to be as low as is consistent with the numbering of the compound and their order conforms to the direction of lettering of the parent component. (…)


Compressibility factor (Z) of a real gas I was wondering whether the compressibility factor of a real gas (given by $Z = V_{\mathrm{real}}/V_{\mathrm{ideal}}$) is supposed to be measured while keeping pressure constant?
I was attempting to answer the question below and got stumped at part (iii) because I had trouble finding out which was the dominating force being that when I tried calculating the compression factor I got a value of 1 (because $V_{\mathrm{real}}$ and $V_{\mathrm{ideal}}$ are the same, but the pressures are significantly different).

3.0 mol $\ce{CO2}$ behaving as a van der Waals gas, when it is confined under the following conditions: at 750 K in $\pu{150 cm^3}$. For $\ce{CO2}$ van der Waals coefficients, a: $\pu{3.610 atm L^{2} mol^{-2}}$, b: $\pu{4.29 \times 10^{-2} Lmol^{-1}}$. The universal gas constant R is $\pu{8.206 * 10^{-2} L atm K^{-1}mol^{-1}}$
i) Calculate the gas pressure (in atm), [5]
ii) What is the molar volume of the gas (in L/mol), [1]
iii) What is the pressure when it is ideal gas and use it to state the dominating force in the above gas?

From the given data, the molar volume $V_\mathrm m$ is $$V_\mathrm m=\frac{150\ \mathrm{cm^3}}{3\ \mathrm{mol}}=50\ \mathrm{cm^3/mol}=0.05\ \mathrm{L/mol}$$

In terms of molar volume, the van der Waals equation reads:
$$\left(p+\frac{a}{V_\mathrm m^2}\right)(V_\mathrm m-b)=RT$$

So, solving for the pressure gives:
$$p=\frac{RT}{V_\mathrm m-b}-\frac{a}{V_\mathrm m^2}$$

So, to get the pressure you don't need to know the compressibility factor.  Which of these two terms dominates numerically in determining the pressure?


Volatile (aroma/scent-defining) compounds from Himalayan Balsam (Impatiens glandulifera)? Has anyone identified the compound(s) that make up the distinctive and intense scent of Himalayan Balsam (Impatiens glandulifera)?

I’ve found a number of compounds associated with Impatiens sp., but it does not look like any of them would be carriers of the scent:

2-Methoxy-1,4-naphthoquinone or "lawsone methyl ether" [an anti-inflammatory, fungicidal agent]
2-Hydroxy-1,4-naphthoquinone (Lawsone) [a dye that is also the principal component of Henna]
Methylene-3,3'-bilawsone (Diphthiocol; 3-[bis(2-hydroxy-1,4-naphthoquinonyl)]methane)
α-Parinaric acid [a polyunsaturated fatty acid]

Neither the naphthoquinones nor the fatty acid are described as having any particular scent, and they are apparently all non-volatile solids, which would rule them out.

I was unable to find any more information online, but maybe someone who is more involved with this kind of research has some pointers. Thank you!

There's seemingly a lot published on the invasion dynamics of Impatiens glandulifera, but little on the scent-giving volatile compounds.

All I could find is a poster from a 2010 symposium on Terpenes - Application, Activity and Analysis, organized by the Phytochemical Society of Europe (PSE).

The authors presented a method to sample volatile compounds from the flowers of the neophytic Impatiens glandulifera and the native Impatiens noli-tangere without cutting them off the plant and analyzed the mixture of compounds by GC-MS.

The scents seem to stem from complex mixtures of monoterpenes.

In the flowers of both plants the following terpenes could be detected:

$\alpha$-pinene
$\beta$-pinene
camphene
1,4-cineole
d-limonene
1,8-cineole
$\alpha$-terpinolene
1-terpineol
$\alpha$-terpineol
$\beta$-terpineol

Only in Impatiens noli-tangere were found

(Z)-ocimene
fenchol
geraniol

Only in Impatiens glandulifera was found

$\beta$-phellandrene


Energy comparison between an atom and its ion Is it possible to compare energy(enthalpy or similar) between an atom and its ion, possibly in the same state(gas)?

For example, $\ce{O}$ and $\ce{O^+}$(intentionally cation), $\ce{Na}$ and $\ce{Na^+}$.

At first I though of ionization energy, but its RHS(products) has electron, so I thought it can't be used to directly compare them.

Ionization energy is, exactly, a comparison of the energy of (i) a given species and (ii) that same species with one electron removed.

Ionization energies can be thought of as a reaction like the following, using the oxygen atom as an example:

$$
\ce{O -> O+} + e^-
$$

When evaluating quantities like the ionization energy, "external" energies (this is my term, not an official/formal one!) like kinetic and gravitational potential energy are disregarded. Further, the reference energy is generally defined to be zero for an isolated electron at "infinite separation".

Thus, the energy of the above reaction is:

$$
E_\mathrm{rxn} = \left(E_{\ce{O+}} + E_{e^-}\right) - E_{\ce O}
$$

But, since the electron is taken as being in its reference state, this is simply:

$$
E_\mathrm{rxn} = E_{\ce{O+}} - E_{\ce O} \equiv IE
$$

PLEASE NOTE that the above energies are only the internal/electronic energies of the $\ce O$ and $\ce{O+}$ species, and not their enthalpies $\left(H_{\ce{O+}},\,H_\ce{O}\right)$, free energies $\left(G_{\ce{O+}},\,G_\ce{O}\right)$, etc.


Will Raman photon counts/Spectral resolution increase if laser beam becomes parallel? Most Raman spectrometer has the laser converging into focus with the sample at center like this.

Usually how many millions of H2O atoms will be covered by one spot size?
What will happen if the laser beam would be made parallel instead of converging. Will it increase the counts of the spectrum? Will it improve the spectral resolution? Or will both become worse?
Water Raman:

Spectral resolution is not dependent on the laser intensity. It is determined by the state of the sample ( gas, liquid or solid) and more importantly by the components of the instrument itself. For example, what is the slit width anf the number of lines on the grating? What type of detector is there? If it is charged coupled device (CCD) then the spectral resolution will also be dependent on it.


According to this equation for glycolysis, what is reduced and where do the electrons go? I know that the glucose is oxidized into pyruvate but I don't understand what is reduced or where the electrons go....

Equation:
$\ce{Glucose + 2 ADP + 2 NAD^+ -> 2 Pyruvate + 2 ATP + 2 NADH}$

NAD+ accepts a hydride ion (proton + 2 electrons) and is reduced to NADH.


Which is more soluble in water and why: K₂HPO₄ or CaHPO₄? If I dissolve $\ce{K_2HPO_4}$ and $\ce{CaHPO_4}$ in water (two different reactions), the latter shows better solubility. Is it because $\ce{Ca^{2+}}$ has ha higher valence of $2$ than $\ce{K^+}$, which has only $1$? Or has it something to do with the lattice enthalpy?

The charge and size of an ion influences both lattice stability and stability of the hydrated ion.

The less stable the lattice the more soluble the ion (less energy required to disrupt the lattice).  The more highly charged and smaller the ion is, the larger are the electrostatic repulsions within the lattice which will act to destabilize the lattice.
The more stable the solvated ion, the more soluble the ion.  The higher the charge on the ion and the smaller its radius, the more stable the solvated ion.

The size of the potassium ion is around 152 pm, the size of the calcium ion is around 114 pm.  So the calcium ion is both the smaller and the more highly charged of the two.  Therefor we would expect the calcium ion to have the less stable lattice and be more highly solvated (stabilized) when in solution.  Both of these factors would lead us to expect $\ce{CaHPO4}$ to be more soluble than $\ce{K2HPO4}$


Why is Cu(II) more stable than Cu(I)? Seeing Copper has an exceptional electron filling in its valence shell(4s1 3d10), the way I see it is that on +1, a d10 configuration should be more stable and hence easier to attain than a d9 which would come on +2. Or is there some reorganisation of electrons? Either way, please help explain. Thank you.

The stability of an ion in an aqueous medium is determined by the tendency of attraction between the ion and the water molecules termed as hydration enthalpy. It, in turn, depends on the charge on the ion and its size.
Smaller the ion greater the hydration power of the ion. It is because when the ion is small the distance between its centre and the molecule that solvates it is less hence providing a stronger attraction, thereby releasing more energy attributing to stability. The example of this fact is provided by Cu2+ which has a smaller cationic radius due to a stronger hold of the nucleus on the remaining 27 electrons after the loss of two as compared to Cu+ where only one electron is lost and the hold of the nucleus is less. The stability gained by energy loss in hydration is more stabilising than the destabilising effect of the removal of 2 electrons.
Also, as mentioned earlier, the greater the charge on the ion greater is its tendency to attract and keep the water molecules solvated to it. Cu2+ having a greater charge has a greater tendency of hydration than Cu+.


What are the main differences of CTAB and DTAB surfactants? And what are practical applications of them? Just doing a little bit of research of both surfactants but I can't seem to find too much online about them. Hopefully you guys can help out!

CTAB, aka Hexadecyltrimethylammonium bromide

DTAB, aka Dodecyltrimethylammonium bromide

Sort of a vague question, don't know where to start. To begin, CTAB has a hydrocarbon chain length of 16 (as its name suggests) whereas DTAB has a -CH2- chain length of 12.

Otherwise they have the same hydrophilic/polar "head"; all this is obvious, you can tell just by the two compounds' names. They also have the aforementioned hydrophobic "tails", which refers to the hydrocarbon chains because they are non-polar.

CTAB and DTAB can be used to create micelles. It's like how a biological cell is separated from its environment with a phospholipid bilayer where the hydrophobic and hydrophilic components of the phospholipids align such that you get an enclosed structure that separates a non-polar "organic" region in a polar environment (e.g. water). The wikipedia link has good diagrams.
The reverse is called a reverse micelle, where the heads create a polar region separated from a non-polar environment (e.g. when the solvent is hexane).

For one example of practical usage, I did a lab just today using hexane as a solvent where I used CTAB to create an environment to limit the formation of CdS quantum dots. This worked by limiting the volumes where conditions aren't non-polar. In this case, reverse micelles were formed; tails pointing out, head surrounding to form a more polar center.

Another: In DNA extraction CTAB is used to pick up insoluble particles (e.g. polysaccharides) and exit the working solution.  You can relate surfactants to soap, which picks up insoluble (i.e. oily) substances and can easily be removed by water due to its hydrophilic outer layer, while containing the oily substances in its hydrophobic center.


What is the reason behind the characteristic colours of elements? I've been trying to find out about this for some time now and when I searched online, I couldn't get any proper reasons.
My question is: What is the reason for the characteristic colours of elements? For eg: Why does Sodium have a characteristic colour of silvery-white?
Is there any solid scientific reason behind this?

Colour is a property of the light reflected or transmitted by any substance. Depending on the substance's characteristics (energy levels of constituent levels, bonding, intermolecular forces), there are certain available energy levels (molecular orbitals, bands etc) for the electrons. Whenever any wide spectrum energy is available to a compound (considering only the visible spectrum), then depending on the various energy gaps between the available to the electrons, different wavelengths will be selectively absorbed and re-emitted, but now in any random direction, therefore the reflected and the transmitted light will be partially deficient n these wavelengths. The absorption and re-emission is very characteristic of all substances, and hence, the wavelengths absorbed and the resultant spectrum (transmitted or reflected) will be characteristic. Depending on this spectrum, various colours can be seen for different compounds.

Sometimes, the energy gap between the electron energy level is very high and therefore the light in the visible spectrum cannot excite the electron. No wavelength will be absorbed and it will act as a transparent glass. If say, the gap corresponds to a energy gap of $E=h(4000A)$, then the blue light will be absorbed and the remaining light will be white-blue=yellow. This is the general way in which colors are produced. The energy gaps can be anything, the band gaps in metals, molecular orbital energy difference in covalent compounds, orbital energy difference in isolated elements, or simply the energy difference between the various vibrational modes in a gas/solid.