Simple distillation & azeotropes In my organic chemistry lab, we recently had an experiment where we were given a mixture of 50:50 ethanol & ethylene glycol. We were supposed to perform a simple distillation to calculate the percent recovery. Based on my readings, I found that the ethanol has a lower boiling point than ethylene glycol and concluded that ethanol would be the compound that would vaporize and transfer over to the cooled flask where it could condense back. At the end of the lab, we just weighed the distillate and the holdup.

My question was, how do you know if the simple distillation process was successful? I assumed that it was successful based on the fact that it produced a distillate, but I realized that can't be the only reason. The difference in boiling point was large enough to be performed as a simple distillation, but how can I conclude that it was overall successful?

Also, based on this, can I figure out if there is an azeotrope for the boiling points of the two compounds? I read that azeotropes "prefer" to be boiled into the gas phase together, but doesn't that imply that both compounds could have condensed, making the simple distillation process unsuccessful?

By definition an azetrope is a distillate that boils at a constant temperature, but is a mixture of two or more compounds. So you can't purify an azetrope by using more distillations.

Pure ethanol has a bp of 78.37 °C and pure ethylene glycol has a bp of 197.3 °C. The bp of the azetrope could be (1) below pp of ethanol, or (2) above bp of ethylene glycol. In theory the azetrope cannot be between or equal to either of the boiling points of the two pure chemicals. (The problem with theory is experimental error....)

I can find no evidence of an azetrope between ethanol and ethylene glycol using Google. So I very confident that no such azetrope exists.

You noted that the distillate temperature was about 67 °C as the pot started to boil. You also noted that you read the thermometer when it began to produce the drops of distillate.

Look at the drawing below.

The glassware setup has two thermometers.

Thermometer-1 is in the pot.
Thermometer-2 is in the pot.

Ideally the pot temperature and the distillate temperature would be the same. However when the solution starts to boil the distillate is just starting to contact the thermometer. The thermometers have some thermal mass, so it takes time to get them up to temperature.

The distillate thermometer is relatively slow to equilibrate. So long as it is dripping distillate solution then it is colder than the distillate and will not give an accurate temperature reading.

So the only explanation for this anomalous result is that you made a small mistake taking the reading.


Resonance in ozone and benzene In inorganic chemistry, I learnt that stability of some atoms like ozone cannot be explained by one structure alone and hence the hybrid of possible structures is taken and assumed to be the structure and this is called resonance structure.

However, the book clearly states that it is wrong to assume that electrons move about back and forth alternating between the two structures, and that resonance is a theoretical concept only.

Here's an example for ozone.

But in organic chemistry I learnt that resonance means electrons moving around and in fact some molecules like benzene are really stable because the moving of electrons causes energy to be continuously lost.
 We also drew the electrons moving around for some structures like nitrobenzene and learnt how these cause ortho para/ meta directing nature.

The two resonances I learnt about don't seem to be the same. My question is, what is the difference? Or is there a concept I misunderstood?

This is because the electrons dont actually jump between the atoms of the molecule, but rather the electron cloud density is even distributed through the resonating bonds.  If atoms had to hold formal charges this would be very unstable for the electron count. In essence all of the resonant structures exist at the same time.  This lack of formal bonding in resonace is why benzene is often drawn as a circle inside a hexagon. This explaination does involve a bit of hand waving, and ignores higher level inorganic, but I think it should provide the needed additional clairity.


Solubility and Evaporation - Would the concentration still be 1 ppm? A certain salt has a solubility of 1 mg/1 L.
I have 1 mg of this certain salt dissolving in 1 L of water, I have a concentration of 1 ppm.

1st Question:
If the solution is left open to evaporation, 0.5 L of water (50 %) evaporated, would I still be getting concentration of 1 ppm while 0.5 mg of this salt became solid again and left at the bottom of the solution?

2nd Question: (if it's a yes to the 1st question)
If I then fill the container back to 1 L mark with water, would the concentration be 1 ppm again after the salts are completely dissolved?

Correct, when the water evaporates you'd have more salt than is soluble in water. The excess would then fall out of solution. Adding the same amount of water that evaporated would leave you with the same kind of solution you started out with, so the concentration of dissolved salt would be the same as when you started (although you might need to stir some to get the salt back in solution).


Is there a way to determine the orientation of a benzene-ring ethyl group in synthesis reactions? On my study guide for synthesis reactions (specifically for the synthesis of disubstituted benzene derivatives), two of the problems look extremely similar.

The question on the guide asks us to synthesize each of these by using benzene as a starting material.

They appear to be the same molecule, just with a different orientation of the ethyl group. And in the solutions section of the guide, both have the same exact synthesis reaction listed.

Is there actually a way to determine the orientation of the substituent that isn't being explained by my study guide? Perhaps something done during the Friedel-Crafts acylation step? Or is this simply an effort to try and trip me up by asking the same question twice?

Unless the temperature is really low, you will not be able to distinguish the two conformation isomers (conformers).

Rotation about single C-C bonds is rarely hindered. It can be, by using bulky groups or a tether, but neither is the case here. I expect the benzyl*-methyl bond and the phenyl*-ethyl bond to rotate freely at reasonable temperatures (say, above 250K).


Why sodium is the worst reducing agent in group 1? Analysing the oxidation enthalpies for $\ce{Li}$, $\ce{Na}$ and $\ce{K}$, we have their respective values:

$\ce{Li} = \pu{162 kJ/mol}$
$\ce{Na} = \pu{202 kJ/mol}$
$\ce{K} = \pu{188kJ/mol}$

Sodium has the largest positive oxidation enthalpy, which means that its oxidation will be less likely to happen according to its Gibbs free energy. But, why does this happens to sodium?

The general trend in the reactivity of the group 1 elements is that it increases down the group. This can be attributed to the decrease in 1st ionisation energy as the atomic radius increases.

Reducing power can be measured by the standard electrode potential E0. This refers to the half - cell:

$$M^+(aq)+e⇌M(s)$$

If you look at the values for group 1 we see a different pattern:

Electrode Potential (V) Li. -3.02 Na. -2.71 K. -2.92 Rb. -2.99 Cs. -3.02

Lithium has an anomalously large and negative electrode potential. After Li the values become more large and negative, indicating, as you rightly say, that sodium has the lowest reducing power.

The reason for the unusual value for lithium is that E0 values are measured in solution whereas ionisation energies are measured in the gaseous phase.

To turn a metal atom in the solid state into an aqueous ion we can think of a 3 step process:

The metal is atomized. $\ce{Li(s)→Li(g)}$

The metal is ionised. $\ce{Li(g)→Li+(g)+e}$

The ion is hydrated. $\ce{Li+(g) +(aq)→Li+(aq)}$

Steps 1 and 2 are both endothermic, i.e. they require energy. Step 3 is exothermic as it is a bond forming process. It is referred to as the enthalpy of hydration.

The small size of the lithium ion gives it a large charge density and so water molecules are strongly attracted giving a large, negative enthalpy of hydration.

This is the main factor responsible for lithium's large, negative electrode potential.

In aqueous conditions it is as reducing as cesium though much less reactive in anhydrous conditions.

This anomaly accounts for sodium having the lowest reducing power, but only in aqueous conditions.


What molecules can be eliminated in condensation polymerization? Is there any specific rule to what kind of molecules can be eliminated in condensation polymerization? From Wikipedia I found that "small molecules" like $\ce{H2O}$ can be eliminated. Is there a more accurate criteria to what these "small moecules" are?

As an abstract answer, to me it would be whatever is commercially viable to manufacture and sell. Commercial polymerisation processes eliminate ethylene glycol (in the manufacture of PET) which is probably recycled,  methanol in transesterification reactions, ethene or propene in metathesis type,  phenol (recycled) in the use of diphenylcarbonate with bisphenol A, etc. In home applications,  acetoxysilicone sealant gives off acetic acid. So, anecdotally, it seems "simple" molecules of MW around 100 or less might meet your criteria.


Dynamic equilibrium of gases $$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$$
  The equilibrium constant for the reaction above is $0.13$ at $\pu{830^\circ C}$.
  In one experiment $\pu{2.00 mol}$ of $\ce{SO2}$ and $\pu{2.00 mol}$ of $\ce{O2}$ were initially present in a flask until it reaches dynamic equilibrium. What must the total pressure of equilibrium be in order to have an 80.0% yield of $\ce{SO3}$?

How do I approach this type of problem?  Is it correct to assume that the total pressure of the equilibrium would be the pressure from the both sides of the equation (pressure of $\ce{SO2}$ + pressure of $\ce{O2}$ + pressure of $\ce{SO3}$)? And how do I find these different pressures?

I am thinking of using the method: $K_\mathrm{p}=0.3=\Large\frac{p^2_\ce{SO3}}{p^2_\ce{SO2}\times p_\ce{O2}}$

Hence, to find the respective pressure of the gasses, I attempted the RICE table, (Reaction, initial mol, change in mol, equilibrium point). How can I proceed further ?

For an ideal gas mixture the total pressure is the sum of the partial pressures of all the gases in the mixture. To make intuitive sense of this, consider that since it is a gas mixture, all the gases essentially have access to the whole container, and so will homogenously fill it. Each gas wil the exert a certain pressure on the container walls. The sum the pressures from all the contributing gases will then equal the total pressure the container walls experience.

We also have the relationship $$p_\text{x}/p_\text{total}=n_\text{x}/n_\text{total}$$

The potential yield for the rxn is 2 mol ($\ce{SO2}$ limiting reactant). To get a 80% yield, we therefore have 1.6 mol $\ce{SO3}$ at equilibrium. Therefore there will be 0.4 mol $\ce{SO2}$ and 1.2 mol $\ce{O2}$ left at equilibrium. $n_\text{total}=3.2$

$n_\ce{SO2}/n_\text{total}=0.4/3.2=0.125$

$n_\ce{SO3}/n_\text{total}=1.6/3.2=0.5$

$n_\ce{O2}/n_\text{total}=1.2/3.2=0.375$

To find the partial pressures of the individual gases, we combine the ratios above with relationship 1

$p_\ce{SO2}=0.125\times p_\text{total}$
$p_\ce{SO3}=0.5\times p_\text{total}$
$p_\ce{O2}=0.375\times p_\text{total}$

As you know, the equilibrium can be expressed with partial pressures, and is constant for all changes in pressure. $K_\mathrm{p}=p_\ce{SO_3}^2/p_\ce{SO_2}^2p_\ce{O_2}=0.13$  at 830 Celsius

Plugging in the partial pressures and solving for p(total):

$(0.5x)^2/((0.125x)^2\times0.375x)=0.13$
$\implies x=328.205$ (The unit can be given by the units of the eq. constant.)


Do molecules with carboxyl groups have properties of carboxyl's sub-functional-groups? We're learning about functional groups of different organic molecules, and there are some I'm confused about: carboxyl, carbonyl, and hydroxyl. I was looking online but I couldn't find a direct answer

I know that the carbonyl group is a "-CO", the hydroxyl group is a "-OH", and the carboxyl group is a "-COOH".

We learned that carbonyl groups make molecules a ketone or aldehyde, the hydroxyl group makes molecules alcohols, and the carboxyl group makes molecules carboxylic acids.

Knowing that that a carbonyl group is just a carbon double-bonded to an oxygen, and a carboxyl group is a carbonyl group bonded to a hydroxyl group, then should carboxyl molecules also have all the properties of hydroxyl and carbonyl groups (such as being an alcohol), on top of its new ones? Or, does it only have the properties associated with themselves?

Thanks.

Carboxyl groups do share some properties with hydroxy and carbonyl groups such as the ability to engage in hydrogen bonding but in terms of chemical reactivity they are very different.

The reason for this lies in the structure of the carboxyl group. Instead of the carbonyl and the hydroxyl groups being completely independent of each other, there is an orbital interaction between the three $p$ orbitals on the carbon and the two oxygens which creates a $\pi$ system across the three atoms that tends to stabilise the group, particularly when it is deprotonated. (I couldn't find any decent pictures of these $\pi$ orbitals but if anyone can please edit them in)

The most obvious property that carboxyl groups have which hydroxyl and carbonyl groups do not display (hydroxyls do slightly but nowehere near as much as carboxyls) is acidity. This high acidity is caused by the fact that the carboxylate ion which forms when carboxyl groups donate a proton is stabilised by the electron delocalisation $\pi$ system. This can also be expressed using resonance structures as shown below.

The strength of the acidity can be seen in the $pK_a$ values of carboxylic acids compared to alcohols.$$Ethanoic~acid~pK_a = 4.74$$$$Ethanol~pK_a = 16$$

Remebering that $pK_a$ has a logarithmic scale we can see that carboxylic acids are over $10^{10}$ times more acidic than their respective alcohols.

An excellent introduction to many other properties and reactions of carboxylic acids can be found here: https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/crbacid1.htm


Detection of Tungsten Hexafluoride I'll be doing some work (as a software engineer) with a vacuum process chamber that consumes various nasty gases.  There are numerous safeguards in the system, but as a cautious type, I decided to read the MSDS of the chemicals involved.  The one that makes me the most nervous is tungsten hexafluoride; one mole apparently reacts with water to form 6 moles of HF, which I distantly remember from chemistry class as being bad news.  Being largely composed of water, I find this troubling.  If there were failures in the multiple safeguards and a leak did somehow occur, how could it be detected?

$\ce{WF6}$ is not an uncommon gas used in industrial manufacturing of semi-conductors. Commercial gas detectors are readily available, and will detect $\ce{WF6}$ down to low ppm range, which is about an order of magnitude less than the $\mathrm{LC}_{50}$ for rats, apparently.

You could google tungsten hexafluoride detector, but some examples include (no affiliation or endorsement):

http://www.newcosmos-europe.com/gas-by-name/144-tungsten-hexafluoride-detection.html
http://www.toxicgasdetection.co.uk/gas/?gas=Tungsten%20hexafluoride&search=Tungsten%20hexafluoride&form=gas&id=416
https://www.gas-find.com/hexafluorure-de-tungstene.html

The most common type of detector is a diffusion monitor with electrochemical sensor.
Some sensors actually use conversion of $\ce{WF6}$ to $\ce{HF}$, and then detect the $\ce{HF}$. Commercial $\ce{HF}$ detectors at low ppm range are also common.


Distribution coeffecient without concentrations From what I understand about distribution coefficient is straight from my book — which does not give any practice examples — is that

$$D = \frac{C_\mathrm{A}(\text{ext})}{C_\mathrm{A}(\text{orig})}$$

where $C_\mathrm{A}(\text{ext})$ and $C_\mathrm{A}(\text{orig})$ represent the total concentration of all analyte species present in the two phases regardless of chemical state. Below is a homework problem that I'm try to solve, but having no luck, since I'm not given any concentrations, only weight and volume.

  In an extraction experiment, it is found that $\pu{0.0376 g}$ of an analyte are extracted into $\pu{50 mL}$ of solvent from $\pu{150 mL}$ of a water sample. If there was originally $\pu{0.192 g}$ of analyte in this volume of the water sample, what is the distribution coefﬁcient?

The answer for this problem is $0.731\%$, but no matter which way I plug in the numbers that are given do I reach this answer. Any ideas on what I'm missing?

Molar concentration can be expressed via mass $m$ and volume $V$ all right:

$$C_i = \frac{n_i}{V_i} = \frac{m_i}{M_iV_i}$$

Analyte doesn't change during the extraction and its molar mass $M$ remains the same $(M_i = \text{const})$ so the distribution coefficient $D$ can be rewritten as such:

$$D = \frac{C_\mathrm{s}}{C_\mathrm{w}} = \frac{m_\mathrm{s}V_\mathrm{w}}{m_\mathrm{w}V_\mathrm{s}}$$

where "s" refers to the solvent phase and "w" to aqueous phase.
At equilibrium

$$m_\mathrm{w} = m_0 - m_\mathrm{s}$$

where $m_0$ is the initial mass of the analyte.
Finally, the distribution coefficient is

$$D = \frac{m_\mathrm{s}V_\mathrm{w}}{(m_0 - m_\mathrm{s})V_\mathrm{s}} = \frac{\pu{0.0376 g}\cdot\pu{150 mL}}{(\pu{0.192 g} - \pu{0.0376 g})\cdot\pu{50 mL}} = 0.731$$


What are unit cells, and how do you determine the number of atoms in one of them? I've read several websites to try to figure out what exactly a unit cell is, but I can't seem to understand it. Could someone explain what a unit cell is, and the different types? Thanks!

The unit cell is the most primitive 3D structure that contains the arrangement of atoms which form the crystal of a compound. The unit cell is probably most meaningful in the study of minerals, so it is more of an inorganic analytical technique rather than an organic analytical technique. There are 7 3D crystal lattice systems which form 14 primitive 3D forms called the Bravais lattices. The 14 Bravais lattices can be further subdivided into 32 point groups.


Intensity of “surface plasmon resonance “ absorption peaks for a nano-rod The SPR absorption spectrum of a nano-rod is seen to have two peaks with the longer wavelength peak corresponding to the absorption by the longitudinal plasmons and the other to the transverse plasmons. When the intensities are compared.. longitudinal plasmon peaks are almost always more intense than transverse ones. Why?

The intensity of a plasmon resonance is related to the magnitude of the transition dipole moment. Longitudinal resonances have a larger transition dipole moment compared to transverse resonances, and so have more intense absorption peaks. You can rationalize this by imagining the change in dipole moment from a longitudinal versus transverse electron density oscillation.


Searching for a solvate that changes its solvation property I'm looking for a solvate that changes its solvation property when being dried.
One example of such a material is gypsum:

at temperatures up to 250°C it becomes γ-anhydrite which will slowly
hydrate to a hemihydrate.
at temperatures above 250°C it becomes
β-anhydrite which will not hydrate again.

However the temperature range is too high to use it as example in my thesis.

http://nvlpubs.nist.gov/nistpubs/jres/086/jresv86n2p181_A1b.pdf
http://www.phasediagram.dk/binary/magnesium_sulfate.htm

Sodium sulfate hydrates,

http://iopscience.iop.org/0022-3727/41/21/212002/fulltext/
http://www.springerimages.com/Images/Chemistry/1-10.1007_s00269-008-0256-0-0

DOI:10.1209/0295-5075/102/28003
http://ej.iop.org/images/0295-5075/102/2/28003/Full/epl15377f5_online.jpg
http://www.phys.tue.nl/nfcmr/PhD-Saidov-2012.pdf

====

http://patentimages.storage.googleapis.com/US7638109B2/US07638109-20091229-D00001.png
http://www.genchem.com/properties.asp
sodium carbonate


Mixtures readily oxidzing organic matter Are there other mixtures than piranha solution with similiar or better ability to completely and readily oxidize organic matter? Perhaps peracetic/performic acid?

In biophysics, when you want to destroy anything organic on a platinum probe(1), you usually use sulfochromic acid (mixture of chromic acid and concentrated sulfuric acid). I do not know anything stronger.

(1) A platinum probe is used just because it can handle being subjected to sulfochromic acid. Most other metals would not.


How does one determine whether a chemical requires hazardous waste disposal? It is pretty easy to find information about the actual hazards of a chemical in the SDS, but I have been struggling to sort out which classifications require that the compound be handled differently, especially when it comes to disposing of waste. I'd like to be able to determine what the restrictions on a chemical are by reading the SDS. For reference, I am in the USA. I know it varies state-by-state, but I am just looking for the "average" protocol, or else just the federal requirements.

For example, I calibrate/check water quality sensors. Some sensors are calibrated with NaCl, which obviously I can pour down the drain. For others I use less common/experimental stuff like p-Toluenesulfonic acid monohydrate as a stable substitute for quinine sulfate. In this case, the SDS clearly says not to pour it down the drain. What I am wondering about is how to interpret the myriad of different hazard classifications shown below, and which of them sets the legal requirement that it not be poured down the drain.

Chemical waste is not common where I work, which is why I need to be able to determine before I buy something whether I'll need to find a disposal service to handle it.

You may find information from the US Federal government relating to this questions at the following sites:
https://www.epa.gov/hwgenerators/steps-complying-regulations-hazardous-waste
https://www.epa.gov/rcra/resource-conservation-and-recovery-act-rcra-regulations

Furthermore, your organization (theoretically) has a local waste discharge permit that it must adhere to as well. If chemical waste is not common where you work, you or the responsible party should review the permit requirements and the operations taking place.


Why is astatine so radioactive? I have heard and read about astatine's radioactivity, and it makes no sense to me. Why is it so unstable?  Uranium has a half life ranging from 70 to 4.5 billion years (natural isotopes), and it has 7 extra protons to deal with. I know that atomic nuclei can be stabler if they are spherical, implying that if you make a nucleus that isn't spherical you could destabilize it, but astatine has a maximum half life of 8 hours. Can someone explain this to me?

Information came from a few sources, including SciShow and Wikipedia.

I don't think there is an actual answer to that. What we know about the radioactivity is based mostly on empiric observation and quantum physics is still not able to give precise answers about the process of radioactive decay.

A similar question could be asked about technetium: why is it unstable, even though it stands in the periodic table between stable elements? Probably it's a matter of their proton and neutron configuration that for some reason are particularly unstable compared to their neighbourhood, but there is still no equation capable of giving the half-life of an isotope.

This has already been discussed and probably we will have to wait until the atom nucleus is better studied before being able to answer this question.


Differentiating esters from glycosidic linkages What exactly is the difference between an ester linkage and a glycosidic linkage? I know both release a molecule of water, but what is the difference? Is it simply which types of monomers are involved in the bond?

In carbohydrates you can find examples of glycosidic linkages; the following example is α-maltose and the image is taken from Wikipedia, where a full list of authors is available.

This following example shows an ester; I chose one of the most simple esters (ethyl acetate) due to boredom. Again, the image is from Wikipedia, where a full list of authors is available.

One side of the molecule — the right-hand side in both of my examples here — always features a typical hydroxy group, i.e. is an alcohol. The difference is on the other side, the coupling partner. Look closely at maltose, and you may spot that the first carbon of the left-hand fragment also contains a hydrogen; its oxidation state is that of an aldehyde. Compare it with ethyl acetate, where the left-hand fragment linking carbon is in the oxidation state of a carboxylic acid (i.e. double bonded to an oxygen).

Technically speaking, glycosidic linkage is just a special case of acetal or ketal linkages; that is the correct name for the functional group $\ce{R^1O-CR^2R^3-OR^4}$ — acetal if either $\ce{R^2}$ or $\ce{R^3}$ is a hydrogen, otherwise ketal if both are carbons. ($\ce{R^1}$ and $\ce{R^4}$ must be carbon-based or we are dealing with hemiacetals/hemiketals/hydrates.) Examples of ketals for glycosidic linkages exist.

(And technically, ortho-esters exist, too, in which the linking carbon is bound to three $\ce{OR}$ groups. But these are very rare and special cases, even though they even exist in natural products such as tetrodotoxin.)


How can we confirm the number of protons in an atom? The periodic table tells us that there are 6 protons in a carbon atom. Is there a way to verify this first-hand? Or are we just expected to believe it unquestioned?

1) While not easy, it is possible to obtain mass-spectrum of full range of ions, including fully ionized. By number of peaks with Z/E corresponding as X, X/2, X/3 ... X/n it is possible to ensure that an element has exactly n electrons and protons. It is still difficult to ensure full ionization of heavier atoms, so the method is not applicable for heavier elements. This, however, is the only direct method I can think of. Should work for carbon, though.

2) Various X-ray-derived spectral methods. While not providing direct evidence, they do provide information on electronic shell structure, that can be compared with theoretical calculations.


Transition metal chalcogenides According to wikipedia, a chalcogenide is a chemical compound consisting of at least one chalcogen anion and at least one more electropositive element. In the same sense does transition metal chalcogenides mean atleast one chalcogen and one transition metal. If I have three classes of materials
$\ce{CrSbX_3}$
$\ce{BaVX_3}$ and
$\ce{Ba_2MnX_3}$ where $\ce{(X}$ = $\ce{S, Se)}$
Can I address them under a common term transition metal chalcogenides?
If not what might be a good term to address them together?
Update:
The common factor between these compounds is that they are all quasi-one dimensional. The $\ce{CrSbX3}$ structure consist of infinite chains of edge sharing $\ce{CrX_6}$. $\ce{BaVX_3}$ is made up of infinite chains of face sharing $\ce{VX_6}$ octahedara. $\ce{Ba_2MnX_3}$ consists of infinte chains of corner sharing $\ce{MnX_4}$ tetrahedra. Also this review has considered $\ce{BaVX_3}$ and $\ce{Ba_2MnX_3}$ among transition metal chalcogenides.

As said by Oscar Lanzi, they are called "ternary metal chalcogenides". However, the name depends on the type of metal present in the compound. From here:

Systems to be studied include the layered intercalates $\ce{AMX2}$ (A=alkali
metal; $\ce{X=S, Se, Te}$) which are potential ionic conductors suitable for
use in solid state batteries: Group 2 ternary chalcogenides of type
$\ce{AMX3}$ and $\ce{AM2X5}$ ($\ce{A=Mg, Ca, Sr, Ba}$ ; $\ce{X=S, Se, Te}$) - some of which are superconducting and ternary mixed transition metal chalcogenides $\ce{MM'X2}$ ($\ce{X=S, Se, Te}$) examples of which have a wide range of magnetic properties including colossal magneto resistance.

You'll find more information on chalogenides of formula $\ce{ABS3}$ in this paper:

The structural diversity of ABS3 compounds with d0 electronic
configuration for the B-cation, John A. Brehma, Joseph W. Bennett,
Michael Rutenberg Schoenberg, Ilya Grinberg, and Andrew M. Rappe, J.
Chem. Phys. 140, 224703 (2014); DOI: 10.1063/1.4879659 (PDF
link)

Also, metal chalcogenides of different composition has also been noted:

($\ce{AB2X4}$): Wold A., Dwight K. (1993) Ternary Transition Metal Chalcogenides AB2X4. In: Solid State Chemistry. Springer, Dordrecht. DOI: 10.1007/978-94-011-1476-9_12

Mixed transition-post transition metal chalogenides of form $\ce{ABS2}$ and $\ce{AB2S4}$

($\ce{CuInS2,AgInS2}$): Ternary Metal Chalcogenides: Into the Exciton
and Biexciton Dynamics, Tushar Debnath and Hirendra N. Ghosh,The Journal of Physical Chemistry Letters 2019 10 (20), 6227-6238, DOI: 10.1021/acs.jpclett.9b01596

($\ce{AgBiS2}$): Effect of different growth temperature on the formation of ternary metal chalcogenides AgBiS2, S.Sugarthia,G.Bakiyaraj, R.Abinayab, M.Navaneethana, J.Archanaa, M.Shimomurad, Materials Science in Semiconductor Processing
Volume 107, 1 March 2020, 104781, DOI: 10.1016/j.mssp.2019.104781

($\ce{ZnIn2S4}$): Ionothermal/hydrothermal synthesis of the ternary metal chalcogenide ZnIn2S4, Cuixia Li, Honghua Lia, Lijun Han, Chunshan Li, Suojiang Zhang, Materials Letters, Volume 65, Issues 15–16, August 2011, Pages 2537-2540, DOI: 10.1016/j.matlet.2011.05.052

Note: if the chalcogen is oxygen i.e $\ce{ABO3}$, then it is called Perovskite.


Is (cycloprop-2-en-1-ylidene)propanedinitrile aromatic or not? Is (cycloprop-2-en-1-ylidene)propanedinitrile ((cycloprop-2-en-1-ylidene)malononitrile) aromatic or not? If yes? How?

At least formally, the target compound is the product of a Knoevenagel reaction of cyclopropenone with malononitrile.

For cyclopropenone, a zwitterionic mesomeric structure is conceivable. This species, a cyclopropenyl cation with a negative charge on the oxygen atom attached, is aromatic. Think that this is the little brother of tropone.

For the compound in question, charge separation is possible as well. The resulting species would again be an aromatic cyclopropenyl cation with a dicyanomethyl substituent over which the negative charge can be delocalized.


Can electrons switch orbitals within a shell? I know that electrons can move from say 2s orbital to an unoccupied 2p orbital, as in Carbon atom which can form 4 bonds this way. But I want to know is it possible for an electron say in orbital 2p to switch its place with an electron in 2s? I know only 2 electrons are allowed in an orbital, but is it possible for one electron to oust another and take its place so the ousted electron would have to take the other one's place? Is this theoretically possible?

The short answer to this is: according to our current understanding of quantum mechanics, we can't tell.

And now for the long answer:

First off, if you haven't already seen Richard Feynman's 60-second primer to the scientific method, you should definitely watch it now. It's short. I promise I won't go anywhere while you're watching it.

...

Alright. So the way science works is to test ideas by experiment. This means that in order to distinguish between which of two ideas is right, we'll need to come up with some difference in their behavior. This can sometimes be problematic in science because the difference in behavior between two models can be incredibly small. To take just one example, it wasn't until the 20th century that an apparent error in the orbit of Mercury was confirmed, which was eventually explained by relativity. Before that, if you had given someone the theory of relativity and the Newtonian theory of gravity and asked them to tell you which one was right, they would have shrugged their shoulders. There simply wasn't enough high-quality evidence to decide between one or the other.

On the other hand, what if there's a nightmare situation in which we simply can't tell the difference no matter how good our instruments are? Imagine that I give you a small cube which lights up when you shake it, but I've used magic so that no matter what experiments you try to use to figure out what's inside of the cube, you can't get any information. You can try to use x-rays to image the inside, or weigh it, or spin it around to see if the insides are unevenly distributed, or look at it with an IR camera...but every single time, the magic renders your results useless.

Now suppose I asked you how the box worked. Without being able to get any info, you're stuck! No matter how clever you are in making your guess or how many guesses you come up with, you can't really say anything about any of them, because you're being magically prevented from getting information about what's in the box.

This is a very important idea, so I want to make sure you really understand it. Without some sort of actual behavioral difference that can be looked at, science has no way of distinguishing between alternate ideas.1

Now, quantum mechanics is sometimes a lot like our magic box (to pretty much anyone who tries to study it, it is a magic box). Our current understanding of QM says that we have this wavefunction that tells us what the system is doing. We can't actually look at the wavefunction. The best we can do is to use some operators on it, which will give us the values of certain properties (don't worry if this all sounds very mysterious, it takes years of study to understand some of this stuff).

So how can we see if the wavefunction matches what we think it is? Well, we can look at what we should get from the operators, and what we actually do get from measuring it. For instance, by applying something called the Hamiltonian operator to the wavefunction, we get the total energy of the system. We can then measure the total energy of the system and see if theory and experiment agree. We can do that a whole bunch of times, measuring the momentum, and kinetic energy, and the average position of the system. If we do it enough times, we can say "eh, it's probably correct" and then move on to the next problem.

Now, to get back to your original question: can electrons ever switch places? Well, our current understanding of quantum mechanics is that, when electrons switch places, the wavefunction changes sign.2 "Aha!" I hear you say. "So we can watch the wavefunction and see if it changes sign, and if we ever see it flip, we can say that the electrons switched places!"

Well...not really. Because we can't look at the wavefunction directly, we can only measure things through the operators. And here's where the nasty trick comes into play: for all operators that we can actually observe, there won't be any change if the wavefunction flips its sign.

We're stuck! Just like our magic box, we can't get any meaningful information about what's going on inside, and so we can't really tell if the electrons switched places. Even if they did, it would look exactly the same to us (no matter how good our instruments are, we cannot see a difference, because there literally isn't one). It's an undecidable problem!

Now, if you're anything like me, this is a really crummy and disappointing answer. It took me a long time to accept the fact that there are some things in science that we just can't decide. So let's get a little creative: what would have to change in science for us to be able to tell?

First, and most directly, we would need an alternate theory to QM that allows us to distinguish between the two cases. Seeing how incredibly complex current theories are, I doubt that will happen anytime soon (though I could be pleasantly surprised). The second is that we would need some way to measure the wavefunction directly, instead of looking at it through operators. I've heard whispers of groups making progress on that front, though I honestly don't know if there's any truth to that or not. One way or the other, I don't expect we'll be able to answer this question tomorrow, or anytime relatively soon.

Some time back, I heard about a very clever paper that had to deal with this problem. There was an effect that could either depend on the temperature or density of water. Of course, you can't control the temperature and density of water independently (the stuff expands when you heat it up), and so everyone had just assumed that temperature was responsible. This group realized that, if you supercool water, it takes on densities similar to those found above 0C. By carefully supercooling the water and measuring this effect, they found that at similar densities of water, the effect was the exact same, even when the temperature differed by over 50C. It was density that was responsible all along!
This is one of the most mysterious things about all of physics. Even after two years of studying quantum mechanics, I have yet to find anyone who can offer a satisfying explanation for why fermions (half-integer spin particles) should flip their wavefunction upon exchange of (ostensibly identical) particles. If you know of a good read on the subject, please put it in the comments and I'll add it here!


Conversion from a PPB value to µg/m3 of Isobutylene I have a quite simple problem, found a lot of information about it, but I am not sure anymore if I do my calculations right.

I own a sensor, which reports measurements in isobutylene units as PPB. So if I understand this right, the measurement would be the number of isobutylene molecules per one billion (x/1'000'000'000).

Therefore to convert this value into the more common µg/m3 unit, I would just use this formula:

$$ \text{Concentration}\ \frac{µg}{m3} = \frac{\text{Concentration}\ \text{PPB}\times\text{Molecular Mass}\ \frac{g}{mol}}{\text{MolarVolume}\ l} $$

Therefore to convert 400 PPM of isobutylene, with a molecular mass of 56.106 g/mol or 0.0005879 g/l, in the molar volume of a gas at STP with 22.4 l, the calculation would be:

$$ \frac{400\times56.106}{22.4} = 1001.9 \frac{µg}{m^3} $$

This seemed very high to me, and the Range of the sensor is 0-1056 PPM.

Is this the correct formula for the conversion?

This is much simpler than you think, even though the value you've obtained looks reasonable to me.
Part per billion ($\pu{1 ppb} = \pu{1e-9}$) on its own is meaningless.
Based on a context, it can refer to anything: mass fraction, mole fraction, particles etc.
The thing is, for gaseous mixtures $\pu{ppb}$ always refers to the volume fraction $\phi_i$:

$$\varphi_i = \frac{V_i}{V}$$

where $V_i$ is the volume of the $i$th gas (here, isobutylene) and $V$ is the total volume.
In order to convert volume fraction expressed in $\pu{ppb}$ to desired mass concentration $\rho_i$ expressed as

$$ρ_i = \frac{m_i}{V}$$

where $m_i$ is the mass of $i$th component in a mixture, you have to use density $d_i$:

$$ρ_i = \frac{m_i}{V} = \frac{m_i}{V_i}\varphi_i = d_i\varphi_i$$

To conform with your units of choice ($\pu{μg m-3}$) we need to juggle the units a bit as the density is usually given in $\pu{kg m-3}$, e.g. for gaseous isobutylene at NTP $d(\ce{C4H8}) = \pu{2.3959 kg m-3}$ (Source):

$$
\begin{align}
\pu{1 g} &= \pu{1e6 μg} \\
d(\ce{C4H8}) &= \pu{2.3959e6 μg m-3}
\end{align}
$$

Now, using this density value you can multiply your volume fraction in $\pu{ppb}$ or $\pu{ppm}$ and get an answer, e.g. for $\pu{400 ppm}$:

$$ρ_i = d(\ce{C4H8})\varphi (\ce{C4H8}) = \pu{2.3959e6 μg m-3}\cdot\pu{400e-6} = \pu{958 μg m-3}$$

Note that all notations involving "part per something" are deprecated and should be avoided.
It also wouldn't hurt if you ask manufacturer whether $\pu{1 ppb}$ the sensor reports is indeed $\pu{1e-9}$ and not $\pu{1e-12}$ (just in case) if you think the reported values are too high.


Which make HCO3- to show two pH values at two scenarios? According to the below two titrations,

Image Reference
If we consider the reactions, at the first reaction (in first figure after adding 1.0), there is $\ce{HCO3-, NaCl,}$ and $\ce{H2O}$ at the first equivalence point. At the second one (after adding 1.0), there is $\ce{HCO3-,}$ and $\ce{H2O, Cl-}$ at the first equivalence point. $\ce{HCO3-}$ only affect to the $\mathrm{pH}$ of the medium. But,

According to the first figure, $\ce{HA- + H2O → H2A + OH-}$    it acts as a base
According to the second figure, $\ce{HA– + H2O → A2– + H3O+}$    it acts as a acid

I am doubtful on why do $\ce{HCO3-}$ perform as two types at these different situations? Do anything in the medium  manipulate $\ce{HCO3-}$ to work as a base or an acid? How to describe these two scenarios? Generally I can't see a theoretical method for describe it.
I searched this in so many resources, but I couldn't find a satisfied answer. So, I hope a theoretical answer, for this problem, not a mathematical description.

The problem is that both titration curves are wrong: at the first equivalence points, the $\mathrm{pH}$ does not equal $\mathrm{p}K_\mathrm{a1}$ and the $\mathrm{pOH}$ does not equal $\mathrm{p}K_\mathrm{b1}$. As well, the two provided dissociation constants for carbonate ion, in the OP's upper figure, are also wrong.
The figure below shows two titration curves: 1) for $\pu{10 mL}$ of $\pu{0.1 M}$ "carbonic acid", with $\pu{0.1 M}$ $\ce{NaOH}$ titrant (black rising curve) and 2) for $\pu{10 mL}$ of $\pu{0.1 M}$ sodium carbonate solution, with $\pu{0.1 M}$ $\ce{HCl}$ titrant (red falling curve):

The $\mathrm{pH}$ of the starting $\pu{0.1 M}$ carbonic acid is 3.67 and the $\mathrm{pH}$ of the starting $\pu{0.1 M}$ sodium carbonate is 11.65. At the first equivalence point, for both titration curves, the $\mathrm{pH}$ is approximately 8.34. The $\mathrm{p}K_\mathrm{a}$ values are as shown in the figure and are from (or derived from) the reference at the bottom.
The next two figures show Excel spreadsheet $\mathrm{pH}$ calculations. The first one shows that $\mathrm{pH} = 8.338$ at the first equivalence point, for the neutralization of sodium carbonate solution with $\ce{HCl}$:

At the first equivalence point, the solution is $\pu{0.05 M}$ sodium bicarbonate plus $\pu{0.05 M}$ $\ce{NaCl}$. Ignoring ionic strength effects, the $\ce{NaCl}$ does nothing to the $\mathrm{pH}$.
The next one shows that $\mathrm{pH} = 8.339$ at the first equivalence point, for the neutralization of carbonic acid with $\ce{NaOH}$:

At the first equivalence point, the solution is $\pu{0.05 M}$ sodium bicarbonate. In both cases, the $\mathrm{pH}$ is approximately 8.34 and is consistent with this answer.
The two $\mathrm{p}K_\mathrm{a}$ values for carbonic acid are from D.G. Harris, In Quantitative Chemical Analysis, 7th Edition; Appendix G; W. H. Freeman and Company, ©2007. The two $\mathrm{p}K_\mathrm{b}$ values for carbonate ion are obtained from the two $\mathrm{p}K_\mathrm{a}$ values for carbonic acid via $\left(\mathrm{p}K_\mathrm{b1} = 14.00 - \mathrm{p}K_\mathrm{a2}\right)$ and $\left(\mathrm{p}K_\mathrm{b2} = 14.00 - \mathrm{p}K_\mathrm{a1}\right)$.


Why should the temperature be maintained at 0–5 °C in a diazotisation? I know that in a diazotisation reaction, the temperature should be carefully maintained within the range of 0–5 °C by use of an ice bath. Why is this so? How does the temperature affect the stability of the aryl diazonium salt?

The temperature at which many reactions are carried out is often arbitrary, with certain values being favoured due to the ease at which cooling baths can be prepared. As an example  -78°C is highly prevalent, and a question about this 'magic temperature' has been asked here previously.

In the case of reactions forming diazonium salts, there has been a common rule of thumb that they must be kept somewhere between 0 - 5°C, which is even quoted on Wikipedia:

  In aqueous solution diazonium salts are unstable at temperatures above 5 °C  (Ref: Wikipedia)

At temperatures above 5 °C, diazonium salts (in aqueous solution) have been known to decompose, often explosively, to liberate nitrogen gas, temperatures too low however can cause crystallisation, which causes issues later on (diazonium salts should never be dried upon isolation - they explode).

Although people often keep to this rule of thumb, theres no real body of evidence for its generality (i.e. different diazonium species will have vastly different decomposition temperatures).

Often the only time people carry out in-depth studies of a particular reaction is when the reaction needs to be scaled up, such as in the the pharmaceutical industry. Process chemists go to extraordinary lengths to study their reactions to avoid any possible safety issues (or even unintended side reactions).  In Organic Process Research & Development 2004, 8, 1059−1064, a team of process chemists at Lundbeck carried out extensive thermal studies on a diasotisation reaction they were doing as part of a Sandmeyer reaction, and they found no issues well above 5 °C (though as a precaution they still kept the reaction below 15 °C).


Why don't we explode after drinking water? It is known that acid should be added to water and not the opposite because it results in an exothermic reaction.

Our stomach contains HCl, so why don't we explode when we drink water?

The hydrochloric acid in the stomach is already quite dilute; its pH is in fact no less than 1.5 so that at the extreme maximum there is only 0.03 molar hydrochloric acid.  And even that small amount is, of course, stabilized by being dissociated into solvated ions.  There is just not enough stuff to react violently.


What is the proton NMR spectrum of p-methoxyphenol? I think there will be two pairs of doublets, since the hydrogens adjacent to the $\ce{OCH3}$ and those adjacent to the $\ce{OH}$ group will be in different environments. The answer, however, is that there'll be a singlet for 4 Hydrogens.

To give some context, I am extremely unused to the proton NMR of benzene derivatives, and any (introductory) links telling me a bit about it would be appreciated. I'm aware of the theory behind NMR spectrosocopy, the 'n+1' rule, and the ways to tell if two Hydrogens are magnetically equivalent.

In theory, this structure is expected to provide four different signals:

a broad singlet, due to the potential exchange of the phenolic OH
a singlet accounting for three protons of the methyl group
and signals for the remaining four aryl protons ... which may be tricky

The chemical shift of the aryl protons is influenced by the two, para standing towards each other, subsitutents; here: OH and methoxy.  With their electronic properties, they influence the electron density of the aromatic ring, and eventually alter the location of the signals of the aryl H's. The centre of theses signals may be predicted by empirical increment rules which many textbooks about NMR include, like Silverstein, for example. Depending on the experimental conditions, accidentally they may simplify to a signal which only looks like a singlet:

(source, recorded in $\ce{CDCl3}$ at a Lamor frequency of 89.56 MHz)

Addition: The distance between the aryl H's and the phenol or the methoxy group (at least three bonds away) and their very similar electronic influence on the aromatic ring may render the aryl H's very similar.  So it takes a spectrometer with higher measurement resolution (measurement points per ppm) and higher magnetic field to discern them.  Rayment et al. published more recently in J. Org. Chem., 2012, 77, 7052–70602012 (doi 10.1021/jo301363h) data recorded again in $\ce{CDCl3}$, on an instrument with stronger magnetic field.  To quote:

  1H NMR (400 MHz, CDCl3) δH 3.78 (3H, s), 4.77 (1H, br s),
  6.76−6.82 (4H, m)

Even here, some discern, yet likely the two doublets overlap a bit, as a prediction by ChemDoodle suggests:

(the other doublet corresponds to the second pair of aryl H's).  Note the spread of the scale, too.  While I imagine even more advanced NMR spectrometer may eventually resolve the doublets (600 MHz is not so uncommon, and a few of about 1 GHz are around, too), I doubt someone will invest the more precious measurement time on these to delve more into detail and publish the then very high resolution 1H NMR, instead of collecting 1H NMR data of proteins and other natural products.

Addendum and correction:  Following the comment by @long, I reread the relevant pages in  Silverstein again.  Stating the for aryl H's would yield two doublets is wrong.

There are three types of coupling each of the aryl protons in p-methoxy phenol participates which are illustrated below:

(Source for the values of coupling constants: Silverstein, loc. cit., 6th edition, 1998, p. 212)

In blue is the proton which signal is of concern.  In the top row, it is the left in ortho position towards the methoxy group.  This proton couples with the close ortho standing proton (only three bonds away, hence $^3J$), typically with a coupling constant of $\pu{9 Hz}$, hence yielding according the $(n + 1)$ rule a doublet.  Simultaneously the proton marked in blue couples with the meta standing one with a typical coupling constant of $\pu{3 Hz}$; hence the previous doublet would become a doublet of doublet (dd).  Even farther distant, the para standing proton, hence coupling with an even lesser coupling constant which not every NMR spectrometer is capable to resolve -- but in principle yielding ddd, a doublet of doublet of doublet -- solely for this proton, up to $2 \times 2 \times 2 = \pu{8 signals}$ may become observable.

Following the theory further, the proton marked in blue of the bottom row is not identical to the one marked in the top row, and consequently, this proton may yield up to 8 observable signals, its own ddd.  This already adds up to 16 lines, which however is not (yet) the end of nitpicking theory as -- according to Silverstein's textbook -- the aryl Hs on the left hand are not equivalent to the ones on the right and a total of 32 lines possibly to observe.

With the graphical example provided by the Silverstein book for p-chloro nitrobenzene

which will serve for a follow-up question by mine, to record the aryl H's in a signal in shape like a singlet was by chance.


Would hydrogen bond form in these scenarios? My notes say that hydrogen bonding occurs between a hydrogen atom covalently bonded to a highly electronegative atom ($\ce{N, O, or F}$) and the lone pair of a highly electronegative atom of another molecule. I am wondering if hydrogen bond would form between a molecule with hydrogen atom covalently bonded to $\ce{N/O/F}$ and a molecule with $\ce{N/O/F}$ that has lone pair but is not bonded to hydrogen. For example, it seems like oxygen, ozone, carbon dioxide are soluble in water so they can form hydrogen bond with water though the $\ce{O}$ atom is not bonded to $\ce{H}$. How about $\ce{CH3CH2COCH2CH3}$, the $\ce{O}$ atom has a lone pair, but it seems like it is not soluble in water and cannot form hydrogen bonding with water.

The molecule your example, 3-pentanone, actually has a water solubility of $\pu{50 g/L}$ at $\pu{20^oC}$. The water solubility of it's pure hydrocarbon analog, n-pentane, is $\pu{40 mg/L}$ at $\pu{20^oC}$. A simplistic interpretation of this comparison is that the addition of a single hydrogen-bond acceptor functionality to n-pentane increases it's water solubility by greater than three orders of magnitude! So the answer to your question is pretty clearly that yes, hydrogen bonding is happening in this scenario.


Plot of Arrhenius Equation with temperature I saw the following plot of the Arrhenius Equation on Wikipedia (cc-by Zivilverteidigung and Matthias M.).

It is a concave upwards graph, but according to the equation
$$k= A \mathrm e^\frac{-E_\mathrm a}{RT},$$
 it is evident that the graph is of the form
$$y= \mathrm e^\frac{-1}{x}.$$

I believe that this should have been the graph (please disregard the negative side of the $x$ axis). It is a concave downwards curve. Why is there a discrepancy?
This maybe also suitable in Mathematics.se, but I have a gut feeling that the solution to this lies in Chemistry and not maths.

I believe that this should have been the graph...(please omit the
  negative side of x axis), so concave downwards. Why is there a
  discrepancy?

Your analysis and your graph are correct.  Let's examine why the discrepancy arises.  I've drawn two graphs below.  The top one is similar to your graph, only I'm examining a smaller range on the x-axis.  My lower graph examines an even smaller range and it is a range close to x=0.  You can see that this lower graph appears to be "concave-up" just like the Wikipedia graph you referenced.

Here's a third graph where I changed the numerator from -1 to -100, notice how it moves the curve further to the right along the x-axis.

Depending upon 1) what numbers we put in the numerator and denominator of our exponential equation and 2) depending upon how we scale our axes, we can dramatically change how our graph appears.  There will always be a relatively flat part to this curve located close to x=0, then there will be a concave-up section, followed by a concave-down section, and finally another relatively flat section further out on the x-axis.  You happened to choose a large range (up to x=60-70) for the x-axis of your graph.  This compressed your graph, so it did not show the first two parts of the curve (the relatively flat part located close to x=0 and the concave-up section) clearly.


Why does helium seem to exhibit an ionization energy affected by the shielding of electrons? According to Wikipedia, the shielding effect only happen in atoms which have more than 1 electron shells. The core electrons repel the electron in the valence shell. However, I have heard some lectures mention the shielding effect on helium atom when they explained why the first ionization energy of $\ce{He}$ is not exactly 2 times more than $\ce{H}$.

Why is this? According to the definition of shielding effect, there shouldn't be one in $\ce{He}$ atom.

The premise is problematic as there is no 'shielding' effects in an isolated Helium atom.

However, Phillipp was correct in explaining why the ionization energy of Helium is not exactly twice that of Hydrogen.  This is because the energies of each electron are higher due to their correlated nature (i.e. they repel each other into higher, non-ground-state-like energies).  Therefore, it takes LESS energy to rip one of those electrons out of the system.


In electrolysis, why does each atom wait to turn into gas until they reach a particular electrode? More specifically why don't bubbles of both oxygen and hydrogen appear everywhere?
If the electricity just breaks apart bonds wouldn’t the hydrogen and oxygen just immediately turn into gas and bubble up wherever they split apart? Why does each atom wait to turn into gas until they reach a particular electrode?

$\ce{H2O}$ is not simply split apart by electricity, as you say. No ! What happens on one electrode is not related to what happens to the second electrode. Let's start by discussing what is happening on the negative electrode, the cathode.
The negative electrode behaves as if it contains plenty of electrons ready to react with anything able to do it. It could be positive ions. But in neutral solutions it may be water that does react with these electrons, according to :
$$\ce{2 H2O  +  2 e^- -> H2 + 2 OH-}$$ So, some bubbles of gaseous Hydrogen $\ce{H2}$ are produced on the cathode and the solution become basic because of the appearance of $\ce{OH-}$ ions in solution.
On the other electrode, the anode, which is the positive electrode, something must occur to produce electrons. Once again, it is water that manages to be decomposed in order to produce electrons. It is done this way : $$\ce{ 2 H2O -> 4 H+ + O2 + 4 e-}$$ Here some oxygen bubbles are produced on the anode. And the solution becomes acidic, if it was neutral in the beginning.
A last consequence of these chemical transformations is the fact that the solution becomes acidic at the anode and basic at the cathode. Of course, by simple mixing the $\ce{H+}$ and the $\ce{OH-}$ ions neutralize one another, producing water. So the only visible effect of the electrolysis is the production of $\ce{H2}$ at the cathode and $\ce{O2}$ at the anode, as if $\ce{H2O}$ has been split. Only "as if".


First order, parallel reaction integration I have parallel equations:
\begin{align}\ce{A &->[k_b] B},&\ce{A &->[k_c] C}.\end{align}
I understand how to determine formation of $\ce{A}$ (with the integration) but I really keep going in circles with integrating $\ce{B}$. If I can have some assistance in how to integrate $\ce{B}$, then I will be fine for integrating $\ce{C}$.

This is what I have:
\begin{align}
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}
&= k_b[\ce{A_0}] \cdot \mathrm{e}^{(-k_b+k_c)t}\\
\int_{[\ce{A_0}]}^{[\ce{A}]}\frac{\mathrm{d}[B]}{\mathrm{d}t}
&= \int k_b\cdot[A] \mathrm{d}t\\
\ln\frac{[\ce{B}]}{[\ce{A_0}]}
&= \int k_b\cdot[\ce{A_0}] \cdot \mathrm{e}^{-(k_b+k_c)\cdot t}\mathrm{d}t\\
\end{align}

To save the embarrassment, once I raise to the "$\mathrm{e}$" of both sides, I get
\begin{align}
\frac{[\ce{B}]}{[\ce{A_0}]}
&= \mathrm{e}^{\frac{k_b\cdot[\ce{A_0}]}{(k_b+k_c)}t} \cdot \mathrm{e}^{\mathrm{e}^ {-(k_b+k_c)t}}\\
[\ce{B}] &=[\ce{A_0}] \mathrm{e}^{\frac{k_b\cdot[\ce{A_0}]}{k_b+k_c}t} \cdot \mathrm{e}^{\mathrm{e}^{-(k_b+k_c)t}}\\
\end{align}

Can someone show me the steps of integrating the $\int k_b\cdot[\ce{A}] \mathrm{d}t$?
Very basic steps are available in solutions, but it still does not fully show the integration/algebra and I am just very lost.

Start by solving the rate equation for $\ce{[A]}$:
\begin{align}
\frac{\mathrm{d}\ce{[A]}}{\mathrm{d}t}
&= -k_\mathrm{B}\ce{[A]} - k_\mathrm{C}\ce{[A]}&
\Longleftrightarrow&&
\ce{[A]}
&= \ce{[A]_0}\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t).
\end{align}
We can then solve the rate equation for $\ce{[B]}$:
\begin{align}
\frac{\mathrm{d}\ce{[B]}}{\mathrm{d}t} &= k_\mathrm{B}\ce{[A]} \\
&= k_\mathrm{B}\ce{[A]_0}\exp\left(-(k_\mathrm{B} + k_\mathrm{C}\right)t)\\
\ce{[B]}-\ce{[B]_0}
&= \int_{\ce{[B]_0}}^{\ce{[B]}}\mathrm{d}\ce{[B]}' \\
&= k_\mathrm{B}\ce{[A]_0}\int_0^t\mathrm{d}t'\,\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t')\\
 &= \frac{k_\mathrm{B}\ce{[A]_0}}{k_\mathrm{B} + k_\mathrm{C}}[1-\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t)].
\end{align}
I use primes to distinguish the variable of integration from the limits of integration.


Difference in pH of water and rainwater In my textbook (Olmsted and Williams 4th ed) it is given that pH of pure water is 7, pH of unpolluted rain water is between 5-6 and pH of acid rain is between 4-5.

It is obvious that pH for acid rain would be low. But how is it acidic, even slightly, for unpolluted rain water? Shouldn't unpolluted mean without any solutes or dust in water? Shouldn't it be similar to pH of pure water (7)?

At first thought, it might seem that rainwater should be clean like distilled water, as it evaporated, condensed, then fell back to earth. It sounds like a condenser or distillation apparatus in a laboratory. But in reality, the process is very different.

To start with, every raindrop must have a particle to condense upon, called a cloud condensation nucleus.  According to this Wikipedia article:

  Cloud condensation nuclei or CCNs (also known as cloud seeds) are
  small particles typically 0.2 µm, or 1/100th the size of a cloud
  droplet on which water vapor condenses. Water requires a
  non-gaseous surface to make the transition from a vapor to a liquid;
  this process is called condensation. In the atmosphere, this surface
  presents itself as tiny solid or liquid particles called CCNs.

Furthermore, the composition of CCN is frequently very acidic, having formed from sulfuric acid or sometimes weak organic acids. These types of CCN form in both polluted as well as pristine environments. Both natural and man-made sources of sulfur emissions result in the photochemical oxidation to sulfuric acid, which forms new CCN which are very hygroscopic (water absorbing).

The type of CCN described above are most commonly formed over land. In marine environments, there are natural sources of sulfur gases, but most CCN form from evaporated sea spray, which is of course much more pH neutral.

However, regardless of how the cloud drop was originally formed, they all have another source of acidity from the atmosphere; they absorb carbon dioxide, which is rapidly converted to carbonic acid when dissolved in a cloud droplet.  It is this absorption of naturally occurring, acidic carbon dioxide that is the primary reason that essentially all rain is somewhat acidic.


Can glass be made with anorthite? The lunar highlands are made mostly of anorthosite (90% anorthite, the rest oxides of magnesium, iron, a little titanium, and traces of others), and pretty pure anorthite ($\ce{CaAl2Si2O8}$) is common. No minerals that are more than about half silica are known. The use of typical batch reactions to purify chemicals is (mostly) not practical on the Moon, as there is no water*, no halogens, and no carbon, nitrogen, or hydrogen.

Human habitats are going to need clear glass, the more the better. If the Moon was being settled, a way to make clear glass locally would be a priority. There are of course many things to be considered about the proper use of glass, but that is a separate issue. This question is about an advanced colony, not the first few habitats. You can assume the infrastructure for an involved process is available.

Could anorthite be processed into decent clear glass? How much impurity could it have? How can I look into the properties it might have and things to consider about the process?

Below is a table from the Lunar Sourcebook, Ch. 5 pg 129, showing
percentages of pure anorthite in various minerals mixed in igneous rocks
from the highlands:

*Actually there is a lot of frozen water and other volatiles at the poles, but getting it in quantity is a big challenge, and it isn't found anywhere else.

In theory, yes it is possible. In practice, not really.

First of all, the anorthite on the moon is never pure anorthite ($\ce{CaAl2Si2O8}$) but rather an anorthite-albite ($\ce{NaAlSi3O8}$) solid solution. That is, the single crystal contains both components. The diagram you added there, where it says "mole %"? The other stuff is albite.

Now, back to glass making. Just about any crystallite solid silicate can be made into a glass. The process is simple: heat it up above the melting point, and then cool it down fast enough so it quenches to glass instead of crystallising. Let's look at these points in details.

Melting the stuff: Anorthite has a very high melting point: 1553 °C. Forsterite (the Mg-rich end member of the olivine group) has an even higher melting point: 1890 °C. To melt the stuff you're going to need a furnace that can go that high, and your energy bill is going to be extremely high. Here's a liquidus phase diagram of the $\ce{CaAl2Si2O8-Mg2SiO4-CaMgSiO6}$ system so you can get an idea of the temperatures required:

Luckily, your material is not pure anorthite. It also has albite! That's good because it takes the melting point down:

However, 1100 °C is still pretty high. The glasses that are commonly used in industry today have melting points of below 1000 °C. This is achieved by having high contents of sodium in the glass, which is perfect in taking the melting point of the material down. However, materials on the lunar surface are not that rich in sodium, and that's a problem.

Adding iron to forsterite (i.e. the fayalite component of the olivine group) will take the melting point of it down, but pure fayalite melts at around 1200 °C which is no good, and the iron will colour the glass green to dark brown (seen as black if thick enough), depending on how much oxygen is around. More about iron later.

Cooling the stuff down: Silicate glasses that we know and love contain abundant silica, commonly above 70%. This makes the glass very viscous because of Si-O polymerisation. A viscous glass can be easily left to cool in air (by the way - not readily available on the moon) and it will remain glass. Cooling a non-viscous glass that has below ~50% silica in air will cause crystals to form in it (anorthite in your case) and it will not be perfectly clear anymore. Therefore, you need to cool it very rapidly, by quenching in water. The problem now is that the glass is very brittle and will most likely crack during cooling or very soon after (this is from my own personal experience in trying to make glasses of these compositions).

So far we have two problems: melting it, and cooling it. There are other problems. You were worried about impurities and rightfully so. Iron is your biggest enemy. Even trace amounts of iron (< 1%) can make your glass dark green or brown. Not only that, if there is oxygen around you might even have magnetite crystals in your glass and it's going to be completely opaque. You need a way to get rid of the iron. Nickel is also a minor problem. As for other "impurities", such as Al, Na, K, Ca, Mg, that's not a problem. They don't give any colour to the glass. It's the transition elements that you should fear.

Yet another problem is that you are not going to get pure minerals. The lunar surface does not contain blocks of anorthite or forsterite or anything else. The regolith is a powder made by weathering down of the minerals, and it contains a horrible mixture of everything. If you want to separate the anorthite or transition-metal-free stuff, you will need to figure out a way to do it.


Why wouldn't you consider the [H+] produce from the weaker acid in a mixture of two acid? Calculate the pH of a solution that contains $\pu{2.0 M}$ $\ce{HF}$ and $\pu{2.0 M}$ $\ce{C6H5OH}$. Also calculate the concentration of $\ce{C6H5O-}$ in this solution at equilibrium.

In these situation, you are told to first find which is the strongest acid. $\ce{HF}$ is.

Knowing this, you can use the ICE table and the $K_a$ of $\ce{HF}$ ($7.2 \times 10^{-4}$) to find $[\ce{H+}]$ in the dissociation of $\ce{HF}$. This $[\ce{H+}]$ is then use to find the $\mathrm{pH}$ of the mixture, but my question is why wouldn't we add the $[\ce{H+}]$ produced from the dissociation of $\ce{C6H5OH}$? Also the $[\ce{H+}]$ that could be made from $\ce{H2O}$ reacting with itself? Is it because the $[\ce{H+}]$ from $\ce{C6H5OH}$ (weaker acid) and $\ce{H2O}$ is so low that it won't really affect the $\ce{pH}$ number?

This question pours over to when you calculate $[\ce{C6H5O-}]$. Why would we just substitute the $[\ce{H+}]$ we found from $\ce{HF}$ for the $[\ce{H+}]$ in the $K_a$ formula of the dissociation of $\ce{C6H5OH}$:

$$K_a = \frac{[\ce{H+}][\ce{C6H5O-}]}{2.0 - x}?$$

Is it because at equilibrium of the mixture we know $[\ce{H+}]$ must be close to the $[\ce{H+}]$ that would be produce at the equilibrium of a $\ce{HF}$ reaction? So it will somehow affect which way the $\ce{C6H5OH}$ reaction will go (Le Chatelier's principle). So because of the addition of HF $[\ce{H+}]$, the reaction will go left and less $\ce{C6H5O-}$ should be made compare if it was just a solution of $\ce{C6H5OH}$?

You don't have to consider the system's sub-components separately, but it makes the algebra much easier.

If pKa of $\ce{HF}$ is 3.14 and pKa of $\ce{PhOH}$ is 9.95 and the problem gives you concentrations to ONE SIGNIFICANT figure, it isn't clear to me why you think the difference between .00072 and .00072000011 is worth worrying about...It can be ignored, clearly. It's not a question of the weaker acid, its a question of whether the values are close enough so that the smaller is significant with regard to the larger. In this case, clearly not. $\ce{PhOH -> PhO- + H+}$ is the chemical equation for the 2nd part, so

$$
\ce{\frac{[PhO][H]}{[PhOH]}} = K_a
$$

is the equation you derive from it (ignoring difference between concentration and activity, and ignoring water).

The fact that $K_a$ for this is $10^{-10}$ and that $\ce{[H+]}$  is on the order of 0.0007 means you get $\frac{X(0.0007)}{(2-X)}=10^{-10}$ which if you have any done a minimal amount of algebra practice/drill immediately suggests that X is going to be on the order of 3E-7 and again is there really a difference between 2 and 2-0.0000003? You'll note I'm not asking the slightly different question about whether the difference between 2.0000000 and 2.0000000-0.0000003 = 1.9999997; 2 is not the same as 2.0000000 when significant figures are in play.


What are the products formed when nitrogen dioxide reacts with water? I've seen two different equations for the reaction of nitrogen dioxide with water:

$$\ce{2NO2 + H2O -> H+ + NO3- + HNO2}$$

and

$$\ce{3NO2 + H2O -> 2H+ + 2NO3- + NO}$$

Are both of these correct to some degree, and is one more prominent over the other?

Both of the reactions are correct. However, the second reaction is written by 'adding' two reaction equations.

When nitrogen dioxide($\ce{NO2}$) is dissolved in water, it produces a 1:1 mixture of nitric acid($\ce{HNO3)}$ and nitrous acid(HNO2). $$\ce{2NO2(g) +H2O(l)->HNO3(aq) +HNO2(aq)}$$
i.e.$$\ce{2NO2(g) +2H2O(l)->H3O+(aq) +NO3-(aq) + HNO2(aq)}$$

However, since nitrous acid is unstable in any environment except very cold solution, it decomposes slowly into $\ce{NO}$ and $\ce{HNO3}$:
$$\ce{3HNO2(aq)->2NO(g) +H3O+(aq) + NO3-(aq)}$$

Your second reaction equation is obtained by adding these two reaction together.

Reference:

Housecroft, C. E.; Constable, E. C. Chemistry, 4th ed.; Pearson, 2010, p 777.

Note:

I wrote the $\ce{H+}$ ions as $\ce{H3O+}$, so there  is an extra water molecule on the left side of the equation, in each case.


Does acetylene give a positive Tollens' test? If not, why is this so? One of the uses of the Tollens' reagent is to confirm the presence of a terminal triple bond in a compound. Does acetylene give a positive Tollens' test? If not, why is this so?

Tollen's reagent (ammoniacal silver nitrate) reacts with terminal acetylenes to form the silver acetylide which precipitates out of solution

$$\ce{RC#CH + AgNO3 + NH4OH -> RC#CAg v + NH4NO3 + H2O}$$

In the case of acetylene itself, since both ends of the molecule are (if you will) terminal acetylenes, a bis-silver acetylide salt is formed and precipitates out of solution (see page 3, item 6b in a PDF from www.sakshieducation.com/ or via the Internet Archive)

$$\ce{HC#CH + 2AgNO3 + 2NH4OH -> AgC#CAg v + 2NH4NO3 + 2H2O}$$

Tollen's reagent is not very stable, therefore it needs to be freshly prepared before use.  The reagent has a number of other uses in chemistry (test for aldehydes and α-hydroxy ketones), anatomical staining, and silver mirroring.  This Wikipedia article provides a nice, concise overview.


Determining the pH of an organic solvent Our work is faced with the issue of determining the charge of compounds that are dissolved in chloroform. One could realistically determine the charge of the relevant moieties knowing the $\mathrm{p}K_\mathrm{a}$ of those groups. However,

How would you determine or calculate the $\mathrm{pH}$ of an organic solvent?

An alternative way of asking this question is how would one calculate the concentration of hydronium ions or free protons in a nonaqueous solution.

I'm not sure why you're wanting to exert so much effort to find the charge of compounds.

If you really want to know the charge, determine it more directly. Electrophoresis can allow you to do so. A charged species is put in a capillary tube that is in between a cathode an an anode. When a charged species is subjected to an electric field, it will experience a force proportional to its charge and the field strength.

Drag forces present in the solvent are also a consideration in this setup and they depend on the "size" of your ion as well as viscosity of your solvent. One also has to worry about Joule heating, as an increase in temperature will cause poor reproducibility. Picking a suitable detector is fun, but hopefully you can get by with the common UV-vis detector.

So if you can find a suitable setup, know the radius of your ion and know the time between injection of your sample and its arrival at a detector, you can at least determine its effective charge.


Stereochemical outcome after elimination followed by addition One of the following diastereomeric chlorides undergoes racemization on heating whilst the other is returned unchanged. Explain.

From the reaction conditions I can see that there must be an elimination of $\ce{HCl}$, presumably via E2 with the base being the amine group on another molecule. Then addition back to the double bond.
However, I have a couple of questions-
Firstly, wouldn't $\ce{HCl}$ add back onto the alkene the other way around so as to form the more stable tertiary heteroatom stabilized carbocation? Which would then be attacked by the chloride ion thereby inducing racemization, and thus it should be impossible to be "returned unchanged".
Secondly, why would one isomer undergo racemization and the other remain the same?

Upon heating, there is an intramolecular $\mathrm{S_N2}$ reaction where the nitrogen lone pair attacks the chlorine-bearing carbon, which gives an aziridinium ion. The stereoelectronic requirement for a "backside attack" results in the formation of a specific diastereomer, illustrated here for the first of the two compounds:

Since there is absolutely no difference between both carbons in terms of sterics or electronics, the free $\ce{Cl-}$ ion can now attack either carbon with equal probability to give a mixture of two products:

The second compound reacts in exactly the same way:

$\hspace{20 mm}$

However, no matter which carbon the $\ce{Cl-}$ ion attacks, the same product is obtained - you can verify this yourself. The product is the same as the starting material.


What is the difference between wet paint, dry paint, and in between those stages? A day after painting a closet, I set a piece of wood on the wall of the closet that was painted a day ago as well. Within minutes, both of their layers of paint fused together (please correct me if there is a better term than 'fused').

I'm assuming that the paint approaches a state of being a solid, but the process is never stopping, (as I've heard:) similar to how concrete is much stronger thousands of years after it is made rather than a couple of years.

So I understand that it started off as a liquid with a high viscosity, and liquids with high viscosity's turn into a solid when spread thin, but why does that even happen. As well, was I right when I explained that it is constantly 'drying', explaining why it was able to fuse with the other layer of the same paint even a day later?

Ullmann’s Encyclopedia of Industrial Chemistry lists numerous components of commercial paints:

dyes, pigments, extenders
organic solvents, water, coalescing agents
binders
plasticizers
paint additives (levelling agents, film-formation promoters, wetting agents, dispersants, antisettling agents, antifoaming agents, catalysts, antifloating and antiflooding agents, antiskinning agents, matting agents, neutralizing agents, thickening agents, preservatives, corrosion inhibitors)

The application method, drying and hardening behaviour of a paint is essentially determined by the binder. Binders are macromolecular products. The low molecular mass binders include alkyd resins, phenolic resins, polyisocyanates, and epoxy resins. These binders must be chemically hardened after application to produce high molecular mass cross-linked macromolecules in order to form acceptable films. The higher molcular mass binders include cellulose nitrate, polyacrylate, and vinyl chloride copolymers, which are readily suitable for film formation.

As the paint dries, a firmly bonded film is formed. Drying and film formation take place physically and chemically. Physical drying involves evaporation of the solvents (organic solvents and/or water). This is the main process for paints with high molecular mass polymer binders. Chemically drying paints contain binder components with a relatively low molecular mass that react together on drying to form cross-linked macromolecules. This process can occur by polymerization, polyaddition, or polycondensation.

In practice, physical drying and the various mechanisms of chemical drying can proceed concurrently or consecutively, depending on the composition of the binder system.


An equilibrium reaction in the presence of other fast reactions -- is it necessary also to consider the individual rate constants? When mathematically describing an equilibrium reaction, we generally treat it as though it is enough just to specify the equilibrium constant, as it relates the concentrations of products and reactants.

If this equilibrium reaction is occurring in a highly reactive environment, however, where the reactants and the products are constantly changing due to other reactions, it seems to me that the equilibrium reaction can no longer be described by equilibrium constant only. Instead, the forward and/or reverse reaction coefficients are needed (in addition to the equilibrium constant) to describe the relationship between the reactants and the products. Is this  correct? If not, why not?

You're correct. If other reactions are occurring that are fast on the time scale of the equilibrium reactions, then you must treat the forward and reverse reactions separately, including by using their separate rate constants.

"Equilibrium" is a mathematical model that never actually perfectly matches reality.  In order for most systems to actually be at "perfect equilibrium" we would have to let them sit and react for an infinite amount of time. Of course, practically speaking, that's not necessary: equilibrium is reached to within measurement precision in quite reasonable amounts of time for many reactions. What's necessary in order for the concept of 'equilibrium' to be useful in analyzing a reaction system is for the time scale of those reactions to be "very short" (or, equivalently, for the rate of those reactions to be "very fast") when compared to all other reactions of interest in the system.

In other words, when we write the following system of reactions:

$$
\ce{A + B <=>[K_1] C + D} \\ ~ \\
\ce{C + E ->[k_2] P}
$$

what we're really writing is:

$$
\ce{A + B \underset{k_{-1}}{\overset{k_1}{\rightleftarrows}} C + D} \\ ~ \\
\ce{C + E ->[k_2] P} \\ ~ \\
\begin{align}
k_1\ce{[A][B]} &\gg k_2\ce{[C][E]} \\
k_{-1}\ce{[C][D]} &\gg k_2\ce{[C][E]} \\
K_1 &= {k_1 \over k_{-1}}
\end{align}
$$

The relative values of the forward and reverse rate constants can vary, with their ratio described by $K_1$, but at all points through the reacting process of interest, both the forward and reverse equilibrium reactions are assumed to be much faster than the final, product-forming reaction. If this assumption doesn't hold, then we can't treat the first reaction as "just" an equilibrium reaction.


Does deuterated water slow down the overall metabolism of a cell? Would deuterated water, being heavier, slow down the metabolic rate of the cell and subsequently the aging process?

edit: lets say I wanted to observe a cellular event, like the formation of the blsatopore or ameboid motion, could I soak a cell in a solution with D2O without killing it or severely damaging it ?

Yes it would, by a few percent. It may or may not be a goal worth pursuing, but there is more to it.

Different reactions would be slowed down to a different extent. Tiny as they are, these discrepancies suffice to disrupt the delicate biochemical machinery of the living cell. No life form more advanced than bacteria is OK with that. Deuterated water in high concentrations is unpleasant for plants and downright lethal for animals, including us.

So much for your slowed aging.


Will cyanide substitute bromine in bromobenzene? I've considered synthesizing benzoic acid from bromobenzene using CN-, and subsequently treating the nitrile with acid and heat. But, is this possible?

Economically, your approach is probably a bad idea.

But in general, the conversion of aryl halides to aryl nitriles does work.

I have prepared large amounts of 2,3,5,6-tetramethylterephthalonitrile from 1,2,4,5-tetramethylbenzene in a two-step procedure

iodination
reaction of the diiodide with $\ce{CuCN}$

back in the days for my PhD thesis. For the cyanation, I followed the procedure described in Synthesis, 1974, 53-55, but used N, N-dimethylformamide (DMF) as a solvent.

The halogen-cyanide exchange is know as the Rosenmund-von Braun reaction and you might find further information in a review on Cyanation of aromatic halides (Chem. Rev., 1987, 87, 779-794).


Special electron configurations in p block on periodic table? Gold and Chromium have weird electron configurations, because electrons are differently divided : eg an electron jumps from a s-orbital to a d-orbital to have a half-filled s orbital and a fully/half filled d-orbital, which is much more stable.
But does this also occur with the p-orbitals? eg why does Fluorine not transfer an electron from its s orbital to its p orbital to have a  half-filled s orbital and a fully filled p-orbital, that would be more stable?

The simplest answer to this problem is related to relative energy differences between the Nd and (N+1)s orbitals in transition metals and Ns and Np orbitals. The Nd orbitals in transition metals are at a slightly lower energy than the (N+1)s orbitals, so for certain atoms there are some unusual changes in the electron configurations, e.g. Cr and Cu. However, the energy differences between the Ns and Np orbitals are quite large, with Np being higher in energy. In these cases, there is a substantial energetic penalty in moving an electron from the Ns to Np orbital.

The electron configuration problem for transition metals is complex, but this J Chem Ed article has a nice description of the theory, although it is somewhat technical.


How to work out equations with period 3 oxides? I know the general equation of reacting an acid with a base to be the salt and water, but I am not able to apply this to the following equations. So in what way can I derive the end product if I have never seen the reactants before?

For example, I can see the water is liberated out in the reaction between silicon dioxide and sodium hydroxide, producing this equation:
$$\ce{SiO2 + 2NaOH -> Na2SiO3 + H2O}.$$

However if the reactants were aluminium oxide and sodium hydroxide, how could I work out the equation:
$$\ce{Al2O3 + 2NaOH + 3H2O-> 2NaAl(OH)4}$$

I would have thought the product formed would be: $$\ce{Al2O3 + 3NaOH ->2Al(OH)3 + 2Na+}$$

The first thing to say is that you final equation doesn't balance. In fact it is impossible to balance your equation using hydroxide because you can never conserve charge with only neutral products.

You can write a reaction that forms $\ce{Al(OH)3}$: $$\ce{Al2O3 + 3H2O -> 2Al(OH)3}$$

However, beyond about pH 6 the $\ce{Al(OH)3}$ will start to dissolve to form $\ce{[Al(OH)4]-}$. Once you get to around pH 7.5 all of the solid has essentially dissolved and you are left with solution of $\ce{Na+}$ and $\ce{[Al(OH)4]-}$ (see footnote). So unless you are very sparing with your addition on $\ce{NaOH}$ you will get the tetrahydroxo product in solution.

$$\ce{Al(OH)3 + OH- -> [Al(OH)4]-}$$

Footnote:
The aluminium species present in aqueous solution are strongly dependent on the pH, particularly around the neutral point. There is a nice graph here showing the different species. The notation is $\ce{Al}$ corresponds to $\ce{[Al(H2O)6]^{3+}}$ and then $\ce{AlH_{-n}}$ denotes the loss of n protons from the hexaaqua complex. This was originally found by @Jan for this question.


Why work done Is 0 and why not a negative value? If a gas expand in vacuum, $P_{ex.} =0$. Then why is the work done 0? Since the gas is expanding, it is doing some work on the piston so the work done must have some negative value.

There is no work done on moving a piston against zero external pressure (not if it is ideal, ie lacks mass and there is no friction in its movement). You might consider the definition of $pV$ work
$$w = -\int_{V_\mathrm{ini}}^{V_\mathrm{fin}}p_\mathrm{ext}
\mathrm{d}V$$
Obviously if the external pressure $p_\mathrm{ext}=0$ then $w=0$.

Conceptually, the point is that a collision by a single gas molecule would suffice to drive the piston to its extended position. In addition, that gas molecule would expend no energy, it would not slow down or change course in pushing the piston, it wouldn't feel it at all (this can get complicated, as explained further below). As absurd as that sounds, this is accurate within the constraints imposed by this conceptual question.

The following schematic contrasts the effect of a gas molecule impinging upon an ideal versus a slightly more realistic piston.

In the real case forces during the collision transfer momentum to the piston and reduce or reverse the forward momentum of the gas molecule. In the case of the ideal piston the collision is best modeled as a perfectly inelastic collision between the massless piston and the gas molecule, the resulting united body having the same momentum (in the direction of the piston axis†) as the original momentum of the gas molecule. There are no forces however and therefore no work is done.

† Components of the molecule's momentum along the orthogonal direction are conserved as well, and the molecule must be reflected once it reaches the chamber wall.


Can wedge and dash bonds on 1, 2 position of cyclohexane ring be used to signify trans orientation of the substituents? I've to make the structural isomers for cyclohexane. On position 1, there's a methane group with wedge bond and on position 2, a methane with dash bond. So I'll have to make the trans isomers only as they are not in the same plane or should I make the cis isomers too?

Or, can wedge and dash bonds on 1, 2 position of cyclohexane ring be used to signify trans orientation of the substituents? Is it actually trans or is it there just to show that the molecules lie in different planes or something?

For reference: as it is the 1,2 position, cis will be axial on 1. equatorial on 2. Trans will be axial on both 1 and 2.

So if you drew a wedge in position 1 and a dash in position 2, it would mean that they are on the same plane - either both axial or both equatorial - assuming that the ring was in a chair conformation.

However, if the objective is to show the structural isomers very clearly, it may be better to use either the chair conformation sketch:

... or the Newman Projection:


Can a copper(II) chloride be added to liquified butane? What would happen to either chemical if the two were mixed together, specifically what would happen if the dust was mixed in when the Butane was in an aqueous state.

I have a limited knowledge of chemistry so I apologize if my terms are incorrect.

If we had butane under pressure in a vessel as liquid and this was to be in contact with a solution of copper chloride in water then I think that nothing will happen. The butane and the water will form two seporate layers.

However if we were to add a copper extraction agent to the butane such as LIX-860 (a phenolic oxime reagent) then it would be possible to convert some of the copper from aqueous copper chloride into an organic soluble copper complex. Then the copper would be extracted into the butane layer.

Aother option would be to add copper chloride powder to the butane solution of LIX-860, I think that this could react but the reaction would be slower. Also I think that the hydrogen chloride formed as side product from the reaction would have a greater inhibiting effect on the reaction.

Another option would be to use a lipophilic chloride ionic liquid such as aliquat 336 to dissolve the copper chloride. This could react (slowly) with powdered copper(II) chloride to form 2(R4N+).[CuCl4]2- which would have some solubility in liquid butane. From my own experience of doing solvent extraction I know that aliquat 336 in alkanes is not a great system. When the metal loading becomes moderate the organic phase can separate into the organic solvent (diluent) and a liquid metal bearing layer.

In short I hold the view that pure butane will not be able to dissolve copper chloride, but with a suitable additive it will be possible to get a solution of copper in butane. The copper will be there as a copper complex.

I think that if we were to place water in a sealed container and put pure butane gas in the head space that the concentration of the butane in the aqueous layer will be very low. The copper chloride is likely to stay in the aqueous layer as hydrated copper ions and chloride anions if the concentration is low. If the concentration is high then it will be able to form chlorocomplexes of copper.


Is Co[Hg(SCN)4] diamagnetic? I was reading a book and it was mentioned that $\ce{Co[Hg(SCN)4]}$ is diamagnetic. I know that it is made up of $\ce{Hg^2+}$ and $\ce{Co^2+}$. $\ce{Co^2+}$ has 7 electrons which means that it should always have an unpaired electron but then why is this compound diamagnetic? Am I missing something?

There was definitely a typo and the book meant to write $\ce{Ca}$ instead of $\ce{Co}$. In the magnetic properties of a coordination complex, both the counter-ions and the coordination sphere metal ion are contributors. Since $\ce{Hg^2+}$ is diamagnetic and the overall compoud is given to be diamagnetic, the counter-ion has to be diamagnetic as well. Hence, it cannot be $\ce{Co^2+}$, which happens to have unpaired electrons.

What the book might have meant to say is that calcium is the counter-ion here. When checking diamagnetism or paramagnetism, you have to instead check the metal ion inside the coordination sphere, which in this case is $\ce{Hg^2+}$. And note that $\ce{Hg^2+}$ will have a $\ce{5d^10}$ electronic configuration.


What is a relationship between Volumetric Analysis and Titration Theory? I believe that Volumetric Analysis and Tetration Theory are not the same subject.
But I need to know what is a main difference between those subjects and what is a main intersection?

I think you mean titration, not “tetration”. “Volumetric analysis” is another (somewhat dated) name for titration, as explained the first paragraph of the Wikipedia page.


How is it possible for a substance to have a high heat of vaporization but a low boiling point? The final paragraph of Dissenter's question here is worthy of standing alone:

  [H]ow does one square a high heat of vaporization with a low boiling point? If it takes a lot of energy to vaporize something, then how can that something have a low boiling point?

(It's similar to this old question by Friend of Kim, but different in that it emphasizes an apparent contradiction.)

This seeming contradiction can be reconciled by examining the thermodynamic quantities involved.  First, per Wikipedia, the enthalpy of vaporization is "the enthalpy change required to transform a given quantity of a substance from a liquid into a gas at a given pressure."  Written symbolically:

$$
\Delta H_{\mathrm{vap}} = H_{\mathrm{gas}} - H_{\mathrm{liq}}
$$

For this situation, it is most helpful to consider $H_i$ as the enthalpy per mole ("molar enthalpy") of the species of interest in phase $i$.

Formally, again per Wikipedia, the boiling point for a substance is "the temperature at which the vapor pressure of the liquid equals the pressure surrounding the liquid and the liquid changes into a vapor."  This is not so helpful for answering this question, however.

Another way of defining the boiling point is the temperature at which the liquid and the vapor of the substance are in equilibrium, for a closed system (no mass can get in or out) containing only the pure substance of interest at a given pressure$^*$.  The equilibrium condition here can be expressed in terms of molar Gibbs free energies (free energy per mole$^{**}$) as:

$$
G_{\mathrm{liq}} = G_{\mathrm{gas}}
$$

By definition, the molar free energy of a phase is equal to the molar enthalpy minus the temperature times the molar entropy:

$$
G = H - TS
$$

So, the boiling point equilibrium condition can be rewritten as ($T_b$ is the boiling point):

$$
H_{\mathrm{liq}} - T_bS_{\mathrm{liq}} = H_{\mathrm{gas}} - T_bS_{\mathrm{gas}}
$$

Subtracting $H_{\mathrm{liq}}$ and $-T_bS_{\mathrm{gas}}$ from both sides gives:

$$
T_b\Delta S_{\mathrm{vap}} = T_b\left(S_{\mathrm{gas}} - S_{\mathrm{liq}}\right) = H_{\mathrm{gas}} - H_{\mathrm{liq}} = \Delta H_{\mathrm{vap}}
$$

Here, $\Delta S_{\mathrm{vap}}$ is the molar entropy of vaporization, which is analogous to $\Delta H_{\mathrm{vap}}$.  The boiling point is thus:

$$
T_b = \frac{\Delta H_{\mathrm{vap}}}{\Delta S_{\mathrm{vap}}}
$$

So, what does this result mean?

Qualitatively, as noted in answers to the linked questions, $\Delta H_{\mathrm{vap}}$ is a measure of how much energy must be added to one mole of molecules in the liquid to free them from their associations with the other liquid molecules.  The physical meaning of $\Delta S_{\mathrm{vap}}$ is somewhat harder to describe, but corresponds roughly to the increase in the 'freedom to wiggle' afforded by the transition to the gas phase from the liquid.

So, in order to have a large $\Delta H_{\mathrm{vap}}$ (lots of energy required to drive molecules into the gas phase) but also a low $T_b$ (low temperature at which molecules "like" to transition into the gas phase) a chemical must have a large $\Delta S_{\mathrm{vap}}$, which means that each molecule gains a significant amount of 'freedom of movement' from entering the gas phase.  This makes sense in the case of ethanol versus water in Dissenter's question: ethanol is bulkier than water, so even though they both have similar values of $\Delta H_{\mathrm{vap}}$, ethanol gains more 'freedom to wiggle' in the gas phase and thus has a lower $T_b$.

This also helps to provide a qualitative explanation for why $T_b$ is proportional to pressure: as the pressure of the gas phase increases, the 'freedom to wiggle' gained by entry to the gas phase decreases, because higher $P$ means more collisions with gas-phase molecules.  Thus, one can expect $\Delta S_{\mathrm{vap}}$ to decrease with increasing pressure ($S_{\mathrm{liq}}$ isn't affected nearly as much by pressure changes as is $S_{\mathrm{gas}}$), which translates directly to the experimentally observed increase in $T_b$.

More details on the relationship between boiling point and pressure can be found in the Wikipedia articles on the saturation temperature and the Clausius-Clapeyron relation.  More information on equilibrium and Gibbs free energy can be found on www.chem1.com.

$^*$ In this model system, the "knob" one has available (see Gibbs' phase rule) to change the equilibrium pressure and temperature is to add to or remove from the system some number of moles of the species of interest.

$^{**}$ As orthocresol rightly notes in a comment, the equilibrium must be expressed in terms of the chemical potentials of each species present. However, in this case there is only one species present and thus examining the total molar Gibbs free energy is sufficient.


Predicting orbital angular momentum effects on magnetic moments For metal complexes with $A_{2}$ or $E$ ground state terms there is angular momentum contribution to the magnetic moment which is generally positive for more than half-filled subshells and negative otherwise, and related to the spin-only magnetic moment by
$$m_\mathrm{eff} = m_\mathrm{SO}\left(1-\frac{\alpha\lambda}{\Delta}\right)$$

where $\Delta$ is the ligand field splitting parameter, $\alpha$ is the orbital angular momentum degeneracy of the state and $\lambda$ is a constant.

My questions:

When do exceptions to this behaviour occur ? For example I have found that $\ce{Co(bipy)_{3}(ClO4)}$ has a lower experimental magnetic moment than the spin only formula predicts despite being a $\ce{Co(I)}$ (d8) compound with a $^{3}\!A_{2g}$ ground state, but I do not understand why.
Is the deviation from the spin only formula always positive for $T$ states ? If not can we predict its sign similarly ?
EDIT: a simple way to do this is via Hund's third rule. If the d-orbitals are half or more full, the highest J state will be lowest in energy, and then the deviation for the T state will be positive ! Exactly why is probably rather subtle.

When do exceptions to this behaviour occur ? For example I have found that $\ce{Co(bipy)_{3}(ClO4)}$ has a lower experimental magnetic moment than the spin only formula predicts despite being a $\ce{Co(I)}$ (d8) compound with a $^{3}\!A_{2g}$ ground state, but I do not understand why.

According to ADVANCES IN INORGANIC CHEMISTRY AND RADIOCHEMISTRY, Volume 12, page 193 "The tris(bipyridal) cobalt(I) cation has a magnetic momment of 3.2-3.4 B.M. which is independent of temperature from 70K to 340K".

Since there are two unpaired electrons, the spin only value is $\sqrt{2(2+2)} =2.83$.  So your experimental data are inconsistent with the literature, as 3.2-3.4 is greater than the spin only value, 2.83.

  Is the deviation from the spin only formula always positive for $T$ states ? If not can we predict its sign similarly ?

No, the deviation from the spin only value is not always positive for complexes with T ground states. The main characteristic of complexes with T ground states is that the magnetic moment is temperature dependent, because the spin-orbital coupling splits the T state into states that differ in energy by an amount close to kT.

For example $\ce{Cs2VCl6}$ has a magnetic moment of 1.4 at 80K and 1.8 at 300K, with the spin only formula predicting 1.73.  So deviation can be negative, and isn't necessarily positive or negative even for the same compound, depending upon temperature.

See section 4.3 of Dr. Mark D. Spicer's magnetism lecture for more information.