Determining the amount of water to create a given pressure in a sealed container I have a sealed container with a volume of 1 liter. I want to go from an initial pressure of $P_{1}=1\ \mathrm{atm}$ and an initial temperature of $T_{1}=298\ \mathrm{K}$ to a final pressure of $P_{2}=9.95\ \mathrm{atm}$ and a final temperature of $T_{2}=453\ \mathrm{K}$..

I want to use water to create the pressure. But I do not know how to determine the amount of water. How would I do that?

As I understand your question, you are limited to a total volume of $\pu{1L}$ and you can only add water (which I assume should remain liquid) and heat to the system, and are able to monitor it's pressure and temperature.

Your limitation is the vapor pressure of water at the desired temperature.

At $\pu{453K}$, the vapor pressure of water is $\pu{9.88 atm}$. At $\pu{9.95 atm}$, water boils at $\pu{456K}$. So, if you heat the water to $\pu{453K}$, you will have the desired temperature and be very close to your desired pressure. Or, you could heat the water to $\pu{456K}$, and you would have the desired pressure.  Or you could heat the water to a temperature between $\pu{453K}$ and $\pu{456K}$ to get some trade off of being even closer to the desired temperature and pressure, but with neither of them being exactly what you want.

Again,this trade-off is a limitation arising from the constraint of using water to control the pressure.  If you selected a liquid with a higher boiling point at $\pu{9.95 atm}$, then you could select a temperature at which the vapor pressure of the liquid would be equal to your desired pressure.


How to use gas laws to calculate the partial pressure of a gas that has partly condensed from a mixture? A mixture of three gases N2, CO2 and O2 has a total pressure of 106000 Pa. The mixture is cooled to -100°C and the resulting pressure is found to have decreased to 82500 Pa. What gas is condensed and what is its partial pressure?
  (BP N2 = -196°C, BP O2 = -183°C and BP CO2 = -78°C)

I've had a look at all the formulas and laws that have been covered in my Chemistry textbook like Dalton's law of partial pressure and the combined gas equation but it appears that there is not enough information to solve it. I mean, if it asked to find the initial temperature, I could do it, but I'm not sure how to find the partial pressure of the gas(es) from the information provided.

You would need the initial temperature to be able to solve this.

You could use the combined gas law (or Gay-Lussac's Law) to determine the total pressure of the gas at -100°C, assuming none had condensed.  This ideal gas pressure would differ from the 82.5kPa by the partial pressure of the condensed gas at -100°C.

The question doesn't specify at which conditions it wants to know the partial pressure of the missing gas.  If it's -100°C, you'd be done at that point.  If it's at the original temperature, you would have to apply the same gas law in reverse to get back to your original conditions.

The boiling point information you added should allow you to determine which of the three gases condensed.  Only one would be a liquid or solid at -100°C.


How to oxidize tantalum? I am interested in learning how to oxidize tantalum. I have seen some tantalum rings that the sellers say is a permanent smokey blackish color. But I've spoken to some people that say tantalum cannot be oxidized, and the rings I saw must have been coated.

So anyone know if and how tantalum can be oxidized?

The smoky black stuff is probably not tantalum oxide.  The only such oxide that's well characterized is $\ce{Ta2O5}$, which is white.  Tantalum itself is a rare element, so if the jewelry contains tantalum at all it's probably a thin coating or cladding on a base metal substrate.


Balancing Redox Equations I was taught to balance redox equations with acid/base considerations. Instead of arbitrarily adding $\ce{H^+}$ and $\ce{HO^-}$ ions and memorizing separate rules for acidic and basic solutions, I was taught simply to keep these four things in mind:

1) Charge must be conserved.

2) Mass is similarly conserved.

3) Do what's thermodynamically more favorable. For example, in strongly basic solution, $\ce{O^2-}$ oxygen ions come from not hydronium ion (because that doesn't exist in significant quantity in basic solution) and not from water (because that requires heterolytic clevage of 2 $\ce{H-O}$ bonds but instead $\ce{HO^-}$ because that only requires clevage of only one $\ce{H-O}$ bond.

4) Strongest base reacts with strongest acid.

On the other hand, here's a gem from Chem Wiki by UC Davis:

My question is:

1) How come I've never seen my professor's method mentioned anywhere? I've Googled, looked at scholarly papers; looked in advanced analytical textbooks, etc.

2) How were you taught?

3) Which way do you think is more advantageous?

1) How come I've never seen my professor's method mentioned anywhere? I've Googled, looked at scholarly papers; looked in advanced analytical textbooks, etc.

It is entirely possible that no one has published that approach! It is certainly not common in introductory chemistry textbooks.

  2) How were you taught?

I was taught (and currently teach) the "standard" algorithm you found on the Chem Wiki.

  3) Which way do you think is more advantageous?

It depends on your goal.

If your goal is to make sure students understand all of the steps and have a good intuition for what is physically happening, then your professor's method makes a lot of sense. This is because you need to already have a clear understanding of what things like "thermodynamically favorable" and "relative acid strength" really mean in order for that approach to work.

On the other hand, if your goal is to make sure students have the functional ability to work through a particular type of problem, regardless of how well they grasp the underlying concepts, then the algorithmic method makes more sense.

And finally, if your goal is to solve a problem quickly without having to think too hard about details, the algorithmic method also makes more sense.


Synthesis of 4-amino-3,5-dimethylphenol from N-(2,6-dimethylphenyl)hydroxylamine using perchloric acid? I come across one reaction in which N-(2,6-dimethylphenyl)hydroxylamine is converted to 4-amino-3,5-dimethylphenol in the presence of strong acid (perchloric acid).

It,s seems rearrangement reaction, but i am unable to understand how does this rearrangement is possible.
What is the reaction mechanism of rearrangement reaction?

N-Phenylhydroxylamines undergo what is known as the Bamberger rearrangement when subjected to aqueous acid. The mechanism is a bit strange, as it involves divalent positively charged nitrogen (very rare - in fact, it is so rare that ChemDraw keeps trying to correct my $\ce{NH}$ to $\ce{NH3}$!), but that's just how it is. Wikipedia already has what I consider to be a very good explanation, but I'll try.

The mechanism begins with protonation at oxygen. Ordinarily, one would expect the nitrogen to be the more basic heteroatom, but protonation at nitrogen is unproductive - you can protonate nitrogen, but it's a dead end. On the other hand, if you protonate on oxygen, you can lose a water molecule.

The cationic intermediate on the right is rather like a benzyl cation, but with a nitrogen instead of a carbon. It has several resonance forms:

and the only decent nucleophile in the system, water, has three choices as to where it wants to attack.

Attack at nitrogen just goes back the way we came.
Attack at the ortho positions in this case would lead to a loss of aromaticity because you can't deprotonate at the end (see the mechanism below).
Attack at the para position. This completes the mechanism; after a proton exchange between N and O, a deprotonation leads to rearomatisation.


Bond-Line Structure for an Alkane My textbook tells me the following structure is 2,2,5-trimethylpentane. I try to change the parent group in different ways so as to maximize the number of substituent groups, but I think the maximum number of substituent groups for any arrangement is 3 - so my answer is 2,2,4-trimethylpentane. Am I wrong in thinking this?

There are 2 methyl groups on the second carbon of the main chain, and 1 methyl group on the fourth carbon of the main chain, so 2,2,4-trimethylpentane is correct! If there was a methyl group on the 5th carbon of a pentane, it would be a hexane, so it would not make sense to name it the way your book did.


Do allenes form rings? I was wondering whether allenes can form rings; geometrically speaking, this seems like it would be energetically unfavorable due to their 180 degree linear geometry.

Yes they do. Such compounds are even found in nature, e.g. vernonallenolides.


Why is iron pentacarbonyl known, but iron(III) hexacarbonyl is not? Why is the neutral complex $\ce{[Fe(CO)5]}$ known, but the cationic carbonyl complex $\ce{[Fe(CO)6]^3+}$ is not known?

There are several things which are needed to be known to be able to deal with this question.

Firstly there is the 18 Valence Electron (18VE) rule (description at libretexts.org). The iron pentacarbonyl is an 18VE complex. It is coordinatively saturated. A low valent transition metal complex with strong ligand field ligands is considered to be coordinatively saturated if the metal has 18 valence electrons (noble gas electron count).

An iron atom has eight valence electrons, each carbon monoxide (carbonyl) ligand adds two electrons to the valence count. Thus we have a total of 18 in total.

Each positive charge which the complex has causes it to have one less valence electron, so the cationic complex will have an iron which will have 17 VE, this alone would make it less stable.

But there is another issue which is more important, the bonding of carbon monoxide to transition metals is dominated by the withdrawal of electron density from the metal atom to the carbon monoxide. So the less electron-rich the metal the less able the carbon monoxides are to bind.

The fact that the iron in the +3 oxidation state will make it less able to bind to $\ce{CO}$ molecules. This is the main reason why it will not be possible to make the complex.

If you consider the series of complexes $\ce{[V(CO)6]-}$, $\ce{[Cr(CO)6]}$, and $\ce{[Mn(CO)6]+}$, the strength of the bonding between the $\ce{CO}$ and the metal decreases along the series $\ce{V, Cr}$ and $\ce{Mn}$ ($\ce{Mn}$ has the weakest bonds).

Secondly, we should only use the 18VE rule when the oxidation state of the metal is low and the ligands are strong ligand field ligands. I do not think that the iron(III) is sufficiently low in oxidation state for the chemistry to be dictated by the 18 VE rule. I would predict that if I was to bubble $\ce{CO}$ through a solution of iron(III) in dilute acid that nothing would happen.

However, if I was to make a solution of an iron(II) complex of a dialkyl glyoxime then it would be able to react with $\ce{CO}$. Here the iron is in a lower oxidation state and has more electron density on it.


Mechansim of reaction between 1‐bromo‐2‐fluorobenzene and furan in the presence of Li/Hg Problem
From FITJEE AITS test series:

What is the mechanism of the following reaction?

Solution

My approach
It looks like Birch reduction with $\ce{Li}$ converting to $\ce{Li+}.$ But I can't make anything much out it.
Maybe there's a typo given in solution because $\ce{Br}$ and $\ce{F}$ should be present initially, or do they get removed by any means?

This is a question based on two reactions.
The first reaction is the formation of benzyne using $\ce{Li/Hg}$. After this a Diels–Alder reaction takes place between furan and benzyne. The reaction mechanism would be as follows [1]:

A more generalised form of this reaction, dealing with the conditions needed and the yield of reaction would be [2]:

Reference

Shashidher, B.; Bajjuri, R.; Guguloath, V. Formation and Trapping of Benzyne. Pharm Anal Acta 2011, 02 (07). DOI: 10.4172/2153-2435.1000137. (PDF)

Comprehensive Organic Synthesis; Wolfgang Oppolzer; Université de Genève, Switzerland 1991. DOI: https://doi.org/10.1016/B978-0-08-052349-1.00128-1


Sulfur hexafluoride creation process and risks/safety I would like to create sulfur hexafluoride and test floating things in it but I am a "chemistry noob" and from what I've read it's non-toxic and generally quite safe but first of all I do not know how to make it (other than that it involves sulfur and fluoride which are both definitely not safe) and the precautions for doing this. Basically: Safe way to make sulfur hexafluoride and how safe is sulfur hexafluoride?

The preparation of sulfur hexafluoride from elemental sulfur and elemental fluorine is not suitable for amateurs or beginners; it is definitely not a home experiment. Fluorine is corrosive and very toxic. It causes severe skin burns and eye damage and it is fatal if inhaled. Furthermore, it is an oxidiser and reacts violently with all combustible materials; thus, it may cause or intensify fire.
Self-made sulfur hexafluoride is not suitable for inhalation or similar experiments since it may contain various toxic impurities.

In the laboratory scale, a nickel boat is filled with sulfur and placed in a reaction tube. The tube is connected to a cold trap that is cooled with liquid nitrogen. The apparatus should be dry and ungreased. A stream of fluorine is then introduced.

The reaction
$$\ce{S + 3F2 -> SF6}$$
is strongly exothermic (standard molar enthalpy of formation at $298.15\ \mathrm K$: $\Delta_\mathrm fH^\circ=-1220.5\ \mathrm{kJ/mol}$[1]). In the fluorine stream, sulfur burns with a blue flame.

Sulfur hexafluoride is collected in the cold trap. The product contains various impurities, such as $\ce{F2}$, $\ce{HF}$, $\ce{SOF2}$, $\ce{SO2F2}$, $\ce{SF2}$, $\ce{S2F2}$, $\ce{SF4}$, and $\ce{S2F10}$. Most of the impurities are easily hydrolysable; therefore the gas is passed through a washing bottle filled with $\ce{KOH}$ solution. However, $\ce{S2F10}$ has to be removed with activated charcoal or decomposed by pyrolysis at $400\ \mathrm{^\circ C}$ according to
$$\ce{S2F10 -> SF4 + SF6}$$

[1] “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


Crystallographic terminology associated with the honeycomb structure I'm trying to quickly learn basic crystallographic principles in 2D under pressure (it's me whose under pressure, not the lattice) so I'm checking many sources to find those that "speak to me" the best (including this excellent answer). Many sources (books, papers, researcher's websites) cite the "honeycomb lattice" as an example of a lattice which is non-Bravais in order to reinforce their explanation of what a Bravais lattice is.

some online examples where honeycomb is a "lattice" but not Bravais: slide 3 and here and slide 7
some online examples where the "lattice" designator is avoided: here and slide 20
ambiguous or noncommittal as to being a "lattice" slide 6

Briefly (I don't have the various texts with me right now to transcribe the paragraphs) it seems to me that a 2D Bravais lattice any set of points $m \vec a + n \vec b$ for all integers $m, n$.

The vertices of a honeycomb don't satisfy this because if you try to use that expression, 1/3 of the points are missing. If you want to use a Bravais lattice, you have to define a unit cell with two "atoms" in it.

Question: Can the vertices of a honeycomb or "honeycomb lattice" really be called a proper lattice (but just not a Bravais lattice), or should we really call it either a net† or a regular complex apeirogons or something else, but not refer to it as a lattice?

From http://www-personal.umich.edu/~sunkai/teaching/Winter_2018/01302018.pdf (click for full size)

†The Five nets

From Elizabeth A. Wood's classic Vocabulary of Surface Crystallography Journal of Applied Physics 35, 1306 (1964); https://doi.org/10.1063/1.1713610

  In triperiodic structures the equivalent points form
  a three-dimensional lattice in which the space units are
  unit cells. In diperiodic structures the equivalent points
  form a two-dimensional net in which the area units are
  unit meshes. The fourteen Bravais space lattices are
  replaced, in the diperiodic case, by five nets, described
  in Table I and diagrammed in Fig. 2.

the concept of lattice is very (too!) often misinterpreted outside crystallography. By definition, the lattice points are all identical, that is each lattice point has identical surroundings to all others. In a honeycomb pattern you can distinguish 2 different sets of points, the red and the blue ones in the figure.
So honeycomb pattern is actually a 2-dimensional structure (not a lattice), resulting from the convolution of a basis (in this casis constituted of one blu + one red point) with a 2D lattice (the hexagonal one in the figure representing the five nets)


Is this textbook bonding picture of hybridization in ethene wrong? Regarding the above picture, why is the $\ce{C-C}$ $\sigma$ bond composed of two $\ce{sp^2}$ orbitals while the $\ce{C-H}$ $\sigma$ bonds composed of $\ce{sp^3}$ hybrid orbitals? I don't see why; I thought the geometry around the carbons was trigonal planar. Why does hybridization differ across atoms in the molecule?

I'm thinking the textbook authors made a mistake; yes, there are four $\ce{C-H}$ bonds, but this does not mean there is $\ce{sp^3}$ hybridization because there are two carbon atoms, each with two $\ce{C-H}$ bonds and one $\ce{C-C}$ bond. Perhaps the authors mistakenly took there to be one carbon and four $\ce{C-H}$ bond (i.e. methane.) That would be $\ce{sp^3}$ hybridization.

Or perhaps I'm wrong altogether. Please enlighten me!

EDIT: Saw this in another text:

You are not mistaken. Your book is wrong. If one desperately wants to involve the hybridisation concept (which makes sense for carbon), then the description should be in accordance with the molecular geometry of ethene:

So the carbons can be described as $\ce{sp_{x}p_{y}=sp^2}$ hybridised, hence there are four $\ce{H_{s}-C_{sp^2}}$ $\sigma$ bonds, one $\ce{C_{sp^2}-C_{sp^2}}$ $\sigma$ bond and one $\ce{C_{p_{z}}-C_{p_{z}}}$ $\pi$ bond.

However the concept of hybridisation is no law, but yet another tool to describe bonding. In case of methane, (ethane,) ethene and ethyne it is quite simple to come to certain $\ce{sp^{x}}$ hybrid orbitals where $\ce{x}\in\mathbb{N}$ - different molecule may lead to fractional $\ce{x}$. (Not a rule, just a guideline)

In response to Dissenters comment:

The coordinate system is chosen arbitrarily to indicate the molecular plane to be $xy$, as indicated in the picture taken from the german article of wikipedia:

As the π orbitals have to be perpendicular to this plane they only can be composed of $\ce{p_{z}}$ orbitals.

The molecular plane can be chosen different, e.g. as the $xz$ plane, the the $\ce{sp^2}$ orbitals would be formed from $\ce{sp_{x}p_{z}}$ and the $\pi$ bond would consist of $\ce{p_{y}}$ orbitals.


Is there any way to determine if the compound is oxide or peroxide or superoxide? I know that in an oxide oxygen's oxidation number is $-2$, in a peroxide oxygen's oxidation number is $-1$ and in a superoxide oxygen's oxidation number is $-\frac12$. So, how will I be able to detect that any compound is oxide or peroxide or superoxide?

For example, to get the oxidation number of $\ce{Mn}$ in $\ce{KMnO4}$ how will I determine whether $\ce{KMnO4}$ is an oxide, superoxide or peroxide? Is there any way?

generally the most universal way to prove this is X-ray crystallography. It allows to obtain crystal structure of target compound directly. It has, however, an inherent flaw: it can't reliably detect position of hydrogen atoms, so in some cases some uncertainty may remain.

In case obtaining crystal is impossible, the second option is NMR, that allows to get various info about atom's neighborhood, but interpreting NMR data is really hard. In case of superoxide, EPR may prove direct evidence.

Generally there is no way to detect such things from brutto-formula. For example, compound $\ce{CH2O2}$ may be either dioxirane (methylene peroxide) or formic acid. In practice, however, peroxides are really unstable, so in most cases if compound is not overloaded with oxygen (like, $\ce{H2SO5}$) and is reasonably stable towards thermal decomposition, it is most likely oxide and not peroxide. Peroxides are usually formed either from another peroxide or electrochemically, or (in rare cases) via cold oxidation with molecular oxygen. And there are exactly three common superoxides.


NaOH being added to Formic Acid (Please catch my error!) I am adding 0.55 mL of 0.1M NaOH to 10 mL of 0.5M Formic Acid.

Here are the values I calculated to use.

Moles H+ : 0.5 * (10mL / 1000) = 0.005 mol

Moles OH- : 0.1 * (.55mL / 1000) = 0.000055 mol

volume : (10mL / 1000) + (0.55mL / 1000) = 0.01055 L

Moles Total : moles H - moles OH = 0.005 - 0.000055 = 0.004945 mol

M Total : moles Total / volume = 0.004945 mol / 0.01055 L = .46872 M

Moles Salt : 0.000055 mol

M Salt : moles salt / volume = 0.000055 mol / 0.01055 L = 0.00521 M

Calculations to find the pH

pH = -log(1.80E-4) = 3.745

3.745 - ( -log(M Salt / M Total) ) = 3.745 - 1.95407 = 1.791

However, this value is too low as the lowest pH of 0.5M Formic Acid is ~2.03.

If the volume of NaOH were to say, 0.5 mL, then the M Salt / M Total value would be less than 1/90, and the pH could then be found by

-log( sqrt( 1.80E-4 * M Total ) ) = 2.036

...which makes more sense, but doesn't apply to the initial problem.

Any idea of where I'm going wrong?

It looks like you misused the acid dissociation constant, which is apparently the $K_a=1.80×10^{-4}$ figure you have.

Properly, you should do the following:

1)  Let $x$ be the hydrogen ion concentration $[\ce{H^+}]$ from the formic acid that actually disdociated.

2)  Since you ended up with $0.469 \text{M}$ total acid the un-dissociated acid $[\ce{HCOOH}]=0.469-x$.

3)  The formate ion comes from both the salt ($0.00521 \text{M}$) and the dissociated acid ($x$), so $[\ce{HCOO^-}]=0.00521+x$.

4)  Plug these into the equilibrium relation

$\frac{[\ce{H^+}][\ce{HCOO^-}]}{[\ce{HCOOH}]}=K_a$

and solve for $x$.  The equation reduces down to a quadratic equation so you should have an easy formula for $x=[\ce{H^+}]$.

5)  Get the pH from the result of Step 4.

Good luck!


Intensive and extensive properties of gases Every chemistry textbook I have ever seen defines intensive and extensive properties the same way: Intensive properties are those which do not depend on how much matter is present, while extensive properties do. I came across a question while looking through sample FE exam questions for which my understanding of these properties by this definition failed. The gist of my question really just revolves around pressure and volume in largely compressible substances (i.e., gases).

It seems that, when considering a gaseous substance, these definitions are not robust enough:

Volume can be shown to be independent of the amount of material present by simply compressing the gas.
Density can be shown to be dependent on the amount of material present by keeping a fixed volume and simply adding more gas molecules.

While I have heard the convincing argument that we should simply consider smaller subsets of the original to determine intensiveness and extensiveness (for instance, if I consider half of the container instead, the volume halves from $V_0$ to $\frac{1}{2}V_0$ and the density remains constant at $\rho$), this doesn't help me to identify why the conventional definition remains in use with what seems like flawed implications. Can anyone help to clarify this?

To give you a more clear explanation of the definition of these properties, it's more easy to remember them this way :

-Extensive properties are those which *are additive * ( mass, volume ) for example 1 L + 1 L = 2 L. On the other hand, intensive properties are not additive ( temperature, pressure, concentration ) for example 2 mol/L + 3 mol/ L does not give you a 5 mol/L solution. These are properties which don't change at any circumstance.

Some other thing to be considered :
The ratio of two extensive properties, gives us an intensive property. For example density= mass/volume. Since it is a ratio it will always be the same regardless of how much of a sample is measured.


Calculation of total pressure given ratio of pressure equilibrium constants The equilibrium constants $K_{\mathrm{p},1}$ and $K_{\mathrm{p},2}$ for the reactions $\ce{X(g) <=> 2 Y(g)}$ and $\ce{Z(g) <=> P(g) + Q(g)}$, respectively, are in the ratio of $1:9$. If the degree of dissociation of $\ce{X}$ and $\ce{Z}$ are equal then the ratio of total pressure at these equilibria are?

My Approach: Let the degree of dissociation be $\alpha$

So $K_\mathrm{c,1}$ for the first reaction is
$K_\mathrm{c,1} = \frac{(2\alpha)^2}{1+\alpha}$

Similarly, $Kc_2$ for the second reaction is $K_\mathrm{c,2} = \frac{(\alpha)(\alpha)}{1+\alpha}$

I don't know how to proceed from here and relate these to the ratio of $K_{\mathrm{p},1}$ and $K_{\mathrm{p},2}$. I need help here.

The equilibrium expressions for these reactions are:

$$ K_{\text{p},1} = \frac{p_{\text{Y}}^2}{p_{\text{X}}}$$
$$ K_{\text{p},2} = \frac{p_{\text{P}}p_{\text{Q}}}{p_{\text{Z}}}$$

Since $p_\text{A} = x_\text{A}p_\text{T}$ for component $\text{A}$, we can rewrite them as:

$$ K_{\text{p},1} = \frac{(x_\text{Y}p_{\text{T},1})^2} {x_\text{X}p_{\text{T},1}} = \frac{x_{\text{Y}}^2p_{\text{T},1}}{x_\text{X}}$$
$$ K_{\text{p},2} = \frac{(x_\text{P}p_{\text{T},2})(x_\text{Q}p_{\text{T},2})}{x_\text{Z}p_{\text{T},2}} = \frac{x_{\text{P}}x_{\text{Q}}p_{\text{T},2}}{x_\text{Z}}$$

Taking the ratios:

$$ \frac{ K_{\text{p},1} }{K_{\text{p},2}}
 = \frac{x_{\text{Y}}^2x_\text{Z}}{x_\text{X}x_\text{P}x_\text{Q}}\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}\label{eqn}\tag{1}$$

We're asked the ratios of the total pressures, and we're given the ratios of the equilibrium constants, so we just need to find the molar fractions ratio in that equation. We know that the degree of dissociation, let's call it $\alpha$, is the same for both reactions.

Assume the first equilibrium starts with $m$ moles of $\text{X}$:

$$\begin{array}{cccc}
  &\ce{X(g) &<=> &2 Y(g)}\\ \hline
\text{I} &m & &0\\
\text{C} &-\alpha m & &+2 \alpha m\\
\text{E} &m - \alpha m & & 2 \alpha m
\end{array}
$$

Then, the molar fractions are:

$$x_\text{X} = \frac{n_\text{X}}{n_\text{T}} = \frac{m - \alpha m}{m - \alpha m + 2 \alpha m} = \frac{1 - \alpha}{1 + \alpha}$$
$$x_\text{Y} = \frac{n_\text{Y}}{n_\text{T}} = \frac{2\alpha m}{m - \alpha m + 2 \alpha m} = \frac{2\alpha}{1 + \alpha}$$

Similarly, for the second equilibrium, assuming we start with $r$ moles of $\text{Z}$:

$$\begin{array}{cccccc}
  &\ce{Z(g) &<=> & P(g) &+ &Q(g)}\\ \hline
\text{I} &r &    &0     &  & 0\\
\text{C} &-\alpha r &    &+\alpha r     &  &+\alpha r\\
\text{E} &r - \alpha r &    &\alpha r     &  & \alpha r
\end{array}
$$

And we calculate the mole fractions of each species:

$$x_\text{Z} = \frac{n_\text{Z}}{n_\text{T}} = \frac{r - \alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{1 - \alpha}{1 + \alpha}$$
$$x_\text{P} = \frac{n_\text{P}}{n_\text{T}} = \frac{\alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{\alpha}{1 + \alpha}$$
$$x_\text{Q} = \frac{n_\text{Q}}{n_\text{T}} = \frac{\alpha r}{r - \alpha r + \alpha r + \alpha r} = \frac{\alpha}{1 + \alpha}$$

Plugging the found molar fractions into $\ref{eqn}$, we have:

$$\frac{ K_{\text{p},1} }{K_{\text{p},2}}
 = \frac{\left(\frac{2\alpha}{1 + \alpha}\right)^2 \left(\frac{1 - \alpha}{1 + \alpha}\right)}{\left(\frac{1 - \alpha}{1 + \alpha}\right)\left(\frac{\alpha}{1 + \alpha}\right)\left(\frac{\alpha}{1 + \alpha}\right)}\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}$$

Which simplifies to:
$$\frac{ K_{\text{p},1} }{K_{\text{p},2}}
 = 4\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}$$

Since $\displaystyle \frac{ K_{\text{p},1} }{K_{\text{p},2}} = \frac{1}{9}$, we get:

$$\frac{1}{9}
 = 4\cdot\frac{p_{\text{T},1}}{p_{\text{T},2}}$$

And finally:

$$\frac{p_{\text{T},1}}{p_{\text{T},2}} = \frac{1}{36}$$

So the total pressures are on a ratio of $1:36$.


How to distinguish between CsCl and NH4Cl by a chemical test? $\ce{CsCl}$ and $\ce{NH4Cl}$ have similar reactions. For example, they can react with $\ce{AgNO3}$ to produce $\ce{AgCl}$ precipitate.
So my question is:

  Can we use a chemical test to distinguish between solid $\ce{CsCl}$ and $\ce{NH4Cl}$?

The $\ce{Cs+}$ ion is very unreactive and I do not know any reactions involving it. I believe the chemical test can somehow convert the $\ce{NH4+}$ ion into $\ce{NH3}$ gas so that we can test it with a moist red litmus paper.

Correct you can distinguish them by their reaction with a base. You should use a base, such as $\ce{NaOH}$. $\ce{Cs^+}$ wouldn't react with $\ce{NaOH}$, while $\ce{NH4+}$ will undergo reaction $$\ce{NH4+ + OH- -> NH3 + H2O}$$ $\ce{NH3}$ is a gas and can be detected with a moist litmus paper.

If you will, you can actually detect $\ce{Cs+}$ using a procedure from this paper:

"Selective Detection of Cs+ in Water Solutions via One-Step Formation of a New Type of Struvite-Like Phosphate". However, this is an exotic reaction (in practice you would use the flame color test) and you don't have to do it since you only need to distinguish samples.


Should chemistry students worry about quarks and such? Even chemistry textbooks less than five years old start by describing atomic structure based only on protons, neutrons and electrons. Is an understanding of chemistry fundamentals likely to become out of date when the textbooks start to include quarks, leptons and the dozens of other sub-atomic particles?

Is an understanding of chemistry fundamentals likely to become out of date when the textbooks start to include quarks, leptons and the dozens of other sub-atomic particles?

Well, the general chemisty books do already include leptons, at least electrons and usually positrons.

Other than protons, neutrons, electrons and positrons (and photons), I think that when discussing beta decay and electron capture, neutrinos and antineutrinos should be included, otherwise the nuclear reactions aren't really balanced (lepton number must be conserved).  Alternatively, the general chemistry books could omit all discussion of nuclear reactions, as being physics.

Other than that, I think the rest of the particles can wait for physics class, rather than general chemistry class.

Beyond general chemistry, other particles are sometimes discussed in more advanced chemistry books, such as quantum chemistry texts, where examples involving muons are common.

For some chemistry topics, such as isotope differences in rotational-vibrational spectra of small molecules distinction between fermions and bosons is critical.  Trying to explain the difference between fermions and bosons (and Fermi-Dirac statisitics vs Bose-Einstein statistics vs Boltzmann statistics) naturally leads to discussion of the standard model and the particles in the standard model.  So by the time a student gets to physical chemistry, around the third year of college, they should have some understanding of the standard model.


How to identify the repeating unit in a polymer? Polyvinyl acetate, PVA, is a useful adhesive for gluing together articles made from wood, paper or cardboard. The monomer of PVA is ethenyl ethanoate.
   PVA is formed from B by the process of addition polymerisation.
  Draw a section of the PVA molecule containing at least 2 monomer molecules and identify clearly the repeating unit.
  (Source: University of Cambridge international Examinations, General Certificate of Education Advanced Level, Chemistry, Paper 4 Structured Questions, May/June 2011, Q4)

Shouldn't the $\ce{C=C}$ double bond break and each of the carbon atoms involved in the double bond are attached to $\ce{CH3OCO}$. so the repeating unit is $\ce{CH3OCO}$.

It's not that the "double bond breaks," it's that the vinyl C=C bond reacts during the radical addition polymerization to create polyvinyl acetate.

I'd disagree that the repeat unit is $\ce{CH3OCO}$ since that ignores the backbone entirely. Instead, if you look at the Wikipedia article (or any other picture of polyvinyl acetate), I'd call the repeat unit something like $\ce{-(CH2CH(OC=OCH3)-)}$ or something like that.


How to identify ionic compounds I have a Chemistry lab in which I must design and carry out an experiment to determine the compound in each of eight numbered vials. Each vial contains one of these ionic compounds:

BaCl2
CaCO3
Ca(OH)2
KI
NaCl
NaHCO3
Na2SO4
Pb(NO3)2

Materials:

Distilled water
Several microwell reaction plates
Tooth picks (to stir)
8x 1mL Beral style pipets
The contents of the lab drawer (beakers, test tubes, etc.)

In addition to the materials above, one of the following four additional reagents may be used:

6M H2SO4
6M HCl
6M NaOH
phenolphthalein indicator solution

I know how to identify a couple of them. For instance, CaCO3 can be identified by its insolubility in water, but as for the rest, I'm not completely sure. I know some can be identified by how and if they react with other compounds, but I don't know exactly how to go about that.

First, you can use phenolphthalein indicator solution, $\ce{Ca(OH)2}$ and $\ce{NaHCO3}$ will turn red. Then use 6M $\ce{HCl}$ and the $\ce{NaHCO3}$ bubbles.

$\ce{NaOH}$ to identify $\ce{Pb(NO3)2}$: it generates precipitations. Use $\ce{Pb(NO3)2}$ to identify $\ce{KI}$: generate golden precipitations.

$\ce{H2SO4}$ with $\ce{BaCl2}$ gives white precipitations, $\ce{Na2SO4}$ with $\ce{BaCl2}$ likewise. So the rest is $\ce{NaCl}$.


Is the recipe for my shaving cream complete nonsense? Preamble
I have only basic education in Chemistry (excelled at Organic chemistry nomenclature, but then subsequently quit when I could not grok the simplest replacement reactions). Also I will note I could not find a tag for retail product labelling.

I bought a lot of shaving cream product that I really like. I have been reading the ingredients and I don't understand why it contains so many acids and bases. Perhaps this is an oversimplification, but I was under the impression any acid and base that were mixed would in time, resolve themselves to a salt and an extra water molecule. Let me share with you the ingredients list verbatum so you can see too:

Aqua (water)
Stearic Acid
Myristic Acid
Potassium Hydroxide
Coconut Acid
Glycerin
Triethanolamine
Theobroma Cacao (Cocoa) Seed Butter
Parfum (Fragrance)
Sodium Hydroxide
Melaleuca Alternifolia (Tea Tree) Leaf Oil
Limonene
Linalool

So it seems to me the major ingredients are acids, bases, water and fats blended together with an emulsifier (Glycerin?).

Question
Wouldn't the acids and bases react with each other shortly after mixing, leaving my product with a different mixture of compounds than in the ingredients list?

Speculation
Here is my attempt at answering, so you can an idea that I did put minimum thought into this.

Are some of these things with "acid" in the name, not really acids as I learned them (pH under 7, with a donor H+ ready to go)? And therefore they do not react?
Considering the vendor spells out cocoa butter with a long latin name, maybe they have arranged the ingredients list to impress the customer? For instance writing out Stearic Acid and KOH instead of, say, "lard" or "wax" or whatever unimpressive name is actually the product of the acid-base reaction?
Perhaps the vendor is legally required to print the reagents rather then the product(s), in some circumstances?
Perhaps there is some process in the mixing of the ingredients, or some catalyst that inhibits the reaction? And to what end? For the reaction to happen on my face when water is mixed into the product?

As @Mithoron commented, these are ingredients, not a chemical breakdown of the result.

And, as @Karl commented, traditionally you mix together some rendered fat or oil -- that's a mixture of fats and oils with the fatty acids and glycerin that make them up -- with potassium or sodium hydroxide to make the potassium or sodium salt (conjugate base) of the fatty acids, which is soap.

Here, you've got:

Solvents:

Water
Limonene

Fats, glycerin, and fatty acids:

Stearic Acid
Myristic Acid
Coconut Acid
Glycerin
Theobroma Cacao (Cocoa) Seed Butter
Melaleuca Alternifolia (Tea Tree) Leaf Oil

Base:

Potassium Hydroxide
Sodium Hydroxide
Triethanolamine

Extra smells

Parfum (Fragrance)
Also Limonene
Linalool
Also tea tree oil

The only reason it has so many is that they've mixed together either a few different soaps or a few different fats before saponification (soap-becoming), and each soap will have been (mostly) neutralised.

Edit: note that the cocoa butter and tea tree oil are almost certainly added after saponification, for fragrance, to alter the material properties of the product, and to be left on the skin.


Why is the equilibrium constant unitless? I read that the equilibrium constant is unitless because the molar activities of each of the species are used in the equilibrium expression, not the actual concentrations themselves. I understand that.

But it is taught that

$$K_c = \frac{k_\mathrm{fwd}}{k_\mathrm{rev}}$$

I’m having some difficulty understanding this equation. Consider the chemical equation

$$\ce{A + B <=>[$k_\mathrm{fwd}$][$k_\mathrm{rev}$] C}$$

The forward rate constant is a second order reaction and has units $\pu{M-1 s-1},$ and the reverse rate constant is a first order reaction and has units $\pu{s-1}.$ Therefore

$$\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}$$

has units $\pu{M-1},$ but it’s supposed to be unitless. Where am I going wrong?

There are a few separate issues here to keep in mind:

$K_c$ (the equilibrium constant in terms of concentrations) is defined as
$$K_c = \prod {c_i}^{\nu_i} \tag{1}$$
Agreement between $K_c$ and the product of forward/back rate constants ($k_{\pu{fwd}}/k_{\pu{rev}}$ in the OP example) is expected only if the mechanism is correct, assuming some proportional rate law, and at equilibrium.
When dealing with gases, $K_c$ cannot generally be equated with $K_p$ (the equilibrium constant in terms of partial pressures). They may be numerically equal under some circumstances. For instance, they are numerically equal in reactions involving only ideal gases, provided the number of gaseous molecular entities remains constant during the reaction.
The (standard) equilibrium constant $K$, which simplifies to $K_a$ (under standard fugacity and some other conditions, agreements) which is defined as $$K_a = \prod a_i^{\nu_i}, \tag{2a}$$ cannot generally be equated with $K_c$. They may be numerically equal under some circumstances. $K$ itself is actually defined as follows:  $$\Delta_\pu{r} G^\circ=-RT\log(K) \tag{3}$$ whereas, strictly speaking, $K_c$ can't be used in this relation (but see below). The equilibrium constant $K$ cannot have units or this equation will not make sense.  In terms of concentrations,$$K_a = \prod \left(\frac{\gamma_ic_i}{c_i^\circ}\right)^{\nu_i} \tag{2b}$$ (To be nitpicky, one should use molality-based activities instead of amount concentration-based $c$ when saying that $K$ reduces to $K_a$).  When activity coefficients are all unity (ideal conditions), it might be that $$K_a \approx \frac{K_c}{K_c^\circ}= \prod \left(\frac{c_i}{c_i^\circ}\right)^{\nu_i} \tag{2c}$$ Strictly speaking, the equilibrium constant $K$ should remain unchanged with a change in the units used to describe concentrations. Units have no place in the constant because they cancel when ratios to reference standard states are computed (see Eq. (2)). $K_c$ and $K$ differ by the product  $$K_c^\circ = \prod ({c_i^\circ})^{\nu_i}\tag{2d}$$ (ignoring, once again, activity coefficients; standard fugacity). This difference is the source of the dangling units leading to the question in the OP.

You should therefore be careful equating $K_c$ with the ratio $k_{\pu{fwd}}/k_{\pu{rev}}$. And, assuming they are equal, the concentration equilibrium constant cannot generally be equated with $K$. However you can divide $K_c$ by $K_c^\circ$, and, assuming activity coefficients are unity, the result might be the thermodynamically correct unitless equilibrium constant (if standard fugacity), equivalent to that which appears in usual thermodynamic derivations.

Sloppy or not, associating units with an equilibrium constant is common practice. Attaching units to the equilibrium constant during reporting is a useful mnemonic when performing some computations, but it should be kept in mind that such a constant may not fulfill the requirements for computation of an associated standard free energy change, unless first modified by a ratio product of the standard concentrations.

Note that some of this is also explained in an answer to this post. IUPAC likes to use the convention standard equilibrium constant when referring to the thermodynamic equilibrium constant.


Boltzmann Distribution in Molecular Dynamics Simulation? The Boltzmann distribution is often used to describe the population of energy states at different temperatures, i.e. something like

$$\dfrac{g_i}{g_j} = \mathrm{exp}\left(-(E_i-E_j)\times \frac{1}{k_B\cdot T}\right).$$

For two given states $i,j$ with energies $E_i,E_j$, we thus have the population given by $g_i$ and $g_j.$

So if I have some observable $\sigma$ calculated for each state $i$, the natural way to calculate the observable of the ensemble would be

$$ \sigma_\mathrm{ensemble} = \left(\sum_i g_i \times \sigma_i\right) / ~\sum_i g_i~.$$

So, if I calculate $n$ ground states, let's say by geometry optimization of several guess structures, I employ the above weighting scheme to calculate the observable for the ensemble.

But here is my problem: If I perform an MD simulation instead, then the states should already be automatically populated according to the Boltzmann population (e.g. higher energy states are less likely to be observed in the MD simulation). So the population of the MD simulation already converges to the Boltzmann population for increasing simulation times $t$

$$ \lim_{t\rightarrow\infty} \frac{g_i^{MD}}{g_j^{MD}} = \frac{g_i}{g_j} $$

Now I am confused, because that would basically mean that there is a continuous transition between e.g. two states that clearly have to be boltzmann averaged up to the "long MD simulation", that is automatically populated accordingly. Where in the latter case, I would accidentally apply the Boltzmann population weighting twice.

Are there any resources on this weird population dichotomy or am I just overlooking something?

Within statistical mechanics (SM) a molecular property $X$ is computed by $$\left<{X}\right>_{SM}=\sum^{states}_i X_i p_i$$where $X_i$ is the value of $X$ for energy state $i$ and $p_i$ is the probability of being in energy state $i$ with energy $E_i$:$$p_i=\frac{e^{-E_i/kT}}{\sum_i e^{-E_i/kT}}$$Within molecular dynamics (MD) the corresponding property is computed by$$\left<{X}\right>_{MD}=\frac{1}{M}\sum^M_i X(t_i)$$where $M$ is the number of time-steps and $X(t_i)$ is the value of $X$ at time $t_i$.

You can also compute $\left<{X_{SM}}\right>$ from an MD by finding all the "unique" states (e.g. conformations) and computing their energy.  "Unique" means that even if a state is observed more than once in the MD you only include it once in the sum over states.

In principle $$\left<X\right>_{SM} = \lim_{t \to \infty} \left<{X}\right>_{MD}$$ but this assumes that the method used to calculate the energy and forces are identical in both calculations.

For example, for MD explicit solvent calculations with periodic boundary conditions it not really possible to calculate the energy of a state ($E_i$).  Also, vibrational effects must be included in $E_i$, which is typically done using the harmonic oscillator approximation, while the MD will include anharmonic effects. Similarly, the harmonic oscillator approximation is quantized (QM) while the vibrations in the MD simulation are not (classical).


Is it possible to calculate the momentum of an electron precisely? In a national level, undergraduate examination (TIFR-2018) following question was asked:

Which of the following is not correct with regard to the uncertainty principle?

An electron in an atom cannot be defined by a well-defined orbit.
The momentum of an electron cannot be measured exactly (correct answer)
A harmonic oscillator possesses a zero-point energy
Measurement of one variable in an atomic system can affect subsequent measurements of certain other variables

In my view answer should be (4). My explanation:
To calculate an electron's momentum precisely, it must have a totally undeterminable position! Since our apparatus have some definite size, you have to confine it in some definite space.
$$\Delta x \Delta p \geq \frac{\hbar}{2}$$
Since we want to make $\Delta p$ to approach zero, $\Delta x$ must approach infinity, which is not possible. So we can't measure the electron's precise momentum. So option (2) shouldn't be the answer.
Moreover, as far as I know, Heisenberg's uncertainty principle restricts us from measuring two quantities(which doesn't commute each other) simultaneously. We can measure them subsequently (one after another). So option (4) should be the answer.
P.S. I know that it is not necessary that we will get the same measurement on repeating the experiment, we can get any of the eigenvalues of that observable. But this is not related to Heisenberg's uncertainty principle. It's related to observer's effect.

To preface this answer I suggest reading a proper QM text which should address many of your issues. Griffiths' Introduction to Quantum Mechanics served me well but you may prefer any of the other excellent texts out there.

Option 4

This depends on your interpretation of quantum mechanics, but in the usual Copenhagen interpretation, the measurement of an observable causes the wavefunction of the system to collapse into the corresponding eigenstate. To give a more concrete example, if we have energy eigenstates $|1\rangle, |2\rangle, |3\rangle, \cdots$ with corresponding eigenvalues $E_1, E_2, E_3, \cdots$ then the wavefunction may be written as a linear combination of energy eigenstates. [This is guaranteed to be true because the Hamiltonian is a hermitian operator, and hermitian operators have complete sets of eigenstates.]

$$|\psi\rangle = c_1|1\rangle + c_2|2\rangle + c_3|3\rangle + \cdots$$

When you measure the energy of this system, the chance of getting the result $E_i$ is equal to $|c_i|^2$. Let's say you measure the energy to be $E_1$. Immediately after the measurement, the wavefunction of the system collapses to $|1\rangle$:

$$|\psi\rangle \overset{\text{collapse}}{\longrightarrow} |1\rangle$$

Now let's say you want to measure some other observable $A$. Again, you can express the original wavefunction $|\psi\rangle$ as a linear combination of eigenstates of $A$.

$$|\psi\rangle = a_1|A_1\rangle + a_2|A_2\rangle + a_3|A_3\rangle + \cdots$$

and the probability of measuring the eigenvalue $A_i$ would be $|a_i|^2$. However, the new wavefunction after collapse will have different coefficients in this expansion:

$$|1\rangle = a_1'|A_1\rangle + a_2'|A_2\rangle + a_3'|A_3\rangle + \cdots; \qquad a_i \neq a_i' \text{ (in general)}$$

and hence the probability of measuring the eigenvalue $A_i$ is now $|a_i'|^2$, which is in general, not the same as $|a_i|^2$.

In the absolute simplest case, we can consider measuring the energy again immediately after measuring the energy and obtaining the value $E_1$. Since the measurement caused the wavefunction of the system to collapse into $|1\rangle$, a second measurement done immediately after this will inevitably return the same value $E_1$ again.

Hence option 4 is true, which means it is not the answer you are looking for. Measurement of an observable causes wavefunction collapse, which in turn causes the probabilities of measuring other eigenvalues to change.

Option 2

First, we note that for a measurement of an observable to have no uncertainty, the system must be in an eigenstate of the relevant operator. This follows from the definition of uncertainty

$$\sigma_A^2 = \left\langle (A - \langle A\rangle)^2 \right\rangle$$

If the wavefunction of an electron is an eigenstate of the momentum operator $\hat{p}$, then a measurement of the momentum will always return the same value. In real life, if we carry out this measurement, will there be some uncertainty due to the precision of the machine? Yes, of course. Is it practical to set up an electron in a momentum eigenstate? Probably not. (By the way, your electron cannot really be constrained within your apparatus, because of quantum tunnelling. That is, unless the walls of your apparatus are an infinitely thick infinite potential energy barrier.)

But, that question is not asking you how you carry something out in the lab. The question is asking purely about theoretical considerations - and there is no fundamental theoretical problem with having an electron in a momentum eigenstate and measuring its momentum to arbitrary precision without changing its state. You are correct that its position will be indeterminate, but that is not a problem in the context of this question.

For interest, an "ideal measurement" of momentum is described in Shankar's Principles of Quantum Mechanics, 2nd ed. (p 123):

  Consider a particle in a momentum eigenstate $|p\rangle$. The postulate tells us that if the momentum in this state is measured we are assured a result $p$, and that the state will be the same after the measurement (since $|\psi\rangle = |p\rangle$ is already an eigenstate of the operator $P$ in question). One way to measure the momentum of the particle is by Compton scattering, in which a photon of definite momentum bounces off the particle.

  Let us assume the particle is forced to move along the $x$-axis and that we send in a right-moving photon of energy $\hbar\omega$ that bounces off the particle and returns as a left-moving photon of energy $\hbar\omega'$. (How do we know what the photon energies are? We assume we have atoms that are known to emit and absorb photons of any given energy.) Using momentum and energy conservation:

  $$\begin{align}
cp' &= cp + \hbar(\omega + \omega') \\
E' &= E + \hbar(\omega - \omega')
\end{align}$$

  it is now possible from this data to reconstruct the initial and final momenta of the particle:

  $$\begin{align}
cp &= -\frac{(\hbar\omega + \hbar\omega')}{2} + \sqrt{1 + \frac{m^2c^4}{\hbar^2\omega\omega'}}\frac{\hbar\omega - \hbar\omega'}{2} \\
cp' &= \frac{(\hbar\omega + \hbar\omega')}{2} + \sqrt{1 + \frac{m^2c^4}{\hbar^2\omega\omega'}}\frac{\hbar\omega - \hbar\omega'}{2} \\
\end{align}$$

  Since the photon always loses energy to the particle (as is clear in the particle rest frame) $\omega' < \omega$ and by sending $\omega \to 0$, we can make the change in momentum $p' - p$ arbitrarily small.


How do you draw (E)-1-phenyl-1,3-butadiene? I'm having trouble drawing this molecule because I don't understand how the functional group Phenyl (by -ene suffix) can also be a substituent (by 1-phenyl). Also, what does the 1,3-butadiene mean?

ManishEarth put together the pieces to understand the connection of the atoms to form the molecule.  Yet since the molecular structure requested in the original post is about (E)-1-phenyl-1,3-butadiene, the answer actually is opposite configuration:

By convention, alkenes posses a reference plane orthogonal to the plane of drawing.  In case the alkene is substituted twice by groups attached at the termini of the alkene -- in the upper example a phenyl and an alkyl chain -- these may be either at the same or the opposite side of this reference plane.

For the ease of convenience, two drawings simplifying the substitutents by "R" demonstrate this. In the first example, they are opposing each other, so their relationship is described either by (E) or (trans). The second example shows them at the same side of the reference plane, therefore, this alkene has a (Z)- or (cis)-configuration.

You might ask about the origin of (E) and (Z).  They are derived from German entgegen and zusammen; similar like the Latin (trans) for across, and (cis) for nearby.  Both ways to describe these configurational isomers are found in the literature for one and the same molecule, yet in an article or report, you should stick to either the (E)/(Z) or (cis)/(trans). (And yes, in italics.)


Where can I get potential energy surface for simple molecules? I'm looking for potential energy functions for simple molecules, such as water or hydrogen sulfide. I only need the region around the equilibrium geometry, preferably as a Taylor series expanded in terms of normal mode coordinates, with mixed terms given to the fourth or higher order. Just the ground electronic state would do, but ideally I need the first excited state as well.

If these data isn't available, I'd appreciate if you could explain how I could obtain them myself (including software recommendations), given that I'm an undergraduate, who understands the theory but doesn't have much practical experience yet.

Extremely accurate state-of-the-art potentials for more than 20 different diatomic molecules (including excited states for many of them!) are given here.

Bob Le Roy has a reputation for building some of the best small molecule potentials in the world, and in recent years he has started to make them available online. But these are not just a Taylor series around equilibrium, they are analytic functions covering the entire landscape of possible geometries, and for diatomics the specific function is given as $V(r)$ here.

You can plug the parameters from the given publications into the $V$ function, and then Taylor expand the global analytic function into a Taylor series if you want, but the original (non-Taylor expanded) function will be more accurate globally.

If the molecule doesn't yet have a potential available, you can build one ab initio (which is just the Latin phrase for "from scratch") using a program such as the free version of MOLPRO which can be found in the row of the linked webpage saying 0.00 EUR.

The input for making potentials in MOLPRO is here.

If you have specific molecules you are interested in, and you can tell us what you wish to do with these potentials, I can try to direct you to the best potentials for the ground and excited states for the particular molecules you ask for. There is no comprehensive database that has all potentials in one place.


Why is it important to remove the used up electrons in oxidative phosphorylation? I get that oxygen is the terminal electron acceptor in oxidative phosphorylation.  But why is it essential to accept these electrons and remove them as $\ce{H2O}?$

Oxygen is essential for cellular respiration, but after learning about cellular respiration it seems oddly unimportant.  Why is having an efficient garbage collector (oxygen picking up the used electrons) so crucial to cellular respiration, and to ATP production?

I get that oxygen is the terminal electron acceptor in oxidative phosphorylation. But why is it essential to accept these electrons and remove them as H2O?

A redox reaction is made of two half reactions. Neither one of them is more or less important; the combination of both determines the energetics of the reaction.

  Why is having an efficient garbage collector (oxygen picking up the used electrons) so crucial to cellular respiration, and to ATP production?

The electrons are not "used" as in an old car or worn-out clothes. There as shiny and new as before. Without oxygen, the reaction would not happen in the first place (electrons are transferred from less tightly bound to more tightly bound). The Gibbs energy of proton transport matches what is necessary for ATP synthesis, which in turn matches the Gibbs energy of the redox reaction. With a different electron acceptor (specificities of the enzymes involved aside), the proton gradient might be too weak to make ATP.


Are there any significant industrial processes that use liquid ammonia as a solvent? There are some interesting laboratory reactions that use liquid ammonia as a solvent (for example, the Birch Reduction where an alkali metal is dissolved in liquid ammonia).

There have also been some high volume industrial processes that use liquid ammonia as a solvent (the weedkiller paraquat has two, quite different, production routes using liquid ammonia producing thousands of tonnes a year). One uses dissolving sodium (like the Birch reduction):

But are there any other significant chemical manufacturing processes that use it?

NB I would include the production of pharmaceuticals or their ingredients as well as any bulk chemicals.

I might be terribly wrong, but I'm  not aware of any other industrial process involving ammonia as a reactant (fertilizers, caprolactam, amino acids) where liquid ammonia is used as a solvent.

As a side note, liquefication at least plays a role over here (Germany) for the transport in rail tank cars.


Any program for conformational searching for sugars? I want to estimate mean Gibbs free energy of formation of glucose and for that purpose I think conformers of ring flip should be taken into account. Do you know if some free program can find those structures automatically? (for example Balloon).

The following image shows the conformers I need to get. I am trying to avoid the possibility of drawing it.

Use RDKit

It sounds like you're only considering free or open source conformer generation tools. In that case, I highly recommend using RDKit, since the distance-geometry (and in turn ETKDG) methods will intrinsically generate all possible ring geometries. That includes not just the two chair forms you indicate, but all possible ring-puckering forms.

There are other free tools which do this, but RDKit has been repeatedly shown to be the best or among the best free conformer generation tools. (Based on current benchmarks.)

"High-Quality Dataset of Protein-Bound Ligand Conformations and Its Application to Benchmarking Conformer Ensemble Generators." J Chem Inf Model 2017 57, 529-539.
"Freely Available Conformer Generation Methods: How Good Are They?" J Chem Inf Model 2012, 52 (5), 1146–1158.

I'd advocate the newer ETKDG method with RDKit, since it seems to produce somewhat better geometries.


How to remove dichlorometane from water? I will have water that have been through dichloromethane. Doing so, the dichlorometane (solubility $2~\%$) will contaminate my water by forming an azeotrop which will have a $38~\mathrm{^\circ C}$ boiling point. Would putting that mixture to $40~\mathrm{^\circ C}$ will make that azeotrop leave? Leaving me dichloromethane free water?

If not I could I extract it?

There are two concepts here which should not be mixed:

The solubility of dichloromethane in water is indeed quite low, around 2% as in the question. But this is not the azeotrope, which is formed when there is 1.5% water - 98.5% dichloromethane mixture. That indeed boils at $38^\rm{o} \rm{C}$. See this. It's worth mentioning that you can't possibly bring the azeotrope to $40^\rm{o} \rm{C}$ - because it will boil at a lower temperature by definition, and the vapour will have exactly the same composition as the liquid.

Because what you have isn't the azeotrope, I'd say boiling your mixture (which will most likely happen near the boiling point of pure water) will remove the dichloromethane content.


Reaction between ammonia and iron I am not a Chemist or did not learn chemistry as a student. I have faced this phenomena from a natural visualization. So this question.

Experiment: What will be happen if we add a solution of ammonia and water over bad quality iron? When it shall be dried I shall pour again. Several day It shall be done regularly in open nature. Some reaction will take place. Rust will be created on iron.

Here is the main question:

Common people will not see anything significant except rust. But As a chemist how do you understand after three to four days after completion of the experiment that I pour ammonia solution on it? Assume that this three or four days it was in open nature. Something will be produced after reaction except rust. How much of them is water soluble?

In fact metallic Iron doesn't react directly with ammonia. However ammonia can influence the corrosion process because tend to increase the pH. From the Wikipedia Pourbaix diagram (below) you can see which are the products of a corrosion at an higher pH.Of course the pH of you solution depends on the ammonia concentration of your solution (a 1.0 M ammonia aqueous solution has a pH of about 11).

Corrosion is in fact inhibited at higher pH, and the products of the corrosion are not specific of the ammonia reaction, so (I think) is hard do identify the presence of ammonia from the corrosion process.
Furthermore ammonia is in fact a gas and many of the compounds it forms tend to became ammonia (e.g. ammonium carbonate) and then flow away. However I think that there are many chances that some ammonium salts are produce by the reaction with other species, (e.g. acids) present in the surface of the metal or in the environment.
In this case you can look for ammonium ions: $\ce{NH_4^{+}}$ you can identify them with a Ion chromatography or with Nessler's reagent (keep in mind that Nessler's reagent contain mercurium $\ce{Hg}$ so is toxic and not very environmental friendly) that detects even ammonia trace.


Why do some periodic tables have 18 groups while others have 8? In school I have been taught about the periodic table with 8 groups, representing the number of electrons in the outer shell (with exceptions). But online almost all I can find is periodic tables with 18 groups. Why do both exist? (Is the one with 18 more modern?) What is the purpose of 18 groups?

Unfortunately, life sometimes is more complicated than school :)

Using 8 groups works well to teach the basic principle, but higher shells have space for more electrons. (And lower shells for fewer - ever wondered why there are only two elements in the first period?)

Hence, you'll get 18 electrons in the next shell (the "transition metals"), then 32 electrons ("Lanthanides/Actinides") - the general rule is $2n^2$, where $n$ is the number of the period. This is unfortunately made even more complex, as the energies of the shells are not quite what one would expect and they are not filled "in order." One probably should stop speaking of "shells" at this point and move towards "orbitals."

Anyway, the community has pretty much settled on using 18 groups nowadays. The idea behind the periodic system is to find regularities (hence the name) between elements.

It turns out, that individual groups and trends within them are similar in the 8 "main" groups (as you have learned). In the transition metals, there are still a fair few regularities, that justify this arrangement - but they are already a lot more similar to each other than say alkali-metals and halides. The Lantanides/Actinides finally are so similar to one another, that they are "grouped" (although they in fact are part of a period) together - they in fact so similar, that they usually only occur in mixtures in nature, and separating them is quite hard.

Finally: no heavier atoms have been discovered so far that would have to go into an even bigger "shell"/orbital. Hence, one could justify up to 32 groups, but for practical reasons, 18 are conventionally used.


Will toluene vapors damage a mirror made from PETG? I'm working on a project for work and trying to avoid using an acrylic mirror because I know that doesn't hold up well against toluene, but I'm wondering if a mirror made out of PETG would?

PETG is also not resistant against toluene.
A month-long contact under ambient conditions results a complete swollen/disintegrated PETG material (PDL rating 0), and an year-long exposure even causes whitening [1, p. 2660]:

References

Chemical Resistance of Thermoplastics; Woishnis, W. A., Ebnesajjad, S., Eds.; PDL handbook series; William Andrew: Norwich, N.Y, 2012. ISBN 978-1-4557-7896-6.


Which is more stable out of the given carbocation? Which is more stable?

$\ce{^+C(CH3)3}$
$\ce{^+C(CD3)3}$

I have read in Solomon's and Fryhle that the +I effect of D is more than that of H but here in this problem it is given that $\ce{(CH3)3C+}$ is more stable.
How is this possible? Since, deuterium enriches the electron density over the central carbon resulting in the diminishing of a positive charge eventually making the second structure more stable.
Second explanation can be on the basis of hyperconjugation which says that $\ce{-CH3}$ is better in showing hyperconjugation than $\ce{-CD3}$, but why is this so?

$\ce{C-D}$ bonds are stronger. Hyperconjugation is thus more easily seen  with hydrogen rather than deuterium.   Hyperconjugation> Inductive effects at stabilization.I believe you can reason the rest out by yourself.  To see why $\ce{C-D}$ bonds are stronger.

See this


Glycoside Formation Reaction Whenever we add $\ce{H+}$/ Ethanol to glucose in its hemiacetal form, why doesn't pinacone/pinacolone rearrangement take place in place of nucleophillic substitution. The product would lead to formation of a carbonyl bond and so, should be the favoured product.

The actual pathway is:

But, why can't the reaction occur like this with ethanol as a polar solvent?:

First of all, there's no pinacole in the first place, so pinacole/pinacolone rearrangement isn't possible. Your rearrangement reaction/hydride shift wouldn't be called pinacole/pinacolone rearrangement.

Second, we have to think about why this would happen. Such rearrangements happen because we produce a more stable carbocation. In case of pinacole/pinacolone, if we do the rearrangement we end up with:

Which is well stabilized and after deprotonation a stable compound.

In case of sugars it's a little bit different, since it's not an alcohol in 1-position, it's a hemiacetale. This is extremly important since it has quite different reactivities.

We can cleave of water under much weaker acidic and mild conditions than in case of alcohols. The produced carbocation is quite stable since we got an oxygen stabilizing the cation, therefore no rearrangement is happening.


Steady-state approach - how to find rate equation for intermediate? The reaction mechanism for the formation of $\ce{NO2}$ is:
\begin{align}
\ce{NO + NO &<=>[$k_1$][$k_1'$] N2O2} & &\text{ (slow)} \\[0.2cm]
\ce{N2O2 + O2 &->[$k_2$] NO2 +NO2} & &\text{ (fast)}
\end{align}

What is the equation for rate of change of the intermediate $\ce{N2O2}$?
My solution is as follows:
The rate of formation of intermediate $\ce{N2O2}$ is given by:

\begin{align}
\frac{\mathrm{d}[\ce{N2O2}]}{\mathrm{d}t} = \color{red}{2} k_1[\ce{NO}]^2 - \color{red}{2}k_1'[\ce{N2O2}] - k_2[\ce{N2O2}][\ce{O2}] \approx 0
\end{align}
But the answer in the text book says:
\begin{align}
\frac{\mathrm{d}[\ce{N2O2}]}{\mathrm{d}t} = \color{red}{2} k_1[\ce{NO}]^2 - k_1'[\ce{N2O2}] - k_2[\ce{N2O2}][\ce{O2}] \approx 0
\end{align}

Can not figure out why my answer is wrong.

Split the scheme into three parts and write down each step. Start by definition with
$$\displaystyle -\frac{1}{2}\frac{d[\ce{NO}]}{dt}=+\frac{d[\ce{N2O2}]}{dt}$$
then the first step $\displaystyle -\frac{1}{2}\frac{d[\ce{NO}]}{dt}=k_1[\ce{NO}]^2\quad$ or $\quad\displaystyle \frac{d[\ce{NO}]}{dt}=-2k_1[\ce{NO}]^2$
and so $\ce{N_2O_2}$ is produced as
$$\displaystyle \frac{d[\ce{N2O2}]}{dt} = k_1[\ce{NO}]^2$$
In the reverse equilibrium step $\ce{N2O2}$ is lost (by dissociatiing) so $\displaystyle \frac{d[\ce{N2O2}]}{dt} = -k_1^{'}[\ce{N2O2}]$
and in the last reaction it is also lost by reaction with oxygen therefore $\displaystyle \frac{d[\ce{N2O2}]}{dt}=-k_2[\ce{N2O2}][\ce{O2}]$.
Adding up all three rates for $\ce{N_2O_2}$ gives
$$\displaystyle \frac{d[\ce{N2O2}]}{dt}=k_1[\ce{NO}]^2-k_1^{'}[\ce{N2O2}]-k_2[\ce{N2O2}][\ce{O2}]$$
You can now find the steady-state conditions.
(In your last equation there is $2k_1$ and it is quite common to see it written this way, all it does is to change $k_1$.)


Enthalpy calculation What is the value of enthalpy change for the following reaction

  $$\ce{X + 2Y -> 2Z}$$

  given

  \begin{align}
\ce{W + X &-> 2Y} &\qquad &\Delta_\mathrm{r}H_1 = \pu{-200 kcal} \tag{1}\\
\ce{2W + 3X &-> 2Z + 2Y} &\qquad & \Delta_\mathrm{r}H_2 = \pu{-150 kcal} \tag{2}
\end{align}

I have the answers, and the correct answer is $\pu{250 kcal}$, so I'm assuming you do $400-150$, but not sure how you would go about it.

$$\ce{X + 2Y -> 2Z}$$

\begin{align}
\ce{W + X &-> 2Y} &\qquad &\Delta_\mathrm{r}H_1 = \pu{-200 kcal} \tag{1}\\
\ce{2W + 3X &-> 2Z + 2Y} &\qquad & \Delta_\mathrm{r}H_2 = \pu{-150 kcal} \tag{2}
\end{align}

$(-1) \times 2$ + $(2)$ gives:

$\ce{4Y +2W + 3X->2W + 2X + 2Z + 2Y} \equiv \ce{2Y +X -> 3Z}$

Using Hess's Law,
the enthalpy change for the required reaction is given by  $-2\times \Delta_rH_1+\Delta_rH_2= 400-150 = \pu{250 kcal}$


Why does dipole must lie parallel to a rotation symmetry axis? While studying group theory, I've got confused by the following statement:

  For a molecule to have a permanent dipole moment, it must have an asymmetric charge distribution. The point group of the molecule not only determines whether the molecule may have a dipole moment but also in which direction(s) it may point. If a molecule has a $C_n$ axis with $n > 1,$ it cannot have a dipole moment perpendicular to the axis of rotation (for example, a $C_2$ rotation would interchange the ends of such a dipole moment and reverse the polarity, which is not allowed — rotations with higher values of $n$ would also change the direction in which the dipole points). Any dipole must lie parallel to a $C_n$ axis.

why does the dipole need to be parallel to the $C_n$ axis? If I put a fluorine atom in just one of the light gray atoms of the following molecule, wouldn't it have a molecular dipole that isn't parallel to the main axis?

No, it would not

A dipole means that there is a difference (in charge) from one "side" to another "side".

If we have determined that something is symmetric, it means the "sides" are equal to each other — and that as you rotate $\left(\frac{1}{n}\right)$ around the symmetry axis, everything stays the same. If there is anything that shifts out of position, that symmetry is not there (it is broken).

Now, your image is a bit hard for me to read, is it supposed to be linear or is it on purpose not? I am going to assume it is not linear, but that the light grey are symmetrically placed… So in your image, there is only a $C_2$ rotational axis, in the approximate "up/down" direction. By placing a fluorine in one end  (and something else at the other end) you would destroy the $C_2$ symmetry. If you rotate it $180$ degrees around (if you could see it through the backside of your monitor…) the fluorine would be on the other side, clearly not identical to $E$, the original. No symmetry.

If it was linear, the light grey and dark gray arranged as a line, then it would still retain a $C_\infty$ rotational axis along the dark grey atoms. Rotating around this in any fraction or direction would still keep everything looking the same.


Optical activity of starch In this German Wikipedia article about optical activity, I found the beautiful picture of starch granules:

Unfortunately, I have no idea what I am looking at. What are these crosses and why are they blue (i.e., same as the background color)?

The figures you see are isogyres and arise in substances that are optically active, which means that polarized light is transmitted through the material in different ways depending on the orientation of the sample. These figures are called interference figures. I highly recommend that you peruse this excellent write-up on the optical properties of minerals, particularly the sections pertaining to uniaxial and biaxial crystals, birefringence, and how a petrographic microscope works.

From the Wikipedia entry on conoscopic interference pattern:

  The figures are produced by optical interference when diverging light rays travel through an optically non-isotropic substance - that is, one in which the substance's refractive index varies in different directions within it.

Starch is uniaxial and the figures you see are characteristic of uniaxial materials when viewed using a petrographic microscope. See this reference (The Principles of Pharmacognocy. F. A. Flückiger and A. Tscirch. William Wood & Company, New York, 1887), pages 118-119.

As for the color seen in the image you reference: that arises from using a mica (or tint) plate which in turn indicates whether the substance is uniaxial positive or uniaxial negative, which refer to the propagation of light along differing optical axes. However, in your figure, the color of the isogyres and the background are both blue, which indicates to me that the blue color is simply a result of polarized light passing through the substance.


How can we say whether or not an organic compound is soluble? In the exam recently conducted by my school, they asked whether 2-hydroxyacetophenone is water soluble or not.

I thought that since there is intramolecular H-bonding in this compund, it won't be soluble. But my teacher marked it wrong. Also, when I checked it on internet, I saw in some websites it was mentioned to be water-soluble. My question is how can we say whether or not an organic compound is soluble?

The traditional rule of thumb has always been that 1 oxygen can dissolve 4 carbons.
In this case, the compound is $\ce{C8H8O2}$ which makes an oxygen:carbon ratio of 1:4, so it should be soluble in water.

The source of this rule of thumb is Dr. Pedro J. Victory, SJ, who taught us Organic Chemistry. That was a long time back but I recall this rule of him.
The context was liquid-liquid extraction.

I do not remember if he added a corollary for the case of esters, where you should count only one of the oxygens of every ester group.
We perform extractions with ethyl acetate all the time and there the ratio is 1:2.  Very probably, he did add that.

To support this rule I have searched for some solubilities at PubChem. On basis of this data, the threshold for soluble (as asked in a comment by @Zhe) would be around 10% (100 mg/mL).

                    mg/mL T (C)
acetone             completely soluble
methyl ethyl ketone 223    25
3-pentanone          43    25
2-pentanone          46    25
isopropanol         completely soluble
1-butanol            63    25
2-butanol           181    25
1-pentanol           22    25
2-pentanol           45    25
3-pentanol           55    25
methyl ethyl ether  230    25
diethyl ether        60    25
methylacetate       243    25
ethylacetate         80    25
n-propylacetate      19    20


Are there any examples of activating meta directors? I have heard about activating ortho-para directors. For example: methoxy group, hydroxy group, alkyl group, et cetera.

I have heard about deactivating ortho-para directors. For example: halogen substituents

I have heard about deactivating meta directors. For example: nitro, trifluromethyl group, carboxylic acid group, et cetera.

So, naturally, I have embarked on a quest to find Activating Meta directors.

I have found one potential candidate, which is dimethylphenylborane:

Boron possesses a vacant $\mathrm{p}_z$ orbital, which can interact with the $\pi$-cloud of the benzene ring.

Considering the fact that boron is treated as a metalloid, it will definitely be less electronegative than carbon, so the dimethylborane group should offer a good +I effect, which could possibly outweigh the -M effects.

Can anyone shed some light on this issue? Are there any other examples of activating meta directors?

Note: This requires an extremely high energy scenario to occur.

This is another example that looks promising, it is essentially an ylide that has enough EWing power to meta direct while giving some density to the ring in another resonance structure. []

The phosphonium ylide derivative (phenylphosphonic acid) may provide the same effect if the hypervalent nature of nitrogen is challenged, or if this substance is unobtainable under any circumstances.


Methylnitrobenzene ring formation Synthetic Route I know the malonic ester is probably used to make propanoic acid or some carboxylic acid, and then joined to the methylnitrobenzene. But I just can't figure out what glues to the methyl. I know the nitro can be reduced to amine and bonded to a carboxylic acid, but then what about the ends?

This is a challenging question as we have so few details.

One method would be to convert the orthonitrotoluene into the benzyl bromide with NBS or another free radical bromination reagent. Then you could react the dimethyl malonate with sodium methoxide in methanol. Next react the sodium salt of dimethyl malonate with 2-nitrobenzyl bromide.

Then filter off the sodium bromide thus formed, next stir it with the product of reacting Merifield's resin with tris-2-aminoethyl amine to scavenge anyunreacted nitrobenzyl bromide. Next subject the mixture to hydrogenation with a palladium on carbon cataylst. This will convert the nitro group into an amine.

Next evapourate it down to form a oil, heat this to make the amine react with the methyl ester to form methanol and an amide. When you are done with this step. Then react it with dilute sodium hydroxide to convert the remaining methyl ester into the sodium carboxylate. Make it acidic again and it will decarboxylate to give you a cyclic amide.

For the next step I would convert the amide to a thioamide with Lawesson's reagent and then treat the thioamide with rainey nickel to convert it into am amine. Between some of the steps it would be a good idea to purify the intermidates, but this is how I would (off the top of my head) try to make the conversion. I suspect that there will be cheaper ways of making the final product.


At home Rochelle's salt synthesis I tried to make Rochelle's salt. It didn't work!

7 oz cream of tartar
1 cup distilled water
heat to 180 F and keep there
add 1/2 tsp sodium carbonate at a time until there is no fizzing
filter and let cool

The solution ended with a strong yellow/brown tint. It took at least an hour of adding sodium carbonate. While cooling it rapidly formed a crystallized honey look and that was it. I reheated and added more distilled water to lighten it up a bit thinking it needed to evaporate slower. This time when it cooled nothing happened.

The sodium carbonate was 100%, and the cream of tartar was McCormick. It's possible that came with some impurities hence the brown/yellow grossness.

Is there as way to save this batch? Should I just try again? Is there a better recipe?

My first instinct is that it didn't work because you don't science with Imperial units.

Tried again with 365 brand (whole foods) Cream of Tartar. Worked much better.

39g Cream of Tartar
50ml distilled water
Sodium Carbonate added slowly until stirring produces clear liquid


Which one is the strongest base among the following compounds? I am confused between option (2) and (4). Both lone pairs of nitrogen are localized, so next, I looked at the hybridization of $\ce{N}$ in each compound. As in (2), the $\ce{N}$ is $\mathrm{sp^2}$ hybridized and therefore is more electronegative, I assumed it would be a weaker base compared to (4).

But the answer is (2). I drew the protonated structures and found that there is resonance in the compound (2).

I was taught that electronegativity is more important in questions like this. Is this a wrong concept? Any help is appreciated!

A good way to compare basicity of basic compounds is to draw their corresponding conjugate acids and compare their stability. As you might figured out from drawing resonance forms of compound (2)'s conjugate acid (its protonated form ), the positive charge is going to be shared by the 3 nitrogens in the compound. On the other hand, in compound (4)'s conjugate acid (protonated form) the positive charge is on a single nitrogen. Of course, this gives compound (2)'s conjugate acid higher stability.

Your reasoning about the nitrogen being sp2 makes it less basic in comparison with sp3 nitrogen is correct,however; it only holds true if you were comparing two nitrogens with the only significant difference being the hybridization state, but in some cases, resonance will stabilize or destabilize the conjugate conjugate acid of the base, this is usually have greater impact than hybridization with regard to ranking baisicity of the compounds as in the question presented.
Electronegativity is an important factor in ranking acids/bases,but as I said, it should be used with compounds where the only significant different in the structures is about it (for example, 4 compounds with same structure but only difference is Cl, I, Br, OH at a specific position), in other cases, other factors such as resonance stabilization and aromaticity should be also considered alongside electronegativity.


If I mix an aqueous solution of hydrogen chloride with an organic solvent, will the organic solvent contain hydrogen ions? For example, let's say I mix $36\ \text{cm}^3$ 2-propanol, $9\ \text{cm}^3$ methyl ethyl ketone and $15\ \text{cm}^3$ hydrochloric acid with a concentration of $1,0\ \text{mol}/\text{dm}^3$.

The hydrochloric acid is solved in water, and since the other compounds are organic and less dense than water they should form two layers. Would the upper layer be "acidic"?

This is an elution I've used in a thin-layer-chromatography lab and I'm trying to figure out what role the hydrochloric acid plays in affecting the compounds' ability to bind to the elution (mobile phase).

$\ce {HCl}$ is so strong a proton donor that bases weaker than water could be protonated.  Alcohols and etherss, if they form a separate phase from water, are likely to be protonated.  Going the other direction, some organic compounds with little solubility in water might become more soluble in a hydrochloric acid solution if they are sufficiently basic.  The basicity of some cyclopropenones can be measured that way.  See (behind a paywall) http://pubs.acs.org/doi/abs/10.1021/cr60288a003?journalCode=chreay.


What does the reaction of hydrogen and iodine to hydrogen iodide look like? The reaction, $\ce{H2 + I2 <=> HI}$ is often used as an example for equilibrium reactions. It is however a difficult reaction to show in a school laboratory: we are not equipped to work with flammable gas at $\pu{450 ^\circ C}$ and high pressure.

It would be great if I could show my students what this process looks like in real life. I have a description, with drawings, by Max Bodenstein, from 1896 -- slightly dated.

Does anyone have pictures or videos of the gases involved and of the equipment used to make $\ce{HI}$ from the elements and to measure reactants and products?

To show an example for equilibrium reactions to your students, you can use a reaction which has visual change as shifting the equilibrium position such as color. My best bet is, use of cobalt(II) chloride equilibrium with water:
$$\ce{[CoCl4]^2-(aq) + 6H2O(l) <=> [Co(H2O)6]^2+(aq) + 4Cl-(aq)}$$

The beauty her is The ion, $\ce{[Co(H2O)6]^2-(aq)}$ is pink while $\ce{[CoCl4]^2-(aq)}$ is blue.  According to the Le Chatelier’s principle, at relatively low chloride concentrations (e.g., diluting the solution or adding $\ce{AgNO3(aq)}$ to remove $\ce{Cl-}$ ions), the equilibrium shifted to the right and, therefore, the solution becomes pink. However, if there is a large concentration of excess chloride (e.g., adding $\ce{Cl-}$ ions by means of adding $\pu{12 M}~\ce{HCl}$ drop wise or adding $\ce{CaCl2}$ beads), the equilibrium tends to shift to the left, making the solution blue.

In addition, similar to its sensitivity to the concentration of solutes, the equilibrium is also sensitive to temperature as well (See picture below). At colder temperatures, the equilibrium tends to shift to the right(pink color at LHS of the picture) while at warmer temperatures, it lies to the left (blue color at RHS of the picture):

Reading:

http://www.rsc.org/learn-chemistry/resource/res00000001/the-equilibrium-between-two-coloured-cobalt-species?cmpid=CMP00005957
https://eic.rsc.org/cpd/understanding-equilibrium-a-delicate-balance/2000012.article
http://research.chem.psu.edu/mallouk/ilab/Equilibrium_and_Le_Chatelier.htm


Overlap matrix (H2 molecule, STO-3G) I'd want to implement SCF method using Modern Quantum Chemistry (Szabo). But when I want to calculate overlap matrix for H2 (STO-3G basis) my results aren't equal to true overlap matrix that there's in the book. However when I compare my results in other basis (STO-1G) I get exact overlap matrix. Below my code for calculation overlap values.

class gauss():

  def __init__(self, alpha, Ra, k = 1):

    self.alpha = alpha
    self.Ra = np.array(Ra)
    self.k = k

  def __mul__(self, other):
    'calculate integral'
    lenght = (self.Ra - other.Ra)
    scalar = np.dot(lenght, lenght)

    return np.exp(-self.alpha*other.alpha*scalar/(self.alpha + other.alpha))*np.pi/(self.alpha + other.alpha)**(3/2)

def overlap(first, second):

  s11 = 0
  for i in first:
    for j in first:

        s11 += i * j * i.k * j.k

  s12 = 0
  for i in first:
    for j in second:

        s12 += i * j * i.k * j.k

  return s11, s12

## gaussian primitives:
s1 = gauss(0.168855404, (0.0, 0.0, 0.0), 0.444634542)
s2 = gauss(0.623913730, (0.0, 0.0, 0.0), 0.535328142)
s3 = gauss(3.42525091, (0.0, 0.0, 0.0), 0.154328967)
s1b = gauss(0.168855404, (1.4, 0.0, 0.0), 0.444634542)
s2b = gauss(0.623913730, (1.4, 0.0, 0.0), 0.535328142)
s3b = gauss(3.42525091, (1.4, 0.0, 0.0), 0.154328967)

first = [s1, s2, s3]
second = [s1b, s2b, s3b]

s11, s12 = overlap(first, second)

print(s12 / s11) # print normalized s12

  0.781180936263708

but exact value from book = 0.6593 (1.4 Bohr)

and if I choose basis as:

s = gauss(0.4166127, (0.0, 0.0, 0.0))
sb = gauss(0.4166127, (1.4, 0.0, 0.0))

I get exact value:

  0.6647924143035879

from book: 0.6648

You have two errors in your code, one minor and one major.
The minor error is in your multiplication function. The value you return should have $\pi^{3/2}$, but the way it is arranged now the power is only being applied to the denominator of that expression.
The major error is how you are normalizing your overlap integrals. You can't simply divide $\langle1|2\rangle$ by $\langle1|1\rangle$ in order to normalize it. You have to explicitly normalize each of the Gaussian basis functions. To do this you can include a factor of ((4*self.alpha*other.alpha)/np.pi**2)**(3/4) in the value you are returning from your multiplication (in terms of readability of your code, it may be wiser to define variables within the function like proportionality_constant, gauss_integral, and normalization and then return the product of these). Or you could simplify and just multiply the exponential by ((4*self.alpha*other.alpha)/(self.alpha+other.alpha)**2)**(3/4) This also means you only need the second part of your overlap function, since the multiplication will take care of the normalization.
The reason your code works for STO-1G is that dividing by $\langle1|1\rangle$ accidentally gives the correct normalization since there is only one normalization constant associated with the one Gaussian function of STO-1G.


How does the application of Schrödinger equation to model a particle in a box explain the origin of degeneracy of atomic orbitals? Is the particle in a box concept analogous to an electron in an orbital? If so, can we apply the equation for the allowed energies of a particle in a box
$$E = \frac{h^2n^2}{8mL^2}$$
to the allowed energies of electrons in orbitals? Are the orbitals degenerate because this equation only includes the quantum number $n$ and not the others and therefore all orbitals in a subshell will roughly be degenerate?
Finally, if this equation can be used to explain the allowed energies of orbitals/electrons, what would $L$ be in this instance?

Is the particle in a box concept analogous to an electron in an orbital?

No, but there is an analogy to the $\pi$ systems of dye molecules, and (for the 3D box) to the band structure of nanodots. Some physical chemistry courses have a lab that explore this relationship, e.g. at Saarland university (sorry, in German, but the figures and the math are universally understandable, I hope).

Finally, if this equation can be used to explain the allowed energies of orbitals/electrons, what would L be in this instance?

It would be the length of the conjugated system (for the dyes) or the radius (for the nanoparticles)

Are the orbitals degenerate because this equation only includes the quantum number n and not the others and therefore all orbitals in a subshell will roughly be degenerate?

For a 1D case with degeneracy, you can explore the particle in a ring, Why do solutions of electron in a box (and in a ring) predict coefficients for LCAO (linear combination of atomic orbitals) in 1D systems?.


Electrolysis of water using 240V AC I remember back in my high school, when we started electrolysis, I became quite interested in it. I decided to do a home experiment on electrolysis of water. I grabbed my graphite pencils, a glass of water made of steel, two wires and some water. I wrapped the wire on my graphite nibs(on one end) and immersed the other end in water. I connected the wires to 240V AC supply(that was the supply which powers our home.)

Observations:

Initially, bubbles formed on both of the electrodes(I think they must be a hydrogen/oxygen mixture since the cathode and anode switches in between due to ac)

The glass of water became very hot with the passage of time(which was due to resistance heating i suppose)

The interesting part came when I added a few grams of table salt to water. Within seconds, the graphite rods turned red and there were fumes of gas(most probably it was steam cause the fumes were white.) There might have been sparks in the plug as well, but I don't remember for sure. I immediately pulled the plug out before it got any worse.

I didnt try anything of that sort again as i knew i could have done worse than that. But it doesnt cause any harm in knowing what actually happened when I added salt.

Why did the graphite turned glowing red?

What was the composition of those fumes?

What more could have gone worse?

What you are trying to do is called a redox reaction (from REDuction-OXydation) in which one molecule will be oxidized (i.e. give one electron) and another will be reduced (i.e. receive one electron). The propensity of two different molecules to exchange electrons is called "redox potential" and measured in Volts.

In the case of water, its dissociation into hydrogen and oxygen has a potential of 1.23V(1), so a 1.5V battery would be enough (however a rechargeable battery usually has a potential of 1.2V which would not be enough).

In your case, you apply a 240V potential, which will not only promote many other reactions but also heat the water up because of the Joule effect and make it boil.

(1) I purposely ignored the sign not to confuse you.


How can anions exist? Lets say we have an nitrogen atom - it should have 7 protons and 7 electrons.

How can an nitrogen anion - lets say $\mathrm{N^{\,1\mathbf{-}}}$- even exist - shouldn't the 7 electrons in valence shell repell the extra one?

What force does hold the extra one electron in the valence shell? There isn't an extra proton in the $\mathrm{N^{\,1\mathbf{-}}}$ that would hold the extra electron.

One reason cations and anions exist is due to the stability of a full or half-full valence shell. The stability from those electronic configurations means that the atom or molecule does not require protons to "hold" the extra electron.

Recall also that nitrogen has three (or five) valence electrons, rather than seven. The 1 s shell is full and is not considered part of its valency. The three 2 p electrons are the valence electrons although they hybridize with the 2 s electrons to produce the trigonal pyramidal structure of ammonia with its lone pair. The single anion $\ce{N^{-1}}$ could exist, but would not be stable because it puts four electrons in the p shell. The p shell would prefer to have three electrons as it does in the nitrogen atom or no electrons as it does in the $\ce{N^{3+}}$ cation.


Testing for asbestos using readily available chemicals and tools? Given that it is difficult to determine who is telling the truth in the recent claim that J&J covered up finding asbestos in their powder, is there any chemical test that we can perform on our own to determine with reasonable confidence whether the particular sample of powder we have bought contains asbestos? (I am aware that one test involves viewing the sample under a sufficiently high-powered polarizing light microscope, but not everyone has such a tool.)
I have not much chemistry background beyond high-school, and it seems that there are numerous kinds of asbestos, apparently characterized more by their large aspect ratio than by their chemical composition. For example, chrysotile (which accounts for ≈95% of asbestos in buildings in the US) has an idealized chemical formula $\ce{Mg3(Si2O5)(OH)4}$, but may not even occur as exactly that. Talc according to wikipedia is either $\ce{H2Mg3(SiO3)4}$ or $\ce{Mg3Si4O10(OH)2}$. These break down as follows:

Chrysotile: $\ce{(Mg^2+)3[Si2O5]^2−([OH]^−)4}$
Talc A: $\ce{(H+)2(Mg^2+)3([SiO3]^2+)4}$
Talc B: $\ce{(Mg^2+)3[Si4O10]^4−([OH]^−)2}$

So we cannot test for $\ce{Mg^2+}$ or $\ce{[OH]^−}$. But is there a chemical test for $\ce{[Si2O5]^2−}$ that can be done at home? And is it safe to rule out chrysotile if no $\ce{[Si2O5]^2−}$ can be detected, or can the amount be too low for chemical detection but still dangerous?

Is there a chemical test for $\ce{[Si2O5]^2−}$ that can be done at home?

First $\ce{[Si2O5]^2−}$ isn't a poly atomic ion like sulfate ($\ce{SO4-}$) or nitrate ($\ce{NO3}$) but represents a stoichiometry in the material.

Unfortunately at this time a test without x-ray diffraction is not possible. Asbestos/chrystolite is not just a chemical formula, but a crystal structure. It's not just what it is made of but also how the atoms are structured that makes it potently carcinogenic. If you were to take asbestos and melt it then cool it sufficiently fast to form a glass (amorphous/non-crystalline substance), it would not have similar carcinogenic properties as the crystalline material.

  I am aware that one test involves viewing the sample under a sufficiently high-powered polarizing light microscope, but not everyone has such a tool.

It is worth noting that using polarized can be done at home with polarized filters but is a presumtive test not a direct method. Essentially all chrystolite will rotate light but not all minerals that rotate light are chrystolite.


Concentration of gas above aqueous ammonia I am a physicist who has run into some chemistry in the lab (and unfortunately, I've completely evaded any formal training in chemistry). I have exposed some samples (I think details of the sample are not important here, but would be happy to provide further information if necessary) to a solution of 30% ammonia and 70% water, and am trying to calculate approximately how much ammonia (in ppm, for example) reached my sample.

My samples are placed in an 3″×3″×1″ (L×W×H) container with the top exposed to the air. Until now, I've just been marking down the volume of the solution that I've placed (about 15 mm) directly below my samples in the container (e.g., 5 μL, 10 μL, …, 200 μL). I think it's good enough to assume room temperature and standard atmospheric pressure for my labs. I've looked into Henry's law and related relationships, but am not sure that these are the appropriate tools. I would greatly appreciate any help or guidance!

Maybe these tables will help (I've stumbled upon them trying to figure this out in the context of Henry's law):

https://www.tannerind.com/PDF/green-aqua-amm.pdf
https://www.atmos-chem-phys.net/15/4399/2015/acp-15-4399-2015.pdf

P.S.
If a closed container is a more tractable problem, I also collected some data with a closed container (same dimensions), and would very much appreciate insight into that problem too (if it is very different, or the only solvable problem). I should also mention that the solution was removed from the container in about 10 seconds in all cases (if that makes a difference).

If the container is open, as shown below, then you have a total mess.

Obviously eventually all the ammonia and water will evaporate into the room eventually. There is absolutely no way to tell how much ammonia gas would be with the volume of the container at any given time, and the ammonia concentration within the container wouldn't ever be homogeneous.

If the container is closed, then you have a somewhat more tractable problem.

However there are two equilibriums to consider. First the gas phase must be saturated with water vapor, and second the ammonia must be distributed between the gas and liquid phases (if any liquid phase remains).

Just out of curiosity, 3x3x1 inchs has a volume of 9 cubic inches, or about 147 mL. So if the air in the lab was at $\pu{20 ^\circ C}$ and completely dry then about $\pu{2.5 \mu L}$ of water would be needed to saturate the volume to 100% humidity.

  Calculation:
  water vapor pressure at $\pu{20 ^\circ C} = \pu{17.5 torr}$
  atmospheric vapor pressure = 760 torr
  molar volume at $\pu{20 ^\circ C}$ and 760 torr = 24.1 liters
  Moles water = $\dfrac{17.5}{760}\cdot \dfrac{0.147}{24.1} = \pu{1.4\cdot10^{-4} moles}$
  water is $\pu{18 g/mole}$ and $\pu{1 g/ml} = \pu{0.001 g/\mu L}$
  $ \pu{\mu L water} = \dfrac{\pu{1.4\cdot10^{-4} moles}\cdot \pu{18 g/mole}}{\pu{0.001 g/\mu L}} = \pu{2.5 \mu L} $

NOTE: If the container is sealed there is still the problem of estimating how long it takes the system to reach equilibrium.


Dissolving Aspartic acid in vinegar? I am trying to dissolve aspartic acid into vinegar as use as a carbon source in saltwater aquariums.   It does not dissolve.

I was told to increase the pH and it will work?  What is the best way to increase pH with minimal negative effects?

Sodium carbonate seems to work at dissolving; however I may be reducing the acetic acid from 5% to something lower?  Might also be creating an acetate?  Calcium hydroxide could be a better choice for the aquarium but I can't seem to get the mixing ratio right?

The good news is there is no good reason to add vinegar to your aquarium.  You are correct that if you did, the addition of hydroxide to maintain proper pH would then neutralize acetic acid to acetate.

You can dissolve about 4.5 g of aspartic acid into a liter of neutral water. I assume you try to keep your saltwater tank at a pH of around 8.3 however, which should significantly increase it's solubility as aspartic acid is acidic, having a pKa of 3.9. This is the reason you could not dissolve the aspartic acid with 5% acetic acid, which still pretty acidic with a pKa of 4.8.

So the solution is to just use the slightly basic aquarium water itself, plus a dose of patience, to dissolve the aspartic acid. You could place it in a tea-bag next to your aerator or something if you wanted to isolate it and see how quickly it dissolves.