How to calculate size of a specific cavity in a protein for every frame in trajectory file? Here is the situation, I have trajecory file of a protein of interest. That protein has 2 main cavities that are participating in interraction with ligands or other proteins and also other cavities. Also residue numbers around this cavity are more or less known. I'm looking for a method or software to get volume of this cavities for every frame in trajectory.

I want to specify a cavity around a residue and  get output for this cavity. What I am looking for is something like this:

Frame Cavity volume
1 1500 A^3
2 1499 A^3
3 1499 A^3
4 1499 A^3
5 1498 A^3

I tried CASTp, it was okey for singe frame if wrote it as pdb but not promising for a 100ns trajectory.

Also tried CAVER. CAVER was able to open a trajectory file and calculate cavity volumes for the frame that I chose but it was not possible to calculate cavity volumes for all of the frames and get output from it.

If anyone has any suggestion I'll be greatful.


Which has a greater dominance: hydration enthalpy or lattice energy? When I was reading my notes I came up with a problem. I read that for the s block element salts, when checking the solubility, we consider the hydration enthalpy so we decide their stability in water. But then there is this exception: hydroxides don't follow this. They depend on lattice enthalpy for their stability. Which has a greater dominance and why? Is it the lattice energy or the hydration enthalpy? How can we differentiate them, and where to use these concepts for deciding stability? As in the case of hydroxides lattice energy plays a dominant role, not the hydration energy: why is it so?

The dominant factor depends on situation to situation.Here is an explanation i found helpful while predicting solubilities

The hydration energy is of Form A÷r1 + B÷r2 .

The lattice energy is of form C÷(r1+r2).
Here r1 and r2 are radius of cation and anion while A,B,C are constants.

When Size of anion is large example Iodine,Lattice energy does not decrease with increase in radius but hydration energy decreases drastically since r2>>r1(See formula)
Therefore in such cases,Solubility of Salts decreases with radius of cation.
Order- LiI is more than NaI is more than KI(in terms of solubility)

Now when anion size is comparable(eg Fluorine),The lattice energy decreases much more rapidly than hydration energy(see formula).Hence Solubility increases with increase in cationic size.Thefore order is LiF is less than NaF is less than  KF


Why won't ethene and ethyne react with halogens? With alkanes, a free-radical halogenation (substitution) is observed.

I was told that there will be no reaction in a similar situation with ethene and ethyne — and I am having trouble understanding why.

Is it because the halogen radical will not be able to abstract a hydrogen atom from carbon because it is influenced by the greater electron density in the carbon-carbon double/triple bonds? Or because the $\ce{C-H}$ bonds are shorter in these molecules compared to alkanes? (resonance?)

I've assumed that the problem arises in the propagation step of the most common reaction mechanism seen with alkanes.

Or is it due to sterics? ($\unicode[Times]{x3c0}$-bonds get in the way of the attacking bromine radical or restricted rotation in double/triple bonds get in way of ideal orientation for reaction to proceed to the transition state)

The necessary heat/$h \nu$ is provided for the reaction. Additionally, no other reactants/catalysts are present.

Please keep your answers on a novice college student level.

Both substance classes have an inherently different reactivity. Alkanes are very unreactive — in fact, save radical reactions, there is not much that will adequately attack them. Alkenes and alkynes have $\unicode[Times]{x3c0}$-bonds which are much less energetically stable than the alkanes’ $\unicode[Times]{x3c3}$-bonds.

$\unicode[Times]{x3c0}$-bonds can be attacked by halogens directly, the mechanism is called electrophilic addition and the usual product is an anti-dihalide with the halogen atoms attached to the carbons previously involved in the double bond. An intermediate species of this reaction is the halonium ion, a three-membered ring consisting of the ex-double-bonded carbons and one positively charged halogen. The other halogen can then attack according to $\mathrm{S_N2}$ from behind.

However, if you try hard enough you will also manage to create halogen radicals in the presence of alkenes. The radical will now not attack a $\ce{C-H}$ bond, but rather the energetically less stable $\ce{C=C}$ bond forming a $\ce{C-Hal}$ bond and a new carbon-centred radical. This carbon radical will then most likely polymerise with further double bonds by the same mechanism, creating a carbon polymer rather than a haloalkene.

I’m not entirely sure how alkynes will react. Their triple bond is more stable than a double bond, so selective halogenation of double bonds is possible in the presence of triple bonds. However, in general there is no reason why they couldn’t react according to the same pathways.


What is the reaction rate constant of an effective reaction? Consider the reactions

$$\ce{A + B ->[$k_1$] C }$$ and

$$\ce{C ->[$k_2$] E + F}$$

with reaction rate constants $k_1$ and $k_2$. I know that the two reactions can be written as follows

$$\ce{A +B->E +F}$$

with a rate constant $k_3$. How does $k_1$ and $k_2$ relate to $k_3$? Does it matter how many species are in the reactants (such as if the first reaction was simply $\ce{A->C}$)?

It is not clear that they are related. It is simpler to consider the equivalent scheme $A \rightarrow C \rightarrow E$ with rate constnats $k_1,\, k_2$ then the decay of A is
$A_t/A_0= \exp(-k_1t)$ the amount of  C rises and falls; it is
$$\displaystyle C_t/A_0 = \frac{k_1}{k_2-k_1}(\exp(-k_1t)-\exp(-k_2t))$$ and
E is
$$E/A_0 = 1-A_t-C_t=1-\exp(-k_1t)-\frac{k_1}{k_2-k_1}(\exp(-k_1t)-\exp(-k_2t))$$

When $k_2 $ is very large and much bigger than $k_1$ then $\exp(-k_2t) \rightarrow 0 $ and $\displaystyle E/A_0= 1-(1+\frac{k_1}{k_2})\exp(-k_1t)$ and in the opposite limit $k_1 \gg k_2$ then $E/A_0= 1-\exp(-k_2t)$.

So you can choose which limit is important. In either case the amount of E rises as time proceeds but with different rate constants. In the intermediate case there seems to be no rate constant but only a slope to the curve which will depend on the time at which the measurement in made. You could for example find the half rise time and try to relate that to rate constants, but it looks messy.


What should be the name of this compound containg both double and triple bond? Which is a more correct name, according to IUPAC nomenclature: pent-2-ene-4-yne, or pent-3-ene-1-yne?

According to R-3.1.1 of IUPAC, Commission on Nomenclature of Organic Chemistry:

Numbers as low as possible are given to double and triple bonds as a set, even though this may at times give "-yne" a lower number than "-ene". If a choice remains, preference for low locants is given to the double bonds. Only the lower locant for a multiple bond is cited except when the numerical difference between the two locants is greater than one, in which case the higher-numbered locant is cited in parentheses (see R-0.1.4)

Numbering from the right side of the molecule as seen in the question figure will assign the locants (2,4) to the double and triple bonds respectively.
On the other hand, following a similar numbering process from the left side of the molecule as seen in the diagram will yield (3,1) as the locants for the double and triple bonds respectively.
Now, as per IUPAC Recommendations 1:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.

Therefore, since (1,3) is a lower set of locants as compared to (2,4) as per the above definition, the IUPAC name of the compound should be pent-3-en-1-yne
Note that this molecule can also exhibit stereoisomerism by the virtue of it's double bond, so depending on whether the hydrogens are cis or trans to each other, the complete name of the molecule could be (3 Z)-pent-3-en-1-yne or (3E)-pent-3-en-1-yne, respectively.


How do you stop a Pharoah's Serpent reaction? I am a novelist and working on a book. I have been searching google for an answer to the above question. How does one stop a Pharoah's Serpent reaction? Thanks!

Just put water on it.  A temperature of $\pu{165^\circ C}$ is required to sustain the decomposition. Adding water will remove heat from the reaction and will dissolve a small amount of the mercury thiocyanate. The heat removed by heating and boiling the water will cool the contents below the decomposition temperature.  Adding soap will help the water penetrate into the powder and more quickly stop the reaction.


IUPAC-Naming of Dioxin In the molecule 2,3,7,8-tetrachlorodibenzo-p-dioxin, the "p" means para--because each chlorine is para to an oxygen?
In IUPAC nomenclature shouldn't "para" be replaced by a number?

In the molecule 2,3,7,8-tetrachlorodibenzo-p-dioxin, the "p" means para because each chlorine is para to an oxygen?

No. The p (for “para”) indicates the dioxine isomer.

  In IUPAC nomenclature, shouldn’t “para” be replaced by a number?

Yes. The preferred IUPAC name for this isomer is 1,4-dioxine.

Therefore, the corresponding name of the parent structure of the compound that is given in the question is 2,3,7,8-tetrachlorodibenzo[b,e][1,4]dioxine. The complete name is 2,3,7,8-tetrachlorodibenzo[b,e][1,4]dioxine.

The preferred IUPAC name, however, is 2,3,7,8-tetrachlorooxanthrene.


I left a bottle of water in the fridge for the past three months. I took it out today. There was a block of ice in it. How? How is this even possible?

I know my fridge is not freezing items. The bottle was sitting at the back and I totally forgot about it. I went to empty it and heard a clank and noticed there was ice in it. How does this make any sense? I don't think that the fridge is disproportionately cooling some regions to below zero. And if so, why would the bottle have a mix of that block of ice and water in it then?

The piping for refrigerant circulation is in the fridge walls, so obviously it's colder near the back as there is no heat from the front door reaching that area and heat exchange with compressor/coolant circuit is at maximum.

As for the partially frozen water, this happens to mineral water and water containing any salts in general. As water freezes out, the salinity increases, resulting in freezing point depression for remaining solution. These processes are the basis for fractional freezing.


Calculating the pH of a dilluted aqueous KOH solution Calculate $\mathrm{p}\ce{H}$ of an aqueous solution of $0.01\:\mathrm{nM}~\ce{KOH}$ in water.

I think that water also has $\ce{OH-}$ ions in concentration of $10^{-7}\:\mathrm{M}$ (since $\mathrm{p}\ce{H}=7$)

So the total concentration of $\ce{OH-}$ after mixing the solution with water would be:

$0.01\:\mathrm{nM} + 10^{-7}\:\mathrm{M} = 10^{-11}\:\mathrm{M} + 10^{-7}\:\mathrm{M} \implies 1.0001 \times 10^{-7}\:\mathrm{M}$

and given that $\mathrm{pH} = 14 - \mathrm{p}\ce{OH}$

$\implies 14 - \log(1.0001\times10^{-7})$

$\implies \mathrm{pH} \approx 7$

Does this make sense?

Yes, that's right, the concentration of $\ce{{}^{-}OH}$ from dissociation of water will be much greater than the amount added from $0.01\:\mathrm{nM}~\ce{KOH}$.

$0.01\:\mathrm{nM}~\ce{KOH}$ will not significantly the affect $\mathrm{p}\ce{H}$ of water.


What happens to the "leftover" Ions in the Mass Analyzer I am very intrigued by what happens to the ions which do not make it to the detector in mass spectrometers, do they just pile up? I have never seen anyone clean our quadrupole, is this done? or do they just stay there indefinitely? I am talking of a triple quadrupole detector in a GC instrument, which deals with very small quantities, but after thousands of injections there must be a considerable mass of analyte on the analyzer. Even more so considering that most ions which go through the quadrupole do not make it to the detector.
 Thanks.

Most would end up being sucked out though the vacuum pump with the other residual gases. I assume that a lot of them collide with the sides of the vacuum chamber and are neutralized before sublimating in the vacuum.

The quads can become contaminated and require cleaning after several years of heavy use. Because there are such small quantities at that stage of the process, it's not uncommon for an MS to reach the end of it's service life before the quads need to be cleaned.


Why does Sodium Hydrogen Sulfate (Sodium Bisulfate) produce moisture? In my chemistry lab, I have a vial of Sodium Hydrogen Sulfate (Sodium Bisulfate). I've noticed that when I take a spoonful using a spatula, the spatula gets covered in many tiny droplets, and the lid of the vial always has a layer of tiny droplets on the inside.

Does the Sodium Hydrogen Sulfate (Sodium Bisulfate) react in any way to produce this moisture? Is this to do with the salt being hydrous or anhydrous?

Does the Sodium Hydrogen Sulfate (Sodium Bisulfate) react in any way to produce this moisture?

This process is called deliquescence (mouthful word). This process is not specific to your salt, even table salt, NaCl, would do the same in a humid environment e.g. on a rainy day. This has nothing to do with the state of hydrated or anhydrous salt. It is the nature of the compound.

The opposite of this phenomenon is efflorescence. The hydrated salt becomes dry when exposed to air.


Correct flask for reacting and distilling on hotplate? I'm running a distillation reaction at under 100 °C with the heat coming from a hotplate (combined with magnetic stirrer).  Is a Florence flask the correct vessel to use on the hotplate?

Given the small contact area of the flask relative to the surface area of the hotplate I'm wondering if it wouldn't be better to use a fully flat-bottomed flask on the hotplate, and if so what best practices are?  Regardless, I need to keep a stirrer running during the process.

(I intend to have a Kjeldahl bulb on top of whatever flask before feeding the condenser.)

At under $100^\circ \mathrm{C}$ The best will be to use water bath and round-bottom flask. Most magnetic stirrers have enough range (magnetic forces do not stop working if vessel is in couple of centimeters from hot plate, though you should use non-magnetic dish for bath itself). Even if magnet in hot plate is weak I would suggest against flasks with flat bottom, because bottom may fall off under heating.


Why can't a balance that can weigh to the nearest milligram be able to weigh 23.8 milligrams of something? I'm having trouble understanding the answer to this question:

  Most chemistry laboratories have balances that can weigh to the nearest milligram, would it be possible to weigh $5.64\times10^{18}$ molecules of octadecane, $\ce{C18H38}$, on such a balance?

I worked through the equation and my end result was $\pu{0.00238 g}$ which is equal to $\pu{2.38 mg}$ so I thought that the answer was that it could, but the book states that the answer is that it cannot. Why can't it? Am I misunderstanding how weighing to the nearest $\pu{mg}$ works?

Couldn't agree less, but check my math.

First $\ce{C18H38}$ has a molecular weight of $254.5~\mathrm{g/mol}$.

Second $\mathrm{5.64\times 10^{18} ÷ 6.02214 \times 10^{23} = 9.3654 \times 10^{-6}}$ and so is $2.4\ \mathrm{mg}$.

A balance with an ability to weigh to the nearest mg would, in a perfect world give you a mass of 2 or possibly 3 mg. This wouldn't be very useful, but it wouldn't give you zero information, either. So, since I don't understand the context in which your book asked the question, I'd have to say it doesn't seem to be completely correct if it states "it cannot" weigh that amount. It certainly cannot weigh that amount with adequate precision for most purposes I can think of. But if, say, you were trying to determine whether there was enough of a sample to use in a mass spec., or to calculate the yield of a reactor-on-a-chip, then it would suffice.


Gel electrophoresis separates proteins on the basis of what property? In SDS-PAGE for separation of proteins, SDS is an ionic detergent that coats the unfolded protein to mask its native charge and give at a uniform charge to length ratio. Since the length of the protein is proportional to its molecular weight, the charge to mass ratio of all the loaded proteins is the same.

To quote this site:

  Since the SDS-coated proteins have the same charge to mass ratio there will be no differential migration based on charge.

Similarly in a DNA agarose gel electrophoresis, the charge to mass ratio remains constant. So what is the separation based upon?

Proteins having larger size will have more SDS bind to it and thus more negative charge than to the smaller proteins. While proteins of higher molecular weight also have lower mobility in the gel. So is this some kind of a trade off between the charge and the size? If charge to mass is constant, what is the property that separation is based on?

What does the above quoted statement mean? Why not differential migration based on charge? Doesn't mobility through gel depend on the quantity of charge?

It is probably a good thing to compare protein electrophoresis with DNA electrophoresis here. For DNA, you know that it has a phosphosugar backbone that comes with a negative charge at ambient $\mathrm{pH}$ per phosphate group. This means, that the charge-to-mass ratio, as the quote implies, is similar, no matter how long the DNA fragment is. (Not identical, because different bases have different masses and DNA typically has a seemingly-random base distribution.) Yet they still travel at different speeds.

This is because the gel is essentially a spider’s web for the DNA fragment. It somehow needs to find the pores of the gel and fit through. These pores have different, randomly distributed sizes and are not all aligned perfectly as you would assume for a crystal structure. Therefore, this process requies some movement perpendicular to the direction of charge. The larger a DNA fragment is, the harder it is to constantly change directions to fit through the misaligned (and potentially small) pores, so the speed of diffusion is slower. Hence, low-molecular weight DNA fragments wander far through the gel while high-molecular weight fragments remain closer to the starting point. Since size and molecular weight nicely correspond to the length in bases, markers are often labelled in terms of $\mathrm{kb}$ — kilo-base pairs.

Similar concepts apply to SDS protein electrophoresis. By denaturing the proteins with the detergent SDS (sodium dodecyl sulphate), they are linearised and equipped with a strong negative charge from the sulphate groups. Much as is the case for DNA, the number of dodecyl sulphate molecules is approximately proportional to the length of the protein in amino acids. And since they are linearised, i.e. their tertiary and quaterneray structures (and to a certain extent also their secondary structures) are destroyed, they very much behave like a DNA fragment when forced through a gel: the shorter they are the better they fit through and the faster they travel.

In the protein case, however, the markers are typically not labelled as a certain number of amino acids but as corresponding to a certain molecular weight (often given in $\mathrm{kDa}$). This is because amino acids vary greatly in size from the very small glycine to the very large tryptophane or arginine. The vicinity of one of those larger amino acids may allow for an extra dodecyl sulphate to be added when compared to the vicinity of small amino acids.

But why don’t larger proteins/DNA fragments that carry a greater negative charge move through the gel faster? Isn’t a greater force being applied onto them?

I’m glad that you asked me that. Yes, a greater force is applied. But, as you may remember from physics classes, $F=m\cdot a$, so consequently $a = \frac{F}{m}$. Therefore, the larger molecule may experience a stronger force, but due to the molecule’s higher mass, the overall acceleration is equivalent (again: not identical but similar). This allows the size-discriminating gel network to do the sorting by size.


What is the the full name of TAPD? Structure provided I have a substituent called TAPD on a molecule I'm working with. I have a hard time locating information about its properties. The bold square in the picture below is supposed to be the rest of the system on which the TAPD is attached. I am a physicist and thus not very good at chemical naming, so I would have good use of some help.

TAPD = Tetraalkyl-p-phenylenediamine

This is because you have a phenyl (benzenoid) ring, with two amino groups (diamine) opposite (para) to each other, and each of those amino groups has two alkyl substituents, giving four (tetra) in total.


What is the correct numbering & order for this molecule? I'm having a hard time understanding why the numbering for this molecule is: 4-bromo-1-tert-butyl-2-methylcycloheptane, instead of: 1-bromo-4-tert-butyl-3methylcycloheptane. Is it just because the overall numbers are lower? I just figured the numbering out by going in alphabetical order, which I now know is wrong. But is that the reason the first numbering is correct?

The parent hydrocarbon chain for this molecule will be the 7 membered ring.

2.There are no functional groups with suffixes.
3.A halogen substituent(Bromine) and two side chains are present. ( methyl and t-butyl or (1,1-dimethylethyl) )
Having identified all these we can proceed to numbering the parent hydrocarbon chain.
After numbering the Parent hydrocarbon chain in all possible ways(ideally) we choose that numbering that satisfies the following criteria:
Numbering of the chain. This is done by first numbering the chain in both directions (left to right and right to left), and then choosing the numbering which follows these rules, in order of precedence.
1.Has the lowest-numbered locant (or locants) for the suffix functional group. Locants are the numbers on the carbons to which the substituent is directly attached.
2.Has the lowest-numbered locants for multiple bonds (The locant of a multiple bond is the number of the adjacent carbon with a lower number).
3.Has the lowest-numbered locants for prefixes.
In the above molecule since there are no suffix functional groups or multiple bonds we can apply criterion 3.
After numbering the parent chain all possible sets of locants are written down in the ascending order.(in this case)
eg: If the set of locants are (2,4,3) and (1,4,3) we write them as (2,3,4) and (1,3,4)
Then choose the one with the lowest possible locants. First compare the first locant, choose the ones with the lowest locant, among these sets of locants compare the second locant, choose the ones with the lowest locant and so on...
eg: If the sets of locants are (1,3,8) and (1,4,5) we choose (1,3,8)
It is only when we get two different numbering for the same compound following all the above mentioned rules that we should choose the set of locants which assign the lowest locant to the substituent or side chain on the basis of alphabetical precedence.


How to calculate the oxidation number of sulfur in H2SO5? I am in class 11th and I am having trouble solving the question.

  Calculate the oxidation number of sulphur in $\ce{H2SO5}$?
  (The answer is given as as $+6$.)

$\ce{H2SO5}$ exists it is named as "peroxy sulfuric acid".

How tried it:
\begin{align}
2(+1) + 1(x) + 5(-2) &= 0\\
2 + x - 10 &= 0\\
x&=\pm8
\end{align}

$\ce {H_2SO_5}$ has an oxygen-oxygen bond. This means that two of the five oxygen atoms have an oxidation number of $-1$. Same case as in $\ce{H2O2}$.

$$2\cdot (+1) + 1\cdot(x) + 3\cdot(-2) + 2\cdot (-1) = 0\\2 + x - 6 - 2 = 0\\
x = +6$$

You have 3 oxygens with oxidation number $-2$ and 2 oxygen atoms with $-1$.


How can I prepare Cu2S powder? How can I prepare $\ce{Cu2S}$ powder?

I have some idea about preparing the ternary composition containing copper, but copper sulfides are difficult due to the large difference between their melting points (Cu: 1084°C, S: 119°C).

It is possible to obtain $\ce{Cu2S}$ by heating a mixture of finely-powdered elemental copper and sulfur in a crucible up to temperatures of above 450 °C. The powdered form of the educts ensures that the reaction starts as soon as the sulfur has melted. Alternatively, the mixture can be milled in a planetary ball mill for 2 hours, which yields $\alpha$-$\ce{Cu2S}$. However, $\ce{Cu2S}$ produced by these methods has limited long-term stability, as it slowly converts into $\ce{CuS}$ even when left in an inert atmosphere (reference).


Time taken in changing surface tension I had read some where that in glasses the surface tension is high, but in a lota the surface tension is low.

My question how much time does the surface tension take to change? I mean that there was water in a glass but I put it into a lota in how much time will the surface tension of water change? Immediately?
This is a lota:

Just as a quick answer, because apparently some people are in doubt so let me try to clean up the confusion.
The 'physics part'
Surface tension that people normally talk about is the liquid-air surface tension, in this case water-air, and I am pretty sure this is the surface tension you are referring to in your question. This water-air surface tension is a property of the water and air and has absolutely nothing to do with the physical shape or material of the container.
For completeness, I have to mention that the contact with the walls that have different chemical properties will change the shape of the water-air interface a bit, which you might misinterpret as a change in water-air surface tension but this is caused by the water-container and container-air surface tensions which indeed DO depend on the container.
The 'chemistry part'
What could * change the water-air surface tension is the exchange of some material out of the container into the water or the other way around. This would mean that you are not really looking at the water-air surface tension anymore, but rather at the (water$+$some solute)-air surface tension. Which might indeed be different, typically lower.
If this exchange is the mechanism then the time over which the surface tension will change is dependent on the kinetics of that process and without knowing what is being exchanged it is impossible to tell you what the time span will be, it could be seconds, it could be hours. Note that the explanation with stuff leaking out of or into the lota would mean that (assuming there will be equilibrium at some point) storing water in the lota and then drinking it from the glass would be the same as drinking it directly from the lota.

*I'm not saying it does I have no clue whether this happens with a lota


The fifth shell of an atom According to the $2n^2$ rule, the fifth shell should accommodate 50 electrons.

However, my chemistry book says that the $2n^2$ rule does not apply to the fifth shell.

Could you explain why this happens?

P.S. I have studied quantum numbers, Heisenberg uncertainty principle, Pauli exclusion principle, and the quantum mechanical model of an atom; please answer accordingly.

Thank You

The $2n^2$ rule is nothing but a mathematical consequence of the mathematical description of electrons around an atom – what chemists call quantum theory and is currently the most accepted description of atomic matter by a mile. Thus, until quantum theory is significantly amended, the fifth shell does, in theory, have the capacity to hold $2\times 5^2=50$ electrons.

I say in theory, because shells aren’t filled all at once bottom to top. In fact, the $n^\text{th}$ shell has $n$ subshells denoted by the letters s, p, d, f, g, h, … which in neutral atoms are (usually; there are exceptions) filled in a specific order known as the aufbau principle. For the first two shells this doesn’t have any effect, but the 4s shell is filled before the 3d shell and the 6s shell before the 4f shell, to name just two examples.

To completely fill the fifth shell (i.e. fill the 5g shell), we would need to also fill the 6s, 6p, 6d, 7s, 7p and 8s shells. The heaviest known (i.e. synthesised according to the standards laid out by IUPAC and IUPAP) element as of 2019 is oganesson (Og) which also happens to be the last 7p element. Two heavier elements (the 8s elements) must be discovered before the third in line will most likely be the first 5g element. The g subshell can hold 18 electrons, so until we have one with a completely filled 5g shell some time will pass.


How to make ammonia from urea? The chemical formula of urea is $\ce{CO(NH2)2}$ or is sometimes written as $\ce{CH4N2O}$. How would I decompose or precipitate the $\ce{CO}$ from urea to get ammonia? I know that in the olden days people used to make gunpowder (potassium nitrate) and ammonium nitrate from human urine, which was made by reacting ammonia and nitric acid, so it should be possible. (Just as a side note: this is for a high school chemistry project, I am not attempting to make explosives out of this).

Short answer: use crushed watermelon seeds (urease)
Combine urea in solution with urease, get some ammonia with carbamate intermediary, which rapidly decomposes to ammonia and carbon dioxide.
$$\ce{(NH2)2CO + H2O ->[urease] NH3 + H2NCOOH -> 2NH3(gas) + CO2(gas)}$$
Of course the ammonia gas dissolves in the water and increases the pH.
Sources:

https://www.hemmaodlat.se/research/citrullus%20lanatus%20seeds%20as%20a%20urine%20catalyst%20for%20anthroponic%20use.pdf
https://en.wikipedia.org/wiki/Urease
https://en.wikipedia.org/wiki/Ammonia_volatilization_from_urea


Anomalous trends in ionization energy I want to address two exceptions in the trend of ionization energy across the period, that are causing me problems:

Taking the second period as an example :

The two exceptions from the general trend are the ionization energies of B lesser than Be and that of O less than N.
My teacher told me the reason to both was that half filled and fully filled orbitals of N and Be are more stable and hence require more energy to pull off an electron. He also said that this affects only the next element in the period for a change in trend( he never said why though).
This place and my textbook tells me that the reasons are:

1) Be,B :

The electron is to be removed from the 2p orbital in B and the 2s orbital in Be. The 2s orbital penetrates more into the nucleus and hence it is more difficult to remove an electron form Be than  from B.

Question: In that case why doesn't this apply to C? Shouldn't it be easier to remove one of the electrons from C's 2p orbital than Be's 2s orbital just like the B's 2p orbital? And so shouldn't the ionization energy of Be must be more than C too?

2) N,O:

N 1s2 2s2 2px1 2py1 2pz1
  O 1s2 2s2 2px2 2py1 2pz1

The reason I have understood here is that there are two electrons in the 2px orbital of O and because of electron-electron repulsion, it is easier to remove an electron from here as opposed to the 2px orbital of N which had only one electron.

Question: Why doesn't this apply to F?

F 1s2 2s2 2px2 2py2 2pz1

F has paired electrons too like O in the px and py subshells. Wouldn't there be repulsion here too? And hence why isn't it easier to remove electrons from F than is it to from N which doesn't have any paired electrons at all? Why is the ionization energy of F more than N instead?

The same problems can be seen in the next period.I hope they can be sorted out with the same logic that i am missing for this period.

Finally, was what my teacher said just a simplification of everything I found in the link and the textbook? Was he trying to convey the same thing? Or are they two complete different explanations?

OK. Lets get a few things cleared up first. The main factor affecting ionizing energy (IE) is the attraction between the nucleus and the electrons. The more the attraction, the more energy is needed to remove an electron from it's orbital and hence higher the IE. The other factors modify this main factor and thus change the expected IE values to the anomalous observed values. Now coming to your questions:

As we go to the right of a row, size of the atoms get smaller. $\ce{C}$ is smaller than $\ce{Be}$ and $\ce{B}$ and hence has higher IE.
The same thing happens in case of $\ce{F}$. The size of $\ce{F}$ is smaller compared to $\ce{N}$ and this factor over powers the repulsion between the paired $p$ orbital electrons.


Can a substance donate a proton without the proton getting accepted by another substance? I have read that the concept of Brønsted and Lowry is based on the fact that acid-base reactions involve the transfer of $\ce{H+}$ ions from one substance to another, and that an acid is a substance that donates a proton to another substance, but:

can a substance donate a proton without the proton getting accepted by another substance? So for example:
$$\ce{AH -> A- + H+}$$
without reacting with another substance?

If yes: Is this a Brønsted-Lowry acid?

The closest thing I can find to this involves a most unlikely proton donor.  Given an appropriate base, sodium hydride donates a proton for encapsulation in the base (rather than forming a localised covalent bond), leaving the valence electrons on sodium (!) in inverse sodium hydride.  The proton is not free (a completely free proton is possible only in plasma), but analogous to complexed metal ion or solvated electron.


Equilibrium: Effect of change in composition and value of Kp, Kc on changing pressure Di-hydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

  $CH_4{_{(g)}}+H_2O_{(g)} \ce{<=>} CO_{(g)}+3H_2{_{(g)}}$

  How will the values of $K_p$, $K_c$ and composition of equilibrium mixture be affected by increasing the pressure ?

Now,

1) According to this answer $K_c$ only changes on changing temperature i.e. unaffected by pressure.

2) According to Wikipedia the article "Pressure dependence" says, "...However, the composition at equilibrium will depend appreciably on pressure when:

1)the pressure is changed by compression or expansion of the gaseous reacting system, and
2)the reaction results in the change of the number of moles of gas in the system....."

here wiki is using term $K_p$ to denote equilibrium constant. Now the "composition of equilibrium" means $K_p$ and not $Q_p$ because its taking at equilibrium, so the $K_p$ changes with pressure.

3) I have studied that on increasing pressure the reaction shifts the side with less number of sum of  gaseous atoms in the reaction formulae. I know from this question that composition of equilibrium is different from "the side at which the reaction shifts". So it seems my study isn't useful here in deciding the composition of equilibrium

So my question is how will the values of $K_p$, $K_c$ and composition of equilibrium mixture be affected by increasing the pressure ?
Does one change and the other remains unaffected and isn't the composition of equilibrium directly related to $K_p$ and $K_c$

Are dependencies of $K_p$ and $K_c$ different for change in pressure.

$K_p$ will not change with pressure, only with temperature. $Q_p$ will instantaneously increase upon applying the pressure for this reaction (why? see below), and will drive the reaction to reactants to re-achieve equilibrium (decreasing $Q_p$) until $Q_p$ = $K_p$ once again. From a LeChatelier approach, you  have more gas molecules in the products, so an increase in pressure will drive the equilibrium composition to the reactants.
As for why $K_p$ won't change with pressure:
$\ln K_p=-\frac{\Delta G^⦵}{RT}$ is at a specified $p^⦵$ (usually 1 bar), so $\left(\frac{\partial {\ln}K_p}{\partial P}\right)_T = 0 $.
And as for why $K_c$ won't change with pressure, see the answer here


What are the effects of Lithium Dioxide on the body? My question is about how much science-fiction and how much chemistry is involved in the the 'Randy's Donuts' scene of Iron Man 2:

  Agent Romanoff injects Lithium Dioxide into Tony Stark's neck to remove the symptoms of Palladium poisoning.

My Question is: Is it safe to inject Lithium Dioxide into our body?
Does it have any side effects?

Is it safe to inject Lithium Dioxide into our body?

No. It will do bad things.

  Does it have any side effects?

Yes. It will do bad things.

As lithium is an alkali metal, it does not usually form covalent bonds, so we should treat whatever it makes as ionic. The common form of lithium oxide is $\ce{Li2O}$, which is not that scary. "Lithium dioxide" would be $\ce{LiO2}$. This implies that that lithium is tetravalent, as $\ce{Li^4+}$. This is not possible, as it requires lithium to give away 4 electrons but it does not have 4, only 3. This means that oxygens are in the superoxide form: is $\ce{O2-}$ instead of the more stable $\ce{O^2-}$. That is definitely not stable in isolation at ambient conditions (unlike other alkali superoxides) and I'm not sure where you will get that stuff.

The result of that is that one molecule (or whatever it is) of $\ce{LiO2}$ will really want to convert itself to $\ce{Li2O}$. The superoxide ion is not going to do fun things. For each $\ce{O2-}$ you're going to need 3 more electrons to make it stable, so that is very strong oxidiser.

As a comparison to what this could do to you, I would guess it would be worse than injecting pure hydrogen peroxide (you have about 5% of that in hair bleach), or pure sodium hypochloride (similar 5% in kitchen bleach). This is definitely not going to treat "palladium poisoning" because it will probably kill you beforehand.

That said, using an oxidiser (such as lithium dioxide) to treat noble metal poisoning (such as palladium) seems dubious to me. I'm not a doctor, but I reckon that there are better method to treat that. So why lithium and palladium? I guess is sounds more credible than "using titanium to treat potassium poisoning" but less credible than "using dysprosium to treat beryllium poisoning", both of which make absolutely no sense as "using lithium to treat palladium poisoning".


Carboxylic acid strength due to the inductive effect Suppose we have a mono carboxylic acid . In that if attach a +I group then sigma electrons will shift towards the compound and O-H bond will become weaker.  Hence H bond can easily break so it should be more acidic but it is not i.e greater the +I effect less the acidic nature

The strength of the carboxylic acid (and any other Bronsted acid, for that matter), is related to the 'stability' of its conjugate base, the carboxylate anion. The carboxylic acid and the carboxylate anion are in equilibrium with one another, and the relative acidity of carboxylic acids depends upon the position of this equilibrium.

One of the main reasons why carboxylic acids are acidic is due to the ability of the charge to be delocalised around the pi-system:

However, in addition to this delocalisation about the pi system, extra stability can be gained by the presence of electron withdrawing groups adjacent to the carboxylate group (i.e. -I groups attached to the R group in the diagram above).

These electron withdrawing groups draw electron density away from the carboxylate anion via the sigma system (the single bonds). This is advantageous as the stability increases with the number of atoms it can be spread across (i.e. how diffuse the charge is).

Electron releasing groups (+I as you call them) destabilise, by increasing the electron density around the carboxylate group, which of-course makes it less favourable for the carboxylic acid to lose a proton and hence makes it less acidic.

The +I groups aren't involved in breaking the O-H bond to any great extent, as you allude to. This is because when the proton leaves, it transfers its electron density to the oxygen it was attached to, cleaving the bond and allowing it to leave as H+. If the +I groups were donating into this bond, you would have to have the H leaving as H-, which is rarely observed.


What properties of later elements are predicted by scientists? Though scientists have predicted up to element 172 (unseptbium) what properties do they predict any elements later on the extended periodic table may have?

Though elements past 92 are quite unstable, so that chemical properties are difficult to determine, heavier elements behave more like metals, since the outer electrons are well shielded from the attraction of the positive nucleus and will therefore be easily lost.

Another predictable property is specific heat, which decreases with increased atomic mass.


Monobromination vs. Dibromination My professor noted that bromination with light and methylene chloride gives you exclusively the monobromination product.

However, bromination without light will give you the multi-brominated product.

I have no found any reference to bromination in my textbook nor online. Is the second statement an accurate statement?

Bromination of alkanes with Br2 will not occur in a dark environment. In the presence of light, multiple halogenations can occur given enough time.

For more information, including the full mechanism, check out: https://en.wikipedia.org/wiki/Free-radical_halogenation


how to separate a mixture of sodium chloride and sodium hydroxide I recently did an double displacement reaction in that I have formed a solution of water with sodium hydroxide and sodium chloride are dissolved in it. now I want to separate both of them

One solution would be to reduce the amount of solvent (under vacuum for instance), then filter out the precipitated NaCl at high temperature (at higher temperatures NaOH has as much as >10x the water solubility of NaCl).
The common ion effect will play in your favour, as the high water solubility of NaOH at higher temperatures will help you reach NaOH concentrations that strongly disfavours NaCl solubility.
The way to go may be to first reduce the solvent mass under vacuum at acceptably safe temperatures, and once the desired mass is reached redissolving most of the precipitated NaOH by pushing the temperature up before filtration to a level at which most NaOH has redissolved.
Use appropriate protection equipment and contact material to handle very hot NaOH.


Why is my NaCl solution seemingly saturated, when I followed the recipe for an isotonic solution? I prepared an isotonic solution of table salt in water following this recipe: 1 leveled coffee spoon of salt in 0,5L of water.

9g/L would be an isotonic solution, a leveled coffee spoon is commonly given as 5g - close enough. I use water from the tap, which was freshly boiled in a water cooker. The water is not chlorinated, I don't know the hardness.

However, even given hours of time, a significant amount of salt doesn't go into solution - visible as turbidity/precipitation. How can this be?

I need the solution to flush my child's nose, so I don't want to use too much salt.

Salt added to food could be sea salt, common table salt or iodized table salt.

Common Table salt typically contains anti-caking agents (desiccants) such as magnesium carbonate, calcium silicate, dicalcium phosphate, tricalcium phosphate, calcium carbonate and sodium alumina silicates.

Iodised table salt may contains added potassium iodide for people who may otherwise have low consumption of this nutrient which is essential for proper thyroid function. Since iodide will slowly oxidize in moist air, stabilizing agents such as sodium thiosulphate or dextrose are also added. Some will also contain a base such as sodium bicarbonate, disodium phosphate or sodium pyrophosphate to also inhibit oxidation of iodide. There may also be added sequestering agents which bond to trace metal ions present as impurities to prevent them oxidizing the iodide.$^{[1]}$

The presence of magnesium and calcium will add "hardness" to the water which could cause precipitates such as calcium carbonate and magnesium carbonate to form.


Why does the image of cyclo[18]carbon look like a nonagon? The $\ce{C18}$ allotrope cyclocarbon has been synthesized and imaged.[1] Science has most details behind a paywall, but this discussion includes an image:

In this octakaidecagonal molecule, each $\ce{\color{blue}{C}}$ is bonded viz. $\ce{C-\color{blue}{C}#C}$. Yet the above image looks like a nonagon. Why does every other carbon atom stand out as an obvious vertex while the others don't? My thoughts:

The internal angles are close to $180^\circ$, making vertices hard to see. However, I wouldn't expect the internal angles of this octakaidecagon to differ much.
I wonder if this molecule has a delocalised ring analogous to the one in benzene. On this idea, each of the nine visible edges could alternate between the states $\ce{C-C#C}$, $\ce{C#C-C}$. But this wouldn't explain why "odd" vertices have one appearance while "even" ones have another. I'm not convinced of this idea anyway, because it would average out to $\ce{C=C=C}$, unlike the $1.5$-bonds in benzene.

References:

Kaiser, K.; Scriven, L. M.; Schulz, F.; Gawel, P.; Gross, L.; Anderson, H. L. An sp-hybridized molecular carbon allotrope, cyclo[18]carbon. Science 2019, eaay1914. DOI: 10.1126/science.aay1914.

The first thing to say is that I'm not sure where that image is taken from; it's neither in the original article nor in the supporting information to the article. Therefore, it appears to be more of an "artist's impression" rather than an actual atomic force microscopy (AFM) image, which is what was reported in the paper.

Nevertheless, the actual AFM images of $\ce{C18}$ are in Figs. 3Q and 3R. They are referred to as "AFM far" and "AFM close" respectively because of the height of the probe ($\Delta z$):

One can indeed see that there is 9-fold symmetry (technically, $D_\mathrm{9h}$). This implies that $\ce{C18}$ has a 'polyyne' structure in which there are two different types of bonds $\ce{-C#C-C#C-\phantom{}}$, rather than a 'cumulene' structure in which every bond is equivalent $\ce{=C=C=C=C=}$ (prior to this, computational studies had been equivocal as to which form was more stable).

The bright spots within the ring do not correspond to carbon atoms, but rather to carbon–carbon triple bonds. This is consistent with the AFM images obtained for other similar intermediates in the synthesis of cyclo[18]carbon. In the authors' own words:

  Assigning the bright features in the “AFM far” images to the location of triple bonds, we observed curved polyyne segments with the expected number of triple bonds: 5 in $\ce{C22O4}$ and 8 in $\ce{C20O2}$. At small tip height, we observed sharp bond-like features with corners at the assigned positions of triple bonds and straight lines in between. This contrast was explained by CO tip relaxation, in that maxima in the potential energy landscape, from which the tip apex was repelled, were located above the triple bonds because of their high electron density. In between these maxima, ridges in the potential landscape led to straight bond-like features.

(The two bright spots outside the ring are due to individual $\ce{CO}$ molecules.)


Can I extract water from diethyl ether to get anhydrous diethyl ether? I am trying to get anhydrous diethyl ether. So, I am considering using a condensor to heat up diethyl ether and extract water. Would this method work?

No it would not work.  Ether boils at a lower temperature than water.  Plus, it forms an azeotrope with water:  a composition of 98.7 % (by weight) ether boils at 34.2 °C and the vapor phase has identical composition as the liquid, meaning that distillation cannot remove the final 1.3% of the water (at atmospheric pressure).

The standard way to remove water is (i) first to use drying agents such as magnesium sulfate to remove most of the water, and then (ii) if needed, eliminating the final traces of water with small amounts of sodium (or potassium) metal.


Identifying coordinate bonds How can we know whether a compound has a coordinate bond or not? I know that I have to draw the Lewis Structure for the compound. But what to do after that? Is there anything to do with formal charges?

The first thing needed for a dative bond is a lone electron pair in one atom and an empty orbital in another atom.
This condition is necessary but enough.

There is also a formula that fails in many cases but may sometimes become handy:
Formal Charge = (Oxidiation State of the centeral atom - 8 + Digit of its group number)/2

e.g.: For SO3 it would be: (6-2)/2 Which means that it has 2 co-ordinate bonds.
However for a compound like NH4+ it won't work.

Of course, safest way is to find out how the compound had formed in the first place.
e.g.: If we know that BF3 + F- made BF4-, we would find out that F- had shared with B two of its electrons, filling the empty orbital belonging to B.


Allotropes of Phosphorus Something I noticed whilst revising about group 15 is that the darker the allotrope of phosphorus, the more stable it is.

white/yellow – very volatile, soluble, pyrophoric
red – air stable, polymer, more stable, one cross link;
black – most stable, polymer, maximum amount of cross link.

What's the reason for this?
Or rather, why would a more stable allotrope have a darker colour?

In phosphorus’ case, the connection between stability of the allotrope and its colour is only accidental.

White phosphorus, i.e. $\ce{P4}$ molecules, has very strained bonds because of the very small tetrahedral angles the atoms need to accomodate for. Going by the edges of a regular tetrahedron, the bond angles are $60^\circ$ which is very strained. This is the reason for its reactivity. Its colour is white, because there are no electron transitions excited by visible light, much like similar compounds of other elements.

Black phosphorus, on the other hand, forms large extensive networks which are well on the way to being metallic. Metallic bonding is typically such a widespread bonding network that every atom contributes to every molecular orbital with more or less the same contribution. You can compare this to large organic π systems. This allotrope is more stable because each phosphorus atom has bond angles that well suit the corresponding angles between atomic orbitals (somewhere just above $90^\circ$) and because that large network is hard to break. And since we have a lot of molecular orbitals with rather small differences in energy, light of the entire visible spectrum can be absorbed. There will always be a corresponding transition. Because of this, black phosphorus appears black.

Red phosphorus is somewhere in-between. The network is not as extensive, it is not as ordered and thus there are not as many molecular orbitals with good energy differences. Therefore, not all visible light is absorbed but green light is preferentially absorbed (giving a red colour). And also therefore, the reactivity is somewhere in-between white and black phosphorus.

(There is also violet phosphorus, which is somewhere between red and black.)


Oxidation state of Mn in MnO₄⁻ I have tried to understand what the oxidation state for Mn in $\ce{MnO4^-}$ is. So far i have understood that O have the oxidation state of -2 so because there is 4 O the number turns to -8. What happens to the minus at O? and what is the oxidation state for Mn?

The sum of the oxidation states of all of the atoms needs to equal the total charge on the ion.

For example, for $\ce{NH4+}$, if each hydrogen atom has an oxidation state of +1, and the overall charge is +1, then we can solve for the oxidation state of nitrogen:

$$\begin{aligned} 4(+1) + OS_\ce{N} &= +1\\
OS_\ce{N} &= +1 -4(+1) \\
OS_\ce{N} &= -3
\end{aligned}$$

Similarly, we can do $\ce{SO4^2-}$. If each oxygen atom has an oxidation state of -2, and the overall charge is -2, we can solve for the oxidation state of sulfur:

$$\begin{aligned} 4(-2) + OS_\ce{S} &= -2\\
OS_\ce{S} &= -2 -4(-2) \\
OS_\ce{S} &= +6
\end{aligned}$$


What is the displayed formula for methylpropene? I am currently studying organic chemistry at school and I came across a question in my textbook which asked me to find the full displayed formula of methylpropene. I knew that there would be three carbons because of the 'prop' and that there would be a one-carbon side chain on the first carbon. However, in my chemistry textbook the one-carbon side chain is on the second carbon (the middle one). I don't know whether this is correct or not. Could someone please explain to me what the correct answer is?

"Propene" is what defines the main chain of three atoms so that the maximum number of carbon atoms in any other chain cannot exceed this number. Hence your proposition that

  there would be a one-carbon side chain on the first carbon

is wrong as in this case there would be a 4-membered chain of either 1-butene or 2‐butene:

In order to comply with the methylpropene name, one can only attach methyl group to the second carbon, and not to the terminal one:

To sum it up, your textbook is correct.


How to calculate bond enthalpies from first principles? Is it possible to calculate bond enthalpies using the periodic table and the molecule's geometric structure? I am interested in a way of calculating this without getting the answer from measurements or pre-existing bond enthalpies.

Firstly the following basic facts can be helpful :

Ionic bond is stronger than covalent bond in most cases .
A metal non-metal bond is more ionic.
Non-metal non-metal bond is covalent in nature.

By considering Valence Bond Theory , we can draw the fact that overlapping of larger orbitals ( usually of large atoms ) with smaller orbitals is less efficient and will result in a weaker bond .

For example ,

HF>HCl>HBr>HI is the order of their bond strengths .

This is because of the fact that as we go down the group , atomic size increases and so does its p-orbital which takes part in the overlapping with small s-orbital of H , making it a weaker bond .

Electronic configurations of various elements ( which is derived from the atomic number provided in the periodic table ) can be used to make the Lewis structures of most molecules and then the number of covalent bonds can tell comparitive enthalpies. ( more the bonds , more is the enthalpy )

Also , some Periodic tables try to be more informative and provide Pauling EN alongside the atomic number and mass . In that case , bond polarity can be measured by calculating the EN difference . Usually , more the polarity of the covalent bonds, more they act like an ionic bond and are more stable .

Though these tips can be of help in most comparisons , relying on actual values cover anomalous behaviors and exceptions which bump along our ways .


How to find chemical market value I would like to retrieve pricing information for specific chemicals. Is there a centralized resource for this?

Edit: I want to clarify that this type of data is to conduct a techno-economic study. I know that for certain markets there are centralized databases which collect prices, and report price statistics, I was wondering if there is something similar for the chemical industry.

Until 2005 or 2006 the best source was Chemical Market Reporter.
It had the form of a newspaper and was edited once a week.
The last pages listed market prices for commodities, some speciality chemicals and also agricultural products.

Hundreds of products were listed and you had some idea of what were the options you had for designing an industrial process.
You could easily decide which solvent to use for an extraction, which metal for a reduction or which raw material for a synthesis.

The publication was on subscription base but it was available in some University libraries or you could get an older copy from a contact in the big chemical industry.

Now this has been converted on a pay per product information scheme.
You can enter in www.icis.com and ask a report for one product.
If you want to compare prices you have to buy the information for every product.
It is almost impossible to have a complete picture of the economic trend in the Chemical Industry.


Why several of isomers of octane such as 2,3-dimethylhexane and 2,4-dimethylhexane doesn't have a value for melting point from any databases? This is a very rough data (as shown below) I have collected from CRC Handbook of Chemistry and Physics 93rd edition in black and ChemSpider in blue. As you can see, I am not quite done, but I was wondering why several of the isomers of octane such as 2,3-dimethylhexane and 2,4-dimethylhexane doesn't have a value for melting point from both databases? (I have tried searching up the melting point for each on a reliable database and have come up with no value given. Wikipedia does give a melting point, but I am very cautious about using that data as it does not a source to prove it.

I recommend to include more than one (secondary) reference and to widen the querry -- if there is access -- to general chemistry data bases like Chemical Abstracts (by the American Chemical Society), or Reaxys (by Elsevier), and Springer Materials, too.

In addition, do not stick to one chemical name, and equally include in your querry unique identifiers like the CAS registry number -- applicable regardless how the organic nomenclature is applied (including databases in other languages) -- rather than one chemical name.  By doing so, you may rapidly find addtional data, for example among chemical suppliers like Sigma-Aldrich, that lists 2,3,4-trimethylpentane (CAS-RN 565-75-3) in its catalog with a melting point of −110 °C and a boiling point of 113-114 °C.

Beside electronic ressources, consider additional ones available in print, like volumes in Landolt-Bornstein dedicated to condensed matter (now included in Springer Materials).


Is Heisenberg uncertainty principle really a demerit of Bohr model? My teacher taught that Heisenberg's uncertainty principle is a demerit of Bohr atomic model.
But as the below paragraph from the article What Is Heisenberg’s Uncertainty Principle? says and as per my thinking Heisenberg's uncertainty principle tells about the uncertainty, if we try to do some practical observation; while what Bohr told us by his formulas was almost fully theoretically derived. The paragraph basically tells that when photon strikes any subatomic particle then change in velocity occurs and uncertainty in position is incurred:

One way to think about the uncertainty principle is as an extension of how we see and measure things in the everyday world. You can read these words because particles of light, photons, have bounced off the screen or paper and reached your eyes. Each photon on that path carries with it some information about the surface it has bounced from, at the speed of light. Seeing a subatomic particle, such as an electron, is not so simple. You might similarly bounce a photon off it and then hope to detect that photon with an instrument. But chances are that the photon will impart some momentum to the electron as it hits it and change the path of the particle you are trying to measure. Or else, given that quantum particles often move so fast, the electron may no longer be in the place it was when the photon originally bounced off it. Either way, your observation of either position or momentum will be inaccurate and, more important, the act of observation affects the particle being observed.

Is Heisenberg's uncertainty principle really a demerit of Bohr model?
Reference

Jha, A. What Is Heisenberg’s Uncertainty Principle? The Guardian. November 10, 2013.

The text is actually describing the observer effect, which is very commonly confused with the Heisenberg uncertainty principle. The latter is a far more fundamental property of the Universe, and has strong theoretical backing (the non-commutativity of certain mathematical operators in quantum mechanics), regardless of experimental setups. The uncertainty relations thus establish a fundamental "fuzz" to any object, precluding its complete and precise characterization even in principle.

The lack of the uncertainty principle is indeed a severe conceptual limitation to the Bohr model (or rather it is better to say that the lack of the groundwork which results in the Heisenberg uncertainty principle is a limitation); whereas the Bohr model still makes predictions in terms of the classical concept of orbits, a more fully-fledged theory of quantum mechanics applied to an atom necessarily predicts the formation of orbitals, where the position and momentum of the electrons is smeared out.

Indeed, the uncertainty principle is ultimately why electrons have stable configurations around atomic nuclei that stay on average around 100 picometers away from the nucleus. The nucleus itself is only around 1-2 femtometers (~0.001 picometers) in diameter, so electrons could have their potential energy greatly reduced if they closely hugged the nucleus at a similar distance. So what's stopping them? If the electrons did scrunch up so tightly against the nucleus, they would essentially be localizing themselves in space very accurately, and thus the fundamental uncertainty in their positions would be very small. However, because of the Heisenberg uncertainty principle, this would cause a very large increase in the uncertainty of the momentum, which in turns causes a very large increase of the average momentum and from there a large increase in their kinetic energy, so that the electrons would ultimately need a higher total energy to stay so close to the nucleus. Again, this behaviour is completely independent of whether the system is being measured or not.

Furthermore, Bohr's model is actually very empirical. In particular, one of its most important considerations is that the angular momentum of the electrons around an atom obey the formula $m_evr=n\hbar$, where $n$ is the positive integer responsible for introducing quantization to his model. However, this is just a postulated equation, with no rigorous derivation. It was chosen because it happened to describe the emission spectrum of hydrogen exceedingly well, so he knew it was right in at least some regime, but had no proof for it.


What is the priority of a phenyl group in a compound such as this? I assumed this:
http://www.chemspider.com/Chemical-Structure.12491.html?rid=222625d4-811c-4b07-b263-8930fafa831a

to be 1-phenylbut-4-ene
Where is the priority here? No source I know of tells me of where the priority of the phenyl  group lies.

It is wrong. The book states it is 4-PhenylBut-1ene.

What is the priority of the phenyl group?

For preferred IUPAC names, the ring is preferred to the chain.
Therefore, the preferred IUPAC name for the given structure is but-3-en-1-ylbenzene.

However, the context may favour the chain, so that, for example, an unsaturated acyclic structure may be recognized.

When one double bond is present, an unbranched alkene is named by changing the ending to ‘ene’. The double bond is assigned the lower locant possible, which is placed immediately in front of the ending ‘ene’. When several structural features appear, low locants are assigned to ‘ene’ endings before detachable alphabetized prefixes.
Therefore, a correct name for the given structure is 4-phenylbut-1-ene.


Is the molar mass of a molecules lowered by the binding energy of the chemical bonds? When calculating the molar mass, is there any measurable difference in total mass of the molecule due to the loss of binding energy?

The difference is the best observed for products of nuclear reactions, related to nucleon binding energy due the strong force.

By comparing the mass of $\pu{2 mol}$ of $\ce{^1H2}$ and $\pu{1 mol}$ of  $\ce{^4He}$, the difference is about $0.7\%$.

The chemical bonds have difference more then million times smaller.

E.g. if there is an exothermic reaction producing $\pu{500 kJ/mol}$, the mass difference per mol is $$\Delta m = \Delta E/c^2 = \pu{-5.56 ng }$$

That means, we have hardly ever scales with the needed accuracy.


Adenine precursor synthesis I am currently exploring the synthesis of xRNA, which contains benzo-expanded purine and pyrimidines. I have found this paper which deals with exactly this subject, and provides the following synthetic pathway for the xA ribonucleotide.

Unfortunately, the first 5 synthetic steps are omitted from the paper, and are likely to be found in one of the following three papers, none of which are publicly available.

1. 2. 3.

Nonetheless, I am interested in any synthetic pathway which will lead me from the first to the second molecule shown above. I understand the presented material, but as an undergraduate cell biologist, I am a little rusty on my synthetic organic chem.

EDIT: The first thing that came to mind was the condensation of a biscarbonyl electrophile and bisamine nucleophile like in pyrimidine synthesis.

Well, you are right, the synthesis of your second compound is described in DOI:10.1021/ja038384r:


How does heat transfer between molecules happen in deep? Inside the container is cold water and outside the container is hot water. B is the microscopic view of container walls .W is water
And A is hot water.
What I have shown is the microscopic view of container walls(Small part )
A is the surrounding and B is the microscopic view of the walls of container ( Since there is some thickness of container walls and we can say that there are molecules present there ).
As you can see , when cold water molecules apply force on B (walls of container at left side) , B walls also apply force on water molecules and same has happened in the part where there is hot water.This is from the Newton’s third law of motion.
Average Kinetic Energy of water molecules will be less since it is cold water.So less force applied by water molecules on the wall.
Average kinetic energy of hot water will be more than cold water.So it applies more force on walls of container than by cold water.
I hope I have written the things clear till here.
Q1 Is it that the walls container will be pushed by B towards right side and not left side ? Because there is more force by Hot water and not cold water.
Q2 Does heat get transferred here because the Average kinetic Energy of the walls of container will increase due to force of A on B? And those will put the molecules inside B into motion which will further put water molecules  into movement and therefore making heat transfer ?
Q3 Does heat transfer happen only because of this displacement?
HAVE I WRITTEN THE MEHTOD OF HEAT TRANSFER?IF NOT, THEN PLEASE HELP IN CORRECTING ME.
I am trying to make my self understand heat transfer in details.
Please tell if how can I improve the quality of my question if not good.

A1: There is no displacement of the wall, forces are in equilibrium with zero net force. There are initiated stronger vibrations on warmer wall side by energetic hot air molecules. And weaker vibrations on colder wall side by energy leaching cold water molecules .
A2a: If the wall initial temperature was the cold water temperature, then yes. Otherwise no, as the mean kinetic energy of wall molecules can generally goes both ways, depending on initial temperatures of air, wall and water, weighted by air and water thermal capacity a conductivity.
A2b: Yes
A3: No, as there is no displacement. Hot air with in average more energetic molecules causes stronger vibrations of wall molecules, which cause faster motion of water molecules.
Imagine the whole scenario in more familiar macro world. 2 boxes of ideally elastic balls in gravityless ISS station, separated by a wall of balls interconnected by elastic strings holding all of them at their place.
Now, balls in one of boxes move faster. What will happen during the time ?

The mean Kinetic energy of molecules is proportional to absolute temperature. Their mean speed is then proportional to the square root of absolute temperature and momentum to the absolute temperature.
Pressure on the walls is proportional to momentum change rate, what is proportional to molecule momentum and the rate of collisions. For gases in constant volume, both is proportional to square root of absolute temperature, so pressure is proportional to absolute temperature of ( ideal ) gas.
But pressure is not essential for molecular mechanism of heat transfer.
Molecules of matter at given temperature do not have the same speed and kinetic energy. During their collisions, it fluctuates all the time. What is constant is their mean value.
For monoatomic ideal gas, the mean kinetic energy of 1 mole of the gas is $E_\mathrm{k}=3/2RT$, where R is the universal gas constant about 8.314 J/K/mol. The speed distribution follows the Maxwell-Boltzmann distribution.
It means that for given kinetic energy and speed of a molecule, we cannot say what is the temperature of matter the molecule belongs.
When 2 different regions of matter have different temperature, they exchange heat, what means molecular kinetic energy.
Molecules from adjacent regions mutually exchange kinetic energy by random way by collisions. As by random collisions, they can both gain or lose the energy.
Molecules from both colder and warmer regions mutually both transfer and accept the energy. If temperatures are equal, their mean kinetic energy (*) is equal and the rates of energy flow are equal in both directions. If one region is warmer, the rate of passing energy is greater than the rate of accepting energy.

(*) It is more complicated, as more exactly, it means the mean kinetic energy per each way the molecule can move, called degree of freedom. But the above is still valid for monoatomic gases.


Why should we be interested in catalyst for electrolysis? I've been reading papers on Phosphided metals (CoP, NiP, FeP) as a catalyst for electrolysis.  They increase the current density of the cell for a voltage.  If it still requires 39.4 kilowatt-hours per kilogram of hydrogen (https://en.wikipedia.org/wiki/Electrolysis_of_water), why does it matter how fast the reaction occurs?

For example, with no catalyst, an electrolyzer runs at 0.6A/cm2 at 2.3V or with a catalyst it runs at 1A/cm2 at 1.4V.  Both cases require about the same amount of power and should produce the same amount of hydrogen, correct?

Catalysts generally decrease the electrode overpotential, which leads to decreasing the needed voltage to perform electrolysis at given current densities toward its minimal theoretical value.

If higher current is possible at the same voltage due the catalyst, the lower voltage is needed for the same current, both leading to better efficiency.

As the energy needed to produce 1 mol of molecular hydrogen is

$$E=2 \cdot U \cdot F$$

where
$U$ is the used voltage
$F$ is the Faraday's constant cca 96490 C/mol

As the consequence, it has direct impact on efficiency of conversion of electricity to hydrogen for hydrogen based engines or just fuel.


What is meant by a 'hydrophilic' sphere? So, I know that a hydrophilic material is defined as those with a special affinity for water, opposite to
hydrophobic, those that naturally repel water.

I am investigating the impact of spheres with water surfaces, and the splashes caused by them. However, one research paper focuses on the difference between the fluid dynamics of hydrophilic spheres and hydrophobic spheres, and the difference in the cavities they make when they sink underwater.

I understand what a hydrophobic sphere is, but what is a hydrophilic sphere? Like a rolled up ball of tissue paper? Or the water absorbing waterbeads/orbeez that get bigger(the ones children play with)?

The answer to your question depends on the size level (microscopic or not). But it sounds like you are referring to spheres at a size that you can hold, or larger.

A hydrophilic sphere could simply be a sphere of some solid material (like plastic), coated with a layer of hydrophilic material/substance.

This page on hydrophilic coatings has some information on what this material may be.


Do lone pairs contribute towards the pi electron count of heterocycles? I'm not able to understand how the following compounds are aromatic.

When should the lone pairs on heteroatoms be taken into consideration when counting the number of π electrons?

Only count the lone pairs/ pi-bonds/ groups which are participating in conjugation and ignore them in all other cases.

For example, in compound 2 (thiophene), there are two lone pairs on sulfur.

One lone pair (brown) is in a p-orbital, and hence participates in conjugation with the two π-bonds. The other lone pair (blue) is pointing outwards from the ring in an $\mathrm{sp^2}$ orbital. This lone pair is orthogonal, or perpendicular, to the π-system and hence cannot take part in conjugation. In total there are six π electrons, and the compound is therefore aromatic.

See also: Conjugated system on Wikipedia.


During phase change in matter, why doesn't the temperature change? I was working on something in school and came across the question:

  Why does the temperature not change much during a phase change?

I'm really not sure why this happens in matter and I couldn't find an answer in my school resources. Does anyone here know?

From Changes of Phase (or State):

...

So, how could there be a change in heat during a state change without a change in temperature?

"During a change in state the heat energy is used to change the bonding between the molecules. In the case of melting, added energy is used to break the bonds between the molecules. In the case of freezing, energy is subtracted as the molecules bond to one another. These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules.

"If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water. Immediately after the molecular bonds in the ice are broken the molecules are moving (vibrating) at the same average speed as before, so their average kinetic energy remains the same, and, thus, their Kelvin temperature remains the same."


Nomenclature priorities with ethers, acyl groups and ketones $\hspace{6cm}$

(2-oxopropionyl)oxy butanoic acid

$\hspace{6.5cm}$

3,4-dioxo 4-methoxy butanoic acid

$\hspace{6.3cm}$

3-(2-oxopropyl),6-oxo heptanoic acid

$\hspace{6.25cm}$

3-propionyl,6-oxo heptanoic acid

$\hspace{6.3cm}$

3-(3-oxobutyl),6-oxo hexanoic acid

Aldehyde takes priority over ketone, yes?

Have I got these right?

PS: With the third molecule, wouldn't it be convenient if we could have a ketonyl group so that instead of 2-oxopropyl , one could say propanonyl?

In principle, your assumptions about the order of priority for classes is correct. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

  (…)
  7. acids
  (…)
  9. esters
  (…)
  15. aldehydes
  16. ketones
  (…)
  42. ethers
  (…)

Thus, it is correct that the names for all compounds given in the question have the suffix ‘oic acid’.

However, the names proposed by you still contain a few errors.

Your name “(2-oxopropionyl)oxy butanoic acid” is ambiguous since the locant for the complex substituent group ‘(2-oxopropionyl)oxy’ is missing.
The complete name is “4-[(2-oxopropionyl)oxy]butanoic acid”.
However, the preferred name is 4-[(2-oxopropanoyl)oxy]butanoic acid.

Your other names include the correct suffixes and locants; however, the suffixes are not cited in alphabetical order. The correct names are

4-methoxy-3,4-dioxobutanoic acid (not “3,4-dioxo 4-methoxy butanoic acid”)
6-oxo-3-(2-oxopropyl)heptanoic acid (not “3-(2-oxopropyl),6-oxo heptanoic acid”)
6-oxo-3-propanoylheptanoic acid (not “3-propionyl,6-oxo heptanoic acid”)
6-oxo-3-(3-oxopropyl)heptanoic acid (not “3-(3-oxobutyl),6-oxo hexanoic acid”)


How toxic chemically is plutonium (Pu), neglecting the radioactive damage? In Rhodes' The Making of the Atomic Bomb, he says that, while Pu is not that radioactive (which is surprising -- maybe he means compared with radium and some other elements), it is very toxic.
I would guess it could not be much more toxic than uranium (U) or lead (Pb) -- I believe thallium is more toxic than U and Pb and so is mercury (Hg). If it is more toxic than lead, do we understand the mechanism?

The toxicity is primarily due to radioactivity and to absorption by the body, where that radioactivity can act internally. There is, "significant deposition of plutonium in the liver and in the "actively metabolizing" portion of bone," according to Miner and Schonfeld.
Many $\ce{Pu}$ isotopes are primarily alpha-emitters, with "high energy radiation with low penetration and thereby requires minimal shielding. A sheet of paper can be used to shield against the alpha particles emitted by plutonium-238," from Wikipedia. The risk of holding a small amount of plutonium, if it is sealed inside almost any container, is negligible, because the the heavy, positively-charged alpha particles are stopped by the container in a very short distance.
By that same token, when absorbed, the plutonium migrates to rapidly dividing liver tissue and bone's hematopoietic cells. The alpha particles are stopped in a very short distance, depositing all their energy in the few nearest cells and massively interfering with cellular reproduction. Small amounts can cause various types of cancer, such as leukemia, a cancer of the previously-mentioned blood cell producing tissues.
That said, plutonium is not quite as toxic as one might think: "The U.S. Department of Energy estimates that the lifetime cancer risk from inhaling 5,000 plutonium particles, each about 3 µm wide, to be 1% over the background U.S. average... no human being is known to have died because of inhaling or ingesting plutonium, and many people have measurable amounts of plutonium in their bodies."
Perhaps what Rhodes means by "not that radioactive" is that the radiation from plutonium is not very penetrating. The isotope used in nuclear weapons, 239Pu, has a half-life of about 24,000 years, comparable to radium, or to 231Pa. If I've done my math  correctly, one mole of 239Pu, about 6e23 atoms, with  half-life of 8e11 seconds, should emit about 400,000,000,000 alpha particles per second. Of course, if it were a sphere, most of those particles would be absorbed inside the mass of metal, releasing energy as heat. However, spread through a human body, each particle has a much greater chance to cause damage. And 238Pu, used in radioisotope thermoelectric generators, is far more radioactive!


Several carbonyl groups, which one does ethylene glycol protect? We know that ethylene glycol is used to protect carbonyl functional groups. But, consider this:

Here, we have two carbonyl groups, but only one equivalent of ethylene glycol. Obviously, only one carbonyl group will get protected in this case. But will it be the $\ce{C}^1$ or $\ce{C}^2$ carbonyl? And why?

Several factors influence the overall rate of a reaction under various conditions. Among the crucial factors are:

structural features of the carbonyl compound;
the role of protons or other Lewis acids in activating the carbonyl group toward nucleophilic attack;
the reactivity of the nucleophilic species and its influence on subsequent steps; and
the stability of the tetrahedral intermediate and the extent to which it proceeds to product rather than reverting to starting material. (Ref.1)

Limiting ourselves to current molecule, aldehydes  are more reactive than ketones. Electron-withdrawing substituents render the carbonyl group more electrophilic.

Image credit: Organic Chemistry, Structure and Function Seventh Ed by Peter Vollhardt, University of California at Berkeley, Neil Schore, University of California at Davis

Because aldehydes form acetals more readily than ketones, we can protect an aldehyde selectively in the presence of a ketone. The following example shows the reduction of a ketone in the presence of a more reactive aldehyde (Ref.2):

With reference to your question, C-2 is more reactive then C-1 carbonyl group for the above reasons.

References

Advanced Organic Chemistry, Fifth Edition, Part A: Structure and Mechanisms by Francis A. Carey and Richard J. Sundberg, University of Virginia
Organic Chemistry, Eighth Edition, by L. G. Wade Jr., Whitman College
Reactivity of Aldehydes & Ketones - Chem LibreTexts


Grapefruit juice to extract herbs for tincture I tried searching my question on google and did not find anything. I am looking to extract herbs with pure 100% grape juice http://www.lakewoodjuices.com/product_detail/id-32/  ( it is pasteurized ) instead of using the old school method of extracting with a 50% alcohol solution.

My theory as of now would be that it would be more beneficial and it would  ferment or extract the same plus the extra beneficial minerals of the juice. I am wonder what your opinion on this is. Do you think it would extract the same and be just as beneficial as a tincture? Would it be more beneficial? How long should I let it sit for? should I let it sit at room temp or in cooler temp say, $45 ^\circ \mathrm{F}$?

Use the old school method and extract with aqueous ethanol. It is more efficient and more healthy:

The solubility of most compounds to be extracted from herbs is likely to be much higher in ethanol than in water (or juice). You get more out of the plants.
The juice was initially pasteurized, but once opened, it will not necessarily stay free from microbes and/or might get mouldy. The risk of infections is definitively lower when staying with aqueous ethanol for extraction.


What is the meaning of the "colon" notation in chemical formulas? For example PEDOT:PSS and P3HT:PCBM.

All it means is that there's an intimate physical mixture of the specified components in a solid or liquid phase (no chemical bonding between the components as happens in a copolymer, for example).

In the context of solar cells, LEDs and related devices, they are often used as part of shorthand device architecture notation. Slashes ( / ) are used to separate different layers of the device, while colons ( : ) are used to enunciate all separate components in a single layer. Do note that every so often a slash is used incorrectly in place of a colon. Checking the experimental procedure should clarify whether two consecutively listed materials are in the same or separate layers.

For example, consider the notation ITO/PEDOT:PSS/P3HT:MEH-PPV:PCBM/Ca/Al for an organic polymer solar cell. There are five layers in total, deposited in sequence from left to right:

A layer of ITO (usually on top of a glass substrate which is omitted).
A layer containing a binary mixture of PEDOT and PSS, usually deposited via spin casting of an aqueous solution with both materials dissolved.
A layer containing a ternary mixture of P3HT, MEH-PPV and PCBM, often deposited by spin casting from a solution of all materials in a chlorinated solvent.
A layer of pure calcium metal, likely deposited by thermal evaporation.
A layer of pure aluminium metal, deposited on top of the calcium layer.

Sometimes the thickness, component proportion and concentration of the spinning solution for each layer is reported inside of parentheses, such as in ITO/PEDOT:PSS (40 nm)/P3HT:PCBM (1:0.8, 150 nm)/Ca (80 nm). Be sure to pay attention to small differences in notation resulting in completely different cell architecture. For example, a solar cell with a ITO/PEDOT:PSS/P3HT:$\ce{C_{60}}$/Al represents a bulk heterojunction solar cell, while ITO/PEDOT:PSS/P3HT/$\ce{C_{60}}$/Al represents a bilayer (or planar heterojunction) solar cell.


Is 2-oxobicyclo[2.2.1]heptan-1-ide resonance stabilized? Is the following compound resonance stabilized?

I think its not resonance stabilized because the p-orbital of carbanion and pure p-orbital of carbonyl carbon are not in same plane. So they can't overlap. (Pure means that the p-orbital is not a part of the $\ce{sp^2}$ orbital. One p-orbital of carbonyl carbon is pure).
Am I correct?

I think, in contrast to cyclohexanone, a resonance stabilization by

is less likely to happen as you already start with a bi-cyclic compound where the cycles involved are small to moderate. The then additional strain introduced by a (partial) double bond at the bridge-head carbon atom would destabilize the molecule too much.  I extrapolate here from Bredt's rule.


Why isn't chlorine reduced or nitrogen oxidised in the reaction between ammonium perchlorate and aluminium? A space shuttle’s upward thrust came from the following reaction between aluminium and ammonium perchlorate.
$$\ce{10 Al + 6 NH4ClO4 → 4 Al2O3 + 2 AlCl3 + 12 H2O + 3 N2}$$
Which statements about this reaction are correct?

Aluminium is oxidised.
Chlorine is reduced.
Nitrogen is oxidised.

I understand why aluminium is oxidized, but why are 2 and 3 wrong?

Here,

The oxidation number of chlorine in the perchlorate radical is +7 whereas, in the product side, chlorine has an oxidation number of  minus 1 in AlCl3. As the oxidation number is reduced by 8 here (gaining 8 electrons), chlorine is, indeed, reduced.

Nitrogen has an oxidation state of minus 3 in NH4ClO4 whereas, it has an oxidation state of 0 in N2 molecule in the product side. Here, the oxidation number of nitrogen has increased by 3 via losing 3 electrons. So, Nitrogen here is oxidised.

Hence, all the options are correct.


Do ligands in ionisation isomerism interchange? The text says:

  This form of isomerism arises when the counter ion in a complex
  salt is itself a potential ligand and can displace ligand which can
  then become the counter ion.

For other structural isomers that I studied in coordination compounds I simply thought that those  compounds (structural isomers) separately exist. In fact I think of structural isomers in coordination chemistry this way only that these are independently existing with different properties but same molecular formula so we study them.

But the text above is confusing it seems that they convert like resonating structures or conformation isomers.

What is true? If they do interchange then isn't this a contradiction to Werner's theory that postulates that secondary valencies are non ionisable.

note:the below problems is related to one of the comments:

    Please Consider this problem-"Give evidences that [Co(NH3)5Cl]So4 and [Co(NH3)5 So4]Cl are ionisation isomers".

  Here we test by dissloving them in water and reacting them separately
  with silver and barium ions  to test for counter ions.One gives
  AgCl and other BaSo4 as ppt. .Now if there is equlibrium like keto-
  enol tautomerism then only one of the two isomers should be stable in water
  then why such kinds of evidences are given?

Actually, in the ionisation isomerism the ligands do not interchange. They get displaced. Thus one ion becomes counter ion for the other. When one has look at the molecular formula it seems the ligands in the coordination sphere are interchanged with the ligands outside the sphere. So, this does not cross out Werner’s theory as there is displacement of ion and not just interchange.


para-dichlorobenzene - number of proton NMR signals I am told that para-dichlorobenzene exhibits two proton NMR signals. This doesn't jibe with me; all the protons on the molecule are equivalent. Here is a book claiming the same thing. Am I wrong, or is the person who told me that para-dichlorobenzene has two signals on its proton NMR wrong?

Be very careful with how you interpret this molecule. Although the 1H NMR spectrum appears as a single line, this is actually an AA'A''A''' system, meaning that, although chemically equivalent,  all four protons are magnetically non-equivalent, but have coincidental chemical shift. The coupling between all protons is different; JA-A''' is different to JA-A''' is different to JA'-A'' is different to JA'-A''' and so on.

So yes, the spectrum is as you thought. p-dichlorobenzene IS a single peak in the 1H NMR spectrum.

Below is a set of simulations for a generic para disubstituted benzene system, which is an AA'BB' system. A is chemically equivalent to A', but magnetically non-equivalent. A is chemically AND magnetically non-equivalent to B. Your system is just a special case example of these types of systems, with all protons being chemically equivalent, but magnetically non-equivalent. These are second order systems, which have a lot of complex additional splitting structure in their spectrum. Like all spin systems, as the chemical shift difference becomes less (the peaks move closer together), the tilting, (or tenting or roofing, depending on which text book you read) becomes more pronounced, and the wings on the main peaks become smaller and smaller. Eventually, they collapse into a single peak at very small chemical shift difference.

The fine text which you can't read (sorry about that), gives the coupling constants for the system, which are bog-standard ortho/meta/para couplings: Jortho=7Hz, Jmeta=3Hz; Jpara=1Hz. The only variable in the simulations is the chemical shift of the downfield peak, which slowly moves towards its upfield partner.


Calculating mass and volume of two different substances required for one solution I want to produce a solution, which contains $0.02~\mathrm{mol~L}^{-1}~\mathrm{Cu}^{2+}$ and $~0.2~\mathrm{mol~L}^{-1}~\mathrm{NH}_3$. The following substances are given:

copper(II) sulfate pentahydrate
$100~\mathrm{mL~H}_2\mathrm{O}$
diluted ammonia solution with a concentration of $2~\mathrm{mol~L}^{-1}$.

I need to determine the required volume ($V$) of $\mathrm{NH}_3$ and the mass ($m$) of $\mathrm{CuSO}_4\cdot 5\mathrm{H}_2\mathrm{O}$.

I have no idea how to calculate $m$. However, to calculate $V$, I solved the following system of equations:

$\dfrac{2~\mathrm{mol}}{\mathrm{L}} = \dfrac{n}{V}$

$\dfrac{0.2~\mathrm{mol}}{\mathrm{L}} = \dfrac{n}{V+0.1~\mathrm{L}}$

Therefore, $V = 0.01111~\mathrm{L}$. Is this correct?

Is $m=555~\mathrm{mg}$ correct?

We have direct formula for calculating the weight of the substance as,
$$\text{Weight}=\frac{\text{Molarity} \times \text {molecular wt} \times \text{volume in mL}}{1000}$$
$$=\frac{0.02\times 250 \times 100}{1000}$$
$$=0.5 \text{ g}$$

So to prepare 0.02 $ \mathrm{mol~L}^{-1}~\mathrm{Cu}^{2+}$ ,0.5 g of copper(II) sulfate pentahydrate should be dissolved in 100 mL water.

To find the volume of ammonia,

Initial molarity = 2 M

Initial volume= ?

Final molarity = 0.2 M

Final volume = 100 mL

We have formula,
 $$ \text {Initial molarity} \times \text{Initial volume} = \text {Final molarity} \times \text{Final volume}$$
$$\text{Initial volume} = \frac{0.2 \times 100}{2}$$
$$= 10 mL$$

Hence, 10 mL of 2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$ should be added in 100 mL water to obtain 0.2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$.


Strange green and gold coloured chemical I have a reagent bottle with an unknown chemical that came with some other lab gear about 30 years ago, I stumbled on it again when clearing out some old stuff.

It is a standard brown glass Merck of about 250ml size.  It has no label and never had any indication of a label and I do not know if it is the original bottle.

The powder is a fine dark free flowing crystalline solid.  It is DARK green with a gold iridescent sheen when in bright light, something like a butterfly wing or some anti-forgery inks on banknotes.

Yet if I wet a very small portion it dissolves completely and creates a vivid dark blue solution that dries blue in a thin film.  If the crystals dry in a clump they form a smutty dark green gold pasty lump that is reminiscent of the original bulk colour.

I had to hand some potassium permanganate crystals and when I mixed a pinch of the two together and added a drop of water a gas evolved and the colour seemed to be somewhere between the permanganate and the mystery chemical blue when spread on paper.  Adding a few drops of acetic acid seemed to have no effect.

Can anyone think of an obvious candidate or some simple experiments that could be used to determine the material?

I worked for a time with squaraine dyes that were often greenish in the crystalline state, but ink-blue when in solution.

However, I would suggest that you take an analytical approach to identifying this compound.  Have you done any NMR, IR or even UV/Vis spectroscopy?  Elemental analysis?

These would be a much safer way to go about identifying the mystery compound.

Good luck, and I'd love to know what you discover!


London Forces and Liquifying Noble Gases From my book, it says that 'without van der Waals (London Dispersion) forces, it would be impossible to liquefy noble gases.'

Why is this the case?

I'd say that because noble gases are non-polar atoms, the only force that applies to them is the London force.

If you want a liquid phase, you need to have forces between the atoms / molecules that constitute this liquid (forces too strong would lead to solid state).

So because London force is the only one that applies on noble gases molecules, they are necessary to liquify to gases.


Can one exclude calibration points using a combination of permutation testing and a non-parametric test statistic? I work in bioanalytical chemistry. I have used an analytical instrument (LC-MS) to measure the response (peak area) of several compounds spiked in to equivalent samples at known concentrations. Each sample, at each concentration, was analysed three times over three separate days. Spike concentrations, for this example, are: 0 (blank), 1, 2, 5, 10, 50, 125, 250, 500, 1000 ng/L. I now wish to construct sensible calibration curves using linear least squares regression (hence, assuming homoscedacity of data, but still acknowledging the potential for outliers) of the response (peak area) versus spike concentration (i.e. the spike concentration).

From visual inspection of data files, it appears that for some compounds there are samples that were spiked at > 0 ng/L that are indistinguishable from blank samples (i.e. those at 0 ng/L). I’d like to test my suspicions statistically and, if significant at a reasonable threshold, would exclude these samples prior to performing the least squares regression.

Given that, for many compounds, the data are often non-‘normal’ around the lower-end of the calibration curve in such dataset, I thought to use permutation testing to generate a sampling distribution. Here, I’d define a null hypothesis that the median peak area of blank samples (0 ng/L) is equal to that of non-blank samples at a defined concentration (e.g. 1 ng/L), i.e. a statistical test of location. Drawing from the world of parametric tests, I’d hoped to use: [median(peak areas of non-blank samples at single concentration) - median(peak areas of blank)] / median absolute deviation(peak areas of non-blank samples at single concentration).

Here, the sampling distribution would be generated by repeatedly randomising class labels (i.e. concentration labels for 0 ng/L and the calibrant level being tested) and calculating the test statistic. The p-value would be derived as the proportion of the permuted test statistics that were equal-to or greater-than the actual measured statistic, i.e. as calculated before randomising labels. If the p-value turns out to be greater than some pre-determined threshold (probably going to be 0.05 or 0.01), I accept the null hypothesis. Given that I would repeat this test for each concentration level, I’d also plan to apply a Benjamini-Hochberg correction to the final list of p-values and reject then exclude concentration levels with adjusted p-values above the cutoff (I’d aim for either 0.05 or 0.01 – yet to decide).

So, feedback on the following would be great:

Does the proposed test statistic seem reasonable as a non-parametric replacement to the t-statistic?
Is it appropriate to exclude samples from the regression, for which the p-value for the test statistic exceeds the defined threshold?

This is more of a comment, but a typical comment doesn't allow a figure...

You have a very wide range (1-1000) for the independent variable. Using a simple linear least squares analysis isn't appropriate.

I grabbed the image below from webpage https://www.mathworks.com/matlabcentral/answers/392636-confidence-band-around-linear-least-squares-line

The point I want to make is shown by the shaded area. Assuming homoscedasticity, you wouldn't weight the errors. Thus the confidence interval for the line is narrowest at the center of the fitted range and gets much wider at either end of the range. So your low samples ( 1, 2, 5, and 10) will essentially be irrelevant.

So you need to assume hetroscedasticity and weight the sample errors somehow.

  1. Does the proposed test statistic seem reasonable as a non-parametric replacement to the t-statistic?

Just because in general non-parametric tests are less powerful than parametric tests, I would try hard to avoid using a non-parametric test on parametric data.

  2. Is it appropriate to exclude samples from the regression, for which the p-value for the test statistic exceeds the defined threshold?

I'll assume that you mean to exclude some of the standards not samples. It is possible to exclude standards, but don't use three standards and then drop the one that you "know" isn't on the line. :-)

Hope this helps a little. I just don't know enough about S/N for LC/MS or the detector system that you're using to tell you what you should do.

Your statement "Given that, for many compounds, the data are often non-‘normal’ around the lower-end of the calibration curve in such dataset,..." is disconcerting.


Determining the exact charge of ions that will be formed by an atom using the periodic table I don't know how to obtain the "exact charge of the ions" that will be formed by each atom mentioned. I also have to figure out whether they are positive or negative.

Hydrogen, for example, can form the ion:

$$\ce{H -> H+ + e-}$$

Now if I were to try to get the exact charge for say, boron using only the periodic table, how would I do so?

Each atom tends to imitate the noble gases which have a full valence shell. Looking at your table you can see that boron is 3 spaces above helium, but 5 spaces behind neon. Boron would "prefer" to have a configuration like the noble gases, and because losing 3 electrons seems easier than gaining 5, boron, as boring as he is, decides to do the obvious thing and three electrons walk the plank... Now he has 5 protons and only 2 electrons therefore imitating helium's electron configuration. This means boron has a charge of $+3$.

Oxidation states are common on most periodic tables, which show you the result of less than obvious behavior of atoms when this trend does not always hold. In the case of hydrogen, it is common than it will bond with another hydrogen to imitate helium, otherwise it will tend to form a $+1$ ion.