Find amount of acid and base required to make buffer I am wondering if it is possible to know exact amount of acid and base that we can put in together to make a buffer.

I want to make a 0.05 M extraction buffer using disodium hydrogen phosphate (DHP) and sodium dihydrogen phosphate (SDP). I want to make 100 mL of buffer so I worked out that I would require 0.78 g SDP and 0.88 g DHP.

The issue is that it sometimes takes a lot of time by mixing one solution into the other and keep it checking with the pH meter, there has to be some mathematical way to get the exact amounts which would save time and my chemicals.

Usually when you are preparing a buffer, you can about the pH and the strength (total concentration). Let's say you want to combine acid $\ce{A}$ and its conjugate base $\ce{B}$ to make a buffer with total concentration $C=[\ce{A}]+[\ce{B}]$ and a set $\text{pH}$.

The Henderson-Hasselbalch equation provides a relationship between the ratio of the two concentrations to the pH:
$$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{[\ce{B}]}{[\ce{A}]}\right)$$

Because $[\ce{B}]=C-[\ce{A}]$, we can write, and solve for $[\ce{A}]$ (and then for $[\ce{B}]$).

$$\ce{pH}=\ce{p}K_a+\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)$$
$$\log_{10} \left(\dfrac{C-[\ce{A}]}{[\ce{A}]}\right)= \ce{pH}-\ce{p}K_a$$
$$\dfrac{C-[\ce{A}]}{[\ce{A}]}=10^{\ce{pH}-\ce{p}K_a}$$
$$C-[\ce{A}]=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]$$
$$C=\left(10^{\ce{pH}-\ce{p}K_a}\right)[\ce{A}]+[\ce{A}]=[\ce{A}]\left(1+10^{\ce{pH}-\ce{p}K_a}\right)$$
$$[\ce{A}]=\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}$$
$$[\ce{B}]=C-\dfrac{C}{1+10^{\ce{pH}-\ce{p}K_a}}=\dfrac{C\left(10^{\ce{pH}-\ce{p}K_a}\right)}{1+10^{\ce{pH}-\ce{p}K_a}}$$

So... if you have a total concentration $C$ and a $\ce{pH}$ in mind for an acid-base pair with a certain $\ce{p}K_a$, just plug them in.


How would I make a basic protein folding algorithm? On the face of it it doesn't seem especially difficult.
Model the amino acids as solid objects linked together. Approximate the 3D force fields around them using a coarse approximation, then set the simulation going in a computer and see what happens.
Why is this such a difficult problem that it has taken decades to crack? Are there subtle difficulties not apparent in the above model?

It is far, far more difficult that a naive approach suggests
The reason why simulating protein folding is extremely hard is the combinatorial complexity of the options you have to explore to get the "right" answer. This exhausts your computers memory or computing capacity far faster than it will for simple molecules.
It is possible to see why this is so from very simple considerations. The backbone of a protein consists of peptide bonds (which can rotate fairly freely) plus two rotating bonds in each amino acid. even ignoring the amino acid side chains (which often have multiple rotatable bonds) even a dipeptide has 5 bonds where you have to explore the possible rotations to find the optimal configuration of the extremely simple molecule. Since proteins consist of between 50 and 2000 amino acids that leaves a gargantuan number of options to consider, even if we can ignore all the other things that matter.
And those other things do matter. A simple example is that the configuration at a specific bond in the backbone may depend on the spatial orientation of other bonds dozens of bonds away so every choice at one site depends on all the other choices on all the other bonds. Optimising one bond at a time just doesn't work. On top of that is the complexity of including the side chains. That makes the problem harger by another gargantuan degree even if you don't consider the specific possible interactions of those side chains with each other (which are know to be important in many real proteins in stabilising specific spatial structures.
On top of those complexities, real proteins fold themselves in the environment of the inside of cells where interactions with water matter. So you have to consider those interactions as well. And, on top of that, the machinery that makes proteins consists of other pieces of molecular machinery designed to force specific shapes to emerge (heat shock proteins or chaperones are known to do this job).
One way to simplify the task so its difficulty is apparent is to realise that a simple protein chain has an extraordinary number of possible ways to fold only one of which is the "right" way in a living system. The problem is not finding a possible structure, but the right structure. Given the almost-impossible-to-count number of options, it is beyond the capacity of computation to find the right one by any known search algorithm.
This is why large teams of professionals have failed over decades to find a good general approach even when using vast amounts of specific biological knowledge from known structures. It is simply not a problem that can be tackled by amateurs.


Why Does Activation Energy not change with Temperature change As The Graph Shows; $E_a$ (Activation Energy) = Energy Of transition State(Threshold Energy) $-$ Energy of Reactants.
So let This be the graph at Temperature $T_1$, Now say We Increase The Temperature to say $T_2$ and so the energy of reactants must increase (as kinetic energy of molecules increases). So at $T_2$, reactants would start with more energy. So Shouldn't the Activation Energy decrease as now less energy is needed to reach Transition State.
Or I can say that since $E_a$ (Activation Energy) = Energy Of transition State(Threshold Energy) $-$ Energy of Reactants, So as Energy of Reactants has changed so should Activation Energy.
I already saw answers on a similar question but couldn't figure out what's wrong with my approach.

You have conceptual misunderstanding. In the activation energy context, energy of reactants, products and transition state are all potential energies, independent on the kinetic energy of molecules.
The kinetic energy of molecules is used to reach the peak of the potential energy, similarly as e.g. a acrobatic aircraft uses its kinetic energy to climb to the apex of an acrobatic loop.
Higher temperature means higher kinetic energy, but it does not mean the potential energy barrier is lower. It means there is higher probability to overcome it, leading to a faster reaction.
There is a famour Arrhenius' equation for the reaction rate:
$$k = A \exp{\left(\frac{-E_\mathrm{A}}{kT}\right)}$$
where $A$ is the frequency factor, describing the rate of molecular collisions, potentially able to take part of the reaction and
$\exp{\left(\frac{-E_\mathrm{A}}{kT}\right)}$ is the term from the Boltzmann distribution,
telling us the probability a molecule at temperature $T$ has sufficient energy, if the activation energy is $E_\mathrm{A}$, and $k$ is the Boltzmann constant.


What is the proper way to prepare a solution from hydrate? I am following a protocol that requires me to prepare $10~ \mathrm{mL}$ $\mathrm{1~M}~ \ce{Ca(NO3)2}$. However, there is only $\ce{Ca(NO3)2*4H2O}$ in the lab.

Originally I thought that all I have to do to figure how much $\ce{Ca(NO3)2}$ to weigh is

and then just add  $10~ \mathrm{mL}$  water. However, that might dilute the solution and make the concentration less than $\mathrm{1~M}$ as there is already water in the hydrate.

So then I found a protocol and made new calculations as followed:

How can I verify that I got $\mathrm{1~M}$ $\ce{Ca(NO3)2}$ as opposed to $\mathrm{1~M}$ $\ce{Ca(NO3)2*4H2O}$?

Which protocol is correct?

Originally I thought that all I have to do to figure how much $\ce{Ca(NO3)2}$ to weigh is

  and then just add  $10~ \mathrm{mL}$  water.

To properly make an aqueous solution of a specified concentration you should dissolve the desired amount of material in about $80\%$ ($8~ \mathrm{mL}$) the desired volume of water. Then dilute to the final solution volume (add water until the solution volume is $10~ \mathrm{mL}$).

  How can I verify that I got $\mathrm{1~M}$ $\ce{Ca(NO3)2}$ as opposed to $\mathrm{1~M}$ $\ce{Ca(NO3)2*4H2O}$?

Hydrates are for the solid state.  You do not have "hydrates" in the aqueous phase.  Therefore the two solutions are identical and indistinguishable.


Liquid nitrogen to cool liquid? Is it possible and safe to use liquid nitrogen to quickly cool a 5 gallon tank of liquid from room teperature to say 34 °F? Or is there another way that can produce really fast results?

I don't know whether liquid nitrogen can be used to cool a volume as large as 5 gallons safely (that is a lot to do at once in a kitchen or restaurant) but it can and has been used in cooking.

Here is an example from youtube.

But it can be dangerous if not done carefully. This BBC story has a warning of what happens if it isn't.


Is there a standard symbol for molar density? Is there a standard symbol for molar density, i.e. $n/V$, where $n$ is the total number of moles of every species in a system?

I realise that there is standard term "molar volume", which is the reciprocal of what I'm asking for, but my equations are going to look pretty silly with $1/V_m$ written everywhere. As a simple example, I want to say that if my system consists of a mixture of $\ce{A}$ and $\ce{B}$ with the reaction $\ce{A <-> B}$, the total concentration of $\ce{A}$ and $\ce{B}$ will be constant and we can write
$$
\ce{[B]} = c - \ce{[A]},
$$
where $c$ is the thing I'm looking for the symbol for.

I didn't find "molar density" in the IUPAC green book, but I don't know my way around it very well, and if it has a different name it might be hiding somewhere I didn't think to look.

To my knowledge, n/V is known as molar concentration and uses symbol $C$ and it seems okay for me to use $C_t$ i.e. total concentration to mark sum of concentrations of all particles in system.


Why is absorbance the negative logarithm of transmittance? What is the physical significance of absorbance? Is it not possible to directly decipher the amount of radiant energy absorbed by the sample from transmittance? Why is that negative log of Transmittance is called absorbance?

Absorbance is useful because it is additive. That is, it's absorbance which is used in Beer's law:

$$A = \epsilon \cdot c \cdot l$$

While you can certainly make a version of this law which uses transmittance instead of absorbance, the math becomes much less straightforward.

The explanation of why absorbance is additive falls out of the derivation of Beer's law. The official derivation relies on calculus, but you can get a sense of things by thinking of the solution not as a single thing, but as a succession of "slabs":

Take a single thin slab of a colored material. This absorbs a certain amount of light at a given frequency. But it doesn't absorb a fixed amount of light. Instead, each photon that passes through has a certain probability of encountering an absorbant molecule and being absorbed. The more light, the more absorbed. For a given slab under typical conditions, we can define the percent transmittance (%T), which is the fraction of light that goes through that material.

But what happens if you start varying the width of the material. Or equivalently, if you stack several of such slabs together. The amount of light going through one slab would be %T. The amount going through two slabs would be %T times %T. (For example, if the slab had a %T of 50%, then the first slab would absorb half the light. The second slab would only "see" half the original light, and would then absorb half of that light. So the amount being transmitted out the back of the second slab would be only 0.5*0.5 = 25% of the total amount.) Three slabs gives you %T * %T * %T. Generally, for N slabs, you'd get only $(\%T)^N$ of the light coming through. (Note the necessity of converting from percentages to fractions to do the calculation here.)

Okay, but how about varying concentration? Well, since the chance that a photon is absorbed is proportional to the chance it interacts with an absorbing molecule, then if you double the concentration, you double the number of interactions that could happen. If you work out the math, it turns out that (in dilute solutions) doubling the concentration is equivalent to just doubling the path length - you effectively have two "slabs", just compressed into the space of a single slab. The effect is the same as increasing the path length. So for a solution at twice the concentration, you's have $(\%T)^2$ the transmittance.

Note the use of exponents. This is all because when you combine two things which absorb, you have to multiply their percent transmittance. Exponents are hard to work with. So instead, we do a trick which can convert mulexponentstiplication into addition: take a logarithm. If we work with logarithms of the transmittance, all of those multiplications are now additions, and instead of using exponents for path length and concentration, we multiply by each. Taking the negative log is just another trick to make things work out reasonably: it's more convenient to have the measurement move in the same direction as concentration and path length (such that more absorbance means a larger concentration), so we stick a negative sign on the result of the logarithm to make things easier.

That's basically it. You certainly could do all your calculations with transmittances, but the equations and reasoning gets more complicated. If you use absorbances, you get the nice Beer-Lambert law, which states the concentration and path length are proportional to absorbance.


Vacuum distillation apparatus selection and configuration I want to assemble a "hobbyist-level" vacuum distillation apparatus for extracting small volumes of "typical" solvents (e.g., water, alcohols, ketones, diols).

I already have standard lab glassware with 24/40 joints, 10mm barbed inlets.

How do I select a suitable pump?

I would want a pump that can run for extended periods to perform vacuum desiccation.
So far my research suggests that I should be looking at diaphragm pumps, but:
I don't know if there are other features or specifications available in a low-three-figure pump that will make life easier.
What kind of tubing do I want?  I know that larger inner diameter is better, but I don't know if something special is used instead of regular tubing to support the vacuum, or what the standard adapters and sizes are to come off the 10mm glass fittings.

What elements need to be part of such an apparatus, and how should they be configured?  E.g., I have seen mention of:

Filters: What are used, and when are they necessary?
Traps: Does this only refer to "cold traps," which I understand to simply be another condenser in front of the pump?  If so, is it correct that this would only help if I can cool the "trap" significantly below the temperature I'm using for the primary distillate condenser?
Manometers
Capillary bubblers: I can't even find anything on these, other than what's mentioned here.

Also, how do these answers change if I'm doing vacuum desiccation of the named solvents (i.e., where the only difference is I don't care to capture the evaporate)?  E.g., can I just let the system exhaust through the pump?

Don't bother with a powered pump: Get a water aspirator (I recommend the $20 Humboldt H-12020, which uses 10mm or 3/8" inner diameter tubing just like your glass), connect to faucet, and using utility water (1-2 gallons/minute) it can pull over 500 torr, which should be adequate for your applications.  Sure, you're flushing away a lot of water, but you don't have to bother with traps or filters.

(Actually, the H-12020 does have a check valve, but to ensure you don't get even a little water sucked into the apparatus when you shut off the aspirator you might want to add a separate valve or simple trap, but that's easy compared to a cold trap.)

You should buy "thick-walled vacuum tubing" for all connections if you want to run full vacuum.

Finally, for safety you do need some bump countermeasure.  A "capillary bubbler" is just a tube that lets a tiny stream of bubbles into the boiling flask.  Yes, that is reducing the vacuum, but a proper capillary tube admits a trivial amount of gas into the system.  When possible, however, I prefer stirring.  Running a magnetic stirrer in the boiling flask also prevents bumping, and is better than a bubbler because it doesn't introduce potentially contaminating gases into the mix (or require the supply of a neutral gas).

A decent reference for practical distillation is Zubrick's Chapter 20 ("Distillation for Dummies") available online here.


Is it possible to make graphite an insulator? Since graphite is anisotropic electrical conductivity, won't it be possible to make it an insulator instead? Like manufacturing it such that the plates of carbon atoms are mostly parallel to one another.

Tempting, but no.  Graphite is of course less conductive perpendicular to the basal plane but still boasts a conductivity of 330 S/m in that direction, https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity.  Electrons can "jump" fairly easily from one pi-electron "layer" to the next.


Is there a way to create natural Red food coloring from Turmeric in the house kitchen? As a non chemist I have read in the Wikipedia article Curcumin that:

  Curcumin becomes bright red when it interacts electrostatically with
  phospholipid film.

This text appeared without source under this image in the article:
https://en.wikipedia.org/wiki/Curcumin#/media/File:Curcumin_thin_lipid_film.jpg

As an amateur of cooking Indian dishes (which often have bright red "Tandoori" color):

Is there a way to create natural Red food coloring from Turmeric in the house kitchen?

That is to ask; what edible supermarket products contain phospholipid films and how should I use at least one together with Turmeric (fresh-grinned or dry-powdery) to get a bright red solution?

Update

I should note that I didn't intend to ask this question as a "how to make Tandoori colored seasoning paste" question;

About six years ago, I was told by the owner of a large Indian restaurant in Israel, what is the secret (or at least one of the secrets) and the following seems to be that secret from what I understood then and did and replicated several times:

1) Take about 500 gram of dried Kashmiri peppers; clean them from pedicel, placenta and seeds.

2) Put the dried Kashmiri peppers in a metal pan and pour boiled water upon them.

3) Let the dried Kashmiri pepper boil and cool ("water-marinate") about 3-6 hours but don't cook them.
Mere boiling, without cooking, enhances pigmentation but doesn't destroy some pigments as in cooking.

4) Remove the water and let the peppers dry.

5) Grind the processed peppers into a paste with a bit of water and cooking oil and add a tablespoon of that paste as a basis for any Tandoori paste you are making (along with salt, lemon, garlic, ginger and possibly other spices).

Curcumin(the active ingredient in Turmeric) can turn red in two ways: Either by acting as an acid-base indicator or by forming a complex with a phospholipid film, as you stated.

Colour change due to Acid-Base action

The curcumin molecule is a tribasic acid in basic solution, where it exists as its enol tautomer. The loss of one or more protons from the phenolic hydroxyl groups convert the benzene rings into their quinonoid form, which have a larger absorption/emission wavelength than the benzenoid form, giving a red colour.

This indeed works experimentally: a paper highlighting this can be found here. A simple home experiment would also suffice

The left one is a suspension of turmeric in water; the right one is the same suspension, but with a few drops of soap added to it, which make the medium alkaline and cause the turmeric to turn red. Unfortunately, around my kitchen, I could not find an edible compound basic enough to cause turmeric to change colour. Lime is a possible compound that I could test with but I don't have any foodgrade lime powder lying around. Another issue with using this colouring is that any acidic medium will revert the colour back to yellow: this means that sprinkling lemon as garnishing on your dish would change the colour from an appetizing red to a dull yellow.

Colour change due to Phospholipid complex

A lot of papers have been released on this topic, such as this one and this one. The second paper mentions the following synthesis method to prepare curcumin liposomes (curcumin surrounded by a spherical membrane of phospholipid)

  The MFGM phospholipids/soybean lecithin was dispersed in chloroform. Curcumin was dissolved in anhydrous ethanol and thoroughly mixed with the MFGM phospholipids/soybean lecithin dispersion in a 250-mL round-bottomed flask. The mixture solution was evaporated on a rotary evaporator at 35°C, until a thin film was formed on the flask. The film was washed with PBS and completely hydrated for 2 h. After the hydration of thin lipid film with PBS, the lipid dispersion was sonicated for 4 min, and finally the liposomes encapsulated with curcumin were obtained.

For cooking food, this above synthesis method seems unreasonable and overkill. I urge you to experiment with a few methods of forming the red curcumin-phospholipid complex yourself, with soy lecithin/eggs. I would have tried some methods out but I don't have these ingredients at home. see edit.

The advantage of having curcumin in a liposome is that it is much more stable: unlike the acid-base action, curcumin in a liposome remains stable across a wider range of pH's, as shown in this graph. This method does show promise, but as I mentioned, you would have to give this a try yourself.

Other practical solutions

Seasoned Advice would probably give you would get a better answer in this field, but I'll drop in my two cents. Generally, chilli powder is used to provide the red colour to Tandoori. More specifically, Kashmiri chilli powder is specifically used for tandoor preparations, such as tandoori chicken and chicken tikka. This powder is applied as a dressing (in combination with oil, rice flour etc after cooking the food) and during marination (chicken is marinated with curd, red chilli powder and turmeric).

Happy cooking!

References:

Kuntal Maiti, Kakali Mukherjee, Arunava Gantait, Bishnu Pada Saha, Pulok K. Mukherjee; Curcumin–phospholipid complex: Preparation, therapeutic evaluation and pharmacokinetic study in rats, International Journal of Pharmaceutics, Volume 330, Issues 1–2,
Sam-ang Supharoek, Kraingkrai Ponhong, Watsaka Siriangkhawut, Kate Grudpan;
Employing natural reagents from turmeric and lime for acetic acid determination in vinegar sample, Journal of Food and Drug Analysis, Volume 26, Issue 2,
Hong-Hao Jin, Qun Lu, Jian-Guo Jiang; Curcumin liposomes prepared with milk fat globule membrane phospholipids and soybean lecithin, Journal of Dairy Science, Volume 99, Issue 3,
Research gate: Why does curcumin have a red colour in the alkaline condition
Wikipedia: Phospholipids, Curcumin

EDIT: I tried to make a phospholipid complex with turmeric and egg yolks in the kitchen. Around a teaspoon of turmeric was mixed with an equal amount of egg yolk and heated in a water bath: An orangish colour was observed, which was in contrast to the bright yellow of the egg yolk. (I scraped the top layer away which had a larger amount of turmeric to reveal the bottom layer, which had more egg yolk)

A possible conclusion that can be drawn from this experiment is that the complex is indeed forming, but the concentration of phospholipids/curcumin is too small to impart the magnitude of red colour required for food colouring. This pretty much conclusively proves that no easily made derivative of turmeric can be used as red food colouring.


Why NH4OH? Why not NH3(aq)? The title says it all.

I had this question when helping someone with their high-school chemistry homework. What are the reasons that someone would use an (from my point of view) obfuscated notation?

The lone pair at the nitrogen is very well suited to form hydrogen bonds. This bond is especially strong, since ammonia has a quite high proton affinity, it is a medium strong base.

Now consider the autoprotolysis reactions in pure water and (liquid) ammonia:
\begin{align}
\ce{2H2O &<=> H3O+ + OH-}\tag1\\
\ce{2NH3 &<=> NH4+ + NH2-}\tag2
\end{align}

Mixing these systems, the reaction of ammonia with water will yield
\begin{align}
      && \ce{2H2O + 2NH3 &<=> H3O+ + OH- + NH4+ + NH2-}\tag3 \\
& (~-~)& \ce{H3O+ + NH2- &-> H2O + NH3}\\\hline
      && \ce{H2O + NH3 &<=> NH4+ + OH-}\tag4\label{eq:ammonia+water}
\end{align}
\eqref{eq:ammonia+water} is incorporated into a huge and strong network of hydrogen bonds. In this case the equilibrium will be very much pushed to the right side, meaning, that almost all ammonia molecules will be protonated.

Therefore it is more appropriate to write:
\begin{align}
\ce{NH4OH (aq) &<=> NH4+ (aq) + OH- (aq)}\tag{4a}
\end{align}

However, it is always advisable to look at the conditions of the whole reaction system.

Note that $\ce{NH4OH}$ is a pretty outdated notation, if you would want to consider one water entity explicitly, $\ce{NH3.H2O}$ would be more appropriate. See the answer by Maurice for some more information: Can acid-base reactions occur in a non-aqueous medium?


Why does Si2H4 readily polymerize at RTP but ethene does not (requires catalyst)? In the case of both Silicon and Oxygen, the double bond is less stable than two single bonds. Consequently, surely both should polymerize, why don't they? What accounts for the difference between Carbon and Silicon in this context?

Why does $\ce{Si2H4}$ readily polymerize at RTP but ethene does not?

$\ce{Si2H4}$, the silicon analogue of ethylene, is referred to as disilene, a member of the silene family.

The silicon atoms in disilene are $\ce{sp^2}$ hybridized and the orbitals  overlap to form both a sigma and pi bond, analogous to the hybridization and bonding in ethylene. However, the $\ce{Si=Si}$ bond length is significantly over 2 angstroms, whereas the double bond length in an ethylene is much shorter at 1.34 angstroms. This much larger double bond length in silenes results in extremely poor p-orbital overlap which yields an extremely weak silicon-silicon pi bond.  While it takes energies on the order of 65-70 Kcal/mol to cause rotational cis-trans isomerization about a carbon-carbon double bond, it only takes 10-20 Kcal/mol to effect similar isomerization about the silene double bond.

Since the pi bond in a silene is so weak it doesn't require much energy to disrupt it making the molecule very reactive, much more so than an olefin.  The molecule can polymerize or react with whatever else may be around such as oxygen, water, etc.

Replacing the 4 hydrogen atoms in disilene with bulky groups stabilizes the molecule and allows the molecule to be isolated and characterized.  The bulky groups make it more difficult for another molecule to get close enough to react with the disilene pi bond.

Still, even these sterically protected molecules react with oxygen and water at temperatures not far above room temperature.  Here is a link to a full paper that discusses the preparation and reactivity of these sterically protected silenes.


HDAC inhibitor and how it works on a cellular level I have been working on a research project on the synthesis of a histone deacetylase inhibitor compound. I just realized that I am still a bit fuzzy on the how it works on a cellular level.

So, how does a histone deacetylase inhibitor work on a cellular level?

Do you have a more specific question? The best I can give you is the mode of action that is generally seen across most biological mechanisms:

An HDAC inhibitor would cause inhibition by binding to HDAC, thus causing a conformational change that disallows the protein from being able to transfer  acetyl-groups to a histone-residue (cysteine I believe?).

Usually, if a protein is something that is abundant in cells, an inhibitors/activators work directly through bonding to the target to cause conformational changes. There is probably also another inhibitor that can bind directly to the HDAC-gene to cease its expression (but I don't see this being beneficial in a growing cell - but maybe in mature cells).


How do α,β-unsaturated acids undergo decarboxylation? I know the concerted mechanism for β-keto acids, but neither could I figure out, nor was I able to find out the mechanism for α,β-unsaturated acids.
Any help is appreciated.

Albeit carboxylic acids and their derivatives lose carbon dioxide under a variety of experimental conditions, the literature on the decarboxylation of $\alpha,\beta$-unsaturated carboxylic acids is not common. The reason for this is all $\alpha,\beta$-unsaturated carboxylic acids seemingly do not undergo decaboxylation under mild conditions, except for those containing substitutions at $\beta$-$\ce{C}$ position, which stabilize the intermediate carbocation.

A wide variety of experimental conditions undoubtedly make the reaction to undergo by many different mechanisms. For example, the role of acid catalysis in the decomposition of cinnamic acid (and related compounds) has been clearly demonstrated in Ref.1.
The mechanism for this acid catalyzed decarboxylation is illustrated in the following figure (see I in the diagram). The structures of some example of these susceptible substrates are also included (Ref.1) in the diagram:

Some carboxylic acids undergo decarboxylation by pyrolysis. It is an experimental fact that the olefins resulting from pyrolysis of simple $\alpha,\beta$-unsaturated acids (note that simple means at least no substitution on $\alpha$-$\ce{C}$) are homogeneous and contain terminal unsaturation (Ref.2). Two different mechanisms leading to the same olefin are formulated in Ref.2. The first one is essentially the same mechanism suggested for the decarboxylation of cinnamic acid and related compounds (see I in the diagram).
Before discuss about the second mechanism, let’s look at the proposed mechanism for $\beta,\gamma$-unsaturated carboxylic acid compounds (see II in the diagram). The decarboxylation mechanism of $\beta,\gamma$-unsaturated carboxylic acid compounds undergo the reaction through a cyclic transition complex to give alkene products. This mechanism mimics the one proposed for the thermal decomposition of $\beta$-keto acids.

As a consequence, an alternative mechanism for thermal decomposition of $\alpha,\beta$-unsaturated carboxylic acid compounds was also proposed if the substrate contains allylic $\mathrm{sp^3}$ carbon to extend the double bond (see III in the diagram). It was assumed that under thermal energy, substrate first isomerizes to the $\beta,\gamma$-isomer, which then decarboxylates through a cyclic transition complex to give the terminal alkene product as given by mechanism I.

References:

W. S. Johnson, W. E. Heinz, “The Acid-Catalyzed Decarboxylation of Cinnamic Acids,” J. Am. Chem. Soc. 1949, 71(8), 2913–2918 (DOI: 10.1021/ja01176a098).
R. T. Arnold, O. C. Elmer, R. M. Dodson, “Thermal Decarboxylation of Unsaturated Acids,” J. Am. Chem. Soc. 1950, 72(10), 4359–4361 (DOI: 10.1021/ja01166a007).


C-F bond lengths of octafluoropropane using Bent's rule I was trying to predict different $\ce{C-F}$ bond lengths of octafluoropropane,  $\ce{CF3-CF2-CF3}$ :

Applying Bent's rule here, we see that

For $\ce{C^2}$, we have $\ce{C^3}$ and $\ce{C^1}$ towards which s-character of $\ce{sp^3}$ orbitals is richer. Hence $\ce{sp^3}$ orbital of $\ce{C^2}$ has lesser s-character directed towards $\ce{F}$.
For $\ce{C^1}$ (or $\ce{C^3}$), we have only  $\ce{C^2}$ toward which  s-character of $\ce{sp^3}$ orbitals is richer. Hence $\ce{sp^3}$ orbital of $\ce{C^1}$ has greater s-character directed towards $\ce{F}$.

Using this I think $\ce{C^2 - F}$ bond length is greater than $\ce{C^1 - F}$.

Q. Is this explanation reasonable?

Q. Can we use Bent's rule to predict bond-lengths here or we should consider steric effects of $\ce{CF3}$ (trifluoromethyl) groups ? (here both lead to same result)

I mean in general on which rule should we rely on? On wikipedia article on Bent's rule its written that often times Bent's rule prediction would contradict predictions based on other classical effects alone (steric effects etc). In such cases how can we justify any prediction?

(Wikipedia) Bent's Rule - Bond lengths

  A prediction based on sterics alone would lead to the opposite trend, as the large chlorine substituents would be more favorable far apart. As the steric explanation contradicts the experimental result, Bent’s rule is likely playing a primary role in structure determination.

It is very much non-trivial to attempt estimations on a perfluorated alkane using Bent’s rule. In Bent’s rule, one should not only consider the effects of the electronegative substituents on the substituted atom itself, but also onto the next atom. Consider for example the following table taken from Bent’s review article:[1]

\begin{array}{lr}\hline
\text{Compound} & \angle(\ce{R^1-O-R^2}) \\ \hline
\ce{Me-O-Me} & 111^\circ \\
\ce{Me-O-H} & 107{-}109^\circ \\
\ce{H-O-H} & 104.5^\circ \\
\ce{F-O-F} & 103.8^\circ \\ \hline\end{array}

This very nicely corresponds to increasing group (!) electronegativities[2] in the series $\ce{Me} > \ce{H} > \ce{F}$. $\ce{CF3}$ should have a higher electronegativity than both $\ce{Me}$ and $\ce{H}$ (compare the acidities of formic acid, acetic acid and trifluoroacetic acid). Yet $\angle(\ce{C-O-C}) = 120^\circ$ in $\ce{F3C-O-CF3}$.[3] This corresponds to a higher $\mathrm s$ contribution than for the previous groups indicating other, more important contributions (here the anomeric effect of oxygen acting on the $\ce{CF3}$ group).[4]

Luckily, there are no such secondary effects in perfluoropropane. However, if you want to estimate all bond lengths compared to propane or 2,2-difluoropropane, these must be considered. Specifically, going from propane to 2,2-difluoropropane should not only reduce the $\ce{C-C}$ bond lengths due to increased $\mathrm s$ contribution but also due to partial double bond character due to a gauche-type effect which can be rationalised in Lewis formalism by the following resonance structure:

$$\ce{H-C-C-F <-> \overset{+}{H}\bond{...}C=C\bond{...}\overset{-}{F}}$$

Of course, the partial $\ce{C-C}$ double bond and the less-than-single $\ce{C-F}$ bonds also mess with the observed bond angles in 2,2-difluoropropane; they don’t only influence bond lengths.

So to attempt an estimation of the differences in bond length of the two $\ce{C-F}$ bonds in perfluoropropane, we would need to know how the group electronegativites of the $\ce{CF3}$ and $\ce{C2F5}$ groups compare to the electronegativity of fluorine and to each other. I don’t see any simple way to do this.

Mack et al. have published gas phase structures of 2,2-difluoropropane and perfluoropropane. These are their results (all values in picometres):[5]

\begin{array}{lcc}\hline
\text{compound} & d(\ce{C-C}) & d(\ce{F2RC-F}) & d(\ce{R2FC-F}) \\ \hline
\ce{Me2CF2} & 151.2 & \text{–/–} & 137.0 \\
\ce{(CF3)2CF2} & 154.6 & 132 & 134 \\
\ce{Me2CH2} & 153.1 & \text{–/–} & \text{–/–} \\ \hline \end{array}

As you can see, the experimental values just confirm your prediction of the central $\ce{CF2}$ having longer bond lengths than the terminal $\ce{CF3}$ groups. As expected, all $\ce{C-F}$ bonds in perfluoropropane are shorter than the bond lengths of 2,2-difluoropropane. And the $\ce{C-C}$ bond lengths increase from 2,2-difluoropropane to propane to perfluoropropane.

The conclusion is that indeed $\ce{CF3}$ is slightly less electronegative than fluorine, but the effect seems minimal.

References and Notes:

[1]: H. A. Bent, Chem. Rev. 1961, 61, 275–311. DOI: 10.1021/cr60211a005.

[2]: Bent explicitly considers group electronegativities. Only by that is the group electronegativity of a methyl group lower than the electronegativity of hydrogen.

[3]: A. H. Lowrey, C. George, P. D'Antanio, J. Karle, J. Mol. Struct. 1980, 63, 243–248. DOI: 10.1016/0022-2860(80)80330-9.

[4]: F. R. Leroux, B. Manteau, J.-P. Vors, S. Pazenok, Beilstein J. Org. Chem. 2008, 4, 13. DOI: 10.3762/bjoc.4.13.

[5]: H.-G. Mack, M. Dakkouri, H. Oberhammer, J. Phys. Chem. 1991, 95, 3136–3138. DOI: 10.1021/j100161a034.


Formation of sulfur colloids What happens when we pass sulfur vapors through cold water? What is the reaction for it? Which type is colloid is this? And why doesn't it form in hot or lukewarm water

What happens when we pass sulfur vapors through cold water?

You get more or less coarse sulfur powder-slurry.

  What is the reaction for it?

Sulfur vapor ( S2 mostly) reacts to form S8
depending on speed of cooling, some S6 might
form temporarily

  Which type is colloid is this?

Sometimes this might result in a "colloid",
but I recommend not to use this outdated word,
because so much nonsense was written in the classical
days of "colloids".

  And why doesn't it form in hot or lukewarm water

Who says/writes this?


What happens if you mix two insoluble liquids but with the same density? Would it make two separate phases? or would it make something like and emulsion?

There is no "or" in the first place. An emulsion is a two-phased system, it is just that one phase comes in a form of small droplets. Small as they are, however, they are pretty big from molecular point of view.

Now to the point. If you stir the liquids vigorously, you'll have an emulsion. Short of that, they will attempt to minimize the surface, and the shape with minimal surface is a sphere. Indeed, that's what you will end up with. It's been done before. Look up the Plateau's oil drop experiment: they put oil in a water/ethanol mixture with ratio tuned so as to match the density of oil, and behold, it just floats there in a huge spherical drop.

If both liquids get in contact with the wall, the geometry might become more complicated. The phase separation surface will develop certain contact angle where it touches the wall, and the rest of it will form some kind of minimal surface, which can be surprisingly diverse.

So it goes.


Synthesis of (1S,2S,4S)-4-ethenyl-5-methylcyclohexane-1,2-diol I'm trying to devise a synthesis for the following question.

I am aware it involves retrosynthetic, Wittig, Grignard and diels alder reactions, but unfortunately I can't really grasp all concepts which make me stuck. If there's any help or further explanation would be great.
I have tried to do retrosynthetic for the alkyne compound, also the Wittig reaction. I think Grignard should be the last step where it will change the oxygen bond to $\ce{OH}$. But I can't get myself to there yet.

Because the target structure contains the motif of a cyclohexane-1,2-diol, I suggest an approach including a Diels-Alder reaction (DA in the illustration below) of 1,3-butadiene and the commercially available diethyl acetal of propargyl aldehyde:

The hydrolysis of the acetal then sets the stage to install the methyl group with the cuprate and 1,4-selectivity in respect to the carbonyl group.  As an intermediate protection of the aldehyde, a cyclic acetal is installed. The Prévost reaction typically yields 1,2-(trans)-diols, thus possibly here, too (brief review).  Eventually, the deprotected aldehyde is transformed into the exocyclic alkene with the Lombardo reagent. Contrasting to the conditions of a Wittig reaction, the conditions of the Takai-Lombardo reaction are milder (absence of a strong Bronsted base).
Alternative installations of the (trans)-diol could be achieved e.g., by epoxidation and subsequent ring opening with a peracid like mCPBA or (in situ generated by) oxone.


How is carbon-halogen bond broken during Finkelstein reaction? Finkelstein reaction is a halogen exchange reaction between haloalkane and a salt of a different halogenide:

$$\ce{R−CH2−Cl + KI -> R−CH2−I + KCl}$$

From an answer of a previous question:

  As per Le chatelliar's principle the forward reaction is favored in
  presence of dry acetone which will dissolve $\ce{KI}$ but not
  $\ce{KCl}$ or $\ce{KBr}$. On account of insolubility of
  $\ce{KCl/KBr}$; they are not available for backward reaction[...]

This reaction proceeds with $\mathrm{S_N2}$ mechanism; i.e. it is a single step reaction. $\ce{C-Cl}$ bond breaks and $\ce{C-I}$ bond forms. Now we know that energy for bond breaking comes from bond making. However we know that
$\ce{C-I}$ bonding cannot supply enough energy to break $\ce{C-Cl}$ bond.

To put in other words: chlorine is a weaker leaving group than iodine.

From where does the energy to break $\ce{C-Cl}$ bond comes then because for $\ce{KCl}$ (precipitate) to form; in first place the $\ce{C-Cl}$ bond must break?

Think of the overall process and you will see where the energy comes from

Looking at just the reaction of iodide with the alkyl chloride doesn't give you the complete picture of the reaction. Sure, in that part of the reaction the bond-breaking and bond formation don't balance out. But in reality this doesn't mean there is no reaction: it just means that the equilibrium of that step strongly favours the alkyl chloride (there is still some dynamic formation of the iodide, perhaps in small quantities).

But the driving force here isn't that initial reaction. The real driving force is the precipitation of the KCl (it is very insoluble in the conditions of the reaction). From a mechanistic point of view this precipitation virtually eliminates the presence of free chloride in solution meaning that, if any is formed, it will rapidly disappear and not be available to drive the equilibrium back to the more stable alkyl chloride (even if the equilibrium of the halide exchange strongly favours the chloride).

From an energy or thermodynamic point of view, the overall reaction gets enough energy from the precipitation of the KCl (which releases energy) to compensate for the energy lost by substituting iodine for chlorine. So, if you look at the overall mechanism, the second step (precipitation) makes up for the apparent unfavorability of the first step.


Why are weak acids weak? What are weak acids? My chemistry and biology classes skimmed over them briefly but nothing in depth. I understand that an acid is weak if it partially dissociates and breaks down into ionic compounds, but they never stated why.

So my question is why do weak acids break down in water? Do strong acids dissociate at all? What if they were in a different solution (or do they just stay themselves all the time?)

What are weak acids?

Examples of weak acids in water solution would be acetic acid (main component of vinegar) and ammonium ion (conjugate acid of ammonia).

  I understand that an acid is weak if it partial dissociates and breaks down into ionic compounds but they never stated why.

An acid is weak if not all of the acid molecules ionize into hydrogen protons and its conjugate base in a particular solvent system. Alternately, if we were to use the broader, Brønsted definition, an acid is weak if it does not completely or nearly completely donate its proton to some base.

Acetic acid is a weak acid in water. Put acetic acid into pure liquid ammonia and now acetic acid will fully dissociate. The proton will be lost from acetic acid to ammonia and this is a favorable reaction. Makes sense, because the only thing that's changed is the solvent that's grabbing the proton - and ammonia is only a few billion times stronger as a base than water.

  That and do the strong acids dissociate at all. Like if they were in a different solution or do they just stay themselves all the time.

Strong acids by definition ionize to an extent of 100% or nearly 100%. Strong acids will give up their protons completely to the strongest base in the system.

Note that in the above text I have used the word "ionize" to describe molecular acids because by definition when these acids react, they ionize. Or form ions. On the other hand salts of acids and bases such as sodium hydroxide are said to "dissociate," because ionic compounds are made of ions, and it wouldn't make much sense to say that ions ionize or become ions.


Why does the BET equation take into consideration a multimolecular-layer, even though it deals with chemisorption? I came across the Brunauer-Emmet-Teller (BET) equation recently, and one thing that I found strange was an implicit reference to the formation of molecular multi-layer.

In the BET equation

$$
\begin{align}
\frac{P}{V(P^0 - P)} &= \frac{1}{CV_m} + \frac{C-1}{CV_m}\cdot \frac{P}{P^0}\, ; \tag{1}\\
C &= e^{(Q_1 - Q_2)/RT}\quad (C:\text{constant})\, ,\tag{2}
\end{align}
$$

where $V_m$ – monolayer adsorption amount, V – adsorption amount at the equilibrium pressure $P$, $Q_\mathrm{L}$ refers to the heat of adsorption of the second and higher layers of adsorbate onto the adsorbent.

Now according to my textbook, chemisorption only results in the formation of a molecular monolayer. Multilayers are formed by physisorption.

The thing is, I'm told the BET equation is an extension of Langmuir's adsorption isotherm for chemisorption at low pressures. If the BET equation applies only to chemisorption, then what is the point of including a term dependent on the formation of multilayers, when chemisorption doesn't even form a multilayer all by itself.

Is there something I'm overlooking here? The internet has not yielded any decent result in that regard.

Brunauer–Emmett–Teller (BET) explains the physical adsorption of gas molecules on a solid surface, and doesn't address chemisorption. Even the wikipedia link that you provided, makes that quite explicit. However, you are correct to think that it is in a extension of the Langmuir adsorption model, which only accounts for monolayers.

The assumption that underscore this model are as follows:

  Adsorptions occur only on well-defined sites of the sample surface (one per molecule)
  The only molecular interaction considered is the following one: a molecule can act as a single adsorption site for a molecule of the upper layer.
  The uppermost molecule layer is in equilibrium with the gas phase, i.e. similar molecule adsorption and desorption rates.
  The desorption is a kinetically-limited process, i.e. a heat of adsorption must be provided:

  these phenomenon are homogeneous, i.e. same heat of adsorption for a given molecule layer.
  it is E1 for the first layer, i.e. the heat of adsorption at the solid sample surface
  the other layers are assumed similar and can be represented as condensed species, i.e. liquid state. Hence, the heat of adsorption is EL is equal to the heat of liquefaction.

  At the saturation pressure, the molecule layer number tends to infinity (i.e. equivalent to the sample being surrounded by a liquid phase)

References: The original article

P.S: Good reading material on BET theory can be found here


Does the boiling point of an aqueous solution depend on what the solute is? Consider a situation where we have two aqueous solutions which have the same concentration (e.g. 1 M), but the solute is different. For example, one is a solution of glucose, and the other is a solution of ethanol.

I am aware that dissolving something in water can lead to changes in boiling point (e.g. elevation). But would the boiling points of these two solutions be affected in the same manner, given that the concentration of the solute is the same? Or will the exact boiling point depend on the chemical species that is dissolved?

No, in water the two solutions will not have the same boiling point.

Ethanol is a volatile liquid with boiling point lower than water, and lowers the boiling point.

Water forms an azeotrope with ethanol.  There are positive and negative azeotropes, so when two or more volatile liquids are dissolved, you cannot easily predict whether boiling point will decrease or increase.

There can also be cases, like adding HCl to water, where as solute is added, first the boiling point increases to a maximum (108 °C around 20% HCl) above the boiling point of water, then decrease to below the boiling point of water (48 °C for 38% HCl).

Glucose is non-volatile and increases the boiling point of water.  The boiling-point elevation formula can be applied to non-volatile solutes in volatile solvent.


What is Vorlander's rule? What is Vorlander's rule? I have got it from a book all of a sudden, but without any further information about it; could you please explain it a bit?

Vorlander's rule discuss about the directing powers of substituent present in benzene ring. Taken from here:

  Saturated groups (containing single bonds) direct predominately 2- and 4-positions.
  Unsaturated groups (containing multiple bonds) direct predominately at 3-positions.

Here are some examples:


Check My Beginner's Guide to Shells, Energy Levels, Subshells, and Orbitals I am a beginner trying to summarize what I've learned about:

Shells
Energy Levels
Subshells and
Orbitals

all at the high school level.
Unlike related threads, I'm going to try to:

Define each word and state a few key properties of each.
Highlight what is most important for beginners.
Summarize all of that in a table.

Please let me know how I've done!
Thanks!

tl;dr Summary: Electrons move within orbitals which are in subshells which are in shells which are in atoms.

Shell aka Energy Level
Below atoms, shells are the largest grouping of electrons beginner chemists should care about, and among those, the innermost two shells are most important to beginners.
They correspond to periods (rows) on the periodic table of the element.

In the 3rd period, all elements have 3 shells.
In the 5th period, all elements have 5 shells.
In the $n$th period, all elements have $n$ shells.

Generally speaking, the larger $n$ means:

The shells are farther away from the nucleus and thus the electrons in it are less attracted to the nucleus.
The shells can hold more electrons, i.e. an atom of that element requires more valence electrons to be stable.

Subshell
Each shell consists of at least one subshell. Below atoms, this is the second largest grouping of electrons beginner chemists should care about.
There are 4 types of subshells: $s$, $p$, $d$, and $f$. For beginners, the $s$ and $p$ subshells are the most important.
The notation for shells and subshells:

$1s^2$ means the 1st shell's $s$ subshell has 2 electrons in it.
$4d^5$ means the 4th shell's $d$ subshell has 5 electrons in it
$AB^X$ means the $A$th shell's $B$th subshell has $X$ electrons in  it.

Note: This notation does not specify which orbital those electrons are in... but that info does not matter much to beginners.
Orbital
An orbital is an actual path (probability distribution function, PDF) that either 1 or 2 electrons can be whizzing through at a time. This is the smallest grouping of electrons beginner chemists should care about. The math underlying that PDF is not critical for beginners.
Each subshell consists of at least one orbital.
THE BIG TABLE
Feel free to edit the table. :)

All looks good at a school level, but I would suggest or convince you that we should not attach any physical pictures or meaning to all these atomic concepts. If you pursue science in future, you will have to unlearn a lot of these stories ingrained by school or even university teachers. It is okay to acknowledge and say that we do not fully understand everything as yet. Nature is very very complex. Our only effort is to make simple models of the phenomena we observe in the Nature.

The shell model is outdated and obsolete yet it is good for educational purposes. It was inspired by the planets rotating around the Sun, so learning about it from a historical perspective is good. For example, you wrote "The shells are farther away from the nucleus and thus the electrons in it are less attracted to the nucleus." It may not be a correct picture, because in reality there are no circular shells per se, like an onion ring around a nucleus. Brilliant physicists spent their life time in understanding the atom so don't be surprised if you find atomic structure overwhelming. Indeed it is.

More importantly just delete this interpretation from your mind "An orbital is an actual path (probability distribution function, PDF) that either 1 or 2 electrons can be whizzing through at a time."

An orbital is just a mathematical construct with no physical reality as textbooks show them as balloon models and so on. The rigorous and correction version of an orbital is "Wavefunction depending explicitly on the spatial coordinates of only one electron" (IUPAC)

Their shape depends on the coordinate system being used to solve the Schrodinger's equation. Can you believe that Schrodinger never showed or drew a single picture of orbitals in his four famous papers?

The typical shapes of s, p, d, f orbital are shown in Cartesian coordinate system, but physicists and mathematicians have plenty of other coordinate systems to solve differential equations. Do the electrons care what coordinate system a human chose to represent them or calculate their energies? The answer is no. Their energy turns out to be the same regardless of the coordinate system. Nobody teaches that and I also recently learned this from a physicist who has published a lot on orbital shapes. That gentleman is a big opponent to the teaching of orbitals and the myths associated with them in undergraduate curriculum. Not all agree with him including Linus Pauling.

The atomic structures "stories" as taught to chemistry students as facts is one of the worst things in modern educational system. In my humble opinion, chemists should take the help of modern physicists and ask them how to represent the latest views of atomic structure right from beginning classes. It does not need to have crazy mathematics, but good scientists can explain and elaborate things in words as well. As someone said, "If you can't explain it simply, you don't understand it well enough."

Book recommendation: Hydrogen: The essential element by J. Rigden (from Harvard). A thin book which is very illuminating. Chemical educators should adapt this style. The last chapter or so talks about anti-hydrogen.


CNT Raman ratio of G band to 2D band? I believe that the G band (~1600 cm-1) shows the stretching mode of the graphene in a carbon nanotube (e.g. chiral, zigzag, armchair) and the 2D band (~2680 cm-1) shows the number of walls in the CNT. I vaguely remember there is some information that can be gleaned from the ratio of the two. Is this the case?

I think there might be a mistake in your question...
Around $\pu{2680 cm-1}$, there is the $S_1$ band (attributed to overtone or combination scattering). I think the $2D$ band you are referring to is the band shouldering the $G$ band (also called $D'$) which is around $\pu{1680 cm-1}$.
About the ratio, there are two ratios that are often used to characterise $\mathrm{sp^2}$ carbon based materials such as the CNT.
$$\text{Intensity ratio} =  R_1 = \frac{I(G)}{I(D)}$$
$$\text{Area ratio} = R_2 = \frac{A(G)}{A(G)+A(D)+A(D')}$$
Both these ratios help characterise the degree of organization of your sample.
For more information or more details, I would recommend checking out Spectrochim. Acta, Part A 2003, 59 (10), 2267–2276


Is hybridization used in ab initio valence bond calculation? Many general chemistry textbooks introduced the concept "hybridization" to construct a symmetry-adapted VB-type wavefunction. In the textbooks, usually the minimal basis is used and without optimizing all resonance forms.

My question is, in the ab initio valence bond calculation with extended basis set and optimizing many resonance forms,

(i) do people still use hybridization to construct the wavefuntion?

(ii) if the answer to question (i) is yes, does the result differ from the textbook description? For instance, the lone pair electron in the $\ce{H2O}$ molecule? (is there any unitary equivalent result like orbital localization in MO-based method?)

I asked the question (ii), since NBO analysis tends to give non-equivalent composition for lone pairs in $\ce{H2O}$, one sp, the other p. I am wondering what is the answer in ab initio valence bond method.

(i) do people still use hybridization to construct the wavefuntion?

Yes and no. No: Only atomic orbitals are included in the base set. Yes: if the wavefunction is optimized, linear combinations of atomic orbitals are allowed, and we all know that one S-orbital + some P-orbitals give hybridized orbitals.

For $\ce{H2O}$, the lowest 3 molecular orbitals are generally these (see below):

At first, this result seems unlike the SP$^3$ bonds between hydrogen and carbon. There is a trick, however. It is called "localization of the orbitals", meaning that we generate linear combinations of molecular orbitals to generate new ones. Below, I'll give you an example in which we linear combinations of molecular orbitals $\varphi_2$ and $\varphi_3$ generate orbitals that look more like the SP$^3$ orbitals from text books.

(ii) if the answer to question (i) is yes, does the result differ from the textbook description? For instance, the lone pair electron in the $\ce{H2O}$ molecule? (is there any unitary equivalent result like orbital localization in MO-based method?)

As you may have noticed above, the results differ slightly, but I can elaborate.

Lone pairs are given by $\varphi_4$ and $\varphi_5$ below.

For $\varphi_4$, note that p$_z$ is along the z-axis, which is in the plane of the oxygen and hydrogen atoms: it is the 2-fold rotation axis.

For $\varphi_5$, note that p$_y$ is along the y-axis, which is perpendicular to the plane of the oxygen and hydrogen atoms: it is out of the plane.

For $\varphi_6$ and $\varphi_7$, note that they are unoccupied orbitals (i.e. lumo and lumo+1)

Now, orbital localization can be applied, linear combinations of $\varphi_4$ and $\varphi_5$ will give 2 lone pair orbitals, that stick slightly out of plane. I haven't tried drawing those yet, as it is not so easy to do so in this representation in 2D, so I'l try to switch to another representation...

As you see, those localized molecular orbitals closely resemble the SP$^3$ lone pair orbitals on oxygen, as predicted by valence bond theory. The only difference is the shape & probability of the small counter-part of the lobe; I included phase, not probability density here.

Does this answer your question?


What is diamagnetic in diamagnetic dilution? I'm reading a paper where the authors study a peptide solution where each peptide may have two spin labels attached to them. They put the peptides on the surface and study them using a diamond sensor.

They perform the experiment in two ways. First, they put only peptides with attached spin labels. Then, they dilute the solution with peptides that do not have spin labels, in a ratio 1:10.

They refer to this dilution as "diamagnetic dilution". I do not know why it's called diamagnetic. I've tried to look for a definition but the term only appears in highly technical papers that I can't understand very well.

Does anyone know why this is called "diamagnetic dilution"?

Ben

Edit: The article I'm referring to is this one:

https://advances.sciencemag.org/content/3/8/e1701116

But my question is general.

The term diamagnetic dilution (in general) implies that the material used for dilution has all its electrons paired, so we will not see its response in an electron spin resonance experiment or any other experiment which can sense unpaired spins.


Discrepancies between smell and taste In my experience, most foods smell like they taste. Fragrant soaps, however, taste quite awful in comparison to their fragrances. How can some things smell and taste identical while others are so vastly different? I have a feeling this may have something to do with the vapor pressures of the molecules that make up the "thing," but I cannot find anything to confirm my hypothesis exactly.

The food that you eat regularly is something that you are used to. You know exactly which smell combines with which taste ... but that does not imply that the chemicals you smell are the same that you taste.

If you had been eating your whole life long fragrant soap with apple smell, you would wonder about the taste of real apples as you wonder right now about the taste of apple soaps.

There are a lot of different substances that are responsible for the special taste of everything you eat. You can find more about flavor or aroma compounds on wikipedia.

In contrast to real fruits, fragrant soaps and other industrial products with a certain smell only contain those substances that are responsible for the characteristic smell of e.g. fruits (often only the main one or two aroma substances or even only synthetic aroma substances that smell similar to the natural products).

But as not all products that shall smell like certain thing also need to taste like those things you only smell the "apple" but you won't taste it.


What are some of the more easily manipulated residues on an enzyme? I am trying to attach a cross-linker to an enzyme and I am looking for candidates for reaction sites. I have several amino acids that are present on the surface of the enzyme (hydrophilic) but I am hoping someone can give some tips for amino acids that are easier to attach linkers to.

I originally was planning on attaching a linker to lysine because it is among the more reactive residues but my enzyme lacks a sufficient number of lysine groups.

Could any one give me some general advice on which amino acids would be better to investigate first? I know the specific residue I choose will depend mostly on the structure of the enzyme, but if you could give me a suggestion of which amino acids are typically easier to manipulate (ie. Lysine > glycine) I would greatly appreciate it.

I'm assuming the OP concerns chemical means of modifying residues. Multiple protein residues other than Lysine may be labeled with an adduct. Here are some examples:

Histidine residues may be labeled using an epoxide attached to a ligand. (Chen et al., 2003)

Cysteine residues may be labeled with a nitrile derivative where the sulhydryl acts as the nucleophile. (Obella et al., 2007)

Arginine residues may labeled using a derivative of the naturally occurring methylglyoxal. (Oya et al., 1999)

Carboxylate functionality such as in Aspartate and Glutamate residues as well as the C-terminus may be labeled via the formation of an amide linker using a carbodiimide such as N-ethyl-3-N',N'-dimethylaminopropylcarbodiimide (EDC). (Gilles et al., 1999)

The above list of residues is not comprehensive, nor are the labeling agents. I recommend checking out the reviews by Basle et al. (2010) and Chen and Wu (2016).
References
Basle, E, Joubert, N, Pucheault, M (2010) "Protein Chemical Modification on Endogenous Amino Acids" Chemistry & Biology 17:213-227. DOI: 10.1016/j.chembiol.2010.02.008
Chen, X and Wu, Y-W (2016) "Selective chemical labeling of proteins" Org. Biomol. Chem. 14:5417-5439. DOI: 10.1039/C6OB00126B
Gilles, MA, Hudson, AQ, Borders, CL (1990) "Stability of water-soluble carbodiimides in aqueous solution" Anal. Biochem. 184:244-248. DOI: 10.1016/0003-2697(90)90675-Y
Gong Chen, Alexander Heim, Doris Riether, Dominic Yee, Yelena Milgrom, Mary Ann Gawinowicz, and Dalibor Sames (2003) "Reactivity of Functional Groups on the Protein Surface: Development of Epoxide Probes for Protein Labeling" J. Am. Chem. Soc. 125:8130–8133. DOI: 10.1021/ja034287m
Oballa, RM, Truchon, JF, Bayly, CI, Chauret, N, Day, S, Crane, S, et al. (2007) "A generally applicable method for assessing the electrophilicity and reactivity of diverse nitrile-containing compounds" Bioorg. Med. Chem. Lett. 17:998–1002. DOI: 10.1016/j.bmcl.2006.11.044
Oya, T, Hattori, N, Mizuno, Y, Miyata, S, Maeda, S, Osawa, T, Uchida, K (1999) "Methylglyoxal modification of protein. Chemical and immunochemical characterization of methylglyoxal-arginine adducts" J. Biol. Chem. 274:18492-18502. DOI: 10.1074/jbc.274.26.18492


What is the use of "pyro" in the naming of organic compound? Recently I asked a question on the naming of polyhydroxy phenols. One of them was pyrogallol (1,2,3-Trihydroxybenzene) which is  synthesized from gallic acid  obtained from oak galls. (BEWARE- not a compound of gallium).

But why "pyro"? Pyro is usen in naming inorganic acids like pyrosulfuric acid ($\ce{H2S2O7}$) or pyrophosphoric acid ($\ce{H4P2O7}$). It is also observed that the carboxylic acid derivative is gallic acid not pyrogallic acid and the hydroxy derivative is pyrogallol not gallol. Why is it so?

So, as @long and @permeakra stated, the name pyrogallol came from the fact that it is obtained by heating gallic acid. The IUPAC goldbook clearly states that:-

  "pyro" is used as a prefix designating compounds formed by heating a compound,
  usually with the elimination of water, carbon dioxide, or other simple
  molecule, e.g. pyroglutamic acid from glutamic acid.

So as per rule, it is pyrogallol and not gallol.


Equation of enthalpy This question arises because: by giving classes in thermodynamics, I have observed that students are often confused between the different definitions (or applications) of the enthalpy concept.

The enthalpy expression is obtained as follows:

From the first law of thermodynamics:
\begin{align*}
 U=Q+W
\end{align*}
Recalling the definition of work:
\begin{align*}
    W=-PV
\end{align*}
Variations added:
\begin{align*}
 \delta U=Q - \delta (PV)
\end{align*}
Its obtained that:
\begin{align*}
 \delta U&=Q-(V\delta P+P\delta V)
\end{align*}
Constant volumen:
\begin{align*}
 \delta U=Q-V\delta P
\end{align*}
Constant pressure:
\begin{align*}
 \delta U=Q-P\delta V
\end{align*}
Variations enlarged:
\begin{align*}
 (U_f-U_i)&=Q-P(V_f-V_i)\\
 (U_f-U_i)&=Q-(PV_f-PV_i)\\
         Q&=(U_f-U_i)+(PV_f-PV_i)\\
         Q&=(U_f+PV_f)-(U_i+PV_i)\\
         Q&=H_f-H_i
\end{align*}
Therefore, the definition of enthalpy is:
$$
    H = U - W
$$

Is this properly proposed?

Definition of internal energy:  U is a function of state, representing the total kinetic and potential energy of the molecules.  U = U(T,V)

First Law of Thermodynamics:$$\Delta U=\delta Q+\delta W$$where the symbol $\Delta$ is used to represent the change in a (path-independent) function of state (like U) between an initial and final thermodynamic equilibrium state of a closed system and the symbol $\delta$ is used to represent the change in a parameter that depends on the process path between and initial and final thermodynamic equilibrium state of a closed system. $\delta Q$ is the heat added to the system over the path, and $\delta W$ is the work done on the system over the path.

Relationship for the Work:$$\delta W=-\int{P_{ext}dV}$$where, for both for reversible and irreversible process paths, $P_{ext}$ is the force per unit area exerted by the gas on the piston face, and, by Newton's 3rd law, the force exerted by the piston face on the gas.  For an irreversible process path, the pressure typically varies with spatial location within the cylinder, so that the average gas pressure does not match $P_{ext}$ at the piston face.  For a reversible process, the gas pressure is uniform within the cylinder, so $P_{ext}=P$ where P is the gas pressure calculated from the equation of state of the gas (such as the ideal gas law), based on the number of moles in the cylinder, the gas pressure in the cylinder, and the gas temperature in the cylinder.

Combining the First Law with the Relationship for Work:$$\Delta U=\delta Q-\int{P_{ext}dV}\tag{rev and irrev processes}$$
$$\Delta U=\delta Q-\int{PdV}\tag{rev processes}$$

Constant Volume ($V_i=V_f)$:$$\delta W=0$$
$$\Delta U=\delta Q$$

Constant Pressure ($P_{ext}=P_i=P_f=P$):$$\delta W=-P_{ext}\Delta V=-P\Delta V$$
$$\Delta U=Q-P\Delta V$$
So,$$\Delta U+P\Delta V=\delta Q$$But, since P is constant,
$$\Delta U+\Delta (PV)=\delta Q$$
The definition of enthalpy is $$H\equiv U+PV$$
Therefore, for a constant pressure process (a process in which $P_{ext}$ over the entire process path is equal to the equilibrium pressures in both the initial and final equilibrium states of the system) $$\Delta H=\delta Q$$


What happens when prop-2-en-1-ol reacts with concentrated hydrogen bromide What happens when prop-2-en-1-ol (allyl alcohol) reacts with concentrated hydrogen bromide?

My Idea: I thought that in the first step protonation of the hydroxyl group will take place followed by formation of a carbocation. Then the attack of bromide ion from hydrogen bromide takes place and that would give me 3-bromoprop-1-ene. Also, I thought since the double bond still remains Markovnikov addition at the double bond might also take place.

But the reaction results into something different
Here is a picture of the above reaction:

The answer has been given as option C. Could someone point out where I might be going wrong or how should the reaction actually proceed?

Rather than trying to rationalise what the question wants to know, I ran a quick SciFinder search. One of the first results was the following reaction, reported for the perdeuterated allyl alcohol by Thiem, Mohn and Heesing in 1985:[1]

Scheme 1: Reaction of perdeuterated allyl alcohol with $\ce{HBr}$. All implicit hydrogens in this scheme are deuterium atoms.[1]

Thiem, Mohn and Heesing do not go into details regarding the mechanism, but as I commented it really doesn’t matter whether the reaction follows a $\mathrm{S_N1}$, $\mathrm{S_N2}$ or $\mathrm{S_N2'}$ mechanism, since the final product will always be the same. In any case, the first step will be protonation of the hydroxy group to turn it into a better leaving group.

Another one of the first results was the following, reported by Karki and Magolan in 2015:[2]

Scheme 2: Reaction of normal allylic alcohol with $48~\%\ \ce{HBr}$ in DMSO/chloroform.[2]

Their proposed mechanism includes the in situ creation of $\ce{Me2S^+-Br}$ which will then react with the double bond giving a bromonium ion; this will then be captured by a second bromide to give 2,3-dibromopropan-1-ol.

I was unable to find single-step reactions of allyl alcohol with hydrogen bromide leading to either 2-bromopropan-1-ol (which would imply a single Markovnikov addition of $\ce{HBr}$ onto the double bond), 1,2-dibromopropane (which would imply Markovnikov hydrobromination of the double bond of allyl bromide) or 1,3-dibromoprop-1-ene (where I have no clue how exactly that would be generated from allyl alcohol in the first place). Funnily enough, you state that this last product, C according to your scheme be the correct answer. I challenge the exam.

References:

[1]: J. Thiem, H. Mohn, A. Heesing, Synthesis 1985, 775. DOI: 10.1055/s-1985-31344.

[2]: M. Karki, J. Magolan, J. Org. Chem. 2015, 80, 3701. DOI: 10.1021/acs.joc.5b00211.


Vapour Pressure and Equilibrium I found this questions on a YouTube lesson and I'm confused with the answers they have given:

"A liquid in its equilibrium with its vapour is in a sealed container at a fixed temperature. The volume of the container in suddenly increased."

Questions:
a) What are the changes in the rate of evaporation and condenation?

b) How will the equilibrium be restored ultimately and what will be the final vapour pressure?

Answers:
a) Rate of evaporation depends on the substance and applied temperature so it will be unchanged and rate of condensation will decrease as vapour pressure is decreased. (Why? Is it true?)

b) They say that the vapour pressure will be restored and the rate of condensation and evaporation will be equal as the system reaches the equilibrium. (But they said that rate of condensation will decrease and rate of evaporation will stay the same. Then how will they be equal?)

Please clarify the whole thing.

The key missing piece is identifying when the system is in equilibrium. When you suddenly increase the volume, the system is temporarily not in equilibrium. In this situation, a few things will happen:

The rate of evaporation won't change for the reason given
The vapor pressure will decrease. As a first approximation, we can use the ideal gas law to see how. $PV=nRT$ can be rearranged as $P=nRT/V$. The number of moles ($n$), gas constant ($R$), and temperature ($T$) will stay the same (this last one in particular isn't really true, but it doesn't change the qualitative result). Then, if the volume increases, the pressure will decrease.
Since the pressure has decreased, the rate at which gaseous molecules bump into the liquid decreases. The ones that do bump into the liquid will condense into the liquid at the same rate. The total condensation rate is the product of those two factors, so the condensation rate will decrease.

Now, since the evaporation rate is the same as it was before, but the condensation rate will go down, we will get a net evaporation. As more liquid becomes gas, the vapor pressure will increase. As the vapor pressure increases, the condensation rate will increase, for the reasons discussed above. Once the vapor pressure is high enough that the condensation rate equals the evaporation rate, then we will have a dynamic equilibrium.


Mercury and 502 glue I do some experiments with liquid mercury and I use a special bottle that was 3d printed (ABS plastic). To prevent leaking of mercury I use 502 glue layered around the bottle.

I want to know, if it is safe to store mercury in this bottle?

Does mercury interact with something in 502 glue?

502 glue: Methylene Chloride , Ethyl acetate, Toluene, Cyclohexane

I am not a professional at chemistry.

Mercury does not react with this compounds. However, I strongly encourage you to

1) use monolithic vial for mercury. 3d-printed pieces are very fragile in my experience. Glass vial should be fine, however I recommend double casing (keep sealed glass vial with mercury in a bigger plastic vial with sulfur dust.)

2) perform all operations with mercury over steel of plastic pan.

3) keep a piece of copper foil/plate and nitric acid at hand. Mercury drops stick to copper, freshly washed in nitric acid, which is used to collect mercury.

4) in case mercury is somewhere you don't have access to, drown it in concentrated iron (III) chloride solution. It will transform metallic mercury into mercury chloride, that does not give vapors.

5) Note, that mercury salts may react with some organic compounds, including aromatics (toluene). While risks are low, organomercury compounds accumulates in living tissues, so caution is required.


Endothermic Conversion of Liquid to Gas When a liquid (in my case I used butane) under high pressure is released into the air, it becomes gaseous, and the container seems to get colder. Why is it that turning a liquid to a gas requires energy when it was simply pressure that was keeping it a liquid? Where does it get this energy? Would the gas formed be colder, or no?

This is probably a bit of a simplistic explanation.

Due to the butane being released from the from the container, the pressure in the container decreases, which in turn, decreases the temperature of the container according to the Pressure-Temperature Law (also known as the Admonton Law), that states for an ideal gas:

$$P  \propto  T$$

The boiling point of butane is normally −1 °C at normal pressures – so in typical use at room temperature, it should have already vapourised. But, due to it being under pressure, it remains a liquid, undergoing evaporation in the container. So, butane has a vapour pressure that increases with temperature, as seen below:

Image source: Wikipedia, who sourced it from the CRC Handbook of Chemistry and Physics 44th ed

Even though butane (like all gases) is not an ideal gas, we can still surmise that if you drop the pressure (by releasing the gas), then the temperature of the gas also drops, especially considering the combined pressure due to that from the container and vapour pressure is rapidly released = rapid depressurisation = rapid loss of temperature, which is felt on the outside of the container.


Osmotic pressure of a solution of a protein? A small protein molecule, code-named "sloth", has a MM of $\mathrm{1.50 x 10^4}$ g/mol. What is the osmotic pressure exerted at $\mathrm{24.0^oC}$ by 25.0 mL of an aqueous solution that contains $\mathrm{3.75 x 10^{10}}$ nanograms of "sloth"? R = 0.08206 (atm L)/(mol K).

The equation my professor gave me is: $\mathrm{O.P. = T*R*M}$. So I did:

$\mathrm{O.P. = (24.0 + 273) K * 0.08206 (atm*L)/(mol*K) * ([(3.75 x 10^{10})x 10^{-9} g] / 1.50 x 10^4 g/mol) / 0.025 L}$

I keep on trying and I get 2.44 atm, but my professor marked the correct answer as $\mathrm{2.44 x 10^{-3}}$ atm. I think he may have divided the moles of sloth by 25 instead of .025 L, which would make his answer $10^3$ smaller than mine. Did I do something wrong?

Ok let's start with $\text{OP} = \text{T}\times\text{R}\times\text{M}$

$\text{T} = (24.0 + 273) \text{K} = 297 \text{K}$

$\text{R} = 0.08206 \dfrac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}}$

$\text{M} = \dfrac{\text{moles}}{\text{L}} = \dfrac{\frac{37.5 \text{ g}}{1.50 \times 10^4 \text{ g/mol}}}{0.025 \text {L}} = \dfrac{2.50\times10^{-3}\text{ mol}}{0.025\text{ L}} = 0.100 \dfrac{\text{mol}}{\text{L}} $

$\text{OP} = 297\text{ K}\times0.08206 \dfrac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}}\times0.100 \dfrac{\text{mol}}{\text{L}} = 2.44 \text{ atm} $

So I get your answer too...


Why is ordinary soap and water adequate to decontaminate possible nerve agents? In the news it was announced that anyone who might have been exposed in the recent Salisbury nerve agent incident should wash any potentially contaminated clothing as normal using regular detergent and water; other personal items can be cleaned with wipes.

The agent in question (identified only as Novichok, a family of chemicals) has been described as the most potent ever made.

How is it that ordinary cleaning can suffice as a method of decontamination? Is it a very unstable molecule? Does water (or detergent, or detergent solution, or alcohol solution) act strongly to decompose such chemicals? If unstable in water, how can it remain stable in the body to act? Could any viable agent remain in the wash waste water (going down the drain) after washing and pose a hazard downstream?

Edited:

The focus of my question is entirely about the apparent paradox of a chemical able to remain stable in an everyday environment, yet readily destroyed/decomposed by everyday means. In the specific case, a chemical which happens to be lethal in trace quantity that can persist, yet simple washing apparently destroys all trace, removing any potential for harm.

Nerve agents are usually susceptible to hydrolysis leading to non-toxic degradation products

Most modern nerve agents are based on fluorophosphonate cores. That is they contain a $\ce{PFO(OR)(X)}$ core where R is some organic radical (-iPr in sarin) and X can be a range of radicals (Me- in sarin, OR or N=R in novichok agents).

Most are not very stable in storage as they are subject to degradation from hydrolysis so have to made not long before use or have to be made ultra pure to avoid any contaminants catalysing their degradation by acid or base hydrolysis. As a result any were deployed as binary agents (where two tables components are mixed inside the weapon to create the active ingredient).

The key to their reactivity is the P-F bond which is easily hydrolysed to give much less harmful products. This hydrolysis is not fast enough to degrade the compounds inside the body before they kill you but is fast enough that they don't hang about for long periods in the environment after use (it is worth bearing in mind that they are extremely powerful inhibitors of acetylcholinesterase, without which your nervous system can't function: it doesn't take a big dose or a long time for them to work). It is also something that can be accelerated by washing. It seems likely, for example, that mildly contaminated clothing can be decontaminated by a standard domestic clothes washing cycle as this would dramatically accelerate any hydrolysis reactions.

It is also worth noting that, if the contamination were large enough, you would notice it very rapidly. So any residual contamination in people not already ill would be of a very low level and easily rendered insignificant by washing.-


How to dissolve a spider with household chemicals? Somehow a spider got into the little window that shows the water level on my electric kettle. The kettle doesn't come apart very easily, and it looks like I can't get this part off without breaking it.

I'd hate buying a  new kettle just because of a little spider but, I also don't want spider juices with my coffee.

Would bleach/some other household chemical dissolve the spider? Is there some other methodology you can recommend?

You're not the old lady that swallowed a fly and trying to avoid your lyrical fate, are you?

"Biological" washing powder might do the trick - plenty of proteolytic enzymes - probably take several days at least.  Then you'd have to wash out the resultant spider soup.

I hope you like coffee with a lot of body to it.


Synthesis of a weinreb amide from an acid The following shows the conversion of an acid to a weinreb amide:

It was taken from the IChO 2015 preparatory problem set problem 23. In this two-step synthesis, I am slighly puzzled by the first step. The rationale behind first forming the ester is that we would like to synthesise an acid derivative which can react with the amine to form the amide since direct conversion of an acid to an amide is rather difficult. However, I wonder why is the t-butyl ester formed instead of say, a methyl ester, since the former sterically-hinders the attack of the amine in the subsequent step. In fact, conversion to an acid chloride first (e.g. using thionyl chloride), instead of an ester would seem to be more optimal than conversion to an ester first since acid chlorides would be more reactive with the amine.

In summary, I would like to know why ester formation is preferred over acid chloride formation in the first step and also why a rather bulky ester is preferred over a much less bulky one. Regarding the former, I believe it could be that the formation of $\ce {HCl}$ by-product may hinder the 2nd reaction by protonating the amines. Regarding the latter, I believe there may be some electronic factors that need to be considered?

Your question is easily answered by the fact that the conditions of Me3COCl + starting acid give a mixed anhydride not an ester. The tButyl group is chosen as the hindrance it brings makes reaction at the desired C=O carbon more likely. Further reading about mixed anhydride coupling is here


Battery acid solution preparation from 98% ACS Sulfuric Acid I'm trying to prepare some battery acid for activating a flooded lead acid battery I had purchased. The battery concentration should be around 36-28% sulfuric acid solution. I have decided to go with 37% acid solution.  I would like to confirm if the volume of acid to be added is correct. So, using a 98% ACS reagent sulfuric acid the volume of acid to make 100mL solution should be 37.57% right? The remaining volume is distilled/deionized water to make up for the 100mL.

The answer begins with the general calculation of mixing solutions with various densities. If you are interested only in the particular numeric results, or if you are scared by algebra, , skip the theoretical part between the horizontal lines.

There is the mixing cross rule, deriving the ratio of mass of 2 mixed liquids from their mass percentage.
$$\frac { m_2 }{m_1} = \frac{|w_1 - w_0|}{|w_2 - w_0|}=k\tag{1}$$
respectively
$$m_2 = k\cdot m_1 \tag{2}$$
and
$$m_0 = m_1 + m_2 = m_1 + k \cdot m_1 =  m_1 \cdot \left(1 + k \right) \tag{3}$$
where
$m_1,m_2, m_0$ are respective masses of mixed solution 1 and 2 and  ofthe final solution 0
$w_1,w_2,w_0$ are respective mass percentages of mixed solutions 1 and 2 and the final solution 0
As we are interested rather in volumes, we involve solution densities:
$$V_1 = \frac{m_1 }{\rho_1}\tag{4}$$
$$V_2 = \frac{m_2 }{\rho_2} = \frac{m_1 }{\rho_2} \cdot k\tag{5}$$
$$V_0 = \frac{m_0 }{\rho_0} = \frac{m_1 \cdot \left(1 + k \right) }{\rho_0}\tag{6}$$
We have the given final volume $V_0$ of the final solution of the density $\rho_0$.
Then, for the solution 1:
$$m_1  = V_0 \cdot \rho_0 \cdot \frac{1}{\left(1 + k \right)} \tag{7}$$
respectively the volume:
$$V_1  = V_0 \cdot \frac{ \rho_0}{\rho_1 } \cdot \frac{1}{1 + k}\tag{8}$$
For the solution 2:
$$m_2  = m_1  \cdot k = V_0 \cdot \rho_0 \cdot \frac {k}{1+k} \tag{9} $$
respectively the volume:
$$V_2  = V_0 \cdot \frac{\rho_0}{\rho_2 } \cdot \frac {k}{1+k} \tag{10} $$

Applying the particular data:
Solution 0: 37% $\ce{H2SO4}$,The density  $\pu{1.267 g/mL}$,  target volume $V_0  = \pu{100 mL}$
Solution 1: 98% $\ce{H2SO4}$.The density  $\pu{1.8361 g/mL}$
Solution 2: water = 0% $\ce{H2SO4}$, the density $\pu{1.000 g/mL}$
From the equation (1):
$k = \frac{|98-37|}{|0-37|}=61/37 \simeq 1.649$
We plug in the data into equations (8) and (10)
For  98% $\ce{H2SO4}$:
$$V_1  = \pu{100 mL} \cdot \frac{ \pu{1.267 g/mL}}{\pu{1.8361 g/mL} } \cdot \frac{1}{1 + 1.649} \simeq \pu{26 mL}\tag{8a}$$
For water:
$$V_2  = \pu{100 mL} \cdot \frac{ \pu{1.267 g/mL}}{ \pu{1.000 g/mL}} \cdot \frac {1.649}{1+1.649} \simeq \pu{79 mL} \tag{10a} $$
Be aware volumes are not additive. After cooling down, the total initial volume shrinks from $\pu{105 mL}$ to $\pu{100 mL}$.
For subsequent dilutions, it is an advantage to keep some volume of the residual 37% solution and to do the dissolution in 2 steps.
First, dilute the calculated portion of 98% H2SO4 in this rest of 37% H2SO4. After cooling, pour this mixture to the calculated volume of water.
The amount of heat per dilution would be divided into these 2 steps.
DO NOT forget to pour acid into water, as not vice versa !! Pouring water into acid would very propably end
by a hot acid splash into your face, as mixing is very exothermic. Mix them carefully, slowly, continually mixing and cool it by a water bath.
Pour the acid along the glass rod and not directly. Wear a lab coat or at least an apron, gloves and shield if possible.
Be aware even tiny drops can cause after some days/weeks holes in your clothes.


Expression of the potential in the hydrogen atom while solving radial part of the wave function I was reading about the derivation of wavefunction of Hydrogen atom from Atkins book. After the separation of variables and writing the  wave function, $\psi_{(r,\theta,\phi)}=R_{(r)}Y_{(\theta,\phi)}$, and further solving the equation, they reach the expression of potential as

$V_{eff}=-\frac{Ze^2}{4\pi\epsilon_0r}+\frac{l(l+1)\hbar^2}{2\mu r^2}$

The first term $(-\frac{Ze^2}{4\pi\epsilon_0r})$ represent the electrostatic potential between the electron and the proton. I am not able to understand the interpretation of the second term $(\frac{l(l+1)\hbar^2}{2\mu r^2})$. In book it is written that it arises from centrifugal effect.
Also why the second term is considered in the potential as potential is defined for a conservative force field? I know that second term in potential comes in picture as a result of the derivation. But the second term doesn't make sense to me as in Bohr's model,  in the expression of the potential we only consider the electrostatic potential. Please explain about the second term in potential.

The 'centrifugal' term appears when the coordinates are changed from $(x,y,z)$ to spherical $(r, \theta, \varphi)$. This is not to say that it is artificial, or a mathematical trick, but is a consequence of the fact that the electron has orbital angular momentum, and this is made clear by changing coordinate systems.
To summarize: the equation is $H\varphi(\pmb{r})=E\varphi(\pmb{r})$ where $\displaystyle H=-\frac{\hbar^2}{2\mu}\Delta +V(r)$. Changing to spherical coordinates produces,
$$\Delta =-\frac{1}{r}\frac{\partial ^2}{\partial r^2}+\frac{1}{r^2}\left(\frac{\partial ^2}{\partial \theta^2}+\frac{1}{\tan(\theta)}\frac{\partial}{\partial \theta}+\frac{1}{\sin^2(\theta)}\frac{\partial^2}{\partial \varphi^2} \right)$$
which means that eigenfunctions of $\varphi(\pmb{r})$ are variables in $r, \theta, \varphi$
The second term is almost the same as that for $\pmb{L^2}$ and substituting for this makes the Hamiltonian
$$\displaystyle H=-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial ^2}{\partial r^2}r+ \frac{1}{2\mu r^2}\pmb{L^2}+V(r)$$
As $\pmb{L^2}\varphi(\pmb{r})=l(l+1)\hbar^2\varphi(\pmb{r})$  then
$$\displaystyle H=-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial ^2}{\partial r^2}r + \frac{l(l+1)\hbar^2}{2\mu r^2}+V(r)$$
The potential is unchanged and still $V(r) = -e^2/r$ and the Hamiltonian is still a function of $r$ but now depends on variable $l$ thus the eigenvalues must depend on $l$.
The centrifugal term is so called because the corresponding force, the negative slope, always repels the electron from the centre and labelled a centrifugal potential or barrier. The term is always positive or zero and can change the shape of the potential the electron experiences and thus forms an effective potential written as  $\displaystyle V_{eff}=\frac{l(l+1)\hbar^2}{2\mu r^2}+V(r)$. When $l>0$ the centrifugal term dominates at small $r$ causing $V_{eff}$ to be repulsive. As larger $r$ the potential becomes attractive.


About purifying ethanol - What type of ethanol? I'm reading this book, trying to learn some basics of chemistry. One of the experiments is about purifying ethanol.

Now, I have some difficulties in deciding what kind of ethanol solution I should try to purify. The drugstores appear to typically sell denatured ethanol (so it could contain e.g. methanol?) or solutions with ethanol and isopropyl alcohol.

Could potable alcohol be used? I think there's glucose in those.

PS. Doing this in my country is legal. It seems that is not the case in all parts of the world (or has not been).

Forget about the denatured alcohol! Additives are typically chosen to render their removal difficult.

Supposed you have decent setup for distillation, analytics still might be a problem.

How will you determine whether your destillation was successful or whether it just was "thermal transfer" from one flask to another?

A hydrometer to measure the density of the liquid could be helpful.

Use some cheap vodka, mix with water and try to distill the vodka off. Measure the density before and after distillation. Note the temperatures, read about strange things that can happen: azeotropes. Cheers!

If you want it more simple and obvious, mix the vodka with some food colourant. Proceed as before and watch the effect. Cheers again.

Note that one usually would not digest anything that went through any lab equipment  for mixed used at a normal lab. But this is your home lab and you know what went through which labware. Anyway, keep your equipment clean and take care. Btw: Cheers!


In an equilibrium, why are partial pressures not affected by the coefficients of the balanced equation? It's probably easiest for me to explain in an example. For the following reaction: $$\ce{2NOBr(g) <=> 2NO(g) + Br2(g)}$$
At equilibrium, if the partial pressure of $\ce{NOBr}$ was 6 atm, the partial pressure of $\ce{NO}$ was 4 atm, and the partial pressure of $\ce{Br2}$ was 3 atm. Why is the equilibrium constant: $$Kp = \frac{3\cdot 4^2}{6^2}$$ instead of the individual pressures multiplied by their coefficients like in molar solubility: $$Kp = \frac{3\cdot \left(2\cdot 4\right)^2}{\left(2\cdot 6\right)^2}$$

One way of thinking about this is writing the original equation
$$\ce{2NOBr(g) <=> 2NO(g) + Br2(g)}$$
as
$$\ce{NOBr(g) + NOBr(g)  <=> NO(g) +NO(g) +  Br2(g)}$$
Now it should be clear you have to multiply, not add concentrations.