Polyethylene terephtalate Biodegradability I'm looking into plastics and how they break down. I have learned 3 ways oxo-biodegradation, photodegradation and thermaldegradation. Does anyone know how PET (polyethylene terephtalate) degrades?

It's oxo-biodegradation* with an asterisk. PET polymers are polyester meaning that the monomer units are link by ester functional groups. This ester group can be hydrolyzed in the presence of water to for the alcohol and acid that formed the ester. For PET the alcohol is ethylene glycol which is readily digestible by bacteria and the acid is terephthalic acid which is not as redily digestable but still digestable.

It is for this same reason that poly(glycolic acid) which is used in dissolvable stitches and poly(lactic acid) which is used in food containers (and 3D printing) are also biodegradable.


Why is phenol more polar than alcohol? Why is phenol more polar than alcohol. I mean the benzene ring seems to be extremely unpolar. Furthermore, is the hydrogen bonding stronger in phenol than in alcohol? Are polarity and hydrogen bonding related (in hydroxy compounds)?

I am doing TLC and I was taught that hydrogen bonding can occur with the alcohol/phenol and that alcohol has stronger hydrogen bonding to the silica gel than it would have despite the lower polarity a lower Rf value than phenol.

But to be honest I am very unsure about those teachings.

The OH group of phenol is indeed more polarized - and a stronger acid - than the OH group of an aliphatic alcohol - maybe that is what you are asking. There are two contributions to this:

(1) inductive: the sp2 hybridized carbon participating in the sigma bond to the OH group is more electron-withdrawing than the sp3-hybridized carbon of an alcohol bonded to OH; and

(2) mesomeric: the phenyl group allows non-bonding electrons on oxygen to partially delocalize into the ring via pi bonding.


Forces of attraction of hydrated potassium ion in aqueous solution I want to study the intermolecular forces (IMFs) in hydrated potassium ion, $\ce{K+ (aq)}$ in an aqueous solution of $\ce{KCl}$. According to my thoughts, among the 4 IMFs I know:

London dispersion forces (LDFs);
Dipole-Dipole interaction;
Hydrogen bonding;
Ion-dipole forces

LDFs are definitely included, but my question is whether dipole-dipole is included as well (because KCl is a polar substance)?

Dipole-dipole interaction, specifically, will probably not happen, or they at least won't contribute significantly to IMFs. The thing with these different chemical bonds is that they are mostly based in identical features, and sometimes calculated differently, taking different things in account.
An ionic bond will be stronger than a polar bond, so that is why dipole-dipole interaction seems irrelevant, in comparison to ion-dipole forces.

All the kinds of interaction mentioned are derivatives of electrostatic interaction, also know as Coulomb interaction. LDFs are a direct derivative of Van der Waals bonds, but if you look more physically at all those bonds you can think of them as electrostatic interaction.


Why is guanidine considered a strong base? I've studied $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15,$ and anything with a $\mathrm{p}K_\mathrm{a}$ lower than should be an acid. Guanidine has a $\mathrm{p}K_\mathrm{a}$ of $13.8$ and is being called by Wikipedia a strong base. Why is it so?
According to Wikipedia, ammonia is a weak base. Then why is its $\mathrm{p}K_\mathrm{a}$ much higher than the $\mathrm{p}K_\mathrm{a}$ of guanidine?

Wikipedia notes that if a base (in water solution) has a conjugate acid whose $\mathrm{p}K_\mathrm{a}$ exceeds $13,$ it's a strong base.  The criterion is (apparently) based on complete or at least predominant dissociation up to $0.1$ molar.  Guanidine makes it, ergo a strong base.


Does the Hydrogen/Chlorine reaction in light proceed instantaneously? When a mixture of Hydrogen and Chlorine is exposed to short wavelength light it reacts explosively. Does the reaction happen simultaneously through the bulk of the mixture?

It depends on how much UV is used, and how it's distributed... from a point source, the flame front would spread as if from an automotive spark plug. In fact, lasers are being tested as a means of ignition in Otto-cycle engines because the flame front can be better controlled than from a spark.

If evenly spread over the surface, as in laser fusion attempts, then the wave-front would propagate from the exterior inwards. Theoretically, using fiber optics inside the container to distribute and delay the beam equally, the explosion could propagate as fast as desired, with the caveat that the delay would still be half the largest space between tips.

This means, for example, that a one-cubic-meter container could be made to explode in far less than 1/3*108 seconds, which would be the minimum if it were to be ignited at the edge. Though not truly simultaneous throughout the container, one can approach any finite speed, even, apparently, greater than light. This is similar to other phenomena that exceed C, such as the crossing point of scissors or the phase velocity of a wave.


Role of the counter/auxiliary electrode in cyclic voltammetry I am doing a school project on cyclic voltammetry and am currently studying the topic. I have learnt that the reaction mostly takes place around the working electrode where the current is measured together with the counter/auxiliary electrode. My question is, how does the counter/auxiliary electrode pass current without oxidising or reducing the analyte?

For example, I were to increase the potential in a solution of $\ce{Fe^3+}$ ions. The working electrode would be the cathode and would reduce the ions, giving them electrons and forming $\ce{Fe^2+}$ ions. In this case, what would happen around the counter electrode? How would the electrode be able to pass current to balance it? Would it oxidise the newly formed $\ce{Fe^2+}$ ions? Wouldn't this result in a never ending process? If someone has already asked this question, I apologize. Please link the page, thank you.

You are right that the counter electrode must balance the redox reactions happening at the working electrode.
For every electron that goes to reduce an $\ce{Fe^3+}$ ion at the working electrode, an electron must be collected from an oxidation reaction at the counter electrode.

What that reaction actually is depends on a lot of things.
First, we must consider what is actually in the solution to oxidize at the counter electrode.
In an $\ce{Fe^3+}$ solution, unless the working electrode is very close to the counter electrode, the newly formed $\ce{Fe^2+}$ ions will not diffuse efficiently to the counter electrode, so something else must balance the bulk of the current.
In many aqueous solutions, this might be the water itself being oxidized, unless there is a more easily oxidized substance in the solution (a supporting electrolyte, for instance).
If the counter electrode material is chosen poorly, it may even be the counter electrode itself oxidizing.

In practice, electrochemists often design cells such that the reactions at the counter electrode are unimportant, perhaps by placing the counter electrode in a separate vessel connected with a salt bridge, or at the least assuring that potential counter electrode products will not interfere with the reactions at the working electrode.


Why does aluminium chloride react with water in 2 different ways? Aluminium chloride has 2 different reactions with water, depending on whether a small amount or an excess of water is added. One produces a solution of hexaaquaaluminium ions whereas the other produces a ppt. of aluminium hydroxide And HCl gas. Does the 2nd reaction occur due to the reaction 1 first occurring followed by vaporisation of the solution due to the heat given out?

Yes, adding a small amount of water to anhydrous AlCl3 is different from an aqueous reaction. Here is an excellent educational institution's answer in the first case, depicting the action of water on anhydrous aluminium chloride, to quote:

If a small amount of water - a few drops - is added to anhydrous aluminium chloride there is a hissing sound and hydrogen chloride gas is given off as steamy acidic fumes. The reaction that occurs will depend on the amount of water added, but is probably
$\ce{AlCl3(s) + H2O(l) → AlCl2OH(s) + HCl(g)}$

In the second case, aluminium chloride in aqueous solution appears to undergo a two-stage reaction, starting with the formation of the hexaaquaaluminium ions, to quote again:

If aluminium chloride is dissolved in a large amount of water the solution is acidic, but this has nothing to do with formation of hydrochloric acid. The solution contains hydrated aluminium ions and chloride ions:
$\ce{AlCl3(s) + aq → [Al(H2O)6]^{3+}(aq) + 3 Cl -(aq) }$

This is followed by an acidic hydrolysis paralleling transition metals:

The hexaqua complex ion behaves exactly like ions of similar type formed from transition metals; the small, highly charged metal ion polarises (withdraws electron density from) the water molecules that are attached to the aluminium ion through dative covalent bonds. This makes the hydrogen atoms d+ and susceptible to attack from solvent water, which is acting as a base. The complex ion is deprotonated, causing the solution to be acidic from the formation of hydroxonium ions H3O+:
$\ce{[Al(H2O)6]^{3+}(aq)  + H2O(l) →  [Al(H2O)5OH]^{2+}(aq) + H3O+(aq)}$


Can you make napalm out of gasoline and orange juice concentrate? According to the novel Fight Club by Chuck Palahniuk:

  The three ways to make napalm: One, you can mix equal parts of
  gasoline and frozen orange juice concentrate. Two, you can mix equal
  parts of gasoline and diet cola. Three, you can dissolve crumbled cat
  litter in gasoline until the mixture is thick.

The napalm is originally called napalm because is mixture of naphthenic acid  and palmitic acid. This makes me very skeptic about it being created by gasoline and juice. Also I see no reason for it to be frozen.

The only thing that sounds realistic to achieve the sticky effect is the use of cat litter - but you could as well just use dry dirt, couldn't you? And dirt doesn't really burn so you're weakening the burning effect.

Here's an excerpt from an interview with the author of Fight Club, Chuck Palhniuk.

  How to make Napalm with Frozen Orange Juice and Gasoline?

  Well, Ed Norton changed one ingredient in every one to make them useless. So, that really pissed me off because I really research those really well. Actually its styrofoam and gasoline - it make the most incredible explosive.

http://www.dvdtalk.com/interviews/chuck_palahniuk.html

So no, Fight Club does not accurately convey any of the recipes. All were clearly modified. And most likely for safety reasons - i.e. you wouldn't want someone to walk out the theatre with a "bright" idea. On the other hand I would still expect a mixture of gasoline and orange juice concentrate to be rather flammable ...


Ipso substitution and its rules I was taught that the following reaction happens:

It is called the "ipso-substitution". My professor had told me that it occurs in the case of $\ce{-SO3H}$ and $\ce{-COOH}$. But he had ended the topic there. My textbook doesn't detail any more either.

My searches landed me up at the fact that ipso-substitution is also employed in the preparation of picric acid from nitrobenzene. But I couldn't find answers all my other questions, which are:

Does the ipso-substitution occur only for these two groups, or are there more? I guess there is a "threshold" value for some parameter, above (or below) which ipso substitution cannot occur, but I am unable to come up with what that parameter is. I guess it's the high stability of the leaving group $\ce{SO3-}$, but I am not sure.
Is ipso-substitution affected by the choice of the electrophile and the solvent? What if, instead of bromine water, I chose bromine in acetic acid? Or something even simpler as $\ce{FeCl3/Cl2}$. Would ipso substitution occur then?
I know that no organic reaction produces 100% of only one compound, as its other isomers are always also formed. Ipso substitution should be no exception. But how do we decide whether the ipso product is a major product or a minor product? I know how to decide major/minor products between ortho/meta/para substitution (electron density/steric hindrance/H-bonding), but I haven't been taught any logic to decide the same for ipso substitution.

The shorthand way that we draw structures in organic chemistry, with implicit hydrogens, leads us to often forget that the hydrogens are there, or to neglect considering them in our analyses. In electrophilic aromatic substitution (EAS) reactions, the generally accepted mechanism involves attack of the aromatic ring on the electrophile to give a Wheland intermediate, which then loses a proton from the $\mathrm{sp^3}$-hybridized ring carbon to restore aromaticity. Ipso substitution, then, is really nothing special - we simply substitute a position on the aromatic ring that bears a non-hydrogen (e.g., $\ce{X}$) group, presumably via an analogous mechanism. For this to be viable, both steps of the mechanism need to be favorable:

$\ce{X}$ needs to enhance, or at least not greatly reduce, the nucleophilicity of the carbon it is attached to. For example, inductive effects would great disfavor ipso attack at groups like $\ce{F}$ or $\ce{CF3}$.
$\ce{X+}$ should be a reasonable species to lose in the second step, to regain aromaticity, or there should be some mechanism by which $\ce{X}$ can be lost, with electrons flowing back towards the aromatic ring.

Sulfonic acids and carboxylic acids are rather electron-withdrawing, but if you drill down into the details of these ipso-substitution reactions, many are actually conducted on the conjugate base (i.e. sulfonate or carboxylate), which also provides a pathway for loss of a neutral group in the second step, i.e. $\ce{SO3}$ or $\ce{CO2}$. Alternatively, harsh conditions are used, e.g., sulfonation/desulfonation.

It turns out that ipso attack is more common than it first appears, because the product of the initial ipso attack can re-arrange. So, for example, nitration of p-cresol involves 40% attack at the ipso carbon bearing the methyl group, followed by acid-catalyzed rearrangement to the expected product.[1] There is more discussion here.

Closer to the original thrust of your question - there are some groups that are even better at being substituted in an EAS reaction than a proton. Trialkylsilyl and trialkylstannyl groups are often used. Both $\ce{Si}$ and $\ce{Sn}$ are less electronegative than carbon and highly polarizable, making the ipso carbon a good nucleophile. These groups also help to stabilize the Wheland intermediate through hyperconjugation. Finally, they both have energetically-favorable pathways to leave with electron density flowing back towards the aromatic ring. For these reasons, ipso substitution at $\ce{SiR3}$ or $\ce{SnR3}$ substituted carbons can often be achieved rapidly, under mild conditions, with high regioselectivity. This is often helpful in specialist applications, like radiolabelling.

I haven't specifically addressed each of the original three numbered questions, but they can be tackled using the principles above. Ipso substitution is possible for any group - it's a competition, $\ce{H}$ vs $\ce{X}$. Is ipso substitution affected by reaction conditions? I'm sure it could be - factors like protonation states, hard vs soft electrophiles, reversibility, etc. would have an influence.


Quantization of particle on a sphere While studying particle on a sphere, I read that boundary condition on $\theta$ and $\phi$ results in quantization.
I understand boundary condition for $\phi$ part which varies from 0 to 2$\pi$. As $\psi(0)$ should be equal to $\psi(2\pi)$ otherwise in standing wave the wave function can have more than one value and can interfere destructively.

But I am not able to figure out boundary condition for $\theta$ part which varies from 0 to $\pi$. In books,the result is directly quoted as "Boundary condition in $\theta$ part resulted in quantization which restricts the quantization in $\phi$ part". Quantization in $\theta$ part is called as azimuthal quantum number$(l)$ and $\phi$ part is called magnetic quantum number($m$) which varies from $-l$ to $l$.

Please tell what would be the boundary condition for $\theta$ part. I am not able to figure out and visualize it.

The idea of boundary conditions is nice and intuitive, but this is one case in which intuition cannot really help us.

In general when you are solving the Schrödinger equation in spherical coordinates, you can separate the radial and angular parts (unless you have some pathological potential energy term, I think). See Griffiths' Introduction to Quantum Mechanics (2nd ed.), Chpt 4 for details. It doesn't matter whether you're solving the H atom, a particle on a sphere, or indeed a free particle: in all cases the potential energy has no angular dependence in the potential energy, so the angular component is the same problem.

When you solve the angular bit, you can (again) separate out the two angles, so you end up with two differential equations: one in $\theta$ and one in $\phi$. The $\phi$ bit is easy enough:

$$\frac{1}{g}\frac{\mathrm d^2 g}{\mathrm d\phi^2} = -m^2$$

where $m$ is unconstrained for now, and $g$ is a function of only $\phi$. This is easy to solve and applying the "common-sense" boundary condition that $g(\phi) = g(\phi + 2\pi)$, you can find that $g = \exp(\mathrm i m\phi)$ with $m$ an integer. So far so good.

The $\theta$ bit is substantially more difficult, but this differential equation is so important that it has been studied for centuries and has its own name, the Legendre equation:

$$(1 - x^2)\frac{\mathrm d^2\! f}{\mathrm dx^2} - 2x\frac{\mathrm df}{\mathrm dx} + \left[l(l + 1) - \frac{m}{1 - x^2}\right]f  = 0$$

where $f$ is again a function of only $x$, and we have made the substitution $x = \cos\theta$. In our case, it turns out that unless $(l, m)$ obey certain constraints (i.e. the now-familiar $l = 0, 1, 2\ldots$ and $m = -l, \ldots, +l$), then the solutions $f(x) = f(\cos\theta)$ go to $\pm\infty$ at $x = \pm 1$, i.e. $\theta = 0$ or $\theta = \pi$. That's not acceptable for a well-behaved wavefunction, so we have to reject all the other cases.

So, you could say that the boundary condition on $\theta$ is that the wavefunction  must be welll-behaved (i.e. not diverge to $\pm\infty$) at $\theta = 0$ and $\theta = \pi$. But that is not something we could say without delving into the maths. A priori, we could certainly say that it is important for the wavefunction to be well-behaved at all points. But we cannot pick out the two particularly crucial points $\theta = 0$ and $\theta = \pi$ just using common sense, or "by observation".

More details can be found in Griffiths (there is a fair bit of mathematical detail in there), or any other book that covers differential equations, really.

Bonus: if you multiply the solutions $f(\theta)$ with $g(\phi)$, then normalise the entire thing, you get what are known as the spherical harmonics, which are also extremely important functions in their own right; that's where the shapes of the atomic orbitals come from.


Analysis project to conduct We have been asked to conduct an analytic project for chemistry. It is important that the project conducted hold social implications and the conclusion achieved help inform people better. Several ideas have come to my mind for such projects, such as determining amount of phosphoric acid in cold drinks, acetic acid in fruits such as guava, amount of insecticides/pesticides in vegetables/fruits in the market. However, I am still yet to find an idea which is striking. I would like to do something unique. Any suggestions?

As you might have guessed: It depends!

What is the educational level of the study (High School, College, etc.)?
What's the time frame - how many man-hours?
What's the equipment of the lab?
If equipment or material has to be bought: What is the budget?
Is it a plain academic study or is it intended to go public with the results in order to change/improve consumer awareness?

"Something unique" might mean that

there are no reference data
it is too "academic" to have social implications (or at least that it will be difficult to explain to the public that it actually has)

What about something as "boring" as sugar content in local soft drink brands as compared to similar products on the international market? I'm sure you'll find lots of references, such as this.

Go for something that is relevant for the average Joe Bloggs. Go for something that can be bought thoughout the while time of the study, just in can you have to repeat measurements or need to cover seasonal variations. If it is a fresh, seasonal product, you might have to plan ahead and stockpile and deep-freeze material.

In the end, it is up to you to decide - planning and outlining the study is part of the project.

Speaking of relevance for the average consumer: Everybody needs excrete - but you do not want to examine that. Fortunately, everybody needs to eat too and seemingly, a lot of people tend to eat too much of the wrong stuff. The recent report of the US Dietary Guidelines Advisory Committee might give some directions what to look for.


How to get a precise reading from pipette? While doing titrations, we are instructed to use pipette and they demonstrated how to use a pipette. But still after all this, we face difficulties of how to take precise reading from pipette.
Example-

Due to the concave water meniscus, until where to take the reading ?
Is it really important to take every single drop of solution from pipette ?
as while putting solution in beaker we see little bit drop left inside pipette.
And what's the best way to hold pipette to avoid breaking it and easy to see the reading which we are taking ?

The bottom of the meniscus (in the middle) is your measurement point and should lie on the line that you are trying to read.
This is completely dependent on the type of pipette that you are using. Some pipettes are designed so that you just leave the last drop and some you are supposed to just touch the tip of the pipette to the surface of what you've dispensed already to get the last bit. TD pipettes stand for "to deliver" (leave the last bit). TC pipettes stand for "to contain" (take the last bit out).
Again, this is dependent on the pipette. If you have a classic bulb pipette then you shouldn't hold it by the bulb as it is the weakest part and that's how you end up with glass in your hand.


How many grams of acetic acid are present in 2.80 L of this vinegar? A vinegar sample is found to have a density of $1.006 \mathrm{\frac{g}{mL}}$ and to contain $8.7~\%$ acetic acid by mass. How many grams of acetic acid are present in $\pu{2.80L}$ of this vinegar?

I multiplied: $$\left(\frac{1.006~\mathrm{g}}{\mathrm{ml}}\right)\cdot \left(\frac{1000~\mathrm{ml}}{1~\mathrm{L}}\right)\cdot 2.80~\mathrm{L} = 2816.8~\mathrm{g}$$

I then converted $8.7~\%$ into $0.087$ and multiplied it into $2816.8$, to get the answer $245~\mathrm{g}$.

Am I correct?

Your math and stoichiometry are correct.
$2.80L*(\frac{1000 ml}{1 L})*(\frac{1.006 g}{ ml})*8.7$%$=245g \checkmark$


How do lone electrons impact the charge of an ion? This is a quite elementary question, but I still can't wrap my head around it.

I know how to calculate the formal charges of an ion or a molecule, but I don't understand what happens when oxygen (when sharing a single covalent bond with sulfur in a sulfate ion) has a lone electron that doesn't bond nor pair. I don't understand why this puts a -1 charge on the atom. What happens?

I've attached an image which explains what I'm trying to figure out:

See those two red dots? What do they mean? I understand that they symbolize the lack of a single electron needed to complete an octet. But why does it give the atom a negative charge?

The two red dots indicate the presence of two electrons, not the absence, if there were no electrons there it would be balanced. The colours of the electrons in the image indicate their origin: blue electrons originate with oxygen, black from sulfur and red from an outside source (eg. Potassium or H2).
Ions have a charge because they have more or fewer electrons than the number of available protons. Sulfate has gained two extra electrons.
The arrows in the second image indicate a 'dative covalent bond' where the electrons in the covalent bond all come from one atom rather than being shared, while the lines indicate a normal covalent bond (ie. an electron from each atom. The extra electrons added to the oxygen atoms to form a full orbital level in the normal covalent bond are responsible for the extra charge.

There's a nice visual depiction here: http://chem-net.blogspot.co.uk/2013/02/electron-dot-structure-of-sulfate-ion.html


Portable library to render 2D structural formulas as vector graphics from SMILES or InChI For a web-based project, I am looking for a piece of software to create a 2D vector graphics structural formula from a textual structure encoding scheme such as SMILES or InChI, running on the server or client, and callable via some kind of API.

I’m imagining a client- or server-side library that you could just use like this:

svg_output = lib.render_from_smiles(smiles_input);

Ideally, it would have the following properties:

Take a SMILES or InChI representation of the compound as input
Compute the structure graph and 2D positional data (positions of atoms, types of bonds, etc.)
Render the structure in some kind of vector graphics representation (SVG would be perfect, but any other open format would do)
Or, alternatively, just output the computed drawing parameters, to be passed on to a separate renderer (which would need to be written, but that is the less challenging part)
Written in a language that is easy to use and deploy in web environments (PHP, JavaScript, Python, Ruby, Go, etc.)
If a renderer is included, allow me to control the rendering style (stroke width/colours, text attributes etc.)
Open source or with license terms that are affordable to a non-profit project :)

The application just needs to render individual structures. No reaction mechanisms or anything fancy. Small organic molecules only.

I know there are various standalone software packages (ChemDraw, MarvinSketch etc.) that do this kind of thing, but I did not find anything that can be used as a library like this.

Any suggestions or ideas? Thank you!

You might want to have a look at Open Babel.

It is licensed under GNU GPLv2
It has bindings for Python
It has bindings for PHP
It has bindings for Ruby
It can read (and write) SMILES
It can read (and write) InChI
It   can write SVG
It can write a list of painter commands


Density of a sodium hydroxide solution The specific gravity of a $50\%$ aqueous solution $\ce{NaOH}$ is $\pu{1.5298 g cm-3}.$
To my understanding, the $50\%$ means $50$ mass fraction $w,$ i.e. $\pu{500 g}$ $\ce{NaOH}$ in $\pu{500 g}$ $\ce{H2O}:$
$$w(\ce{NaOH}) = \frac{m(\ce{NaOH})}{m(\ce{NaOH}) + m(\ce{H2O})} = \frac{\pu{500 g}}{\pu{500 g} + \pu{500 g}} = 0.5.\tag{1}$$
However, when I try to calculate the density of that solution I end up with
$$
\begin{align}
\rho(\ce{NaOH}) &= \pu{2.13 g cm-3}; \\
\rho(\ce{H2O}) &= \pu{1.00 g cm-3}.
\end{align}
$$
$$V = \frac{m(\ce{NaOH})}{\rho(\ce{NaOH})} + \frac{m(\ce{H2O})}{\rho(\ce{H2O})} = \pu{234.74 cm^3} + \pu{500 cm^3} = \pu{734.74 cm^3},\tag{2}$$
$$\rho_\text{sln} = \frac{\pu{1000 g}}{\pu{734.74 cm^3}} = \pu{1.361 g cm-3}.\tag{3}$$
What am I missing here?

What's your first miss is the definition of $50\%(w/w)$ $\ce{NaOH}$ solution (although it doesn't matter here). Actual definition is $\pu{50 g}$ of $\ce{NaOH}$ in $\pu{100 g}$ of solution. Since water is the solvent, your interpretation is correct.
Your second mistake is the lethal one. You have considered water and solid $\ce{NaOH}$ are additive. Assuming $V_\text{total} = V_\ce{NaOH} + V_\ce{H2O}$ is a mistake.
Using a given density of $\ce{NaOH}$ ($\pu{1.5298 g}$),we can calculate the volume of $\pu{1000 g}$ of solution: $\frac{\pu{1000 g}}{\pu{1.5298 g/mL}} = \pu{653.68 mL}$. Therefore, dissolving $\pu{500 g}$ of $\ce{NaOH}$ in $\pu{500 mL}$ of water (assuming temperature is $\pu{25 ^\circ C}$, hence density of water is $\pu{1.00 g/mL}$), the volume increased by only $\pu{153.68 mL}$. Do you still think volumes are additive?


Why should space whisky taste different to Earth whisky? I recently read an article about a vial of whisky being stored on the International Space Station for 3 years, and having its taste compared to a control sample stored on Earth.

What could cause this difference in taste / chemical composition?

Theories so far

I'd assumed that the effects of (absence of) gravity would have altered the taste by affecting how the liquid's molecules interact with one another (i.e. being free to move in space; whilst being separated into layers on Earth), though I've been told that gravity doesn't affect diffusion within a liquid.

Other possibilities that myself & my father came up with:

Effect of UV if the vial's unprotected
Effect of oxidation (due to increased surface area of the liquid in the zero-g environment, if there's any air in the vial)
Effects of increased G during take off / reentry possibly causing increased interactions between the molecules (though this would be short lived).

From the press release from Suntory, the Japanese company which sent 6 vials of their whisky into space, which details their study:

  Tokyo, Japan, 30 July -- Suntory Global Innovation Center is about to embark upon space experiments on the “development of mellowness in alcoholic beverage through the use of a microgravity environment.” This research will be conducted in the International Space Station’s Japanese Experiment Module (nicknamed “Kibo”), with the cooperation of the Japan Aerospace Exploration Agency (JAXA).
  H-II Transfer Vehicle No. 5, commonly known as “Kounotori5” or HTV5, is scheduled to be launched from JAXA’s Tanegashima Space Center on August 16 (Sunday) carrying alcohol beverages produced by Suntory to the Japanese Experiment Module aboard the International Space Station, where experiments on the “development of mellowness” will be conducted for a period of about one year in Group 1 and for two or more years (undecided) in Group 2.

  With the exception of some items like beer, alcoholic beverages are widely known to develop a mellow flavor when aged for a long time. Although researchers have taken a variety of scientific approaches to elucidating the underlying mechanism, we still do not have a full picture of how this occurs.

  Our company has hypothesized that “the formation of high-dimensional molecular structure consisting of water, ethanol, and other ingredients in alcoholic beverages contributes to the development of mellowness,” and we have been conducting collaborative researches on this topic with research groups of Professor Shigenao Maruyama of the Institute of Fluid Science, Tohoku University and Professor Mitsuhiro Shibayama of the Institute for Solid State Physics, the University of Tokyo, the Japan Synchrotron Radiation Research Institute and Suntory Foundation for Life Sciences. The results of these collaborative researches have suggested the probability that mellowness develops by promoted formation of the high-dimensional molecular structure in the alcoholic beverage in environments where liquid convection is suppressed.

  On the basis of these results, the space experiments will be conducted to verify the effect of the convection-free state created by a microgravity environment to the mellowing of alcoholic beverage.

  The Suntory Group aims to use these experiments to help find a scientific explanation for the “mechanism that makes alcohol mellow.”

It appears that the working hypothesis is that the lack of gravity in space precludes liquid convection and thus alters the characteristics of liquids aged in that environment ${\it vs.}$ that on Earth. As you indicate, it's allegedly one way that promotes the formation of compounds which are otherwise not formed when gravity and convection within the ageing liquid are present.

From the same press release, details about the study:

  Title of study
          Elucidating the Mechanism Mellowing Alcoholic Beverage

  Experimental period
          Group 1: August 16, 2015 (Sunday) (planned) to September 2016 (planned)
  Group 2: August 16, 2015 (Sunday) (planned) to undecided date two or more years later

  Content of experiment
          One set of samples consisting of various alcoholic beverages will be stored in a convection-free state in Japanese Experiment Module “Kibo” on the International Space Station, and another set of the identical samples will be stored in Japan for the same period of time. The following methods will be then used to analyze and compare the two sets of samples.

  Measurement of substance diffusion coefficient with the use of a phase shifting interferometer$^{1}$ in cooperation with the Institute of Fluid Science, Tohoku University.
  Detection of high-dimensional structure by small angle X-ray scattering$^{2}$ using SPring-8, in cooperation with the Japan Synchrotron Radiation Research Institute and the Institute for Solid State Physics, the University of Tokyo.
  Measurement of substance diffusion with the use of the NMR method$^{3}$ in cooperation with Suntory Foundation for Life Sciences.

  [1] Phase shifting interferometer: Uses the properties of light (visible light) waves to visualize temperature and concentration distributions, etc. that cannot be seen by the eyes, making it possible to detect slight changes in the ingredients of alcoholic beverages and determine how they are distributed.

  [2] Small angle X-ray scattering: Can be used to obtain nanoscale (1 millionth of a meter) structure by irradiating a substance with X-rays and observing the scattered X-rays in an angular region of a few degrees or less.

  [3] NMR method: Uses the phenomenon of nuclear magnetic resonance (NMR) to extract information from the organic compounds that are the object of measurement. Can be used to obtain information on the molecular structure of a compound and its dynamic properties.

  Samples
          Five types of distilled spirits differing aging periods and 40% ethanol: Total of six samples

Suntory is not alone. Ardbeg scotch made a similar trip into space, and you can read about the experiment and results here. They used GC, HPLC, GC-MS, and organoleptic assessment to analyze their samples.

As stated in their summary:

  The key aims of the experiment were as follows:-

  (i) To determine if conditions of micro-gravity would have an impact on the range of naturally occurring terpene compounds and on the extraction
  of key flavour-active compounds in oak wood by Scotch malt distillate.

  (ii) To assess the impact of maturation in conditions of micro-gravity on the flavour profile of oak matured distillate.

  (iii) To investigate the presence of novel compounds and flavours in model maturation systems for Scotch malt distillate, as a result of conditions
  of micro-gravity.

  The key findings of the above three aims would then be considered in conjunction with actual maturation conditions on Earth to investigate
  techniques for developing novel flavours in Ardbeg Islay Single Malt Scotch Whisky.

In the Results and Discussion section (I do encourage interested parties to jump to the link, and there are several tables and too much information to post here), they note:

  GC analysis of major volatile congeners demonstrated no significant differences between the International Space Station samples and the control
  samples on Earth. Since this group of compounds comprising primarily alcohols, aldehydes, ketones and esters, are produced during fermentation
  and concentrated up, or in some cases, catalysed during distillation, their presence and quantities should have been identical in each set of samples
  to start off with. The conditions of micro-gravity on the ISS have not impacted on their levels, as demonstrated by the similar levels in the Earth
  based samples (Table 1.)

  A more detailed investigation of these compounds, along with phenolic compounds, was carried out on our behalf by external whisky experts and
  scientists, using GC-MS. In the resulting chromatograms, the areas of 75 separate peaks were integrated, most of which were identified as alcohols,
  esters, acetals and phenols. Again, no major differences were found between the ISS and Earth samples. From an Ardbeg perspective, this was
  particularly interesting as the conditions on the ISS have not impacted on the absolute quantities of the various phenolic compounds. These are
  the compounds which contribute the powerful, smoky, medicinal flavours to Ardbeg spirit.

  The most significant variable between the ISS and Earth samples was discovered when the results of the HPLC analysis of key maturation related
  congeners (wood extractives) were considered. The absolute concentration of these compounds was far higher than would normally be expected
  in standard, barrel-matured spirit, almost certainly as a result of the much higher surface area of woody material that the spirit was exposed to in
  the MixStix. However, when the two samples, ISS and Earth, were compared to each other, the lignin breakdown products - which are the major
  flavour active derivatives from oak wood - were present in higher concentrations in the control samples than in the space samples. This indicates
  that conditions of micro-gravity have inhibited their extraction from oak. On the other hand, ellagic acid and gallic acid, which are tannin breakdown
  products, both appeared to be higher in the ISS samples. The levels of these compounds are detailed in Table 2.

  By drilling down further into these results, and subjecting them to some statistical analysis, the relative abundance of the main, easily extractable
  lignin breakdown products (vanillin, coniferaldehyde and sinapaldehyde) expressed as a percentage of the total wood extractives, is broadly
  similar in each sample. However, other extractives, such as the syringyl compounds, are very different in their total abundance, with much lower
  percentages in the ISS samples. This result means that while the overall extraction of wood-derived compounds is negatively impacted by micro-gravity,
  the impact is not equal; the compounds can be subdivided into ‘readily extractable’ and ‘less readily extractable’.

  While it is clear that micro-gravity is inhibiting the release of wood extractives, not all compounds are affected equally, as would be expected
  in normal conditions. This has resulted in an unusual ratio between the readily extractable compounds and the less readily extractable ones in
  the ISS samples. This offers the intriguing possibility of using the ratios of such compounds as a potential marker to identify ‘unusual’ maturation
  characteristics, and thus a means of identifying for example, spurious age claims in whiskies. This possibility may be further investigated by the Scotch
  Whisky Research Institute.

  For studies such as this, where we are examining the impact of differing conditions on either model systems, or on the full scale in actual barrels,
  the key question is always ‘What does this actually mean for the flavour?’, so the organoleptic analysis was of great interest to us. The results from
  the key volatile congeners, wood extractives and phenols analysis did not lead us to believe that from an organoleptic perspective, there would
  be large differences between the ISS and control samples. However, when we finally nosed and tasted the samples in the sensory laboratory,
  the results were unexpected.

  Triangle tests gave a high number of correct responses, with almost all participants picking out a difference between the ISS and control samples.
  When we examined the actual aroma and flavour profiles, the differences between the two were as remarkable as they were unexpected; all tasters
  described the Earth control samples as being of the general Ardbeg whisky style, while the ISS samples displayed a quite different set of flavours.
  These have been summarised in the detailed descriptions below.

The above-mentioned MixStix are described in the write-up:

  MixStix™ technology, which provides a convenient platform for small scale sample experimentation, is a trade-marked piece of technology used
  by NanoRacks. Essentially, they are small glass vials which can hold a range of materials; in our case these materials were 6ml of Ardbeg new spirit
  distillate, along with a small quantity of charred oak wood shavings (from a once used American oak Bourbon barrel), c. 1mm x 1mm x 6cm in
  size. The two materials are separated from each other by a glass dividing partition, which is broken by bending the Teflon outside covering of the
  MixStix™ at time of experiment activation, thus allowing the two separate components to mix together.


Why is the smell of car exhaust much stronger in winter? I noticed that in winter, the smell of cars (to be precise the smell of the exhaust gases) is much stronger in winter. I just can't figure out why. Has this got something to do with the air temperature or with the humidity?

While car engines might look simple there is a vast amount of technology installed to ensure as complete a combustion as possible. Complete combustion according to the equation

$$\ce{C_{$m$}H_{$n$} + \frac{2$m$ + $n$}{2} O2 -> $m$~CO2 + \frac{$n$}{2} H2O}$$

(not accounting for oxygen, nitrogen or other elements present in the fuel) requires a precise amount of oxygen added. Too little oxygen will result in uncombusted hydrocarbons while too much oxygen will result in $\ce{NO_{$x$}}$. A small oxygen sensor is thus installed to improve the oxygen regulation. (It is a very interesing compound because it functions as an oxide ion conductor.†)

The thing about this oxygen sensor is that its function is temperature dependent: It needs to be heated to $\approx 500~\mathrm{^\circ C}$ to function properly. This is achieved quickly due to the heat of the exhaust, but does take a few minutes from starting the car. So throughout the first few minutes, the oxygen sensor does not function and incomplete combustion results in $\ce{NO_{$x$}}$ and uncombusted hydrocarbons that cause the odour. Once working temperature has been reached, the smell disappears mostly.

In cold weather it will take longer for the oxygen sensor to function properly and thus more incomplete combustion is observed. This becomes more pronunced the colder it gets.

†: Its German name is Lambdasonde from the Greek letter $\lambda$ which is used as a variable to determine if the correct amount is present or not. At the correct oxygen concentration, $\lambda = 1$.


Counterintuitive Reaction of Aluminium On Steel Balls There is a certain video which I just watched about 2 large cast iron balls. If the balls are struck together, no reaction happens, but if one is covered in aluminium foil and then struck, a large quantity of sparks are produced. This flies in the face of my metalworking experiences with these metals. When cut, steel reacts with oxygen in the atmosphere producing large quantities of sparks, but aluminium will not produce any sparks, despite being cut with an abrasive wheel or grinded with an abrasive wheel.

So why does the aluminium spark when the balls are grinded together? Could the temperature be higher forming a reaction with oxygen? Is there pure aluminium in the center of the foil which does not have the typical Alumina composition on the exterior and is being exposed to the atmosphere and is being oxidized?

Video: (CrazyRussianHacker): https://www.youtube.com/watch?v=w7yWoO7vqvI

This is actually related to the thermite reaction, where the rust on the balls reacts with the pure aluminum in the foil. Here is a pdf detailing the experiment.

The activation energy for the reaction is reached through the friction and pressure, and the rust on the balls is actually very important, since it is the exchange of oxygen between the iron oxide and aluminum that releases the heat as iron and aluminum oxide form.


Inductive effect vs electronegativity Why is the inductive effect of chlorine, bromine, and iodine stronger than that of hydroxy group, even though oxygen is more electronegative?

With hydroxyl groups, the oxygen is donating electron density into the ring (resonance). Additionally, the overlap between oxygen and carbon pi orbitals is sufficient enough to make up for the electronegativity of oxygen.
Halogens also donate electron density, however, the orbitals do not overlap as well with halogens and carbon, causing less electron donation then oxygen. Thus, electronegativity trumps resonance.


Freezing all the water on earth with a seed crystal I was recently reading the satirical, fictional book Cat's Cradle by Kurt Vonnegut. In this book, one of the major subplots is the invention of a mythical polymorph of ice called ice-9 in the book. In the book, it is described that if this polymorph of ice were to touch actual water, all the water, inevitably touching through oceans, rivers, lakes, etc. would freeze into this polymorph of ice which is purported to melt at a temperature over 100 degrees Fahrenheit.

This raises two questions for me:

First, would something like this actually happen? I know, for instance, a method people use for purifying crystals is to create a supersaturated solution and disturb the substance and it will crystallize quite dramatically. Could something similar happen here in the sense that the mythical ice-9 structure would be introduced into liquid water (which to ice-9 is a supersaturated solution of water) and create a massive crystal out of all water?

Second, and more in reality, is there any solid that is like this? A solid which has two stable crystal structure that exist at temperature reachable in everyday life at about atmospheric pressure? And if so, can you actually convert one crystal structure to another easily? And would one of the liquids behave as a supersaturated solution of the crystal?

Sorry if that is a little unclear. I am sincerely interested in this and can clarify if there is any confusion.

Anything interesting along those lines would be considered relevant in my mind.

A precise physical argument against the existence of an Ice-9-like state of water is probably not trivial (ab initio methods required?). However, there is a simple but powerful argument against it; we haven't found anything like it yet. Water is a very common substance, and humans have been applying the scientific method to it for hundreds of years, humans have manipulated it for hundreds of thousands of years, but most importantly, the entire Earth has been churning massive amounts of water around for billions of years. If an Ice-9-like state of water did exist, we should have seen it by now.

To be clear, superficially this argument may seem to be a fallacious argument from ignorance. However, we can qualitatively back it up with a bit more knowledge. Assuming that Ice-9 is the true ground state of water, and that all liquid water on Earth is currently in a false ground state, separated only by a kinetic barrier, and realising that the conversion of liquid water to Ice-9 is autocatalytic, we must ask ourselves, if it only had to happen once, why hasn't it happened yet?. The only reasonable conclusion is that, if it can happen, then the kinetic barrier to conversion must be enormous. The interconversion of diamond to graphite is a chemical process with a famously large kinetic barrier (around $\mathrm{540\ kJ\ mol^{-1}}$), yet we have plentiful evidence of it happening many times in geological time scales (and we can do it in the lab too!). Thus the kinetic barrier for conversion between liquid water and Ice-9 would have to be even larger, which is unlikely. There's also a limit to how high we can expect chemical kinetic barriers to be, which we can put very generously at $\mathrm{100\ eV \approx 10\ MJ\ mol^{-1}}$. Water molecules on Earth have had many, many, many interactions at energies way higher than this, as they have been incessantly exposed to cosmic radiation, nuclear decays, lightning strikes, high temperatures, etc. In short, the collective history of all water molecules on Earth has explored a vast region of the space of allowed states, and not once has an Ice-9-like state been found. It is thus fair to assume it simply does not exist.

Regarding your second question, I just remembered a great example: pure tin. At ambient pressure, solid tin can exist in two different polymorphs, and the transition between the most stable form is conveniently close to room temperature. Above 13.2 ºC, tin is most stable as a ductile metallic material, called white tin, whereas below 13.2 ºC, the most stable form is a brittle non-metal called grey tin. It is possible to sustain white tin below 13.2 ºC for a period of time, creating a metastable system, much like a supersaturated solution. However, an adequate disturbance will cause the white tin to fall to its ground state. Like Vonnegut's Ice-9, this transition is autocatalytic; once conversion begins, there's no stopping it below 13.2 ºC, creating the so-called tin pest. There are actually several great videos showing the process, which you can easily find on YouTube, such as this one.


Finding final temperature for adiabatic expansion An ideal gas, molar heat capacity $C_V=\frac{5}{2}R,$ is expanded adiabatically against a constant pressure of $1\ \mathrm{atm}$ until it doubles in volume. Given initial temperature of $25\ \mathrm{^\circ C}$ and initial pressure of $5\ \mathrm{atm}$, calculate final $T$ (etc.).*

The author uses the First Law to calculate $T_2$ as $274\ \mathrm K$.

My question is, why can't we use the relation $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$?

It appears we cannot, since $\gamma =C_p/C_V$ and so, since $V_2=2V_1$, $T_2=274\ \mathrm K$, $T_1=298\ \mathrm K$,

$$\log_2\frac{T_1}{T_2}+1 =\gamma = \frac{C_p}{5/2}\rightarrow C_p\approx 2.8, $$

so assuming $C_p - C_V = R$, giving $C_p = \frac{7}{2}R$, does not seem to work.

Can someone explain why the highlighted equation doesn't apply?

*This is adapted from an example in Castellan, Physical Chemistry.

The equations you are proposing to use are valid strictly for an adiabatic reversible expansion. In the present adiabatic expansion, the external force per unit area is dropped suddenly from 5 atm to 1 atm, and held at that value until the volume doubles.  This takes place rapidly and spontaneously, and is not reversible, so the equations for an adiabatic reversible expansion do not hold.  For an irreversible expansion, the ideal gas law (or other equation of state) cannot be used to describe the force per unit area acting on the piston face, because the equation of state only applies to a system in thermodynamic equilibrium (or, since a reversible expansion consists of a continuous sequence of thermodynamic equilibrium states, to a reversible expansion).  In an irreversible expansion, the force at the piston face (where work is occurring) is determined not only by the amount of volume change, but by the rate at which the volume is changing.  This is because of viscous stresses that are important in rapid expansions and compressions.  In addition, the mass (inertia) of the gas comes into play, causing the pressure and temperature to be spatially non-uniform within the gas during its deformation.  So there is no single uniform value of the pressure or the temperature that can be assigned to the gas.


Volatile nature due to hydrogen bonding Why does intramolecular hydrogen bonding make an organic (or possibly inorganic as well(?)) compound volatile.

What I think is that it might be due to decrease in the solubility of the compound as intermolecular hydrogen bonding cannot be done after that.

"Volatile" usually refers to ease of evaporation, high vapor pressure, so it is a property of a pure substance while solubility is a property of the combination of two or more substances.

An intramolecular hydrogen bond makes a compound more volatile because the charge imbalances are offset internally instead of by forming an interaction with another molecule.

Interactions between different molecules of the substance would lower volatility, but intramolecular hydrogen bonds prevent the hydrogen bond donor and acceptor of one molecule from interacting with another molecule.


Will this series of reactions work to produce potassium phosphate? I want to end up with a solution or anhydrous sample of potassium phosphate with which I can grow larger crystals. I want a challenge and something to do so I don't want to buy potassium phosphate, I would rather make it myself.

I don't have access to a lot of chemicals but I do have a fair amount of monoammonium phosphate, (or ammonium dihydrogen phosphate), from a crystal growing kit. I plan to heat this until the ammonia is released to leave behind phosphoric acid to which I can then add water to have an acidic solution. I will then dissolve potassium carbonate from corroded alkali batteries into water and add this to the phosphoric acid. The products will then be carbon dioxide, water and potassium phosphate.

My question is will these reactions work and how easy/difficult will it be. I am sorry if this makes no sense at all because I have done my research for this on Wikipedia and it isn't exactly a 100% reliable source. Thanks.

Getting phosphoric acid by heating ammonium phosphate probably won't work since it requires temperatures above 200°C where phosphoric acid will start to vaporize and decompose.

A better approach might be the reaction of ammonium phosphate with potassium carbonate. The ammonium carbonate formed by this reaction decomposes at much lower temperatures (> 60°C). However, heating will produce corrosive, irritant and toxic products (ammonia, phosphoric acid, phosphorus oxide), so I do not recommend doing such experiments.


Why do you need a strong base for a titration For a titration, if you want to know the concentration of an acid you use a strong base such as $\ce{NaOH}$. My question is; why you can't use a weak base for this purpose? Is it because the calculation would be tougher? Is it because you'd need more of the base?

There is no reason why you cannot use a weak base to titrate a strong acid.  The calculation is probably considered a bit more complex but I think the real reason is a practical one.  For the sake of simplicity, I swapped your question around just a bit, and the figure below shows the titration of a strong base (blue) weak base (reddish) and very weak base (dashed line) with a strong acid (say, for example 1 M $\ce{HCl}$).

In each case the concentration of the acid is the same (1.0 M) and the concentration of the base is 0.100 M.  The volume of the base titrated is 100 mL.  For the standard strong acid/strong base scenario, one finds a well defined equivalence point.  When you transition to a weak base, however, there is additional information presented in the titration curve, namely the pKb of the base, which is a measure of its strength.  The pKb of the base can be obtained by finding the pH at the half-way mark between no acid added and the equivalence point (and if you know a bit of calculus, this is the point of minimum slope prior to the equivalence point). I've marked the pKb of the reddish titration curve with a dot, and the pKb is 9.  As the base gets weaker (less willing to accept a proton) the region before the equivalence point becomes less well defined.  In an extreme example, I have shown a simulated titration curve of acetic acid (a very poor base indeed!) with a strong acid.  Notice how the equivalence point in this curve would be very hard to pick out if we didn't know where we were looking.

So the bottom line is this: the calculations aren't harder (one uses the Henderson–Hasselbalch equation and this is a standard exercise in a first year analytical chemistry class) but the data one obtains from the experiment are harder to analyze.  And in practice, we do perform weak/strong acid/base titrations when we are interested in determining the strength of the acid or base under consideration.


Why is glycine hydrochloride more acidic than chloroacetic acid? I cannot figure out why glycine hydrochloride is more acidic than chloroacetic acid. Could I have figured this out from a $\mathrm pK_\mathrm a$ table and maybe am just not seeing it? On the table I have in my book (Organic Chemistry, Seyhan Ege, 3rd edition), neither acid nor its conjugate base are there.

Glycine hydrochloride is the amino acid glycine adsorving a molecule of Hydrochloric acid, wich is a strong acid. I think this is the main reason of the acidicity difference, since the pKa's are similar (comparing glycine and chloroacetic acid).

Update: glycine is amphoteric: it can either act as a base or an acid.
In water the hydrochloric acid molecule of the glycine hydrochloride will dissociate: $$\ce{H2O +HCl -> H3O+ +Cl-}$$
Then:
$$\ce{Gly +H+ -> GlyH+}$$
$$\ce{GlyH+ +Cl- <-> Gly + H+ + Cl-}$$
You can conclude, therefore, that Glycine hydrochloride is more acid than glycine (and also the chloroacetic acid). Hope it helps, and sorry for the english mistakes!


Can you heat something at alcohol bath? I have a mixture of $\ce{Et2O},$ $\ce{EtOH}$ and water, and I want to distill off the ether. I am using a water bath so not much water is distilled over, as the boiling water is 100 °C, and the solvent has like 20 °C lower temp than the water bath working temp, I believe.

So, my point is, is it safe and reasonable to make a bath with 95% ethyl alcohol to prevent much $\ce{EtOH}$ evaporation without needing to closely keep track of the temperature?

It would be extremely dangerous to use a flammable solvent for a heat bath because any spill on the hot plate (or even worse, a Bunsen burner) could cause a fire.

If you want to distill off ether, it is by far better to just use warm water from a tap or a kettle, and replace your hot bath regularly.

Trust me, ether can spontaneously catch fire VERY easily ! I have been there!


Which of the following compounds were involved in the titration? We have a sample that weighs $\pu{0.3148 g}$.This sample may contain one,two or even three of the following compounds: $\ce{NaOH, Na2CO3, NaHCO3}$. We have as a clue, that the sample was titrated with $\pu{20 mL}$ $\ce{HCl}$ $\pu{0.09 M}$ using phenolphthalein as an indicator. Later, the sample was titrated with $\pu{50 mL}$ of the same $\ce{HCl}$ using orange methyl as an indicator.

The question is how do we understand which of the compounds were involved? Should we set for example mass of $\ce{NaOH} = x$, mass of $\ce{Na2CO3} = y$ and mass of $\ce{Na2CO3} = z$, then write down the reactions between all of the compounds with $\ce{HCl}$ and then extract some equations from the equivalent point? One equation comes from the total mass.

Yeah,  you just gotta do that.

Assume the variables $x$, $y$, $z$. You need three equations to get the values.

First equation: Add all three and equate to $\pu{0.3148g}$.

Second equation: In the titration involving methyl orange, due to it's pH range being 3.1 to 4.4, changes color as soon as carbonates and bicarbonates are neutralized. That will give you the second equation.

Third equation: Consider all three bases for phenolphthalein titration, since phenolphthalein changes color at 8.3 to 10.0.This gives the third equation.

Solve all the three, and enjoy the efforts of your hard work.


How to find the lifetime of sodium atoms in the 3p using information on the spectral line width? I am trying to find the lifetime of sodium atoms at the 3p state using the relationship between the spectral linewidth and the uncertainty principle:
$$ \Delta E \times \Delta t = h / 4\pi $$
The information I have is the full-width-half-maximum (FWHM) linewidth of the sodium atomic absorption spectrum $\lambda = \pu{9.2E-14 m}$. How would I use this equation to find the lifetime?

The atomic transition from ground 3s to excited 3p states has a wavelength $\lambda = \pu{589.0 nm}$.

I used the relationship
$$ \Delta E = \frac{h c \Delta \lambda}{\lambda^{2}} $$

Plugging in $\Delta \lambda = \pu{9.200E-14 m}$, then
$$ \Delta E = \frac{h \cdot \pu{3E8  m} \cdot \pu{9.200E-14 m}} {(\pu{589.0E-9 m})^{2}} $$

So using the uncertainty relationship, we have that
$$ \Delta t = \frac{h}{ 4 \pi \Delta E} = \frac{(\pu{589.0E-9 m})^{2}}{4 \pi (\pu{3E8 m s^-1} \cdot \pu{9.200E-14 m})} = \pu{1.000E9 s} $$

My attempt to convert the linewidth from unit of $\pu{nm}$ to $\pu{eV}$:
$$ E = \frac{hc}{\lambda} = \frac{\pu{6.626E-34 J s} \cdot \pu{3E8 ms^-1}}{\pu{9.200E-14 m}} = \pu{2.161E-12 J} = \pu{1.348E7 eV} $$

Half the fwhm has to be added/subtracted  to the central wavelength and then this change in wavelength converted to an energy. Thus the limits on the wavelength are $\lambda \pm \Delta \lambda /2) $ and the energy change $\Delta E=hc/(\lambda - \Delta \lambda /2)-hc/(\lambda + \Delta \lambda /2)$ Joules with $c$ in m/s and $h$ in J s. Plugging in the numbers gives a lifetime of $\approx 10^{-9}$, i.e. $1$ ns.

As a rule of thumb, lifetimes of excited electronic states are rarely less than $10^{-12} $s and if the transition is allowed, as in this case often of the order of nanoseconds to microseconds. If transitions are forbidden then the lifetimes can extend from microseconds to seconds.


Derive expression for internal energy of mixing and entropy of mixing using statistical thermodynamics I want to derive an expression for the internal energy of mixing, $\Delta_\mathrm{mix}U$, and entropy of mixing, $\Delta_\mathrm{mix}S$. The framework for this is the Lattice Theory of Ideal Solutions. Specifically, I am mixing $N_A$ molecules of type A with $N_B$ molecules of type B together. I am supposed to find the following two relations:

$$
\Delta_\mathrm{mix} U = 0 \\
\Delta_\mathrm{mix} S = -kM (x_A \ln x_A + x_B\ln x_B)
$$

My attempted solution

I have the canonical partition function for an ideal solution of two different components:

$$
Q = \left( q_A e^{-cw_{AA}/2kT} \right)^{N_A} \left( q_B e^{-cw_{BB}/2kT} \right)^{N_B} \frac{(M)!}{N_A!N_B!}
$$

Taking the logarithm

$$
\ln Q = N_A\ln q_A - N_A\frac{cw_{AA}}{2kT} + N_B\ln q_B - N_B\frac{cw_{BB}}{2kT} + M \ln M - N_A\ln N_A - N_B \ln N_B
$$

where $w_{xx}$ is the pairwise interaction energy, $M=N_A+N_B$ is the total number of molecules = number of lattice sites.

Further, some thermodynamical relations for this system

$$
U = kT^2\left( \frac{\partial \ln Q}{\partial T} \right)_{N,M}
$$

$$
S = \frac{U}{T} + k\ln Q
$$

The internal energy of mixing and entropy of mixing should be

$$
\Delta_\mathrm{mix}U = \Delta U_A + \Delta U_B \\
\Delta_\mathrm{mix}S = \Delta S_A + \Delta S_B
$$

Generally we have that

$$
\Delta U = U_\mathrm{final} - U_\mathrm{initial} \\
\Delta S = S_\mathrm{final} - S_\mathrm{initial}
$$

I am a bit confused about how to compare the final and initial conditions of the system. I assume that we have initially two separated and equal volumes, one containing $N_A$ of A and one containing $N_B$ of B. Then we allow flow between the two volumes. Hence, the final volume is twice the initial volume. $N_A$ and $N_B$ are constant during the mixing. So I should have something like this

$$
\Delta U_A = U_{final} - U_{initial} = kT^2\left( \frac{\partial \ln Q}{\partial T} \right)_{N_A,M} - kT^2\left( \frac{\partial \ln Q}{\partial T} \right)_{N_A,M/2}
$$

and similarly for the other component. I just replace $M$ with $M/2$ for the expression of the initial state for both components? Most of the terms will be zero when doing the derivations, and the non-zero terms will actually just cancel. This gives that $\Delta_\mathrm{mix} U = 0$, which is the expected result for an ideal solution.

Looking at $\Delta_\mathrm{mix} S$ we have

$$
\Delta_\mathrm{mix} S = \Delta S_A + \Delta S_B = [S_{f,A} - S_{i,A}] + [S_{f,B} - S_{i,B}] \\
= \left[ \left(\frac{U_{f,A}}{T} + k\ln Q_f \right)- \left( \frac{U_{i,A}}{T} + k\ln Q_i \right) \right] + \left[ \left( \frac{U_{f,B}}{T} + k\ln Q_f \right) - \left( \frac{U_{i,B}}{T} + k\ln Q_i \right) \right] \\
=\frac{1}{T} (U_{f,A} - U_{i,A} + U_{f,B} - U_{i,B}) + 2k(\ln Q_f - \ln Q_i) \\
= \frac{\Delta_\mathrm{mix}U}{T} + 2k(\ln Q_f - \ln Q_i) \\
= 2k(\ln Q_f - \ln Q_i)
$$

where I used in the last step that $\Delta_\mathrm{mix} U=0$ for an ideal solution, as explained above. However, using this last expression to calculate the entropy of mixing, I find that (most of the terms cancel directly)

$$
\Delta_\mathrm{mix}S = kM \ln (2N_A + 2N_B)
$$

which is quite different from the correct expression. Is there something wrong with my procedure? (there likely is!)

Following the commented suggestion:

$$
S = k\ln W \Rightarrow \Delta_\mathrm{mix}S = k\ln \frac{W_\mathrm{mix}}{W_AW_B} = k\ln W_\mathrm{mix} = k\ln \frac{M!}{N_A!N_B!} \\
= kM [\ln M - x_A\ln N_A - x_B \ln N_B]
$$

which is almost correct. If I could divide all logarithmic arguments by $M$, I would get the correct answer.

In the solution there are two types of molecules $N_1$ and $N_2$. Assume that they do not interact with one another but simply occupy particular 'lattice' sites by blocking them. The total number occupied sites is $N=N_1 + N_2$. The first molecule can be placed at any of the $N$ sites, the second at $N-1$ empty sites and so on. The total number of possibilities is $N(N-1)(N-2)\cdots = (N_1 + N_2)!$ . Now, molecules of the same type are not distinguishable from one another thus this number must be divided by $N_1$ for molecules of the first type and similarly for the other type. Thus the number of distinguishable arrangements is

$$W_\mathrm{mix}= \frac{(N_1 + N_2)!}{N_1!N_2!} $$

The pure compounds have only one type of molecule thus for them $W_1=N_1!/N_1! = 1$ etc.

The entropy is defined as $S=k\ln(\Omega)$ where $\Omega$ is the number of arrangements $W_\mathrm{mix}/(W_1W_2)$ thus
$$ \Delta S=k\ln\left(\frac{(N_1 + N_2)!}{N_1!N_2!}\right)$$

To simplify this it is convenient to use Stirling's approximation as $N$ is large:

$$\ln(N!) \approx N\ln(N)-N+1$$ and also as $N$ is large the $1$ can be ignored, giving $$\ln(N!) \approx N\ln(N)-N$$

Expanding gives
$$ \Delta S = k[(N_1 + N_2)\ln(N_1 + N_2)-N_1\ln(N_1)-N_2\ln(N_2)]$$
expanding, multiplying by $-1$ and simplify gives

$$ \Delta S = -k\left[N_1\ln (\frac{N_1}{N_1 + N_2} )+ N_2\ln(\frac{N_1}{N_1 + N_2})\right]  $$

which can be expressed as mole fractions
$$ \Delta S= -k[N_1\ln(x_1) + N_2\ln(x_2)]$$

in molar terms, $ \Delta S= -R[n_1\ln(x_1) + n_2\ln(x_2)]$. This is the ideal solution entropy of mixing.

If the molecules do interact then this is a complicated problem in regular solutions, the free energy is
$$\Delta G = RT[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2w$$

where $w \approx (2E_{12}- E_{11} - E_{22})$ and $E_{12}$ is the interaction energy between molecules of type $1$ and $2$ when they are adjacent and similarly for $E_{11}$ and $E_{22}$ all of which are negative. From a consideration of the energies involved $w$ is expected to be positive.


Determining structure from C-NMR spectrum, is my answer correct? Determine the structure of the compound whose spectral data is given below and explain your reasoning (5 marks for structure, 15 marks for reasoning). The peak at 7.27 ppm does not belong to the compound. What is the source of this peak?
  (2.5 marks) Looking at the full spectrum you will see small peaks at about 4.2 and 3.7 ppm. What is the origin of these peaks? (2.5 marks)

  Formula: $\ce{C9H10O3}$

  IR: small peaks observed between 3010 and 2853. The largest peak is at 1693. All peaks are in $\pu{cm-1}.$

  $\ce{^{13}C}$ NMR (DEPT-135 result in brackets): δ 190 (+), 153 (no peak), 149 (no peak), 131 (no peak), 127 (+), 111 (+), 109 (+), 58 (+), 57 (+) ppm.

  $\ce{^1H}$ NMR (300 MHz. See expansions on next page):

I mainly looked at the $\ce{^{13}C}$-$\mathrm{NMR}$ aspect of this question. Given that there is a peak at $\pu{190ppm}$, it is probably an aldehyde with a $\ce{CH}$ (which explains the +DEPT sign). The molecular formula also showed a degree of unsaturation of 5, which could be explained by a benzene ring and a double bond (which links up nicely with the carbonyl group). But in order to come up with no peaks in an aromatic ring I added some methyl groups. I eventually came up with the structure 2,3-dimethylbenzaldehyde. The +signs indicated $\ce{CH}$ or $\ce{CH3}$, the 5 $\ce{CH3/CH}$ groups could be explained by the methyl groups and the rest of the carbons in the benzene ring. Do you think this structure is correct?

It is not always a good idea to focus only on $\ce{^{13}C}$-$\mathrm{NMR}$ when you are doing structure elucidation on unknown compound using spectral data. Yes, $\ce{^{13}C}$-$\mathrm{NMR}$ gives you valuable info, so does FTIR to some extent. Yet, $\ce{^{1}H}$-$\mathrm{NMR}$ gives the most info about the compound as I depicted in following diagram:

All your analysis is correct except for the final conclusion on the compound. It is not 2,3-dimethoxybenzaldehyde as you concluded. But it is 3,4-dimethoxybenzaldehyde. How do I know that? Look at the diagram and analyze the coupling pattern (disregard the small but significant meta-coupling, which give close to $\pu{2 ppm}$ here).

If the compound is 2,3-dimethoxybenzaldehyde, the $\ce{^{1}H}$-$\mathrm{NMR}$ spectrum of it should displays a doublet, a triplet (or dd), and a doublet in aromatic region (~$\pu{6.7-8.5 ppm}$), in that order going from upfield to downfield. However, what given $\ce{^{1}H}$-$\mathrm{NMR}$ exhibit is a doublet, a singlet, and a doublet in the aromatic region (disregarding the meta-coupling). The singlet indicates the presence of isolated aromatic ring proton, and two separate doubelets also indicate the presence of two isolated but neighboring aromatic ring protons. Since three substitution groups are one formaldehylde (-$\ce{CHO}$) and two methoxy groups (-$\ce{OCH3}$), there are three possibilities: The compound would be 2,5-dimethoxybenzaldehyde, or 2,4-dimethoxybenzaldehyde, or 3,4-dimethoxybenzaldehyde:

The possibility of been 2,4-dimethoxybenzaldehyde can easily be ruled out, because it would show the upfield shift of the chemical shift of the singlet corresponding to the proton on position 3 of the ring (been ortho to two -$\ce{OCH3}$ groups), even shielded than the doublet at the $\pu{6.99 ppm}$. This is justifiable conclusion since the resonance structures of 2,4-dimethoxybenzaldehyde (using only 2-methoxy group) indicate that the ortho ($\ce{C}$-3) and para ($\ce{C}$-5) positions will be more shielded than meta positions. Also, with respect to 4-methoxy group, the ortho ($\ce{C}$-3 and $\ce{C}$-5) positions will be even more shielded than corresponding ortho positions  in 3,4-dimethoxybenzaldehyde.

However, distingushing 3,4-dimethoxybenzaldehyde from 2,5-dimethoxybenzaldehyde is tricky. Both would have a doublet at $\pu{6.99 ppm}$ (for the proton at $\ce{C}$-5 for the former and the proton at $\ce{C}$-3 for the latter) since substitution effects are the same for both. Same argument is valid for the singlet (with meta-coupling) at $\pu{7.43 ppm}$ (for the proton at $\ce{C}$-2 for 3,4-dimethoxybenzaldehyde and the proton at $\ce{C}$-6 for 2,5-dimethoxybenzaldehyde) since, again, the substitution effects are identical for both. Yet, chemical shift for the proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde and that of the proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde would not be identical. That is because ortho-effect of a substitution group is somewhat larger than para-effect of the same group in general (there are few examples you may find here), regardless of its sheilding or deshielding effect. Thus, the chemical shift for the proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde should be more shielded than that for the proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde. My reasoning:

The proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde should be deshielded by para-$\ce{CHO}$ group (blue shift), while shielded by ortho-$\ce{OCH3}$ group (larger red shift).
The proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde should be deshielded by ortho-$\ce{CHO}$ group (larger blue shift), while shielded by para-$\ce{OCH3}$ group (red shift).
The both protons at $\ce{C}$-6 position of 2,5-dimethoxybenzaldehyde and at $\ce{C}$-2 position of 3,4-dimethoxybenzaldehyde are deshielded by ortho-$\ce{CHO}$ group (larger blue shift), while shielded by ortho-$\ce{OCH3}$ group (larger red shift).
Comparition of reason 1 & 3 clearly shows these two chemical shifts should be switched (instead of singlet followed by doublet, it should be doublet followed by singlet).

Thus correct resonance order would be given by 3,4-dimethoxybenzaldehyde.


Measuring volume change from change in mass I want to measure the expansion (by change in volume) of a material in liquid. However the shapes are so complex that I can only measure their mass. Is it possible to compute the volume of the expanded shape by measuring its mass alone?

Given mass $m_0$ at $t_0$ and mass $m_1$  at $t_1$ and the density $\rho_\mathrm m$ of the material and $\rho_\mathrm l$ of the liquid, can I find the volume change of the object from $t_0$ to $t_1$?

One could assume that $(m_0-m_1)/\rho_\mathrm l$ is the correct answer, but I am not sure if this is the right way to go about it?

You can't compute the volume of a solid from the mass alone.

You can measure the specific gravity as a function of the temperature of the liquid. You'd need to know how the density of the liquid changes with temperature to convert the specific gravity data at each temperature to actual density value which could then be used to calculate the effective volume of the solid at each temperate.


When will peroxide form? A very typical example of a compound of peroxide is $\ce{H2O2}$. When like the title says, will such compounds form, despite usually oxygen will have a oxidation number of 2-?

From Greenwood and Earnshaw's excellent book Chemistry of the Elements:

  When the metals are burned in a free supply of air the predominant product depends on the mtal: $\ce{Li}$ forms the oxide $\ce{Li2O}$ (plus some $\ce{Li2O2}$), $\ce{Na}$ forms the peroxide $\ce{Na2O2}$ (plus some $\ce{Na2O}$) [...]

In general, formation of inorganic peroxide salts in this fashion doesn't appear to be especially favorable for other metals, at least by comparison to the variety of other oxides possible. Sodium's peculiar preference seems anomalous to me, especially given that the lattice energy for the oxide is higher (albeit slightly). This may be a function of the availability of oxygen, which is roughly $21\%$ in air, and is certainly temperature-dependent as well.

As for organic substances, formation of peroxide linkages is possible from a variety of functional groups and using various reagents. Ordinary $\ce{O2}$ will form peroxide bonds with various substrates via radical mechanisms, especially under photochemical conditions. Carboxylic acids react with $\ce{H2O2}$ to form peroxyacids. These are just two examples that come to mind, and numerous others exist.


Finding x and y in Pt(NH3)xCly Questions 8 and 9 refer to the analysis of a salt whose formula is $\ce{Pt(NH3)_xCl_y}$.

  $0.02$ moles [sic!] of this salt required $40.0~\mathrm{cm^3}$ of $2.0~\mathrm{mol\,l^{-1}}$ nitric acid for exact neutralisation.

  The number of moles of $\ce{NH3}$ per mole of salt is

  A 2

  B 4

  C 6

  D 8
  $0.02$ moles [sic!] of the salt were dissolved in nitric acid and excess silver(I) nitrate solution was added. The precipitate formed was filtered, washed and dried. It weighed $5.74~\mathrm{g}$.

  The number of moles of chloride per mole of the salt is

  A 1

  B 2

  C 3

  D 4

For question 8, I first found the no. of moles nitric acid to be $0.08\ \mathrm{mol}$, which is four times the number of moles of the salt. However, I just can't see how I would get $x$ from this. (The answer to this question is somehow B which makes even less sense since that's the number of moles of nitric acid for each mole of salt.)

For question 9, I tried to write a balanced equation for the formation of the precipitate but I don't think that's possible seeing that we don't know $x$ and $y$. Now, I know that the number of moles of the precipitate (silver chloride) is $0.04\ \mathrm{mol}$ which is twice than that of the salt. Yet again, I don't know how I would get $y$ from this. (The answer to this is also B, which is also the number of moles of precipitate for each mole of salt.)
 It would be appreciated if someone could explain how one can obtain the values lf $x$ and $y$ here.

You're almost there!

For 8:  For every mole of acid added, it reacts with a mole of the base included in the Pt(NH3)xCly, which is the ammonia.  So you can figure out how many moles of ammonia have reacted, and compare this with the moles of salt you started with.

For 9: For every mole of AgCl you have formed, you have used up one mole of Cl from the Pt(NH3)xCly.  So you can figure out how many moles of Cl you have captured, and compare this with the moles of salt you started with.


Can we prevent metal corrosion by creating a vacuum around it? I have a device in a box which has parts of metal. So can I prevent corrosion by creating a vacuum around the metallic object?

The word corrosion has a range of meanings, but it is generally taken to indicate the chemical reaction of a material with its environment. For example iron corrodes by reacting with oxygen in the air (or water if it's underwater).

If you put the metal in a vacuum then it doesn't have anything around it to react with, so in this case yes putting the metal in a vacuum will stop it corroding.


Why was rhenium used as an igniter in flash bulbs? It looks like it typically had a current passed though it to ignite zirconium threads.

What properties of rhenium made it the best metal to use for this?

New tungsten alloys containing rhenium were developed as a filament material in flashbulbs for the purpose of improving their reliability. Weak batteries (anything but rare in the carbon-zinc era) could easily result is a slightly-delayed flash or no flash at all, spoiling the photo in either case.
An important characteristic of batteries is their internal resistance. This impedance is small in a fresh dry-cell battery but increases as the battery wears out, eventually reaching a value high enough to make the battery unusable. The rhenium acts to "increase the cold resistance of tungsten wire filament so as to permit sufficient power to be obtained over large variations in impedance."
As the prior post suggested, anti-oxidation properties of the rhenium also helped greatly  reduce the chance of a filament burn-out and consequent flash lamp ignition failure.
Because photography wasn't cheap back in the 50's and 60's, the vast majority of cameras and flash units sat in closets at least nine-tenths of the time. Even unused, their carbon-zinc batteries could go bad in three years. Considering the many millions of flash  pictures attempted with nearly-dead batteries in that era, this reliability feature looks like an excellent idea.
For details, please see United States Patent No. 3,123,993 granted 3/10/1964 to George W. Cressman, Chagrin Falls, and Louis A. Demchock, Jr., Cleveland, Ohio, assignors to General Electric Company.


How to determine the concentration after a dilution with Beer's law? The molar extinction coefficient of pure antibody is $1.4$ absorbance units per milligram antibody at $\pu{280 nm}$. You have a concentrated antibody solution, so you make a $5$-fold dilution and measure at $\pu{280 nm}$, getting an absorbance of $0.845$. What is the concentration of that solution?

Using Beer's Law this is my work:
$$0.845 = (1.4)\cdot (1)\cdot (c) \\
\therefore c = \pu{0.604 mol L^-1}$$

Then I think we need to factor in the dilution, therefore $c = (5) \cdot (0.604) = \pu{3.02 mol L^-1} $. Is this correct?

Beers law is

$$ E= \epsilon_{\lambda}\ \cdot c\ \cdot d$$

Where $c$ is the concentration of the solution. Then, you've found your answer once you've found $c$.

As Jerepierre points out, if you were to account for units in the answer, the question is actually missing some information. The pathlength $d$ is unaccounted for (but we're assuming its $1$), however if it was say $\mathrm{1\ cm}$, then the units for $\epsilon$ would need to be $\mathrm{AU\cdot mg^{-1} \cdot cm^{-1}}$. Which would mean that the units for your concentration would be in $\mathrm{mg}$. But, for concentration we would need to know antibody per some amount of volume. So taking units into account shows that something is missing.

What you're adding at the end is a step you would take to find the original concentration of the undiluted solution. Given that they diluted the concentration 5 times, but we don't know to what factor so we would have trouble solving for that.


What is 'power law' kinetics? I've come across a lot of instances where 'power law' description of the kinetics was mentioned, but I cannot think of any integrated rate law that follows something like:

$$A(t)=bt^n+c$$

I'm pretty sure the 'power law' refers to an integrated rate law instead of rate equations like:

$$r = k\prod_i [A_i]^i$$

So what is power law kinetics anyway? Does this 'power law' have a physical meaning or just a polynomial approximation that tells nothing about the mechanism?

Usually these type of complex kinetics arise when there is an inhomogeneous distribution of properties in a system. For example reaction over barriers which have a range of sizes, this can occur in proteins or in quantum dots but not in 'normal' kinetics. Fluorescence quenching in micelles can follow non-single exponential decays because there is usually a distribution of quenchers  solubilised in micelles. The fluorescence has the form $\ln(I)\sim -At+B\exp(-Ct)+D$ where , $t$ is time and $A,B,C,D$ constants.

In diffusion controlled quenching or energy transfer quenching in solution the rate constant will depend on time as $k=k_0+k_1/\sqrt{At}$. So there are many situations in which 'strange' kinetics can be observed.

However, vey often 'power law' kinetics is just a name given to an empirical method of fitting data to a function, say of the form $\exp(-kt^n)$ where $k, n$ are variables used to fit the data. In this form the decay function has no physical meaning.

Addendum. The first equation above was calculated as shown below.

As a model we suppose that the fluorophore and quencher are both in a micelle, as example of a restricted geometry, and that effectively no fluorophore exists outside. There is always an excess of quencher over fluorophore whose concentration is small so that no more than one will exist in any micelle. It is possible that zero, one or  several quenchers can be in any micelle.

We suppose that the rate constant for quenching is $k_i=k_0+ik_q$  where $k_0$ is the fluorescence decay rate constant value without quencher and $i$ the number of quenchers in any micelle, $i=0,\,1,\,2\cdots$ and $k_q$ the constant for quenching of the fuorophore.

The measured fluorescence signal is the sum of contributions from all micelles and is therefore

$$I(t)=I_0\sum_{i=0}^\infty p_i\mathrm e^{-k_it}$$

where $p_i$ is the probability of there being $i$ quenchers in a micelle. It seems reasonable to suppose that this has a Poisson distribution, and in that case $\displaystyle p_i= \frac{m^i\mathrm e^{-m}}{i!} $ where $m$ is the mean number of quenchers in a  micelle. Substituting and simplifying the intensity then becomes

$$ I(t)=I_0\mathrm e^{-k_0t-m}\sum_{i=0}\frac{(m\mathrm e^{-k_qt})^i}{i!}$$

The summation term has the form $\sum x^i/i!$ which is the expansion of the exponential thus

$$ I(t)=I_0\mathrm e^{-k_0t-m}e^{m\mathrm e^{-k_qt}} = I_0\mathrm e^{-k_0t +m(\mathrm e^{-k_qt} -1)      }    $$

which when logs are taken is the expression above.


Visualizing rotation about a double bond So I'm having trouble visualizing rotation about a double bond without using a model kit. The 2nd image has a red dot where I rotated it to make the molecule look exactly like the second image with my head and then my model kit. This is also consistent with my answer key that says these molecules are identical.

Now, the next two molecules:

I visualize rotating the left molecule at the red dot carbon and in my head, it matches up with the right molecule. However, when I pull out my model kit, it does not rotate like that and does not match up. This is also consistent with my answer key that says these molecules are diastereomers. So my question is, are there any rules that I could use or a way to visualize that will help me realize if a bond can be rotated to match a molecule without using a model kit.

Also, when people talk about it is not possible to rotate a double bond without breaking it, what exactly are they talking about that makes it different from this?

Adding one more element to help you long-term: I think the underlying reason you are having a problem with this is that the visualization you are using is a little misleading.

You're selecting an atom with your red dot.  You can't rotate around an atom. You can rotate around the bonds between the atoms, though, assuming that the constraints of that bond allow it.

So in the first example, there are two possible carbon-carbon bonds you can rotate around that would both correspond to that red dot. Your dot should be on the sigma bond, based on your diagram.

In the second, you have three carbon-carbon bonds connected to your red dot. The one you are rotating is the double bond, not either of the bonds in the ring.  So the dot should instead be on that bond.

If you change the way you're marking the connection in your head, it could help you think about the nature of that connection, and that could help you with the rest of the information everyone is providing.


At equilibrium are the rates of the forward and backward reactions equal to zero? At equilibrium, the concentrations do not change with time. So, is it true that the rates of the chemical reactions are zero at equilibrium? Wikipedia says that they are not zero. Why is this?

In equilibrium, the forward and backward rates are equal to each other. The net is zero, but the individual rates are not zero.

Consider something as simple as water — we know that the water molecule can dissociate in to an $\ce{OH-}$ and $\ce{H+}$, and the pH is a direct measure of the amount of $\ce{H+}$. Once equilibrium is achieved, it isn't like $\ce{H2O}$ stops splitting, or $\ce{H+}$ and $\ce{OH-}$ get back together — it is just at a macro level nothing changes.


Electron shell, how do they know it? I started to learn Chemistry with something like Hydrogen having 1 electron in the first shell, Oxygen having 2 electron in the first shell and 6 electron in the second shell... But I don't really understand how scientists know it? How do they know there are how many shells and how many electron per shell? Is that something visible? Or just an imagination theory or they using a method to measure?

Electrons are not actually separated in shells. That is a simplified model to explain the atom's structure when people first start approaching Chemistry or Physics.

In reality, electrons' movements around the nucleus are governed by complex equations (some of them not even resolvable so approximations are needed) to form what is called an electron cloud. This cloud is confined in proximity of the nucleus by the electromagnetic force. Every electron is defined by its wave equation, which tells us the probability that the electron appears to be at a particular location when its position is measured. These equations allows us to define regions in space where the electron spend the most part of its time around the nucleus, and they can be visualised as atomic orbitals, as shown below. Orbitals can have one or more ring or node structures, and they differ from each other in size, shape and orientation.

Each orbital (which is defined by a different equation) has different associated parameters, called quantum numbers, and a particular energy. These properties allow us to divide them into levels and sublevels, to form a structure that resembles multiple shells one inside each other. But as you can see from what I explained the structure is in fact far more complex.

This knowledge has been gained over time both from the mathematical and physical analysis of different equations (like Schröndiger equation) and by experimental data collected by the means of different techniques. Starting from Thomson and Rutherford the model of the atomic structure has seen many changes (Bohr, Moseley, Van der Broek, de Broglie, etc.) and it has come closer to a realistic representation of the actual atom. Take a look at their experiments to understand how it has been studied. However, they often involve particle physics and are not really straightforward to understand.


Why, during aspirin synthesis in acidic conditions, anydride carbonyl oxygen is protonated rather than salicylic acid carbonyl oxygen ? The common route for aspirin synthesis (described here) employs acetic anhydride and salicylic acid, in acidic conditions.

The first step of such a reaction mechanism is the protonation of the carbonyl oxygen of acetic anhydride.

Why, instead, protonation of carbonyl oxygen in salycilic acid should not occur ? Of course, such a protonation is not desired, since it won't lead to aspirin: however, can it be neglected and why ?

Thanks for any hints.

Mechanisms are explanative, and not predictive.

The first step of the reaction mechanism is protonation of the anhydride because that is the first step of the reaction.  That just is.  If the anhydride is not protonated, step 2 simply won't happen.

This does not mean that the salicyclic acid is not protonated.  It just means that the protonation of salicyclic acid is not necessary to explain how acetylsalicyclate forms.  Proton exchanges are at equilibrium.  It is likely that the salycilic acid is protonated during this process.  However, that protonated salicyclic acid may then protonate acetic anhydride and proceed as indicated in your link.  For the purposes of describing the mechanism, it is not necessary to distinguish where the proton that protonates the acetic anhydride comes from.

So your question really is self-answered, where you say, roughly, 'Such a protonation does not lead to product directly'.


How to calculate the equilibrium composition after a change in partial pressure of one component? Question
  Ammonium carbamate dissociates as shown below.
  $$\ce{NH2COONH4 (s) -> 2NH3 (g) + CO2 (g)}$$
  In a closed vessel containing ammonium carbamate in equilibrium with ammonia and carbon dioxide, ammonia is added such that partial pressure of $\ce{NH3}$ now equals to the original total pressure. What is the ratio of the new total pressure to the original pressure?

I tried solving the problem and I know I am quite close to the answer which is $\frac{31}{27}$.
Firstly I took out $K_p$ for equilibrium assuming total pressure to be $p$.
So,

$K_\mathrm{p} = \left(\frac{2p}{3}\right)^2 \times \frac{p}{3} = \frac{4p^3}{27}$

After this the par pressure of ammonia becomes $p$ so let's say that of $\ce{CO2}$ is $\frac{p}{3} + x$.

Now since $K_\mathrm{p}$ doesn't change so:

$\frac{p}{3} + x = \frac{4p^2}{27}$ (cancelling one of the $p$)

But solving further and finding new pressure and dividing by the old one doesn't give the answer

As follows:

Just to simplify the math a little bit, assume pressure of $\ce{CO2_{(g)}}$ is p. Thus, pressure of $\ce{NH3_{(g)}}$ is 2p at equilibrium. Total pressure in this case is 3p.

Let 3p equal new pressure of $\ce{NH3_{(g)}}$.

Added pressure of $\ce{NH3_{(g)}}$ would shift the equilibrium to the left, so say we lose 2x pressure of $\ce{NH3_{(g)}}$ and x pressure of $\ce{CO2_{(g)}}$. New equilibrium pressures will be 3p-2x for $\ce{NH3_{(g)}}$ and p-x for $\ce{CO2_{(g)}}$. Thus, ratio of new to old is $\frac{4p-3x}{3p}$. You can solve for x in terms of p using $K_p$.

Solve for x and find the ratio, which should be $\frac{31}{27}$


How to find trace methanol content in fermented fruit/distillate? Suppose we have some fermented (optionally also distilled) fruit/fruit juice. The sample therefore contains water, sugars, ethanol etc. Which analysis method is the simplest to find out whether the toxic methanol is also present?

Methanol, formed by degradation of pectines from the skin of fruits, as well as higher alcohols is volatile and does not show significant absorption in the UV.

Consequently, gas chromatography is the method of choice to determine these components in the fermentation mix.


Why is mass gained at the cathode? I'm dealing with a voltaic cell constructed like this:

(From “Electrochemistry”, chapter 19 from the book Principles of General Chemistry (v. 1.0).)

I understand that as oxidation occurs at the $\ce{Zn}$ electrode, the electrode loses mass as $\ce{Zn}$ dissolves into $\ce{Zn^{2+}}$ in the solution and eventually will bond with $\ce{Cl-}$ coming from the salt bridge. But on the $\ce{Cu}$ cathode side, how could it gain mass if initially the $\ce{Cu}$ electrode lost mass after the $\ce{Cu^{2+}}$ ions dissolved in the solution? I get that the $\ce{Cu^{(2+)}}$ ions are re-attracted to the $\ce{Cu}$ electrode once electrons flow from anode to cathode, but wouldn't that mean that the $\ce{Cu}$ will retain the same mass once the cell is complete?

As you have figured out, the zinc electrode is undergoing oxidation since zinc is a stronger reducing agent than copper.

$$\ce{Zn -> Zn^2+ + 2e-} $$

Thus, the zinc electrode becomes thinner as shown in your image. It is oxidizing into soluble $\ce{Zn^2+}$ ions and thus its mass decreases.

In the right solution, $\ce{Cu^2+}$ ions are being reduced to metallic copper which deposits on the copper cathode

$$\ce{Cu^2+ + 2e^- -> Cu} $$

The $\ce{Cu^2+}$ ions are already there to begin with. The copper electrode isn't dissolving to form $\ce{Cu^2+}$ ions to begin with (the copper electrode does not react with the $\ce{Cu(NO3)2}$ solution), so there isn't any loss in mass of the copper electrode initially. Rather, during the operation of the cell, the reduction of copper(II) ions forms metallic copper, which causes the copper electrode to increase in mass and gets thicker, as seen in the image.


Applying Avagadro's law: the relationship between moles and volume of a gas If $\pu{2.4 mol}$ of gas fill a volume of $\pu{175 mL}$, what volume, in mL, will $5.4$ additional moles of gas fill?

This was a quiz question which I got wrong the first attempt, and I think I got it right this time however I would like some confirmation that it is/is not. The unnamed gas is assumed to be ideal.
$$\frac{\pu{2.4 mol}}{\pu{175 mL}} = \frac{\pu{7.8 mol}}{X}, \quad \text{or} \quad X = \frac{\pu{7.8 mol} \times \pu{175 mL}}{\pu{2.4 mol}} = \pu{568.75 mL}$$
Is this accurate?

You are correct in your approach and answer. Assuming the gas to be ideal, one can apply Avogadro's Law which states that for an ideal gas $V \propto n$.

This implies $V_1/n_1 = V_2/n_2$.

Using the given values one does obtain the volume as $568.75\ \mathrm{mL}$


Why the bond strength of HCl is more than HBr? Most people say that it is because the 3p orbital of chlorine when overlapping with the s orbital of hydrogen covers more area than when 4p orbital of bromine overlaps with the s orbital of hydrogen which makes bond between H and Cl more stronger than the bond between H and Br. But how can we say that the area of overlapping is more for 3p and s orbital compared to overlapping of 4p and s orbital?

You can think of the overlapping volume between the electron clouds, but a simpler form to show why this happens is when you consider the electrical potential between the atoms. Even tough this isn't an ionic compound, it has many characteristics that are similar.

If I remember correctly, this is the expression for the potential energy between charged particles: $E_p = k\frac{Q_1 \cdot Q_2}{d}$

Since the radius of $\ce{Br}$ is greater than the radius of $\ce{Cl}$, the potential energy involved in $\ce{H-Br}$ bond will be smaller than $\ce{H-Cl}$ bond.

Another thing that came to my mind, and I'm not sure, is that as the 4p cloud of $\ce{Br}$ is bigger than that of $\ce{Cl}$, the relative amount of the volume that is overlapped with the 1s cloud of $\ce{H}$ gets smaller with $\ce{Br}$ than with $\ce{Cl}$, which means the connection is weaker.


Determining the Equivalence Ratio Determine the equivalence ratio of the following mixture: 4g of Butane, $\ce{C4H10}$, and 75 grams of air.

Answer is 0.82

Ok so I try converting everything into mols

$4~\mathrm{g}~(\ce{C4H10}) \times \frac{1~\mathrm{mol}}{58~\mathrm{g}} = \frac{4}{58}~\mathrm{mol}$

$75~\mathrm{g~(Air)} \times \frac{1~\mathrm{mol}}{28.966~\mathrm{g}} = \frac{75}{28.966}~\mathrm{mol}$

Air-fuel ratio = $\frac{\frac{75}{28.966}}{\frac{4}{58}} = 37.54$

This is just a guess that the air-fuel ratio is some sort of equivalence ratio.

I saw that equivalence ratio =  actual air-to-fuel ratio divided by the theoretical. I don't know how to get either actual or theoretical? Is the data in the problem sufficient enough to get both?

I also see a formula in the reviewer: Fuel + Air = Products of combustion and the appropriate reaction, but it only works for finding the exact amount of mols like: $\ce{C4H10 + aO2 + 3.76aN2 = bCO2 + CH2O + 3.76aN2}$

This is for a mechanical engineering board exam review; unfortunately, university only tapped briefly on the chemistry side of fuels.

Any hint on what I should be doing? Point me to the right direction?

$$\begin{align}
\text{AFR (Air to Fuel ratio)} &= \frac{n_{\text{air}}}{n_{\text{fuel}}} \\
\text{Equivalence ratio} &= \frac{\text{AFR}}{\text{AFR}_{\text{stoich}}} \\
\end{align}$$

According to the problem statement, if you did everything right,
$$\begin{align}
\text{AFR} &= 37.54 \\
\text{AFR}_{\text{stoich}} &= \frac{\nu_{\text{oxygen}}}{\nu_{\text{fuel}}}\times 4.76 \\
\ce{C4H10 + $6.5$\,O2 &-> 4 CO2 + 5 H2O} \\
\text{AFR}_{\text{stoich}} &= 4.76 \times 6.5 = 30.94 \\
\text{Equivalence ratio} &= \frac {37.54}{30.94}=1.21
\end{align}$$
Now there are two types of equivalence ratio. One is air to fuel and another is fuel to air. I suspect in the answer they meant fuel to air equivalence ratio:
$$\text{FAR equivalence ratio} = \frac{1}{\text{AFR equivalence ratio}} = \frac{1}{1.21}=0.82$$