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Does chemistry need machine learning?
In many fields of science (e.g. biology, medicine, psychology, statistics, physics), machine learning and artificial intelligence techniques are becoming more and more popular to analyze data. Is it also so in chemistry?

On https://www.chemistryworld.com there is a nice series of articles on machine learning in chemistry.

The main challenge here is the paucity of large open datasets in Chemistry. When you can get the data, machine learning techniques often work.

@chrishmorris Welcome to Chemistry.SE! Please take the [tour] to get familiar with this site. I have converted your answer to a comment, because in this state it does not really fit the philosophy of this site.

Try out Chembrows (https://www.chembrows.com/website/) in parallel to your usual literature search, and you may see how well (predictive / suggestive) AI works for you (or not).

For one thing, ML is already used in drug development and similar R&D situations.

The short answer is yes. Machine learning, data mining, AI and other techniques are highly useful in chemistry.
I completely agree with Fred's answer that lots of machine learning, expert systems and statistical analysis in chemistry goes back a long time. This is particularly true in analytical chemistry - match a mass spec or NMR or IR against a library of known compounds.
Now I saw your tag "computational chemistry" and there are some newer applications here. Basically the question is whether you can predict properties (e.g., the heat of formation, dipole moment, etc.) that would normally require quantum mechanics.
There are a few groups attempting this, but one of the more successful attempts so far has come from Anatole von Lilienfeld.

The researchers found that with a landscape of more than 5000 molecules, the error for predicting atomisation energies of new molecules drops below 10kcal/mol, approaching the 5kcal/mol accuracy of hybrid DFT. 'Calculating a molecule's atomisation energy using hybrid DFT would take on average one hour on a single CPU,' says von Lilienfeld. 'With machine learning, it's milliseconds.'

There are more conventional approaches to use machine learning, evolutionary algorithms and the like to perform parameterization of force fields, semiempirical quantum mechanics, etc.
Basically, if you have a lot of data, machine learning techniques can be effective ways to analyze the data and use it for other purposes.

Yes. It's not a new thing at all in chemistry; you'll find papers on chemical AI applications going back four decades or more.
We have massive amounts of data that must be processed quickly. For example, a diode array detector in HPLC can collect large numbers of spectra per minute at hundreds of wavelengths, and we need to use that data to distinguish and possibly identify two closely eluting compounds coming off the column. Intelligent, automated data analysis and pattern recognition is a must. A lot of "artificial nose" sensors have similar requirements; they can use neural nets to answer questions like "are these potatoes infected with dry rot?" We also use expert systems in various areas of analytical chemistry.
Here are some nice overviews by Hugh Cartwright:
Using Artificial Intelligence in Chemistry and Biology: A Practical Guide (CRC Press)
Applications of Artificial Intelligence in Chemistry (Oxford University Press)
Development and Uses of Artificial Intelligence in Chemistry (Reviews in Computational Chemistry, Volume 25, Wiley)

I know I'm a little late to this party, but in the last few years there have been some potentially very important developments in the application of machine learning to chemistry. Both of which apply to molecular dynamics.
First, one potentially obvious observation is that when performing a molecular dynamics simulation, each of the time steps is highly correlated with the previous time step, assuming the time step employed is small enough to calculate any meaningful property. This is a perfect situation for machine learning because it means that after using some of these time steps as a training phase, the subsequent time steps can be simulated with minimal loss of accuracy extremely quickly. This idea, and its application to ab initio molecular dynamics, is discussed in ref. [1]. This is huge because it means with a clever use of machine learning, ab initio MD may be possible on much larger time scales than have been done before.
As an interesting point, one of the keys is that MD simulations are often done because eventually some abnormal, but probably important configuration is sampled. This means that a so-called decision engine must be implemented so that the machine learning can switch off and start learning again from the unusual configuration. This is also described in ref. [1].
The second important development is in the construction of molecular force fields which are very commonly used in MD simulations. The purpose of these force fields is that they are much faster than solving the Schrodinger equation approximately. Force fields developed from machine learning potentially solve two problems which normal force fields face. First, normal force fields are only valid in a very specific context for which they were developed. Mainly, they are limited by the functional form upon which they were built. Machine learning also has the problem of needing a training set, but machine learning force fields are adaptive and thus able to become more robust upon visiting configurations not previously encountered. Second, machine learning force fields can be extended to new atoms and types of molecules without starting completely from scratch. After all, the same laws of physics apply to different elements, so the force field just needs to learn, approximately, how these same laws apply to different elements. Basically, it's faster than starting over from zero. Considerations such as these can be found in ref. [2].
I have also seen a paper at some point which discussed machine learning as a method for providing better guesses of the optimized geometry of a system, which would require fewer iterations using a QM method. Also, I think I have seen a paper about using machine learning for finding new minima on complicated potential energy surfaces such as those of large water clusters. This is related to the optimization point.
Basically, the possibilities are endless.

[1] Botu, V., & Ramprasad, R. (2015). Adaptive machine learning framework to accelerate ab initio molecular dynamics. International Journal of Quantum Chemistry, 115(16), 1074-1083.
[2] Botu, V., Batra, R., Chapman, J., & Ramprasad, R. (2016). Machine learning force fields: construction, validation, and outlook. The Journal of Physical Chemistry C, 121(1), 511-522.

If we don't use artificial intelligence in chemistry , then chemistry will be the only field amongst all , not to have used that .
But the truth is that lot of artificial intelligence has already been depolyed in chemistry , without us being known that it is being used.
Artificial inteligence can solve almost all the problems in the chemistry.
An interesting article for this can be found in wikipedia
https://en.wikipedia.org/wiki/Artificial_intelligence_in_chemistry
Many chemist think that artificial intelligence is tough , but in fact it is very easy. A chemistry person with basic computer knowledge can learn it in 6-7 days , and can deploy it in millions of ways in Chemistry.
An application of artificial intelligence has been developed by Wiley .It is a software called chemplanner .It uses is that it can synthesize any organic molecule in the world that a human can think of , in matter of seconds , using literature of already existing  reactions in the world.
So , with the help of this artificial intelligence software by wiley , you can synthesize any organic molecule in the world .

Very naive analysis. Think of AI as being a car. So yes if you are given a working car then you can learn to drive it in a few days. But if you have to build the car, then it is a much more difficult proposition. Also there is a nasty hidden feature in most AI. The AI makes a decision, but explaining that decision so that a human can understand why that decision was made is an even harder problem.


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Splitting across a double bond in ¹H NMR
Why do the Hs in (E)-1-bromo-2-chloroethene split into doublets instead of staying as a singlet. Wouldn't they be isolated from each other since the double bond does not rotate?

The hydrogens are not 'isolated' - they are still vicinal protons (3-bonds away) and so the J-coupling between them is strong. The strength of a splitting interaction is primarily determined by the number of bonds in between the hydrogens involved.
The angle between them also plays a role but to a lesser extent. The Karplus equation describes the three-bond J-coupling between two protons as a function of their dihedral angle.

The hydrogens are not identical, one H is connected to a carbon directly connected to a bromine and the other to a chlorine. Since chlorine is more electronegative, the H on the carbon with Cl will be more unshielded.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


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Why is chloridric acid used in salt bridges?
I've been planning an experiment on how to make galvanic cells and during my research I came across an article in which chloridric acid was used, along with a soluble salt and agar-agar,  in salt bridges.
Since then, I've been wondering why chloridric acid was added to agar-agar.
Thanks in advence.

Any ionic solution can function as a salt bridge. Having it acid based helps with the fast conduction paths of H+ in water.
The critical question is the leak rate. Any salt bridge will leak some amount of the ions into the solution. Good ones leak less, bad ones leak more.
So if you cannot tolerate any acidity in your solution, don't use an acid. Similarly, if you cannot tolerate any chloride in your solution, this would be a bad choice.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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Chromatography in geology
I just started reading about the technique of chromatography and learning how important it is.
Now I'm trying to focus on whether is possible to use this technique in the geological fields (fossils and dating for example.)
I'm just brainstorming so any thoughts will help here.

Welcome to chem.SE. This is a rather vague question. A fair bit of chromatography is used for various problems in geology. Perhaps you could be more specific in what you want to know?

Gas chromatography is commonly used in the earth sciences.

chromatography is routinely used to characterize oil and gas. However, since geology typically works with crystalline materials, XRD and various elemental analysis (say, atomic emission spectroscopy) are more common.

A typical example for the application of models and experimental methods of chromatography for geologic materials (e.g. soil, sediment) is the $K_\mathrm d$ model, which can be used for understanding the long-term behaviour of contaminants in the subsurface environment (e.g. the rooting zone of crops).
The parameter $K_\mathrm d$ is the partition coefficient, i.e. the ratio of the concentration of a contaminant sorbed on a specified solid phase (e.g. soil) to the concentration in a specified liquid phase (e.g. groundwater):
$$K_\mathrm d=\frac{\text{concentration in solid phase (per kg)}}{\text{concentration in liquid phase (per l)}}$$
The partition coefficient $K_\mathrm d$ can be used to calculate the retention factor $k$:
$$k=\frac\rho\varepsilon K_\mathrm d$$
where
$\rho$ is the dry bulk density of the solid phase, and
$\varepsilon$ is the total porosity of the solid phase.
Typical values for soil are $\rho=1.4\ \mathrm{kg\ l^{-1}}$ and $\varepsilon=0.3$.
This parameter is related to the retardation factor $R_\mathrm F$:
$$R_\mathrm F=\frac1{k+1}$$
The retardation factor $R_\mathrm F$ corresponds to the ratio of the distance travelled by the contaminant to the distance simultaneously travelled by the mobile liquid phase:
$$R_\mathrm F=\frac{v_\text{contaminant}}{v_\text{mobile phase}}$$
For example, if the annual precipitation exceeds the evaporation by $300\ \mathrm{mm}$ and the porosity of the soil is about $\varepsilon=0.3$, the drainage water travels with an average velocity of $v_\text{mobile phase}=100\ \mathrm{cm\ a^{-1}}$ downward through the soil.
Assuming a mean value for the partition coefficient of plutonium in all soils of
$$K_\mathrm d=7.4\times10^2\ \mathrm{l\ kg^{-1}}$$
the corresponding retention factor is
$$\begin{align}
k&=\frac\rho\varepsilon K_\mathrm d\\[6pt]
&=\frac{1.4\ \mathrm{kg\ l^{-1}}}{0.3}7.4\times10^2\ \mathrm{l\ kg^{-1}}\\[6pt]
&=3.5\times10^3
\end{align}$$
and the retardation factor is
$$\begin{align}
R_\mathrm F&=\frac1{k+1}\\[6pt]
&=\frac1{\frac\rho\varepsilon K_\mathrm d+1}\\[6pt]
&=\frac1{\frac{1.4\ \mathrm{kg\ l^{-1}}}{0.3}7.4\times10^2\ \mathrm{l\ kg^{-1}}+1}\\[6pt]
&=2.9\times10^{-4}
\end{align}$$
Therefore, the migration rate is
$$\begin{align}
v_\text{contaminant}&=R_\mathrm F\cdot v_\text{mobile phase}\\[6pt]
&=2.9\times10^{-4}\times100\ \mathrm{cm\ a^{-1}}\\[6pt]
&=0.029\ \mathrm{cm\ a^{-1}}
\end{align}$$
i.e. a downward migration of $1\ \mathrm{cm}$ takes about 35 years.


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Interpreting X-ray diffraction data
So I've got my data for some x-ray crystals of small molecules of Ag, Cu,... I was curios to know what the best way to determine what type of lattice each of these atoms may have?

You'll have to be more specific.  What kind of diffraction do you have?  Is this a homework problem?

Specifically, is this data from an area detector where the crystal is rotated about one axis only, is it from a simple detector (no spatial resolution) attached to a 3 or 4 circle goniostat, or is it film data from a precession camera?

Your question has no sense – sorry.
You talk about a few small molecules (with coinage metals) but ask about lattice type for atoms. Atoms in molecular crystal have no their own lattice. Lattice describes how your molecule are aligned in 3D-space.
Assuming that you have a single crystal data then in the beginning of experiment your X-minion (that's the person that took X-ray data for you) has already found the cell parameters and then usually also Bravais cell. To be sure that selected lattice was right whole data set need to be collected, the data have to be integrated and then using Lattice exceptions (see International Tables for Crystalography Volume A,  2.2.13. Reflection conditions) you can tell what lattice type you have. If you do the same for glide planes and screw axis exceptions then eventually you will find space group.

X-minion... That is the best term ever


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Dissolving/denaturing bitterant residue
In what seemed like a good idea at the time, I used a can of compressed air to clean out my Magic Mill flour grinder.
Picture of one: http://ic.pics.livejournal.com/ccdesan/4606041/68177/68177_320.jpg
It did a great job of cleaning out all the extra flour stuck to the bottom of the top part. However, what I didn't realize at the time was it also left bitterant residue, ruining a batch of flour I just made!
The MSDS for the compressed air (Compucessory Power Duster) just lists diflouroethane, but it says right on the can it contains a bitterant!
The bitterant is likely Denatonium (http://en.wikipedia.org/wiki/Bitterant). Is there any household chemical that will dissolve it and the other common bitterants, or better yet, break them down into non-bitter compounds?

Since probably only ppm levels of denatonium benzoate are in the compressed air, there are ppm levels or lower of denatonium in your flour grinder.  At that concentration, water should dissolve denatonium benzoate.  Is that a viable way to clean your grinder?
According to this solubility table, the best solvent for denatonium benzoate is methanol.  However I would not recommend using methanol on anything you are using for food.  Ethanol is only slightly less good.  Maybe you can wash your grinder in some cheap vodka a few times?
Here is a random thread from Google Groups where someone claims to have identified a vendor for non-bitterized air duster cans.

Thanks. Well, I just tried wiping it several times with everclear, and it did help, but there's still a noticeable bitter flavor on the plastic. Is there a certain soap that would help, or another chemical that would actually break down the Denatonium?

On some time scale and at some temperature, I'm sure bleach or hydrogen peroxide would oxidize/degrade denatonium.  However I'm not sure how long & hot it would need to be.  If you are comfortable wiping your grinder with bleach or hydrogen peroxide, then you could try that.

After a bit of trial and error, this formula seems to work:
1/2 tablespoon bleach
7.75 oz mouthwash (enough to make a cup)


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Why is acetone and diethyl ether miscible, but not water and diethyl ether?
Acetone and Water are miscible, and water and diethyl ether are not. So why are acetone and diethyl ether miscible if water and acetone seem to have similar polarities?

Acetone's dipole moment is 2.91D while that of ether is only 1.15D.  Water is a very polar substance, so acetone will dissolve in it while ether will not.
Two substances are miscible when their intermolecular forces (IMFs) are similar enough such that the forces of attraction between molecules of different substances are similar in strength to the forces of attraction between molecules of the same substance.  If we look at the miscibility of A and B as an example, if A attracts B about as strongly as A attracts A and B attracts B the substances should be miscible.  If A and A have much stronger attraction than A and B the substances are less likely to be miscible.
In your example it's not that some molecules of ether will not dissolves in water and vice versa, it's just that if the two liquids are in contact the water will stay with water and the ether with ether for the most part.  This is because water's strongest IMF is hydrogen bonding while ether's is dispersion with some dipole-dipole forces.


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Fenton's reagent application to leachate
What is the experimental procedure for applying fenton's reagent to leachate? What doses of hydrogen peroxide and ferrous sulphate required? How much contact time needs to be provided?

Welcome to chemistry.se? Lern more about this site in the [help] and taking the [tour]. I am not very certain if this paper contains the information you are looking for, but it might get you started.

@Martin maybe the OP doesn't have access to the Researchgate. I myself couldn't make an account because I don't have an institution or anything.


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About purifying ethanol - What type of ethanol?
I'm reading this book, trying to learn some basics of chemistry. One of the experiments is about purifying ethanol.
Now, I have some difficulties in deciding what kind of ethanol solution I should try to purify. The drugstores appear to typically sell denatured ethanol (so it could contain e.g. methanol?) or solutions with ethanol and isopropyl alcohol.
Could potable alcohol be used? I think there's glucose in those.
PS. Doing this in my country is legal. It seems that is not the case in all parts of the world (or has not been).

Forget about the denatured alcohol! Additives are typically chosen to render their removal difficult.
Supposed you have decent setup for distillation, analytics still might be a problem.
How will you determine whether your destillation was successful or whether it just was "thermal transfer" from one flask to another?
A hydrometer to measure the density of the liquid could be helpful.
Use some cheap vodka, mix with water and try to distill the vodka off. Measure the density before and after distillation. Note the temperatures, read about strange things that can happen: azeotropes. Cheers!
If you want it more simple and obvious, mix the vodka with some food colourant. Proceed as before and watch the effect. Cheers again.
Note that one usually would not digest anything that went through any lab equipment  for mixed used at a normal lab. But this is your home lab and you know what went through which labware. Anyway, keep your equipment clean and take care. Btw: Cheers!

Thanks for the input! I don't plan to digest anything :)


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Atomic mass ratios - Is the problem providing enough info?

One gram of Hydrogen reacts with exactly (almost) 8 grams of oxygen to produce $\ce{H2O}$. Another single gram of Hydrogen reacts with 16 grams of Oxygen to produce $\ce{H2O2}$.

Can the ratio $\frac{m_O}{m_H}$ be determined from this information? If not, what else do chemists need to know to find this ratio?

This appears to be a homework question, with no effort shown on how you would approach it.  What is the concept that you are finding difficult?

It is not actually a homework question. My question is how these mass ratios in general are deduced from reactions. I couldn't derive this ratio from these particular reactions.

If you know that 1 gram of Hydrogen (well, H2) reacts with 8 grams of Oxygen (fortunately O2) to produce H2O, and you know it is H2O, that is kind of all the info you need to get the mass ratio of H:O.  The tricky part is, without knowing that you start with H2 and O2, and end with H2O, how would you get there...

yes exactly assumption is that you don't know. It seems to me the information is not enough and the first derivation is due to Avogadro.

Well, it is kind of like the classic high school chemistry experiment mixing various salts to see what precipitates - looking at enough reactant and product combinations eventually gives you a pattern that gets you to integer ratios. But, yes, a key is being able to figure out moles of reactants and products.

Are you asking whether we can figure out the ratio of the mass of an oxygen atom to a hydrogen atom from this information? If so, then the answer is yes, assuming that you know that $\ce{H2O}$ and $\ce{H2O2}$ are the formulas for each compound, and that atoms exist and are indivisible in chemical reactions.
$$2n\space \rm{atoms \space H} = 1 \space g \space H$$
$$1n \space \rm{atoms \space O} = 8 \space g \space O$$
Divide the second equation by the first:
$$\frac{1n \space \rm{atoms \space O}}{2n\space \rm{atoms \space H}} = \frac{8 \space g \space O}{1 \space g \space H}$$
Cross multiply to get a 16:1 mass ratio. You can confirm this by looking at $\ce{H2O2}$, where the ratio of atoms is 1:1 and mass is again 16:1.
If you didn't know the formulae, then you could assume that the ratio was some integer multiple of 8:1, but without more data you wouldn't know which. This is because you would not know whether water had a 1:1 atom (or mole) ratio of $\ce{H}$ to $\ce{O}$, or if it was something else.
Incidentally, what you are describing is the law of multiple proportions, and it is one of the key findings that led to the development of atomic theory.

Incidentally, John Dalton (developer of the atomic theory) got all of his atomic masses and many of his formulas wrong because he assumed that water has hydrogen and oxygen in a 1:1 ratio of atoms (per his book A New System of Chemical Philosophy). You can't really fault him though, since the very existence of atoms was, at the time, both revolutionary and controversial.


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Why does 1 cubic meter = 1000000 cubic centimeters?
I have the SI chart in front of me. It states that a centi is 10^-2 and since they are already designated cubic meters why does 1 cubic meter equal 1000000 cubic centimeters?

Take a look at the Wikipedia article for SI prefixes, in particular, the following excerpt:

When units occur in exponentiation, for example, in square and cubic forms, the multiplication prefix must be considered part of the unit, and thus included in the exponentiation.

$1\ km^2$ means one square kilometre, or the area of a square of $1000\ m$ by $1000\ m$ and not 1000 square metres.

$2\ Mm^3$ means two cubic megametres, or the volume of two cubes of $1000000\ m$ by $1000000\ m$ by $1000000\ m$ or $2\times10^{18}\ m^3$, and not 2000000 cubic metres ($2\times10^{6}\ m^3$).

Thus, the relationship between the cubic metre and the cubic centimetre is as follows:
$$1\ m^3 = (1\ m)^3 = (100\ cm)^3 = 100\ cm \times 100\ cm \times 100\ cm =  1000000\ cm^3 = 10^6\ cm^3$$
$$1\ cm^3 = (1\ cm)^3 = (0.01\ m)^3 =  0.01\ m \times 0.01\ m \times 0.01\ m = 0.000001\ m^3 = 10^{-6}\ m^3$$
For a real case, converting $243.7\ cm^3$ to $m^3$ step by step:
$$243.7\ cm^3 = 243.7\times 1\ cm^3 = 243.7\times (1\ cm)^3 =243.7\times (0.01\ m)^3 = 243.7\times 0.01\ m \times 0.01\ m \times 0.01\ m = 243.7\times 0.000001\ m^3 = 243.7\times 10^{-6}\ m^3 = 2.437 \times 10^{-4}\ m^3$$
This is simply the way it was defined, and it is the standard everyone adheres to. More examples of SI prefix usage can be found on this page from one of the organizations which created the SI units, the BIPM.

In short, by $cm^3$ we mean $(cm)^3$, not $c(m^3)$

If I were you, I would have only written: In short... ;)

$1\ \texttt{m}^3$ is equal to $(\texttt{100 cm})^{3}$ because $1\ \texttt{m}$ is equal to $100 \ \texttt{cm} = 10^2\ \texttt{cm}$.

Forgive me, a few hours ago I read your answer improperly and mistakenly downvoted you, even though it is correct. The site doesn't allow me take it back after two hours unless you edit your answer somehow.

Though not explained, it isn't incorrect. I don't vote correct answers down. Now. That'll wash it a bit away.

1 cm = $10^{-2}$ meter
i.e 1 m = $10^2$ cm
So 1 $m^3$ = $(10^2)^3 cm^3$
i.e 1 cubic meter = $10^6$ cubic centimeter

You can use MathJax to better format your posts.


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Why do molecules have to have a change in dipole moment in order for them to be IR active?
My chemistry textbook keeps saying that in order for a molecule to be able to absorb infrared radiation, it has to have a change in dipole moment when the bond vibrates.
I don't understand why that is. Some clarification would be appreciated.

The actual quantum mechanical justification for this is the "Fermi Golden rule", and it can be used to justify selection rules across spectroscopy.
I can post a derivation/explanation here if you like, but it is very involved (by chemistry standards) so you might not find it useful unless you've got some background in quantum.

Thank you for the information, please do post an explanation!

Seconded, do please post!

As a molecule vibrates, if there is a fluctuation in its dipole moment, then this induces an electric field that interacts with the electric field associated with the infra red radiation. If there is a match in frequency of the radiation and the natural vibration of the molecule, absorption occurs.
For more details, please see this page on the UC Davis ChemWiki website.

There is an intuitive reason why a vibration is IR-active only if it involves a change in the dipole moment.
Recall that the typical wavelength of IR radiation ($\sim 10\mu\mathrm{m}$) is much larger than the typical size of a molecule ($\sim 1\mathrm{nm}$). Hence, to a very good approximation, the (time-varying) electric field of the IR radiation is spatially uniform within a molecule.
Now observe that under a spatially uniform electric field, all positive charges, regardless of their positions, are pushed to a common direction, and all negative charges are pushed to exactly the opposite direction. In this case, can the change in the total dipole moment, defined as $\Delta \vec{\mu} = \sum_{i} q_{i} \Delta \vec{r}_{i}$, be equal to zero?
Clearly, the answer is "No." The displacements $\{\Delta \vec{r}_{i}\}$ are parallel between like charges and antiparallel between opposite charges. Therefore, $\{q_{i}\Delta\vec{r}_{i}\}$ are all parallel to one another, and $\vec{\Delta\mu}$ cannot be equal to zero.
So far, I have argued that a motion induced by a spatially uniform electric field always involves a change in the dipole moment. A corollary is that any motion that does not involve a change in the dipole moment cannot be induced by a spatially uniform electric field (of IR radiation).
As a simple example, imagine applying a spatially uniform electric field to a $\ce{CO_{2}}$ molecule such that the direction of the field is parallel to the length of the molecule. Can this field ever induce a symmetric stretch, in which the two oxygen atoms move in opposite directions? Certainly no. Both oxygen atoms are negatively charged by exactly the same amount. Under a spatially uniform electric field, they are pushed to the same direction by the same magnitude, and there is no room for a symmetric stretch.

Good argument, and it is true even for shorter wavelengths.

@Greg Thanks. Indeed, that the electric field is spatially uniform within a molecule (which is equivalent to the dipole approximation) is more or less true for visible light and probably even for UV.


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Why does decreasing the pressure of the system increase the relative volatility of a binary solution?
In our teaching lab, we were posed with the following question as an exersize:

If the boiling points of two compounds differ by $\pu{50 ^\circ C}$ at atmospheric pressure, what will be the effect on the relationship by decreasing the pressure to $\pu{1 mm}$ (no number, just a general process)?

I did not find much information on the subject in our lab manual or book but, but based off of some reading I've done around the internet (see below), it seems that for most binary solutions, decreasing the pressure increases the relative volatility.  However, asking some of my peers, they seems to believe that while both boiling points would decrease, the difference in their new boiling points would remain about the same (i.e. $\pu{\approx 50 ^\circ C}$), To make matters slightly more confusing, using a pressure-temperature nomograph, picking arbitrary numbers, I found that the relative volatility decreased (see Figure 1).

Who is more correct?  Does decreasing the system pressure increase the relative volatility or does it stay the same?
If the relative volatility increases when system pressure is decreased, why is this the case?  I would actually guess the opposite, because as you decrease the pressure towards $0$ (a vacuum), it would seem that all boiling points would approach $\pu{0 K}$, thus be getting closer together.
If the relative volatility increases when system pressure is decreased, does that mean it is always more efficient to perform fractional distillation under a partial vacuum?

Figure 1  Two nomographs measured via Pressure-Temperature Nomograph Interactive Tool by Sigma-Aldrich decreasing the system pressure to $\pu{1 mm}$ for two arbitrary substances with $\pu{50 ^\circ C}$ boiling point disparity at $\pu{1 atm}$.
References:

Effect of temperature or pressure on relative volatility chemengineering.wikispaces.com via the Internet Archive (under "Effect of temperature or pressure on relative volatility")
Industrial-scale applications of vacuum distillation at Wikipedia

Let's assume that the vapor pressure of both components can be modeled with the Antoine equation.  This gives us a convenient way to address the question algebraically.  The vapor pressure of most chemicals is excellently modeled by the correct Antoine equation for that chemical.  The equation is:
$$\log_{10}{p} = A-\frac{B}{C+T}$$
where the the parameters $A$, $B$, and $C$ are compound specific parameters.  The equation can be solved for temperature explicitly yielding $T = \frac{B}{A-\log_{10}\, p} - C$.
So at $\pu{1 atm}$, the equation for pure species $i$ and also a different pure species $j$ that is higher boiling by $\pu{50 ^\circ C}$ is:
\begin{align}
T_1 &= \frac{B_i}{A_i-\log_{10}\, p_{\pu{1 atm}}} - C_i\\
T_1 &= \frac{B_j}{A_j-\log_{10}\, p_{\pu{1 atm}}} - C_j - \pu{50 ^\circ C}
\end{align}
where $T_1$ is the boiling point of $i$ at $\pu{1 atm}$.  To make the math easier, let's suppose the equation is in units of atmospheres, in which case $\log_{10}\, p_{\pu{1 atm}} = 0$.
\begin{align}
T_1 &= \frac{B_i}{A_i} - C_i\\
T_1 &= \frac{B_j}{A_j} - C_j - \pu{50 ^\circ C}
\end{align}
The question is, what happens when $p$ changes, in this case, when it is lowered by ~760-fold, let's call that 1000-fold to keep the math easy.
\begin{align}
T_{0,i} &= \frac{B_i}{A_i-\log_{10}0.001} - C_i = \frac{B_i}{A_i+3}-C_i\\
T_{0,j} &= \frac{B_j}{A_j-\log_{10}0.001} - C_j = \frac{B_j}{A_j+3}-C_j
\end{align}
Now we have four equations but six unknown parameters, $A_i$, $A_j$, $B_i$, $B_j$, $C_i$, and $C_j$.  So in general, there is no generally valid constraint for the temperature difference $T_{0,i}-T_{0,j}$:  It could be higher, lower, or about the same as the difference at room temperature.
However, we could make some further assumptions.  A good one here could be that both species in question have a constant heat of vaporization, which means that $C_i=C_j=0$.  If that assumption is true, then the four equations become:
\begin{align}
T_1 &= \frac{B_i}{A_i}\\
T_1 &= \frac{B_j}{A_j} - \pu{50 ^\circ C}\\
T_{0,i} &= \frac{B_i}{A_i+3}
 =\frac{\frac{B_i}{A_i}}{1+\frac{3}{A_i}}
 =\frac{T_1}{1+\frac{3}{A_i}}\\
T_{0,j} &= \frac{B_j}{A_j+3}
 =\frac{T_1+50 °C}{1+\frac{3}{A_j}}
\end{align}
Now, the temperature difference of interest $T_{0,j}-T_{0,i}$ is
$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{1+\frac{3}{A_j}} - \frac{T_1}{1+\frac{3}{A_i}}$$
To keep the formulas easy let me define new parameters $a_i$ and $a_j$ such that $a_x=1+\frac{3}{A_x}$.  Then the temperature difference is
$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{a_j} - \frac{T_1}{a_i}= \frac{a_i(T_1+\pu{50 ^\circ C})-a_j T_1}{a_i a_j}=\frac{(a_i-a_j)T_1 + a_i(\pu{50 ^\circ C}) }{a_i a_j}$$
Now we must introduce further assumptions about the Antoine $A$ parameters.  They are positive and thus the $a$ parameters must also be positive, and further positive $A$ implies  $a>1$.  If we further assume $a_j=a_i$, then
$$T_{0,j}-T_{0,i} =\frac{(\pu{50 ^\circ C}) }{a_j}$$
Since we know already that $a_j>1$, then the temperature difference (i.e. the difference in boiling points) at the lower pressure is less than the temperature difference of $\pu{50 ^\circ C}$ at \pu{1 atm}.  Now, even if $a_i$ is not precisely equal to $a_j$, then the temperature difference will still be less than $\pu{50 ^\circ C}$ as long as the $a_i$ and $a_j$ parameters are similar enough to keep $(a_i-a_j)T_1 << a_i(\pu{50 ^\circ C})$.  It would take a very unusual choice of chemicals $i$ and $j$ for that to happen.  I don't think I could identify such a pair.
So, in summary, to get to the result that I think your instructor was asking for, we required many assumptions:

Both chemicals obey the Antoine equation in the temperature range of interest (probably an OK assumption).
Both chemicals have approximately constant enthalpy of vaporization in the T range of interest so that we can neglect the Antoine $C$ parameters.
The Antoine $A$ parameters for both chemicals must be positive.  (A good assumption but it still must be made.)
The chemicals have Antoine $A$ parameters (and thus also $a$ parameters as I defined them here) that are not too different from each other.

It might be possible to weaken some of the assumptions and still find that the lower-pressure boiling point difference must be lower than $\pu{50 ^\circ C}$, but I'm confident that the result is not necessarily generally true.  It isn't a thermodynamic law or anything.  It definitely depends on several key assumptions about how vapor pressure "usually" or "typically" behaves for most chemicals.

Thank you so much for this well explained answer!  Could you postulate why in the references I provided above that they say "decreased pressure increases relative volatility?"  It seems that most of the literature that says this is aimed at an engineering audience.  Is this a theory vs. practice scenario?

The difference in boiling point temperatures is only part of what people mean by "relative volatility".  What is important practically is the concentration ratio of the compounds in the vapor phase.  Consider 8 variables: {P, T, yi, yj, xi, xj, pi, and pj} where xi and xj are liquid mole fractions, yi and yj are vapor mole fractions, pi and pj are vapor pressures, T is temp., and P is absolute pressure.  There are 7 equations: P = yipi + yjpj, 1=yi+yj, 1=xi+xj, and (for ideal solutions) yipi = xi, yjpj=xj.  There are also two Antoine equations for pi and pj.

With 8 variables and 7 equations, there is one degree of freedom.  If you solve the equations the right way we can let the pressure P be that degree of freedom, i.e. the independent variable.  Then you can solve for the ratio yi/yj as a function only of P.  If you do all that math you should see that for reasonable choices of Antoine parameters, yi/yj is an decreasing function of P, so that as P goes down, the ratio of concentrations in the vapor phase goes up.

OK, the general idea I had in the above comments is OK but I've botched the equations quite badly.  I'll write up a separate answer.

My first answer focused on the boiling point difference.  But that's not what's practically important when separating compounds.  The concentration ratio in the vapor phase of the two compounds to be separated is what is important.
Let's define some variables.  The total overall pressure is $P$ and the temperature is $T$.  The partial pressures of components $i$ and $j$ are $p_i$ and $p_j$.  The mole fractions of $i$ and $j$ in the liquid phase are $x_i$ and $x_j$ and in the vapor phase are $y_i$ and $y_j$.  Lastly, the vapor pressure of pure $i$ is $p_i^*(T)$ and of pure $j$ is $p_j^*(T)$.  Those vapor pressures are functions of temperature, we could choose the Antoine equation for example.  I won't write the $(T)$ but keep in mind vapor pressures of pure species are functions of temperature (only).
For simplicity let's focus on ideal solutions, which mean's the simplest form of Raoult's law applies:
\begin{align}
\tag1 p_i&=p_i^*x_i\\
\tag2 p_j&=p_j^*x_j
\end{align}
The sum of mole fractions must be $1$ in both vapor and liquid:
\begin{align}
\tag3 1&=x_i+x_j\\
\tag4 1&=y_i+y_j
\end{align}
There is also the definition of partial pressure:
\begin{align}
\tag5 p_i &= y_i P\\
\tag6 p_j &= y_j P
\end{align}
And there are vapor pressure relationships, e.g. Antoine equations for each of $i$ and $j$:
\begin{align}
\tag7 p_i^* &= p_i^*(T) = \text{some equation with parameters and T only}\\
\tag8 p_j^* &= p_j^*(T) = \text{some equation with diff. parameters and T only}
\end{align}
That is eight independent equations but there are ten variables.  So there are two degrees of freedom.  Let's assume that we have an equimolar liquid, i.e. that $x_i=0.5$.  From equation 3 that means $x_j=0.5$ too.  Now there is only one degree of freedom left.  The general question is what pressure $P$ maximizes the relative volatility $\frac{y_i}{y_j}$.  The math is tough because you have to substitute one Antoine equation into the other to eliminate temperature.  I assumed $C_i=C_j=0$ again for Antoine parameters, and got that $p_j^*=K (p_i^*)^{\frac{B_j}{B_i}}$, where K is a proportionality constant that depends in a complex way on Antoine $A_i$, $A_j$, $B_i$, and $B_j$ parameters.  The other non-Antoine equations reduce to:
\begin{align}
y_i P &= p_i^*/2\\
(1-y_i)P &= p_j^*/2
\end{align}
Subbing the Antoine result in for $p_j^*$ and then subbing the first equation into the result gives
$$2^{d-1}P^{d-1} = \frac{K y_i^d}{1-y_i} $$
That expression doesn't have an analytical solution for $y_i$ as a function of $P$ but in theory you could find the solution numerically if you knew all the Antoine parameters.  I graphed the equation for a few different values of $d$ and y_i was always increasing as P got lower and lower.
So that's a long-winded way of saying that "relative volatility" is different than the difference in boiling point temperatures.

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text:
A good book for learning paper and thin-layer chromatography
What book would you recommend for learning paper and thin-layer chromatography? I don't have access to fancy equipment (just to these two basic techniques that are relatively cheap), so it would be nice if the book focused on them in addition to being easy to understand etc. It does not matter if advanced techniques are discussed in a chapter or two but if the book is solely about them, it would not offer me much.

I would be interested in this as well, in the form of a complete guide. Maybe, an introductory book on biochemistry techniques could contain what you are looking for.

I think an entire book on TLC/paper chromatography would be quite overkill and your best bet is on a more general practical organic chemistry book. I can recommend The Organic Chem Lab Survival Manual: A Student's Guide to Techniques by Zubrick if you are only interested in technique or need a introductory text. Else Vogel is of course the book for practical organic chemistry. TLC is covered in detail.
I'm not so sure about paper chromatography, I don't believe it is often used in the lab (at least I have never worked with it). TLC is much more common.


text:
Open data in chemistry?
I am a student in computer science, which is a field where sharing research data and source code is very very much recommended. I am looking for sites where data from chemistry experiments could be openly distributed. Do such exist? I would like to test some algorithms for analyzing certain kinds of experiments, without having to do them myself (which I easily can't). The motivation is that I think (.. or maybe hope) that I would have some insights to offer.
I did not find many suitable tags, sorry.

There's all kinds of different chemistry data. What kind of experiments are you looking to analyze? There are many efforts to provide open access data, but it is in disparate locations based on discipline.

For a given experiment, one ideally would share all records. In some fields, it is not possible before anonymization (e.g. neuroscience; removing the subject's face from magnetic resonance images) but all essential data is still given. The ideal site I'm looking for would not be dedicated to just one class of experiments. However, if no such exist, I'm happy to hear about the narrow-scope ones too.

There are a variety of "open data" sources in chemistry. In your case, you probably want "open notebook science.".
There are a few practitioners in chemistry, although Jean-Claude Bradley sadly passed this last year.

JC Bradley
Andrew Lang
Open Source Malaria

I believe there will be a collection of articles published in the open access Journal of Cheminformatics as a memorial to Jean-Claude, but I don't think these have cleared review yet.

The types of databases you are thinking of probably don't exist for most types of chemistry work. The reason is that interpreting data typically isn't the most difficult and expensive part, obtaining the data is. There are exceptions:

Bioinformatics: http://www.ebi.ac.uk/services
Protein structure: http://www.rcsb.org/pdb/home/home.do
Basic properties of small molecules: http://cccbdb.nist.gov/
Material properties: http://www.matweb.com/search/search.aspx

These are just example sites, there are many others out there. What they all have in common is that they touch on areas where theory is not very well developed, but we have the ability to generate large amounts of data at relatively low cost. I'm sure there are other areas like this as well, but you aren't likely to find things like tables of reactants and product concentrations for reactions. You mention "certain kinds of experiments" - if you can be more specific, someone can probably point you in the right direction.
If not, check out the NIST website - it lists a lot of free online scientific databases. Not all are chemistry, but maybe you will find what you are looking for there.

I know it is a bit necromancy to post this so late after the question, however you might have a look at the OpenPHACTS system, that brings a lot of medchem data together.
https://www.openphacts.org/


text:
How to convert sodium aluminate back to sodium hydroxide?
I had dissolved a bunch of aluminium foil in an aqeous solution of sodium hydroxide. The resulting solution consists mainly of sodium aluminate with very little excess sodium hydroxide.
What I want to do is convert this back to sodium hydroxide. I am thinking electrolysis should do the trick. But, I am not completely sure if this is right. If it is, what kind of electrodes should I use? If not, why not? (What will be the product?)
Thank you for your time.

I don't think you will have much luck recovering the hydroxide.  To do so you would need to reverse the reaction that happened when the aluminum dissolved, which was
$$\ce{2 Al + 2 NaOH + 2 H2O -> 2 NaAlO2 + 3 H2}$$
Your idea is essentially to substitute reduction at an electrode for reduction with hydrogen gas.  Unfortunately electrochemical reduction of $\ce{Al(III)}$ is very difficult.  If your idea was feasible, it would revolutionize aluminum manufacture!  Instead of doing electrolysis of molten cryolite at ~1000 degress with sacraficial carbon electrodes, aluminum manufacturers could use your process.  So if you could get it to work, you'd be rich...but I am not hopeful.
Note however that you don't necessarily need to do electrolysis to recover sodium hydroxide.    There is hope!  You could just boil down your solution so that it is supersatured with respect to aluminum hydroxide, and then wait for the aluminum hydroxide to precipitate, which would happen faster if you add seed crystals of aluminum hydroxide $\ce{Al(OH)3}$) to initiate precipitation.
$$\ce{2H2O + NaAlO2_{(aq)} -> Al(OH)3_{(s)} + NaOH_{(aq)}}$$
This step is part of the Bayer process for commercial refining of aluminum ore.

I guess You can do electrolysis of sodium alluminate.
The result is Al and Na ions.
But the aluminum formed, converts to aluminum hydroxide and aluminum oxide very fast.
And some of the aluminum compounds can react with NaOH again. And this can creat a loop and waist energy. But for your application, I think you can do a trick to minimize waist.
Add sodium aluminate solution to a bigger container and put the cathode there. Then put a smaller pot in the bigger one, or make some small holes on a plastic bottle (not near bottom not near solution's surface, in middle), this is where the anode goes in.
This way it can work but Bayer process is easier.
Also NaOH formed can reacts with CO2 in air (if concentration of NaOH is lower than 50 percent), and this also waists some energy.

This comment needs more credit. A clay pot seperator between anode chamber and cathode chamber  would help minimize the mixing.

Electrolysis is good, and it is also helpful if you want aluminum powder too, the electrolysis of Sodium Aluminate will yield Sodium Hydroxide in solution and aluminum metal as sponge which crumbles into a powder when dry, this is not efficient though because the Hydroxide will react with a little bit of the aluminum that had formed, so you will have a solution of Sodium Hydroxide with a little bit of Sodium Aluminate in it, this method also works with sodium zincate as well, NurdRage has a video on that topic, the Sodium Aluminate process is very similar to the Sodium zincate process he uses in his video.
Link to his video:
https://youtu.be/3X9c6epL7HQ

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text:
Is there a relationship between bond strength and polarity?
I believe that a more polar bond would be stronger, since it would be similar to an ionic bond.
However, this seems to be wrong when applied to the study of acidity of some compounds.

What compounds? And, you can't say that for example ionic bond is stronger than covalent bonds, since no specific molecule is under study.

In terms of acidity, ionic(polar) bonds dissociate more easily than covalent(polar) because the electrons are shared less equally in the ionic bond leading to easier breaking apart of the molecule. An acid for with an ionic bond will dissociate more easily leading to a stronger pH than an acid with a covalent bond.


text:
Methylene blue used as funguscide in cichlid reproduction - concentration measurement?
I have an aqueous solution of methylene blue. How do I determine the concentration of this solution, so I can duplicate it?

Do you have access to a UV-VIS spectrometer?

As @KlausWarzecha said, what kind of equipment/instruments do you have access to?

The concentration of an aqueous solution of methylene blue can be determined using a UV-VIS spectrometer.
Measure the absorption at $\lambda = 664\, \mathrm{nm}$, use the molar extiction coefficient of $95000\, \mathrm{dm^{3}\cdot mol^{-1}\cdot cm^{-1}}$
(Reference) and calculate the concentration using the Lambert-Beer equation.

If in addition to a UV-VIS spectrometer you also have access to a fluorimeter, I would recommend using fluorescence instead of absorbance.  It will be more sensitive and less prone to interference.  However, fluorimeters are more expensive and harder to come by than UV-VIS spectrometers which measure absorbance only.
When I measured methylene blue by fluorescence, I used 630 nm for excitation and 700 nm for emission.  I found fluorescence to be much better than absorbance at 670 nm for the same samples, but the absorbance measurement also worked.  The fluorescence properties of methylene blue are less well-known than the absorbance.  Other papers often use different values for excitation/emission than what I used, but if you have a fluorimeter you can determine what's best for you.


text:
How do I utilize Henry's Law Constant?
I have tried for several hours to answer this question with no luck. I believe I may be missing a couple of steps or I am not utilizing the given information properly. Nonetheless, some guidance will be greatly appreciated.
The concentration of $\ce{N2}$ in the ocean at 25C is 445 $\ce{\mu}$M. The Henry's Law constant for $\ce{N2}$ is 0.61 x $\ce 10^{-3}$ mol $\ce L^{-1}$ atm $\ce L^{-1}$.
Part A:
Calculate the mass of $\ce{N2}$ in a liter of ocean water.
4.45 x $\ce 10^{-4}$ mol/kg = molality
$\ce{N2}$ in moles = 28.014 g/mol
Using Molality Equation
Molality = $\frac{amount of solute (mols)}{mass of solvent(kg)}$
I want kilograms, thus:
4.45 x $\ce 10^{-4}$ $\frac{mol}{kg}$ = $\frac{28.014 g}{mol}$
Where do I go from here?
Part B:
Calculate the partial pressure of $\ce{N2}$ in the atmosphere
Use C = K x $P_(gas)$
4.45 x $\ce 10^{-4}$ = (0.61 x $\ce 10^{-3}$ mol $\ce L^{-1}$ atm $\ce L^{-1}$)x
x = 0.73 atm

No one can help me with this question?

I formatted your question using MathJax http://meta.chemistry.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here/88#88
This makes your question computer readable and typing out your whole is generally the way to go, so it can potentially help others (and in increases the chance of a response.) If you hit edit on your question, you can see some of the functions I used to format it. I kept your image in case I made an error. Best of luck finding the answer to your question, I'm studying something fairly similar right now.

Part A: Why do you want to use the molality?! You have the concentration i.e the number of moles in one liter. So the mass of nitrogen in one liter is simply:
Number of moles in one liter $\times$ molar mass.
Part B: It's correct.

Thnaks. I do not understand the definition well enough.


text:
What property of zinc chromate makes it a good anticorrosion paint
What properties of zinc chromate make it a good anticorrosion paint?  I know that stainless steel is protected by a layer of chromium oxide.  Is it because the zinc chromate doesn't flake off the metal it is protecting?

Long story short, if it's scratched in the presence of water(humidity) the chromium ion inside the chromate will get reduced to Cr 3+ from Cr 6+. This 3+ ion will then form a Chromium hydroxide compound,which is insoluble in water and corrosion resistant, inside the scratch. This reduction of the chromium ion first is what inhibits the oxidation of the underlying metal.
Here's the full article http://www.electrochem.org/dl/interface/wtr/wtr01/IF12-01-Pages34-38.pdf
The answer: Being water resistant in the first place and being capable of producing ions that can be reduced forming another corrosion resistant compound( chromium(III)oxy-hydroxide.)


text:
What Can Dissolve Polypyrrole?
I am looking for a solvent for polypyrrole. I read M-cresol and chloroform can be used, but I am not entirely sure.

Unmodified or (partly) protonated and with counterions, such as arylalkyl sulfonates or polystyrene sulfonate?

Without reacting with polypyrrole it is not possible to dissolve it. Whoever is demanding in paper that they have succeeded in dissolving the pure polypyrrole, without transforming to its derivative, its wrong.

Polypyrrole is quite a rigid polymer (one of the reasons it's a conductive polymer) and it's generally not considered particularly soluble without being derivatized. It seems to be possible to dissolve it somewhat in polar organic solvents if it's been polymerized with a large anion like dodecylbenzenesulfonate, basically reducing the interactions of the chains of polymer, but it loses some of its conductivity when cast into a film. One group has apparently managed to make it water soluble by polymerizing oxidatively with horseradish peroxidase in the presence of a sulfonated polystyrene counterion, which keeps the chain growing linearly along the length of the polystyrene, apparently allowing it to be soluble.


text:
How to quantify amount of pigments in plant leaves?
Using thin-layer chromatography, it is possible to see what pigments plant leaves contain. For example, I can test whether there are chlorophyll a or b, carotenoids, or anthocyanins.
What kind of technique I would need to quantify the amount of these pigments in the leaves?

Nice question, and welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

Once you've found a suitable mobile phase in analytical TLC, you could upscale the separation and transfer the results to preparative TLC or column chromatography.
In the case of column chromatography, collect the eluted material in test tubes, check the content by analytical TLC and/or UV-VIS spectroscopy, unite identical fractions, evaporate the solvent and determine the weight of the respective fractions.
In the case of preparative TLC, scratch the separated bands off the plate, elute with a solvent of your choice and proceed as above.
If necessary, repeat the chromatographic separations for the different compound classes with other solvent mixtures.

How should the results from spectroscopy be interpreted in this case? Is it a confirmatory step or does it serve some other purpose?

@mmh One spot on tlc indicates at least one compound. A UV spectrum gives additional Information on which fractions to pool and which not.


text:
Double anode in simple cell
What would happen if I kept two copper plates and one zinc plate in a solution of dilute sulfuric acid, and joined two copper plates with wires?
(The wires would connect to a single wire and be joined from that wire to the single zinc plate.)
What would be the voltage? Will there be something wrong?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. || You lost me! :)

The two copper plates will be at the same voltage, they have the same potential.

It would be like connecting the positive of a battery to a wire and the end of that wire to the positive and then connecting that wire to a single wire, kinda useless...

Neglecting any resistance in the connecting wires, this situation is no different than having one large copper plate. If everything is connected together, you can't measure any voltage because the fermi levels of all the metals will be equal (at equilibrium). This is essentially shorting the terminals of a battery and the zinc will start to dissolve in the electrolyte and hydrogen will be produced at the copper electrodes. You can measure a potential difference between the copper plates and the zinc plate, if they're not shorted, but it won't be any different with one copper plate or two for an open circuit.
If you're actually using the cell to do work, you may get more work out of the cell with more copper area, as the larger surface area will result in less solution resistance and better diffusion.


text:
What is the smell of a glass of water in contact with fresh air?
When you pour water into a clean glass in a small confined room with no or little fresh air circulation, the only smells you perceive are those of the glass and the water (or the impurities from both of them).
However, expose that very glass to a bit of fresh air, and voilà, it starts to smell different, a bit like the (wet) shells of mollusks in the sea or from a sweetwater source (it is not a "fishy" smell in the sense that is smells like fish but rather one reminding of the sea, lakes or rivers).
I found this smell with wet glass, ceramics, porcelain, but apparently not really or not at all with plastic or metal (maybe the smell of metal somehow obstructs perception, or maybe it alters or destroys the chemical compounds).
So what is this smell actually?
What chemicals are there causing this smell?
Why does it apparently need contact with fresh air (maybe oxygen radicals or something) to emerge? And it seems to be volatile so that the smell weakens with time (or maybe you just get used to it).
I hope at least some here know what I am talking about... if not, maybe try it out.
Some people are apparently more susceptible to this smell than others (a friend of mine is very susceptible to it while others have no idea what this fuss is all about).
It also works when you get the glass hot and fresh from a dishwasher, though it is recommended to try with a non-scented detergent.
And you absolutely have to try it with non-chlorinated water.
Update:
I've come across two substances that may be related to the phenomenon, Trimethylamine and Dimethyl sulfide, however, I don't know how they smell, so I can't really say if any of these are responsible for the odors, but from the Wikipedia articles, trimethylamine sounds more fishy than "mollusky" while dimethyl sulfide is described to be cabbage-like.

I am searching for the chemicals involved as well. The two you have mentioned are per se totally fishy and rotten eggs respectively. It might be that in mixture and at extremely low level they smell so. The smell we perceived is well known here but it happened to me that I saw puzzled people elsewhere. In my dialect has a name (raffrescume) but I am not sure if it does exist in my mother tongue. I try to look for lakes and ponds since it might be easier to find out. Also handling eggs improves the chance to get that smell around the kitchen and tools

And it should be related to bacterial metabolism.  I think is mostly low level of the pair you have mentioned. ...

I do have a very similar experience! What the OP calls the mollusk smell, I call it slightly rotting "seaweedy" smell. It has been a very unpleasant experience for us in our new home.
Similar to OP's experience, this seems to be more intense with glasses and ceramics right after the newly washed dishes get exposed to air when the windows are open. But I also detect it with plastic cups as noted below.
In order to eliminate other variables such as dishwasher problems and tap water quality, I bought bottled water and disposable plastic cups. I did the following experiment.
WITH CLOSED WINDOWS
Water from bottle poured in a previously unused plastic cup, sitting for 15 minutes : No smell
Water from bottle poured in a cup and couple of sips taken, sitting for for 15 minutes : No smell
WITH OPEN WINDOWS
Water from bottle in a previously unused plastic cup, sitting for 15 minutes : No smell
Water from bottle in a cup and couple of sips taken, sitting for 5 minutes : Seaweedy smell
So I'm about to conclude that something in the outside air combines with the saliva (and/or chemicals/enyzmes from the detergents) to produce the seeweedy smell! The same thing happens with the newly washed dishes out of the dishwasher, and hand washed dishes after they get exposed to air with the windows open.
Yes, I know this doesn't make any sense to people that haven't experienced it before! It sounds like nonsense, but it's quite annoying every time I take sips from water cup.
I'm almost on the verge of calling up the EPA about it. There must be something in the air, but what...

Nice findings! I think it's related to microbial metabolism and not restricted to saliva/detergents. You may find a similar smell on wet roads sometimes. Also, it appears to diminish with time, so leaving stuff in the open for some time ahead may improve your situation.

I'm not 100% sure yet but I tested this several times. With open windows, water from a bottle in a plastic cup doesn't smell.  But after I take 2-3 sips, it starts smelling. With closed windows, there's no smell. Yesterday I ran the dishwasher, and when I unloaded it (closed windows), the glasses had only lemony smell from the detergent. I put one glass (still lukewarm from the dishwasher) outside for two minutes. The rotten seaweed smell was back in full force! I'm really perplexed about this. Maybe there are different mechanisms here...

The only smell you're perceiving is evaporated water molecules. Since higher temperature increases the rate of evaporation it makes sense the smell of water vapor is more prevalent when the glass is hot. Fresh air has a much larger affect on your ability to smell than it does on the rate at which water evaporates.
So really it's the fresh air that let's you smell the water vapor more easily not the water actually changing composition and smelling differently.

I seriously doubt that all water smells "mollusky". And that would contradict that the odor hardly seems to develop in contact with plastics. What I'd consider a true "watery" smell is perceivable with air moisteners.  They smell like fog.

It's not all water. It is sometimes.


text:
Kinetics vs Thermodynamics
The way kinetics is taught at the undergraduate level (Arrhenius and collision theory) chemical equilibrium is determined governed immensely by activation energy of the reaction. According to thermodynamics, however equilibrium is a function of free energy change. In a way, thermodynamics and kinetics seem to contradict each other. What insight am I missing. Is the energy in energy vs reaction coordinates supposed to be Gibbs free energy instead of enthalpy? Would this imply activation energy changes as a reaction proceeds? Is the way kinetics is taught wrong?

I suggest you add a bit more background to your question - most people will have no idea how kinetics is taught at your university.

I'm a high schooler actually, using an undergrad textbook. Its your typical kinetics lesson. rate is proportional to reactant concentration depending on mechanism (1st order, 2nd order etc), there is a frequency factor (temperature based), steric factor, percentage of molecules with activation energy exp(-Ea/RT). I'm wondering if one is supposed to calculate activation energy wrt to free energy or enthalpy. if it is free energy, free energy change keeps changing so your simplistic university approach doesn't work

"...chemical equilibrium is governed immensely by activation energy of the reaction."

This sentence is incorrect. The relative amounts of species in a chemical equilibrium are not at all influenced by the kinetics; if anything, kinetics are influenced by the chemical equilibrium (such as in the Marcus theory of electron transfer reactions). A rigorous derivation of the expression for an equilibrium constant can be made with no direct reference to reaction kinetics.
Chemical equilibria are what happens to systems after they've been left without interference for an infinite time; any system must reach a steady state in finite time, otherwise it would essentially be a perpetual motion device. In other words, for a chemical equilibrium to be achieved, the system is given more than enough time for any finite activation energy barrier to be overcome.

I reasoned that acidic strength is a thermodynamic measurement from this (since it's determined using pKa). Is such an analysis correct? Also, clear and concise answer.

@HarryHolmes That is correct. Though it is possible to talk about the kinetics of proton transfer or Lewis adduct formation ("kinetic acidity" or "kinetic basicity") using other measurements, pKa and pKb values (and therefore acidic and basic strength) are a purely thermodynamic property

A great deal of chemistry is determined by the interaction of kinetics and thermodynamics. The world would be a dull place if we didn't have both.
While thermodynamics determines what directions a reaction can go, the kinetics often determines whether the reaction can happen. Take a simple example: diamond is not the stable from of the element carbon at room temperature; graphite is. If the world were purely determined by thermodynamics all carbon would be graphite and men would be unable to impress women with the expensive shiny jewellery they buy them. But this reaction doesn't happen.
The reason it doesn't is because to convert diamond into graphite we would have to have to radically restructure the chemical bonding in diamond from a tetrahedral network breaking at least one bond for every carbon atom. Breaking all those bonds requires a huge input of energy to kick off the process (thermodynamics says you will get that energy plus a little more back if you can make this happen). You can think of diamond with a whole bunch of broken carbon-carbon bonds as an intermediate structure on the path from diamond to graphite. But that intermediate structure needs a great deal of energy to create and there simply isn't enough energy at room temperature to get there. The reaction will proceed at very high temperatures (and, if the pressure is high enough, diamond will be the thermodynamically favoured form). This is how the earth (or industry) creates diamonds: at high temperatures there is enough energy to break the carbon-carbon bonds and the equilibrium can be established (and, at high pressures, will favour diamond and at lower pressures graphite).
The point of all this is that, in chemistry, what happens involves the interplay of thermodynamics and kinetics. If it takes a lot of energy to get to an intermediate, the thermodynamic products cannot be reached. So being able to achieve an equilibrium depends not just on the start and end products of a reaction but the intermediate structures or compounds that you have to pass though to interconvert them. If those structures are hard to get to because you don't have enough energy around then thermodynamics is irrelevant to what will happen. you always have to think through the mechanism behind the reaction and understand the energy required to create the intermediate states along the reaction path.

The energy referred to on an energy diagram is enthalpy.  Think of the reaction A = B, an exothermic reaction involving two gases.  The forward reaction has lower activation energy so it is faster.  In order for the reaction to be in equilibrium there has to be more B than A so that the forward and reverse rates can be equal.
Look at the reaction A = B + C, an exothermic reaction where all the species involved are gases.  The reaction will produce more products for 2 reasons.  One, the forward reaction is easier in terms of activation energy than the reverse reaction.  Two, the reverse reaction requires 2 species coming together rather than 1.  Both have to be at the correct angle etc.  That could be thought of as a way that entropy affects the equilibrium position.
I am not sure how to deal with the entropy change in a reaction like A = B.  I suppose if B has greater entropy than A this will slow the reverse reaction due to a lower frequency factor A as it is more difficult for B to be in the correct configuration to react.


text:
Reduction of amides
The reduction of an amide using LAH yields a primary amine . I looked up the mechanism over here.
The mechanism suggests that the final main product (amine) is produced before giving the products a water bath (hydrolyze) .

So why do we need to add water at the end of the reaction?

This page's  last paragraph suggests that 4[H] are required for the reaction.

Two [H] are used up to make the amine (according to the mechanism), so what happens to the other two nascent hydrogens?

4 [H] in the second reference are general one-electron reducing agent equivalents. They are not 'literal' hydrogens. If you need water for the reaction is somewhat debatable, but since primary amines are very weak acids and hydrides are extremely strong bases, the primary amine product likely will form a complex with aluminium, that can be destroyed by water.

@permeakra This(http://www.curvedarrowpress.com/partd/aminefamide.html) provides the reason for the workup but with a completely different mechanism . Also check out my comment on J.LS's answer . Which is the correct mechanism ?

@DelPate Likely, you won't find proven correct mechanism for the reaction. It certainly includes oxygen coordination by aluminium and hydrogen attack on carbon, but proving anything else is hard and unneeded.

@permeakra Why do you say unneeded ? If the mechanism liberates H2 gas , wouldnt it be dangerous or explosive ?

Technically speaking, LAH liberates hydrogen on contact with acids, so yeah, unsubstituted amides may produce some hydrogen (this is unrelated to the reduction itself, though). However, LAH is so active it is always used under fume hood and in (weak) argon stream, so accumulation of any significant amount of hydrogen is not an issue.

You don't need to add water to produce the amine, but it's the simplest way of destroying any remaining hydride reducing agent (to give aluminium hydroxide, lithium hydroxide and hydrogen). Presuming you're doing the reaction in ether, the $\ce{LiOH}$ will all go to the aqueous layer, whereas $\ce{Al(OH)3}$ is pretty much insoluble in any solvent you're likely to use in organic synthesis, so it's easy to extract the amine from the aqeuous solvent.
The "chemguide" page is a bit confusing. The reaction they write out isn't a great description of what's happening to the oxygen in the amide (it's probably going to be incorporated into $\ce{Al(OH)3}$ but the other two "$\ce{[H]}$"s just go to making the carbonyl oxygen into water.

Thank you . I think the mechanism of that link(that I mentioned) is wrong . In the 1st step itself , a hydride is given out for attack and AlH3 must remain but they seem to have converted the H to CH3 . This (https://m.youtube.com/watch?v=CJU4NIFac9k) provides an altogether different mechanism which involves H2 release . And this mechanism (http://www.curvedarrowpress.com/partd/aminefamide.html) actually provides the use of the workup . Which is the correct mechanism ?

You'd really have to look at the research literature to check what people think at the moment. I suspecct that as permeakra said, there's no "established" route because while AlH3 is pretty reactive, it is deactivated by oxygen coordination so it's hard to tell exactly what processes are going to dominate.


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Maltose Nomenclature
Maltose has a 1,4 glycosidic linkage between two molecules of α-D-Glucopyranose.
My textbook suggests the final name of maltose as

4-O-(α-D-Glucopyranosyl)-D-glucopyranose

I wanted to know what does "4-O-" stands for ?

I dont think so it provides information about the atoms which are linked and moreover feel that Wikipedia's nomenclature (not IUPAC) :

α-D-Glucopyranosyl-(1→4)-β-D-glucopyranose

more detailized .

I found 6-O-(whatever) not 4-O-(Whatever) :). Seriously though. I never found any such molecule nomenclature in the whole web.

There is this statement in the wiki link I provided "Appropriate symbols are used to indicate the stereochemistry of the glycosidic bonds (α or β), the configuration of the monosaccharide residue (D orL), and the substitutions at oxygen atoms (O)." The last 3 words could be the partial answer for the "O" in the name . But I cant understand what it means .

And also ,is maltose is made up of two alpha-glucose molecules? Since the wiki name has beta attached to the latter part of its name . I am confused .

Maltose is 4-O-α-D-glucopyranosyl-D-glucose.  The "O" in "4-O" indicates the attachment of atom of the substituent.  The substituent in this case is "-α-D-glucopyranosyl".  It is attached to the oxygen atom that is attached to carbon 4 in the base molecule.  The base molecule in this case is "-D-glucose".  "4-α-D-glucopyranosyl-D-glucose" could be interpreted as a different structure where the α-D-glucopyranosyl substituent is bound directly to carbon atom number 4.
The "base molecule" of "-D-glucose" still has an unsubstituted anomeric OH group.  Thus, in aqueous solution, maltose will be an equilibrium mixture of at least these three isomers:

4-O-α-D-glucopyranosyl-α-D-glucopyranose
4-O-α-D-glucopyranosyl-β-D-glucopyranose
4-O-α-D-glucopyranosyl-D-glucose (open chain on the right-most "glucose" moiety in the name)

Thus, it doesn't really make sense to disambiguate α vs. β vs. open-chain forms of the "right-most" glucose ring in maltose, unless you need to refer to a particular conformation of maltose that matters for a certain reaction or condition.  However, the "left" ring is locked (in maltose) to an α-D-glucopyranosyl conformation because it does not have an anomeric OH group and rapid equilibration in water does not take place.

The latter name that you describe as Wikipedia-not-IUPAC is actually the preferred name variant. From IUPAC Nomenclature of Organic Chemistry (Preferred names 2013), the “Blue Book”:

P-102.7.1.2 Disaccharides with a free hemiacetal group
Disaccharides which can be regarded as formed by elimination of one molecule of water from one glycosidic (anomeric) hydroxy group and one alcoholic hydroxy group, are named as glycosylglycoses. Locants and anomeric descriptors must be cited in the full name.
There are two established methods for citing locants:
(1) in parentheses between the components with an arrow going from the locant of the glycosyl component to that of the glycose component;
(2) at the front of the glycosyl component.
Example:

(1) α-ᴅ-glucopyranosyl-(1→4)-β-ᴅ-glucopyranose
  (2) 4-O-α-ᴅ-glucopyranosyl-β-ᴅ-glucopyranose
     β-maltose (trivial name; not β-ᴅ-maltose)

Here it's “preferred” just as in “listed first”; but really explicitly preferred (on the other hand, not in a strict sense, like PINs in the Blue Book) in IUPAC Carbohydrate nomenclature (Recommendations 1996) itself (which the portion above was based on, but not completely literally):

2-Carb-36.3. Disaccharides with a free hemiacetal group
(…)
There are two established methods in use for citing locants: either in parentheses between the glycosyl and glycose terms, or in front of the glycosyl prefix, as in the names of glycosides. The former (preferred) method is derived from that used to designate residues in oligosaccharides (see 2-Carb-37 and 2-Carb-38).
Note. The latter method is that used by Chemical Abstracts Service for disaccharides.
The O-locants used for the former method in the previous recommendations [1] are omitted here.

The bottom note refers to previous Carbohydrate nomenclature (1969), that included the O- also in the “arrow based name”,

Example:
  α-Lactose: O-β-ᴅ-galactopyranosyl-(1→4)-α-ᴅ-glucopyranose

As already mentioned in Curt F.'s answer, the O- prefix is to specify that the glycosyl (carbohydrate residue, with formal free valence after anomeric $\ce{OH}$ group removal) is bound to the oxygen rather than on the carbon atom (which is automatically assumed in the “arrow style” names), just as for carbohydrate general O-substitution, e.g.

5-O-methyl-ᴅ-galactitol

or

2,3,4,6-tetra-O-methyl-β-ᴅ-glucopyranose


text:
Is ozone an electrophile in the ozonolysis reaction?

From the above mechanism it is clear to see that the electrons from the double bond attack one of the terminal oxygens in ozone, thus meaning it's an electrophile. How can this be? If you draw the resonance structures, the negative charge is concentrated on the terminal oxygens too. I am OK with the rest of the mechanism but I would never think to draw oxygen as an electrophile.

This type of reaction is more appropriately described as a pericyclic reaction.

Electrons are concentrated on the two terminal oxygens in ozone but remember, ozone exhibits resonance. The electrons can be delocalized across each oxygen through p-orbitals.

So yes, at first if you think about the reaction in purely Coulombic terms then it doesn't make much sense. Why would electrons go to a region that already has a relative surfeit of electrons?
But what happens is that by attacking the terminal oxygen, the double bond effectively displaces electrons to the central, electron-deficient oxygen atom.
I'm sure someone can explain this in much greater detail with reference to molecular orbital theory but I'll leave that to someone else.

As Ben Norris mentioned in his comment, ozonolysis is better described as a pericyclic reaction. Pericyclic reactions fall outside of the usual nucleophile-electrophile paradigm. This reaction is an example of a dipolar cycloaddition, because ozone is a 1,3-dipole. Chemists have observed that electron-rich alkenes react faster with ozone than electron-poor alkenes. This implies that the alkene is providing the HOMO to the reaction and ozone is providing the LUMO. In nucleophile-electrophile reactions, the nucleophile reacts through its HOMO, and the electrophile reacts through its LUMO. So if you want to make an analogy between a dipolar cycloaddition and a nucleophile-electrophile reaction, you could say that the alkene is the 'nucleophile' and ozone is the 'electrophile.'
It's not always obvious that the 1,3-dipole is going to provide the LUMO in a dipolar cycloaddition, but we can speculate as to why ozone does. Oxygen is a very electronegative atom, so its atomic orbitals are lower energy than carbons. Since ozone is made up of three oxygens (with low energy atomic orbitals orbitals), its pi-system will be low energy with respect to an alkene's pi-system. Furthermore, the pi-system will be even lower energy due to placing a cation on an electronegative atom. These factors contribute to the LUMO being low energy relative to an alkene.
To a first approximation, if there's a positively charged oxygen in a compound, it can be assumed that compound will be electrophilic.

I'm sure someone on this site could run calculations to support these rationalizations.


text:
pKa differences and reactivity
I know a lot of chemists like to quote $\mathrm{p}K_\mathrm{a}$ values and subtract them to reinforce some notion about some acid-base reaction being highly favorable or highly unfavorable or whatnot.
For example, the $\mathrm{p}K_\mathrm{a}$ of acetic acid is about 5 and the $\mathrm{p}K_\mathrm{a}$ of LDA's conjugate acid is about 36. 36 minus 5 gives us 31 "zeros" of favorability.
However, $\mathrm{p}K_\mathrm{a}$ values themselves don't tell you what solvent you are using … can using this method blindly lead to problems when working across solvents? And as far as I know, a lot of $\mathrm{p}K_\mathrm{a}$ values are measured in water … but in organic chemistry we don't just use water as a solvent. So does this method even have general validity?

An important case of acidity constants changing with solvent, with consequences in organic chemistry, can be found for compounds with hydroxyls, especially water. It is possible to deduce that hydroxyl compounds are more acidic in protic solvents compared to aprotic solvents, which is attributed to an abnormally high stabilization of the resulting conjugate base via strong $\ce{O^{-}\ ^...\  H-O}$ hydrogen bonds.
The difference is most clear for water. Water molecules solvated by water itself have a $\mathrm{p}K_\mathrm{a}$ of 15.7 in ambient conditions, but if the unusual stabilization from hydrogen bonding were removed, they would have a $\mathrm{p}K_\mathrm{a}$ of around 28. In other words, $\ce{H2O}$ is a weaker acid by some 10 orders of magnitude in non-protic solvents compared to water, and correspondingly, its conjugate base $\ce{OH^{-}}$ is a stronger base by the same factor. This means hydroxide ions can effect quantitative deprotonation of even rather weak acids, including the hydrocarbons cyclopentadiene, indene and fluorene, in solvents such as 1,2-dimethoxyethane.
Unfortunately, compounds containing hydroxide ions have somewhat limited use as superbases in non-protic solvents because the solubility of common hydroxide bases such as $\ce{NaOH}$ or $\ce{KOH}$ is rather small. Other hydroxides such as quaternary ammonium hydroxides have appreciable solubility in non-protic solvents, but they are always found in the form of hydrates. Attempts to prepare anhydrous $\ce{R_4N^+OH^{-}}$ inevitably result in decomposition, because the crystallized water solvates the hydroxide ion and tames its reactivity; once the water is removed, the bare hydroxide ion immediately attacks the quaternary ammonium cation creating an alcohol and a tertiary amine.
As mentioned previously, other compounds containing hydroxyls also show changes in $\mathrm{p}K_\mathrm{a}$ comparing water and non-protic solvents, though the difference tends to be smaller. For ethanol, the difference is at most about 4 units of $\mathrm{p}K$.

Doing anything blindly is, of course, counter productive. If you were to make the mistake of comparing pka values in different solvents then there would be issues ,however, if you compare pka values while maintaining a constant solvent the assumption is fine for qualitative reasoning. The only counter example I could find is a 1,8-Bis(dimethylamino)naphthalene. It's pka was significantly lower in ACN and DMSO than expected. Again, just make sure you compare things with constant solvents and you're fine.
I think the chemists you speak of are just saying " for a given solvent blah dissociates more so than blah" which lets them see where the reaction lies at equilibrium.

"Doing anything blindly is, of course, counter productive." - Hey, Helen Keller objects! Also, isn't that how a lot of things were discovered - by accident or "blindly"?

All of the best discoveries are by people who already have an absolute mastery of their discipline. When they say blindly they don't really mean it. It's years of well educated reasoning and experimentation. They call it blind or accident when it's really not.

pkA Values are measured in water, so it's only an approach to think that in other solvents they behave in a similar way. The pKa values are also depending on the temperature.
But a good approach is better than nothing.
If the organic solvent is a protic solvent like alcohol (an alcohol is chemically a bit like water) this approach is quit o.k.
Remember that the pKa tells something about the equilibrium and the energy that is needed to move a proton to another molecule (and so to charge a molecule). If a charged molecule is stabilized as it is in a protic environment the equilibrium allows more protons to move around.
In contrast a "naked" small positive charge like a proton without any stabilizing environment need so much energy that it can't exist in a normal laboratory - this is already chemistry of a plasma.

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

I don't think pKa values are exclusively measured in water.


text:
How to buy 1g of enzyme that is sold in units/mg?
I am trying to buy 1g of horseradish peroxidase. On sigma-aldrich, it is sold as units/mg. I am confused as to what this unit is. It comes in 50-150, 150-250, and 250-330 u/mg. The way I currently understand it is if I buy 25 KU (kilo-units?) of 50-150 u/mg, than I should have  at minimum 167 mg of the peroxidase. Is this a correct understanding? The price of 25 KU of horseradish peroxidase is $95, so it seems ridiculous to pay that for only 1 2/3g.

Sigma Aldrich defines their units as:

One pyrogallol unit will form 1.0 mg purpurogallin from pyrogallol in 20 sec at pH 6.0 at 20 °C.

Assuming you're doing a biological imaging experiment, these activity values tell you much more about what you're getting than mass.
From a casual internet search it appears that HRP is pretty expensive. I would consider whether you actually need so much of it for your research.

Indeed. 1 g is a lot of HRP for most things. 10 mg of ~150 U/mg HRP is enough to do several well plates' worth of peroxide assays for me. I usually aliquot out a stock solution and freeze the extras.

I am using this for synthesis of polypyrrole. To do so, I need to be able to measure out masses of HRP.

@jarred: no, you probably don't. What you need to know is the activity of a certain amount of enzyme, which is what U/mg gives you.

And I hope you are aware that you will have to work in aqueous buffer solution when using enzymes, which might be a bit limiting when working with pyrroles.

If you're talking about this paper, they use 1 mg of ~170 U/mg HRP in 10 mL of solution (probably because they didn't want to do serial dilutions and 1 mg is about the lowest that's practical to weigh out), so you just need to make up a solution that's around 17 U/mL. How much HRP you need depends on the activity level of the stuff you buy.

Ok, that makes sense @Michael

The enzyme is sold by activity, not by weight.  The different descriptions from Sigma refer to the amount of active enzyme (in units according to their definition) in 1 mg of the powder or lyophilized residue they sell you.  You will not be buying pure HRP.  "Pure" HRP is a bit of a misnomer anyway, since it comes it at least seven isoforms, differing slightly in MW and glycosylation patterns, etc.  So if the U/mg number is higher, the amount of active enzyme per mg of residue is higher.
Most HRPs seem to have a molecular weight of around 40,000 g/mol.  If you knew the rate constant $k_{cat}$ of HRP for your substrate in your assay conditions, you could use the rate constant and the MW to calculate a "theoretical" activity in terms of U/mg.  My guess would be that this theoretical activity would be higher even than 250 - 330 U/mg.
The ratio of the U/mg that you buy to this "theoretical" activity is the weight fraction of what you buy that is active enzyme.  Probably for the cheapest preparation, more than 90% of the weight is not active enzyme.  There are probably salts, water, and inactive/misfolded HRP molecules, and other protein contaminants in there to varying degrees.


text:
Why doesn't the atomic number change by two in the process of production of Americium?

Looking at the bomardment of uranium-238 this is what I see.
First, I see a uranium-238 atom with 92 protons bombarded with a neutron. The neutron then undergoes beta decay into a proton. So we then get Neptunium as the neutron decayed into a proton and hence, atomic and mass number increase by one. An electron is also released from the atom.
That makes sense.
Looking at the second equation of the bombardment of Plutonium-239.
Plutonium-239 with 94 protons gets bombardment with 2 neutrons. These neutrons undergo beta decay to produce protons. Thus, the atomic and mass number must increase by two. The mass number, indeed, does increase by two. The atomic number does not. An electron is also released? Not sure why aswell.

Only one neutron decays.

What might be confusing is that it's actually two reactions. Two neutrons are captured followed by a single beta decay:
$$\ce{^239_94Pu ->[2(n,\gamma)]^241_94Pu->[\beta^-]^241_95Am}$$
The first reaction leaves the new nucleus in an excited state, so a gamma ray is emitted. The second reaction is actually quite slow with a halflife of ~14 years. It's a beta minus decay so that's why an electron is emitted.


text:
Calculating moles of Hydrogen gas in multiple gas container
If there was a 500 mL container filled with oxygen and hydrogen so it burns completely (2H:1O), how many moles of hydrogen gas would there be if there if pressure is 101.3 kPa and the temperature is 293 K?

$PV = nRT$ is your friend in this case.

I know this would work if it was pure hydrogen, but what about mixed with oxygen?

Calculate total moles of both gases with the equation and divide those moles between gases in mole ratio (2:1) that you have given in your question.

Using pv = nrt$$1atm\times(0.5L) = (total moles)(.08206\frac{atm\cdot L}{mol\cdot k})(293K) \\ \text{total moles} = .02079mol $$ There are 3 total moles of gas with 2 of those being hydrogen and 1 being oxygen. This means 66% of the total moles is hydrogen and 33% is oxygen. Soooooo 66% times .02079 is the amount of hydrogen moles and 33% times .02079 is the amount of oxygen. TADA!


text:
Iodate ion and Octet rule
'm trying to figure out why the oxidation state of the iodate ion is +5. The iodine atom has 7 electrons in its outermost shell (comprised of s and p sub-shells). Two oxygen atoms receive 2 electrons from iodine to obtain full valence shells (s2 p6) and only one electron goes to the other oxygen atom. This leaves iodine with a full sub-shell s2. Is this a deviation from the Octet rule?
And how is the charge on the iodate ion negative when the third oxygen requires an electron?

First of all I strongly advise you to go through the following link as it will make my answer more clear to you .
Iodate ion

Two oxygen atoms receive 2 electrons from iodine to obtain full valence shells (s2 p6) and only one electron goes to the other oxygen atom.

The bond between the iodine atom and oxygen atom is a covalent bond . The electrons from the iodine atom are not given but shared .

This leaves iodine with a full sub-shell s2. Is this a deviation from the Octet rule?

The two double bonded oxygens ( refer the structure in my link ) share in all 4 electrons and the 3rd oxygen shares a single electron . Iodine hence has more electrons than octet rule specified 8. This is why it violates the rule .

And how is the charge on the iodate ion negative when the third oxygen requires an electron?

The third oxygen shares a single electron and hence has a negative charge on it which makes the overall molecule an anion with charge -1.


text:
Does symmetry lower the energy of molecules?
Why does nature "prefer" symmetry? Why is symmetry energy-lowering? I keep hearing statements along the lines of this tossed around by chemists. Here are some examples:
https://physics.stackexchange.com/questions/135671/nature-favours-symmetry
https://books.google.com/books?id=WkRy8b0F2HwC&pg=PA153&dq=nature+favors+symmetry&hl=en&sa=X&ei=V__fVPvpGsuxggSptIG4AQ&ved=0CDAQ6AEwAw
IBrF2 - van der Waals repulsions - and number of unique bond angles (The more symmetric conformation of the molecule is probably preferred).
So:
1) Are any of the above statements true? If so, can you provide example(s)?
2) I think the statement is false at least to some extent. Consider cyclobutadiene ... it's not a square; it's a rectangle.

$H_2S_2O_5$ is not symmetric. Myth busted!

@Rohinb97 what's the significance of the molecule you named?

Well symmetry isn't favored so much. Nature does like it, but it's not everywhere.

Why does nature "prefer" symmetry?
Why is symmetry energy-lowering?
Are any of the above statements true?

Things are either "true" or "false".  The condition is binary, there is no "somewhat true".  In that sense, the above statements are false.  There are many, many non-symmetric things in nature.

The human heart is not symmetrically positioned within the human body.
The narwhal's tusk is helically grooved and projects from the left side of the jaw.
You are not symmetric with the person standing next to you.
Like people, most snowflakes are different one from another (see here).
The staggered conformation ($\ce{D_{3d}}$, lower symmetry) of ethane is lower in energy than the eclipsed conformation ($\ce{D_{3h}}$, higher symmetry).
Cyclooctatetraene exists in a tub shaped conformation with 4-fold symmetry, rather than planar with 8-fold symmetry.
Cyclobutane exists in a puckered $\ce{D_{2d}}$ geometry rather than a planar geometry with 4-fold symmetry.
Cyclohexane exists in a chair conformation with a 3-fold axis of symmetry rather than a planar configuration with 6-fold symmetry.

From these examples we see that higher symmetry is not always preferred.  In those situations where a symmetric arrangement is preferred it is likely due to

It requires less (DNA) code to create identical arms or arrangements on a living structure
Electrostatic repulsions are often minimized by creating structures with higher symmetry (e.g. methane is tetrahedral rather than planar).

In those situations where the symmetric arrangement is not preferred evolutionary advantages or steric\electrostatic repulsions, hyperconjugation (ethane conformation), etc., outweigh other factors.

"The staggered conformation (lower symmetry) of ethane is lower in energy than the eclipsed conformation (higher symmetry)." - actually, the symmetry group is just as big in either case. There is a 3-fold rotational symmetry with a mirror symmetry parallel to the axis in either case, and while the eclipsed conformation has an extra mirror symmetry that the staggered conformation doesn't have, the staggered conformation has an extra point symmetry. Moreover point symmetries are usually better at distributing points along a sphere - of all platonic solids, only the tetrahedron doesn't have one.

@JanDvorak  staggered ethane is $\ce{D_{3d}~ (C_3 ~+~ 3C_2~+~3\sigma_{\mathrm{v}})}$; eclipsed ethane is $\ce{D_{3h} ~ (C_3 ~+~ 3C_2~+~3\sigma_{\mathrm{v}}~+~ \sigma_{h})}$, the latter containing the extra $\ce{\sigma_{h}}$ symmetry plane - because of this, I've always seen $\ce{D_{nh}}$ listed as higher symmetry than $\ce{D_{nv}}$

D3d also has i (inversion)

Symmetry is preferred if it lowers energy, but it does not always.
My high school chemistry teacher had a good rule of thumb. A chemical system will [almost] always tend towards the lowest energy configuration. (I insert "almost," because there are some cases where this isn't true - usually because of entropy.)
So yes, many things are symmetric and this usually lowers the energy.
You raise a good counter-example. Cyclobutadiene is not square. A square cyclobutadiene would be a di-radical, so it undergoes symmetry breaking (i.e., a Jahn-Teller distortion) and becomes a rectangle.

We see here that the rectangular distortion has lower total energy - one orbital goes down in energy and becomes doubly-occupied. The other goes up in energy and is now empty.
So what about benzene? Why is it symmetric, vs. a double-single-double.. bond alternation?
Well, in this case, the symmetric (aromatic) version is lower in energy.
(Figure, Chemwiki)

The Jahn-Teller theorem says almost opposite:

any non-linear molecular system in a degenerate electronic state will be unstable and will undergo distortion to form a system of lower symmetry and lower energy, thereby removing the degeneracy


text:
How has iodine gained electrons?
$$\ce{Cr2O7^2-(aq) + 14H+(aq) + 6I-(aq) -> 2Cr^3+(aq) + 3I2(aq) + 7H2O(l)}$$
Two chromium atoms gain 3 electrons each. They go from $0$ to $3+$
Similarily, 6 iodine atoms go from $-1$ to $0$, thus gaining one electron. However, my textbook says that iodine has lost electrons. How?
It went from negative to positive, i.e. from $-1$ to $0$, just like chromium from $0$ to $+3$.

Are you sure those chromium charges are correct? 2) If you take a neutral species and add an electron, what is its resulting charge?

yes
+1?..

Check that $\ce{Cr2O7^{2-}}$ ion again. What are the charges on the chromium and the oxygen atoms? Remember the charges of all atoms in an ion must sum to the ion's net charge. 2) Electrons have negative charge. If you add a negatively charged electron to a neutral species, is the resulting charge positive or negative?

Two points before we begin:

+ charge means deficiency in electrons.
- charge means excess of electrons.

Now lets come to the reaction you are talking about.

Two chromium atoms gain 3 electrons each. They go from 0 to 3+

Moving from neutral to +3 charge means it just lost 3 electrons. (Electrons have negative charge.)

Similarily, 6 iodine atoms go from -1 to 0, thus gaining one electron. However my textbook says that Iodine has lost electrons. How?

Your textbook is right and you're wrong here. Moving from -1 charge to 0 means that an atom just lost an electron (in other words, it lost the excess negative charge it had and now it is neutral).

The mistake here is understanding what happens to the dichromate ion. Chromium has gone from the +6 to the +3 oxidation state. Its oxidation state has decreased by three therefore it has gained three electrons in the process: $$\ce{Cr2O7^{2-} + 14H+ + 6e- -> 2Cr^{3+} + 7H2O}$$
Iodine has gone from the -1 to the 0 oxidation state. Its oxidation state has increased by one, losing one electron in the process. $$\ce{2I- -> I2 + 2e-}$$
Putting these two half equations together gives you the overall equation which you started with. $$\ce{Cr2O7^{2-} + 14H+ + 6I- -> 2Cr^{3+} + 7H2O + 3I2}$$


text:
Green Potassium Manganate and Sodium Hydroxide Equation
When a sample of green potassium manganate (VI), $\ce{K2MnO4}$, is added to aqueous sodium hydroxide, a brown black solid A and a purple solution of B are obtained.
I know that the brown black solid A is $\ce{MnO2}$, but I can't seem to write out the entire equation.
Can someone help me with this?
Thanks! Really appreciated.

What are your thoughts on the identity of B?

definitely a Mno4 ion, possibly NaMno4

Great, that's correct. To make balancing simpler, I suggest removing the spectator ions, or if you find that difficult, imagine for now that all alkali metal ions are the same type, for example using $\ce{KOH}$ instead of $\ce{NaOH}$. So put $\ce{K2MnO4}$ in the reagents, and $\ce{KMnO4}$ and $\ce{MnO2}$ in the products. On which side does the $\ce{KOH}$ go? Wherever you put it, notice that only one side will have hydrogen, and the equation cannot be balanced. What do you think is a reasonable compound to put on the other side that also contains hydrogen?

i managed to use half equations in basic sln to solve this question. Removed all the spectators

Showing your work by answering your own question is highly encouraged!

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$$\ce{ 3K2MnO4 + 2NaOH + 2H2O -> MnO2 +2NaMnO4 + 6KOH }$$
As you said the brown solid A is $\ce{MnO2}$ while the purple solution is due to the permanganate ion .