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How to identify a compound as chiral or achiral
I read my book and learned everything about enantiomers, racemization, $\mathrm{S_N1}$ and $\mathrm{S_N2}$ reactions, but when I'm trying to identify molecules as chiral or achiral, I get confused.
Consider the two molecules below:
1.CH3-CH-OH
      |
      Br

2.CH3-CH-Br
      |
      Br

How can I tell if a compound is chiral?

Look at your molecule, if you see any plane of symmetry the molecule must be achiral. If you don't see a plane of symmetry that doesn't mean the molecule is chiral. But checking for a plane is an easy first step and may often make the time consuming step of drawing and comparing structures unnecessary.

@ron why does the lack of planes of symmetry not imply chirality?

Compounds that possess any $S_n$ axis are achiral and a plane of symmetry is only one example of such an axis.

Chirality can be determined by visualising the molecule in space. Once you do that, you can check to see if there exists a non-superimposable mirror image. If this is indeed the case, then your molecule is chiral.
You can look for an asymmetric carbon atom--because this often results in chirality.
An asymmetric carbon (or chiral carbon) is, very simply,  a carbon atom attached to four different groups.
With this in mind, consider your examples: 1) should thus be chiral and 2) is achiral.
Hope this makes sense.

I drew out the wedge and dash representations as per your steps. It turns out that 1st one is Chiral and 2nd is achiral. Thanks

Good job. The same reasoning can be applied to more complicated molecules (structurally). With some practice, the process of identification becomes almost intuitive.

@getafix I downvoted this question because it could mislead the OP into thinking that the presence of chiral carbons implies chirality.

@DHMO Well that's up to you. But I don't think that's the case. The answer opens with, "Chirality can be determined by visualising the molecule in space. Once you do that, you can check to see if there exists a non-superimposable mirror image. If this is indeed the case, then your molecule is chiral." Which is general. "You can look for an asymmetric carbon atom--because this often results in chirality." Which is also true. Nowhere do I say asymmetric carbons automatically imply chirality in all cases.

@getafix That is why I said "could" and "mislead". These words mean that you did not say directly that the presence of chiral carbons implies chirality. I made that judgment because right after you define chirality on the first paragraph, the second paragraph starts with "[y]ou can look for an asymmetric carbon". You did use the word "often", but I feel that the point "not always" has not been emphasized enough, albeit not the main topic of the question.

Look for carbons with four different groups attached to identify potential chiral centers.
Draw your molecule with wedges and dashes and then draw a mirror image of the molecule. If the molecule in the mirror image is the same molecule, it is achiral. If they are different molecules, then it is chiral.
Here you're particularly interested in the C in the CH. However, take a look at the C in CH3. Why isn't it a chiral carbon?

I don't think CH3 Groups can be chiral centers. Am I wrong?

Nope, you're correct. :P It was just a hint toward why the carbon attached to the two bromines wasn't chiral.

Draw both molecules. Don't draw them flat. Alternatively, model them using Avogadro.
Exchange the positions of $\ce{Br}$ and $\ce{OH}$ in the first case. Is the new molecule identical to the original one?
Do the same for the dibromo-substituted compound. How is the result here?

Thanks for pointing out on using avagadro. I did exchange the postitions of OH and Br  and the new molecule was indeed non-superimposable to the first one and hence chiral.

The first method I used to understand chirality was to grab some different colored zip ties and make a couple small tetrahedral structures with them, assigning a color to each different radical (say blue for $\ce{CH3}$, white for $\ce{H}$, red for $\ce{OH}$ and yellow for $\ce{Br}$.)
Then look at it by rotating the first to see if I can get the second molecule, if you can't then that molecule is chiral.


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Can bone mineral be obtained in pure, crystalline form?
Can bone mineral be obtained in pure, crystalline form outside a living organism? Had this been done ever?

What do you mean by bone mineral? Its calcium carbonate / phosphate content?

Bone has a complex structure and formation. There has been studies which shows a relationship between the formation of the "mineral" ie hydroxyapatite  and protein (collegen). I doubt that any materials similar to bone is seen outside a living organism (except exoskeleton)

Apatite is the chemical name for the mineral found in bone and teeth. It is occasionally found in the form of large crystals, such as this:

BTW, just as bone is made of the mineral apatite, there are critters with skeletons of silica, another common mineral, such as Venus' flower basket!

Thank you. I have already found that bones are mostly built from hydroxylapatite [img]https://www.mineralauctions.com/public/multimedia/images/auctions/items/thumbs/111506_465xV.jpg [/img] It is mostly mined but also used in medicine for prosthetics.

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Relative atomic mass
Is it possible for an element to have a calculated relative atomic mass that is higher than the relative isotopic mass of its most massive component isotope?
The equation is:
$\mathrm{Relative~Atomic~Mass}=\sum[{\mathrm{(Isotopic~Mass) \cdot (Relative~Abundance)]}} \times100\%$

No. Mathematically, that would not make sense.
Consider an imaginary element with only one known isotope: imaginarium-100.
The percent abundance of imaginarium-100 is $100\%$, so its relative atomic mass is: $$100 \cdot 1=100~\mathrm{amu}$$
Now imagine a group of scientists were to discover that in fact imaginarium-99 consists of $0.0001\%$ of all imaginarium samples.
The new relative atomic mass would be:
$$100 \cdot 0.999999+99 \cdot 0.000001=99.999999~\mathrm{amu}$$
The relative atomic mass goes down with the discovery of a less massive isotope. So the closest that the relative atomic mass of an element can be to the mass of its most massive isotope is if the most massive isotope is also the only known isotope, but the relative atomic mass can never go over this.

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Change in current in an electrolytic cell over time
Why does the current increase over time in an electrolytic cell with a $\ce{CuNO3}$ electrolyte. The current increased about $2~\mathrm{A}$, and the temperature increased during the experiment to around $55\ \mathrm{^\circ C}$.
I’m not sure whether the formation of hydrogen and oxygen gas at the electrodes would decrease the internal cell resistance causing the current to increase. Or is there something else causing this?

The electrolyte conductivity is dependent on the temperature of the electrolyte. Higher temperature increases ions mobility, thereby increasing electrolyte conductivity. The conductivity increases around 2% per degree (assuming linear increase of conductivity). More info here.
As Burak Ulgut mentioned, temperature increase is due to ohmic (Joule) heating. This can be estimated accordingly: $Q=V \cdot I \cdot t$

Temperature will increase due to resistance.

As the reaction proceeds, your electrodes will get rougher since the surfaces will be etched. This will in turn result in an increased electroactive surface area increasing the current.
The temperature effect mentioned by Jaroslav Kotowski is there, but the growing electrode area is also a contributing factor.

Thanks this clears it up because there were some anomie in my data where a higher temperature didn't always result in a higher current.


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Nomenclature of organic compounds containing complex side chains
How do we decide the sequence of substituents in a organic compound if one of the substituents is complex. Should the complex chain must always be written first while writing the IUPAC name or are there any other rules governing it?
The following IUPAC names got me confused:
7-(1,2-dimethylpentyl)-5-ethyltridecane
AND
5,5-dimethyl-6-(1,1-dimethylbutyl)-6-pentyltridecane
In the first compound the complex substituent is written first although p(pentyl) comes after e(ethyl).
In the second compound, the complex substituent is written later although m(methyl) comes after b(butyl).

Simple prefixes are arranged alphabetically disregarding any multiplicative prefixes. The multiplicative prefixes are inserted later and do not alter the alphabetical order.
For example, ‘1,2-dibromo-’ is considered to begin with ‘b’.
However, the name of a compound substituent is considered to begin with the first letter of its complete name.
For example, ‘1,2-dibromobutyl-’ is considered to begin with ‘d’.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.
P-14.5.2 The name of a prefix for a substituent is considered to begin with the first letter of its complete name.

Therefore, the correct alphanumerical order corresponds to the name 7-(1,2-dimethylpentyl)-5-ethyltridecane (not 5-ethyl-7-(1,2-dimethylpentyl)tridecane) since the compound substituent name ‘dimethylpentyl’ starts with ‘d’, and ‘d’ is earlier alphabetically than ‘e’.
(Nevertheless, the preferred IUPAC name is 5-ethyl-7-(3-methylhexan-2-yl)tridecane.)

In your second example, the correct alphanumerical order corresponds to the name 6-(1,1-dimethylbutyl)-5,5-dimethyl-6-pentyltridecane (not 5,5-dimethyl-6-(1,1-dimethylbutyl)-6-pentyltridecane) since the compound substituent name ‘dimethylbutyl’ starts with ‘d’, and ‘d’ is earlier alphabetically than ‘m’.
(Nevertheless, the preferred IUPAC name is 5,5-dimethyl-6-(2-methylpentan-2-yl)-6-pentyltridecane.)


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What is the world's finest powder substance?
What is the world's finest powder substance that has the smallest granules but remains solid?

Welcome to Chemistry SE! Please define, in your terms, what finest is.

Clusters of just a few atoms, such as gold atomic clusters, and quantum dot nanoparticles could be considered very fine powders, though they usually are suspended in another substance to keep the clusters from agglomerating. Particles this tiny have different properties from the bulk material, both because of quantum effects (there are few particles, so there is not a continuous valence band) and because more atoms are at the surface interface than are surrounded by similar atoms in the interior.
See Quantum dot for more information.
I hope this fits your definition of a "fine powder". You might also be interested in pyrophoric iron.
BTW, ruby glass is made using colloidal, i.e. super-finely divided, gold.

While not quantum dots, there are some pretty small nanoparticles which can be bought in bulk as a powder. The type that come to mind the most are titanium dioxide nanoparticles, which are regularly sold in kilogram bags with particle diameters as small as $\rm{21\ nm}$.

Maybe fullerens and similar nanocarbons qualify as one of the finest powder stable without any solubilizer.

@Greg this is interesting question, can a substance consisting of individual molecules be a powder rather than a liquid or gas?

@Anixx There is no substantial difference between large single molecules and nanparticles/fine powders. If you take a look at C60 or nanotubes, there is nothing in there physical appearance that would make them gas or liquid. see http://commons.wikimedia.org/wiki/File:C60_powder.jpg

@Greg the powder on the photo may be composed of particles greater than single molecules

@Anixx True, but the stability of a single molecule (eg placed on a surface after sonification) is similar if not better than the aforementioned metal or semiconductor nano-particles.


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Is there an otherwise liquid material that has angle of repose?
I wonder whether there is a material which consists from individual molecules (that is, fluid), which has its angle of repose thus exhibiting propertes of a powder?

A liquid by definition has no fixed shape and therefore can have no angle of repose. I have noticed you have asked a similar question on Physics.SE; the presence or absence of this kind of property can be used as a means to classify powders versus fluids.
Note also that the conical "fixed shape" observed in powders is due to frictional forces between the particles, whereas any surface curvature in liquids from surface tension (e.g., a meniscus) is due to intermolecular forces between individual molecules.
Furthermore, take into account the melting point of the substance: it's a powder if the temperature of interest is below the melting point, and a liquid if it is above.

Such materials are called Bingham plastics.  An angle of repose means that the material must withstand a non-zero shear stress (from its own weight) while still exhibiting zero shear rate.
Examples include toothpaste, ketchup, mayonnaise, latex paint, and concentrated suspensions of coal particles in a otherwise newtonian liquid.  I'm not sure if any of these materials meets your requirement that it consist "of individual molecules".  I don't think a single-component, single-phase, non-polymeric substance will ever exhibit Bingham plastic behavior, although (I might be wrong and) I can't cite any theoretical reason why this is so.

Do they really have angle of repose?

Imagine a large tank of ketchup.  You stick a one end of a long pipe straight into the surface of the ketchup, and let the other end be a large height above the bulk surface.  Fill the pipe with ketchup, and after it is full, remove the pipe.  The large column of ketchup will collapse into the bulk, but not completely.  There will be a small pile left that will have some maximum angle w.r.t. the bulk surface normal.  Isn't that an angle of repose?


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degradation of sertraline

The picture above is a drug called Setraline. Suggest a likely mode of degradation of sertraline hydrochloride alone in aqueous solution, giving likely structures of the degradation products.

From my knowledge, degradation in aqueous solution can only mean hydrolysis and sertraline does not contain hydrolysable groups like amide or ester. What are the other forms of degradation besides hydrolysis? in aqueous solution.

I agree.  I don't see anything happening quickly when sertraline is in a neutral aqueous solution.  That is, as long as the solution is protected from light.  Zoloft is photolabile and will decompose in the presence of light, see here.


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What is the meaning of cmmol in chemistry?
I came across a unit called $\text{cmmol dm}^{-3}$ in buffer solutions. What does this unit mean?

I guess it simply lack space between C and m - it's concentration in mmol per dm3

Can you give us more context? Copy and paste here the paragraph where you found it.

@NicolauSakerNeto: I'm sorry can't paste it because it was a handwritten one. But I've found some other references online. Such as http://www.thenakedscientists.com/forum/index.php?topic=35027.0  and  http://www.labfert.agr.br/ferramentas.php

That first reference doesn't seem clear to me, and the second one uses $\rm{cmol}$ rather than $\rm{Cmmol}$, in which case it simply means centimol, i.e. a hundredth of a mole ($\rm{1\ cmol} = \rm{10\  mmol}$).

@NicolauSakerNeto I agree that the references are not relevant. The original text I referred mentions $cmmol,dm^{-3}$ and not $cmol,dm^{-3}$. And I have changed 'C' into 'c'.

The unit $cmmol\,dm^{-3}$ is the same as $mmol_{c}\,dm^{-3}$ which denotes the millimoles of charge per litre. I think the former notation is not well used anymore. On the other hand, latter is used often and a google search would return a number of references. For example,
http://micromaintain.ucanr.edu/Prediction/Source/Groundwater/Potential_for_clogging/CP/Water_analysis_for_hazard/Levels_of_concern/
http://www.slidefinder.net/c/converting_various_units_mmol_mmolc/32208520
http://www.scielo.br/scielo.php?pid=S0103-90162011000400012&script=sci_arttext

Upon researching this, I found that this a measurement of charge:
$$\mathrm{cmol=10~\mathrm{meq}}$$
and
$$1~\mathrm{eq}= \mathrm{F}$$
where $\mathrm{F}$ is Faraday's constant and is:
$$\mathrm{F}=96 485.3365~\mathrm{C}$$
So:
$$\mathrm{cmmol}=\frac{\mathrm{cmol}}{1000}=\frac{\mathrm{eq}}{100000}=.96 4853365~\mathrm{C}$$
Sources:
http://en.wikipedia.org/wiki/Equivalent_%28chemistry%29
https://www.agronomy.org/files/publications/jnrlse/pdfs/jnr014/014-02-0084.pdf

Do you mean that $C,mmol,dm^{-3}$ refers to milli moles of charge per litre ?

@chemkatku actually, since Faraday's number is 1 mole of charge, which is equal to 1 equivalent, $1~\mathrm{cmmol}~\mathrm{dm^{-3}}$ is 10 micro moles of charge (per liter).

Do you have any reference to back this up? One mole is similar but not the same as one equivalent, and I've never heard anything relating $\mathrm{eq}$ with $\mathrm{F}$. I agree with @Mithoron and I think this is simply a typo. It was supposed to be either $\mathrm{cmol}$ or $\mathrm{mmol}$ (or the $\mathrm{C}$ is from concentration and a space is missing) per $\mathrm{dm^{-3}}$, which is a litter, and this sounds a lot like a buffer concentration, which in turn can be related to it's "capacity" as the full question mentions on the link provided above.

Also, the reference given uses lower case $\mathrm{c}$ and not $\mathrm{C}$, which is another point against the Coulomb thing.

@Molx sources cited

OK I changed Cmmol to cmmol. My bad, the original text mentions $cmmol,dm^{-3}$ with a simple 'c'

@ringo the agronomy paper talks about $\mathrm{cmol}$ which is centimole. They really like using that in agronomy, as well as equivalents, which is related to Normality. I still think it's a typo but I don't see how one can post an answer saying the question is "wrong".


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Lambert-Beer's law coefficient
While the Beer's law equation is $A=a \cdot b \cdot c$, according to my book, $a$ is a coefficient which is:

a characteristic property of the analyte in a specific solvent, and its values will depend on the wavelength of absorption radiation.

This is quite unclear to me.
If we have for example  $a= 1.5$, what does that exactly tell us ? What do we know when this coefficient is given ? Or is this just a constant that helps us to solve the equation?

A large value for $a$ indicates a strong interaction between light and the chromophore.  In other words, the larger $a$ the stronger the absorption of light by the chromophore.

Think of a shooting gallery at a carnival, where you fire pellets (‘photons of light’) at targets (‘molecules’) randomly moving around in a deep shooting gallery. Then b is how deep the gallery is, c is how many targets there are in the ‘volume’ of the gallery (that the moving targets move around in) and a is the size of the target, i.e., its cross-sectional area. Only a small fraction of fired pellets will be transmitted entirely through the shooting gallery if there are many large targets and the depth is large. Conversely, a large fraction of fired pellets will be transmitted entirely

through the shooting gallery if the targets are small, there are not many of them in the gallery and the gallery is short. In other words, a, c, and b, respectively, are small. The transmittance is the number of transmitted pellets divided by the total fired pellets. Then A, the absorbance, is defined as the negative logarithm, base 10, of the transmittance.

To start with, I recommend to reformulate Beer-Lambert's law in a way more frequently met in (analytical) chemistry, i.e.
$\mbox{ABS} = c \cdot{} l \cdot{} \varepsilon = \log_{10} \frac{I_0}{I}$
because then it more obvious that the (dimensionless) absorption (ABS) recorded is the product of analyte concentration ($c$), optical path length ($l$) across the sample investigated, and the molecular extinction coefficient ($\varepsilon{}$).  Absorption equally is defined as the logarithm (to base of ten) of the fraction of incoming light intensity ($I_0$) and light intensity that is transmitted across the sample ($I$).
Now to your question.  While concentration $c$ and optical path length $l$ may be changed (the later, for example, by exchanging a measurement cell of 1 cm path length to one of 2 mm or 5 cm -- depending on the setup of the spectrometer), the molecular extinction coefficient $\varepsilon$ is indeed more valuable than just assistance to solve the equation.  It relates to the ease the analyte may absorb irradiation.
UV-Vis spectroscopy investigates electronic transitions, prominently of $\pi{}$-systems (like of alkene double, and aromatic bonds), yet investigation of electronic transitions of $n$- (non-bonded, like of carbonyl-O electrons), $d$- (for example in transition metal complexes) and $\sigma{}$-electrons (in the short wavelength UV-range).  In principle, these transitions are discrete, i.e. only photons of this attuned energy ($h\nu{}$) that fits the energy gap between the electronic ground state and the excited state are absorbed.
An analyte molecule may possess more than one allowed electronic transitions.  They may occur simultaneously, with different efficiency.  For example, many organic dyes have a strong absorption in the visible range, like indigo has one around 600 nm.  Yet if you tune the wavelength of investigation to short wavelengths, below 250 nm, for example, absorption may still occur, too, yet to a lesser extent.  (This eventually contributes to processes that indigo stained cloths fade over time, for example.)

Side note:  The molecular extinction coefficient $\varepsilon$ depends on other, additional parameters than the "solvent" and "wavelength of investigation".  Varying the pH-value, ionic strength of the solution, presence/absence of other dissolved molecules than of your analyte and solvent may alter $\varepsilon{}$, too.  Do not forget, the concentration $c$ of your analyte may influence $\varepsilon{}$, too -- pointing to (concentration dependent)  potential dimerisation / aggregation of your analyte (for example charge transfer complexes).

Thanks but i find this a little bit off topic ( not helpful ), I asked only for the coefficient meaning and not for the transitions that may occur. Yet the only thing that seemed to be related is ' it relates to the ease the analyte may absorb irradiation ', which is also hard to understand (not explained well ).

The absorption observed is result of changes of the quantum state. There are selection rules, according to them some transitions are "allowed" (more probable), and others "forbidden (less probable to occur).  Wavefunction, symmetry, and kind of experiment come into play here.  This is an other avenue to answer to your question.

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Fehling's Test meaning for partial formation of a precipitate?
Whilst doing tests on two unknown solutions, I used Fehling's test. The solutions could be a mixture of compounds or just one compound dissolved in water/ another solvent - I don't know. Upon adding Fehling's solution both solutions (which were originally colourless) turned a blue colour. I then heated the solutions very gently for about 5 minutes and after this time one had completely formed a red precipitate, whereas the other still had the blue solution left in it (while forming some red precipitate). After waiting for about 20 minutes both solutions had completely formed a brick red precipitate and were the same. I assume this is to do with rate of the oxidation reaction occurring. Is there any explanation as to the rate of reaction in each solution or an explanation which is not to do with rate at all perhaps?

Is that 20 minutes of heating? or just sitting on the bench?

It's 20 minutes of warming at around 80 degrees Celsius.

The reason is thus:
For aldoses (like glucose), the copper sulfate in the reagent easily oxidizes it into the carboxylic acid forming copper oxide (a precipitate).
For ketoses (like fructose), the ketone cannot be oxidized, so there is no color change. Over time at elevated temperature, the α-hydroxy ketone isomerizes to a α-hydroxy aldehyde. This aldehyde is then oxidized to the carboxylic acid, producing a red precipitate.


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Is there an alternative wet chemical method or electrochemical method to obtain precipitation of iron, nickel and silicon powder?
There are many methods available to obtain powdered metals and alloys. Wikipedia show some of them. Another form is grinding with posterior removal of wheel particles. All these methods produce metal powder with the help of specific machinery. I am asking for an alternative wet chemical method (water solution) or electrochemical method (water or ionic liquid electrolyte) to obtain metal powder (specifically iron, nickel and silicon, in pure or controlled alloy form).

I want to obtain powder of those metals (pure, not alloyed) without an atomizing machine.

I need power of separate element or mixed in known proportion. Grinding will introduce impurities. ¿Will they be easy to remove? I need a small amount, big enough to be overpriced if I buy it from a laboratory supplier and small enough so that I can not import from a wholesaler.

Even after reading all the comments, I still don't understand what you are asking.  Could you revise the question?  Otherwise it might get closed as too vague.

Thanks for updating!  I'm still a bit confused by "obtain metal powder (specifically iron, nickel and silicon, in pure or controlled alloy form)".  Does that mean you are asking for (a) three separate wet electrochemical methods, one for Fe, one for Ni, and one for Si; OR (b) a single method to generate a Fe-Ni-Si alloy; OR (c) both?

@CurtF. (a) or (b) or both. As fewer restrictions are imposed, more likely to get a answer.

Both metal powders can be preapred electrochemically.
Iron dust is prepared using solution containing $\ce{FeCl3}$ with addition of $\ce{AlCl3}$ to make the solution more conductive. The process occurs at elevated temperature 60°C. Unfortunately exact process parameters are hard to find.  But following graph could help.

Nickel could be prepared electrochemically too and can be prepared by dissolving of Ni-Al alloy in hydroxide -Raney_nickel. The description of patent is here.
This is link to a study book PRÁŠKOVÁ METALURGIE, K. Skotnicová, M. Kursa
, unfortunately only in czech.

Nickel powder can be made by electrolysis as well. Raney nickel is a speciality for use as a hydrogenating catalyst, and is very pyrophoric!

Did you even read my answer? I am saying on 2 positions that the electrolysis is possible. Altough the powder is pyrophoric, it is a way how to prepare the nickel dust. What was actually the question.

I am asking for an alternative wet chemical method (water solution) or electrochemical method (water or ionic liquid electrolyte)

Water will only be a efficient electrolyte to generate metals that are more weakly reducing than hydrogen gas.  Otherwise, attempts to produce the metal will just result in mostly water hydrolysis.  The chemwiki page at UC Davis has a good overview of this particular topic.  The standard reduction potential for iron is too low (too reducing; too negative) to be made efficiently via an aqueous electrochemical process.  Update: I softened the wording of this paragraph based on advice from Jaroslav Kotowski who noted that the original wording I used was wrong.  Iron dust can apparently be made via electrolysis, but efficiency is very low.
Nickel and silicon are less powerful reducing agents than iron, but are still slightly more powerful than water under standard conditions.  Some combination of pH and concentration adjustment could make it theoretically possible to generate these metals in a wet (aqueous) electrochemical process, but it would not be very efficient.  I dug up a 1908 paper that reports that nickel films (not powders) can be electrodeposited from a (presumably aqueous) solution of nickel ammonium sulfate.
Non-aqueous electrolytes such as ionic liquids do not necessarily have this problem.  You should read a review article on metal deposition from non-aqueous media such as this one.

Thanks Jaroslav -- I'll edit the answer to reflect this.


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What is SPDF configuration?
Recently in my chemistry classes, the teacher spoke about SPDF configuration and then said that we'll be taught about it in higher classes.
But I'm sorta curious to know that what is SPDF configuration and is there something like - obtaining electronic configuration in SPDF format?

Sounds like your teacher was talking about the electron configuration, though, I've never heard anyone calling it "SPDF configuration".

Without knowing the context (what level of chemistry you are in now), this could relate to term symbols.

s, p, d, f and so on are the names given to the orbitals that hold the electrons in atoms.  These orbitals have different shapes (e.g. electron density distributions in space) and energies (e.g. 1s is lower energy than 2s which is lower energy than 3s; 2s is lower energy than 2p).

(image source)
So for example,

a hydrogen atom with one electron would be denoted as $\ce{1s^1}$ - it has one electron in its 1s orbital
a lithium atom with 3 electrons would be $\ce{1s^2~2s^1}$
fluorine has 9 electrons which would be $\ce{1s^2~2s^2~2p^5}$

Again, as we read from left to right the energy of the orbital increases and the superscript shows the number of electrons in the orbital. You can read more here.

By SPDF configuration, he meant orbital configuration. Now the basic of this concept is from very fundamental quantum chemistry formulation but I don't think you need to know that now (It is usually taught in 4th year of UG or 1st year of graduate study in Engineering discipline but i am not sure about pure science discipline). Whatever I would like to think about it from a different perspective. Are you familiar with quantum numbers? There are 4 different quantum numbers:

Principal quantum number (This is like the city you live in). There are lots of electrons in an atoms. Now if we want to distinguish between those electrons than we need to name them or there should be something to distinguish. Imagine you are living in a circular states and the cities are named as the radius it has. Like sector 1 is the city which has a mean radius of 1 unit and so on. By principal quantum number, we actually mean that the probability of finding that electron is high within that particular radius. It is named as n=1,2,3...
Azimuthal quantum number: This is like the building you live in. Now if you are living in a city that is very small and has only one building, then you don't need to specify each building of that city differently. Like for n=1, l=0 (Here l=azimuthal quantum number/building number) but if n=3, then l=0 to (n-1), that means in Sector 3 city there are 3 building named 0,1 & 2.
Magnetic quantum number: This is like apt number on that building. m=0 to (+/-)l. So, if you are living in building no. 3, you can either live in apt -3,-2,-1,0,1,2 or 3.
Spin quantum number: Each apt has two rooms (room A and room B) (This is actually unrestricted Hartree-Fock case) but if you are living with spouse than you can have a large room by breaking the wall between those two rooms (Only room A or restricted Hartree-Fock case)

Now, The building name can be rephrased as spdf orbital. If you are living in building no. 0 that means you are living in s orbital. Similarly
building no.1= p orbital
building no.2= d orbital
building no.3= f orbital
So, in your building no.0 (s orbital),
total number of room = 1apt * 2rooms/apt = 2rooms or 2 electrons
In building no.1 (p orbital),
total number of electron/room = 3apt * 2rooms/apt = 6 rooms or 6 electrons
Now if you want to know more you can read:

Hund's rule
Pauli's exclusion principle
Aufbau principle

But they all are superficial theory, they can say what is going on but they can't say why. But quantum chemistry method will give you mathematical understanding of why there is 2 rooms/apt or why building 2 has 5 apt etc.

excellent answer ! is helpful in getting to know some basics, except for the total number of electron/room concept , which is somewhat difficult to grasp

First of all, I must appreciate your eagerness to learn new stuff (though it makes it harder to explain). I will try my best to explain in a way you can understand.
You might have been taught about eletron filling order like: 2 electrons in K shell, 8 electrons in L shell and so on. However, that works only up to a certain level.
Its a fact that each shell itself is composed of subshells (experiments involving spectra have shown this). The number of subshells each shell has depends on the number of the shell (like 1st shell, 2nd shell; a.k.a principal quantum number). These subshells are called as s, p, d, or f. The s-subshell can fit 2 electrons, p-subshell can fit a maximum of 6 electrons, d-subshell can fit a maximum of 10 electrons, and f-subshell can fit a maximum of 14 electrons. The first shell has only an s orbital, so its called as 1s. Since it can have either one or two electrons, its called as $1s^1$ and $1s^2$ respectively. They are also the respective "SPDF" configurations of hydrogen and helium. In this way, you would consider the electronic configuration of oxygen to be $1s^22s^22p^4$.
Another important point to note, is that the filling of electrons in subshells doenst actually fill from low to high. There is a special rule called aufbau principle (german word for 'building up'). Here is a diagrammatic representation of aufbau principle:

This infact, is the actual way of writing an electronic configurations. Schools teach to primary classes the 'configuration-per-shell' method simply because its easier and they usually dont come across geniuses like you. Now, I think you can understand the 'SPDF' configuration much better.

Ya. It's good to know.Where the finding of an electron is maximum is known as orbital.The first shell contain s orbital in which two electron can be filled .The second shell contain s & p orbital ,the p orbital can have maximum 6 electrons.The third shell contain S , p & d orbital ,d orbital can have maximum 10 electrons .The forth shell contain s , p , d & f orbital, f can have maximum 14 electrons .
The p orbital have dumbbell shape , it contain subshell  known as px which lies on x axis , py which lies on y axis and pz which lies on z axis . The d orbital have double dumbbell shape . it contain subshells known as dxy lies between x and y axis , dyz lies between y and z axis, dzx lies between z and x axis , dx2-y2 lies on the x and y axis & finally dz2 which lies on z axis . Here are the images of subshell

"the p orbital can have maximum 6 electrons." It hurts me to see this. The p subshell is composed of three p orbitals, each of which can hold two electrons, allowing the p subshell to hold up to 6. No orbital itself holds more than two electrons (at least not the standard hydrogen atomic orbitals, or any molecular orbital I've encountered.)

They refer to the secondary quantum numbers (L) :
The S holds one orbit which contains 2 electrons
The P holds three orbits which means 2×3=6 electrons
D holds five orbits 2×5=10 electrons
F holds seven orbits 2×7=14 electrons

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Photometric titration
If we have  $\ce{A + T -> P}$
where $\ce{A}$ is the analyte, $\ce{T}$ is the titrant, and $\ce{P}$ is the product,
photometric titration happens only in one condition, that one of the following substances will absorb the electromagnetic radiation.
When stable complexes formed such as $\ce{Bi}$-EDTA, they will not absorb the radiation (or in small amounts), but when unstable complexes formed such as $\ce{Cu}$-EDTA, they will absorb highly. The same goes also for the acids. Strong acids will absorb much less the electromagnetic radiation than the weak acids.
The reason for this is that strong acids are completely dissociated.
Why does that indicate the photometric titration ?

This statement seems incorrect: "Strong acids will absorb much less the electromagnetic radiation than the weak acids. The reason for this is that strong acids are completely dissociated". Sulphuric acid is a strong acid which is colourless whether dissociated or not. Trinitromethane is a relatively strong acid which is colourless when undissociated, but strongly yellow when dissociated. Furthermore, many pH indicators are weak acids which are strongly coloured both when dissociated or not. I don't see how the proposed relation between colour and acid strength works.

photometric titration happens only in one condition, that one of the following substances will absorb the electromagnetic radiation.

Almost ;-)
Regardless whether you monitor a complexometric titration by eye or with the help of a spectrometer, it is a good idea to use an indicator, such as Eriochrome Black T. Then, it doesn't matter, if or how  strong the EDTA complex of the metal absorbs!
Instead, you monitor the absorption of the indicator, that is liberated from the metal-indicator complex upon addition of EDTA.

Yes but i didn't mean classical methods by that

@NdrinaLimani so what exactly did you mean? I find your question quite unclear as it stands.

Well i think you should read it more carefully, because I re-read it and I dont know any other way to explain it


text:
Difference between a dimethyl and ethyl substituent groups
I'm teaching myself organic chemistry and am struggling with differentiating between dimethyl and ethyl substituent groups.
Example: tert-butyl (1,1-dimethylethyl)
This chapter follows the one on bond-line structures, and I realize one of the rules for making a valid bond-line structure for straight-chain alkanes involves zig-zagging carbon. But I am confused on how to represent the substituent groups - is the ethyl group the V, and the methyl groups are each of the branches from the middle carbon atom of the substituent group?

Just the opposite of your color scheme, the ethyl group is red and the dimethyl (meaning 2 methyl groups) are your 2 blue lines.  BTW, we typically draw hydrocarbon structures using zig-zag lines like you've done above, but these are not resonance structures.

Sorry, I meant bond-line structures. How do we know the ethyl group is the straight line and the two methyl groups are the 'branches'? (without being given the name of the substituent group)

the longest straight chain (coming off the backbone of your main chain) in your example turns out to be 2 carbons or ethyl.  You can select it 3 different ways (center red line, bottom of center red line plus right blue line, bottom of center red line plus left blue line) but it doesn't matter, in this case you get the same result.  Whichever of the three 2 carbon fragments you select, it has two methyl groups branching off of it.

So, we use the 'central' atom of the substituent group as the reference point, and connect it to the parent group (to obtain ethyl), then 'add up' any remaining carbon bonds (in this case, only one - methyl), to properly name the substituent groups. I applied this method and successfully identified (2-methylpropyl). Thank you!

@imaginov 2-methylpropyl is a different structure, this is 2-methylpropan-2-yl, however tert-butyl is the preferred name.

If an organic compound has dimethyl in its name than that means that it has 2 methyl groups. If they are on the same carbon they form a V in the bond line structure. An ethyl group on the other hand  forms a 2 carbon unbranched zigzag in the bond line structure instead of the V that a dimethyl group forms in the bond line structure.
For the ethyl group it is standard on all bond line structures to represent it as a 2 carbon unbranched zigzag. For the dimethyl group it is standard to represent it as a V in all bond line structures.
As to how we know this it is because an ethyl group coming off of a carbon looks like propane if you just look at that carbon and the ethyl group. In fact some people in the early years of organic chemistry might have called propane ethylmethane before the prop prefix was used for alkanes. However if this was the case some of them would call propane dimethylmethane. This was because they were looking at different carbons as the central atom. If you look from one of the ends it looks like an ethyl group sticking off of methane. If you look at the center it looks like a dimethyl group sticking off of methane.

Thank you for clarifying. But I don't understand why this is the case - is this standard for all bond-line structures? How do we know this?


text:
Different elution profiles of carvacrol and thymol
I have identified carvacrol and thymol via thin-layer-chromatography and via gas chromatography
On a normal silica plate the thymol has a higher Rf-value then carvacrol. The explanation for this would be the position of the hydroxyl-group and the difference size of the nearby methyl or isopropyl group. I think thymol should be more hydrophobic then carvacrol, explaining the less interaction with the silica gel resulting in a higher Rf-value.
The problem is, that the analysis by gas chromatography with a non-polar column shows a different elution profile. I would expect that thymol, according to the higher hydrophobicity, should have a higher retention time then carvacrol. This isn't the case. Carvacrol (27.08 min) elutes after thymol (26.51 min).
Is there any explanation I'am not aware or is my thinking simple wrong?

Thymol has a lower boiling point (232 °C) than Carvacrol (237.7 °C, both from Wikipedia), I'd bet this is the reason because AFAIK the volatility has a more important effect on the separation than the column polarity in gas chromatography.

@Molx You are right, gas chromatography's separation is more influenced by volatility (related to boiling point) of the constituents of a mixture than their adsorption to the stationary phase.  In addition -- as already stated in the original question -- the column deployed in GC was a non-polar one.  Except explicitly stated differently, thin layer chromatography more often uses a polar stationary phase (silica, acid/neutral/basic alumina, for example) where different strength of adsorption is important.  On contrast, HPLC more often deploys reverse phase (less/apolar) stationary phases.

For gas chromatography, boiling point is one of the most important determinants of elution time.  For thin-layer chromtagraphy, that isn't true.
The comments from @Molx and @Buttonwood on this question expound on this point quite well.
@Molx:

Thymol has a lower boiling point (232 °C) than Carvacrol (237.7 °C, both from Wikipedia), I'd bet this is the reason because AFAIK the volatility has a more important effect on the separation than the column polarity in gas chromatography.

@Buttonwood:

... gas chromatography's separation is more influenced by volatility (related to boiling point) of the constituents of a mixture than their adsorption to the stationary phase. In addition -- as already stated in the original question -- the column deployed in GC was a non-polar one. Except explicitly stated differently, thin layer chromatography more often uses a polar stationary phase (silica, acid/neutral/basic alumina, for example) where different strength of adsorption is important. [...]


text:
Nomenclature of an Alkane
Is the correct name for the following structure 3-tert-butylhexane or 3-ethyl-2,2-dimethylhexane?
I think it is 3-tert-butylhexane, but the textbook says it is 3-ethyl-2,2-dimethylhexane. Can someone please explain?

According to IUPAC (source):

2.6 - If chains of equal length are competing for selection as main chain in a saturated branched acyclic hydrocarbon, then the choice goes in series to:
  (a) The chain which has the greatest number of side chains.

Therefore the correct name is 3-ethyl-2,2-dimethylhexane as opposed to 3-tert-butylhexane because the former has more side chains.

So straight chain parent groups aren't necessarily favored and this rule always dominates?

The parent name is always a straight chain (hexane in this example). The source I gave has the full list of rules for naming if you are still confused.

That makes sense. I did more practice problems and have understood nomenclature a little better. Thanks for your help.


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

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


text:
Why does Pb(II) promote ligand substitution?
I have heard of an article where divalent lead $\ce{PbL2}$ ($\ce{L}$ is ligand) reacts with something (again not sure, as I dont know the article) and what it reacts with gains a ligand.
In less confusing words:
$\ce{PbL2 + MXn}$ ($\ce{X}$ is another ligand, not necessarily halogen) ==> $\ce{PbL(n-1)X + MX(n-1)L}$
Or something similar. That is what I got from hearing about the article. I was wondering if you guys can share your thoughts as to why happens?

It is maybe a better question if you yourself try to clean up some of the "not sure", "or something similar" points and clean up your reaction equation (there is 2L on the left, n L on the right). Also, if you can be a little bit more specific than "why happens?"... It seems like a vaguely documented ligand exchange, which can happen in any equilibrium.


text:
Phosphorus and Boron groups
I know some sulfur groups such as sulfones and disulfides and sulfonic acids. I also know of 1 phosphorus group. That is the phosphate ion.
I don't know of any boron groups.
What is the phosphorus equivalent of aldehydes, ketones, and carboxylic acids?
Phosphals, Phosphones, and Phosphonic acids I suppose?
And what about the boron equivalent of aldehydes, ketones, and carboxylic acids?

For phosphorus, one analogy between carbon and phosphorus compounds is the following:
Carbonate -> carboxylate -> acetal* -> alcohol -> alkane
Phosphate -> phosphonate -> phosphonite -> phosphinite -> phosphine
Note that I put acetal instead of ketone/aldehyde because it is a better analogy to phosphonite.  A few cool P-containing functional groups like phosphine oxides and phosphorus ylides don't map cleanly into that analogy though, so don't think the list is exhaustive.

Please don't try to summarise your question with tags. For more information, see meta.se and the [help].

Boron-oxygen double bonds are not very stable so I doubt any direct boron analogues have been prepared.

I didn't think I was summarizing with tags. I just thought that putting in a phosphorus tag and a boron tag along with the organic chemistry tag would attract people who know about phosphorus and boron functional groups so that I could get an answer faster.

There is no such thing like phosphals, phosphones and its boron equivalent but phosphonic acid and boronic acid do exist.


text:
What is the color and structure of CrO3?
What is the structure and color of $\ce{CrO3}$?
Also, is it possible to produce $\ce{CrO3}$ when dichromate is heated with sulphuric acid?

According to its Wikipedia page:
Color
$\ce{CrO3}$ is a dark-purple solid under anhydrous conditions, that turns bright orange when wet.
Production
$\ce{CrO3}$ is generated, as you said, by the following reaction:
$$\ce{H2SO4 + Na2Cr2O7 -> 2 CrO3 + Na2SO4 + H2O}$$
Structure
Its monomeric structure is predicted to be pyramidal like so:

Its crystalline structure is:

Each chromium center shares two oxygen centers with neighbors, but two oxygen atoms are not shared, giving the overall stoichiometric ratio of 1:3.


text:
Why does raising the temperature affect the mass of dissolved solute?
Consider a solution of $\ce{KOH}$ which is in equilibrium with undissolved $\ce{KOH}$ at $25\ \mathrm{^\circ C}$.
After raising the temperature to $50\ \mathrm{^\circ C}$, the mass of undissolved $\ce{KOH}$ decreases. Why is this?

For many (though not all) solids dissolved in water, solubility increases with temperature. The increase in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the intermolecular forces of the solute molecules.
For many salts and some gasses there are visual representations of solubility variation with temperature called a solubility curve, for which the units are $\frac{\mathrm{g~solute}}{\mathrm{100~g~solvent}}$. I could not find one for $\ce{KOH}$, but the Wikipedia page for $\ce{KOH}$ lists the solubility of $\ce{KOH}$ at various temperatures, and it is clear the trend is that it is more soluble at higher temperatures.
As a result of Le Châtelier's principle, substances that release heat upon being dissolved tend to dissolve better in colder water. This accounts for gasses and some salts.
You should note, however, that although $\ce{KOH}$ does evolve a large amount of heat upon being dissolved, as the concentration of $\ce{KOH}$ increases, the process becomes less exothermic, and eventually near the saturation point it becomes endothermic (it becomes less favorable to dissolve more $\ce{KOH}$), which explains why it is more soluble at higher temperatures.

How can I relate the solubility with decreasing the mass of dissolved solute?

@user16139 edited to reflect your question

Apparently $\ce{Ca(OH)2}$ has an exothermic heat of dissolution, yet its solubility is very low at room temperature and decreases further with increasing temperature. This seems to contrast with the explanation for the temperature behaviour of $\ce{KOH}$. Would you happen to know why?

I would suggest that $\ce{Ca(OH)2}$, unlike $\ce{KOH}$, becomes more exothermic or remains as exothermic at higher temperatures, and is therefore affected by Le Châtelier's principle in a way that makes it less soluble at higher temperatures.


text:
Stability constant of complexes
I know that this constant tells us the strength of the interaction between the metal and the ligand,  so should it also mean that the higher the constant, the stable the complex ?
I need to make sure of it because in photometric titration, all that matters is the stability of complex, and if the constant is given, one should know which complex is formed earlier.

As $K$ is a thermodynamic constant, and if you define your reaction as
$A + B \rightarrow{} C + D$
leading to
$K = \frac{c_C \cdot{} c_D}{c_A \cdot{}c_B}$
then -- thermodynamically -- yes.  Especially for chelate complexes (like with EDTA), entropically favoured. This answer excludes (macroscopic) kinetics


text:
Can two metals bond covalently?
Can metal ions be engineered to behave like non-metals in terms of chemical properties (by supplying or removing electrons), would the altered metal ions bond covalently?
Would the ions be too unstable and begin to decay?

There are quite a lot covalently bound metals to metals already, and I do not understand what you mean with engineering metal ions; Could you exlain a little more?

some examples of these molecules and/or covalent networks would be appreciated. also by 'engineering' i mean creating an ion that behaves in a specific chemical manner due to the number of valence electrons it possesses by removing or adding electrons from the valence shell.

The most ready answer to this question is $\ce{Hg_2^{2+}}$ and other polycations of mercury.


text:
How do I create a spreadsheet containing the coordinates and predicted bond angles, molecular properties, torsion properties, etc of a molecule?
On a previous answer here (https://chemistry.stackexchange.com/a/30841/876) I was informed in the comments that OpenBabel & Python terminal could be used to print the data contained in the property matrices of Avogadro (more info can be found here How do I print property matrices in Avogadro?). I should be clear, however, most of the files I want to be able to do this for are CIF files where the bond angles and lengths are not explicitly mentioned in the files, so these properties may need to be predicted by the software. So I felt I couldn't be the only person interested in this so I thought I'd ask this question on this site.
I would like any answer to be suitable for those operating on Windows machines, if possible. I have Python & OpenBabel installed on my 64 bit Windows 7 PC, so I believe this shouldn't be a major issue, hopefully.

You want the Open Babel "report" format, e.g.:
obabel benzene.cml -oreport

FILENAME: benzene.g03
FORMULA: C6H6
MASS: 78.1118
EXACT MASS: 78.0469502
INTERATOMIC DISTANCES

              C   1      C   2      C   3      C   4      C   5      C   6
              ------------------------------------------------------------------
   C   1    0.0000
   C   2    1.3975     0.0000
   C   3    2.4207     1.3975     0.0000
   C   4    2.7952     2.4205     1.3976     0.0000
   C   5    2.4205     2.7948     2.4205     1.3975     0.0000
   C   6    1.3976     2.4205     2.7952     2.4207     1.3975     0.0000
   H   7    1.0841     2.1549     3.4030     3.8793     3.4029     2.1549
   H   8    2.1548     1.0841     2.1548     3.4028     3.8789     3.4028
   H   9    3.4030     2.1549     1.0841     2.1549     3.4029     3.8793
   H  10    3.8793     3.4029     2.1549     1.0841     2.1549     3.4030
   H  11    3.4028     3.8789     3.4028     2.1548     1.0841     2.1548
   H  12    2.1549     3.4029     3.8793     3.4030     2.1549     1.0841

              H   7      H   8      H   9      H  10      H  11      H  12
              ------------------------------------------------------------------
   H   7    0.0000
   H   8    2.4816     0.0000
   H   9    4.2984     2.4816     0.0000
   H  10    4.9634     4.2982     2.4817     0.0000
   H  11    4.2982     4.9630     4.2982     2.4816     0.0000
   H  12    2.4817     4.2982     4.9634     4.2984     2.4816     0.0000

ATOMIC CHARGES
   C   1   -0.1860370000
   C   2   -0.1859930000
   C   3   -0.1860370000
   C   4   -0.1860370000
   C   5   -0.1859930000
   C   6   -0.1860370000
   H   7    0.1860240000
   H   8    0.1860180000
   H   9    0.1860240000
   H  10    0.1860240000
   H  11    0.1860180000
   H  12    0.1860240000

BOND ANGLES
   6    1    7  Car  Car    H    120.003
   2    1    6  Car  Car  Car    119.993
   2    1    7  Car  Car    H    120.004
   1    2    3  Car  Car  Car    120.014
   1    2    8  Car  Car    H    119.993
   3    2    8  Car  Car    H    119.993
   4    3    9  Car  Car    H    120.003
   2    3    4  Car  Car  Car    119.993
   2    3    9  Car  Car    H    120.004
   5    4   10  Car  Car    H    120.004
   3    4    5  Car  Car  Car    119.993
   3    4   10  Car  Car    H    120.003
   4    5   11  Car  Car    H    119.993
   6    5   11  Car  Car    H    119.993
   4    5    6  Car  Car  Car    120.014
   5    6   12  Car  Car    H    120.004
   1    6    5  Car  Car  Car    119.993
   1    6   12  Car  Car    H    120.003

TORSION ANGLES
  11    5    4   10     -0.000
  11    5    4    3   -180.000
   6    5    4   10   -180.000
   6    5    4    3     -0.000
  11    5    6   12      0.000
  11    5    6    1   -180.000
   4    5    6   12   -180.000
   4    5    6    1     -0.000
   5    4    3    9   -180.000
   5    4    3    2     -0.000
  10    4    3    9     -0.000
  10    4    3    2   -180.000
   5    6    1    7   -180.000
   5    6    1    2     -0.000
  12    6    1    7     -0.000
  12    6    1    2   -180.000
   6    1    2    3     -0.000
   6    1    2    8   -180.000
   7    1    2    3   -180.000
   7    1    2    8     -0.000
   4    3    2    1      0.000
   4    3    2    8   -180.000
   9    3    2    1   -180.000
   9    3    2    8     -0.000

UPDATE:
You mentioned wanting a CSV-parsable output, e.g., from Python.
I wrote up a fairly quick tool for this molcsv.py
It should be pretty obvious about modifying as needed in the different for loops.

Thanks, one thing though is there a way to create a CSV file of this, in an automated manner (i.e., without me having to copy-paste everything)?

Yes, I can probably whip up a script, it'll probably be tonight though.

@BrentonHorne OK, give it a shot. It's not entirely obvious how you want the output but with a bit of Python it's easy for you to modify as needed.

OK, please forgive my ignorance when it comes to programming & Avogadro but how do I use this to create a csv? Do I like open a CIF in Avogadro & run the file from the Python terminal therein? If so where do I place the file such that the avogadro python terminal will recognize it?

This is not an Avogadro script. If you have Open Babel and Python/Pybel support installed, you'd run  python molcsv.py file.cif from the command-line. If you have other questions, it's probably easier to handle by e-mail.

Load your CIF file in PLATON and run CALC-GEO
In the *.LIS file generated, you'll find all the tables that you need, namely

Bond Lengths (Angstrom)
Bond/Valence Angles (Degrees)
Torsion/Dihedral Angles (Deg.)

You can still use the regex module of Python to parse the data,\s*(?P<atoms>([A-Z][a-z]?\(\d*\w\)\s*){4})\s*(?P<value>-?\d*\.\d\(\d+\)) will split entries in the torsion table into two named groups, which you can further process and write to a CSV file.

EDIT
I think that OpenBabel is an excellent tool! In this case, however, I suggest to use the approach above, rather than parsing the OpenBabel report file for the following reasons:

In a couple of cases, I noticed that the report generator of OpenBabel had some problems with CIF files that were checked both using the internal check routines of PLATON, as well as  CheckCIF, an online tool provided by the IUCr, and were found to be valid.
In the table for the dihedral angles, the OpenBabel report does not use descriptors like C(5), H(10A) to denote atoms in a dihedral, but uses the atom indices. If you would want to write the decriptors to a CSV file, you would have to look them up at another table.

If you can live with the indices, the following Python snippet should suffice:
#!/usr/bin/env python3

import csv
import argparse

def get_torsions_from_report(report_path):
    '''Read all rows with he indices to four consecutive atoms and
       the corresponding dihedral angle from an OpenBabel report file.'''
    try:
        torsion_list = []
        with open(report_path) as infile:
            for line in infile:
                if line.startswith('TORSION ANGLES'):
                    break

            for line in infile:
                if line.strip():
                    torsion_list.append(line.split())

        return torsion_list
    except IOError:
        print(e)

def write_torsions_to_csv(csv_path, torsion_list):
    '''Write torsion data to CSV.'''
    with open(csv_path, 'wt') as fout:
        writer = csv.writer(fout)
        writer.writerow(('Atom1', 'Atom 2', 'Atom 3', 'Atom 4', 'Torsion'))
        writer.writerows(torsion_list)

if __name__ == '__main__':
    parser = argparse.ArgumentParser(description='Read torsions
    from OpenBabel report and write them to a CSV file.')
    parser.add_argument('-i', dest="infile", required=True,
    help="OpenBabel report file")
    parser.add_argument('-o', dest="outfile", required=True,
    help="CSV file")

    args = parser.parse_args()
    torsions = get_torsions_from_report(args.infile)
    write_torsions_to_csv(args.outfile, torsions)

An outstanding answer; isn't it CALC GEOM PLATON command instead of CALC-GEO?


text:
Mechanism of photolysis of HX
What is the mechanism for the photolysis of $\ce{HX}$ to dihydrogen and dihalogen ? From the low quantum yield (0.3% for $\ce{HI}$ according to one paper) it seems unlikely to be a chain process but I am not sure why it would not be.

Can you reference the paper? I am absolutely uncertain about this, but my guess would be, that there are just not enough right pairings to form dihalogen and dihydrogen.

the question is pretty much based on an incorrect premise as it turns out the actual quantum yield is closer to one - I was incorrectly reading the quantum yield for triodide formation. I've found a mechanism here: https://books.google.co.uk/books?id=o24m1CLMR4cC&pg=PA253#v=onepage&q&f=false


text:
How to calculate the percentage of calcium carbonate in seashells in an experiment?
We have a investigation in my chemistry class in which we must calculate the amount of calcium carbonate in seashells. We can have as many sea shells as we wish, lab equipment, and hydrochloric acid of varying concentration from $\pu{0.5 mol dm-3}$ to $\pu{2 mol dm-3}$. How is it to be done?

$\ce{CaCO3}$ reacts with $\ce{HCl}$ in the following reaction:
$$\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) +CO2(g)}$$
In my opinion, the easiest way to measure the results of this reaction would be to collect the $\ce{CO2}$. Grind up the seashell into a powder, record the mass of this. Assemble an apparatus like this, and react the seashell with an excess of $\ce{HCl}$ in the reaction chamber to collect and measure the volume of gas evolved (docbrown.info):

As this is a gas law calculation, you will need the atmospheric pressure and ambient air temperature at the time you perform the reaction. After obtaining the final volume of the gas in the reaction, you will then need to account for the partial pressure due to water vapor, and subtract this from the atmospheric pressure accordingly. Now this is simply a matter of stoichiometry, calculating how much $\ce{CaCO3}$ would be required to obtain this volume of gas at this temperature and pressure.
You can then use the equation RobChem stated:
$$\left(\frac{m({\ce{CaCO3})}}{m(\mathrm{shells})}\right) \times 100\% = \mathrm{Percent}~\ce{CaCO3}$$
to find the percent $\ce{CaCO3}$ by mass in the seashell.

$\ce{CaCO3}$ is basic so it will neutralize the acid.
$$\ce{CaCO3 + 2HCl -> CaCl2 + H2O +CO2}$$
In your experiment you need to devise a way of finding out what the end point of the reaction is (using a $\mathrm{pH}$ indicator to find when the mixture is neutral). You need to be able to measure the amount of hydrochloric acid used in moles  (you should measure the volume and you know the concentration. The number of moles of water is twice the amount of moles of the calcium carbonate (from the chemical equation above).
You should measure the mass of the amount of shell you use at the start. You will be able to figure out the mass of the calcium carbonate in that sample of shells that you use (since you now know the number of moles of calcium carbonate). Then:
$$\left(\frac{m({\ce{CaCO3})}}{m(\mathrm{shells})}\right) \times 100\% = \mathrm{Percent}~\ce{CaCO3}$$
You should be able to do it with just one concentration but your teacher probably wants to talk to you about reliability and experimental errors.


text:
Predicting melting points of substances based on their molecular/atomic structure
Wouldn't it be great to be able to predict the melting point of a metal based on the atomic number and not by looking in some table for the value that was measured experimentally by someone else? Or to predict the same melting point for a simple salt like NaCl or KI by applying some formula to the atomic numbers of the 2 atoms involved?
Can you help me understand why is it so difficult to build a theoretical model for this particular physical property? Were there attempts made in the scientific world for such a model?
--EDIT--
Yes, this is part of my question:
I want to understand if there is a connection between the theoretical explanations of properties like melting point and the current theoretical model of atomic structure. Can the electronic structure of $\ce{Fe 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2}$ tell us anything about it's probable melting point without measuring that point in the first place?

Melting point depends not only on the structure of the molecule itself, but also on the structure of the solid (crystalline) phase and molten/liquid state of the substance. That is a much bigger computational problem than calculating energetics for a small molecule or ensemble. It is not easy to predict how a molecule (especially a complex one) will crystallize, and even if you do know that, a big computation to calculate the energy of the crystalline state.

I suspected that much, but can you help me understand why? "It's difficult" does not enlighten me one bit. What does "calculating energetics for a small molecule or ensemble" mean?

One way to think about melting point is in terms of the energy difference between the solid and liquid phases of a substance. There are a range of ways to calculate the energy of a chemical system, but for a bulk material, even the simplest of these requires an energy calculation for each of a long series of geometrical configurations of the atoms and molecules in your system. The complexity of the computation increases, the bigger the molecule and the more of them that are present in the system you are calculating the energy of. You might want to check out "molecular mechanics" on Wikipedia.

Thanks. "Molecular mechanics" was what I needed to know :) .

With regard to organic compounds, you might be interested in group contribution methods, e.g. the Joback method.
These methods divide a molecule into functional groups or similar small building blocks and estimate the melting point (and other thermodynamic properties) based on the contributions of the individual groups.

do you know how this model was built? Was it statistically inferred from a smaller database and and then verified on a bigger data set? Or did they propose some theoretical model (of quantum level interactions, let's say) and then calculated the individual group contributions and built the formula?

@Liviu-AurelianRau-Neacsu These are statistical methods. It is way more easy to do that way than create universal theoretical models from ab initio or dft theories.


text:
For a given element do heavier isotopes form stronger bonds than lighter isotopes?
I know that $\ce{C-D}$ and $\ce{C-T}$ bonds are stronger than $\ce{C-H}$ bonds, but is this generally true for isotopes of all elements? (By isotopes I mean stable ones with large half lives.)

Isovalent isotopes will have the same force constant. However the different masses of the isotope will affect the position of the vibrational state in its potential well. You can rationalise the difference in well depth briefly using the vibrational frequencies of a classical oscillator as the harmonic approximation to the asymmetric well for low lying states. If the stretching frequency is given by;
\begin{equation}
\nu \propto \frac{1}{\sqrt{\mu}}, \ \ \ \ \ \ \ \ where \ \ \ \ \ \ \ \mu =\frac{m_1\cdot m_2}{m_1+m_2}
\end{equation}
Then as the reduced mass increases the frequency decreases and the vibrational energy decreases $E=\hbar\omega $ with  $\omega\cdot 2\pi=\nu$. We can see that if we were to double the reduced mass then the energy would decrease by root 2.
So we need to examine the reduced masses of some simple systems, lets say the diatomics you suggest. If we let the masses of $H=1$, $D=2$, and $T=3$ then we calculate the reduced masses to be $1/2$, $2/3$ and $3/4$. The reduced mass is increasing as we go along the series and hence we expect the successive molecules to have stationary states lower in the well.
In vibrational spectroscopy this is best seen by the decrease in zero point energies for HCl and DCl. It is often easiest to take the ratios of the frequencies $\omega _{HCl}/\omega _{DCl}$, where using $m_{Cl}=35$ and H and D as above we obtain 1.395. indicating that the deuterated molecule is deeper in its well.
Edit
I have been advised to explain zero point energy. In short it is a irremovable quantity of energy from a quantum oscillator. The classical oscillator has no parallel to this. It is a manifestation of the uncertainty principle in position and momentum. If you want the truth of where it comes from you will need to consult the physics SE, but I will do my best :)
The quantum oscillator has a Hamiltonian of the classical oscillator except just replace the classical position and momentum variables with their quantum operator analogues $p\mapsto \hat p$ and $x\mapsto \hat x$.
The classical Hamiltonian (or total energy) is learnt at school to be kinetic + potential,
\begin{equation}
H=\frac{p^2}{2\mu}+\frac 12 kx^2
\end{equation}
For the reduced two body system. It is a postulate of quantum mechanics that in the position representation (just a way of formulating things) that wherever you see a $p$ and an $x$ you replace them by $\hat p$ and $\hat x$ where,
\begin{equation}
\hat p =-i\hbar \frac{\partial }{\partial x}, \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \hat x =\cdot x
\end{equation}
Therefore with these changes in the above classical expression we generate the quantum Hamiltonian of the quantum harmonic oscillator,
\begin{equation}
\hat H=\frac{\hbar^2}{2\mu}\frac{\partial ^2}{\partial x^2}+\frac 12 kx^2
\end{equation}
Where $i=\sqrt -1$ so $i^2=-1$ and $--1=1$.
It is another principle of quantum mechanics that the Schrodinger equation is the equation to solve for quantum systems, just like in classical mechanics you use Newton's second law, $F=ma$, This equation has a bit of a different shape but in essence (decoupled from time) takes the following form.
\begin{equation}
\hat H\psi =E\psi
\end{equation}
Substituting the above expression for $\hat H$ in and doing a bit of maths (wayyyy to advanced for us here) we can solve for the energy $E$ in the Schrodinger equation, when we do this we get,
\begin{equation}
E_n=(n+\frac 12)\hbar \omega
\end{equation}
What this tells us is that the vibrational levels are not continuous like the classical oscillator (above) but can only take discrete values. We see in the maths of the solution for $E$ the need for discrete integers $n$ which we call quantum numbers. These may take in this case values of $0,1,2,...$ etc etc. Each corresponding to the next adjacent energy solution.
Something interesting happens if we choose $n=0$ however. You would think that the lowest energy would be zero right? well no... quantum mechanics says we can never remove a little bit of energy from the system. (try substituting $n=0$ in the above expression and we see that the answer is non zero (1/2 $\hbar \omega$).
This is a zero point energy.
As promised here is the home-made graph of the molecular potential energy. Im sorry its not very good but honestly the best I could do with the available tools! Apologies.
The main points to follow are:

At the base of the well the graph looks like a parabola, i.e. the harmonic oscillator approximation to the molecule works best at low lying states since the anharmonicity term grows with excitation (well increasing quantum number).
Where the harmonic oscillator is valid you can see that the energy levels of are evenly spaced, each one corresponds to the quantum numbers $n$ as given above. You can calculate $\Delta E=E_{n+1}-E_{n}$ to get the difference in energy as $\frac 12\hbar \omega$. In this approximation the equilibrium bond length does not change with excitation. However we can see it does in the anharmonic case.
To answer your question, the energy levels corresponding to the heavier isotope are the blue ones since they sit deeper in the well. So you could imagine the blue ones as C-D while the red ones would be C-H. (normally you would just be looking at one species per graph, so only one set of coloured lines.

Maybe it'd be useful if you could expand slightly on zero point energy for a harmonic oscillator (add the formula?), and perhaps add one of those graphs with potential energy versus internuclear separation, with the vibrational levels shown for bonds involving different isotopes.

Sure thing, are we allowed to use images off the web in answers? :)

Why wouldn't we? If the image isn't generic, you can always cite the source!

I strongly suggest not using images off the web due to copyright concerns; the major exceptions are images on Wikimedia Commons which are freely licenced. "Generic" images may well still be copyrightable.

@HritikNarayan I have updated the answer with the requested graph, I feel it goes slightly off topic but it is just background information to help you, the main part of the answer is above the edit section. Sorry the graph is so shoddy but I'm terrible at art! hope that helps :)

@AngusTheMan I'm not sure if this is helpful, but do check out Desmos Graphing Calculator, for graphing functions.

When comparing isotopes, the different nuclear masses control the bond length and bond strength.
The radius of the $n^\text{th}$ Bohr orbit is given by
$$r_{n} = {n^2\hbar^2\over Zk_\text{c} e^2 m_\text{e}}$$
where $Z$ is the atom’s atomic number, $k_\text{c}$ is Coulomb’s constant, $e$ is the electron charge, and $m_\text{e}$ is the mass of the electron. In this equation it is assumed that the electron orbits the nucleus and the nucleus remains stationary. Given the mass difference between the electron and nucleus, this is generally a reasonable assumption. However, in reality the nucleus does move too. It is relatively straightforward to remove this assumption and make the equation more accurate by replacing $m_\text{e}$ with the electron’s reduced mass, $\mu_\text{e}$
$$\mu_\text{e} = \frac{m_\text{e} \cdot m_\text{nucleus}}{m_\text{e} + m_\text{nucleus}}$$
Now the equation for the Bohr radius becomes
$$r_n = {n^2\hbar^2\over Zk_\text{c} e^2 \mu_\text{e}}$$
Since the reduced mass of an electron orbiting a heavy nucleus is always larger than the reduced mass of an electron orbiting a lighter nucleus
$$r_\text{heavy} < r_\text{light}$$
and consequently an electron will orbit closer to the nucleus of a heavier isotope than it will orbit the nucleus of a lighter isotope. In turn, this means that the heavy isotope will have shorter, stronger bonds.

Does the heavy isotope necessarily have a shorter bond length? Taking HCl [http://webbook.nist.gov/cgi/cbook.cgi?ID=C7647010&Mask=1000#Diatomic] and DCl[http://webbook.nist.gov/cgi/inchi?ID=C7698057&Mask=1000#Diatomic] as the common examples, NIST has their bond lengths as nearly equal despite the large difference in reduced mass. In fact, DCl actually has a slightly longer equilibrium bond length.

@Tyberius Both answers indicate that the compound with the heavy isotope will be lower in the potential well and hence have a shorter bond length.  The NIST data is pretty old.

But lower in the potential well  the equilibrium bond length shouldn't change significantly. Its effectively a harmonic potential when you are deep in the well, so shifting it slightly down shouldn't change anything. I also found another link with more recent values that also includes $\ce{H^37Cl}$ and $\ce{D^37Cl}$ and it too shows larger bond lengths for the cases with larger isotopes.

@Tyberius Look at the primary isotope effect with carbon, $\ce{K_H > K_D}$, the $\ce{C-D}$ bond is shorter and stronger.  This second reference is pushing 30+ years from when the data was actually collected.  Also, the effect is just harder to see (smaller) in $\ce{HCl}$ because the effect of the reduced mass is much less than with lighter nuclei such as carbon.

Even 30 years ago, this level of quantum mechanics would have been known and if I were them I wouldn't bother publishing results that not only don't reflect what you say should be the observed trend, but actually show the exact opposite. I'm fine with the idea $\ce{C-D}$ is shorter, but does the heavier isotope have to be shorter? Alternatively, I know it should usually be stronger, but does stronger = shorter always? It seems like there are very few cases where the observed $\ce{X-D}$ bond length is measurably shorter.

In particular from the link, I think the $\ce{O-D}$ and $\ce{O-H}$ bond lengths and bond energies give an interesting example of a case where the bond energy is different, but the bond length is not.  In my mind, this makes sense as the molecules could have the same equilibrium bond length while having different vibrational zero point energies, leading to equal bond lengths, but a stronger bond for $\ce{O-D}$.

@Tyberius  Symmetry tells us that if things are not the identical then they must be different.  Therefore an $\ce{X-H}$ bond cannot have the same length, strength or any other property as an $\ce{X-D}$ bond.  If an experimental measurement suggests the bond lengths are the same, then the measurement must be made with greater precision - they cannot be the same.

Let us continue this discussion in chat.


text:
Calculating new volume during expansion of an ideal gas

Question
  $\mathrm{5~mol}$ of a gas is confined in a piston (volume $\mathrm{100~cm^3}$) at $\mathrm{400~K}$. $\mathrm{10~kJ}$ of heat energy is supplied, causing a reversible and isothermal expansion of the gas. Calculate the volume of the expanded, heated, gas. You may use the equation below (assume ideal gas behaviour).
$$W = -nRT(\ln V_2-\ln V_1)$$

I substituted the values (after converting to SI units) into the equation and attempted to rearrange to find $V_2$, but work was unknown. I tried to use the enthalpy value (10 kJ) given but I was unsure of what to do with it.

I dont understand where you are getting stuck up since everything is already given to you and you just have to substitute stuff using:
$\Delta U= W+Q$, the basic conservation equation.
$Q$ is the heat supplied. Think about the change in internal energy ($U$) during an isothermal process.


text:
What is the enthalpy of formation of 1 mol of sulfuric acid?

In the contact process, the commercial process of producing $\ce{H2SO4}$, several reactions occur before the final product is produced:
\begin{align}
\ce{S8(s) + 8O2(g) &-> 8SO2(g)} &\Delta H_\mathrm{f} &= \pu{-70.9 kcal mol^{-1}}\tag{1} \\
\ce{2SO2(g) + O2(g) &->[V2O5] 2SO3(g)} &\Delta H_\mathrm{f} &= \pu{-94.5 kcal mol^{-1}}\tag{2} \\
\ce{SO3(g) + H2O(g) &-> H2SO4(g)} &\Delta H_\mathrm{f} &= \pu{-193.9 kcal mol^{-1}}\tag{3}
\end{align}
Given that $\Delta H_\mathrm{f}$ for $\ce{H2O(g)}$ is $\pu{-68.3 kcal mol^{-1}}$, what is the enthalpy of formation of $\pu{1 mol}$ of $\ce{H2SO4}$?

What I've tried to do:

Summing all the enthalpies, I arrived at to $\Delta H_\mathrm{f} = \pu{-427.6 kcal mol^{-1}}$.
Using the data from equation $(3)$, I conclude that the answer is $\Delta H_\mathrm{f} = \pu{-193.9 kcal}$.

The second point in what you have tried that is not $d$H$_f but  only dH

@DSinghvi actually the question stated like this  (given that the standard formation enthalpy of SO2, SO3, H2O, and H2SO4 consecutively are -70,9 kkal/mol, -94,5 kkal/mol, -68,3 kkal/mol, and -193,9 kkal/mol) I moved the enthalpy to the reaction.

@DSinghvi By the way, from your response, that means I have to conclude every steps In order for me to have the ΔHf of H2SO4?

what do you understand by $/Delta H_f/$

The enthalpy of the second and third reaction should not be labeled formation enthalpy because the reactants at not elements (and in the second reaction, the stoichiometric coefficient of the product is different from one).

Since enthalpy is a state function, all that is needed is the $\Delta H_{\mathrm{f}}$ of each reactants that form $\ce{H2SO4}$. Multiply the equations through by the correct fractions (including the enthalpies) to result in the correct coefficients and enthalpy for the desired reaction:
$$\ce{SO3{(g)} +H2O{(g)} -> H2SO4{(g)}}~~~~~\Delta H_{\mathrm{f}}=-193.9~\mathrm{kcal~mol^{-1}}$$
$$\ce{H2{(g)} + \frac{1}{2}O2{(g)}->H2O{(g)}}~~~~~\Delta H_{\mathrm{f}}=-68.3~\mathrm{kcal~mol^{-1}}$$
$$\ce{SO2{(g)} + \frac{1}{2}O2{(g)} \xrightarrow{\ce{V2O5}} SO3{(g)}}~~~~~\Delta H_{\mathrm{f}}=-47.25~\mathrm{kcal~mol^{-1}}$$
$$\ce{\frac{1}{8}S8{(s)} +O2{(g)} -> SO2{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-8.8625~\mathrm{kcal~mol^{-1}}$$
Combining these equations yields:
$$\ce{\frac{1}{8}S8{(s)} +2O2{(g)} +H2{(g)} -> H2SO4{(g)}}~~~~~~~~~~\Delta H_{\mathrm{f}}=-318~\mathrm{kcal~mol^{-1}}$$
(according to the equations and numerical values provided)


text:
What is formed when you leave iron(II) sulfate in plain air?
What is formed when you leave iron(II) sulfate in plain air?
Knowing that iron(II) is easily oxidised to iron(III), and assuming that the reactive component of air is oxygen, I solved it this way:
$$\ce{FeSO4 + O2 -> Fe2O3 + SO2}$$
But in my textbook it's given that iron(III) sulfate, $\ce{Fe2(SO4)3}$, is formed. Is this because $\ce{Fe2O3 + SO2}$ react to form $\ce{Fe2(SO4)3}$?

Usually, the original iron(II) sulfate is present as green $\ce{FeSO4.7H2O}$.
Oxygen can oxidize $\ce{Fe(II)}$ salts to $\ce{Fe(III)}$ salts; for example, at $\mathrm{pH}=0$:
$$\begin{alignat}{2}
\ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}\\
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad &&E^\circ = +1.229\ \mathrm{V}
\end{alignat}$$
$\ce{Fe(II)}$ is even easier oxidized under alkaline conditions; for example, at $\mathrm{pH}=14$:
$$\begin{alignat}{2}
\ce{FeO(OH) + H2O + e- \;&<=> Fe(OH)2 + OH-}\quad &&E^\circ = -0.69\ \mathrm{V}\\
\ce{O2 + 2H2O + 4e- \;&<=> 4OH-}\quad &&E^\circ = +0.401\ \mathrm{V}
\end{alignat}$$
However, you cannot simply oxidize iron(II) sulfate $\left(\ce{FeSO4}\right)$ to iron(III) sulfate $\left(\ce{Fe2(SO4)3}\right)$ in dry air since you would need additional sulfate to balance the equation:
$$\ce{2FeSO4 + SO4^2- -> Fe2(SO4)3 + 2e-}$$
Nevertheless, $\ce{Fe(II)}$ can be oxidized to $\ce{Fe(III)}$.
The resulting $\ce{Fe(III)}$ is subject to hydrolysis. By way of comparison, the ion $\ce{[Fe(H2O)6]^3+}$ is only stable under strong acidic conditions. Already at $\mathrm{pH}=0{-}2$, it turns into yellow $\ce{[Fe(OH)(H2O)5]^2+}$ and begins to form $\ce{[Fe(OH)2(H2O)4]+}$:
$$\begin{align}
\ce{[Fe(H2O)6]^3+ \;&<=> [Fe(OH)(H2O)5]^2+ + H+}\\
\ce{[Fe(OH)(H2O)5]^2+ \;&<=> [Fe(OH)2(H2O)5]+ + H+}\\
\end{align}$$
Further addition of base causes precipitation of amorphous iron(III) hydroxide.
Accordingly, the likely product when iron(II) sulfate is oxidized is basic iron(III) sulfate, i.e. approximately:
$$\ce{4FeSO4 + O2 + 2H2O -> 4Fe(OH)SO4}$$
However, the real weathering and aging of iron(II) sulfate in dry air actually yields a mixture of various compounds, including iron(III) sulfate and iron(III) oxide-hydroxide.
You may observe this reaction in some iron fertilizers for lawns. The fresh product typically contains green iron(II) sulfate, which gradually becomes yellow.

From wikipedia:

Upon exposure to air, it oxidizes to form a corrosive brown-yellow coating of basic ferric sulfate, which is an adduct of ferric oxide and ferric sulfate: $$\ce{12FeSO4 + 3O2 -> 4Fe2(SO4)3 + 2Fe2O3}$$

The Pubchem page for ferrous sulphate says that this reaction occurs rapidly in moist air and that the rate of oxidation is increased by increasing the pH, temperature, or by exposure to light. However, I don't have access to the original sources.
As @Greg pointed out, oxygen appears to be acting as the oxidiser here and it forms $\ce{Fe2O3}$. There is not enough sulphate present in the starting materials to form only $\ce{Fe2(SO4)3}$ and hence the oxide is formed as well.


text:
Energy of Eclipsed Interactions
I am asked to identify the highest and lowest energy conformations for the following bond-line structure (I think it is 2,2-dimethylbutane). I drew a Newman projection (for all 6 conformations) and was able to identify the lowest energy conformation correctly but the textbook solution for the highest energy conformation does not match what I drew. It seems as though the textbook solution is missing a methyl group, and has 3 $\ce{H}$ bonds instead. Can someone please clarify?

Well methyl groups don't fly away, so if that is truly what the book says it must have an error in it. Could you copy the actual text from the book please?

My camera makes the pictures in the textbook very blurry, so I've tried to draw the images to the best of my ability.

Can you then at least refer us to the textbook and the pages your question is concerned with?

Sure, the book is Organic Chemistry 2e by Klein and the question is on pg 165.

You are correct in drawing this. The "answer" molecule depicted by your book is different from the one it first depicted, so it cannot be right.
You correctly depicted the high energy conformation of the molecule as a rotation of the low energy that shows the methyl groups aligned. This is high energy because of the intramolecular repulsion of the hydrogen atoms and the electron clouds.


text:
Reference request for the number of solvents around a solute
I want to know if there is a way to find number of solvent molecules that are directly embracing a solute molecule (something like a reference or table that shows waters of hydration).
Particularly I am interested in solvents such as water or ethanol, and the solutes acetone, acetaldehyde, 1-butanol, methylene blue, benzene, anthracene, and phenanthrene.

This is too broad. For a single atom ion this would sort of be answerable. But there are large molecules like proteins or polymers where the solute molecule could contact hundreds (or more) of solvent molecules.


text:
Why is it the middle ring of anthracene which reacts in a Diels–Alder?
I'm wondering why maleic anhydride adds to the middle cycle of anthracene, and not the outer two.

I would have expected that a Diels–Alder with the outer ring would be better, because I expected a naphtalene part to be lower in energy than two benzene parts (more resonance stabilisation).
I guess it has to do with reactant based arguments that the atomic coefficients for the two center carbon atoms (C-9 and C-10) are higher than from the outer cycle (C-1 and C-4). I ran a calculation using http://www.chem.ucalgary.ca/SHMO and the coefficients on C-9 and C-10 were 0.44, whereas those on C-1 and C-4 were only 0.31.

Clar rule, see https://en.wikipedia.org/wiki/Polycyclic_aromatic_hydrocarbon#aromacity

To provide a reason for the observed regioselectivity, it is helpful to draw anthracene's aromatic π-electron system in alternance of single and double bonds.  In this instance, it is more beneficial than "the ring" symbolizing the delocalised electron system, as this helps you to account for the precise number of π-electrons before the reaction (starting materials), during the reaction (the mechanism), and after the reaction (the product).

For the Diels–Alder reaction, you may imagine two different pathways.  I invite you to draw the mechanisms by yourself:

A reaction that involves carbon atoms #1 and #4 (or #5 and #8).  Possible, by mechanism.  There are five double bonds remaining in conjugation, and you count one six-membered ring in the state of "a benzene ring" (the very left one).  In the very right six-membered ring, there is only a single double bond, too.
Alternatively, a Diels–Alder reaction with carbon atoms #9 and #10.  This is more favourable then the former example, because both the very left as well as the very right ring of the product are in the state of "a benzene ring", each of these rings contain 6 π-electrons, the prerequisite of a Hückel aromatic compound.  It is so much favourable to the former, that this is the reaction observed.

It may be helpful to add that benzene, naphthalene and anthracene are of course Hückel-aromatic compounds; with 6, 10 or 14 π-electrons they fit into the rule of $(4n + 2)$.  Yet gradually, as experimentally found, in this group of three, benzene is the most, anthracene the least aromatic compound.

In general, the reactions of anthracene almost always happen on the middle ring: https://en.wikipedia.org/wiki/Anthracene#Reactions


text:
Andersen thermostat in MD with rigid bond constraints
I have been writing a custom MD code for some time. I have been using the Andersen thermostat, mainly because its ease of use, and the fact that I am not concerned with gathering dynamical data, but rather free energies. I have implemented the SHAKE algorithm found in LAMMPS. However, the thermostat seems to be undercooling now, when constraints are present. For instance, if I make the set point temperature 350 K, my system will equilibrate at around 330 K.
I am calculating the number of degrees of freedom for the temperature calculation as $3N-3-$number of bond constraints. Has anyone else had a similar experience with the Andersen thermostat? Would a Langevin thermostat work better here? Why is the Andersen thermostat not playing well with the SHAKE algorithm?

Have you checked out paper by Feller, Zhang, Pastor & Brooks, JCP, 103, 4613 (1995)?

Hünenberger’s Thermostat algorithms for molecular dynamics simulations (PDF available here) is a generally concise and accurate reference on the topic. On stochastical thermostats such as Langevin, it does not use the number of degrees of freedom (i.e. reduced by the number of constraints) but actually uses the number of atoms.
One would have to follow the proper derivation of the thermostat's target ensemble (from the original Andersen paper) to make sure that is correct, but it seems likely to fix your problem of having too low kinetic energy.