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What is the empirical formula of Haemoglobin?
Everywhere I look on the internet tells me a different formula. Are there genuinely different formulas for Haemoglobin, or are some of my internet sources incorrect?
(I know there are different formulas for oxyhaemoglobin, deoxyhaemoglobin, carboxyhaemoglobin, etc but the formulas I am seeing are very varied and do not account for this... I don't think)

A reliable way to get the formula would be to search for human hemoglobin in the Protein Data Bank, pick the type you want, and then open the file in a protein visualization program or extract the aminoacid sequence + sidechains/intercalated molecules and calculate the formula from that.

I've gotten to this (http://www.rcsb.org/pdb/explore/jmol.do?structureId=5EE4&opt=3&bionumber=1) page and can see the ball and stick model for Haemoglobin subunit α, how do I find the empirical formula from here?

Unfortunately I don't work with proteins, so I don't know how to process that data. I don't know if you can get the formula by viewing the molecule in your browser. I think you may be able to get the empirical formula by opening the .pdb file in the appropriate software. Also, I'm not sure that's the structure you're looking for; something like this is more likely what you want. You can also download the FASTA sequence and open it as a .txt to retrieve the aminoacid sequence for each heme subunit and maybe do something with that.

Okay, thanks, I'll try to interpret the PDB file.

Aha! Go to the "Experimental Data & Validation" section in the protein page, then look at "View Full Validation Report". In the PDF file, go to section 2, "Entry composition", and you can find the molecular formulae for each subunit and the ligands.

Argh, I just realized that the reported composition doesn't take into account any hydrogen atoms. You'll need to insert those with a program. Not sure how else to do it.

Oh. Well, you've given me a lot to work on, and I'm very thankful for that. Perhaps I can try and cross reference the data with other data about Haemoglobin subunit α which includes the hydrogen atoms.

Unfortunately this is only half an answer. I'd be thankful if someone could chip in with the rest.
When you want structural information about proteins, the first place to look is the Protein Data Bank, a freely accessible repository of protein structures determined by techniques such as X-ray crystallography, protein NMR and neutron diffraction, among others.
The problem now is being sure about what you want to find. A basic search for human hemoglobin currently provides 190 hits, so you need to browse a bit to find the most adequate entry. To keep things simple, I looked for a structure for adult hemoglobin, with no mutations, ligands or associations with other proteins. There are some database entries which seem to be right on target, namely 2DN2 for deoxyhemoglobin and 2DN1 for oxyhemoglobin.
There is a lot of information in each entry page. The whole structure can be downloaded as a .pdb and opened in appropriate structure visualization programs, of which there are probably several free options. Depending on the program used, retrieving the empirical formula can be trivial.
Forgoing the use of molecular visualization software, there is an "Experimental Data & Validation" section in the protein page, containing a "Full Validation Report" link to a PDF file. There, you can find a section called "Entry Composition" with useful information. For example, in the deoxyhemoglobin structure there are 4777 non-hydrogen atoms. The protein is composed of two copies each of two unique subunits (the alpha and beta subunits), totalling four subunits. The alpha subunit contains 1069 non-hydrogen atoms (685 $\ce{C}$, 187 $\ce{N}$, 194 $\ce{O}$ and 3 $\ce{S}$), while the beta subunit contains 1123 non-hydrogen atoms (724 $\ce{C}$, 195 $\ce{N}$, 201 $\ce{O}$ and 3 $\ce{S}$). Additionally, each subunit also contains one molecule of protoporphyrin IX coordinated to an iron atom. One important thing to note is that the structure also contains solvent molecules, in this case a total of 221 molecules of water, which must be removed from the atom count.
The only flaw is that hydrogen atoms are not shown. This is because X-ray crystallography measurements generally cannot detect hydrogen atoms, and so they are usually added separately to the backbone of the protein, in such a way to complete all atom valencies. I don't know how this can be done other than using a program.
A last thing to keep in mind is that no matter how good the measurement or the data processing, you can very well end up a few atoms off, especially regarding hydrogen atoms. Hopefully perfect precision is not required.


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Materials with solid volume greater than liquid volume
Water has a solid volume greater than its liquid volume.
This question is in two parts.
1) Are there other materials that have solid volumes greater than liquid volumes? 2) What about the structure of a molecule makes its solid volume greater than its liquid volume?

There is a bunch of such materials; among the elemental compounds, they include silicon, gallium, germanium, and bismuth. As for the properties... well, it just so happens that their crystal structure contains a little too much of empty space. Crystal packing is a tricky thing, it can't be predicted with certainty even with the most sophisticated methods.

Antimony as well in the elements.

Yeah, that too. Oh, and then there is gray tin (though it does not survive until melting).

Indeed. The water being unique is a meme that really needs to die...

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Nomenclature of a branched alkane

Here I need to give the systematic name of this molecule so I decided to first find the longuest chain of carbons which seems to be the one I put in red. Then I have 2 ethyls at 3 and 4 and I have an isobutyl at 6 so would the name be
3,4-diethyl-6-isobutyldecane ?

The name ‘3,4-diethyl-6-isobutyldecane’ that is proposed in the question is unambiguous and describes the correct structure. However, it corresponds to obsolete IUPAC recommendations.
The retained prefix ‘isobutyl’ was still contained in the 1993 recommendations but is explicitly no longer recommended in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).
According to current IUPAC recommendations, the preferred prefix for the group $\ce{(CH3)2CH-CH2-{}}$ is ‘2-methylpropyl’.
Therefore, the preferred IUPAC name (PIN) for the compound given in the question is 3,4-diethyl-6-(2-methylpropyl)decane.

Isobuytl is a trivial name (Although it is very clear what is means, even if not formally correct).
Try:
3,4-diethyl-6-(2-methylpropyl)-decane

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Relationship between Lewis Theory and Redox theory?
First of all, thank you ahead of time for taking the time to read my question and answer!
Let us take the reaction $\ce{SnI4 + I^{-} -> SnI2 + I3^{-}}$ as an example.
Here we see one obvious redox reaction, $\ce{Sn(IV)}$ gets reduced to $\ce{Sn(II)}$.
Less obviously, the iodide gets oxidized to $\ce{I(I)}$ in the triiodide anion. The other 4 iodines maintain their negative 1 oxidation state.
Is there also a lewis theory acid/base reaction here as well?
The tin acts as a lewis acid and accepts an electron from each of the two of the iodines (lewis bases)? Those iodines, now electron deficient (act as lewis acids), bond with the iodide (acting as a lewis base) to create $\ce{I3^{-}}$ product.
In this case, Tin accepts one electron from two atoms rather than a "pair of electrons" from one atom like you see in traditional examples of lewis acids.
So that would be my first question, can Tin act as a "lewis acid" even though it is accepting one electron from two different sources rather than a pair from one source? Or in other words, can the iodines be considered "lewis bases" even though they only donate one electron?
Or is the whole paragraph above a moot point because there is no acid/base reaction here because the Tin is simply regaining "ownership" of the electrons it let iodine "borrow" to create a covalent bond and thus its not actually getting any "new" electrons.
My second question is: Can we make a general statement about the relationship between lewis acids and their roles in a redox reaction? For example, can I say that ALL lewis acids are oxidants and ALL lewis bases are reductants? Or are these concepts mutually exclusive? In other words, sometimes lewis acids can act oxidants, sometimes as reductants, and sometimes as neither, like in the example below?:
$\ce{Al(III)Cl3 + Cl- -> Al(III)Cl4-}$ (no redox here)

Under the Usanovich definition of acids and bases, any substance which loses electrons is a base while any substance which gains electrons is an acid. Here, any redox reaction is also an acid-base reaction, whereas Lewis theory has nothing to state in these cases.
Regarding your second question, blanket statements are dangerous. Depending on exactly what you mean, I can provide a counter-example. Here is a strong oxidant ($\ce{HNO3}$) reacting as a Lewis base:
$$\ce{HNO3(l) + H2SO4(l) -> H2NO3^{+} + HSO4^{-}}$$

Is Tin acting as a "usanovich acid" here?

I think $\ce{SnI4}$ is acting as an Usanovich acid in your first reaction, yes.


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Recrystallization of salts with similar solubilites
If you had two solutions, for example, 100 ml each of 0.5 M $\ce{RbCl}$ and 0.5 M $\ce{NaBr}$, at 20 °C, where the salts have very similar solubility in water, and you mixed the solutions, and allowed the water to evaporate. Would you theoretically end up with equal number of moles of $\ce{RbCl}$, $\ce{RbBr}$, $\ce{NaCl}$, $\ce{NaBr}$? If not, is there an equation or methodology you can use to determine how much of each salt you'd end up with?

I would be looking at the free energies of formation and work it out from there. The presence of water is a red herring in the crystallization question.

You would end up with an equimolar mixture of all four salts if they all would have exactly the same solubility, which just can't be so. In reality, some salt would reach its solubility limit before the others, and would start to crystallize. Then the concentration of its ions would stay the same, but the other two would continue to grow until something else happens. So it goes. Oh, and then there is a possibility of solid solutions which makes the whole story a lot more complicated.

Not only solubility is important, but crystal structure as well: due to differences in ionic radius, the crystals of one salt may exclude those of another.


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How to calculate theoretical yield if there are minor products?
Calculating theoretical yield for a reaction with a single product is pretty trivial: Multiply the amount of moles of limiting reagent to the molar ratio of the limiting reagent and product to the molecular weight of the product.
However, if the reaction is also capable of producing side product, this method no longer works because some portion of that limiting reagent would be used to create the minor product. Therefore, the ONLY way to calculate theoretical yield for a reaction with multiple products would be to know the ratio between them. In other words, if the ratio of major to minor product is unknown, then it would be impossible to calculate theoretical yield for that reaction. Is this correct?
We can use an example to illustrate this. Take 1-methyl,1-bromo-cyclohexane. When mixed with a strong base, we get two E1 products, a major trisubstituted alkene and minor disubstituted alkene. In order to calculate the theoretical yield of the major product, we would have to know its molar ratio to the minor product?
If I did know the ratio, was lets say, 80 to 20. All I would have to do is multiply the mass of the limiting reagent by 0.8 because only 80% of the mass would be used to create major product?
Thanks!

The theoretical (not actual)  yield will be 100%, irrespective of the reaction selectivity.

Ok, so I was overthinking it! This was very illuminating. Thank you!


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Nomenclature of a cyclic ketone

Why isn't this molecule named 2-fluorocyclopent-1-one?

The 1-specification for the ketone function is superfluous, there is no information obscured by omitting the '1'.  It is the most important functional group in molecule, and therefore establishes the 1-position.
Also, cyclopentone is not a proper name.  You are missing the 'ane' suffix to indicate saturation.  It would be a cyclopentanone.
Final name:
2-fluorocyclopentanone.

@Lighthart Nitpicking: according to IUPAC nomenclature, ‘ane’ is an ending, not a suffix.

True.  I think the OP got the point

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), names of cyclic ketones are formed substitutively by using the suffix ‘one’.
The terminal letter ‘e’ in names of parent hydrides (here: cyclopentane) is systematically elided when followed by a suffix beginning with ‘a’, ‘e’, ‘i’, ‘o’, ‘u’, or ‘y’.
Therefore, the name of the unsubstituted cyclic ketone given in the question is cyclopentanone.

Note that the locant ‘1’ is omitted in monosubstituted homogeneous monocyclic rings.
However, if any locants are essential for defining the structure, then all locants must be cited in preferred IUPAC names (PINs).
Therefore, the omission of the locant ‘1’ in 2-fluorocyclopentanone, while permissible in general usage, is not allowed in PINs, thus the name 2-fluorocyclopentan-1-one is the PIN.


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Mechanism: explanation of final step in synthesis of the thyroid hormones (T4, T3)
In the following mechanism for the synthesis of T3 (triiodothyronine) on thyroglobulin, what are the factors causing the last step to occur? Does it have something to do with the electron density of the intermediate and/or bond strength? (NB thyroperoxidase abstracts the electrons in the first step so I don't know if it's involved much in the last step)

Aromatization might be the driving force? In that case, wouldn't the arrows point in the opposite direction to give the final methylene group on the enzyme?

I would think so but this is just how they've drawn it. Could you explain how aromatisation drives this change?


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Synthesising Complex Aromatic Compound

Synthesize the following compound using only toluene as the carbon source-

I have never done something like this before and I have no clue on how to start.
I am even more confused by the fact that the top two phenyl groups are meta substituted despite having a methyl group which is suppose to be ortho and para directing.

From an answer below it states I need to use Friedel–Crafts acylation. However I am not entirely sure how to do this. This is my guess:
Convert toluene into benzoic acid using $\ce{KMnO4}$ and then react half of it with $\ce{SOCl2}$ to form $\ce{C6H5COCl}$. Then react that with the other half of the benzoic acid followed by Wolff–Kishner reduction to form the below molecule:

After this, I am not sure how to chlorinate it so that I do another acylation.

An alternative approach might be something like this. The tertiary alcohol product with two identical groups suggests that a Grignard reaction with an ester might be a useful step.

Looks good, except technically you were supposed to use toluene as the only carbon source so maybe you lose points esterifying the benzoic acid with methanol.  Form an acyl chloride from the acid + SOCl2 instead, maybe?

You'll obviously need meta-directors, so convert those methyl groups into a different function.
I'd start with a benzylic oxidation to get benzoic acid, and work some Friedels-Craft acylation magic.
Then you can reduce the resulting ketone and chlorinate, and go at an acylation again.
Finally a DIBAL and then Wolf-Kichner Reduction to get rid of your carbonyls.
Probably you need to flesh this out some.

thanks for you answer. For the acylation step, do you mean that I need to convert toluene into benzoic acid using KMnO4  and then react half of it with SOCl2 to form C6H5COCl. Then react that with the other half of the benzoic acid?

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Alphabetical order in nomenclature
Shouldn't 4-isopropyl-2-cyclohexyl-3,5-dimethylheptane be written as 2-cyclohexyl-3,5-dimethyl-4-isopropylheptane, by following alphabetical order?
Or maybe you don't look at the cyclo or the di prefix when you look for alphabetical order.

Related: Nomenclature of an organic compound

Both names that are proposed in the question, ‘4-isopropyl-2-cyclohexyl-3,5-dimethylheptane’ and ‘2-cyclohexyl-3,5-dimethyl-4-isopropylheptane’, are unambiguous and describe the correct structure. However, they are not in accordance with IUPAC recommendations.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl, cyclohexyl, and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. For example, ‘3,5-dimethyl’ is considered to begin with ‘m’. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.

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.

Therefore, the correct alphanumerical order for the example given in the question corresponds to the name 2-cyclohexyl-4-isopropyl-3,5-dimethylheptane since ‘cyclohexyl’ begins with ‘c’, isopropyl begins with ‘i’, and methyl begins with ‘m’.
However, note that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred prefix is ‘propan-2-yl’. When using this prefix, the correct alphanumerical order  corresponds to the name 2-cyclohexyl-3,5-dimethyl-4-(propan-2-yl)heptane.
Anyway, to construct a preferred IUPAC name (PIN), the ring is selected as the parent hydride. Therefore, the PIN for the example given in the question is [3,5-dimethyl-4-(propan-2-yl)heptan-2-yl]cyclohexane. (In general nomenclature, the context may favour the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.)


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How does iceskating work?
I was willing to understand a little bit better the physics behind iceskating.
Messy thoughts
From New Scientist (1964) (probably not the most up-to-date reference $\ddot \smile$), I found a diagram showing the relationship between both the static and kinetic coefficients of friction and the temperature (x-axis is reversed) for ice.

When we ice skate, we put a big pressure on the ice. As the ice is less dense than liquid water, this pressure may eventually be high enough to turn ice into liquid water. Intuitively I would think that high pressure would yield to high temperature but I suppose that this would hold true exclusively for gazes ($PV=nRT$). In any case, I don't quite have a good intuition for why warmer (or more pressurized eventually) ice would decrease the coefficient of friction of ice.
Questions

What is the intuition behind the fact the coefficient of friction of ice decreases as temperature increases?
Does the coefficient of friction of ice decreases with higher pressure too? For the same reason as above? As a correlated-side effect of increasing temperature?
Why do ice skates slide so well? Because it puts a high pressure on the contact surface or because it puts a high pressure on the contact surface which in turn increases the temperature?

Possible duplicate of What makes ice slippery?

I don't think this is a duplicate. The linked question is related but this one is specifically about the effects of temperature and pressure, which the linked question does not address.

Maybe it has something to do with the effect of pressure on the melting point of ice.  Check the phase diagram.  As the pressure increases, the melting temperature decreases.  I just don't know if the pressures under the blade of a skate are high enough to allow the ice to melt so that a layer of water could provide lubriction.  You might check this using an estimate of the contact area and the weight of the person.

Similar question on Physics: https://physics.stackexchange.com/q/1720

Note that the topic is much trickier than it appears, consisting from several levels.  The first level is obvious reason of decreasing melting point under pressure, working down to about -20 deg C. Next one is heating up by friction. But the last levels are not understood at all.

There is an excellent article by Robert Rosenberg in Dec 2005 Physics Today magazine that discusses the effect of pressure, friction and a molecular layer of water on the ice surface as initially suggested by Michael Faraday which seems to be essential.

Calderon & Mohazzabi give an excellent summation of the various theories proposed through the years to explain why ice is so slippery in their 2018 paper "Premelting, Pressure Melting, and Regelation of ice revisited" in the Journal of Applied Mathematics  and Physics.
They  offer both theoretical and experimental evidence that neither pressure melting nor friction melting explain the phenomenon and conclude from atomic force microscopy, among other evidence, that there is a pre-melting Quasi-Liquid surface layer with special properties that make ice skating possible.
References

Calderon, C. and Mohazzabi, P. (2018) Premelting, Pressure Melting, and Regelation of Ice Revisited. Journal of Applied Mathematics and Physics, 6, 2181-2191. doi: [10.4236/jamp.2018.611183.][1]

Preview/read online at :  https://www.researchgate.net/publication/328766489_Premelting_Pressure_Melting_and_Regelation_of_Ice_Revisited

There's a very nice—and fairly accessible—discussion of your third question in Nature Magazine's News and Views Section[1]. In this article, Bonn discusses the results of a fairly recent work that appeared in Physical Review X[2]. Bonn summarizes the key points as follows:

The idea that a thin film of meltwater wets the surface of ice has
been accepted since the nineteenth century...[but]...Water is not a
good lubricant, because its low viscosity means that it is easily
squeezed out of gaps. The idea that a layer of water is sufficient to
lubricate a skate on ice is therefore strange. It doesn’t even make
intuitive sense, given that it is impossible to skate on a road or a
kitchen floor with a layer of water on it...[The explanation thus
seems to be that water and ice together]... form a viscoelastic,
liquid–solid third body in response to friction and wear. (Emphasis
mine)

References

Bonn, Daniel. The physics of ice skating. Nature 577, 173-174 (2020). link: https://www.nature.com/articles/d41586-019-03833-5

Canale, L., Comtet, J., Niguès, A., Cohen, C., Clanet, C., Siria, A. and Bocquet, L. Nanorheology of interfacial water during ice gliding. Physical Review X, 9(4), p.041025 (2019). link: https://journals.aps.org/prx/abstract/10.1103/PhysRevX.9.041025

@Nilay Ghosh:  Thanks for making my references pretty :).  However, the phrase "and fairly accessible" is parenthetical, and thus needs to be separated by parentheses, commas, or dashes.  Hence I've restored the dashes you removed.   The test for whether a phrase is parenthetical is whether you can remove it without changing the essential meaning of a sentence.  For more information, see:  https://www.englishclub.com/grammar/sentence/parenthetical-expression.htm

My understanding of ice skating is that it has to do with the pressure of the skate, melting the ice and you skate on the melted water.  Look at the phase diagram of the water.  At 1 atm, the water is solid, but at higher pressures (keeping the temp. the same) the water changes to a liquid.

The effect is called regelation. You can cut ice with a thin steel wire and weights. Since ice expands when it freezes, pressure at the freezing point can cause thawing.

This question has more physics into it than chemistry but the ice melting to provide a slippery surface (and guess subsequently re-freeze) makes sense. We can also try asking someone to glide over dry-ice made out of carbon-dioxide instead of regular ice and see if there is any increase or decrease in the friction .
Two possibilities -

(a) If water contributes then the idea is that dry-ice under pressure would turn directly to gas and as such not contribute to being slippery
or
(b) The process of solid turning to gas makes the surface even more slippery than normal - but the process would be inherently destructive since the gas will not re-solidify back into ice.

Alternatively we can try out an experiment using ice-cubes out of water and dry-ice and run it over a butter knife and apply some pressure to simulate a person skating on ice.

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Determining Rate Law and Reaction Mechanism from Method of Initial Rates

The data collected from using the method of initial rates is tabulated above.
Suggest a rate law and a mechanism consistent with the data. State any assumptions.

My Attempt
Obviously B and D has no impact on the rate of the reaction. From the first table it appears that the reaction is first order in A, however this doesn't make sense when you examine the last two rows of the second table. Since the rate law is first order in A and is independent of B, decreasing the concentration of A by 4 and increasing the concentration of B by 4 should result in a fourfold decrease in the rate. However the rate of the reaction decrease by 2.5 times. So then my logic is somehow wrong.
Furthermore I am even more confused about the order of C. I know that concentration of C is inversely proportional to the rate of the reaction. However the reaction still occurs without the presence of C (evident by the first table). Doesn't this imply that the rate law is independent of the concentration of C?
From the comments below, I have realized that the rate order of B is in fact not $0$. However now I am even more confused. How can it be that the rate is not dependent on B when there is no C, but it is dependent on B when there is C present? What would its rate law be like?

It looks like B does affect the rate.  Compare cases 2 and 6 in the lower table.

@ChesterMiller thanks for that, I didn't see that. However this just confuses me even more. I don't see how it can be possible that has B has no impact on the rate, but when in the presence of C, it does. I have no idea what the order of B is suppose to be now.

Empirical rate law is:
$$\mathrm{Rate} = \frac{[\ce{A}][\ce{B}]}{[\ce{B}] + [\ce{C}]}$$
In the first block of data you can notice that $\ce{C}$ and $\ce{D}$ are not necessarily required for reaction to proceed because otherwise the rate would be zero. You can also deduce that their concentration must appear as a sum in the rate law for the same reason. The sum will appear in denominator because $\ce{C}$ slows down the overall reaction.
Next step is to guess elementary reactions. From the law's denominator sum you can conclude that intermediate will competitively react with $\ce{B}$ and $\ce{C}$.
$\ce{A ->[k1] I}$ (Probably some kind of excitation)
$\ce{I + C ->[k2] D}$
$\ce{I + B ->[k3]Product}$
Then you derive the overall rate via steady state approximation (assumption).
$$\mathrm{Rate} = \frac{k_1k_3[\ce{A}][\ce{B}]}{k_3[\ce{B}] + k_2[\ce{C}]}$$
and ${k_1}={k_2}={k_3}$

I think you should consider some sequence of reactions involving mass action kinetics.  The rate they give in the table must be the rate of disappearance of A (I guess). So the rate might be represented by something like $R=C_1A+C_2B+C_3C+C_4A^2+C_5B^2+C_6C^2+C_7AB+C_8BC+C_9AC$.  Not all the coefficients would be positive because of reverse reactions.  There is not enough data in the table to determine all the coefficients even for this set up to 2nd order.  So you might have to play with the coefficients to determine the simplest version which matches the table   I hope this helps and I hope it works.

Thanks for you thoughts. But I think that there must be a simpler way of doing it since we have never been taught mass action kinetics and it is expected that we should be able to do this question within 7-8 minutes.


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what is the difference between excitation and emission spectroscopy?
I know about absorption spectroscopy but really confused between excitation and emission spectroscopy. If they are different, then how?

http://chemistry.stackexchange.com/questions/9816/fluorescence-spectra-emission-spectrum/9818 might be helpful.

Absorption:

Light with a specific frequency is selected
This frequency is shined (radiated) onto a molecule (sample)
The amount of light that shines through the molecule is measured, ie. the detector is behind the sample (180 degrees).
Step 1 is repeated with a different frequency.

When the sample absorbs a specific frequency, less light of that frequency will shine through to the detector. The machine records this and a peak appears in the spectrum for that frequency.
Frequency is usually measured in wavelength, and a wide range of wavelengths are applicable. In UV-VIS, this can be eg. 300-800 nm, so 800 nm is measured, then 799 nm, then 798 nm...down to 300 nm. In IR, you will go into micrometers. In X-ray, you will be around 10 nm I think.
Emission:

A light with a specific frequency is selected
The sample is irradiated with this frequency.
The amount of light that shines from the molecule is measured, ie. the detector is to the side of the sample (90 degrees).
The emitted light is measured for a range of frequencies
Step 1 is repeated with a different frequency.

In both processes, the sample absorbs light. The major difference here is the detector being located to the side of the sample. This allows it to ignore the light being shined onto the molecule, and instead measure light that is emitted from the sample. Think of the incident light as a column. In absorption spectroscopy, that column is absorbed by the sample and continues to the other side, where the detector can measure it. In emission spectroscopy, you want to avoid the column of light. Think of the sample like a light bulb that shines light in all directions. The detector in fluorescence is placed to the side, avoiding the column of light and picking up the light from the sample instead. The advantage to fluorescence is you can often selectively measure a moiety (part of the molecule) with superior sensitivity, whereas in absorption impurities or stray light are more prone to interfere with the signal and also leads to higher concentrations usually being necessary.
The reason the emission wavelength detected is different than the frequency used to irradiate it is due to Stoke's Shifts.
Stoke's Shift
Stoke's Shift requires basic knowledge of the difference between electronic and vibrational states. A molecule has electronic states, S=0, 1, 2, 3...etc. Within each electronic state is a set of vibrational states, v=1,2,3.... For example, two electrons can be in the 0 electronic state (termed ground state), but one electron can be in the v=2 vibrational state, while the other electron in v=4.
This is easier to picture if you look up Stoke's Shift, or specifically the diagram on Wikipedia's page for Fluorescence.
For example, in emission spectroscopy, an electron absorbs light, being excited electronically from S=0, v=1, to S=1, v=2. Vibrational relaxations will occur before electronic relaxations. That means the electron will first relax from S=1, v=2 to S=1, v=0. After this, it relaxes to S=0, emitting a photon in the process for detection. Because it relaxed its vibrational state first, energy was lost not in the form of light (thermodynamically). As a result, the photon finally emitted is at a lower energy than the photon that had been absorbed. That's the rudimentary explanation.

Thanks!! But is the position of detector makes all the difference between them? can we still calculate the emission from the sample if we keep the detector just behind the sample?

No, for 2 reasons:

The emission wavelength is not known.
The incident light (the light used to excite the sample) shines like a column through and behind the sample.

The first reason means the detector has to scan every possible wavelength to test for the emission wavelength. This is a problem for the second reason. While scanning every wavelength, it will also be detecting the massive signal from the incident wavelength which will not only overlap with the emission signal, but it will drown it out with its intensity. You need to avoid detecting the incident light, hence 90 degrees.

In order to appreciate the difference between UV-VIS emission and excitation spectroscopy it is crucial to have a closer look at the instrument that is used to carry out both techniques, i.e. the Fluorometer.
The fundamental optical components of a Fluorometer are: a lamp that provides the light excitation, a monochromator of excitation (Mex) to select a monochromatic radiation (the radiation has always a certain bandwidth so it is never monochromatic) used to excite the sample, a monochromator of emission (Mem) that selects the radiadion coming out from the sample, this can be luminescence (fluorescence or phosphorescence), raman or scattered light (with a 90 degree geometry used in the experiment the scattered light is greatly reduced but not completely eliminated). Finally a detector, usually a photomultiplier.
In EMISSION spectroscopy the Mex is fixed at the wavelenght of excitation and the Mem scans, i.e. it lets pass different monochromatic radiation sequentially, in a selected wavelenght range. The final spectrum (emission) results in emission-intensity vs wavelength.
In EXCITATION spectroscopy the Mex scans in a wavelength range equal to that used in absorption spectroscopy. The Mem is fixed at a certain wavelength. What you want to have is a spectrum (excitation) that is proportional to the absorption spectrum. If that is the case it means that all the excited states (those that have been populated by exciting the sample in a Mex scanning mode) contribute to populate the emitting state (the one that is monitored by fixing the Mem). If that is not the case it means that not all the excited states contribute to populate the emitting state, i.e. there are other deactivation paths for the excited states located at higher energy compared to the emitting state.
In other words, with emission spectroscopy you have information on the capability of your sample to emit light. On the other hand, with the excitation spectroscopy you want to investigate whether high-energy excited states communicate with the emitting state, i.e. if their population deactivate towards the emitting state or not. This also implies that to perform an excitation spectrum you have to have an emitting compound. Thus, first you run an emission experiment and then an excitation.

There are three types of spectra here, absorption, emission (fluorescence, phosphorescence, in general 'luminescence') and excitation. (a) Absorption as you know is a transition from the ground state to the first, second or higher electronically excited states. These states will decay back to the ground  state, assuming no chemistry from an excited state occurs, by radiationless and radiative transitions. Radiationless transitions occur from a singlet excited state to the ground state (internal conversion) or to a triplet state (intersystem crossing). (b) In competition with these transitions is fluorescence from the singlet and phosphorescence from the triplet to ground state. This is the second type of spectrum and is obtained by fixing the excitation wavelength and scanning the emission monochromator over different wavelengths and recording the signal vs. emission wavelength. (c) The third type of spectrum is excitation. To obtain this the detector is set at the  required wavelength in the emission spectrum and the excitation wavelength is scanned and the signal recorded vs. excitation wavelength. This spectrum is related to the absorption spectrum, but differs from it in intensity because the Franck-Condon factors are different in absorption and emission. Additionally, suppose that radiationless transitions are greater from one excited state than another then less emission will be observed, compared to that expected from the absorption spectrum, and so a weak excitation spectrum will now be observed at these wavelengths. Thus the excitation spectrum is like an 'action' spectrum. In general excitation spectra are difficult to interpret accurately.


text:
Is tungsten gas visible?
I apologize if this isn't a sufficiently technical question to be posed in this forum.
I'm planning a short story, and unfortunately I've worked in a somewhat sience-y opening. Most of the information I need can be found online, but so far I haven't been able to find any comprehensive information about the properties of some of the less common gases, namely tungsten.
Would pure tungsten gas be visible to the human eye? Does it have any other unique properties that I should know before writing about it?

You realize that tungsten is a metal, and a pretty heat-resistant one, right? To make it a gas, you have to heat it way above... well, pretty much above everything. So yes, it would be glowing hot and thus quite visible.

You can easily see it: "burn out" the filament of an incandescent lamp from a flashlight (torch) with over-voltage (e.g. put a 1.5 volt lamp across a 12 volt battery). The bluish flash you see is vaporizing tungsten (W), but to the unaided human eye, it looks little different from, say, the "spark" in xenon flash lamps.
Since the filament is so hot, it radiates heat rapidly and cools quickly, so is visible only as a flash. If you were to vaporize a few kg of W, it would be visible longer as a blindingly-bright object.
BTW, notice the film on the "dead" lamp: it is condensed W. If the lamp has a vacuum, the film is silvery, but collision with argon or other inert filler makes it rougher, or grayish.


text:
How do Cryocoolers that use Helium as Coolant work?
I was unable to find elaborate information on helium based Cryocoolers. I like to know how they use helium to cool samples to cryogenic temperatures and why helium is the best option as a coolant.

A helium based cryocooler works just like an air conditioner, but helium is the gas rather than a halocarbon compound. The boiling point of helium is −268.928 °C, so such a cryocooler could get really cold.
Helium is a rare gas and relatively expensive. So to cool a small experimental apparatus it is much cheaper to buy one helium based cryocooler rather than dewars and dewars of liquid helium.

Could you tell me how it cools the samples?

http://inspirehep.net/record/852153/plots
The actual design depends on the experiment you want to perform, as you usually cool the sample for some reason, not just for fun.


text:
Number of stereoisomers of a polyene
Find the number of stereoisomers of the following compound:

I think the answer is 16:

The double bond in the ring cannot show geometrical isomerism.
The first double bond on the top substituent can show cis-trans isomerism and the one on the end will not show cis-trans isomerism because of two same groups on one carbon.
The substituent on the bottom that is: $\ce{-CH=C=C=CHMe}$ can also show cis-trans isomerism.
Then, there are two chiral centres and the number of optical isomers is $2^2=4$. The total number of stereoisomers is 16.

But the answer given is 4. Am I correct or not?

A picture is worth a thousand words so I've redrawn your compound and have numbered the areas to be considered.

Double bond "1" is not capable of displaying cis-trans
stereoisomers since there are two identical substituents (hydrogens)
on one end of the double bond.
Double bond "2" is capable of displaying cis and trans stereoisomers since there are two different substituents on each end of the double bond.
Double bond "3" can only exist as a cis isomer due to the geometric constraints imposed by the 6-membered ring.
Double bond "4" is a 1,2,3-triene (allene would be a 1,2-diene); such trienes  display the same stereochemical cis-trans possibilities as a double bond.
The substituents on the two cyclohexene carbons contained in circle "5" also have cis-trans arrangements; these are further illustrated in the cyclohexene drawings pictured below.
Finally, each of the carbons contained in circle "5" are chiral (they each have 4 different substituents) and can exist as R and S stereoisomers.

Considering only the circles that can have stereoisomers (2, 4 and 5) we can have the following spatial arrangements:

2-cis, 4-cis, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-cis, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-cis, 4-trans, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-trans, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-cis, 4-cis, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-cis, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-cis, 4-trans, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-trans, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.

Therefore, there are a total of 16 stereoisomers for your molecule, 8 diastereomers, each of which can exist as an enantiomeric pair.
Thanks to Loong for pointing out that the cis and trans geometry in circle "5" limits the R and S possibilities at the two chiral carbon atoms in circle "5".  That is, these carbons in (for example) the cis isomer cannot be R,S R,R S,R and S,S, - in the cis isomer they can only be either R,S and S,R or R,R and S,S.


text:
What should be included in the equilibrium constant expression?
Some writes that pure solids and pure liquids and also solvent should not appear in the Kc, yet a book writes that an carbonyl compound, insoluble in the water, the Kc for this reaction finally includes the water as the density of water is not a constant according to the notes.
Can anyone explain to me about this and what should be included into the Kc expression? And are there any special case for me to memorise in the exam?


text:
Why does calcium oxide react with sulfur dioxide?

Which of the following compounds reacts with calcium oxide, $\ce{CaO}$?
  (a) $\ce{K2O}$
  (b) $\ce{Na2O}$
  (c) $\ce{SO2}$
  (d) $\ce{MgO}$

I thought that since sodium is higher in the reactivity series than Ca, then it would be B. However, the answer is C, and I'm not quite sure why. Can anyone enlighten me?

This will happen: Why is the reaction of calcium oxide with sulfur dioxide to form calcium sulfite not a combination redox reaction?

If a reaction takes place there should be a reasonable chemical equation.
Calcium oxide and sulfur dioxide may react to give calcium sulfite:
$$\ce{CaO + SO2 -> CaSO3}$$
On the other hand alkali metal oxides and alkaline earth metal oxides do not form mixed oxides.
The reactivity series gives us information about the feasibility of redox reactions. Since no redox reactions can occur here, the reduction potentials are not relevant.

One way of thinking about it is that calcium is a metal and therefore forms a basic oxide, $\ce{CaO}$. On the other hand, sulfur is a non-metal and forms acidic oxides $\ce{SO2}$ (or $\ce{SO3}$). So, the reaction between $\ce{CaO}$ and $\ce{SO2}$ can be considered an acid-base reaction.
There is a somewhat obscure acid-base theory, the Lux–Flood theory, which is applicable to such reactions. In this context, bases are oxide donors (like $\ce{CaO}$) and acids are oxide acceptors (like $\ce{SO2}$).


text:
How to come up with overall equation of a galvanic cell
How are overall equations of the galvanic cell written? In a zinc-copper cell, the overall equation is
$$ \ce{Cu^2+(aq) + Zn(s) -> Zn^2+(aq) + Cu(s)} $$
So in order to come up with this equation, do you take the net ionic equation that occurs when $ \ce{Zn} $ is placed into $ \ce{CuSO4} $ solution (the electrolyte for copper half cell)? In other words, do you calculate the overall equation of the galvanic cell by coming up with the net ionic equation of the single displacement reaction that occurs between the anode electrode and the cathode electrolyte?
In this example, it would be
$$ \ce{Zn(s) + Cu^2+(aq) + SO_4^2- (aq) -> Zn^2+(aq) + Cu(s) + SO4^2- (aq)} $$
where $ \ce{Zn(s)} $ would be the anode electrode and $ \ce{CuSO_4} $ would be the cathode electrolyte.
And the $ \ce{SO_4^{2-}} $ will cancel out since it is a spectator ion to give the overall equation of the galvanic cell.
Am I doing this right?

Since $ \ce{Zn} $ never touches any $ \ce{Cu^2+} $ ions in a galvanic cell - the salt bridge/porous plate is preventing this - it would not be helpful to take this chemical equation as a starting point. As the aim of such cells is to separate the reduction and oxidation process, this is where the equations should originate.

Direction of the Reaction
To start with, one should find out which of the 2 electrode materials is more noble, hence will be the oxidizing agent aka cathode material, and will get itself reduced during the process through receiving electrons from the anode over the external circuit. A higher standard electrode potential ($E^\circ$) correlates with higher nobility. Information on different $E^\circ$ values can be found in galvanic series (Wikipedia: Standard electrode potential (data page)). A lookup of $\ce{Cu}$ and $\ce{Zn}$ gives:
$$
\begin{array}{rclc}
\hline
\text{Oxidant} & & \text{Reductant} & E^\circ/\mathrm{V}\\
\hline
\ce{Zn^2+(aq) + 2 e− & <=> & Zn(s)} & −0.7618\\
\ce{Cu^2+(aq) + 2 e− & <=> & Cu(s)} & +0.337\\
\hline
\end{array}
$$
Since $\ce{Cu}$ has the higher $E^\circ$ (is more noble), $\ce{Cu}$ will act as cathode, and $\ce{Zn}$ as anode, to provide the reactions spontaneity.

Setting up the Reaction
Looking at the electrochemical process, first the $ \ce{Zn} $ from the zink electrode gets oxidized according to:
$$ \text{Ox - Anode half equation:}\qquad\ce{Zn(s) -> Zn^2+(aq) + 2 e-} $$
After passing through the external circuit the electrons get to reduce the $ \ce{Cu^2+} $ ions according to:
$$ \text{Red - Cathode half equation:}\qquad\ce{Cu^2+(aq) + 2 e- -> Cu(s)} $$
Which shows, that the anode is going to lose mass through corrosion, while the cathode is going to gain mass through the formation of solid copper on its surface.
After having these two half equation we can sum them up and cancel out the $ \ce{2 e-} $ respectively to yield:
$$ \text{Redox - Total equation:}\qquad\ce{Cu^2+(aq) + Zn(s) -> Cu(s) + Zn^2+(aq)} $$
While these processes happen, the salt bridge/porous plate is allowing the balancing of charges in both half cells.

Note: Of course someone could begin with the reduction, rather than with the oxidation, but it would anyway end in the same total equation, and it's harder then to imagine the electrons wandering through the external circuit.

Edit: As requested in the comments: The reactions with spectator ions:
$$\begin{align}
\text{Ox - Anode:}&& \ce{Zn(s) + SO4^{2-}(aq) &-> Zn^2+(aq) + SO4^{2-}(aq) + 2 e-}\\
\text{Red - Cathode:}&& \ce{Cu^2+(aq) + SO4^{2-}(aq) + 2 e- &-> Cu(s) + SO4^{2-}(aq)}\\
\text{Redox - Cell:}&& \ce{Cu^2+(aq) + Zn(s) + 2SO4^{2-}(aq) &-> Cu(s) + Zn^2+(aq) + 2SO4^{2-}(aq)}
\end{align}$$
Edit 2: Added Direction of the Reaction

but how do you know zinc is getting organized. What is the equation in half cell (i.e. are there spectator ions that cancel out and just leave Zn(s)⟶Zn2+(aq)+2e−. How do they come up with these half cell reactions.

Zinc is oxidized, since it is less noble than copper. This information, including the half equations can be found in the galvanic series (Sry for the german wiki, but in the english version there is no table). Of course you could write the anions in both half equations, I omitted them, since they are spectator ions and would cancel out anyways.

could you please show them with it, I am a little confused.

Show what with it?

the reaction with spectator ions

added also equations with spectator ions

@user510: How to find out which is the cathode, and which is the anode half equation has already been answered in one of your questions.

why does Zn(s) dissociate into Zinc ion in the anode half reaction?

What else would you expect to happen?

but what is the theory behind expecting the zinc solid to immediately turn into zinc ions?


text:
How do I find the final concentration after an additional dilution?
I took poor notes and didn't record what the final concentration of an amylase solution I  made last year. I recorded that I took $\rm 0.2\ g/10\ ml$ of the amylase solution, and then took $\rm 1\ ml$  of that and diluted to $\pu{35 ml}$.
How would I solve for the final concentration? Would I use
$$
\begin{align}
c_1V_1 &= c_2V_2\tag{1}\\
(\pu{0.2 g}/\pu{10 ml})(\pu{1 ml}) &= c_2(\pu{35 ml})\tag{2}\\
c_2 &= \frac{\pu{0.02 g}}{\pu{35 ml}}?\tag{3}
\end{align}
$$
Therefore, I would take $\pu{0.02 g}$ of amalyase powder and add up to $\pu{35 ml}$ with distilled $\ce{H2O}$ to achieve the final concentration that I would have previously had after dilution?

Check your math: It looks like you forgot to divide by $ 10\ \mathrm{ml} $ in finding $ c_2 $.
Also: the process you are describing in the first paragraph will not produced the same concentration that you started with. If you want $ 35\ \mathrm{ml} $ at the same concentration ($ 0.02\ \mathrm{g/ml} $) by taking some fraction of your original solution and diluting it, you will need to add more solute.
e.g.
$$ \rm(0.02\ g/ml) \cdot (35\ ml) = 0.07\ g $$
$$ \rm(0.07\ g + \mathit{m}) / 35\ ml = (0.02\ g/ml) $$
$$ m \rm = (0.02\ g/ml) \cdot (35\ ml) - 0.07\ g $$
The first expression is the concentration of your fraction (which doesn't change from the original solution, assuming a homogenous solution).
The second expression is equating the concentration that you want (conc. of the original solution, for example, with the concentration of your diluted fraction with $ m\ \mathrm{g} $ of solute added (using the mass we just found).
Then I rearranged this to express the mass of solute needed to obtain this concentration.


text:
How is adenosine synthesized in biology?
I understand the constituent parts of adenosine (adenine and ribose) and that it can be broken down from ATP anabolically. Can adenosine also be formed catabolically? If so, what is the mechanism?
What's the mechanism for combining ribose and adenine in the cell?
A general explanation or citation would work. No need for someone to go out of their way to draw pictures. Thank you very much!

https://en.wikipedia.org/wiki/Purine_metabolism?wprov=sfla1

I saw this page, but it only refers to the degradation of AMP to adenosine. I was looking for its formation from adenine and ribose. Sorry, but am I missing something there on the page?
I am starting to wonder whether adenosine does not biologically form at all via ribose and adenine.

Where it says "Biosynthesis", i.e. the first half of the page. It goes all the way to IMP - for the last steps to AMP you'll have to click the respective link.

Adenosine is synthesised from Inosine Monophosphate (IMP) as the nucleotide form (adenosine monophosphate).
IMP does not accumulate in the cell but is rapidly converted to AMP and GMP. AMP, which differs from IMP only in the replacement of its 6-keto group by an amino is synthesized in a two-reaction pathway.

In 1948, John Buchanan obtained the first clues as to how nucleic
  acids and their components are synthesized by describing the synthesis
  of purine ribonucleotides.

Buchanan investigated how this process occurs de novo by feeding a variety of isotopically labelled compounds to pigeons and chemically determining the positions of the labelled atoms in their excreted uric acid (a purine).

The biosynthetic origins of purine ring atoms. Note that C4, C5, and N7 come from a single glycine molecule but each of the other atoms is derived from an independent precursor.
The actual pathway by which these precursors are incorporated into the purine ring, was elucidated in subsequent investigations performed largely by Buchanan and by G. Robert Greenberg. These investigations showed that the initially synthesized purine derivative is inosine monophosphate (IMP),
(Here I am not going to explain the biosynthesis of IMP but focus on the former)
IMP is converted to AMP (or GMP) in separate two-reaction pathways:

In the first reaction, aspartate’s amino group is linked to IMP in a reaction driven by the hydrolysis of GTP to GDP + Pi to yield adenylosuccinate.
In the second reaction, adenylosuccinate lyase eliminates fumarate from adenylosuccinate to form AMP (adenosine monophosphate).

It is important to note that free adenine (and guanine) can be reconverted to corresponding ribonucleotides through salvage pathways requiring the enzyme Adenine phosphoribosyltransferase (APRT) (mediates AMP formation)
References
Voet and Voet Biochemistry: Section 28-1. Synthesis of Purine Ribonucleotides


text:
How to find the time elapsed in a second order reaction given the final concentration?
This is an example in my textbook that I am trying to figure out
$$1/[\ce{A}] = 1/[\ce{A}]_0 + kt$$
In this case:
\begin{align}
[\ce{A}] &= \pu{4.62E-2 M}\\
[\ce{A}]_0 &= \pu{0.217 M}\\
k &= \pu{0.540 M-1 s-1}
\end{align}
so I get this
$$1/(\pu{4.62E-2 M}) = 1/(\pu{0.217 M}) + (\pu{0.540 M-1 s-1})(t).$$
The answer should be $\pu{31.5 s}$.
I can’t get this answer I must be missing something
in the order that it should be.

What is the answer you get, then?

When you manipulate the equation to solve for t you'll want to subtract 1/[A]0 from 1/[A] and then divide by k.

It must be a math error.  Show us the next step in your calculation please.

((1/0.0462) - (1/0.217)) / 0.540 = 31.5 s

Solving the equation

\begin{align}
  \frac{1}{[A]_t} &= \frac{1}{[A]_0} + k \cdot t\\
  \frac{1}{[A]_t} - \frac{1}{[A]_0} &= k \cdot t\\
  \frac{1}{[0.0462 \ M]_t} - \frac{1}{[0.217 \ M]_0} &= 0.540 \ M^{-1} \cdot s^{−1} \cdot t\\
  21.64 \ M^{-1} - 4.61 \ M^{-1} &= 0.540 \ M^{-1} \cdot s^{−1} \cdot t\\
  17.03 \ {\color{red}{M^{-1}}} &= 0.540 \ \color{red}{M^{-1}} \cdot s^{−1} \cdot t\\
\\ \text{Molarity is simplified} \\ \\
  \frac{17.03}{0.540\ s^{-1}} &=t\\ \\
  \pu{31.53 s} &= t
\end{align}

Unsig Wxmaxima (or Maple)

Please always carry the units in every step.

I have added the units of measurement. Thanks a lot


text:
Is there Any Advantage to Using Multiple Steel Plates in an Electrolyzer?
If I have an electrolyzer (For water) is there any advantage to having additional steel plates in between the electrodes in solution? For some reason this appears to be a popular design and I'm not sure why. I'd think that at worst the voltage would have to be slightly higher to conduct through the extra solution in a 2 electrode system.

As the reactions in electrolysis take place on the electrodes surface, it can therefore be concluded, that the more electrode surface (contact surface) exists, the more water molecules can be electrolyzed in one instance of time. Therefore, more steelplates result in more surface, which increases the productivity of the electrolyzer.


text:
Plasma, Fourth State of Matter, and Magnetic Fields and Electric Fields
Plasma is the fourth state of matter consisting of positive atoms, and freed negative electrons.  Does plasma have an electric field, and magnetic field?

You know there's Physics SE? Why you posted it here? And what do you mean by "having" fields?

I'm voting to close this question as off-topic because it's better suited for physics.SE.

There is not necessarily a preexisting magnetic or electric field in a plasma, since the charges can be randomly distributed. Think of plasma as similar to a metallic state conductor -- a sea of mobile electrons (and, of course, ions in plasma, so perhaps it better approximates a semiconductor, with electrons and holes). A plasma in thermal equilibrium, not exposed to an external electromagnetic field, should be uncharged.
That said, unbalances occur for a number of reasons. For example, electrons are thousands of times less massive and therefore more mobile than ions, so charge separation occurs, leading to a plasma potential.. Lasers can generate a charged plasma, accelerating particles with the wakefield effect.


text:
How do I assign R/S configuration?

So I have this molecule shown up here and I need to give its systematic name according to the UIPAC and need to give its R/S configuration in the name. I suspect the name to be 3-tert-butyl-2-methylheptan-4-one but I don't know the R/S thing however.

This answer may give you some insight on the rules for assigning R or S. http://chemistry.stackexchange.com/a/23988/402
Which group has the highest priority? Why?

Ok but where is my asymmetric carbon? Would it be at 3?

then it would be (3S)-3-tert-butyl-2-methylheptan-4-one ?

Yes, this would be correct.


text:
How can I make a black precipitate without a sulfide?
I'm supposed to make a black precipitate using only the following chemicals: copper sulfate(II), lead nitrate, acetic acid, barium chloride, sodium hydroxide, copper(II) carbonate, methanol, potassium permanganate, hydrochloric acid, hydrogen peroxide, magnesium sulfate, and potassium dichromate.
I know that solid sulfides (used in qualitative analysis) are typically black in colour, such as copper(II) sulfide, but I don't have any source of the sulfide ion here. Is it actually possible to make a black precipitate?

The most common black pigment is charcoal. The chemical actions are interesting, but one might ask how color film was able to depict black without using a black pigment? All that is needed is a mix that absorbs all the visible frequencies. Please use care these are a bunch of nasty materials to play with. An interesting experiment: do paper chromatography on black ballpoint ink.  There is much more to chemistry than just mixing chemicals.

Hint:  Manganese dioxide is pretty black.
Look up potassium permanganate and note its reactions especially the reduction in neutral solutions.

It is possible to obtain a black precipitate of copper(II) oxide from an aqueous solution of $\ce{Cu(II)}$. Adding a solution of $\ce{NaOH}$ to an aqueous solution of $\ce{CuSO4}$ will initially give a blue precipitate of copper(II) hydroxide. This redissolves with an excess of $\ce{OH-}$ to form a deep blue solution of tetrahydroxocuprate(II). In strongly alkaline medium at room temperature, $\ce{CuO}$ can precipitate from this complex by means of a condensation reaction.
$\ce{Cu^2+ + 2OH- -> Cu(OH)2(s)}$
$\ce{Cu(OH)2(s) + 2OH- -> [Cu(OH)4]^2-}$
$\ce{[Cu(OH)4]^2- <=> CuO(s) + 2OH- + H2O}$
Alternatively, the precipitated $\ce{Cu(OH)2}$ can be converted into the oxide by boiling the mixture.
$\ce{Cu(OH)2(s) ->[\Delta] CuO(s) + H2O}$
Reference:
Yannick Cudennec, André Lecerf. The transformation of Cu(OH)2 into CuO, revisited. Solid State
Sciences, 2003, 5 (11-12), pp.1471-1474. DOI: ff10.1016/j.solidstatesciences.2003.09.009ff. (Open Access Link)

In boiling water, the blue precipitate $\ce{Cu(OH)2}$ is quickly transformed into black $\ce{CuO}$ according to : $$\ce{Cu(OH)2 -> CuO + H2O}$$

Disclaimer
This answer is somewhat an addition to the existing ones. Here, I will try to present a practically feasible method for obtaining lead(IV) oxide $\ce{PbO2}$ from the available reagents. It is a dark brown precipitate, almost black, as required in the question.
Required reagents
The following reagents from the list are needed in order to get the desired black precipitate of lead dioxide: lead(II) nitrate $\ce{Pb(NO3)2}$, hydrogen peroxide $\ce{H2O2}$ (the higher the concentration, the better) and sodium hydroxide $\ce{NaOH}$. All solids have to be dissolved in water while sodium hydroxide solution should not be too strong.
Procedure
At first, it is necessary to mix the solutions of sodium hydroxide and hydrogen peroxide, obtaining an alkaline $\ce{H2O2}$ solution. Then, when the alkaline solution of hydrogen peroxide is added to a $\ce{Pb(NO3)2}$ solution, a precipitation of dark brown solid occurs according to the following reaction equation:
$$\ce{Pb(NO3)2 (aq) + H2O2 (aq) -> PbO2 (s) + 2HNO3 (aq)}$$
Heating can help to speed up the reaction, but too much heating can dissolve the precipitate:
$$\ce{PbO2 (s) + 2OH- (aq) + 2H2O (l) \xrightarrow{\textrm{heating}} [Pb(OH)6]^2- (aq)}$$
Reference

Ronald L. Rich. Inorganic Reactions in Water (1st ed.): "Peroxide", p. 356.

A very good answer, however I doubt OP should use lead salts :D


text:
Identifying chiral carbons
Looking at this structure, I would say this molecule has five chiral carbons. However, its IUPAC name (3R,4R,5S,6R)-3-amino-6-(hydroxymethyl)oxane-2,4,5-triol suggests that there are only four. Which is it?

It is the hemiacetal carbon atoms. It is chiral but racemization can occur through the open form of the molecules.

The diagram you drew doesn't indicate the stereochemistry, so I am curious as to how you came up with that IUPAC name.

The structure as drawn indeed contains 5 chiral carbon atoms.  However, the one carbon that is bound to two oxygen atoms is called the anomeric carbon.  It is part of a hemiacetal moiety in the structure you have drawn.  That means that at room temperature and in aqueous solution, it will equilibrate reasonably quickly between:

One enantiomer of hemiacetal
An open-chain non-cyclic molecule where the anomeric carbon is an aldehyde moiety and is thus not chiral.
The other enantiomer of hemiacetal.

Thus, for many applications it is probably better to think of this molecule as having only four "fixed" stereocenters.


text:
Calculations on an irreversible adiabatic expansion

A $\pu{5.00 L}$ sample of $\ce{CO_2}$ at $800\ \mathrm{kPa}$ underwent a one-step irreversible adiabatic expansion against a constant external pressure of $\pu{100 kPa}$. The initial temperature of the gas was $300\ \mathrm{K}$
An alternative path between the initial and final states consists of a reversible isothermal expansion
  from $5.00\ \mathrm{L}$ to the final volume $V$, followed by (reversible) constant volume cooling to the final temperature $T$.
a) Give the equations for $\Delta U$, $q$ and $w$ for all three processes in terms of $V$ and $T$
b) Calculate the final volume and temperature of the gas

My Attempt
Adiabatic Expansion
$q = 0$ since it is adiabatic. $w = -100 \times (V-5)$. $\Delta U = q + w = -100 \times (V-5)$
Isothermal Expansion
$\Delta U = 0$. $w = -q = -\mathrm{nRT}\times \ln\frac V5$ = $-4000 \times \ln\frac{V}{5}$
$q = 4000 \times \ln\frac{V}{5}$
Reversible Cooling
$w = 0$. $\Delta U = \mathrm{c_v \Delta T = c_v \times (T -3oo)}$
$q = \mathrm{c_v \Delta T = c_v \times (T -3oo)}$
Part b)
$-100 \times (V-5) = \mathrm{c_v \Delta T = c_v \times (T -3oo)}$ since $\Delta U$ is a state function
$\mathrm{V = \frac{nRT}{p}} = 0.133T$  Now I am not sure if this step is correct. I used that $p = \pu{100 kPa}$. Is that right? From there I can solve for $T$ and hence for $V$.
Could someone please check if above working out is correct?

I didn't check you arithmetic, but your methodology is correct.

@ChesterMiller thanks, that is fine. Just to make sure, it is correct to use p as 100 kPa for finding the volume?

Yes.  The implication of the problem statement is that you allow the gas to continue to expand until it equilibrates at its final state.


text:
Free chemical database to search for physical properties?
I am looking for an Open-access tool that can search "backwards", eg. give me all the structures corresponding to a given heat of formation, critical pressure, etc. It would be even better if the database canbe automatically searched by API or by downloading it. Currently my best go is ChemExper, where I can search for some of the properties, but it is challenging to make it automatic. Is there anything similar in the field?

Related: http://chemistry.stackexchange.com/questions/22285/public-api-for-physical-properties

As I've mentioned on other related questions, this is very much an open topic. There aren't many open-access databases in chemistry, and far fewer that have APIs.

As suggested in the other answer, there are multiple ways to query / scrape Wikipedia.
ChemSpider offers some property search features.
Reaxys offers some property searching, including spectra features. It's not open access.

Indeed, what I see is open access chemistry databases have a long way to go. The reason I asked this question is exactly this: there might be some databases that slipped my knowledge. On an open forum like stackechange, there might be someone who knows something.

@Ezze I think it's a good question. It's just not that different from the last question

Yes, you are right. I didn't see the question. However, I am hoping for a better answer this time.

@Ezze - Nothing has changed since then. In a few months, we'll be rolling out API-based property search for PQR but these would be computed values.

You might search the German Wikipedia using SPARQL via DBpedia.
Go to http://de.dbpedia.org/sparql and run your query, e.g.
PREFIX dbo:<http://dbpedia.org/ontology/>

SELECT ?chem ?bp
{
  ?chem a dbo:ChemicalSubstance .
  ?chem dbo:boilingPoint ?bp .
  FILTER ((?bp >= 372) && (?bp <= 374))
}
ORDER BY ?bp

I've just discovered this website where you can find a tremendous amount of data, especially about inorganic chemistry with things like E-pH diagrams, phase-diagrams, things about alloys of different metals, ultra pure silicone, salts, elements, nuclear data and so on.
And if you prefer sugar derivatives here is a great database with online calculators about many sugar things, syrops etc.
Enjoy !!!


text:
What solvents work on aloe vera?
I'm looking for a solvent which can fully dissolve dried aloe vera gel. I assume the main problem is the constituent polysaccharides.

Hot water, the "universal solvent", should work nicely.

Care to elaborate? This looks more like a comment than an answer at the current moment.

Aloe vera gel is almost entirely made of water-soluble  components, or at least contains a large enough fraction of water-solubles that hot water will work nicely to clean it.


text:
How to approximate the pH of a box of water molecules with one proton?
I have a cubic box of length $L = 12.42\ \mathrm{\overset{\circ}A}$, containing $60\ \ce{H2O}$ molecules and $1\ \ce{H+}$ ion. How can I approximate the $\mathrm{pH}$ of the system?
Following the basic definition, $\mathrm{pH}=-\log\left([\ce{H+}]\right)$, I tried to calculate it like the following two ways:
\begin{align}
  \mathrm{pH} &= -\log\left(1/60\right) = 1.78 \tag1\\
  \mathrm{pH} &=
    -\log\left(
      \frac{1/6.022\times 10^{23}}{12.42^3\times 10^{-30}\times 10^3}
    \right)
    = 0.0621 \tag2
\end{align}
However, I am not sure whether these approximations are valid or not in this regime.

Chemistry is not about individual molecules at all. But if you insist, well, then your second way must be right.

For a rough, back of the envelope, sanity check sort of calculation, your second approach is more appropriate than the first. (Though check your value for Avogadro's number.) This is something that should jump out at you if you keep track of units in your calculation. The $\ce{[H+]}$ in the equation should be measured in moles of hydrogen ion per liter, so your calculation needs to result in an inserted value that's in moles of hydrogen ion per liter. You can also rule out the first equation by performing a thought experiment: if the size of the box doesn't factor into the equation, does it make sense that the pH wouldn't change if the box size changes? (No matter how much the size changes?)
However, the statement that pH is equal to the negative log of the hydrogen ion concentration is strictly speaking only true for dilute aqueous solutions near standard temperature and pressure. For concentrated acids/bases, solutions which are not primarily water, or for pressure and temperatures well outside standard, things get more contentious. There's a number of different interpretations of what "pH" means, and each have their advantages/disadvantages, depending on the situation.
One of the big considerations here is that "concentration" is not really really the thermodynamically relevant parameter when figuring out how chemical systems behave. Instead, it's the compound activity which is typically the more important quantity to keep track of. As it happens, the thermodynamic activity of a compound is normally proportional to the concentration in dilute solutions, so by convention we set unit activity to be equivalent to 1M solution.* This means the distinction between concentration and activity is normally ignored/glossed over in most introductory chemistry courses, but make no mistake, it's the activity of the compounds which should be used in things like the Henderson–Hasselbalch equation, not the concentration.
*(This normalization by 1 mol/L is the reason we can take the log and not run up against any strange results with dimensional analysis - it's technically a dimensionless value inside the log: the actual concentation, divided by the reference concentration of 1 M.)
The reason this matters for you is that you're dealing with a situation which isn't quite the same as a dilute aqueous solution. Nominally, it sounds like you're close. Your water density is around standard water density, and the amount of hydrogen ions are close to 1 mole per litre. But there are other considerations to take in to account. For example, what temperature and pressure are you using? Extreme values can change the hydrogen ion activity. Also, you didn't mention anything about the presence of a counter-ion. The unbalanced electrical charge might affect the predicted activity as well. The other issue is that just a single hydrogen ion will result in some pretty extensive fluctuations in the system as the system changes state. The "pH" (or hydrogen ion activity) of the system may actually exhibit time dependence, which might be important for the system you're modeling.
If you're truly interested in what the equivalent pH of the solution is, you should look into simulation techniques which attempt to actually gauge the chemical activity of the hydrogen ion under the conditions you're interested in. -- That is, figure out why you want to know the "pH" of the system, and look for more direct measures of that.

Thank you for the answer. To be more be specific, I have been trying to see the activity of 1 $H_3O^+$/2 $H_3O^+$ ions in presence of oxyanions, $XO_4^{2-}$. The systems have been modelled at NTP having overall density of about 1.0-1.1. Can you please help me with some more insights?

And thank you very much for pointing out the mistake in putting Avogadro's number, which now been corrected!

@SangkhaBorah I've actually never done such simulations myself, so I'm a little short on practical help. What I would highly recommend is an extensive literature search for people which are doing similar things, (looking at hydrogen ion activity in simulations) and attempting to replicate the approaches they used.


text:
What are some technologies/biosensors for real time kinetic analysis of reactions?
I am in a lab that has used surface plasmon resonance (SPR) and ELISA very heavily over the past few years, and we are thinking of getting a new machine. I'm wondering what other technologies exist for making similar measurements; particularly protein -protein/antibody-antiggen binding.
Octet in particular looks very different because rather than passing a flow of analyte over the ligand, it puts them together in a chamber, which I'm guessing allows the user to make much longer measurements, which could be really useful for measuring low affinities and low concentrations.
Also what are peoples experiences of different Biosensors?

This question is broad as many types of reactions exists.  Which specific reaction is of interest?

You need to specify what analytes you are interested and the level of accuracy you need.  Also mention what is the scale of the reactions.

Protein -protein interactions; particularly antibody -antigen. Small volumes of reactants are preferable to large, as some of the reactants are expensive, but I'd be interested in reading about similar technology for all scales


text:
Numbering in chlordiazepoxide
The IUPAC name of chlordiazepoxide is given as : 7-chloro-2-methylamino-5-phenyl-3H-1,4-benzodiazepine-4-oxide, but when I started numbering the structure, I am unable to find chlorine on 7th carbon, it is on 10th. Then why 7-chloro... is given is it wrong? Or I am wrong? Please help by numbering the structure in a diagram even on a paper an upload the photo.

The parent structure of chlordiazepoxide consist of the heteromonocyclic component 2H-1,4-diazepine that is fused to a benzene ring.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the name of the fused parent structure is generated as follows:

P-25.2.2.4 Heteromonocyclic components fused to a benzene ring
Unless listed as a retained name (…), a benzene ring fused to a heteromonocycle of five or more members (a benzoheterocycle) is named by placing the locant(s) indicating the position(s) of the heteroatom(s) at the front of the name consisting of the fusion prefix ‘benzo’ followed by a retained name, a Hantzsch-Widman systematic name, or a name formed by skeletal replacement (‘a’) nomenclature (…). The locants cited correspond to the full bicyclic structure. As in Hantzsch-Widman names, locants are placed in the order corresponding to the order of citation of the heteroatoms in the heterocyclic component. The locant ‘1’ is always assigned to the atom of the heterocyclic component next to a fusion atom. Heteroatoms are allocated lowest locants as a set, without regard to kind; if there is a choice, lowest locants are assigned in accordance with the seniority of the ‘a’ prefixes (…). In general nomenclature locants may be omitted when the name is unambiguous; for preferred IUPAC names locants must be cited. The letter ‘o’ of the ‘benzo’ prefix is elided when followed by a vowel. Indicated hydrogen is placed at the front of the name, when required.
  (…)

Accordingly, the name ‘benzodiazepine’ consists of the prefix ‘benzo’ followed by the name ‘diazepine’. The locant ‘1’ is assigned to the atom of the heterocyclic component next to a fusion atom and the heteroatoms are allocated lowest locants as a set (here: 1,4):

The same numbering of locants applies to the complete structure of 7-chloro-2-(methylamino)-5-phenyl-3H-1,4-benzodiazepine 4-oxide:

So is it correct that we should not number the fused carbons?

@JM97 That is correct. If you need to number the fusion carbon atoms, they are given the same number as the immediately preceding nonfusion skeletal atom, modified by a letter ‘a’, ‘b’, ‘c’, ‘d’, etc.


text:
Why is the change in entropy is different for different interpretations?
Forgive me if the title is a bit vague. Read on and I'll clarify.
So, today in class, my teacher was explaining entropy. He gave various examples explaining how the degree of randomness changes. One example in particular stuck out to me. Here it goes:
Imagine there is a container with a hypothetical boundary in the middle. Neon gas is on the left of the boundary, and Argon on the right. As long as the boundary exists, the gases can't cross it to intermix.
Now the boundary is removed, and the gases mix. This increases the randomness in the system and thus entropy increases. We get $\Delta S \gt 0$. Thus the process is spontaneous.

My Question:
How can we definitively say that entropy has increased (or that randomness has increased)?

My Reasoning:
So my teacher says that entropy increases. These following cases show that change in randomness depends on your interpretation.

Interpretation 1:
Before mixing (boundary exists):
If we take some volume of gas, say $1 \text{ cm}^3$, at a random position in the container, we might get:

Pure $\ce{Ne}$
Pure $\ce{Ar}$
A mixture of the two gases

The point is that we aren't sure what composition we get, because the distribution of gases isn't homogeneous.
After Mixing (boundary removed):
The gases diffuse and intermix to form a homogeneous mixture.
Now if we take $1 \text{ cm} ^3$ from a random position, we can be $100\%$ sure of the composition of the gases, because the mixture is homogeneous.
This means there is more uniformity now. Or, we can say that there is less randomness. This tells us that $\Delta S \lt 0$.

Interpretation 2:
Using a new approach now.
Before Mixing (boundary exists):
Take any single atom from either side of the boundary at random. My chances of correctly predicting what atom I get (either $\ce{Ne}$ or $\ce{Ar}$) are $50\%.$
After mixing (boundary removed):
I again pick any atom at random from a random point in the container. My chances of correctly predicting what atom I get is again $50\%.$
This tells us that since I still have the same chance of correctly predicting, the randomness of the system has not changed.
This gives us $\Delta S = 0$.

Same physical process, but we get 3 different values of degree of randomness ($\Delta S$) depending on our interpretation. We know that diffusion does take place, so the process is spontaneous, and hence $\Delta S \gt 0$.
My Questions:

How to accurately know whether the degree of randomness increases or not?
How is the exact value of $\Delta S$ measured in actual experiments?

Note:

Assume the gases to be ideal.
Assume the container to be sufficiently large.

Entropy is not exactly degree of randomness. see this: http://www.chem1.com/acad/webtut/thermo/entropy.html

@adianadiadi Yeah, user36790 pointed that out in the comments of his answer. The fault lies with the author of the book I am reading, who has managed to totally confound me. He uses "randomness" but doesn't explained what exactly it is and how to measure it.

From the perspective of Statistical Mechanics:
Multiplicities do depend on the external parameters that specify the energy level of the system.
For ideal gas of $N$ indistinguishable molecules, the multiplicity is given by $$\Omega= \frac{1}{N!}\frac{V^N}{h^{3N}}\,\frac{2\pi^{3N/2}}{(3N/2- 1)!}(\sqrt{2mE})^{3N-1}$$
So, indeed the multiplicity increases when the position space increases.
When the external parameter like volume of the container is increased, the energy levels come closer and then the particles can have more microstates for the same $E$.
Prior to the volume increase, the microstates of the system are $\Omega_{N,E,V_i}\;_;$ when the volume is increased, not will be there the microstates $\Omega_{N,E,V_i}$ still be available after the removal of the constraint, but also, there be addition of microstates. In general, $$\Omega_{V_f}\geq \Omega_{V_i}\;.$$
So, as the volume increases, the multiplicity increases and so does the entropy.
As requested by OP:
From the perspective of Thermodynamics:
Suppose the two dissimilar indistinguishable gases $A$ and $B$ having the same number of particles, energy, volume are partitioned from each other.
Now, when the partition is removed, the energy of the gas $A$ confined to a localised area gets dispersed over a greater volume.
Thus the entropy of gas $A$ increases and that is given by \begin{align}\Delta S_A &= Nk\,\ln \frac{V_f}{V_i}\\ &= Nk\,\ln \frac{2V}{V}\\ &= Nk\,\ln 2\;.\end{align}
Same happens for gas $B$ also.
Therefore, the change in entropy of the system is given by \begin{align}\Delta S_\textrm{system}&= \Delta S_A + \Delta S_B\\ &= 2Nk\ln 2 \gt 0\end{align}
Therefore, entropy increases when the partition is removed.

Um... I'm sorry, but I can't understand what you said. I'm just starting with entropy. So, dumb your answer down a bit please. Thanks

@Pratyush Yadav: Edited.

Thanks. So just to comfirm, "degree of randomness" is not an accurate way to describe entropy, right?

@PratyushYadav: What according to you is degree of randomness? IMO, you seem to be confused in those interpretations, I notice now actually.

It is rather confusing. My textbook says, and I quote "Entropy is the measure of the degree of randomness in a system". That's why I am confused in the first place. There is no further elaboration about what "randomness" means and how to precisely measure it. When I first read that, I just assumed that since the gases mix, obviously "randomness" increases. I started getting confused when I tried to analyze that further.

Dumb such books which maintain anonymity; follow some good texts if you want to get these topics; they are subtle for novice but equally interesting.

Randomness is actually the number of microstates accessible to a system for a given constraint. There are many words like disorder, randomness which spoil the party for a beginner (at least it did annoy me a lot).

A non-mathematical look at entropy
Now while this doesn't directly answer the OP's questions, it provides another way of looking at the concept of entropy, something that's not the dreaded ‘disorder’[1]. An addendum to @user36790's answer perhaps.
Rather than disorder or chaos entropy is probably better described as the number of ways I can do something at the individual particle level and still get the same overall result on the large scale.
Entropy is a measure of the number of micro-states[2] a system can potentialy occupy and still give the same macro-state. A micro-state is the way the particles (in your case Ar and Ne atoms) are arranged and the energy of the system is partitioned between these particles. A macro-state is a particular arrangement of a group of particles characterised by temperature, volume (sometimes pressure), number of particles etc. it is a collection of microstates. (See [1] for a some technical definitions)
Entropy compares the number of microstates that make up different macrostates. So given one macrostate we can look at the number of ways we can make that state, and then look at the number of ways we can make a different macrostate. That number we are comparing is analogous to the entropy. If one of the macrostates has more ways it can be made than the other – it has a higher entropy.
Examining the OP's System
Let's start with just looking at one gas, neon. (Why Ne? All the others are gone ;)
Before Mixing (boundary exists):
The N particles of neon occupies a volume V at temperature T. There are a number of different ways you can arrange the particles and partition the energy to give you that N, V and T.
After mixing (boundary removed):
Now lets remove the partition and let the particles equilibrate. It doesn't really matter if the partition was halfway, one third, all that matters is that there is now more volume for the particles to occupy. There are more ways to arrange the neon atoms in your container (it's bigger).[3]
You need to be careful. The macrostate your microstates can be combined to give before the partition is removed is different to the one after the partition is removed. What entropy compares is macrostates. It does this my enumerating microstates.
You can do the same for argon. The process and thinking are the same, increase in space give an increase in the number of ways to arrange the particles to give the same (new) outcome.
An analogy: You have 10 books and one shelf in your study. There are a certain number of ways you can arrange the books on your shelf (microstates) to obtain the same end result, books on the shelf (macrostate).
Now, you buy a second shelf. There is now more space (and ways) to arrange your books and obtain a macrostate.
Be careful though, this is a different macrostate to before, your shelf space (volume), and pressure (roughly how close the books are packed) have changed. There are still more ways to arrange the books on the shelf in the second state than the first though, there's more space, so the second state is considered to have a higher entropy.
($S$ is larger in the second state, going from state one [one shelf] to state two [two shelves] results in an increase in entropy of the system $\Delta S \gt 0$)

[1] ‘Disorder—A Cracked Crutch for Supporting Entropy Discussions’ Frank L. Lambert, J. Chem. Educ., 2002, 79 (2), p. 187, DOI: 10.1021/ed079p187 (Behind a paywall I'm afraid.)
Quote from the above paper:

A macrostate is measured by its temperature, volume, and number of molecules; a group of molecules in microstates (“molecular configurations”, a microcanonical ensemble) by their energy, volume, and number of molecules. In a microcanonical ensemble the entropy is found simply by counting: one counts the number $W$ of microstates that correspond to the given macrostate and computes the entropy of that macrostate by Boltzmann’s relationship, $S = k_B \ln{W}$, where $k_B$ is Boltzmann’s constant.

[2] When I see microstates I think ‘ways the system can be arranged’, every time I see microstates I just substitute in ‘ways the system can be arranged’. You have to be careful though, the way a system can be arranged includes both the positions of the particles and how the energy could be distributed through out the system. For an preliminary understanding though you can just look at the positions.
[3] You can actually look at the the box before and after, as one box, it just gets bigger. Removing the partition is just a way of visualising transitioning from one volume to another, the mixing has very little to do with it. The gasses don't interact anyway, and they occupy negligible space compared to the volume of the box, they're assumed to be ideal.

Thanks a lot for this answer. Really helpful. Wish I could upvote, but don't have enough rep.


text:
Naming an Ion with respect to its oxidation state
When identifying compounds, where certain elements are in various oxidation states we write the oxidation state in Roman numbers after it. Does the same go for ions? What would the correct way of writing Cu$^{2+}$ in words be?

Does the same go for ions? What would the correct way of writing $\ce{Cu^2+}$ in words be?

The most common way of saying $\ce{Cu^{2+}} $ would be "copper(II) ion".  However keep in mind that in aqueous solution, various ligands bind to naked copper(II) ions forming various complexes.  So for example "hexaaquacopper(II)" ions are very common.  In these names the charge is inferred from (a) the oxidation state of the metal and (b) the ligands on the copper.  So copper(II) with four negative charged chloride ligands would be "tetrachlorocopper(II)", although since the latter is an anion, it is often called "tetrachlorocuprate(II)", e.g. on Wikipedia.


text:
Why selective ring opening of the compound (attached image) by acid catalyst through results in a wide variety of branched iso-paraffins?

Basically, I want to know that why through beta scission, there is a possibility of formation of so many products, for example;
2-Ethyl-4-methylheptane;
2,4,6-Trimethylheptane;
2,2,5-Trimethylhexane;
3-Ethyl(2-methyl)hexane;
2,4-Dimethylhexane;
2,4-Dimethylpentane.

typo error- there's no 'through' in the question statement


text:
Tools for investigating hydrogen bonding
What would be the best tool to investigate local molecular organization in liquid mixtures due to hydrogen bonding?
For example, I might want to look at how water molecules hydrogen bond to PNIPAM vs with itself.
Qualitative analysis would be good (does one type of site preferentially hydrogen bond with a specific other site) but quantitative would be even better (enthalpy of h bonds and entropy  reduction due to local organization).
What is a good way to look into this?

What is PNIPAM?

Do you want a computational tool? It's a bit unclear when you say "qualitative analysis."

@ssavec I added a link to PNIPAM's wikipedia page which explains a bit about how it interacts with water. The short story though is that its a molecule that interrupts water's hydrogen bonding with itself.

I'd prefer a computational tool but I didn't want to restrict the answers if an experimental option is far superior.

The polymer is obviously very sensitive to the enthalpy/entropy balance. This would be very difficult to describe correctly using theory. You need MD simulations to have at least semi-qualitative approximation. Experimentally, THz spectroscopy is good for detecting the solvation. Check the Resolv cluster, which tries to study solvation by all possible means: http://www.ruhr-uni-bochum.de/solvation/


text:
Can we recycle garbage with the principles of mass spectrometry?
'Mass Spectrometry (MS) is an analytical technique that sorts ions based on their mass (or "weight"). Mass spectrometry is used for many chemical analyses, ranging from the analysis of a complex mixture of petroleum, to the products of genetic engineering.'
Is it possible to atomise some garbage by applying a very high voltage and a very high current across it, thereby obtaining high-temperature sparks, and then apply the principles of mass spectrometry, to separate out the individual elements for use in manufacturing?

Theoretically yes but it is highly uneconomical.

The garbage part of question makes it look weird. Besides that, I consider it quite interesting.

That's exactly how uranium enrichment worked as well. An even for uranium enrichment, the process was considered to be too expensive, so many effort was spend on developing other (chromatography based) methods that are used today. If it is already too expensive for nuclear weapons, I guess you wouldn't consider using it for waste. Bacteria do a much more efficient and inexpensive job at recycling waste for a long time now.

So what precisely makes the process so uneconomical? Besides, the target is not uranium, but common lighter elements. Doesn't that save some electricity?

Think of it like this: You'd have to evaporate megatons of material. Does that sound like a clever idea? And actually evaporation is not quite enough.

If we can get heaps of purified elements in the end, why not?  Much electricity can be derived from renewable sources nowadays anyway.

Yes it is possible, but is very expensive and would be orders of magnitude more costly than what people are willing to pay for recycled materials.  Let me give two data points to explain why it is expensive.

Some commercial material is manufactured by mass spectrometry.  The starting material isn't garbage, but reasonably chemically purified metals of various types.  The product(s) are isotopically separated metals.  Iron 57 ($\ce{^{57}Fe}$) is an isotope that is naturally present in terrestrial iron supplies at an abundance of about 2%.  It is useful to researchers that study iron materials using various spectroscopic techniques.  Pure iron 57 costs hundreds of dollars per milligram.
Mass spectrometry requires very high voltage, a very high vacuum, and large amounts of purified gases; additionally, the throughput of material separated by mass is very low.  I think that for commercial isotope purification, magnetic sector-type instruments are used.  I'm not too familiar with them, so let me go through some rough numbers for a different type of mass spectrometry: time-of-flight: Several thousand volts are applied to packets of ions which leads to their separation. Total ion currents are as high as say $10^9$ particles per second, which corresponds to about 0.1 nanoamperes or $10^{-15}$ moles per second.  So to purify one mole of iron would require $10^{15}$ seconds, or several million years of continuous operation of very expensive machinery!  Even if other types of mass spectrometry do better than time-of-flight by three orders of magnitude, it will still be cost-prohibitive.$%edit$

The Calutrons at Oak Ridge in WW2 used about 1/7th of the US electricity to mak 10s of kg of enriched uranium. That is a data point to consider.

One mole of iron-57 would be 57000 milligrams; assuming $1000 per milligram, that means you could make a mole of iron-57 for less than $60,000,000. How does that match the $10^{15}$ machine-seconds needed?

Excellent point immibis.  I think that the types of mass spectrometry that I am not familiar with are cheaper than TOF.  Even if it is a few orders of magnitude cheaper it still isn't close to economical, unfortunately.

Is this a theoretical limitation, or merely a limitation of human ingenuity? I understand bond breaking takes energy, but has someone performed an in-depth analysis on the energy cycle involved? Why precisely does the process require so much electricity? We can burn the garbage first, which I think will release some heat. The burning should be able to provide part of the energy required. To increase the throughput, can't we simply build a bigger recycling plant? As hydrogen, oxygen etc. are separated out, they can also be used to fuel the process.

@ChongLipPhang, if you burn something, energy is released because you are forming stronger bonds than before. These stronger bonds take more energy to break, so how do you propose to break even? Furthermore, MS typically does not break all bonds, so in any case you will be left with a soup of random (molecular) ions, not a collection of atomic ions nicely sorted by mass.

Who knows? By forming some stronger bonds on some less useful elements, you get to separate out the more useful elements. For instance, releasing energy in the process, O2 reacts with H2 to give H2O, which is not entirely useless. Being typical doesn't mean it is impossible. I think all bonds break if the temperature is high enough. I suggested in my original post electrical sparks as a possible means.

The preparative mass spectrometry is field of active research,
trying to overcome many obstacles. One of them is capturing the species (not only atoms, but almost anyting, undestroyed), with the hot candidate being  soft landing
Another problem is producing high enough flow of ions and removing the strict vacuum requirements of the traditional MS setup. This resulted in high-flux electrospray ionization source, where they report the deposition rate of $\ce{\approx 1 \mu g/day}$.
You can see that it is usable for preparation of very small amounts of very valuable substances, but definitely not for garbage recycling.

This is a very interesting answer.  +1 !  But the question did say it is focused on elemental separation, not necessarily of intact ions.  That said, if the deposition rate for electrospray-based preparative MS is comparable to the deposition rate for high-flux magnetic sector instruments, it goes some way to answering @immibis's comment to my own answer.

@CurtF. I intentionally ignored the elements, because that renders it more of a mining issue, extraction, separation, etc...  Regarding the flux, using the estimate of 1 ug/day, for one mole of iron it is around 150 000 years, so much faster than your estimate, but I'd better postpone the start of project by 10 years, waiting for better technology ;)

So far the assumption seems to be that the yield is one of a few valuable substances needed in a small amount. I am not sure if the same can be said about the large-scale elemental separation of garbage.


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Precipitation of Calcium ions in basic Sodium Hydroxide solution?
I recently conducted an EDTA titration of milk to find its Calcium contents. I got a value significantly smaller than the theoretical value. (≈10 times smaller). I made $50ml$ samples to which I added $2ml$ 8M NaOH to precipitate any Magnesium ions. However, assuming milk is neutral, this would make the pH of the sample equal to $13.5$. Would this cause precipitation of Calcium? If so, how could I calculate the extent of this? I know it has to do with solubility, but it is not a part of the curriculum at my school, and after a long research, I am still lost. I would really appreciate the help.