text:
Copper pyrite representation
What is the oxidation state of iron in $\ce{CuFeS2}$? And also could you please tell it's proper representation: $\ce{Cu2S.FeS.FeS2}$ or $\ce{Cu(I)FeS2}$?

$\ce{Cu2S.FeS.FeS2}$ and $\ce{CuFeS2}$ are fine, but $\ce{Cu(I)FeS2}$ is wrong. Oxidation state in parentheses used in written names; if you want to denote monovalent copper, then $\ce{Cu^IFeS2}$

Can you please also comment on their oxidation state ie. Is  iron in +2 and copper in +1?

But you've already answered this part by typing $\ce{Cu^I}$, haven't you? :)  The structure contains  $\ce{Cu+}$ and $\ce{Fe^3+}$ cations.

In the formula Cu2S.FeS.FeS2, the 1st iron is in +2 and the second one is also in +2 by me, ie it is polysulphide ion (  S2 (2- ) ion ), so their avg will also be 2 right?

No. $\ce{Cu2S.FeS.FeS2}$ is just a convenient representation, nothing else. In reality the cubic  structure consists of array of vertex-linked $\ce{CuS4}$ and $\ce{FeS4}$ tetrahedra, that's it. There is no separate standalone metal sulfides as one would imagine.

No prob, there were debates in the past whether it's iron(II) or iron(III), but these days Mössbauer and photoelectron-emission spectroscopies alongside with magnetic measurements suggest it's iron(III).

If you are really interested in details, there is an open-access article on what's going on with copper and iron in chalcopyrites.


text:
Hacking a pH meter and digital burette to partially automate a titration
This is a chemistry question about experimental techniques and technology.
I would like to connect a digital burette (Brand Titrette) and a pH meter to a computer and automate data acquisition to Excel. The pH meter would be connected to a platinum electrode with which I am measuring electrical potential during a titration. The objective is streamlined data collection to generate a titration curve (x-axis: volume, y-axis: potential) that would be used to identify the equivalence point of the titration.
Can anyone recommend an open source software that can be used to link excel to instruments?
Thank you,
Jeremy
Thank you,
JS

I suspect you are asking about a hardware hack, so if you are at the university, I'd talk to the guys at the workshop.

I'm voting to close this question as off-topic because it's about technical aspects of chemistry tools.

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text:
Validity of Law of Equivalence
Is law of equivalence valid for partial displacement reactions?
【I found this in my textbooks:
According to the law of equivalence, whenever two substances react, the equivalents of one will be equal to the equivalents of other and the equivalents of any product will also be equal to that of the reactant.】
If I consider this reaction, then the law of equivalence holds good:
$$\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}$$
Here, number of gram equivalents of each of the reactants and products are equal.
But, if I consider this reaction:
$$\ce{H2SO4 + NaOH -> NaHSO4 + H2O}$$
Here, 2 equivalents of $\ce{H2SO4}$
react with 1 equivalent of $\ce{NaOH}$ to produce 1 equivalent of $\ce{NaHSO4}$ and 1 equivalent of $\ce{H2O}$ .
Here, the law of equivalence does not hold true.
Similarly,
$$\ce{HC#CH ->[1eq  of NaNH2] NaC#CH}$$
The law of equivalence does not hold true for this.
Please explain the limitations of law of equivalence, if any.

"number equivalents" depends on reaction, that's one of reasons they are obsolete and I wonder why they were introduced in first place.

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text:
Does a polyethene shrinking signify negative or positive entropy?
When I think of something shrinking, I would intuitively think it decreases in entropy as there is less movement. E.g. if a gas was to reduce in volume, there would be less movement of the particles and hence less entropy.
How about for as solid?
I have read if polyethene shrinks, it INCREASES in entropy.
Why is this the case?

As polyethene shrinks, it doesn't necessarily reduce in volume, to begin with.

Sorry, could you please elaborate? Doesn't shrink imply a decrease in volume?

It shrinks in one direction and expands in the others.

When stretched, the polymer chains elongate and are held taut. This means there are more crosslinks and fewer opportunities for internal degrees of freedom (movement). A relaxed polymer is not held taut and therefore can freely move, having more degrees of freedom, and an associated increase in entropy.


text:
Chemistry of toilet bowl mineral stains on porcelain
I live in Massachusetts, my house has 3 bathrooms and I live alone.  If I don't scrub my toilet bowls once a week they build up brownish-yellow mineral stains.  These seem independent of whether I've actually used that toilet bowl since the previous cleaning (since I live alone some toilets are seldom used).
Looking on the web there's lots of advice from "folksy" sources like "frugal blonde" or "professional mom" about how to clean toilets -  usually involving vinegar, Clorox, pumice-sponges, etc, and lots of elbow grease, rubber gloves, bathroom fans and time.   One guy uses WD-40!     But being "folksy" none of these involve any discussion of the chemistry.
I thought if I understood the science I could save work and maybe even prevent the stains but I can't find ANYTHING on the web about this!  What are the minerals in my water and what form are they in?   Are they ionized or just some sort of suspension?  What are the principal forms -  I assume calcium, magnesium (and silicate?) compounds -  and what form do they precipitate out in on the porcelain?   What IS porcelain (chemically-speaking) anyway, and what makes such tight bonds at the interface between the minerals from the water and the porcelain surface?
If I knew the answers to these maybe I could even prevent these porcelain-mineral bonds from forming

Is the water drinkable, I.e. the same you could use as tap water?  If so, it can only be calcium carbonate. Perhaps the water is pretty hard in your area. I would made a test using a small amount of a product against limestone in the toilet reservoir as well as directly in the toilet. Then I

...then I would use only one bathroom for a while. Finally compare the porcelain of the two spare :) bathroom, the treated and the untreated one. It may work.  Congratulations for the number of bathrooms, by the way.

Addendum limestone can accumulate every kind of dirt and therefore assume tints

What do you mean " a product against limestone "?   Limestone is CaCO3, so what is the chemical reaction we're trying to obtain with such a product?   Say I use vinegar (acetic acid) - I get CaCO3 + 2CH3COOH --> Ca(CH3COO)2 +H2O +CO2.   But what, in colloquial terms, is Ca(CH3COO)2 - is it something I want in my plumbing  system?

I mean you buy a product (they are existing in liquid form too) normally use to prevent limestone (not the rock, the deposit, CaCO3 anyway) in washing machine. In your case you just poured it to the toilet not in use and you let there. It is prevention and it should work. The aftermath treatment you refer to  require to solubilze the already formed trace and thus is more difficult. Obviously you have to replace the product after flushing the bowl.

The treatment you mention is ok as Ca and acetate ions do not harm your plunbers. Just vinegar is not strong enough . Acetic acid will be more effective. Thought it can harm you depending on concentrion. It won't be as effective as prevention as it is volatile.  Better look for a product nicely dosable and that comes with a nice laundry like perfume. Hope thus help and especially works.

As I said before colour could be due to other things (see Answer) but it is likely limestone that accumulates all. Pouring store available HCl (conc ca. 8%) in  the toilet can be done periodically. Let it there and you should get a bright porcelain with minimum amount of scratching ( AFTER FLUSHING first).

Many places have water with dissolved iron and manganese, which precipitates out as a brown or black stain when exposed to air.
Massachusetts has areas of podzol, where the fine sandy glacial till had been covered by pine forests. This makes for acid soil, which in anaerobic conditions readily dissolves iron in the ferrous form, $\ce{Fe++}$ from the ground, which is a pale green color, almost unnoticeable. If you dig into a podzol, exposing it to air, the soil turns brownish or orange in a matter of minutes as the iron is oxidized to its ferric, $\ce{Fe+++}$, form. That's what is happening on your porcelain fixtures.
It's hard to dissolve the brown ferric iron stain, but there are two ways to do so:

Reduce the iron (and manganese) to the lower-valence state. Sodium metabisulfite $\ce{Na2S2O5}$ is sold for this purpose under names such as Red-B-Gone. It works well on clothing and light stains on porcelain. Use this with caution, since it can generate noxious sulfur dioxide fumes if it comes in contact with an acid.
Dissolve the oxides in a strong acid, such as $\ce{HCl}$, hydrochloric acid. Though more dangerous than reducing agents, 10%-30% $\ce{HCl}$ can remove thicker layers of stain. Use with great caution, including eye protection and rubber gloves. Rinse immediately with plenty of water if any gets on you or your clothing! $\ce{HCl}$ is available in hardware stores and in plumbing supply shops.

You can also treat the water to remove iron and manganese or to sequester it with a polyphosphate additive.
Do not use alkaline drain cleaners (e.g. $\ce{NaOH}$ and Drano®) on vitreous china, as it etches the glaze and allows stains to soak into the porcelain, making the stains difficult or impossible to remove.


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Calculate work done by burning benzoic acid exposed to the free air
I'm self-studying and can't solve this problem in Ball's Physical Chemistry. Now, the errata shows that the answer in the book is not correct; still, I can not make sense of it.

$(2.55)$ Benzoic acid, $\ce{C_6H_5COOH}$, is a common standard used in bomb calorimeters, which maintain a constant volume. If $1.20\text{g}$ of benzoic acid gives off $31,723\text{J}$ of energy when burned in the presence of excess oxygen at a constant temperature of $24.6^\circ\text{C}$, calculate $q, w, \Delta H,$ and $\Delta U$ for the reaction.
$(2.56)$ $1.20\text{g}$ of benzoic acid, $\ce{C_6H_5COOH}$, is burned in a porcelain dish exposed to the air. If $31,723\text{J}$ of energy is given off and the system temperature is $24.6^\circ\text{C}$, calculate $q, w, \Delta H, $ and $\Delta U$.

The answers(corrected from errata) are,

$(2.55)$ $q = -31,723\text{J}, \Delta U = -31,723\text{J}, w = 0, \Delta H = -31,735\text{J}$
$(2.56)$ $q = -31,723\text{J}, \Delta H = -31,723\text{J}, w = 12, \Delta U = -31,711\text{J}$

The book never talks about heats of combustion prior. Computing using a value I found on the internet of $26.4 \,\text{kJ}/\text{g}$, I get a value for $q$, I get that $\Delta U = q + w$ then and we can solve for work. This seems to be an incorrect approach since the energy given off as heat seems to already be given.
Why is the enthalpy the same as the heat in the second case? I would assume it is because the pressure is constant, but only the external is constant, which is not what is used to determine that enthalpy is the same as heat.
Where does this work come from? The volume changes in the non-calorimeter experiment, so work on the acid is done. But how does on obtain this change?
How do I solve this problem correctly?

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text:
Why does it yield iron(II) chloride instead of iron when copper meets iron(III) chloride?

Complete the equation:
$$\ce{Cu + FeCl3 -> ? + ?}$$

I wrote $\ce{CuCl2}$ and $\ce{Fe}$, but it turned out to be $\ce{Cu}$ and $\ce{FeCl2}$. I really wonder why.

Reading this q it looks like a typo was made.  Should the answer be $\ce{CuCl_2}$ and $\ce{FeCl_2}$?

Take a look at standard reduction potentials at $\pu{25 ^\circ C}$:
$$
\begin{array}{clr}
\hline
\text{Reaction} & E^\circ~(\pu{V}) \\
\hline
  \begin{align}
    \ce{Fe^3+ + e- &-> Fe^2+} \\
    \ce{Cu^2+ + 2e- &-> Cu} \\
    \ce{Fe^3+ + 3e- &-> Fe} \\
    \ce{Fe^2+ + 2e- &-> Fe}
  \end{align}
&
  \begin{array}{r}
    +0.77 \\
    +0.34 \\
    -0.04 \\
    -0.41
  \end{array} \\
\hline
\end{array}
$$
A more positive potential means a greater tendency to be reduced, so $\ce{Cu(II)}$ is easier to reduce to copper metal than either $\ce{Fe(II)}$ or $\ce{Fe(III)}$ being reduced to iron metal. So, copper metal can't force the formation of iron metal.
But $\ce{Fe(III)}$ can be reduced just to $\ce{Fe(II)}$ at a relatively high potential, high enough to allow the copper to be in its oxidized form as $\ce{Cu(II)}$. So, we can have $\ce{Cu(II)}$ and $\ce{Fe(II)}$ at equilibrium and thus we can get the corresponding chlorides $\ce{CuCl2}$, $\ce{FeCl2}$.

How to get the table in this answer properly aligned?  I tried to get the Reaction and potential columns aligned with spaces but this did not display with my intended alignment.  Thanks.

Are you comfortable with the edit?

It should be $E^0$ in table. Otherwise very nice.

@NilayGhosh Usually the reactions are aligned around arrows $\ce{->}$, and there is no need to use bold math font in header and \mathrm{...} for numbers inside array environment. Modern tendencies also discourage unnecessary vertical lines in tables.

@OscarLanzi Would you please explain your edit when you changed the notation $E^\circ$ back to $E^0$?

@OscarLanzi According to IUPAC Green Book, p. 57, only $^⦵$ or $^\circ$ are allowed to denote standard state. Since plimsoll is tricky to get right with $\rm \LaTeX$, $^\circ$ is commonly used. Putting zero ("$^0$") there is just wrong.

I was misled by @MaxW.

Yup my mistake. But geeze, super picky with the markup.

@MaxW Yes, I am. Ordnung muss sein :)

@andselisk - Nope. If you believe in entropy, then disorder rules. ;-)

@OscarLanzi (sorry for following up on an old post!) When you compare the reduction potential of Cl- ion and Cu, Cl- has a more negative reduction potential, implying Cl- is more readily oxidized to be Cl2.
I was wondering why we already knew that Cl2 would not be a product?

Iron (III) is reduced to iron(II) at a potential well below the 1.36 V listed for the chlorine/chloride equilibrium.  The chlorine thus remains reduced.

@OscarLanzi Why does that happen? Specifically, why does the iron(III) reduction occur if chlorine has a greater reduction potential?


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Any side effects to using calcium chloride as a desiccant?
Based on cost, availability, and moisture absorption capacity, calcium chloride seems to be too good to be true. Are there any side effects to using calcium chloride as a desiccant?
I'm interested in effects that are directly attributed to the desiccant itself, not so interested in the effects of maintaining an extremely low moisture level in an enclosed environment.
For example, if I were to put a container of it in my vehicle, am I exposing the inside of the vehicle to corrosive fumes? Is there a danger of the calcium-chlorine bond breaking and releasing chlorine gas? I realize that the upholstery may dry up and crack if the moisture level is too low, but that can be managed. Rust build-up in the vehicle frame, not so much.

Is your goal with this to reduce the humidity inside your car or was that just an example of an enclosed space? I think you'll find that space isn't as enclosed as you might think, at least if the car is being used.

I agree. Even if it's not used, given how fast the interior cools down at the winter or autumn, the air exchange rate is probably high. The large amount of dessicant and forced airflow through it may turn out a major challenge.

The only issue with $\ce{CaCl2}$ I've encountered is that eventually it pulls in so much water that it dissolves.
To keep dust down and to prevent the $\ce{CaCl2}$ solution from sloshing around, place it in a plastic container with a tightly-fitting sponge atop.
$\ce{CaCl2}$ is relatively non-toxic (it's used to make pickles), is inexpensive and is readily available, in cold climates, for melting snow and ice.

...if I were to put a container of it in my vehicle, am I exposing the inside of the vehicle to corrosive fumes?
Not so much fumes as dust particles from calcium chloride.
Inside the car the calcium chloride will get jostled about such that the granules will grind on each other and create dust particles. The dust particles will go everywhere. Calcium chloride will attract enough water so that it dissolves. Thus it would create a highly corrosive salt solution which could cause problems.
With all the modern electronics with connectors and PCB boards of all sorts the last thing I'd want in the car is something that could cause corrosion problems.


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Lewis Structure For H2SO4

When I was drawing the lewis structure for $\ce{H2SO4}$ I got this configuration, now I know this isn't the best configuration but I don't see whats wrong with it. The number of valence electrons is right, the free electrons are on the most electronegative element and the formal charges are all 0. I'm not sure if there's a rule I'm missing.

related https://chemistry.stackexchange.com/questions/42675/different-possible-lewis-structures-of-hso5

Obviously, what is wrong with it is that sulphur is exceeding its octet.

BTW, I think this is a good question because I see many students asking similar things - how do I know what is and what is not a valid Lewis structure?

As Jan mentions in the other answer, in general, a Lewis structure is preferred if it retains the octet rule. There can be more complicated bonding, and sulfur and phosphorous are two common examples (e.g., $\ce{SF6}$).
I find that many students can come up with "non-traditional" Lewis structures that, like yours, satisfy the number of valence electrons and minimize the formal charge. I usually consider these correct responses on an exam.
The "missing piece" is ring strain, which is not typically discussed until organic chemistry courses.
Note that your diagram as two O-S-O three-membered rings. These are extremely high in energy, because the O-O-S angle won't be anywhere near the expected 109.5°.
I tried a few quick calculations using the PM7 method and MOPAC. The best ring structure I could find looks like this (a distorted octahedral shape) and is estimated to have a $\Delta H_f^0$ = +33.05 kcal/mol. The O-O-S ring angles are ~65°.

The lowest energy geometry is the traditional $\ce{H2SO4}$ Lewis structure, estimated at $\Delta H_f^0$ = -177.88 kcal/mol. While PM7 isn't a highly accurate method, the difference in stability isn't close.

Sir so here the problem is just the ring strain, we can have hypervalecy shown in lewis structures right
?

Personally, I don't like the term "hypervalency" because the real picture is less clear but typically we draw Lewis structures as I depict, and the problem with the structure in the question is of ring strain. When I see logical (albeit incorrect) Lewis structures, I typically give full marks.

You are missing the octet rule. In your proposed structure, sulphur would have a total of twelve valence electrons which just doesn’t happen.

Except, there are plenty of "hypervalent" sulfur compounds - it's one of the more common examples, e.g. octahedral $\ce{SF6}$

Agree with Geoff. The traditional Lewis structure of sulfuric acid also has 12 electrons. You can argue you draw that with formal charge on S and O and whatnot, but you could easily do the same with OP's structure - just break two of the S–O single bonds and tack on + and – charges appropriately. Heck, you can even have resonance forms.

@GeoffHutchison Hypervalent =/= violating the octet rule. We know about 4e3c bonds and the likes and we should teach students to follow these principles rather than outdated and wrong ways such as d orbital participation.

@orthocresol True, the short answer I gave does not go into all those details which I consider not relevant for this question. But I could add a note on all those other possibilities.

@Jan - I used quotes around "hypervalent" very intentionally. If students are aware that $\ce{SF6}$ is a known, stable, sulfur compound, they can and will come up with this possibility for $\ce{H2SO4}$. Yes, my preferred structure for $\ce{H2SO4}$ satisfies the octet. But IMHO, the main problem with this structure is ring strain.


text:
Stoichiometry : dilution questions
I need 500 ml of 0.2 mol/L  H2SO4 for a tutration lab. What volume in mL  of the 18 mol/L stock solution should you use?
I think this is a dilution question asking me to use C1V1=C2V2. I am confused because it sounds like s dilution question, but its asking for volume so i could possibly be using the (v/v) formula. If someone could help me out that would be greatly apprecaiated, trying to study for a test.

This is a homework-type of question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your work for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

I gotchu.
It's definitely a dilution question. Just plug it into the equation you have, C1V1=C2V2.
$C_1V_1=C_2V_2$
$(0.2\: M)(500\: mL)=(18\: M)(x\: mL)$
Solve for x:
$x=5.56\: mL$
Therefore, you want to use 5.56 mL of 18 M stock solution. To make your 500 mL of 0.2 M sulfuric acid, just take the 5.56 mL of the stock solution, then fill it up with water until you have 500 mL total solution.
The idea behind the equation, by the way, is to preserve the number of moles in the solution while changing the total volume of the solution. Here, we find out that (0.2 M)(0.5 L) = 0.1 mol of sulfuric acid is needed in the final solution. So, we find the amount of stock solution needed to provide that 0.1 mol of sulfuric acid, by dividing 0.1 mol by the concentration of the stock solution.


text:
How to be able to tell if a molecule will react via SN1 or SN2?
I've been having a lot of issues with a question.
2 compounds, (1-chloro-2,2-dimethylpropyl)benzene and (1-chloro-2-methylpropyl)benzene react with EtOH in water.
It asks to classify the reaction mechanism for each and to then go on and explain why (1-chloro-2-methylpropyl)benzene has a much faster relative rate.
I am having major issues with the question, because I just cannot tell where there are any major differences between molecules which could make a difference. Both of them have a benzene ring, but this stabilises both the transition state and the carbocation intermediate so will favour both SN1 and SN2. Both reactions are done in the exact same solvent with the same nucleophile.
The only differences I can spot is that in (1-chloro-2,2-dimethylpropyl)benzene, there will be more stabilisation due to the inductive effect and hence it will favour SN1. But this does not help with determining the mechanism for (1-chloro-2-methylpropyl)benzene.
Any help will be much appreciated. Many thanks :)

I think you need to consider steric factors when thinking about the relative reaction rates

Factors affecting SN1 mechanism:

Weak nucleophile / base and Good leaving group.
More steric hindrance.
Polar protic solvent should be used.

Factors affecting SN2 mechanism:

Strong nucleophile / base and Good leaving group.
Less steric hindrance.
Polar aprotic solvent should be used.

Coming back to your problem. Actually, both of the compounds can go through SN1 or SN2 mechanism as the carbon linked with $\ce{Cl}$ is a secondary carbon. But, as you mentioned that you're using $\ce{EtOH}$, and you should know that $\ce{EtOH}$ is little basic, almost neutral, therefore, we can say that now, both of the compounds will go through SN1 mechanism (Remember, the first point of SN1 mechanism, i.e., weak base).
Now, the question is if both the compounds will follow SN1 mechanism, then which will react faster. For this, we will have a look at the SN1 mechanism:

Step 1: Leaving of the (-X) group
Step 2: Weak nucleophile attacks from either side
Step 3: Substitution product formed (Inversion or retention, based on
  the step 2)

So, our concern is only the step 2.
Note that, rate of reaction in step 2 depends on steric hindrance in your case. So, let's have a look at the structure of (1-chloro-2,2-dimethylpropyl)benzene and (1-chloro-2-methylpropyl)benzene:

The Carbon with red dot is the point where $\ce{EtOH}$ will attack after the (-Cl) will leave.
Now, on comparing the both the compounds, you will notice that in the first compounds there is greater steric hindrance as compared to the second compound. And, that's the reason why second compound will react faster than the first one.
Conclusion:

(1-chloro-2-methylpropyl)benzene will react faster than
  (1-chloro-2,2-dimethylpropyl)benzene due to less steric hindrance
  around the carbon that is directly attached to aromatic ring (i.e.,
  the carbon with red dot).
And, both of the compounds will go through SN1
  mechanism due to presence of $\ce{EtOH}$.

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text:
Inductive effect in cyclic compounds vs open-chained compounds

Consider the two carbocations. Which one is more stable? I think it should be the open-chained one as it has 2 more H atoms than the cyclic one, so polarity will be greater; but I suspect that there might be some other factors at play  here due to the ring that I might have missed.


text:
How good is aromaticity at stabilization?
I've been taught that aromaticity is often a very good stabilization factor. Then I came across this question in which another compound was said to have more stability.

Upon deprotonation arrange in the order of stability.

Why is S more stable than P?
(the answer given is (a); from Himanshu Pandey Organic Chemistry chapter General Organic Chemistry)

Indeed the comparison is between totally different molecules and/or their different carbocations. I can even push further saying that in principle one doesn't know without observation or calculations. Of course an answer is possible and that given is correct. Consider the effect of the two C=O in S. It is stronger than the moderate effect in Cyclopentadyenil cation.

I agree that this is not answerable without recourse to either (1) memorising pKa ranges; or (2) experiment (either in the lab, or in silico). I would recommend the former. If you look up the pKas, a typical 1,3-dicarbonyl compound has pKa 10. Cyclopentadiene has pKa 16 ish, so it is pretty close, actually. 1,3-Dithiane has pKa... I’m not sure, but I’d guess around 30 or 35. Of course, a lower pKa means stronger acid means more stable conjugate base.

I am not sure lab experimentation would work either.  How do you work with (P) when it's prone to self-dimerize?  It looks like the student is supposed to know pKa values by rote.  Sucks to be given that kind of question.

Where can i get the pka ranges? Not all of them i just want to get a rough idea.

@OscarLanzi, memorising a few pKa ranges is not that hard and I'd venture to say that anybody half-serious about organic chemistry should have a rough idea of how acidic different molecules are. Just like how an transition metal organometallic chemist worth their salt should know the layout of the d-block.

@AvatarShiny A very extensive listing of pKa values is here; obviously it makes no sense to memorise all of them, but they are conveniently categorised based on functional groups, so you should be able to get a general feeling of relative pKas as well as how much they can be affected by substituents. You will also probably see more of these as you do more chemistry.

@Orthocresol the problem was you had to decide between a ring made inherently electronegative by aromatic stabilization versus conjugation to highly electronegative oxygen atoms.  At an intuitive level that is a much tougher call than, let us say, comparing the carbonyl groups to sulfur atoms.

The situation in S and P is very similar.
Essentially you have a negative charge $\alpha$ to two double bonds which will delocalize these charges.
In the case of cyclopentadiene, this will be stabilized because the anion is aromatic.
But still, there will be a charge which is supported by carbon atoms.
In the case of the 1,3-diketone, the charge will also be delocalized by two double bonds, but instead of an aromatic anion of five members, the charge will be supported by only three atoms.
However, two of them are oxygens which can hold a negative charge much better than a carbon.
The electronegativity of carbon is 2.55 while for oxygen it is 3.44.
Note that the electronegativity scale ranges from 0.7 for $\ce{Fr}$ to 4.0 for $\ce{F}$.
This ability of supporting negative charge is much bigger than the small contribution of aromaticity to the stabilization of a carbanion.

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text:
Can a molecule contain a linear series of conjugated pi bonds twisted through 180 degrees?
Imagine a long chain of carbon atoms connected by single and double bonds in conjugated fashion. All of the p orbital pi bonds would normally be aligned to either side of the carbon atoms forcing the molecule to adopt a flat structure.
Would it be possible to take this structure and twist it slightly so as to form a long helical shape?
The bond overlap would not be as good but if the chain were long enough I would have thought the bond weakening would not be excessive.
Obviously both ends would need to be firmly bonded to some other inflexible linking components such as a large protein molecule or a long chain of aromatic rings.
Has anything like this ever been synthesised? Is it even possible?
Here is an example either end of this structure would be connected to a larger molecular segment that would hold it in place and prevent it from untwisting. I want to give this normaly flat structure a 180 degree twist.

Is this what you mean? https://www.chemistryworld.com/news/acene-synthesis-sets-new-record-for-twisted-molecules/3008304.article

@Zhe I think it might be, although it’s not quite clear what the structure is between the two 5 membered rings as they are both side on, is that just a double bond or is it a ring? Hard to be certain from side on, although I assume a ring wouldn't fit?

There are examples of molecular Möbius strips where the ends of the chain connect with each other to form a loop where the pi orbitals rotate 180° across the chain.  See this earlier answer for such an example.

@Slarty : There are no 5-membered rings in that image.  The side-on component is another of the 4-ring structures seen at the left and right ends.  (As the person looking over my shoulder attests, the "little red ends" give this away.)

@Slarty what are you actually looking for as crystalline polymers can adopt helical structures, with the number of repeat units in a single "twist" depending on the side chain. E.g. Polypropylene has more repeat units occurring in a single helix twist than polystyrene.

The material you imagine has been a target for Natta, best known together with Ziegler for they work on isotactic polymers.
Natta was interested in what you describe - polyacetylene - as the extended conjugation would have likely led to unusual electronic properties.
Calculations and models were around theoretically, before and in  parallel to the actual synthesis.
The material prepared by Natta was however insoluble and basically no work up was possible.  This is not totally surprising if we look at the structure.
After decades, the fortuitous use of thousand time the amount of catalyst led to the formation of nice shiny films of polyacetylene. Still insoluble but so different from commons residues to attract sufficient attention to cause the gathering of brilliant scientists and finally the emergence of an entire field, that of conducting (or better, conjugated) polymers and so called plastic electronics.
https://www.nobelprize.org/nobel_prizes/chemistry/laureates/2000/
https://en.m.wikipedia.org/wiki/Polyacetylene
This is to answer the final part of the question.
Concernig the first part, if I understand correctly your picture, I can tentatively answer in an apparently convoluted way.
As far a chain of alternating single and double bonds is considered, and without accounting for delocalisation, we have basically free rotation around the single bound. As such, any statistical coil can be obtained.
Conversely, if one imagine a "metal like" structure with full electron delocalisation, the chain is than stiff.
The reality is somewhere in the middle. Real conjugated polymers should be regarded as delocalized pi systems (which at the end is the reason of the huge interest on them) inerhently not much flexible and not very soluble.
Still, a long chain is susceptible to adopt a variety of elical or coiled conformations as a result of collective little departures from planarity. Ideally, you can turn very rigid rod on themselves providing they are long enough.
One cannot apply Flory theory as in polyethylene, but adaptations exist to account for rigidity due to order bond > 1 in conjugated polymers.
Also let me note that independent of the extent of delocalisation (also known as effective conjugation) a linear series of sp2 hybridized C atoms can close to a ring - no need for examples here).

Additional notes:
1- Solubility in common solvents is a major challenge towards real  application of conjugated polymers. A common strategy is the side substitution with moderately long alkyl chains. This in generak results in improved solubility via an extra entropy term.
2- polyanilyne far preceded the interest on polyacetylene. This systems is more complicated and in a way interesting, but less as far organic electronics is in scope.
3*- a detailed and chemically exhaustive paper appeared well before the paper mentioned in the linked Nobel prize. It has been published in the Journal of the  Australian chemical society and it details the preparation and properties of polythiophene.
*(Making products and selling them are two different tasks. This is the lecture I got)

The longest conjugated chain with well-determined crystal structure that would allow to determine the twist angle is cyano-substituted oligoene with 13 all-trans conjugated $\ce{C=C}$ bonds 2,5-bis(11-phenylundeca-2,4,6,8,10-pentaen-1-ylidene)hex-3-enedinitrile (CCDC 847947) [1]. It is also discussed in the article what might lead to a twist in the chain:

The longer oligomers, DPDC7–DPDC13, co-crystallize with one solvent molecule per unit cell and are not fully planar. The central dicyanobutene
  moiety is planar and to either side out-of-plane bending of the conjugated backbone is accompanied by a slight twist. DFT calculations at the B3LYP/6-31G** level predict fully planar conformations, however the incorporation of solvent as well as bending is common in crystal structures of oligoenes having
  more than six $\ce{C=C}$ bonds.
We were concerned that the cyano groups would disturb the local structure of the oligoene backbone and minimize the degree of conjugation. Despite the observed bending, the central alkene unit (that having doubly allylic cyano groups) remains planar with a dihedral angle of $180.0^\circ$. This suggests that deviations from planarity may be attributed to co-crystallization with
  solvent molecules and not the result of cyano-functionalization.

Twist angle between mean planes of terminal phenyl groups (red) and cyano groups in the middle (green) is $20.6(3)^\circ$ (each plane is denoted with an arrow(normal); both terminal phenyl groups are parallel):

So, the twist can be activated by external factors; however, there is still no known examples of $180^\circ$-twisted trans-conjugated oligoenes.
Reference

Meisner, J. S.; Sedbrook, D. F.; Krikorian, M.; Chen, J.; Sattler, A.; Carnes, M. E.; Murray, C. B.; Steigerwald, M.; Nuckolls, C. Chem. Sci. 2012, 3 (4), 1007–1014. DOI: 10.1039/C2SC00770C.

I just add that in crystals a rather coiled/random  conformation is obviously unlike. Your example however is a very good one to point to the strength of effective conjugation. I will look at the paper as I am curious about Vis abs changes between crystals and a more amorphous sample.

This does not quite fit your description because it is not a chain but rather a ring, but take a look at Möbius aromaticity. The concept revolves around building large rings of conjugated p orbitals in such a way that following the ring will make you end up on the opposite phase of the same carbon atom’s p orbital. This is an to Möbius band that you can create in real life and are mathematically somewhat fascinating.
A good introduction into Möbius aromaticity, its history as a concept in chemistry and the first attempts to calculate and synthesise such molecules is presented in a review article by Rzepa.[1]
In recent literature, a number of polyporphyrin structures which can adopt Möbius, figure-8 or even more twisted structures have been synthesised; an overview is provided by Stępień et al.[2] However, I also want to highlight one specific example which I personally consider to be most interesting.

Figure 1: the octaporphyrin system synthesised by Tanaka et al.[3]
In 2008, Tanaka et al. published a figure-8 type octaporphyrin.[3] When drawn traditionally (figure 1), it seems like there are two porphyrin-like coordination sites side by side; however, when viewed three-dimensionally it is evident that these are not planar due to the crossover at the centre of the 8 and there is no easy way for one metal to be coordinated by four nitrogen atoms in each of these. Instead, when palladium(II) acetate is added, two palladium atoms are coordinated in an NNNC fashion with one $\ce{C-H}$ bond being activated by each. Two such products were isolated in $\pu{51\%}$ and $\pu{20\%}$ yield, respectively; both gave crystals suitable for diffraction analysis.

Figure 2: major and minor products corresponding to the Hückel (A) and Möbius (B) isomers of $\ce{[Pd2(\mathbf{2}\text{-}H_{-4})]}$ with the π path highlighted by colours.[3]
The major product was identified to be the figure-8 twice-twisted and thus Hückel (anti-) aromatic compound depicted in figure 2 A. The minor product was identified to be the remarkably stable Möbius aromatic compound depicted in figure 2 B. Therein, you can also see the π conjugation pathway: in compound A it can be clearly separated into a white path and a blue path while in compound B the white and green path meet in a phase transition.
The $\ce{^1H}$-NMR shifts of protons on the inside of the Möbius ring — $\pu{-1.77ppm}, \pu{-2.93ppm}$, and $\pu{0.24 ppm}$ — clearly indicate the presence of a ring current which is further confirmed by the downfield shift of outer protons. The aromaticity of the 36π electron system is best explained by Möbius aromaticity, the authors claim. (Likewise, for the figure-8 system, a Hückel antiaromatic system is expected by the same reasoning.)
Finally, the authors conclude that:

In summary, metal complexes [B and others] are stable, conformationally locked Möbius aromatic molecules that display distinct aromaticity. The formation of such Möbius aromatic complexes is not only spontaneous, but is also surprisingly general upon metalation with Group 10 metals, as seen in the case of octaphyrin, heptaphyrin, and hexaphyrin. While requiring further study, we currently believe that the formation of Möbius-type aromatic systems under the conditions of metalation is driven by a need to relieve the internal strain that would accumulate upon the formation of complexes in the absence of twisting, with the energetic stabilization provided by Möbius-type aromatization. These results unambiguously validate the concept of Möbius aromaticity predicted by Heilbronner more than 40 years ago. Moreover, the ready availability of Möbius aromatic complexes from expanded porphyrins is expected to stimulate further studies in this area, including their applications in development of functional materials.

References:
[1]: H. S. Rzepa, Chem. Rev. 2005, 105, 3697−3715. DOI: 10.1021/cr030092l.
[2]: M. Stępień, N. Sprutta, L. Latos-Grażyński, Angew. Chem. Int. Ed. 2011, 50, 4288–4340. DOI: 10.1002/anie.201003353.
[3]: Y. Tanaka, S. Saito, S. Mori, N. Aratani, H. Shinokubo, N. Shibata, Y. Higuchi, Z. S. Yoon, K. S. Kim, S. B. Noh, J. K. Park, D. Kim, A. Osuka, Angew. Chem. Int. Ed. 2008, 47, 681–684. DOI: 10.1002/anie.200704407.

Very interesting. I wonder if the "single sided" p orbital pi bond has any unusual effects over the usual double sided versions?

The example of this I've seen contains silicon as the d- orbital is in phase with one side of the ring and out of phase on the other side.

@Beerhunter … Seriously? Silicon’s d orbitals are, of course, far too far removed to take part in conjugation to any meaningful extent. (Wish I could downvote a comment.)

The molecule was cyclic but not fully conjugated


text:
How does % purity on household products work?
For example, toilet bowl cleaner is 9.5% HCl. Does this mean that for every 90.5 moles of H2O (or other stuff) there is 9.5 moles of HCl? Does it mean that for every 90.5 g of H2O there are 9.5 g of HCl?

Do you see a little "(w/v)" next to those figures? ;-)

@paracetamol no, it just says 9.5% active ingredients and 90.5% other ingredients edit: here's an img https://media-services.digital-rb.com/s3/live-productcatalogue/sys-master/images/hfe/ha4/8814543568926/110573.jpg?width=1280&height=1280 It's right in the bottom, you can see 9.5% and 90.5%.

Well, I'd expect it to be $x$ grams of $\ce{HCl}$ in $y$ units of water. Whether the water's weight or volume is considered would depend on the regulations for such products (which I'm unfamiliar with). If we're talking about pharmaceuticals, the the percentage is often a w/v thing (number of grams of something in 100ml of solvent).

Thanks, I guess I'll keep looking. Kinda weird that we don't know if this is weight or molarity lol

Amusingly (not really), the MSDS also lists things as a simple "percentage", and I can't find a standard as to whether this is v/v, w/v, w/w, or whatever the manufacturer's degenerate heart desires that day.

For example, toilet bowl cleaner is 9.5% HCl.

First off, do you see "w/v" mentioned after those figures, i.e- $\ce{HCl}$ 9.5% (w/v)?

Does this mean that for every 90.5 moles of H2O (or other stuff) there is 9.5 moles of HCl? Does it mean that for every 90.5 g of H2O there are 9.5 g of HCl?

Even if you don't, I'd still expect the concentration to be expressed in the form: $x$ grams of $\ce{HCl}$ for every $y$ units of water. As for whether water's mass or volume's considered, it would depend on the regulations for such products (which I'm unfamiliar with).
However, I wouldn't be surprised if it did turn out to be a w/v quantity (which is quite common in pharmaceutics).

Kinda weird that we don't know if this is weight or molarity.

Like I said, I'm going to throw my weight behind "weight", since it's far more common than molarity outside of a lab (come to think of it, I've never seen a household product's ingredient concentration expressed in molars).
Not exactly important, but since you mentioned "molarity", here's a handy formula to convert something in % w/v to molars:
$$\text{Molarity} = \frac{{10}{(w/v)}{(d)}}{M}$$
Where $\ce{d}$ is the density of the solution, $\ce{M}$ is the molar mass of the solute and $\ce{w/v}$ is the mass concentration of the solute.

I would take it here and in general  as mass percentage. Expressing conc as % always leads to doubts although it is originally intended as mass to mass ratio. The good practice should be stating m/v when is the case.
Household HCl per se (not other ingredients) should be at about the percentage you mention in m/m.
Note that as always with solution in water, the concentration expressed in the two ways tend to the same value as lower it is. That is the reason people started to neglect the difference.
I am sensitive to this because in my domain (interdisciplinary) you face concentration given as % with no more info  and the solvents are even organic !
So that the actual concentration or two supposedly equally concentrated samples can be very different. This leads to ambiguity in protocol but can highly interfere with data interpretation and relationships. Fortunately it is nothing about quantitative analysis.
Basically you are let in dark and (I do, based on experience) guess they mean 1 mg for 1 ml solvent !
I mostly write this to sensitised the readers to adopt the good practice. 1 mg of matter in 1 ml of dichloromethane is 1 mg/ml, not 1%! Alternatively it does not cost much to add m/v.
Edit: not to mention that v should refer to the final volume of the solution. ...


text:
Am I Using the Correct Number of Significant Figures (Heat Problem)?
The question is:

A patient arrives in the emergency room with a burn caused by steam. Calculate the heat, in kilocalories, that is released when $\pu{18.0 g}$ of steam at $\pu{100 °C}$ hits the skin, condenses, and cools to body temperature of $\pu{37.0 °C}$.

1) I solve for the steam as it condenses:
Heat of vaporization: $$\Delta H_\text{vap}(\ce{H2O}) = \pu{2260 J/g}$$
Heat: $$\begin{align}Q_1 &= \pu{18.0 g} \times (\pu{2260 J/g}) \times (\pu{1 cal/4.184 J}) \times (\pu{1 kcal/ 10^3 cal})\\ &= \pu{9.722753346 kcal} \to \text{rounded to 3 SF} = \pu{9.72 kcal}\end{align}$$
2) I solve for the cooling of the $\ce{H2O}$:
Specific heat capacity: $$c(\ce{H2O}) = 4.184\ \mathrm{J/(g\ ^\circ C)}$$
Temperature change: $$\Delta T = \pu{100 °C} - \pu{37.0 °C} = \pu{63 °C}$$ (2 SF)
Heat: $$\begin{align}Q_2&=\pu{18.0 g} \times \pu{63 °C} \times 4.184\ \mathrm{J/(g\ ^\circ C)} \times (\pu{1 cal/4.184 J}) \times (\pu{1 kcal/10^3 cal})\\&= \pu{1.134 kcal} \to \text{rounded to 2 SF} = \pu{1.1 kcal}\end{align}$$
I then add the two results together and get:
$$\begin{align}Q&=Q_1+Q_2\\&=\pu{9.72 kcal} + \pu{1.1  kcal} = \pu{10.82 kcal}\end{align}$$
$\to \text{rounded to 3 SF} = \pu{10.8 kcal}$ (using the addition/subtraction rule for SFs)
The textbook's answer is $\pu{11 kcal}$.
Is my answer right, according to significant figure rules, or is the textbook's provided answer correct?

Todd, I have a quick question: in an equation, such as specific heat, is the temperature change (Delta T) irrelevant as a significant figure? It's causing me a ton of grief. And, do you know of any texts or resources that may help me understand multi-step problems, in regards to significant figure amounts? Thanks a lot.

Actually 18.0 has 3 significant figures. 18 would have two

Nice job of clearly showing all your work so someone could find your mistake and give an answer. That is what we ask of homework type of problems, so I have no idea why the downvotes. Also, welcome to Chemistry.SE.

@Raven - Thanks for the catch. Ignore my previous comments.

This is sort of tricky, but I'd say that neither 10.8 nor 11 is right!
You got
$$Q_1 = 9.7\bar22753346\ \mathrm{kcal}$$
which you rounded to $9.72\ \mathrm{ kcal}$. The bar over the first two indicates that is the significant figure and that the excess numbers can be carried in intermediate calculations but would be trimmed for the answer.
Now for the second part you got
$$Q_2 = 1.\bar134\ \mathrm{ kcal}$$
which you rounded to  $1.1\ \mathrm{ kcal}$.
Using the intermediate calculations you are calculating the sum.
$$\begin{align}
9&.7\bar22753346\\
1&.\bar134\\
Q_1+Q_2 =10&.\bar856753346 \end{align}$$
Since 9.72+ answer is good to hundredths, but 1.1134 answer is only good to the tenths, thus the answer must be rounded to the tenths. So the "best" answer using significant figures is:
$$Q_1+Q_2 = 10.9\ \mathrm{ kcal}$$
You should not round intermediate results to their "correct" number of significant figures since that leads to rounding errors in the intermediate calculations.
$$\begin{align}
9&.7\bar2\\
1&.\bar1\\
Q_1+Q_2 =10&.\bar82 \end{align}$$
Obviously carrying the ten digits in $9.7\bar22753346$ is somewhat ridiculous, so a good rule of thumb with a calculator would be to carry two extra digits in intermediate calculations. So round the intermediate number to $9.7\bar228$. The value $1.\bar134$ has two extra digits so its fine as shown.
Notice that this is exactly the same rule you used in calculating the temperature difference.
$$\begin{align}
10\bar0&.\\
-37&.\bar0\\
=6\bar3&.\end{align}$$
Had it been 37.5 degrees for body temperature, then you should have used $6\bar2.5$ in the intermediate calculations.
$$\begin{align}
10\bar0&.\\
-37&.\bar5\\
=6\bar2&.5\end{align}$$
As a matter of curiosity I used the rnd() function in Excel to simulate uniform distribution between all the values. So:

17.95 grams to 18.04999 grams of water
99.5 to 100.49999 degrees C
36.95 to 37.04999 degrees C

$\text{=(2.26)*(18+0.1*(RAND()-0.5))/4.184+(18+0.1*(RAND()-0.5))*(100+(RAND()-0.5)-(37+0.1*(RAND()-0.5)))/1000}$
The 4.184 and 2260 values were assume to be exact. Averaging over a thousand such values I got 10.85703268
with a standard deviation of 0.016370732.
So the gist is that using significant figures doesn't give as good a result as possible. The problem is in the use of 2 significant figures in the calculation of $Q_2$. For the difference between the uniform differences of 100 and 37.0 the relative standard deviation is 0.46%. However the uniform distribution of 1.05 to 1.15 has a relative standard deviation of 2.6%, much worse than what is really happening.

Thanks for all the responses! I had just delved deeper into this problem as well, and after deciding that I didn't think 11 kJ was the correct answer, I was also able to find that 10.9 kJ would be the most "precise" answer. I have to say, this problem really helped me to develop stronger SF skills! I'm taking CHEM 101 next semester, so I'm just trying to be as prepared as I can.

I'd wonder if any chemistry book used SF correctly though out the book. So the book would have problems like "5 grams of X". Is that really just 5 grams, or 5.0 grams, or 5.00 grams etc... back in the day I used a slide rule to solve the problems. 3 SF was pretty much it, even on intermediate calculations. Rounding errors are the bane of calculations. You can't totally avoid the problem, you can just try to minimize it.

Dually noted, MaxW! Thanks for your help. This definitely taught me not to round my intermediates, and implement guard digits (like 2 more SF than needed), while doing intermediate calculations. It's amazing how it affects the end result. I joined this website now, so maybe I can contribute a bit, myself.


text:
How much additional hydrochloric acid should be added to arrive at ‘excess acid’?
I am an IB student. And I am trying to find the mass percentage of $\ce{NaHCO3}$ in antacid. I am dissolving the antacid in $\pu{0.1 M}$ $\ce{HCl}$ solution. So, first of all I conducted some preliminary experiments with pure $\ce{NaHCO3}$ powder to optimize the experiment, get a hang of it etc.
For this preliminary experiment, where I used $\pu{0.5 g}$ $\ce{NaHCO3}$, I calculated the amount of $\ce{HCl}$ solution required as being approx. $\pu{60 ml}$. Hence I added $\pu{100 ml}$ in the flask so that I have an excess of the acid as required.
Is adding $\pu{40 ml}$ as the supposed excess amount reasonable? Should it be more or less? I am confused because my teacher suggested that by even putting $\pu{100}$ ml of $\ce{HCl}$ in $\pu{0.5 g}$ of antacid tablets, I still don't have the acid in excess? But do I not? Doesn't it only take $\pu{60 ml}$ to neutralize $\pu{0.5 g}$ of $\ce{NaHCO3}$? So $\pu{100 ml}$ is definitely in excess, isn't it?

The volume you've established seems correct, as well as your assumption that extra $\pu{40 mL}$ of $\ce{HCl}$ would be a good excess:
$$\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}$$
$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{C(\ce{HCl})} = \frac{m(\ce{NaHCO3})}{M(\ce{NaHCO3}) \cdot C(\ce{HCl})} = \frac{\pu{0.5 g}}{\pu{84.0 g mol-1} \cdot \pu{0.1 mol L-1}} = \pu{59.5 mL}$$
I suspect your teacher either prepared a series of various antacids, or this one contains additional components, such as $\ce{CaCO3}$, $\ce{Mg(OH)2}$ of $\ce{Al(OH)3}$. Note that for the neutralization of the same mass of, say, aluminium(III) hydroxide, over than 3 times more $\ce{HCl}$ is required:
$$\ce{Al(OH)3 + 3 HCl -> AlCl3 + 3 H2O}$$
$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{C(\ce{HCl})} = \frac{3m(\ce{Al(OH)3})}{M(\ce{Al(OH)3}) \cdot C(\ce{HCl})} = \frac{3 \cdot \pu{0.5 g}}{\pu{78.0 g mol-1} \cdot \pu{0.1 mol L-1}} = \pu{192 mL}$$
Either way, it looks like you understand how to properly find out the amount of acid for the neutralization. I think if you demonstrate your detailed attempt to the teacher, it should clarify the situation.


text:
Energy absorbed by the gas from a PV-diagram

The $PV$-graph for a monoatomic gas is shown in the figure. Find the energy absorbed by the gas during this process.

I’m having trouble with this question because both pressure and volume are changing and I cannot calculate a value for $T$. So far, I ended up with an answer using the ratio of $C_p$ and $C_v$, and got $6.25P_0V_0$. The answer is $8.5P_0V_0$.

Using the equation you learned for calculating the amount of work, how much work is done on the surroundings n this process?  From the ideal gas law, what is the change in temperature in this process?

I suspect you forgot to take the work done by the gas into account. From the first law of thermodynamics:
$$Q = \Delta U - W \label{eq:1}\tag{1}$$
For the monoatomic gas $U=\frac{3}{2}PV$, so
$$\Delta U = U_2 - U_1 = \frac{3}{2}(P_2V_2 - P_1V_1) = \frac{3}{2}(3P_0 \cdot 2V_0 - 2P_0 \cdot V_0) = 6P_0V_0 \tag{2}$$
On the other hand, since the gas expanded ($V_2 > V_1$), the work is negative and can be found geometrically as the area under the line:

$$W = -(\color{blue}{2P_0(2V_0-V_0)} + \color{orange}{0.5 \cdot (3P_0 - 2P_0)(2V_0 - V_0)}) = -2.5P_0V_0 \tag{3}$$
Now back to \eqref{eq:1}:
$$Q = 6P_0V_0 -(-2.5P_0V_0) = 8.5P_0V_0$$

@Sanjana No prob:) Try to understand the concepts behind these problems, in reality they are all quite similar.

I actually interpreted "energy absorbed" to mean $\Delta U$. Obviously, the answer shows that I was wrong. Was there a reason why you interpreted it to be $q$?

@orthocresol Same here. I did the math for $\Delta U$ and realized that's not what the author expects from the reader:)


text:
Organic Structure Challenge
Recently we had this challenge in our Organic Chemistry Department.
I was not involved in it but I thought it might be interesting for the SE Chemistry site, so here it is.
Propose a structure for C20H18 with the longest IUPAC name.
Following rules apply:

Names must be generated with MarvinSketch: Structure>Generate Name…>Preferred IUPAC Name (https://www.chemaxon.com/products/marvin).
No isotopic markings can be used.
The structures must be possible (no cycles with too much unsaturations nor systems with impossible tension).
All the characters will be counted: numbers, letters, points, colons, dashes, everything.
Longest name wins.

Here is the example that was given:

and the name provided by MarvinSketch:
(1'R,4'Z,6'Z,8'Z,10'Z,12'S,15'R,16'R,17'R)-spiro[cyclopropane-1,18'-tetracyclo[13.2.1.04,17.012,16]octadecane]-2,2',4',6',8',10',13'-heptaene
141 characters
Good luck and prosperous year 2018!
EDIT
Well, what is a challenge without a prize?
So, I have set a bounty worth +50 reputation for the longest name.
Sorry, I can not spend much more on it.
Remember that you can post more than one structure.
If you have proposed one and after some days you find a structure with a longer name, you can post it too.

How is longest determined? By character count (e.g. using a mono-spaced font) or by the actual physical length using a proportional font. If a proportional font is used, which one?

@mhchem You just edited a formula with HTML tags instead of using own package. Should we be worried?

@MaxW  Just the character count without spaces. Any text editor will do.

@andselisk No. Don't worry. If its just 2 subscripts, <sub> does do its job quite well. I'd prefer the MathJax font would not look different to the text font, then I would.

$\ce{C20H18}$, 246 characters

InChI=1S/C20H18/c1-2-13(1)5-15(13)7-17(15)9-19(17)11-12(19)20(11)10-18(20)8-16(18)6-14(16)3-4-14/h1-4,11-12H,5-10H2/t11-,12+,15-,16-,17-,18-,19-,20-/m1/s1
PIN: (1'R,1''R,1'''R,1''''R,1'''''R,1''''''R,2''''R,3''''S)-octaspiro[tetrakis(cyclopropane)-1,2':1',2'':1'',2''':1''',4''''-bicyclo[1.1.0]butane-2'''',2''''':1''''',2'''''':1'''''',2''''''':1''''''',1''''''''-tetrakis(cyclopropane)]-2,2''''''''-diene

This is very good. Your approach has given me some further ideas, so you're not safe yet! :<

$\ce{C20H18}$, 153 characters
(1'S,2'S,3'R,6'S,8'S)-3'-(cycloprop-2-en-1-yl)-11'-[(cycloprop-2-en-1-yl)methylidene]spiro[cyclopropane-1,7'-tricyclo[6.3.0.02,6]undecane]-2,4',9'-triene

P. S. I guess someone should just start, so I simply used the slightly modified structure from the example. I'm pretty sure this one is going to be criticized a lot by experienced organic chemists, so it's more like a kick-starter.
I found it very useful to start with Avogadro first to sketch (and optimize, Ctrl+Alt+O) the initial geometry (very handy for the geometries which are distorted or resembling platonic solids), then import cml file to MarvinSketch and do the rest.
For those who would like to check the names, copy it from here, the press Ctrl+Shift+N, Ctrl+V and click Import in MarvinSketch. For the record, currently MarvinSketch 17.29.0 is being used.

Thanks for participating!!! We normally use ChemDraw but since MarvinSketch is free, I have set that one so that more people can participate.

@RaoulKessels I agree that MarvinSketch might be not that advanced, but it gets the job done, free and is available virtually anywhere, so I fully support your choice:)

Let's start nitpicking ;)  Bonds directing over plane aren't placed properly, it seems.

$\ce{C18H20}$ : 145 characters
I was curious as to the isomer with the longest name that had actually been reported in the literature rather than hypothetically proposed (so we at least know it exists).
The IUPAC name, whilst not quite as long as some of the examples, is a respectable 145 characters, largely due to the representation of stereochemistry and connectivity.

(2aR,4aR,4bS,6aS,8aR,8bS)-1,2,7,8-Tetramethyl-2a,4a,4b,6a-tetrahydrocyclobuta[a]cyclobuta[2',3']cyclopenta[1',2':3,4]cyclobuta[1,2-b]cyclopentene

The molecule was reported in a 2002 paper (J. Org. Chem. 2002, 67 4436–4440) on the rearrangements of a set of spiro[2.4]hepta-1,4,6-trienes

Are you certain that this PIN is correct? Also, this is $\ce{C18H20}$, not $\ce{C20H18}$ as in the OP. Still interesting, just bad luck.

Here is my input with character count increasing up to down. I have added InChI codes because it seems MarvinSketch is less capable of reading names than it is in their generation.
1. $\ce{C20H18}$: 164 characters

InChI=1/C20H18/c1-11-3-4-12(2)17-16-9-15-13(11)5-6-14(15)18(17)20(16)10-19(20)7-8-19/h3-6,9,13,15H,1,7-8,10H2,2H3/b4-3-,17-12+/t13-,15+,20-/s2
PIN: (2'R,4''Z,7''S,11''S)‐3''‐methyl‐6''‐methylidenedispiro[bis(cyclopropane)‐1,1':2',14''‐tetracyclo[8.4.0.02,13.07,11]tetradecane]‐1''(10''),2'',4'',8'',12''‐pentaene

2. $\ce{C20H18}$: 167 characters

InChI=1/C20H18/c1-11-3-4-12(2)17-16-9-15-13(11)5-6-14(15)18(17)20(16)10-19(20)7-8-19/h3-8,14-15,18H,2,9-10H2,1H3/b4-3-,13-11-/t14-,15+,18+,20+/s2
PIN: (2'S,2''R,3''S,7''R,11''R)‐3''‐methyl‐6''‐methylidenedispiro[bis(cyclopropane)‐1,1':2',14''‐tetracyclo[8.4.0.02,13.07,11]tetradecane]‐1''(10''),2,4'',8'',12''‐pentaene

3. $\ce{C20H18}$: 169 characters

InChI=1/C20H18/c1-11-3-4-12(2)17-16-9-15-13(11)5-6-14(15)18(17)20(16)10-19(20)7-8-19/h3-6,9,14,18H,2,7-8,10H2,1H3/b4-3-,13-11-/t14-,18+,20+/s2
PIN: (1''S,2'S,4''Z,6''Z,10''S)‐6''‐methyl‐3''‐methylidenedispiro[bis(cyclopropane)‐1,1':2',14''‐tetracyclo[8.4.0.02,13.07,11]tetradecane]‐2''(13''),4'',6'',8'',11''‐pentaene

4. $\ce{C20H18}$: 171 characters

InChI=1/C20H18/c1-11-3-4-12(2)17-16-9-15-13(11)5-6-14(15)18(17)20(16)10-19(20)7-8-19/h3-6,9,13,15H,1,7-8,10H2,2H3/b4-3-,17-12+/t13-,15+,20-/s2
PIN: (1''S,2'S,4''Z,6''Z,10''R,11''R)‐6''‐methyl‐3''‐methylidenedispiro[bis(cyclopropane)‐1,1':2',14''tetracyclo[8.4.0.02,13.07,11]tetradecane]‐2,2''(13''),4'',6'',8''‐pentaene

5. $\ce{C20H18}$: 172 characters

InChI=1/C20H18/c1-11-3-4-12(2)17-16-9-15-13(11)5-6-14(15)18(17)20(16)10-19(20)7-8-19/h3-9,14-15,17-18H,2,10H2,1H3/b4-3-,13-11-/t14-,15+,17+,18+,20+/s2
PIN: (1''S,2'R,2''S,4''Z,6''Z,10''R,11''S)‐6''‐methyl‐3''‐methylidenedispiro[bis(cyclopropane)‐1,1':2',14''‐tetracyclo[8.4.0.02,13.07,11]tetradecane]‐2,4'',6'',8'',12''‐pentaene

$\ce{C20H18}$, 290 characters
Heavily based on Daniel Bonniot de Ruisselet's idea, the two "spiraned" cyclopropane pairs next to ending cyclopropenes are fused to bicyclobutanes as well, allowing remaining methyls attachment to the ending cyclopropenes, thus also allowing more chiral centers.

PIN: (1S,1′R,1′′′S,1′′′′′R,2′′S,2′′′′S,3′S,3′′′S,3′′′′′S,4′R,4′′′R,4′′′′′R)‐2,3′′′′′′‐dimethylhexaspiro[cyclopropane‐1,2′‐bicyclo[1.1.0]butane‐4′,1′′‐cyclopropane‐2′′,2′′′‐bicyclo[1.1.0]butane‐4′′′,1′′′′‐cyclopropane‐2′′′′,2′′′′′‐bicyclo[1.1.0]butane‐4′′′′′,1′′′′′′‐cyclopropane]‐2,2′′′′′′‐diene

InChI=1S/C20H18/c1-7-3-15(7)9-10(15)17(9)5-19(17)13-14(19)20(13)6-18(20)11-12(18)16(11)4-8(16)2/h3-4,9-14H,5-6H2,1-2H3/t9-,10+,11-,12+,13-,14+,15+,16+,17-,18-,19+,20-

CXSMILES: [H][C@]12C@([C@@]11C=C1C)[C@]21C[C@]11[C@]2([H])[C@@]1([H])[C@]21C[C@@]11[C@@]2([H])[C@]1([H])[C@@]21C=C1C |c:6,32|

gzipped and base64 encoded molfile:
H4sIAGimq1sCA6WWTW7CMBCF9znFXKDW/PlvXSqxocsepIuev7YjGhFPGrBRhODl5WPmeWyxANy+
fyixAGYOlChy5gv8vZYFOIB4AGwXtfftlXOGL0bEpX5Tp4EywJsXlxVT1dCVuwjv8Iiwr0bxjmLQ
SlEnXsSk8Bllq0VdDH66lvopySgl5ZpGpfh8kAs/VYtUSijPHuSCZ5TgAno56Yifp/xTy2ku0Qnl
s1xOKcmhKM92tFFm0t0oM7ObXNQUZ2d3o8zsRimU6FeK5zSYS92NjCuF6KAjfKWjcUoutQj/daSj
HaWoLRd2iqSjHYkEnqW0NbJPqevzFO+yIK5ThwUzRml72p666ytr5JHvp9RjLtdXzjofmO67sa7W
GGU76xKPpcttp9FOFVPV3r562SRor3pTDUfe0P9abDf23mSqudkNr61mi0BoqtRVViTDWyTqMyMx
vXr34s5rq31mFID87hgvagKKvVq8fcfFyEbH1Lj7taAKMQhGZWUeeH08PE4fc6eWdll6tRC0J5R6
fa+m+uewI2iNYtfFDeDj87L8Autmps9wCgAA

(All data according to MarvinSketch 18.23, image postprocessed in a vector image editor; name postprocessed chem.SE and other text editor - RS italicization and apostrophe to prime replacement.
Marvin obviously has some issues, importing the produced InChI string produces some unspecified centers, so I have generated ChemAxon SMILES, this when pasted produces different stereochemistry, so I have added a compressed MDL molfile as well)

Can someone beat 300 characters, by further modifications?

 **162 Characters

PIN:(1'S,2'S,6'S,7'R,12'S,14'S)‐7'‐methyldispiro[cyclopropane‐1,8'‐pentacyclo[7.5.1.02,6.02,14.012,15]pentadecane‐
13',1''‐cyclopropane]‐2,2'',3',9'(15'),10'‐
pentaene
I have used MarvinSketch 17.28.0 for generating pin.


text:
Solubility of Calcium Oxalate in Aqueous Alcohol?
In searching through the literature for the solubility of calcium oxalate, I came across this post https://www.researchgate.net/post/Are_there_any_organic_inorganic_solvents_for_dissolving_Ca-oxalate I am not sure where the people answering this question have gotten the information that calcium oxalate is soluble in an aqueous solution of alcohol, but I can't view the only article that I could find agreeing with them past its abstract (I can't see the full text).  https://www.ncbi.nlm.nih.gov/pubmed/641826  Does anyone know if this is true?

I think it depends on what you define as "soluble." Wikipedia states that the solubility in water is 0.67 mg/L (20 °C) which should be more than the solubility would be in pure ethanol.

Actually calcium oxalate is a lot more soluble in ethanol.


text:
How to rationalise the drop in pH when excess aluminium nitrate is added into an aqueous ammonium solution?

Consider the flowing reaction
  $$\ce{NH3 + H2O <=> NH4+ + OH-}$$
  Explain “when excess powdered aluminium nitrate is added to the reaction mixture and the $\mathrm{pH}$ value of the solution is lowered.”

There are 3 questions that confuse me:

When $\ce{OH-}$ is removed by aluminium(III) ions, $\ce{NH4+}$ is produced, which is acidic. In that case, is the decrease in $\mathrm{pH}$ due to the formation of $\ce{NH4+}$ ions or because of the removal of $\ce{OH-}$ ions?
When $\ce{OH-}$ is removed, according to $K_\mathrm{c} = [\ce{H+}][\ce{OH-}]$, the $\ce{H+}$ ions will increase since $K_\mathrm{c}$ is a constant value. Hence, the pH decreases. Is is a correct explanation to the drop in $\mathrm{pH}$ when $\ce{OH-}$ ions are removed?
Does addition of excess aluminium nitrate mean that the equilibrium reaction will go to complete and there is no any ammonia at the end?

Note that there is also hydrolysis of $\ce{Al(NO3)3}$ (when excess is added), which also significantly drops pH.

As andselisk has already pointed out in a comment, the hexaquo complex of aluminium is a known "cation acid"
$$\ce{[Al(H2O)6]^{3+} + H2O<=> [Al(H2O)5(OH^-)]^{2+} +H3O+}$$
A practical application is the use of solid $\ce{KAl(SO4)2·12 H2O}$ (alaun) for blood clotting during wet shave.


text:
How do they ensure there is one electron on an oil drop in Millikan's oil drop experiment?
Since air is ionized using X-rays, in the 'observation chamber', there must be several electrons in the chamber, so the oil drops are exposed to several electrons so it seems intuitive that a single oil drop should catch more than one electron.
Then why is it that there's just one electron on a droplet or do they selectively choose those droplets which have a single electron residing on them?

https://physics.stackexchange.com/questions/311522/question-about-milikans-oil-drop-experiment

Oil drops would surely catch various numbers of electrons. It is just that we select those with one electron.

And, more to the point, they can only catch an integer number of electrons.

Perhaps even more to the point, did Millikan ignore droplets which apparently had fractional charges?

@matt_black Where would a fractional charge come from?

@DavidRicherby the point is that, unless you know the answer, you can’t use this experiment to prove it. And even when you do know the answer, there are experimental errors. Questions have been raised about whether Millikan ignored some actual observations to make his results look better.

You can overcome experimental errors with a sufficient body of data, using statistical techniques to show a tendency in favor of integer multiples of some factor.  @andselisk in his reference also suggests that Millikan effectively got more than one data point per drop by observing drops over time.

If only it were somehow possible to reproduce the experiment and verify the results... Oh wait.

Now we know exactly what we should look for in such a verification experiment.

There does not have to be just one electron per drop.  Say you have a drop which, in reality, picked up four electrons, another with five and a third drop with seven.  None has one electronic charge, but when you measure charges you find they have a common factor; the first drop has four times that factor, the second has five times that factor and the third drop shows a multiplier of seven.  It was this common factor, not necessarily the charge on any specific drop, that was recognized as one electronic charge.

The experiment showed all the drops had discrete amounts of charge. That means, the charges weren't all over the place (any random value). They only had specific values. Some had $2$ or $3$ or $4$ times the charge of others, but it was always some specific value that they had multiples of.
The conclusion was that oil drops didn't seem to pick up "any random amount" of charge, and the reason seemed likely to be because electric charge couldn't be just "any value". There seemed to be some basic unit of a "single electric charge", the smallest value that was found. Some oil drops had $1\times$ or $2\times$ or $5\times$ that charge, but no oil drops had (say) $3.77\times$ or $1.628\times$ that value.

Actually, the initial formula $(1)$ that was employed in the experiment $$e'= 9\unicode[Times]{x3c0}\frac{v_0+v_E}{E}\sqrt{\frac{2\eta^3v_0}{g(\rho-\rho_0)}}\tag1$$ failed to give a constant common divisor, and Stokes law was consequently corrected for radii near the mean free path of a gas particle. $|$https://doi.org/10.1103/PhysRev.2.109 (Slightly different notation, here I used subscript '$0$' to denote having to do with air, and dropped subscript $n$ in the paper for an apostrophe.)

This is my absolute favorite physics experiment; I actually wrote a miniature essay on my lab report containing, in essence, the following:
Millikan's experiment is amazingly indirect.  You don't directly measure the charge of the electron, nor do you compute it from an equation (like, say, the one relating electrostatic force to charge, though you do use this equation to deduce the non-unit charge on each oil drop).  You don't even deduce it by fitting a curve to a set of data points.  You do deduce it from a frequency graph, but it is, again, not any kind of fitting of the actual frequency data; rather, you identify spikes in the frequency counts and mark off the corresponding charges.  Then you find their greatest common divisor and argue that your data was random enough that this must be the charge of one electron (since it is vanishingly unlikely after enough droplets that each one contained only, say, even numbers of electrons).  It is actually a simple form of image recognition.
You don't need to luck into one-electron droplets.  You just need to keep measuring until you have an unambiguous set of frequency spikes at sufficiently many different charges and apply (what?!) a number-theoretic computation.
Addendum
The business of taking the greatest common divisor is a bit tricky in the presence of measurement errors: after all, the charges you find are not only subject to error from your lab equipment but also from your identification of the center of each spike.  You can assume the charges are all integers by converting your limited-precision floating point numbers to fixed point, but those integers are almost certain to have a GCD of 1.  For example, if you measure charges of 201 and 302, you'll find that the fundamental charge is not 100 (which is the obviously correct answer) but rather 1.
You can, of course, eyeball it: say, you can take various ratios and fit them to nearby rational numbers with a small common denominator (in the above example, the ratio is approximately 1.5025, so you easily find 1.5 = 3/2 as a likely "correct" ratio). A better way is to use an "error-tolerant" version of Euclid's algorthm.  In short, proceed as usual by dividing (with remainder) the smallest number into all the others and repeating, except that rather than waiting for all the remainders to be 0 (indicating that the last remainder was the GCD), you wait for them all to be "small" in some sense.  Say, an order of magnitude smaller than the previous one.
Take the above example: Euclid's algorithm gives you the following sequence of remainders: 302, 201, 101, 100, 1 (each one is the remainder of the division of the previous two).  This suggests that 100 is the correct GCD, as indeed it is.  Amazingly, the algorithm actually wiped out the measurement errors and got the exact correct GCD; I don't know if this kind of "focusing" effect is typical or if I just happened to use the right numbers.
This only increases my love for this experiment.

I actually think 101 is a better fit for your initial values. Euclid's algorithm gives progressively "better" numbers on average (not necessarily on every step), so you will find a "good" but not exact fit before you complete the sequence, unless the sequence has a low true GCD. Try it with Fibonacci numbers: 21, 13, 8, 5, 3, 2, 1. 21 and 13 are both roughly divisible by 5, for example, and 3 divides them even better because it is smaller. Note: the result of each division is very close to a Lucas number: 1, 3, 4, 7, 11...

I agree, and thanks for the discussion. One can make precise one's concept of "correct values" by putting error bars on the measurements, which may help you find a range of approximate gcds, and is better science anyway. I'm on my phone, so I don't want to work out how that would affect this example. I'm actually not entirely sure you could deduce a range just knowing the error bars, without trying every value in between.

I was taught in high school that the experiment had to be reran for hours and hours to get enough ones to be reasonably sure there wasn't a smaller number, and the GCD was used only to cross-check afterwords.

You definitely have to go for hours :). I suppose that method has the advantage of actually directly observing a single election's charge, but I'm not sure how reasonable it is. Being a mathematician, I probably like the abstract existence proof because it is more elegant and, statistically, airtight.


text:
Which is the correct mechanism for the ozonolysis of an alkyne?
According to Organic Chemistry by Jonathan Clayden, Nick Greeves and Stuart Warren (2nd edition), the mechanism for alkene ozonolysis is given as

In Advanced Organic  Chemistry, Part B: Reactions and Synthesis, Francis A. Carey and Richard J. Sundberg (5th edition), the ozonolysis of alkene is given as:

But mechanism of ozonolysis of alkynes is not given.

What is the correct mechanism of the ozonolysis of alkynes?

See this: https://chemistry.stackexchange.com/questions/25190/preparation-of-oxalic-acid-from-acetylene

Basically, ozonolysis of alkyne forms different types of compounds, and thus the reaction is too complicated and thus exact mechanism is not completely known. See this: http://pubs.acs.org/doi/abs/10.1021/ja010166f

As to your 2 examples of alkene ozonolysis, if a hydroxylic solvent is present (MeOH), then the zwitterion is trapped as the methoxy hydroperoxide  [R(MeO)C(H)OOH] and it is this species that is reduced. In non-nucleophilic solvent, the ozonide is formed and then reduced.


text:
Time until completion for $n$-th order reaction
For an $n$-th order reaction, how to determine the time until all the reactants run out?
My approach is the following. The reaction kinetics is modeled by the differential equation
$$\frac{\mathrm d[A]}{\mathrm dt}= - k{[A]^n}$$
where $n$ is the reaction order, which can take values in the nonnegative integers and even in the rational and negative numbers. Suppose we have a chemical reaction similar to the following one
$$\ce{A -> B}$$
In this problem, I'll take an example $5$ equation with similar form with the equation above, and different order $n$. Assuming the $n$ value of the $5$ example is $1,2,0,-1$ and $\frac{1}{2}$, we will get these equations for the five example after we input the $n$ value and integrate the kinetic differential equation.
(reaction with 1st order) $\ln[A] = -kt + \ln[A]_0 $
(reaction with 2nd order) $\frac{1}{[A]} = \frac{1}{[A]_0} +kt$
(reaction with 0th order) $[A] = [A]_0 -kt$
(reaction with $\frac{1}{2}$ order) $2\left(\sqrt {[A]} - \sqrt {[A]_0}\right) = -kt$
(reaction with $-1$ order) $\frac{1}{2}[A]^2 = -kt + \frac{1}{2}\left([A]_0\right)^2$
what I was thinking is, if I want to determine the time until the reaction complete, then I will have to define $t$ when $[A]=0$ (note: I will call the time that needed to complete the reaction $t_\text{complete}$) which is not defined for reaction with 1st and 2nd order because it will involve $\infty$ value, in fact as much as I know, $t_\text{complete}$ will not be defined for $n \ge 1$, but we can calculate $t_\text{complete}$ for $n \lt 1$.
so for 0th order, the $t_\text{complete} = \frac{[A]_0}{k}$
for $\frac{1}{2}$ order, $t_\text{complete} = \frac{2\sqrt{[A]_0}}{k}$
for $-1$ order, $t_\text{complete} = \frac{([A]_0)^2}{2k}$.
my question. is all the reaction with $n \ge 1$ never complete? and is all the reaction with $n \lt 1$ is the only reaction that can complete?
and I know what I've done is just manipulating the rate law equation, but if this is correct, can the $t_\text{complete}$ that I got above determine the time that needed for order $n \lt 1$ to complete?

These laws are made by observing large amount of molecules and using statistics. Once you go down to only a few you cannot (or should not) use statistics any more. You might want to look into explanations about this exact problem for radioactive decay which follows a first order rate law.

The math is right, but it hardly matters. Time until completion is not a thing at all.

As @Ivan Neretin correctly points out the completion time is not very useful. Instead we define a time for the population to be halved or more usefully when it has reached 1/e of the initial value which is called the lifetime and its reciprocal is the rate constant $k$.


text:
Image of crystal structure
Is there any websites or resources where I can get a good image of a certain crystal structure? In my case I'm looking for lead(II) tungstate $\ce{PbWO4}$ and I can't find a vivid graphic on google (except of the structural formula).
I've seen modeling structure of crystals, but unfortunately the software is commercial.

I recommend to fetch the crystal structures of inorganic compounds via specific Google search pattern site:crystallography.net <keyword(s)>, e.g. in your case
site:crystallography.net PbWO4
yields in this COD page where you can use JSmol to do basic visualization right in your web browser. Alternatively, you can download CIF and work further offline with numerous free tools such as Mercury or Olex2. Both are free and available for Windows, Linux and MacOS.
Mercury

Olex2

Thanks! The image provided there doesn't look that good. The aim is to have a high quality image for my thesis. Is there something that matches my interests?

@Ben The image you are referring to is the one generated by JSmol script, which is very configurable in terms of visualization. Do read the JSmol manual if you want to know more. Same for the desktop apps, there is no easy way around.

Ah, sorry, the other two tools provide nice graphics. Will have a look on them! Thank you!

@Ben No prob, but as I said, you got to read the manuals first, and it's going to take some time (an evening, maybe a couple). The "beauty" of JSmol is that you can easily see right in you browser whether that's the structure you want or not without tedious downloading and opening CIF files on the computer:)

Thanks, can you maybe recommend a certain drawing style ? My thesis is about particle physics and I only want to "show" the material we're using to detect particles. This means it's indeed more about (simple) visualization and less about specific information.

@Ben That's a broad one and there is no good answer. I'd recommend to check out recent related publications in your field and stick to what others did to show their structures. IMO in inorganic chemistry it makes sense to show the filled unit cell and draw coordination polyhedra for the atoms of interest  if you wish (Mercury cannot do that BTW, Olex2 can), and then show few projections. Images must be simple, clear and support your thesis; they are not there to be pretty, they are to be informative in the first place.

The tools seem to need too much effort for my actual purposes. But for instance, Diamond creates a nice image directly.
Sure information is more important than the graphic itself but we're no chemists (or something similar).

@Ben Diamond needs a lot tweaking as well, there is no program which would give you exactly what you want because there is no telepathic interface yet. Yes, Diamond is somewhat easier, but it's probably the only advantage; it only runs on Windows, stores structures in own proprietary format (diamdoc), crashes a lot, and costs a lot of money.

Yes, that's right. I got a perfect image right after importing, then I clicked somewhere and now I cannot use it anymore..  but Vesta seems to match my interests. Mercury and oxys are probably also fine but I only want an image and two weeks effort for that is not worth it (for my purposes).

I follow a similar methods as mentioned above. First, locate the CIF file as described above.
I use VESTA which is a free, multiplatform molecular visualization program. Its relatively easy to get a decent image out of it (File>Export Raster Image):

You can even render vector images! You have relatively little control over the style of the image, but it works well for basic images. I usually only find myself tweaking the shininess of the atoms and bonds(Objects>properties>Atoms).

Oh, I actually forgot about this gem, very nice answer! Probably you might want to add that VESTA is also free (as a beer) and is cross-platform:)


text:
Volume of nitrogen gas produced during taking off

The fuel in the buster rockets of the Space Shuttle is constituted of a mixture of ammonium perchlorate, $\ce{NH4ClO4}$, and aluminum powder. One of the reactions taking place during taking off is given by
  $$\ce{6 NH4ClO4(s) + 10 Al(s) -> 5 Al2O3(s) + 3 N2(g) + 6 HCl(g) + 9 H2O(g)}$$
  The indication (s) and (g) signify solid and gaseous phase. Suppose that these rockets are loaded with $\pu{11.75 ton}$ of ammonium perchlorate and $\pu{2.70 ton}$ of aluminum.
Calculate the volume of nitrogen gas produced during taking off under normal
  conditions ($\pu{0^\circ C}$ and atmospheric pressure, so that $\pu{1 mol}$ corresponds to $\pu{22.4 L}$ of gas).

I got to the point where it's $\pu{10^5 mol}$ for each of the compounds.
How do you get to the answer $\pu{6.72e5 L}$?

LOL - Had to run through calculations twice. I used English tons rather than metric tons.

I ran right over this reading the problem statement but the statement (0 ∘C and atmospheric pressure, so that 1 L corresponds to 22.4 mol of gas). is backwards. 1 mole of gas is 22.4 liters.

https://chemistry.stackexchange.com/questions/80780/what-volume-does-one-mole-of-an-ideal-gas-occupy

@MaxW I have no idea how I missed it. Those sneaky 22.4 liters.

Your next step is to identify the limiting reagent.
Observe that six moles of ammonium perchlorate react with ten moles of aluminum.  So by taking ratios you find that:
1)  The $10^5$ moles of ammonium perchlorate you started with react with approximately $1.7×10^5$ moles of aluminum.
2)  The $10^5$ moles of aluminum you started with react with $6×10^4$ moles of ammonium perchlorate.
Now there is a problem with (1) -- you have only $1.0×10^5$ moles of aluminum available so not all of the ammonium perchlorate can react as indicated in (1).  So we now look at (2) -- it indicates that all of the aluminum can react and you still have $4×10^4$ moles of ammonium perchlorate remaining.
Thus we can realize condition (2) rather than (1) because the amount of reaction is limited by the aluminum you have available.  Thus aluminum is the limiting reagent.
Now that you know the limiting reagent you can infer how much product is formed.  With $10^5$ moles of aluminum all being consumed, you should be able to see that $3×10^4$ moles of nitrogen molecules are formed as gas.  The volume under normal conditions should then follow.


text:
detecting/tracking nano particle
I am new to nano-sized particles and conducting a very simple experiment using nano particles. In my experiment I have nanomagnetic particles inside water and they are moving due to applied magnetic field. I am looking for a method to track or detect the motion of nano particles. Is there any easy way?

Umm, with your eyes would be simplest... Otherwise there's lots of instrumental methods, but it would depend on particles.

One particle? Mass flow??

it is superparamagnetic iron oxide nano powder. Actually I want to study morion if a single particle. But if they dont interact for better observation, I can use cluster too.. I dint know that ine can see nano partivles with naked eye.

There are a few ways to do nanoparticle tracking, although I'm not sure they will exactly apply to your system.
Dynamic light scattering or DLS has been used to get the distribution of nanoparticle sizes in solution/suspension over time. Generally, the data gives information on the ensemble, not single-particle tracking.
Nanoparticle tracking analysis or NTA also relies on scattered light, but gives frame-to-frame analysis of each particle's trajectories.
There are other related techniques... Another approach is to create an Anti-Brownian ELectrokinetic (ABEL) trap - to monitor the particle or protein dynamics and apply forces to push it back. This technique was developed by the Moerner lab at Stanford, for example Acc. Chem. Res., 2012, 45 (11), pp 1955–1964


text:
Why is potassium bitartrate so insoluble?
The solubility of potassium bitartrate in water at 20°C is given by Wikipedia as 0.57 g per 100 ml, which is considerably lower than that of sodium bitartrate. Generally, a potassium salt is more soluble than the corresponding salt of most other metals—in particular, more soluble than the sodium salt. For example, potassium hydrogen oxalate is more soluble than the corresponding sodium salt (in H₂O at 20°C). Tartaric acid itself is highly soluble.* Is there something special about tartaric acid, or are there other potassium salts of highly soluble organic acids that have markedly low solubility?
*133 g per 100 ml for either optically active acid. (Curiously, and probably irrelevantly, the racemic mixture has much lower—albeit still high—solubility: 21 g per 100 ml.)

Well, as you can see it's not so simple; also yeah it's not unique behavior, there's KClO4 etc.

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Thanks, @Mithoron: that's a good example, and an inorganic one to boot.

Potassium hexachloroplatinate is also insoluble.


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How to write an entire trajectory into a single PDB file?
I have a trajectory (few frames) that I want to stack together and build a single PDB. I thought either to use VMD or MDAnalysis library.
With VMD if I do atomselect it takes a frame identifier as an argument. With MDAnalysis I saw this:
with MDAnalysis.Writer("all.pdb", multiframe=True) as pdb:
    for ts in u.trajectory:
        pdb.write(u)

However, I am not sure how to use it in a complete script.

Not only is your edit asking a different question, it's not the right question. That PDB file doesn't have multiple "trajectories" or frames. It's a protein with quaternary structure, where each subunit is separated by a TER card. You should ask a separate question about how to split apart PDB files.

@pentavalentcarbon I see, that was my original problem. I created a new question.

#!/usr/bin/env python2

import MDAnalysis

# file formats are automatically deduced; this is a standard topology and binary
# (NetCDF) trajectory from AMBER
topology = 'peptide_1.prmtop'
trajectory = '02_peptide_1_equil_NVT.nc'
u = MDAnalysis.Universe(topology, trajectory)

with MDAnalysis.Writer("all.pdb", multiframe=True) as pdb:
    for ts in u.trajectory:
        pdb.write(u)

One thing that may be important: I see lots of warnings when trying to use Python 3, so 2.7 is probably better at this point.


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Does Brønsted-Lowry theory apply to neutralization reactions?
Does the Brønsted-Lowry theory apply to neutralization reactions such as:
$$\ce{HCl + NaOH -> H2O + NaCl}$$
and would there be a conjugate acid/base?

The Brønsted-Lowry acid/base theory is based on a proton donor (acid), and a proton acceptor (base).
In the reaction: $$\ce{HCl + NaOH -> H2O + NaCl}$$
the $\ce{HCl}$ will dissolve into $\ce{H^+}$ and $\ce{Cl^-}$ because it is a strong acid, and likewise $\ce{NaOH}$ will dissolve into $\ce{Na^+}$ and $\ce{OH^-}$.
Considering that $\ce{H^+}$ is a free proton (neglecting the hydrogen isotopes), then in the left side of the reaction, $\ce{HCl}$ is an acid because it donating a proton in form of $\ce{H^+}$ to $\ce{NaOH}$ according to the following mechanism:
$$\ce{HCl -> H^+ + Cl^-}$$
$$\ce{H^+ + NaOH -> Na^+ + H_2O}$$
$$\ce{Na^+ + Cl^- -> NaCl}$$
This means that $\ce{HCl}$ is an acid because it's donating a proton, while $\ce{NaOH}$ is a base because it's accepting the proton.
The same is true for the conjugate: $\ce{H_2O}$ is the conjugate acid while $\ce{NaCl}$ is the conjugate base.
$$\ce{H_2O -> H^+ + OH^-}$$
$$\ce{H^+ + NaCl -> HCl + Na^+}$$
$$\ce{Na^+ + Cl^- -> NaCl}$$

Please read about mhchem for formatting chemical equations.

The Brønsted-Lowry theory describes acid-base reactions as proton transfers. The acid is the species which donates a proton while the base is the species that accepts a proton.
Usually, $\ce{HCl}$ and $\ce{NaOH}$ are considered dissolved in water in the context of acid base reactions. In the case of sodium hydroxide, this means that it is better considered as $\ce{Na+ (aq) + OH- (aq)}$, while in the case of $\ce{HCl}$ both $\ce{HCl (aq)}$ and $\ce{H+ (aq) + Cl- (aq)}$ are valid descriptions; in the Brønsted-Lowry context things are easier to understand if $\ce{H+ (aq)}$ is replaced by $\ce{H3O+ (aq)}$ although the latter is neither more nor less correct than the former when dealing with acidic aqueous solutions. I will restrict myself to discussing the problem assuming $\ce{HCl (aq)}$; but note that hydrogen chloride is a much stronger acid than hydronium and thus will be practically completely dissociated.
To fully analyse the reaction in question, we need to determine where a proton has been transferred. I will write it slightly differently to emphasise the fact that the products remain in aqueous solution; compare $(1)$.
$$\ce{HCl (aq) + Na+ + OH- -> H2O + Na+ + Cl-}\tag{1}$$
If we look at the non-hydrogen elements, we notice that chlorine is bound to hydrogen on the reactant side but no longer on the product side. Likewise, oxygen is bound to one hydrogen on the reactant side but to two on the product side. We have thus determined the species that will donate and accept a proton; the reaction can be classified as a proton transfer reaction.
$$\ce{\underset{acid}{HCl (aq)} + Na+ + \underset{base}{OH-} -> \underset{{conj. acid}}{H2O} + Na+ + \underset{{conj. base}}{Cl-}}\tag{1'}$$
As expected, the neutralisation can be considered a Brønsted-Lowry acid-base reaction. The corresponding proof in the case of dissociated hydrochloric acid ($\ce{H3O+ (aq) + Cl- (aq)}$) is left to the reader.


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Why are alpha carbonyl halides most reactive towards SN2 reactions?
Relative rates of some compounds towards SN2 is given below

The reason for reactivity of alpha carbonyl  halides is given as

It’s not clear what exactly is happening.

I have asked a question previously which concerns the same electrophilic system with the exception that nucleophilic attack happens on the carbonyl carbon and not on the α-carbon.

From the organic chemistry textbook by Clayden, Warren, Wothers and Greeves on pp. 890-891:

The $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$ orbitals add together to form a new, lower-energy molecular orbital, more susceptible to nucleophilic attack. But, if $\ce{X}$ is not a leaving group, attack on this orbital will result not in nucleophilic substitution but in addition to the carbonyl group. Again, this effect will operate only when the $\ce{C-X}$ and $\ce{C=O}$ bonds are perpendicular so that the orbitals align correctly.

This interaction when considered in the context of an attack on the carbonyl is the principle of the polar Felkin-Anh model used to describe a certain diastereoselectivity. However it can likewise be used to explain the lowering of the $\unicode{x3c3}^*(\ce{C-Br})$ orbital which is attacked in a nucleophilic substitution reaction. It may seem as though these statements are contradictory, but the orbital interaction will create a three-centre π-type double-antibonding orbital with only the contribution on oxygen being neglegible due to oxygen’s electronegativity being highest and thus its contribution to high-energy orbitals being lowest.
It will depend on your nucleophile what will actually happen. It can attack both on the carbonyl carbon and on the α-carbon. If the attack on the carbonyl carbon is reversible, only the attack on the α carbon leading to an observed $\mathrm{S_N2}$ process will be observed after the reaction — even if the carbonyl attack is faster. Only if the attack on the carbonyl is not reversible as is the case e.g. with Grignard reagents, a mixture of two products will be observed.
Naturally, if $\ce{X}$ is a very poor leaving group — e.g. $\ce{OTBS}$ — the attack on the carbonyl will dominate this way or that.

To sum up:
The reaction rate is increased because the $\unicode{x3c3}^*(\ce{C-X})$ orbital and the $\unicode{x3c0}^*(\ce{C=O})$ orbital linear combine to give a lower-energy orbital. Since this is the LUMO, the LUMO energy is lowered and nucleophilic attacks are facilitated.

Why does the presence of carbonyl on α-carbon of a substrate restrict SN1?

@Neha Draw mesomeric structures and the answer will become apparent.

Most of the time, a $\text{S}_\text{N}2$ reaction will involve a negatively charged nucleophile attacking an uncharged substrate (as is depicted in the second picture of the question). This negative charge can be stabilized effectively by the carbonyl group through conjugation. Note that in the example given in the first picture, the stabilization probably also involves the phenyl ring. Of course, the molecular structure must allow for alignment of the orbitals involved.
This mechanism is valid for most examples given in the table of the first picture bar the methoxy-methylchloride (MOM chloride, an old-fashioned protection agent for alcohols). In this case, I would think that the oxygen lowers the $\sigma^\ast$ orbital energy such that it may serve better as a point of attack.

Why does the presence of carbonyl on α-carbon of a substrate restrict SN1?

@Neha It makes the $\text{S}\text{N}2$ so much faster that $\text{S}\text{N}1$ does not stand a chance.


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Electrolytic refining of copper
In refining copper, an aqueous solution of copper(II) sulphate is electrolyzed, using impure copper as the anode and pure copper as the cathode.
At the anode, metal impurities higher than copper in the electrochemical series such as zinc, iron and lead are oxidized first, then copper loses its electrons to form copper(II) ions. Metal impurities lower than copper in the electrochemical series are hardly oxidized and deposit at the bottom of the tank as sludge. So, I have two questions:

If ions from metals more reactive than copper exist in the solution, why aren't they preferentially discharged first to form zinc or iron metal on the other copper electrode first rather than copper?
Why can the pure copper cathode be thickened with pure copper only?

More reactive metals aren't preferentially discharged precisely because they are more reactive. As for copper cathode, well, what choice does it have? There is current running through it and there is copper which is going to be deposited, no matter what.

What do you mean by the last question, "Why can pure copper cathode be thickened with pure copper?". It doesn't make any sense. How can you thicken a pure copper cathode by pure copper? I think you should explain that question by a few more words.

To your first question: Consider what happens if you put an iron nail or a zinc pellet into a copper(II) solution: the less noble metals will oxidize and reduce the copper ions to copper(0). So even if some of the less noble metal ions are reduced and deposited at the cathode, they would react with the copper ions (that are the most numerous in this scenario).
To your second question: if ions of more noble metals than copper were in the solution, they would be deposited on the cathode. However, by carefully tuning the voltage and current, no more noble metal ions are produced at the anode. As stated by Ivan Neretin in the comments, the cathode will reduce what is available to be reduced: copper ions (which are most numerous and the potential fits).


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Work done by a gas
An ideal gas changes from state $a$ to state $b$ as shown in the figure. What is the work done by the gas in the process?

This is how I reasoned it:
As the pressure in the gas increases, its temperature also increases. This the gas was not allowed to exchange heat with the surrounding so we can say it’s an adiabatic process. Now the work done on the gas would be negative (since temperature increased in increasing the pressure so we can assume that gas was compressed) and the work done by the gas would be zero (since the system was not allowed to exchange heat with the surroundings). Is my reasoning correct? What would be a more mathematical way to solve the problem?

The work done on the gas and the work done by the gas is the same thing, only with different sign.

@Ivan Oh right! So where did I go wrong ?

Who said the work done on the gas is negative? True, the pressure increased, so what? It seems to have increased thanks to temperature alone. There was no work.

@Ivan sorry , my bad ,I thought the PV diagrams were drawn for constant temperature

Maybe (though then again, maybe not), but this is not PV.

At constant volume, there is no work done by the gas or on the gas.

Your reasoning seems correct, but you shouldn't assume the nature of processes unless stated. I find that very uncertain, and unnecessary as in your question.
From the given diagram it is clear that there is a liner relation between $T$ and $P$,
$T=kP$    $\space $(Identical to the standard equation of a straight line, $y=mx$)
The questions says that the gas is ideal, which enables you to use ideal gas equation.
$PV=nRT$,
$V=\frac{T}{P}nR$
$V=knR=constant$
Now according to you, when the temperature increases, the pressure increases but the work is zero. The flaw is this reasoning could  could be understood by considering the ideal gas equation. This reasoning only holds when the change in $T$ with respect to $P$ is constant (which is the definition of the slope of an line, $k$) which leaves you with a volume, but there can be cases where the graph between $T$ and $P$ is not a straight line, and in those cases work by the gas won't be zero.
As @Ivan added in his comment, you second statement which states that the work is zero, contradicts with your first one. This is not a very clear way to look at the process indicated by the diagram.
In case of adiabatic process, it is not necessary that work will be nil if no heat exchange takes place. It just means that heat exchange is zero, it does not tell you anything about the work done by (or on) the system.
What you need here is a change of perspective, you are thinking the temperature increases when the pressure increases (which indicates compression), but instead, you can increase temperature at constant volume which will increase the system's pressure.