text:
How do I take chemistry notes on Mac?
I am learning Chemistry now, and I want to take some notes that involve huge amount of organic chemistry on Mac. How can I do that? Pages does not seem support scientific use.

Please elaborate a little bit more what you want to accomplish. For example tanking notes would be easily accomplished by any word processing program. More difficult is setting mathematical formulas or chemical reaction equations. Going to organic chemistry and explaining reaction mechanisms is maybe also something you might want to consider. There is a huge variety of programs on the market - most likely you will have to use a combination of more than one. (Unfortunately I have little experience with apple.)

@Martin Advices taken. Thanks for your wisdom!

@AveMaleficum Simply lay a sheet of paper on top of your Mac and apply a pencil.

You might also consider one of those digital pens. You write on a piece of paper, but the pen tracks its own motions and whatever you write is transferred to your computer in real time. You still have to write on paper, but you get the digital note immediately. I know nothing of how well these pens actually work, though, but it might be worth checking.

If you know LaTeX and you want to see what you are writing you can use lyx, and some TeX package like mhchem and chemfig. So you can write all the things you want and add comments. This is what I have done with my Mac just now:

Here is the code, you can make these lines as a default template:
\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\usepackage{chemfig}
\usepackage{tikz}
\usepackage[version=4]{mhchem}
\setlewis{4pt}{0.8em}{black}

From here you have to write! The date is automatically added!
\title{Introduction to chemistry notes}
\begin{document}
\maketitle
\section{mhchem}
You know that \ce{ CH4 + 2O2 -> CO2 + 2H2O}     this is a simple equation\ldots
Okay maybe you can do this with word, but if you are familiar with Chemistry S.E. you will be very fast with this notation.

\ce{Zn^2+
<=>[\ce{+ 2OH-}][\ce{+ 2H+}]
$\underset{\text{amphoteres Hydroxid}}{\ce{Zn(OH)2 v}}$
<=>C[+2OH-][{+ 2H+}]
$\underset{\text{Hydroxozikat}}{\ce{[Zn(OH)4]^2-}}$
}

\section{Chemfig}
\begin{figure}[h]
\setatomsep{1.8em}
\schemestart
\subscheme[90]{\chemfig{*6((-[@{a1}]H_2@{a2}\Lewis{26,N})=[@{r1}]-[@{r2}]=[@{b2}](-[@{b1}]N =[@{c1}] N -[@{c2}]*6(=[@{d1}]-[@{d2}] =[@{e1}](-[@{e2}] OH)-=-))-=-=)}\chemrel{<->}
\chemmove[-stealth,shorten <=1pt, shorten >=1pt]{%
    \draw(a2)..controls +(90:5mm) and +(135:5mm)..(a1);% first arrow
    \draw(r1)..controls +(225:12mm) and +(300:12mm)..(r2);% ring arrow
    \draw(b2)..controls +(10:3mm)..(b1);% ring arrow
    \draw(c1)..controls +(340:8mm)..(c2);% ring arrow
    \draw(d1)..controls +(225:12mm) and +(300:12mm)..(d2);% ring arrow
    \draw(e1)..controls +(20:4mm)..(e2);% ring arrow
    }
\chemfig{*6((=[@{a1}]H_2@{a2}\chemabove{N}{\oplus})-[@{r1}]=[@{r2}]-[@{b2}](=[@{b1}]N-[@{c1}] N =[@{c2}]*6(-[@{d1}]=[@{d2}] -[@{e1}](=[@{e2}] \chemabove{O}{\ominus}H)-=-))-=--)}
}
\schemestop
\caption{This is a coniguated system\ldots}
\label{fig:con}
\end{figure}
Okay the lesson will end and in one hour you'll draw this figure only, maybe you should use an external software\ldots %what the professor means whit this last phrase?!
\end{document}

Sincerely I think that is best to take handwrite notes because I think that writing and drawing chemical formula allow you to memorize better but of course this is only a thought.

Pencil and paper is the god, but I think taking notes using LaTeX also have some benefits.You will be more familiar with LaTeX.

@AveMaleficum Yes, furthermore you can collaborate with other students to create lecture notes...

@AveMaleficum If you're able to 'keep-up' with typesetting your notes in real-time then I must bow before you.  That would be an amazing feat.

@GM Once you've become well versed in LaTeX, you don't even see the code anymore.  All you see is a benzene molecule here, a water dimer there, and a substitution reaction over there.

@LordStryker yes, but I think to draw chemical structure with LaTeX in "real-time" you need at least one year of training!

@GM I have troubles drawing a scheme in ChemOffice in "real time" ... I am very much impressed how this all can be accomplished by now. I also seriously have to consider switching to mhchem. Is there a good tutorial for chemfig?

@Martin chemfig has a good documentation you can scroll down until you see what you are interested in copy the example and edit it according to your needs. Recently I'm trying to create an Avogadro plugin to export directly from 3d structure 2d chemfig structures using python terminal but I've just begin to explore the functions I don't know if it is feasible...

My wife used ChemDraw when she was in grad school and teaching. http://www.cambridgesoft.com/Ensemble_for_Chemistry/ChemDraw/Default.aspx

As one answer already recommends, ChemDraw is pretty good. But you should also look at ChemDoodle (which also has tablet/phone versions): https://www.chemdoodle.com/
Whether packages like this are good enough for live notes is up to you to decide, but they are clearly good enough to produce good, well formatted, publication-quality output.

Try Avogadro http://avogadro.cc/wiki/Get_Avogadro
It's an advanced molecule editor and visualizer designed for cross-platform use in computational chemistry, molecular modeling, bioinformatics, materials science, and related areas. It offers flexible high quality rendering and a powerful plugin architecture.

I don't see how Avogadro is going to help someone take notes in a lecture...


text:
Nitration of naphthalene and anthracene
In the nitration of compounds such as naphthalene and anthracene, how can you determine which would be the major product? I tried drawing resonance structures for the intermediate formed because stability of intermediate determines which product will be major, but this became too hard for me as there were too many resonance structures and I couldn't really compare them. Can someone please help?

In the electrophilic substitution of polycyclic aromatics, when drawing resonance structures keeping as many benezene rings intact as possible is important.  Note too that a naphthalene ring isn't as "good" as two separate benzene rings.
Look at the structures below, electrophilic attack at the 1 position in naphthalene (top row of drawing) allows you to draw 2 resonance structures with a benzene ring remaining intact (4 if we count structures with the double bonds simply shifted in the intact benzene ring); attack at the 2 position (middle row in figure) only has 1 resonance structure (2 if we count structures with the double bonds simply shifted in the intact benzene ring) with a full benzene ring intact.  Hence, electrophilic attack at the 1-poisition is preferred in naphthalene.
Applying the same considerations to anthracene, we see that only attack at the central ring position allows 2 full benzene rings to remain intact in the possible resonance structures. Consequently electrophilic attack at anthracene's 9-position is preferred.

When you compare the stability of the carbocation intermediates of polycyclic aromatic compounds, you have to consider if the aromaticity of one of the rings is sacrificed in order to delocalize the positive charge. If this is the case, the intermediate is less stable.
For naphthalene, for example, substitution at C1 is favored, because the cation is stabilized by allylic resonance and the aromatic character of the second ring is maintained. After attack of the electrophile E at C2, the charge in the resulting cation is delocalized without retaining the aromaticity of the second ring (it no longer has 6 π electrons). Hence, this C2-substituted intermediate is less stable (source).

You can make analogous considerations for anthracene, except that there is a third position for substitution, namely C9 in the middle ring. Delocalization of the charge on this position creates a cation with two intact benzene rings, which favors substitution at this position.

To supplement Jannis Andreska's excellent answer, you can consider the total number of resonance structures available to both carbocation intermediates.

Nitration at position 1 produces a carbocation that has 7 total resonance structures, 4 of which appear to preserve the aromaticity of the second ring.

Nitration at position 2 produces a carbocation that has 6 total resonance structures, 2 of which appear to preserve the aromaticity of the second ring.

You can do the same analysis for anthracene, and you will probably find that nitration at position 9 (on the middle ring) is favored.

source: Stack Exchange

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sort: votes


text:
Different Equivalent Weight for the same compound

Let us say that we have the following two reactions:
$$\ce{A + O2 -> B}$$ where $\ce A \text{ and } \ce B $ are different oxides of metal $\ce M$.
Weights of $\ce A$ and $\ce B$ are $\pu{5.72 g}$ and $\pu{6.36 g}$.
$$\ce{B + H2->M + H2O}$$
The weights of $\ce B$ and $\ce M$ are $\pu{4.77 g}$ and $3.81$ g, respectively.

So I was trying to infer some information from the given data and this is what I did:
In the first reaction we can calculate the equivalent weight of $\ce A$ and $\ce B$ to be $\pu{71.5 g}$ and $\pu{79.5 g}$ respectively.
In the second reaction we can calculate the equivalent weight of $\ce B$ and $\ce M$ to be $\pu{39.75 g}$ and $\pu{31.7 g}$ respectively.
If we compare the equivalent weight of $\ce B$ in the two reactions, the equivalent weight of $\ce B$ in the first reaction is twice the equivalent weight in second reaction. What exactly does this mean?

Should the second equation be $\ce{B + 2nH+ -> M + nH2O}$? Otherwise, the second equation will have mass imbalance.

@BenNorris Since I'm working with equivalents, I don't need to balance the equation. Although the edited question has H2 instead of H2O.

Equivalent weights are considering a redox equation with one mole equivalent of electrons. But some redox reactions require more than one mole equivalent of electrons to be equilibrated.
Thus, if you have the second reaction that leads to a weight equivalent of $1/2$ of that for the first reaction, it means that the reaction 2 has consumed $1/2$ of  $B$ compared to reaction 1 for one mole equivalent of electrons.
You still cannot proceed to the molecular weights on these basis, but you can now try to consider that the most likely case is that the first one involved only one electron and the second one two electrons. Thus you can write that this is consistent with:
First equation:
$$\ce{A \rightarrow B + e-}$$
(equilibrated by $\ce{\frac 14 O_{2} + e- \rightarrow {1\over 2} O^2-}$).
And for the second equation:
$$\ce{B + 2 e- -> M}$$
(equilibrated with $\ce{H2 + O^2- -> H2O + 2 e-}$).
One this is said, one should consider the molar weight of $\ce M$ to be twice the equivalent weight of $\ce M$ calculated from the second equation, leading to a molar weight of 63.5 g/mol. Quite unsurprisingly, this is the molar weight of copper, so $M = Cu$. And going backwards, $\ce{B = CuO}$ with copper in its second oxidation state. We can check this, as the molecular weight of CuO is 76.5 g/mol, exactly the equivalent weight calculated for reaction 1. And thus $\ce{A = Cu2O}$ with copper in its first oxidation state.

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text:
Resonance stabilized carbanion
I am sure that the answer is (b) because I am familiar with the other rings and I think they are showing resonance (and it is a single choice question so it was easy to guess)
I wish to know why is (b) not showing resonance (because I think it can as far as I know) and if I am wrong about the answer then please correct me.

(b) does not show resonance because the second resonance structure (with negative charge at oxygen and double bond in the ring) violates Bredt's rule. The carbon $\ce{sp^3}$ orbital with the negative charge and the $\ce{p_z}$ orbital of the adjacent carbonyl C cannot overlap because they are not in plane, so the charge cannot be delocalized into the carbonyl group.

You are right about compound B.  Try to build a model of the resonance form for B where charge is delocalized to the adjacent carbonyl.  Here is a picture of the situation.

Note what is called a "bridgehead double bond" in the resonance form on the right.  Such a structure would be a violation of Bredt's Rule, which states that double bonds can't be placed in bridgehead positions if there would be a lot of ring strain as a result.  Again, try to build a model of the molecule on the right and see how strained that bridgehead double bond makes the system.

For any cyclic compound to exhibit resonance, the following rules have to be obeyed:
1) $(4n+2)\pi$ electrons.
2) Conjugation. Conjugation is not possible for either b) or d).
3) Planarity. Compound b) is not planar.
And of course, compound b) violates Bredt's rule as well.
Hence, neither b) nor d) can be resonance stabilised.

These are rules for aromaticity not conjugation.


text:
Taking Castor Oil from room temperature to 300 degrees Celsius
I am doing a project where I heat up a container (around 500ml) filed with castor oil at room temperature. I want to add a pressure/safety valve that opens up when the castor oil reaches around 300 degrees Celsius (boiling temperature). In order to get the right valve I have to know how much the pressure increases with in that process. How can I calculate this? And were can I find the required data?
Is there something else I need to consider?
Thanks,
EDIT: Like I have experienced, it seems impossible to find the required data. What if I used mineral oil instead?

If castor oil boils at 300 °C, then adding a pressure would increase the boiling point. What you need is a thermometer that signals when a defined temperature is reached. The boiling point of mineral oil depends on the refinement. Although I would think most would be above 300 °C.

http://pubs.acs.org/doi/abs/10.1021/je049898o

Can you be more precise about the intended function of the "safety" valve?  If it's just intended to release pressure to avoid rupturing the container, you can just use what is called a "check valve", set at the appropriate pressure.  In that case, if the pressure rises too high in the vessel, it will release pressure until it drops back down below the limit of the check valve (within some specified tolerance).  Does that sound like what you need?  I'd also point out that castor oil vapor may not be compatible with the designs of many standard check valves .. you need to check with the vendor.

Meh .. just noticed the date on this question.  Oh well, perhaps the OP is still interested in this project.


text:
Why oxygen is enclosed in parenthesis in some chemical reaction?
While going through my chemistry text book I came through the following reaction.
$\ce{RCH2OH + (O) -> RCHO + H2O}$
In the above reaction why oxygen atom is enclosed in parenthesis ?

It represents oxygen from an oxidising agent, as opposed to from diatomic oxygen. It's normally done in square brackets like this:
$$\ce{CH3CH2OH + 2[O] → CH3COOH + H2O}$$
Is the reaction between ethanol and potassium manganate (VII) to form ethanoic acid. The oxygen in the reaction is from the permanganate, not from diatomic oxygen. It kind of simplifies the reaction, so you only see the reaction of what you want -the ethanol and oxygen - instead of the whole reaction with the manganate there. (Manganate is an oxidising agent).
In your reaction the alcohol:
$\ce{RCH2OH}$ is reduced by the oxygen to form an aldehyde and water, using oxygen from an oxidising agent - the [O].

$\ce{(O)}$ means nascent oxygen. When you use an oxidising agent, fresh Oxygen is produced which is called nascent oxygen.


text:
Probability of measuring a particle in the ground state: having trouble with the integration
I am working on a problem in the context of an infinite symmetric potential barrier around the origin, with barriers at $x=- \frac{a}{2}$ and $x=\frac{a}{2}$. The wave function is a symmetric triangle-function:
$$
\Psi (x) = \begin{cases} 0 & x < - \frac{a}{2} \\ \sqrt{\frac{12}{a^3}} \left( \frac{a}{2} - \lvert x \rvert \right) & -\frac{a}{2} \leq x \leq \frac{a}{2} \\ 0 & x > \frac{a}{2} \end{cases}
$$
I must calculate the probability of measuring a particle in the above eigenstate in the ground state (when $n=1$). The way to attack this problem is to calculate the coefficient $c_1$, and then square it. Here follows my (wrong) attempt.
$$
c_n = \int\limits_{-\infty}^{\infty} \psi_n^*(x)\Psi (x)\,\mathrm dx
$$
I understand from the above expression that the $\psi_n^* (x)$ is just the "regular" solution to the particle in a box. Given this symmetric potential, and that $n=1$ (odd $n$, even function), I must set
$$
\psi _1^* (x) = \sqrt{\frac{2}{a}} \cos\frac{\pi x}{a}
$$
The expression for $c_n$ becomes
$$
c_n = \int\limits_{-a/2}^{a/2}  \sqrt{\frac{2}{a}} \cos\frac{\pi x}{a} \sqrt{\frac{12}{a^3}} \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx \\
$$
Simplifying gives
$$
c_n = \frac{2 \sqrt{6}}{a^2} \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx
$$
This can be solved by partial integration, setting $u'=\cos \left( \frac{\pi x}{a} \right)$ and $v= \left( \frac{a}{2} - \lvert x \rvert \right)$. Hence, $u=\frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) $ and $v'=-1$. The integral itself then becomes
$$
\int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx = \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right)\left( \frac{a}{2} - \lvert x \rvert \right) - \int \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) (-1)\,\mathrm dx
$$
Cleaning up
$$
\mbox{Integral} = \frac{a^2}{2 \pi} \sin \left( \frac{\pi x}{a} \right) - \lvert x \rvert \frac{a}{\pi} \sin \left( \frac{\pi x}{a} \right) - \frac{a^2}{\pi ^2} \cos \left( \frac{\pi x}{a} \right)
$$
Now we just need to insert the limits. However, we see quite easily from the above expression that upon inserting $x=\pm \frac{a}{2}$, the last term vanishes, since $\cos ( \pm\pi / 2 = 0)$. The sine terms will equal $1$ when the limit is positive and $-1$ when the limit is negative. Hence, we are left with
$$
\mbox{Integral} = \left[ \frac{a^2}{2 \pi} - \frac{a}{2} \frac{a}{\pi} \right] - \left[ \frac{a^2}{2 \pi}(-1) - \lvert \frac{-a}{2} \rvert \frac{a}{\pi}(-1)  \right] = \frac{a^2}{2 \pi} + \frac{a^2}{2 \pi} = \frac{a^2}{\pi}
$$
Plugging this into the original expression
$$
c_n = \frac{2 \sqrt{6}}{a^2} \cdot \frac{a^2}{\pi} = \frac{2 \sqrt{6}}{\pi} \approx 1.56
$$
So, you see the problem. My $c_n$ is larger than one, which will give me a rather un-physical probability. Am I wrong in this derivation in my assumption for $\psi _n^* (x)$? If not, then where?
Wolfram Alpha gives me that
$$
\mbox{Integral} = \frac{2a^2}{\pi ^2} \\
\Rightarrow c_1 = \frac{2 \sqrt{6}}{a^2} \cdot \frac{2a^2}{\pi ^2} = \frac{4 \sqrt{6}}{\pi ^2}
$$
Using this gives me a probability of $c_1^2 \approx 0.986$, which at least has physical meaning. Also, logic tells me that there is a very high probability of finding the particle in the ground state, so this further backs up WA's answer.

How do you get from Wolfram Alpha's expression $c_1 = \frac{2a^2}{\pi ^2}$ to $c_1^2 = 0.986$ - what value of $a$ are you using and where does this $a$ come from?

$a$ is the length of the potential well. I have made an error; I see that now! The WA expression for $c_1$ should be multiplied by the constant that was factored out from the integral. I will update my question!

I think your tactics using partial integration from the start is the problem because $\lvert x \rvert$ is not differentiable at $x=0$ so working with its derivative is not quite kosher (and $v' = -1$ is certainly wrong since for $x < 0$ the gradient of $-\lvert x \rvert$ is $1$).
I suggest you go about it step by step:
\begin{align}
  c_1 &= \frac{2 \sqrt{6}}{a^2} \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right) dx \\
      &= \frac{2 \sqrt{6}}{a^2} \biggl( \frac{a}{2} \underbrace{\int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) dx}_{ = \, I_{1} } - \underbrace{\int\limits_{-a/2}^{a/2} \lvert x \rvert \cos \left( \frac{\pi x}{a} \right) dx}_{ = \, I_{2} } \biggr)
\end{align}
Now, integral $I_1$ is easy:
\begin{align}
  I_{1} &= \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) dx \\
   &= \left[ \frac{a}{\pi} \sin \left( \frac{\pi x}{a} \right)  \right]_{-a/2}^{a/2} = \frac{a}{\pi} \sin \left( \frac{\pi}{2} \right) + \frac{a}{\pi} \sin \left( \frac{\pi}{2} \right) = \frac{2a}{\pi}
\end{align}
For integral $I_2$ I suggest that you use that
\begin{equation}
 \lvert x \rvert = \begin{cases} x &  x \geq 0 \\ -x & x \leq 0 \end{cases}
\end{equation}
and split up the integral into two integrals
\begin{align}
  I_{2} &= \int\limits_{-a/2}^{a/2} \lvert x \rvert \cos \left( \frac{\pi x}{a} \right) dx \\
   &= \int\limits_{0}^{a/2} x \cos \left( \frac{\pi x}{a} \right) dx - \int\limits_{-a/2}^{0} x \cos \left( \frac{\pi x}{a} \right) dx
\end{align}
Now use integration by parts with $u'=\cos \left( \frac{\pi x}{a} \right)$ ($\Rightarrow u=\frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) $) and $v= x$ ($\Rightarrow v'= 1 $) to solve the integral
\begin{align}
  \int x \cos \left( \frac{\pi x}{a} \right) dx &= \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) - \int \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) dx \\
  &= \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \ ,
\end{align}
so
\begin{align}
  I_{2} &= \int\limits_{0}^{a/2} x \cos \left( \frac{\pi x}{a} \right) dx - \int\limits_{-a/2}^{0} x \cos \left( \frac{\pi x}{a} \right) dx \\
        &= \left[ \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \right]_{0}^{a/2} - \left[ \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \right]_{-a/2}^{0} \\
        &= \frac{a^2}{2 \pi} - \frac{a^2}{\pi^2} - \frac{a^2}{\pi^2} + \frac{a^2}{2 \pi} = \frac{a^2}{\pi} - \frac{2 a^2}{\pi^2}
\end{align}
Plugging this into the equation for $c_1$ gives
\begin{align}
  c_1 &= \frac{2 \sqrt{6}}{a^2} \left( \frac{a}{2} I_{1} - I_{2} \right) \\
      &= \frac{2 \sqrt{6}}{a^2} \left( \frac{a^{2}}{\pi} - \frac{a^2}{\pi} + \frac{2 a^2}{\pi^2} \right) \\
      &= \frac{4 \sqrt{6}}{\pi^2}
\end{align}
and that is indeed what Wolfram Alpha gives you.

Hmm. The text does not explicitly say that the function is normalized; I just assumed that it was. That square root fooled, because it looks like a normalizing constant. But still, WA gets another answer than mine, and that one looks (to me) to be the more likely correct answer.

@AndersMB I checked it again and have to correct my first comment: the wavefunction is indeed normalized.

I just redid my integration based on your notes. Thanks a lot! I never thought that I had to treat $\lvert x \rvert$ in a special way!

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text:
Which of the following has highest solubility in water?
Which of the following has highest solubility in water?
A. $ Ca(OH)_2 $
B. $ Fe(OH)_3 $
C. $ Cr(OH)_3 $
D. $ Zn(OH)_2 $
And also please explain how can we predict their solubilities when we are not provided with their solubility products.
I know that calcium hydroxide is sparingly soluble. So it cannot be the right option. I think Iron hydroxide is also soluble to a good extent. I haven't got any idea about option C. I think D may be the right answer.

Hi SkyDreamer welcome to ChemistryS.E.! Can you show us your efforts and your attempts, we can't do your homework see our homework policy! :-)

I know that calcium hydroxide is sparingly soluble. So it cannot be the right option. I think Iron hydroxide is also soluble to a good extent. This is evident as Iron rusts in presence of water vapours, but it is a slow process. I haven't got any idea about option C. I think D may be the right answer.

Please help! I have got my final exam tomorrow.

What are your thoughts on what the answer is? We can't solve problems for you; we only guide you through them.
In addition, iron hydroxide is a gel-like substance and is hydrated. I'm not aware of how this gel-like precipitate may rust.

Ohh. I think I was wrong about Iron hydroxide. I just have got a rough idea that the answer is D.

I've add to your question your thoughts, and I've retract my close vote!

OK.. Thank you! But what do you think could be the most probable answer?

@SkyDreamer9 you are welcome here is a video to learn it quickly good luck for your exam!

You can't predict solubility reasoning, you have to remember solubility rules in your case this one:

Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and $\ce{Al^3+}$ are insoluble. Thus, $\ce{Fe(OH)3, Al(OH)3, Co(OH)2}$ are not soluble.

So in this case $\ce{Ca(OH)2}$ even if slightly soluble theoretically should be the right answer.


text:
is water a non-metallic oxide? and hence acidic in nature?
I just read about non-metallic oxides in my textbook which says that non-metallic oxides are a compound formed of an oxide ion and a non-metal. It also said that they are acidic in nature. It gave examples of carbon dioxide and sulfur dioxide.
But then wouldn't hydrogen oxide (water) also be a non-metallic oxide and hence, acidic in nature. But most textbooks say that water is neutral. So how's that possible?

Water is neutral by convention. Life on an ammonia planet with would consider water an acid.

When textbooks say that water is "neutral," textbooks mean that the water meets this condition:
$\ce{[H_3O^+]} = [\ce{HO^-}]$.
That's all.
Water can be neutral, in that its hydronium ion concentration equals its hydroxide ion concentration. Water can also be acidic if there is carbon dioxide dissolved in the water. Carbon dioxide can dissolve in water simply by letting the water come into contact with the atmosphere. Water can be basic too if you put a basic solute in water.
Water itself can also act as a weak Brønsted acid by donating a proton. This however is not a favorable process.
Water can also act as a Lewis acid because the oxygen in water withdraws electron density from the hydrogens, giving each hydrogen a partial positive charge. This in turns makes the hydrogens electrophilic - fitting the definition of Lewis acidity.
I think what your textbook is getting at is Lewis acidity. Consider carbon dioxide. This doesn't fit the Brønsted definition of acidity because it has no hydrogen protons to donate. So consider the partial charges on the carbon dioxide molecule. There is a positive partial charge on the carbon because the oxygens are more electronegative than carbon and have withdrawn electron density from the carbon. This makes the carbon electrophilic. And hence carbon dioxide as a whole can be considered a Lewis acid (but do understand that it's the carbon specifically in carbon dioxide acting as a Lewis acid).
So in general, your textbook says that non-metallic oxides tend to be acidic (in the Lewis sense) because the oxygens withdraw electron density from the non-metal (remember that oxygen is one of the most electronegative elements). This in turns creates partial positive charge(s) and makes the centers of these partial positive charges electrophilic.
Also, you could say that non-metallic oxides are precursors to Brønsted acids; consider this Lewis acid/base reaction. Remember that the central conceit of Lewis acid/base chemistry (coordination chemistry) is electron flow; nucleophiles attack electrophiles (not the other way around; that would be the fatal conceit).
So consider this reaction:
$\ce{SO_3 + H_2O -> H_2SO_4}$
We have sulfur with an electronegativity (EN) of 2.6. Oxygen has an electronegativity of 3.5. The difference is 0.9. The bond is clearly polar covalent. In addition, we have not one but three oxygens isolating electron density from the sulfur. Therefore, the sulfur most definitely has a partial positive charge and the oxygens partial negative charges. We say the oxygens are nucleophilic. And back to water, we have already concluded the hydrogen atoms in water are electrophilic due to their positive partial chage. The sulfur trioxide will attack the hydrogens and form the sulfuric acid molecule - a strong Brønsted acid (its first ionization is complete in water solution).

Hydroxonium(H3O+)=hydroxide(OH-) one acts as a strong acid and the other a strong base- the instant they come to be, we have water, which results in a neutral PH in isolation; the PH is altered, in solution, by changing relative concentrations of these ions caused by exposure to molecules such as atmospheric carbon dioxide(distilled water can absorb more CO2 and hence tends to be slightly acidic) which creates carbonic acid in solution with water. This equilibrium is also part of the fundamental nature of pure water.  H2O (the molecule) is an oxide of hydrogen mathematically. Water is bound H2O plus a relatively tiny amount ofhydrogen; hydroxide ions(oxides are  binary compounds of oxygen with another element).   Additionally "water", under various circumstances can have oxygen dissolved at varying levels(from very low dissolved oxygen (fish breathe) in the open underwater areas of the sea to relatively high oxygen in babbling brooks and rivers) and carbon dioxide (added by contact with air)- so there are many forms of "water" but "pure water in isolation" is without these common additions. Water is in a category of it's own. There is much yet to be learned about this wonderful substance, with all of it's unique properties, including long held beliefs regarding bonds and bond strength in my opinion, but that is another discussion. The bound molecule, H2O, has been categorized as a neutral, non-metal oxide albeit with significant unique aspects as water which have not been fully explored.   The molecule of H2O is existant, in our reality, in the form  of water.  Water is not H2O; H2O is not water.  H2O, should it be isolated, is not water and does not have the properties of water. Additionally, all H2O molecules are not the same, as a small percentage of H2O molecules, in water, are deuterium, as opposed to the aneutronic hydrogen(protium), which is far more common(with the definition that "pure water" is not "heavy water"(water with a high amount of deuterium)). Both protium (most H2O) and deuterium are stable non-radioactive isotopes of hydrogen.
Water Ph, in solution, changes up or down depending upon the relative concentrations of the two ions, hydroxonium (acid) and hydroxide(base). When the acid ions outnumber the base there is an acid and when the base ions outnumber the acid there is base. That is, when there are more hydroxonium ions than hydroxide ions, we have an acid and when there is more hydroxide we have a base- pure water has an equal number of these 2 ions and hence is neutral, by default. The confusing part of the process is the value of "neutral PH" which changes as the temperature of water changes, matching the value of the PH of water at a given temperature.  Water, on it's own, always maintains a perfectly neutral PH value, even upon change in temperature, albeit it water can react amphoterically- as an acid or a base when in solution with other substances.  The equilibrium point of water shifts when the temperature is raised (an effort, at reattaining balance, by effectively moving against change). Thus, water responds with an opposite reaction to the change in temperature. The counterbalance, to heating, is an effort to move in the direction of cooling. Heating pure water also causes both of water's ions to increase in numbers(equally); this causes a seeming decrease in the numerical value of the PH of water. However, this doesn't cause the water to become any more acidic as temperature increases. Water, in solution, is only acidic with an excess of hydrogen ions, as i have already said above. Pure H2O, is PH neutral, even if it's PH value changes due to the shifting of it's equilibrium point by increasing temperature because the new PH value instantly becomes the neutral PH value for water at that particular temperature.  As another way of looking at it, the neutral PH value of water at room temperature, PH 7, would actually be alkaline for water tested at higher temperatures. Effectively, the neutral PH value, of water, lowers, as water raises in temperature. The "real PH" effectively remains unchanged within the instantly stable equalibria of water at a given temperature. The same result is seen, but in the other direction, as the temperature of water is lowered from room temperature. The PH value of water rises as the temperature of water falls from room temperature but so to does the neutral PH value of water at the new temperature. The unaffected PH state of pure water, at any temperature, is a neutral PH- meaning that water is not acidic and not base in addition to meaning an equal number of hydroxonium and hydroxide ions.

Hydrogen(H+)=hydroxide(OH-) results in a neutral PH of 7- the PH is altered by changing their relative concentrations. This is wrong. pH is solely dependent on the concentration of hydrogen ions. Water has a pH of 7 at 25$^\circ$C but if you heat it then it's pH will decrease because of increasing self ionization. Water is neither a compound nor an oxide Water is definitely a compound, and most people would consider it an oxide as well (dihydrogen oxide).

Sorry, I think it should be part of the original answer.  I thank you for telling me how the site works and I will be very careful to separate things as you advised in future.  Water is not a compound in my opinion.  Yes many people believe it is but that is there right as it is yours.  I did not go into any response on this question because it went outside the bounds of the original answer.  I was simply trying to say that there are people that think it is does not change my conclusion.  I don't change my conclusions based upon the number of people believe a particular way.

I am only pointing out the technical aspects of the post. 0. Please take some time and learn about this site by taking the [tour]. Also consult [Answer] 1. Please do not excessively edit your post. Even in its current stage, after 16 edits, the wall of text is barely readable. Consult this page to learn more. 2. Do not respond to comments directly. Use comments to make your post more clear, but address comments by commenting. You can even reply to a comment with typing @ and the name of the commenter you want to reply to.

It is also not a good practice to repost the same answer, after it has been deleted with good reason. 4. Your answer reads like an opinion only, it would be appropriate to back up your claims with scientific facts and not only mere statements.

Thank you for your help.  As I said, i felt it was important to alter the text to meet the editorial format of  the site which appears not to be inclined to fluid exchange of ideas, as I had wrongly assumed.  I have only been on this site for a day and was trying to answer the original question which should include whatever length of answer I feel is needed.  I have altered the response to a more mainstream paradigm whilst isolating simpler versions of my views. I answered the question with a new response following deletion. I answered the commenter as an extension to my answer to the question

‘H2O, should it be isolated, is not water’ stopped reading there.

Gosh it's awfully difficult to read- I cannot imagine how you get through an entire book. The response I made is all of 2 little paragraphs.  H2O is not water - Water is H2O-OH-H3O (and more yet) and water is the interactivity - connected mass, of it's components, not the singular molecule H2O, which produce the properties of water. One cannot think of water as being the simple molecule H2O.

Let me just say that books follow the rules of typography and punctuation — they alone facilitate reading a lot.

You will understand more as you go beyond basic Chemistry. Do not get so easily frustrated- it is all within the realm of possibility, with lots of hard work. I also recommend a strong, focus on Physics, to gain a more complete understanding of Chemistry.  Best regards.

@James Truth , can you please explain how H2O doesn't have the properties of water ?


text:
How to do this reaction - adding 2 different groups to an alkyne with 2 terminal hydrogens?

How to add a methyl group and an ethyl group to an alkyne with 2 terminal hydrogens separately?

No catalytic hydrogenation. Doing a sequential alkylation of the starting acetylene using sodamide and liquid ammonia with the appropriate alkyl bromide will do the trick.  Be sure to only use one equivalent of the first alkylating agent.  It doesn't matter whether you use $\ce{MeBr}$ or $\ce{EtBr}$ first.


text:
Percentage of carbon in stainless steel to avoid brittleness
What is the maximum percentage of carbon allowed in stainless steel 202 so that it does not cause brittleness?

All  standardised grades of alloys have definite limits of composition.
If its common, then a quick search will find it. 0.15% is the specified limit for SS202.
http://www.azom.com/article.aspx?ArticleID=8209
Whether its brittle or not depends on what you mean by brittle and its thermal processing. Its a martensitic stainless steel so note the details in azom and that it will be more brittle when hardened as described.

The specified maximum carbon content in stainless steel 202 is 0.15 %.

Could you back up your claims with a reference?

http://www.azom.com/article.aspx?ArticleID=8209

Type AISI 202 stainless is very tough regardless of carbon content ( in specification). It is an austenitic stainless with roughly half the nickel replaced by manganese. It becomes stronger with less ductility when it is cold worked , regardless of carbon . As an austenitic , it can not be hardened by heat-treatment. It is relatively uncommon so little data is available: However, the mechanical properties of 202 are similar to AISI 301 , a much more common grade.

Depends on the environment and mechanical stress (+ speed etc.)
The brittleness of a piece of steel highly depends on it's composition (<10% Cr can make it super brittle, too) it's 'temperature history' (e.g. hardened, welded) and the temperature where it is exposed.
So this is found in table-works. You start with the temperature of your environment: (for details see charpy impact test)

Note that transition temperature. I guess, that's the line, your question refers to. So now you look up the table-works and find an adequate composition for your steel.
An example:
room temperature, welded/hardened → less than 0,25 % C (e.g. S235) are always fine; 0,25…0,5 % may work too
not welded/hardened → the same steel (0,25…0,5 C) can be ductile until -50°C
I couldn't find a reference in a quick search, but up to 0,8 % C shall be fine for most things you have in mind.


text:
Why 1-octanol is immiscible with water?
Octan-1-ol is polar due to the presence of the $\ce{-OH}$ group, but it is immiscible with water. How is it possible?

I just thought I'd add a visualisation of this concept. Electrostatic maps show the electrostatic potential at a given point of the molecule. We know water is very polar, and this is evident from its electrostatic map:

If we look at octanol though, we see that the vast majority of the molecule is non-polar (evenly distributed electrostatic charge):

The hydroxy-group of octanol is indeed hydrophilic (polar), but the majority of the molecule wants to stay in the non-polar phase.
Molecules with strongly hydrophilic "heads" and non-polar tails have unique properties in aqueous solutions, including some very important for life. Take phospholipids. In water, they self assemble into the basic structure of the cell membrane, the so-called phospholipid bilayer:

You can see how the phospholipids arrange themselves to have their polar portion in contact with water (hence "hydrophilic" - "loving water"), while their non-polar portion is in contact only with the non-polar portions of the surrounding phospholipids (hence "hydrophobic" - "fearing water").

Sometimes chemistry is a balancing act.  In the case of 1-octanol the hydroxyl group contributes 17 units towards the molecular weight, while the 8-carbon tail contributes 113 units.  In this case the molecule is more non-polar (hydrocarbon like) than polar (water like).  As you make the chain shorter and shorter you continually increase the solubility of the alcohol in water.  For the very short chains, methanol and ethanol, they are completely miscible in water.  Take a look at the chart here...
chart

It is not soluble because the head (OH) is slightly polar and the whole chain is fat. 1-octanol is used to assess partition coefficient in environmental literature because it mimics 'well' animal fat tissue absorption.
$$K_\mathrm{partition} = \frac{[X_\ce{octanol}]}{[X_\ce{water}]}$$

Technically, fats are triglycerides - a triester of glycerol and fatty acids. But yes, they tail is a hydrocarbon; the same feature that makes fatty acids immiscible with water.


text:
Identifying protons in NMR using the J values
When looking over an old organic midterm I came across this question.

As you can see I got half of it right, but honestly I'm not sure how I got that one right or how you would answer this in general.  Looking back at this now it appears that the first signal was split into a doublet and each peak was then split into a triplet, which makes sense for proton c (first split by b and then by d).
The second one seems trickier.  It looks as though the signal was split into a quartet, and then each peak into a triplet.  To me that looks like it should be d - split into the quartet by e and then split into the triplets by c and b.  So why is the answer b?  These J values wouldn't have been supplied for now reason - is there a way I could have used them?  More generally, is it possible (and if so how) to identify the type of proton and the structure of a molecule if you just had access to the NMR and the J values like this?

The proton(s) in question absorbs at 5.95 ppm, the olefinic region - so it cannot be D which is expected to absorb upfield in the aliphatic region.  B is the correct answer.  Proton B would be split by proton C into a doublet with a 15.5 Hz separation (the coupling constant $\ce{J_{BC}}$).  Each line in the doublet would be further split into another doublet with a J=8 Hz spacing by proton A.  So now we have 4 lines, a doublet of doublets.  Each of these four lines will be further split into a triplet with 1.5 Hz spacing by the two (hence the triplet) D protons.
Using the chemical shift of a proton and the observed splitting (e.g. the coupling constants, or J values) can go a long way in helping to determine a structure, particularly in relatively simple molecules.  As molecules get more complicated you can do advanced experiments to help with the structure determination.  One example would be spin decoupling experiments.  For example, if we "know" which signal is due to the D protons, we could irradiate them with energy during our nmr experiment.  If they are coupled to proton B (they are), then the triplet in the B signal should diappear and we would just see our doublet of doublets.  This type of experiment confirms which protons are coupled to which other protons.  There are many other advanced nmr techniques that can be used to help in structure elucidation.

How were you able to identify the split separation in Hz when the spectra was in ppm? Is there a way to do that without the carrier frequency or was it just based on your experience and ability to visually analyze the spectra?

2D experiments are much more common now, typically you'd simply record a COSY, HSQC, HMBC or similar experiments if a 1D is not enough. That give a lot more information than selective irradiation, and is probably easier to set up in most cases.

@Dannnno  No, I wasn't able to read then from the spectrum because, as Mad Scientist pointed out, I don't know your magnet strength.  I just read them from the coupling constants provided in your top figure.

You can read the coupling constant directly from the NMR spectrum. Typically you would use processing software for this which makes it much easier, but you can do it from the spectra you have available as long as you know the field strength of the spectrometer. Your scale is in ppm in this case, to convert it to Hz you have to divide by the spectrometer frequency in MHz, e.g. 0.1 ppm on a 300 MHz spectrometer are 30 Hz.
I've marked where you would measure the coupling in your first example:

This is a nice example of the typical pattern you get for a proton on a double bond that couples once via the double bond and once via a single bond.
The second example is more complicated, the most important thing you should recognize is that it is not a real quartet. The four peaks are all roughly at equal height, if this was a true quartet they should follow Pascal's triangle and be at a ratio of 1:3:3:1.
This means there are more than two couplings involved, and if the couplings are identical or multiples of each other they can shift the peaks onto each other. In this case you have the big 15.5 Hz coupling, that signal is then split up again by the 6 Hz coupling which results in four roughly equally spaced signals that are again split up by a small 1.5 Hz coupling.
In this example the absolute chemical shift is a rather big help, both signals are around 6 ppm which pretty much tells you that those two are the two protons directly at the double bond.

What about if we don't know the field strength (like in this question)?  I suppose we could find that from the information given but to do that wouldn't we have found the answer?

@Dannnno Without the field strength you cannot directly read the coupling constants out of this spectrum. You can rather easily guess them in this case as there are three very different couplings here, and those you can see in the spectra.


text:
Can the amount of shielding in proton NMR caused by ring currents be predicted?
My textbook (Organic Chemistry, 2nd Ed. Hornback, Joseph) states that if a hydrogen is located near the center of an aromatic ring it will experience an (often quite sizable) upfield shift, as seen in these images (borrowed from aforementioned textbook):

Theoretically it all makes sense - the ring current of the aromatic system shields these protons rather strongly. But is there a way to estimate what the actual upfield shift will be? For example, if provided with structure 1 and asked to estimate the chemical shift of the red hydrogen, how could you do so? Or do you just run the NMR and find out? My book doesn't give any advice on how to figure this out, just that it happens.

In principle it is possible to calculate shifts using Density Functional Theory, for example using Gaussian. In practice, such calculations are highly demanding and (due to the necessity to approximate) are prone to substantial errors.

Johnson and Bovey first proposed a theory describing the ring current effect,[1] and in 1972 Haigh and Mallion published their often-cited paper on ring current shielding.[2] A later review by these authors in 1979 describes the mathematical prediction of ring currents very well.[3] You may often see this graphic describing ring current contributions to chemical shifts in textbooks.

So, in answer to your question, it is possible in theory to make chemical shift predictions for nuclei within the shielded regions of ring currents, however in practice, it is very difficult and prone to large errors. Calculations require a very thorough knowledge of distances and angles between the aromatic plane and the nucleus of interest. And a reasonable computer.
Very few chemical shift prediction tables in texts or literature give examples of upfield-shifted aromatics, and none that I'm aware of give enough tabulated data to extend an even ballpark prediction. This makes it very, very difficult to make a prediction unless you have experience with the particular type of molecules you are dealing with. For example, I work with one research group who makes quite good estimations of chemical shifts for the inner ring proton within porphyrin systems, but this has evolved over many years of experience of dealing with this particular class of molecule. They would really be at a loss trying to predict any of the shifts for the 4 compounds you have shown.
Most desktop chemical shift predictors use an additive model based on combined chemical shift contributions, and do not do a full calculation of ring current properties. In fact, many desktop predictors give very poor estimations for these types of protons for this very reason, as they do not distinguish shielded and deshielded regions arising from ring current. For small organic molecules the variables dictating the ring current contribution are difficult to predict, especially for those signals experiencing increased shielding, such as overhanging methyl groups. They require knowledge of distance from the centre of the aromatic ring and angle from the plane. This information is often used qualitatively in spectral analysis to construct a 3-dimensional structure of a molecule, in explaining upfielded shifts of certain nuclei.
In proteins, NMR shift prediction models are more inclusive, as the molecule conformations are often very highly constrained and well predicted. Many shift predictors for protein NMR includes a ring current contribution.

References

Johnson, C. E.; Bovey, F. A. Calculation of Nuclear Magnetic Resonance Spectra of Aromatic Hydrocarbons. J. Chem. Phys. 1958, 29 (5), 1012–1014. DOI: 10.1063/1.1744645.
Haigh, C. W.; Mallion, R. B. New tables of ‘ring current’ shielding in proton magnetic resonance. Org. Magn. Reson. 1972, 4 (2), 203–228. DOI: 10.1002/mrc.1270040203.
Haigh, C.; Mallion, R. Ring current theories in nuclear magnetic resonance. Prog. Nucl. Magn. Reson. Spectrosc. 1979, 13 (4), 303–344. DOI: 10.1016/0079-6565(79)80010-2.

Magnetic anisotropy effects are very difficult to calculate. As such I don't believe there is a table that help you calculate the value so an experiment or a similar example is required. A look at Pretsch shows only one example of this system (structure 1 in your book). Running one of the structures in a NMR predictor gives results that are quite far from the experimental values and obviously don't take this effect into consideration


text:
Mechanism of sulfur trioxide reaction with water to make sulfuric acid
This question touches on whether sulfur is hypervalent (whether it expands its octet when bonding).
The first picture shows both sulfurs with an expanded octet.
The problem is that in the top half of the below picture, we are supposed to end up with sulfuric acid, but we actually end up with sulfuric acid plus two lone pairs (if I drew the mechanism correctly). Did I? (I actually separated out the top part of the picture into two parts - first, hydrogen attacking oxygen; second, oxygen attacking sulfur - so the arrow pushing wouldn't get confusing).
The second picture follows updated chemical fact. According to Wikipedia, sulfur trioxide has one double bond and two dative single bonds. Sulfuric acid is also a charge separated species with four single bonds, not two single bonds and two double bonds. I seem to end up with the proper product here without any hand-waving.

Wikipedia says the bonds to all oxygen atoms in sulphur trioxide are equal double bonds. In section 2 you seem to have incorrectly started off with the sulphite anion, instead. With that correction, does the hydration reaction make more sense to you now?

This is what Wikipedia says to me: "The Lewis structure consists of an S=O double bond and two S–O dative bonds without utilizing d-orbitals."

Also did I push my electrons correctly in section 2? In the middle of my reaction - right in part 2 of section 2 - I have 36 electrons on the reactant side but only 32 on the product side. My electron pushing isn't up to snuff yet.

You are right however that I did start off with the sulfate anion. Will fix.

I hadn't noticed that sentence about the bonding in Wikipedia. Oddly, the structure shown in the top-right of the article is the usual representation for the molecule. I'll have to read the article referenced in the section you pasted.

The structure I'm seeing is one with three triple bonds. What are you seeing?

This is the most used representation for the structure of sulphur trioxide. Also, I was mistaken, you didn't draw the sulphite anion in section 2 after all, otherwise there would be an additional non-bonding pair on the central sulphur atom. Furthermore, the article that mentions the correct structure should have one double and two single bonds comes to that conclusion based on quantum chemistry calculations, which are far deeper than the basic qualitative covalent bond/octet completion theory you're dealing with.

Makes sense. Looks like we both saw the negative formal charges on the oxygens in section 2 but forgot to consider the +2 formal charge on the sulfur. My problem is that if we consider sulfur to be hypervalent we can't seem to achieve the product of sulfuric acid.

let us continue this discussion in chat

I've drawn a more correct mechanism for the reaction of dilute $\ce{SO3}$ with water in the liquid phase:

$$\ce{SO3 (aq) + 3H2O (l) -> SO4^2- (aq) + 2H3O+ (aq)}$$
$\ce{SO3}$ is a strong electrophile, enough to react quickly with water, which is a relatively weak nucleophile. A water molecule is added to the structure, facilitated by the dislocation of a $\ce{S=O}$ bonding electron pair in $\ce{SO3}$ to the oxygen atom, forming the intermediate structure shown. Additional water molecules then quickly abstract hydrogen atoms from the intermediate (actually, the first abstraction is far easier than the second, so it might have been more proper to show each proton abstraction separately in sequence). The final product of the reaction is sulfate anions and hydronium cations ($\ce{H2SO4}$ is a strong acid in water, so it doesn't really exist as a molecule. The second proton isn't nearly as acidic as the first, but in dilute solution both are ionized). The formal global reaction is shown below the mechanism. Notice that at every point the formal charge on the central sulfur atom is exactly the same, zero.
Now, with regards to our little discussion about the actual structure of $\ce{SO3}$. Classical valence bond theory, enhanced with the concept of hypervalency and VSEPR theory, attributes the structure shown, with three double bonds. The Wikipedia article for $\ce{SO3}$ mentions another structure, based on modern valence bond theory computational calculations, with one double and two single bonds. This structure is quite interesting as it does not invoke hypervalency. Indeed, hypervalency is thought to be an old concept, and not really necessary in more modern theoretical treatments of chemical bonding.
However, if one opts to display the structure according to modern valence theory, they would do well to make it explicit. Almost all structural formulae today are drawn in the context of classical valence bond theory. Certain assumptions from the classical theory do not carry over to the modern theory. For example, the authors of the article "Modern valence bond theory", Chem. Soc. Rev., 1997,26, 87-100 (free access mirror) find that the bonding in $\ce{N2O}$, described classically as a mixture of the canonical forms $\ce{^{-}N= N^{+}=O}$ and $\ce{N≡N^{+}-O^{-}}$, is best described as $\ce{N≡N=O}$. Even with five bonds, the central nitrogen atom has a calculated occupancy close to 7, i.e. it is not hypervalent. Since the structural construction rules of classical and modern valence bond theory aren't the same, the structures generated from both do not mesh and cannot be compared directly. The simplest thing to do, then, is to draw $\ce{SO3}$ according to classical valence bond theory, with three double bonds.

It really troubles my eyes seeing the double bonds there. It pays no respect to the actual charges in the molecule. The other bonding picture does not pay tribute to the symmetry. Lewis is a nice concept, but when it comes to hypercoordinate species it is just not applicable any more. The bonding might be better described as $\ce{S^{+}(\bond{~-}O^{-\frac13})3}$ as there is a $A_{2}^{"}$ MO.

@Martin Indeed, I could have added further canonical forms and established resonance to make the picture more accurate, but since my main point was to show the mechanism, I just let it slide.


text:
How to calculate the rotational constants for different states from a stick spectrum?
Given a stick spectrum for an unknown diatomic molecule, how do you determine the rotational constants $B_\nu$ for both the excited and ground states given that
$$E_{j}=B_{\nu}j(j+1)$$
\begin{array}{cc}
\text{Line position}\, \mathrm{(cm^{-1})} & \text{relative intensity} \\
\hline
27128.20 & 1.47 \\
27171.74 & 2.35 \\
27211.40 & 3.13 \\
27247.18 & 3.34 \\
27279.08 & 2.62 \\
27331.24 & 1.00 \\
27351.50 & 2.62 \\
27367.88 & 3.34 \\
27380.38 & 3.13 \\
27389.00 & 2.35 \\
\hline
\end{array}

I noticed you added a line list, but I don't see how it contributes to the question. Was there some particular reason for it?

In your question, it's not clear whether the quantum number assignments for the transitions are known in advance, so for generality I'll assume that you don't know what they are, and help you find them.
Combination-differences is the most common paper-based method to determine line assignments (for more detailed analysis, a computer-aided fitting program such as PGopher is best). From the line assignments, it's easy to find the $B$-values.
Rotational energy levels are given by
$$F_\nu(J)\approx B_\nu J(J+1)+D_\nu [J(J+1)]^2$$
where $B_\nu$ and $D_\nu$ are the rotational constant and centrifugal distortion constant associated with the manifold $\nu$.
Denoting the initial and final states by $1$ and $2$ respectively, the $P,R$ transition energies are given by
$$P(J)=F_2(J-1)-F_1(J)
\\R(J)=F_2(J+1)-F_1(J).$$
Combination-differences is can be done in two ways:

If your rovibronic transition originates from an initial manifold with well-documented spectral characteristics (including $B_1$), note that
$$R(J-1)-P(J+1)=2 (2 J+1) \left(B_1-2 D_1 \left(J^2+J+1\right)\right)\approx2B_1(2J+1)$$
when $D_1$ is small, and so then you can just try to find pairs of $P,R$ peaks with spacing $2B_1(2J+1)$ for various small values of $J$. This allows you to number the peaks as to their identity.
If instead the upper state has well-known $B_2$, then note that
$$R(J)-P(J)=2 (2 J+1) \left(B_2-2 D_2 \left(J^2+J+1\right)\right)\approx2 B_2 (2 J+1)$$
and so then you can just try to find pairs of $P,R$ peaks with spacing $2B_1(2J+1)$ for various small values of $J$. Again, this allows you to number the peaks as to their identity.

Once the quantum numbers are assigned, finding both $B_1$ and $B_2$ is easy. For example, $R(J)-R(J-1)=2 \left(B_2 (J+1)-B_1 J\right)$, so if you measure it for two values of $J$, then you have two unknowns and two datapoints, so just solve for $B_1$ and $B_2$.
Fun Pictures
Just to give a simple idea of how spectra are simulated, the following Mathematica code does a basic rotational spectrum assuming Boltzmann statistics at a reasonable temperature (hit Cell>Convert To>StandardForm to get it to render the subscripts correctly):
Subscript[F, k_][J_] :=
  Subscript[B, k] J (J + 1) - Subscript[D, k] J^2 (J + 1)^2;
R[J_] := Subscript[F, 2][J + 1] - Subscript[F, 1][J];
P[J_] := Subscript[F, 2][J - 1] - Subscript[F, 1][J];
ListPlot[Transpose[
  Table[{{R[J], (2 J + 1) Exp[-(1/30) Subscript[F, 1][J]]}, {P[
       J], (2 J + 1) Exp[-(1/30) Subscript[F, 1][J]]}} /. {Subscript[
      D, k_] :> 0, Subscript[B, 1] -> 1, Subscript[B, 2] -> 0.9}, {J,
    0, 15}], {2, 1, 3}], Filling -> Axis]

I think a simple way is 2B is the interval between two spectral lines so subtract line position 2 from line position 1 and then divide by 2 you will get B i.e. rotational constant as $21.77\rm~cm^{-1}$.


text:
how to speed up Ferric Chloride and copper reaction
im doing pcb etching process (for those who don't know pcb is that thing with electronics on it lmao) using Ferric Chloride and it's taking forever
is there any chemicals i can find that will speed up this process (could it also be found at home, salt or acid for instance)
thanks in advance

Chemicals are probably not the good way to go. What is the temperature of the solution? Increasing the temperature speeds up the reaction. Btw: take the proper safety precautions when etching pcb's!


text:
Dipole moment of symmetrical molecules
While studying dipole moment in organic chemistry, I got stuck at a particular statement written in the book :

Symmetrical molecules without lone pairs of electrons will have $\mu = 0.$

The main problem was, why would they have $\mu =0$ ?
What type of Symmetrical molecules are possible?
The book gave the example of $\ce{BH3}$. How is this a symmetrical molecule?

I'd rather start any answer with a reminder that BH3 does not quite exist as such; B2H6 does.

Yes, the statement in your book is very vague.  Molecules are often classified by the symmetry elements they contain.  For example, $\ce{BH3}$ contains a $C_3$ axis perpendicular to the plane containing the $\ce{BH3}$ molecule.  "$C_3$" means that if you rotate the molecule around this axis by $360^\circ/3 = 120^\circ$, you'll get a molecule that is indistinguishable from the molecule you started with.  $\ce{BH3}$ also contains 3 $\sigma_\mathrm{v}$ planes of symmetry.  These planes contain the $C_3$ axis and one of the $\ce{B-H}$ bonds.  $\ce{BH3}$ also contains one $\sigma_\mathrm{h}$ symmetry plane that is perpendicular to the $C_3$ axis and contains the $\ce{BH3}$ molecule.  In addition to these symmetry elements, a number of other symmetry elements exist as well.
It turns out that only molecules that belong to symmetry classes (point groups)

$C_n$ (the molecule only contains a $C_n$ axis
$C_{n\mathrm v}$ (the molecule contains a $C_n$ axis and $n$ total $\sigma_\mathrm{v}$ planes)
$C_{\mathrm s}$ (the molecule only has a plane of symmetry that contains
the entire molecule)

can have a permanent dipole moment.  If the molecule belongs to any other point group it cannot have a dipole moment.  Since $\ce{BH3}$ contains a $\sigma_\mathrm{h}$ plane of symmetry ($\ce{BH3}$ belongs to point group $D_{3\mathrm h}$) - it therefore does not belong to groups $C_n$, $C_{n\mathrm v}$ or $C_{\mathrm s}$ - and cannot have a dipole moment.

$\ce{BH3}$ is a flat triangle molecule, so it has mirror symmetry. Since the dipoles cancel out, the molecule does not have a dipole moment.


text:
Strongest B/L Acid Which Can Exist (Leveling Effect)
Got a few issues with this:

Here is the Lewis Structure:

As we can see, we have five lone pairs on which a hydrogen proton can land and form the solvent's conjugate acid. Let's consider them left to right.
1) Oxygen lone pair. Adding a hydrogen proton here would make the oxygen have a formal charge of +1. Not exactly ideal for an atom which is so electronegative (even more EN than nitrogen is).
2) sp3 nitrogen. This seems like a good site; it parallels that the ammonia molecule. However, we would be creating a +1 formal charge.
3) sp2 nitrogen. We again would be creating a +1 formal charge.
4) sp nitrogen. And finally, we would be creating a +1 formal charge.
I know that the more s-character, the more able is the hybridized atom able to stabilize negative charge. It isn't called electronegativity for nothing.
That being said, is electronegativity inversely related to electropositivity?
So should the sp nitrogen be least able to stabilize a positive formal charge?

From a Lewis acid/base consideration, we would say that each of these lone pair sites are pretty nucleophilic. So which is the most nucleophilic. We need the strongest electron donating tendency. And that appears to be the sp3 nitrogen because in an sp3 nitrogen, the electrons are further from the nucleus on a time-average basis than the electrons in an sp2 or sp nitrogen.

Nucleophilicity does not necessarily correlate with basicity. And by adding a proton you always have to create a charge, and this charge is real not formal.

Your reasoning is not entirely correct, as you consider that the charge is only located on the atom which bears the proton upon protonation. But nearby atoms, bonds and molecules will also participate in the stabilization of the positive charge, through several factors.
In the gas phase (so considering no solution phase effects), you should at least consider that, in addition to the instrinsic capability to bear a proton of the atom (which goes as the inverse of electronegativity, as you state in your discussion), one has also to consider the following factors:

Resonance effects (delocalization of the charge on other atoms, that may be less electronegative than the atom directly bearing the proton). In the molecule given above, this would lead to a stabilization of the $sp$ and $sp^{2}$ hybridized nitrogens, as there would be a partial positive charge on the neighbouring carbon atom, as: $$\ce{RCH=N+H-R' <-> RC+H-NH-R'} $$ or $$\ce{RC#N+H <-> RC^{+}=NH}$$
Inductive effects (or if you prefer availability of electrons in nearby atoms that can be displaced towards the charged site). In your constrained textbook example, this would be hard to indicate: there are a lot of heteroatoms (nitrogens and oxygens) and electron withdrawing groups (CN) which are likely to remove electrons from the carbon atoms, and in the central part of the molecule, these are likely to contribute to destabilize the molecule.
Intramolecular hydrogen bonds. Although sometimes neglected, these can contribute quite a lot to the stabilization of the charge when mutiple protonation sites involving heteroatoms are present. Just as a reminder, in the gas phase, a hydrogen bond in a protonated molecule can reach up to 120 kJ/mol. In your example they would be hard to predict without modelling the complete structure, but one could consider a possible H-bond between the protonated imine ($sp^{2}$ nitrogen) and the alcohol oxygen.

Then if we were to consider solution phase, the interactions between neighbouring molecules and the propensity to form hydrogen bonds with the protonated molecule will also play a role on the possible protonation sites.
Based on the above considerations, one should consider that the most likely stable forms would be protonation on either the nitrile group ($sp$ nitrogen) or the imine group ($sp^{2}$ nitrogen), as both allow to put a partial positive charge on the neighbouting carbon atom. The protonated nitrile group is a protonation on an $sp$ nitrogen, with charge delocalization on an $sp$ carbon which means that the charge stabilization will be less efficient than for an imonnium ion.
If one considers purely gas phase consideration, one can use the proton affinity scale to rank these various contributions.
If we look for proton affinities in the NIST database we have:

for a secondary amine $\ce{H3C-NH-CH3}$ 929 kJ/mol
for a nitrile $\ce{H3C-C#N}$ 779 kJ/mol

Unfortunately little experimental data is available for imines. But some theoretical data is available in this article by Hammerum et al.. If one considers the methylimine $\ce{Z-CH3-CH=N-CH3}$ proton affinity is calculated at 943 kJ/mol, higher than dimethylamine. But on the other hand if one considers $\ce{H2C=N-CH3}$, its proton affinity is of 880 (exp) or 902 (theo), lower than dimethylamine.
So inductive effects clearly play a very important role in the order between amine protonation and imine protonation in the gas phase. In a really constrained molecule as above, they are likely to play a major role, as is the possibility to form hydrogen bonds in the solution.
If one now puts in the possible formation of intermolecular hydrogen bonds, protonated secondary amines can make two hydrogen bonds with two hydrogen bond acceptors, whereas protonated secondary imines can only make one hydrogen bond. This will likely tip the balance towards amines and would make answer (4) most likely in solution, whereas I would be quite at loss to give an answer in the gas phase.


text:
How to calculate the concentration after a certain timespan from the specific reaction constant?

A container at $30~^\circ\pu{C}$ contains $\ce{HNC_{(g)}}$ reacting to form $\ce{HCN_{(g)}}$. Knowing that the specific reaction constant is $4.4\cdot10^{-4}~\pu{s-1}$ and that you initially have $\pu{1 g}$ of $\ce{HNC}$: how many moles of each gas are in the container after one hour and a half?

I'm pretty sure that the reaction is reversible here. However, I have no idea how to calculate how many moles of each gas will there be after $\pu{1.5 hrs}$ (even if it was not reversible).
I have many doubts:

Did they skip the concentration because there's only one reactant?
Does $K$ take into account the $30~^\circ\pu{C}$?
What's the equation for "change of concentration over time" here?
How can I figure out how many moles of $\ce{HNC}$ and $\ce{HCN}$ the container will have after 90 minutes, ignoring the fact that it's a reversible reaction?

Where did you get the idea that the reaction formed an equilibrium at equal amounts of HNC and HCN? In truth, the equilibrium is quite favoured towards products; hydrogen isocyanide is a rather unstable molecule. For this problem, you don't really need to think about chemical equilibrium at all (or equivalently, assume the equilibrium constant is infinite).

I think that because it is such a slow reaction that you don't reach equilibrium in 90 minutes. The provided rate is for the specified temperature of 30°C. For a first approximation, I would treat it as only a forward reaction to see what the concentrations would be. If the HCN concentration is still small (in comparison), I would go with that.

Okay thanks a lot! I understand. However, how can I figure out how many moles of HCN I have after 90 minutes?

Because the rate constant is $4.4\times 10^{-4} s^{-1}$, it tells us that it is a first order reaction. If you look at this section in Wikipedia, you should see the direction to go. http://en.wikipedia.org/wiki/Rate_equation#First-order_reactions

To answer this, you should know what is the concentration of HNC after 90 minutes. First, you should know the initial concentration of HNC.
After that, look at the rate constant. It tell you if the reaction is a first order. So, you can apply the data in integrated first-order rate law equation:

Don't forget if you must change 90 minutes to seconds :)
Then, now you find the concentration of HNC after 90 minutes. To figure out the moles of HCN after reaction, we know that the initial moles of HNC is the same as the moles of HNC after reaction plus the moles of HNC that changed to HCN.
Then after submit the initial moles of HNC and moles of HNC after reaction, you get the moles of HNC that changed to HCN. This is the same of moles of HCN that
present in the container.


text:
How to explain the concept of mol and reactions involving mol quantities
From time to time I do little pro bono lessons for second grade in math, physics, and chemistry. Recently I came across this question.

In a container is $2~\mathrm{mol}$ of $\ce{SO3}$ (sulfur trioxide). This compound under certain conditions breaks down into $\ce{SO2}$ (sulfur dioxide) and $\ce{O2}$ (oxygen). At the same those two resulting compounds react together back to form $\ce{SO3}$. In equilibrium we have $1.52~\mathrm{mol}$ of $\ce{SO3}$.
  How much of remaining resulting compounds do we get?

The answer: It is obvious that $0.48~\mathrm{mol}$ of $\ce{SO3}$ broke down into $0.48~\mathrm{mol}$  of  $\ce{SO2}$ and $0.24~\mathrm{mol}$ of $\ce{O2}$.
The problem: A few of students cannot grasp how could we start with $2~\mathrm{mol}$ of some substance and end up with $2.24~\mathrm{mol}$ of other substances. I have tried to explain to them that $\mathrm{mol}$ is not a quantity as they are used to as in weight or volume, but that a $\mathrm{mol}$ is a number of molecules. I have tried to simplify an example using just 2 molecules of $\ce{SO3}$ and showed that the break down to 2 molecules $\ce{SO2}$ and 1 molecule of $\ce{O2}$. That was understandable. But as soon as we got back to moles: "How can we get more than we started with?"
What is the best way to explain this? Any experience? To me, it seems so obvious.

Try showing them with legos maybe, and use the dissolution of NaCl as an example.

This is an exceptionally well asked questions! Ball-and-Stick modelling by hand may be an option. If you do not want to purchase an available set you can use plasticine and matches. You can go from single molecules and double and double [and double]$_n$ and maybe they understand better this way.

Before getting into the less trivial combustion-like reactions, have your students mastered displacement equilibrium?

I should point out that I am not a teacher. I teach this in my sparetime to kids who have problems but are willing to do something about it. Therefore I am bound to the issue at hand before exams.

The fundamental misunderstandings here are related to two concepts:

Conservation (atoms are conserved, molecules are not)
A mole is just a number (like the chemist's dozen)

I would start with conservation, then move on to moles.
Explaining why atoms are conserved, but not moles
For second-graders, hands-on activities are probably best, so molecular models or something similar would be a good approach. I like your example using two molecules - if you can get some models, you could try having groups of students start with two (or more - 4+ would make it easier to see) molecules of $\ce{SO3}$. Have them count the number of molecules and the number of each atom. Write the results down where everyone can see them. Then, have them do the "reaction" - let them take apart the molecules and put them back together to make$\ce{O2}$ and $\ce{SO2}$. Again, have them count the molecules and the atoms. Show them that the number of atoms stays the same (for each element and in total), but the number of molecules can change.
Once your students are comfortable with this idea, introduce the concept of moles as a way of counting very large numbers. This is the more difficult part, and in my experience is the "make or break" concept for most chemistry students.
A mole is just a very large number
Goal: The number of atoms in macroscopic samples is huge
You can motivate the discussion by calculating the number of molecules/atoms in very small samples. In my classes (college level) I use the number of molecules in one mL of hydrogen gas at STP, and show that if you counted one molecule per second, it would take longer than the age of the universe (a lot longer) to count them all. Sometimes I use the number of silicon atoms in a bag of sand and show them that it is larger (much larger) than the number of stars in the universe. Other examples would work as well - the idea is to get them thinking about the largest quantity that they can possibly imagine and show them that the number of atoms in even a small amount of matter is much larger than that.
Goal: A mole is just a number
Now you want to link this to moles. I start with the concept of a dozen. I ask things like "how many donuts are in a dozen?" "Ok, how many cookies?" "How many cats?" "How many people" until someone points out that it's always 12. Then I ask them why we say a dozen instead of 12. If they don't figure it out quickly on their own, I tell them that for bakers and their customers, it is a convenient way to count stuff. Then I tell them chemists also have a convenient way to count atoms and molecules, which is good because there are a lot of them! The number they use is called a mole, and similarly to how a dozen = 12 of anything, a mole = 6.022 x 10^23 of anything.
Goal: A mole is a HUGE number
Depending on how good the students are with scientific notation, you might want to actually write out that number of zeroes - it's an impressive demonstration! When there is time, I also like to demonstrate the scale by comparing it to different things as the number of zeroes grows. For example, I say (or have them say) "thousand," "million," "billion," "trillion," for every three zeroes. At around 11 zeroes I start comparing it to unimaginably large things. As examples:

The number of stars in the Milky Way galaxy is about 100 billion ($1 \times 10^{11}$).
The number of sheets of paper that would fit in a stack between Earth and the moon is about 4 trillion ($4 \times 10^{12}$).
The number that would fit between Earth and Mars (at their average distance) is about $2 \times 10^{15}$.
Age of the universe in seconds is $4 \times 10^{17}$.
There are $10^{22}$ stars in the universe.

I think that this is effective because it connects the concept of moles being huge and the number of atoms being huge to a bunch of astonishing facts - people tend to remember weird little surprising facts, and because of the way memories form, the moles stuff gets stored along with it, leading to an artificially created "intuitive" understanding of the size scale of moles and their relationship to counting atoms.
Now you need to tie it all together.
Goal: Moles are used to count atoms and molecules
I connect the concept of a mole being a very large "convenience" number used for counting to the stoichiometry and conservation concepts with a simple exercise. I write out a chemical equation, which in your case would be:
$$
\ce{2SO3 <=> 2SO2 + O2}
$$
Underneath, I write the number of molecules of each, starting with the smallest. So, I would write...
$$
\begin{array}{|c|c|c|c|c|}\hline
\ce{2SO3} & \ce{<=>} & \ce{2SO2} & \ce{O2} \\ \hline
2 \space \rm{molecules} &  &  2 \space \rm{molecules} & 1 \space \rm{molecules} \\ \hline
\end{array}
$$
... and say "If we have two molecules of $\ce{SO3}$, we will end up with two molecules of $\ce{SO2}$ and one of $\ce{O2}$." Then I ask them: "What if I start with 10 molecules $\ce{SO3}$?" and let them figure it out, writing on the board as I go. Next, I ask

"What if I start with 12?"
"What about a dozen?"
"How about 100?"
"A billion?"
"$6.022 \times 10^{23}$?"
"A mole?" - at this point, most of them will have the idea
"10 moles?"

At the end, the board looks like this:
$$
\begin{array}{|c|c|c|c|c|}\hline
\ce{2SO3} & \ce{<=>} & \ce{2SO2} & \ce{O2} \\ \hline
2 \space \rm{molecules} &  &  2 \space \rm{molecules} & 1 \space \rm{molecules} \\ \hline
10\space \rm{molecules} &  &  10\space \rm{molecules} & 5 \space \rm{molecules} \\ \hline
12\space \rm{molecules} &  &  12\space \rm{molecules} & 6 \space \rm{molecules} \\ \hline
1\space \rm{dozen} &  &  1\space \rm{dozen} & 0.5 \space \rm{dozen} \\ \hline
100 \space \rm{molecules} &  &  100 \space \rm{molecules} & 50 \space \rm{molecules} \\ \hline
1 \times 10^9 \space \rm{molecules} &  &  1 \times 10^9\space \rm{molecules} & 0.5 \times 10^9 \space \rm{molecules} \\ \hline
6.022 \times 10^{23} \space \rm{molecules} &  &  6.022 \times 10^{23}\space \rm{molecules} & 3.011 \times 10^{23} \space \rm{molecules} \\ \hline
1 \space \rm{mole} &  &  1 \space \rm{mole} & 0.5 \space \rm{mole} \\ \hline
10 \space \rm{mole} &  &  10 \space \rm{mole} & 5 \space \rm{mole} \\ \hline
\end{array}
$$
At this point most everyone will "get it" - and the advantage of this method is that they sort of figure it out for themselves along the way, which means they will remember it more clearly. If anyone still struggles with the conservation part, you can have them calculate the number of each atom and each molecule for all or some of the examples above, in the same manner as at the beginning of the lesson.

Nice answer! However sometimes some professors may think that is wrong to say that a mole is just a number: Avogadro number is just a number, a mole is an Avogadro number of "atoms, molecules, ions, electrons, other particles, or specified groups of such particles."

@GM, I know that IUPAC specifies elementary particles, and that in practice that is what moles are used for, but I would argue that "elementary entities" can be defined for anything, and therefore the definition is needlessly strict (and by trying to draw line, actually becomes more ambiguous). Also, I agree that it is important to understand that a mole is an abstract concept, while Avogadro's number is the constant derived from the definition of a mole - but it's not necessary to understand that at this stage, and it usually just confuses people when it is introduced too early.

@thomij Yes preservation of atoms. I knew that, I said that at the moment but did not emphasize it enought. Great tip for future.

This is indeed a advanced exercise so maybe first you should made more easier exercises. In this case the student need to know mole concept and equilibrium concept that are not directly linked, the problem here is more related to equilibrium concept. For the future he/she should begin to reasoning following these steps:
Write the balanced equation:
$$\ce{2SO3 <=> 2SO2 + O2}$$
Then for simplify you could write it so, this is more easier to understand, always ask Can I write it so?:
$$\ce{SO3 <=> SO2 + \frac{1}{2}O2}$$
Then you should ask him: Which is the difference between this notation: $\ce{SO3 -> SO2 + \frac{1}{2}O2}$ [eq. 1]  and this $\ce{SO3 <=> SO2 + \frac{1}{2}O2}$ ? You should focus  that in the first reaction all the reagent are transformed in the products  in the second only a part. So you can show him what happen if this is not an equilibrium. For the first case you can write this table:
\begin{array}{|c|c|c|c|c|}\hline
 &\mathrm{SO}_{3}  & \mathrm{SO}_{2} & \mathrm{O}_2 \\ \hline
\mathrm{Initial}              & 2\ \mathrm{mol}   &  0 &0    \\ \hline
\mathrm{Variation} & \small{(0-2=)}-~2 & (2-0=)+2&(1-0=)+1 \\ \hline
\mathrm{At\ the\ end}         & 0~\mathrm{mol}& 2~\mathrm{mol}& 1~\mathrm{mol}\\ \hline
\end{array}
(Maybe write the initial row and let the student write the "At the end" row)
At this point you can talk about equilibrium: (the variation row may lead to confusion so maybe you could write a 3 row table). You can tell him to fill something like this:
\begin{array}{|c|c|c|c|c|}\hline
 &\mathrm{SO}_{3}  & \mathrm{SO}_{2} & \mathrm{O}_2 \\ \hline
\mathrm{Initial}               & 2\ \mathrm{mol}   &  0 &0    \\ \hline
\mathrm{Variation} & ? & ?&? \\ \hline
\mathrm{At\ the\ end\ (equilibrium)}   & 1.52\ \mathrm{mol}      & ? & ?\\ \hline
\end{array}
At the end:
\begin{array}{|c|c|c|c|c|}\hline
 &\mathrm{SO}_{3}  & \mathrm{SO}_{2} & \mathrm{O}_2 \\ \hline
\mathrm{Initial}              & 2\ \mathrm{mol}&  0 &0    \\ \hline
\mathrm{Variation} & -~(2-1.52\ \mathrm{mol}) & +~(2-1.52\ \mathrm{mol}) & +~\frac{(2-1.52\ \mathrm{mol})}{2} \\ \hline
\mathrm{At\ the\ end\ (equilibrium)}           & 1.52\ \mathrm{mol} & 2-1.52\ \mathrm{mol} & \frac{(2-1.52\ \mathrm{mol})}{2}\\ \hline
\end{array}
So the core concept: you have a fixed number of molecules one part is transformed in products one part is still reagent.
So starting form 2 mol of reagents only a part of its can be transform in products. The goal is simply find this part (with a simple subtraction) and calculate the products using eq. 1.

Could be useful use metaphors, some times I use dozen of eggs. This could be dangerous please note dozen can be compare to Avogadro's number but not to mole in fact a mole is not simply a number is a "number of" because only refers to elementary entities  (may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles, so not only molecules but neither orange or ping pong balls). With these exercise maybe you should NOT use metaphors: one misconception you should avoid is that it is a static equilibrium, tell him that as $SO_{3}$ molecules decompose they are formed but at equilibrium the rate is constant. This is a dynamic equilibrium!

Say you have a piece of bread. Lets refer to that piece of bread as one large crumb. Then you split this large crumb into 10 smaller crumbs.
Start: one large crumb
End: 10 smaller crumbs
More advanced example:
Assume you have a piece of bread, consisting of three different kinds of bread (three pieces of each type of bread) that you just squeezed together in your hand so that they would stick. Then, by some magical force, the large multicrumb is split into its constituents (the three different kinds of bread).
Start: 1 large multicrumb
End: $\ce{3Crumb_{type 1} + 3Crumb_{type 2} + 3Crumb_{type 3}}$

Currency transactions or making change can be viewed as examples of an equilibrium process.  Maybe this example might help the students see your point.  For example, $$\ce{4Q <=> 2Q + 3D + 4N}$$ $$\ce{4Q <=> 3Q + 25P}$$ $$\ce{4Q <=> 1DB}$$ $$\ce{Q=quarter, D=dime, P=penny, DB=dollar bill}$$

Very nice analogy, I'll take it.


text:
Water-water vapour equilibrium above the boiling point

Consider $\ce{H2O_{(g)} -> H2O_{(l)}}$ at $\pu{105^{\circ} C}$. Under these conditions it is false that:
(1) $T\Delta S_r$ is negative.
(2) Relative humidity is 100%.
(3) $|\Delta H_r|>|T\Delta S_r|$
(4) $P_{\ce{H2O_{(g)}}}>\pu{1 atm}$
(5) $|{\Delta G_f}_{\ce{H2O_{(g)}}}|=|{\Delta G_f}_{\ce{H2O_{(l)}}}|$

Been trying to understand this question. Here is my reasoning:

True, because $T$ is positive $\pu{(105 + 273)K}$ while $ΔS$ should obviously be negative because the final entropy is lower than the initial entropy. We are after all going from water's gaseous state to its liquid state. There are fewer degrees of freedom in water's liquid state.

If there is an equilibrium between water vapor and water liquid, then by definition relative humidity is 100%. The air is saturated with water vapor to the point where it can't even hold the water vapor and water vapor exists in a dynamic equilibrium with liquid water. From Wikipedia:

Saturated air is really just air which contains water vapor in equilibrium with a liquid water source.

False, because $ΔG$, not $ΔG^{\circ}$. $ΔG$ for a system at equilibrium is always $0$. $ΔG^{\circ}$, not necessarily $0$. So we have $0 = ΔG = ΔH - TΔS$. This implies $ΔH = TΔS$.

True. For liquid water to boil, vapor pressure must equal atmospheric pressure. Similarly for water to condense, atmospheric pressure must equal vapor pressure. Also, vapor pressure is solely dependent on temperature of the pure liquid. The below statement is from UC Davis' ChemWiki. I would think that adding solute would change the vapor pressure (and yes it does, and it's discussed further down the page of the site).

Vapor pressures are dependent only on temperature and nothing else.
Taking into consideration that water boils at $\pu{100^{\circ}C}$, we may conclude that at $\pu{105^{\circ}C}$, the vapor pressure of liquid water is greater than $\pu{1 atm}$. And therefore, for this equilibrium to exist, both the vapor pressure of liquid water and the atmospheric pressure of gaseous water must be equal and thus greater than $\pu{1 atm}$.

This is true. This would make $ΔG = 0$ since $ΔG = G_{\text{(products)}} - G_{\text{(reactants)}}$.

I would like to know if my reasoning is correct.

All the arguments are perfect except the fourth.
The statement is true, but the premise that the vapour pressure of water must equal the atmospheric pressure is incorrect. A phase equilibrium between the gaseous and the liquid phase can be established at a particular temperature without the liquid boiling. Boiling is a special case of this equilibrium when the atmospheric pressure is entirely constituted by water vapour in the vicinity of the liquid surface and hence the vapour pressure can no longer rise at that pressure. In normal cases, as is possible in this case also, vapour pressure of water is just one component of the atmospheric pressure and the liquid need not be boiling.
A more accurate sentence would be $P_{atm}\geq P_{vap}$. This would still mean that both vapour pressure and the atmospheric are higher than 1 atm but not necessarily equal.


text:
Can a volume of gas be measured by change in PSI?
I had a visit from an air conditioner repairman who brought with him a tank of refrigerant. He measured the system and determined that the pressure was about 20 psi too low. So he added refrigerant to the system until it was right.
When it came time to pay him, he said, "well I had to raise your system by 20 psi so it's going to be [20 x price for 1 psi]."
The price was fine, but I am still wondering: would it have taken the same amount of gas to raise my system from 0 to 20 psi as it would to go from 80 to 100, say?
Or, in other words, is it absurd to charge by pressure delta instead of some more reliable measure of what was dispensed from the tank?

Volume is inversely proportional to pressure only when:
1) the number of moles of gas is held constant. We can say this is true in your situation because the repairman measured the pressure in the tank and I don’t think a significant amount of gas disappeared from the tank in the time it took for him to measure the pressure.
2) The temperature is held constant. Again, I don’t think temperature dropped 20 kelvin in the time it took for him to measure the pressure in the tank.
This follows from the ideal gas law equation – a good approximation most of the time:
$PV=nRT$.
So I say your repairman still has his wits about him and isn’t trying to rip anyone off.

Sure PV = nRT works... but I think the poster's original comment is also valid.  A delta pressure as a unit of "refridgerant" seems a bit odd since different units (heat pumps) will have different volumes.  He may not have been trying to "rip off" the customer... but unless all heat pumps have the same internal volume (and they don't) a pressure delta is a strange / semi-bogus way to measure things.  The flip side is that there probably are a small number of common volumes for commercial heat pumps, but he didn't say "delta pressure for this size"...

Dan, that's a good point. In Dissenter's defense, though, most of the ACs here are of a very predictable size so his experience may have led him to charge by pressure delta as a rough indication of the material he's dispensed.

In fact $PV=nRT$ works better when you deal with gas at high temperature and low pressure. In your case it depends on the pressure of your conditioner and the compressibility factor of your refrigerant. If you know the refrigerant and his compressibility factor you can evaluate it.


text:
Cis and Trans Isomerism in Resonance-Stabilized Compounds
Would it be valid to describe $N_2O_5$ isomers as cis or trans? As a matter of fact, does this molecule even have isomers, given that all its outer N-O bonds are resonance stabilized and therefore equal in length and strength?
Also for non-resonance stabilized compounds, would it be okay to describe say the N to O double bonds as cis or trans with respect to each other? I've just never heard of cis/trans being used this way.

$\ce{N2O5}$ doesn't have cis-trans isomers.  Can you add a picture to clarify what you mean by "N to O double bonds as cis or trans with respect to each other"?

They were trying to say something like this: http://www.shodor.org/Master/environmental/general/decomposition/n2o5.gif
has its double bonds trans to each other.

@Dissenter The "outer" $\ce{N=O}$ double bonds are delocalized within their respective $\ce{NO2}$ groups (like it is shown here). The picture you link to in your comment only shows one of the possible mesomeric structures and not a different isomere.

Right, that's what I thought. In a resonance stabilized compound - as I wrote in my original post - all the bonds are equal. However, my question is, if there were no resonance, could we describe molecules in this fashion? I.e. N=O is cis/trans to the other N=O?

Thanks for the picture, now I understand your question. All of the atoms in $\ce{N2O5}$ don't need to lie in a plane as is shown in the picture in your link.  At room temperature there would be rapid rotation about the 2 bonds connecting the central oxygen to the two nitrogen atoms, so there is no "fixed" geometry from which you could use the cis and trans references.  For example, imagine the conformation with one $\ce{NO2}$ group in the plane of the screen and the other perpendicular to the screen, even if each $\ce{NO2}$ group had one $\ce{N=O}$ double bond and one $\ce{N-O}$ single bond, the terms cis and trans would still have no meaning in this conformation. Further, as you noticed, all 4 bonds between nitrogen and the exterior oxygens are equivalent due to resonance, so again terms like cis and trans would have no meaning.  Finally, I can't imagine a situation where an $\ce{N=O}$ double bond (e.g. where the oxygen is unsubstituted) could have cis-trans isomers because the 2 lone pairs on the oxygen are equivalent.

Ah, thank you. So we can only use cis-trans to describe molecules in a plane? Wait, I think I know the answer. We don't use cis-trans to describe 3D molecules that rotate around a sigma bond. And we can use cis-trans in 3D - e.g. with a substituted cyclohexane ring.

Correct, the cyclohexane was a good example!  Here is a list of terms from Wikipedia to describe different conformations that can result when we consider rotation around single bonds:  a torsion angle of ±60° is called gauche; a torsion angle between 0° and ± 90° is called syn (s); a torsion angle between ± 90° and 180° is called anti (a); a torsion angle between 30° and 150° or between –30° and –150° is called clinal; a torsion angle between 0° and ± 30° or ± 150° and 180° is called periplanar (p);

Thank you! Only knew about cyclohexane because I'm a beginning organic chem student ;)!

(cont); a torsion angle between 0° and ± 30° is called synperiplanar or syn- or cis-conformation (sp); a torsion angle between 30° to 90° and –30° to –90° is called synclinal or gauche or skew (sc); a torsion angle between 90° and 150° or –90° and –150° is called anticlinal (ac); a torsion angle between ± 150° and 180° is called antiperiplanar or anti or trans (ap)


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Liquid crystal structure: Equivalent effect of alkoxy group?
Consider a structure like this,
R - Cy - Cy - Ph - OR'
where Cy represents a cyclohexane ring, Ph is a phenyl ring, and R, OR' is alkyl group and oxygen bonded to different alkyl group.
If 2 linearly bonded fluorines on top of the phenyl ring are moved to either cyclohexane ring, do you get the same effect?
What "effect" means I'm not sure, but this is related to liquid crystals having  $\Delta\varepsilon=\varepsilon_{\parallel}-\varepsilon_{\perp}<0$.
My initial guess is that because cyclohexane rings have $\sigma\rightarrow\sigma^*$ transitions, resulting in lower birefringence $\Delta n=n_e-n_0$, the effect is the not the same. I'm also assuming the dipole will be different? This is an organic chemistry question but sort of physics related so i wasn't sure where to put this question.


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Identifying Nucleophiles and Electrophiles
Are all polar molecules both nucleophilic and electrophilic, depending on which atom you are talking about? Take water for example. Oxygen bears the partial negative charge. Oxygen is nucleophilic. Because oxygen isolates electron density from the hydrogens, the hydrogens bear partial positive charges. So would the hydrogens be electrophilic?

Nucleophilicity and Electrophilicity are no absolute measurable quantities - they are based on an per entry basis and usually in reference to some kind of reaction. So basically you are right in assuming that polar molecules have both properties. In the special case of water these properties also stem from $$\ce{H2O <=> {}^{-}OH + H3+O}$$
But let us have a look at some molecules:

$\ce{CO}$ pretty good nucleophile especially towards transition metals. It has quite a small dipole towards the carbon, but a massive overload of electron density due to the $\sigma$-type HOMO
$\ce{CO2}$ also quite polar bonds, but no (permanent) dipole - still quite a potent electrophile given the right nucleophile
$\ce{SO3}$ has no permanent dipole (but a transitional one) although quite polar bonds. However, it is only a strong electrophile.
$\ce{NH3}$ quite polar, hence a reasonably good nucleophile. But it is also almost always way too strong a base to act as an electrophile
$\ce{NH(Et)2}$ not so different from ammonia, quite polar, but yet a terrible nucleophile, because it is too big
$\ce{BH3}$ not very polar (no permanent dipole) but quite a good electrophile
$\ce{I-}$ a monopol, very polarisable, good nucleophile, good leaving group.
...

The consideration of the strength of nucleophiles and electrophiles lies in many variables. It might be polarity, local electron density, molecular orbitals, polarisabilities. It is often highly dependent on the solvent and the other reactants. If you add $\ce{I-}$ to a acyl chloride you will certainly have a substitution to some extend:
$$\ce{R-COCl + I- <=> R-COI + Cl-}$$
If you add hydroxide ions, then you will certainly end up with the acid:
$$\ce{R-COCl + I- + {}^{-}OH <=> R-COOH + I- + Cl-}$$
Look at it nucleophilicity or electrophilicity from a case to case basis.