text:
What is the simplest approach to balance a complex reaction equation?
Consider the redox reaction:

$$\begin{multline}
\ce{ a K4Fe(CN)6 + b Ce(NO3)4 + c KOH ->\\ d Ce(OH)3 + e Fe(OH)3 + f H2O + g K2CO3 + h KNO3}
\end{multline}$$

The book's answer for the coefficients $a, b, c, d, e, f, g,$ and $h$ is $1, 61, 258, 61, 1, 36, 6$, and $250$. The final answer for the problem is the sum of the coefficients which is $674$.
I have tried balancing it by using the general methods such as oxidation number method, ion-electron method, etc. However, they seem to be giving me incorrect equations. The answer given by the book is perfectly correct. I am sure of it, as I multiplied and checked each atom on both the sides, and they checked out completely. Also, the coefficients don't have a common factor other than 1.
What is the simplest way to solve this problem?

I answered a related question there.

$\ce{ a K4Fe(CN)6 + b Ce(NO3)4 + c KOH ->\\ d Ce(OH)3 + e Fe(OH)3 + f H2O + g K2CO3 + h KNO3}$

Because there is one $\ce{Fe}$ in $a$ and one in $e$, we know that $a=e$.
To better see things, I rewrite the formulae.

For $\ce{Fe}$: $a=e$
For $\ce{Ce}$: $b=d$
For $\ce{C}$: $6a=g$
For $\ce{N}$: $6a+4b=h$
For $\ce{K}$: $4a+c=2g+h$
For $\ce{H}$: $c=3d+3e+2f$
For $\ce{O}$: $12b+c=3d+3e+f+3g+3h$

Now I replace $e, d, g, h$ with their counterpart, leaving just $a, b, c, f$. I get

For $\ce{K}$: $4a+c=2(6a)+(6a+4b) \Rightarrow c=14a+4b $
For $\ce{H}$: $c=3b+3a+2f$
For $\ce{O}$: $12b+c=3b+3a+f+3(6a)+3(6a+4b) \Rightarrow c=39a+3b+f$

Replacing $c$.

For $\ce{H}$: $14a+4b=3b+3a+2f \Rightarrow 11a+b=2f$
For $\ce{O}$: $14a+4b=39a+3b+f \Rightarrow -25a+b=f$

Combining at $f$.

$11a+b=-50a+2b \Rightarrow 61a = b$

Now I assume $a=1$ and get

$a=1$
$b=61a=61$
$c=14a+4b=258$
$d=b=61$
$e=a=1$
$f=-25a+b=36$
$g=6a=6$
$h=6a+4b=250$

And, finally

$a+b+c+d+e+f+g+h=674$

The last step could also be achieved by taking $a+b+c+d+e+f+g+h$ and replacing $c, d, e, f, g, h$, then replacing $b$, which should eventually lead to $674a$.

Yea, so using the coefficients, this can be done in this way. However, can we do it by using the oxidation no. method, the ion-electron method etc. If we can, then pls show how, as I tried to balance it using those methods, but got wrong answers.

I would also suggest using the coefficients, but if you can use a graphing calculator or a computer spreadsheet, you can automate the process very simply. We can construct a composition matrix for your problem that looks like this:
$$
        \begin{pmatrix}
        4 & 0 & 1 & 0 & 0 & 0 & 2 & 1 \\
        1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
        6 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\
        6 & 4 & 0 & 0 & 0 & 0 & 2 & 1 \\
        0 & 1 & 0 & 1 & 0 & 0 & 2 & 1 \\
        0 & 12 & 1 & 3 & 3 & 1 & 3 & 3 \\
        0 & 0 & 1 & 3 & 3 & 2 & 0 & 0 \\
        0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
        \end{pmatrix}
$$
The columns of this matrix are your compounds and the rows are the number of each element in them. The  rows, from top to bottom, are $\ce{K,Fe,C,N,Ce,O,H}$). I also added one row with just a one in the last column. This is done because a square matrix is needed for the next step
Now, all you have to do is obtain the inverse of your matrix, which can be done quickly on a graphing calculator(or by hand if you are looking to kill time). We ignore most of this inverted matrix and focus in on the last column. This will give you:
$$
        \begin{pmatrix}
 -.004 \\
 -.244 \\
 -1.032 \\
 .244 \\
 .004 \\
 .144 \\
 .024 \\
 1 \\
        \end{pmatrix}
$$
When scaled to whole numbers, this column gives the coefficients for the equation from top to bottom in the same order as the original columns. The sign of the coefficient tells you whether it is a reactant($-$) or product($+$). For more information on how this method can be applied in general, take a look at this article from the Chemical Educator. This method can also tell you if an equation can not be balanced or if it has an infinite number of independent solutions.

+1 Wow! Using matrices to balance chemical equations!! This is something I had never heard of before. Thanks!

A nice method for it was only something I found relatively recently. Essentially, it is the same approach as the other two answers, but puts it in a context where the calculations, even for large/complex systems, can be done much more easily.

Solve for the coefficients $a,b,c,d,e,f,g,$ and $h$.
$\ce{aK4Fe(CN)6 + b Ce(NO3)4 + cKOH -> dCe(OH)3 + eFe(OH)3 + fH2O + gK2CO3 + hKNO3}$
(1) The only Fe on the left is in $\ce{K4Fe(CN)6}$ and on the right in $\ce{Fe(OH)3}$.
Therefore $e$ = $a$
(2) The only Ce on the left is in $\ce{Ce(NO3)4}$ and on the right in $\ce{Fe(OH)3}$.
Therefore $d$ = $b$
(3) The only C on the left is in $\ce{K4Fe(CN)6}$ and on the right in $\ce{K2CO3}$.
Therefore $g$ = $6a$
Rewriting the chemical equation we now have:
$\ce{aK4Fe(CN)6 + b Ce(NO3)4 + cKOH -> bCe(OH)3 + aFe(OH)3 + fH2O + 6aK2CO3 + hKNO3}$
and we're down to 5 variables starting from 8.
I can see that $\ce{2KOH + H2CO3 -> H2O + K2CO3}$. Also the $\ce{OH-}$ in both $\ce{Ce(OH)3}$ and $\ce{Fe(OH)3}$ must have come from $\ce{KOH}$. But none of this really breaks the problem open. So let's slog it out from here...
Looking at K we have $4a + c = 12a + h$ or $c = 8a + h\tag{1}$
Looking at N we have
$6a + 4b = h\tag{2}$
Looking at H we have
$c = 3b + 3a + 2f\tag{3}$
Looking at O we have $12b + c = 3b + 3a + f + 18a +3h$ or
$9b + c = 21a + f +3h\tag{4}$
The fifth restriction is that the the equations above are Diophantine equations and that the GCD of $a,b,c,f$ and $h$ is 1.
Using equation (2) we can eliminate $h$ in equations (1) and (4)
From equation (1) we get $c = 8a + (6a + 4b)$ or $c = 14a + 4b \tag{5}$
From equation (4) $9b + c = 21a + f +3(6a + 4b)$ we get
$c = 39a + 3b + f \tag{6}$
So our set of equations is now:
$c = 3a + 3b + 2f\tag{3}$
$c = 14a + 4b \tag{5}$
$c = 39a + 3b + f \tag{6}$
and the restriction that the GCD of $a,b,c$ and $f$ is 1.
We can now use equation (5) to eliminate $c$.
From equation (3) we get $(14a + 4b) = 3a + 3b + 2f$ or  $11a + b = 2f\tag{7}$
From equation (6) we get $(14a + 4b) = 39a + 3b + f$ or $b = 25a + f\tag{8}$
We can now use equation (8) to eliminate $b$.
From equation 7 we have $11a + (25a + f) = 2f$ or $36a = f\tag{9}$
So we only have to worry about $a$ and $f$ having a GCD of 1 and $a=1$ works, so $f=36$.
I'll leave it up to you st solve for the rest of the coefficients.
So this isn't hard, it is just a lot of bookkeeping.

The approach I would take is to regard all of the $\ce{Fe(CN)_6^{4-}}$ ion as the reducing agent.  We then have the oxidation reaction
$\ce{Fe(CN)_6^{4-} \rightarrow Fe^{3+} + 6C^{4+} + 6N^{5+} + 61e^-}$
The charges on the products represent the oxidation states of the elements in the actual products, such as $+4$ for carbon in the carbonate.  The oxidation states on the oxidized atoms add up to $+57$ so to balance that with the initial four negative charges the $\ce{Fe(CN)_6^{4-}}$ must give off $61$ electrons.  We then need $61$ ceric ions to take them up, therefore $a=1, b=61$ possibly multiplied by a common fraction to eliminate fractions that appear later on, if needed.
With $a=1, b=61$ inserted you can then balance elements sequentially -- cerium, then iron, then carbon, nitrogen, potassium, finally hydrogen or oxygen.  You do not need an overall multiplier after all, as there are no fractions.  Everything falls into place without dealing with any coupled linear equations!


text:
Why does the AraC arabinose sensing system only respond to the presence of L-arabinose?
I understand that when arabinose is present it interacts with araC and changes it shape, promoting the binding of RNA polymerase with the promoter and GFP is produced. However, I don't understand why the sensing system in E.coli only responds to L-arabinose and no other pentose sugar nor their D-isomer.
I would truly appreciate it if someone could explain why this is.

Hello and welcome to Chemistry.SE! I suggest you take the short [tour] to better familiarize yourself with the site. This also seems kind of like a homework-type of question, so you might want to read through this discussion.

This could probably be a good question for http://biology.stackexchange.com/, but I am not sure.

Your question is really about the origin of specificity in molecular recognition.
The simple answer to this is that two molecules (like the AraC protein and arabinose) bind to each other because they make favorable interactions with each other. ("Favorable" here means that free energy is released in the "reaction" of going from separate molecules to the complex.) There's a number of different types of interactions, but the ones you typically talk about with non-covalent protein binding are van der Waals,  hydrogen bonding, and electrostatics.
The reason AraC binds to arabinose is that it has functional groups in the binding pockets which exactly match three dimensional organization of corresponding atoms in the arabinose molecule (see the experimental structure of the AraC/arabinose complex). You can see a schematic representation of these interactions here.
The thing about these interactions is that they only happen if the three dimensional geometry matches up -- we might write molecules in a 2D schematic form, but molecules are intrinsically 3D. So when you have the hydrogen bond between, for example, Tyr82 and the C4 hydroxyl of arabinose, it can only form that favorable interaction if the C4 hydroxyl ends up in the correct 3D position.
If instead of L-arabinose you have D-arabinose (or L-ribose), then you have a different spatial orientation of atoms. Your C4 hydroxyl no longer goes "down", but instead goes "up". Since Tyr82 is still in the same place, it can no longer form a favorable hydrogen bond to the C4 hydroxyl - the distance is too far. There's no way to put the 3D structure of D-arabinose into the binding pocket of AraC and keep all the interactions. If you lose enough of these interactions (and sometimes one is enough), the total energy of binding is no longer favorable, and there's no driving force for binding.
Keep in mind that it's not always an all-or-nothing thing. It all depends on which interactions are destroyed by changing the chemical structure of the ligand. For example, AraC has no difficulty binding D-fucose. The difference between L-arabinose and D-fucose is just an addition of a C6 methyl group. For AraC this addition doesn't disturb the existing interactions enough to significantly perturb them. Therefore, you still get a favorable energy of interaction when the AraC binds to fucose.

source: Stack Exchange

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

sort: votes


text:
Why do strong acid-weak base reactions "essentially go to completion"?
This is what my textbook says. I understand that the strong acid would readily donate a hydrogen ion, but I can't see how all the weak base will readily accept it (at least not to "essential completion").

This is a matter of equilibrium. Let's consider the  acid-base reaction of $\ce{HCl}$ and $\ce{NH3}$:
$$\ce{NH3 + HCl <=> NH4+ + Cl-}$$
Which can be broken apart into:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
$$\ce{HCl + H2O <=> H3O+ + Cl-}$$
For these reactions at equilibrium:
$$K_\mathrm{eq}=\ce{\frac{[NH4+][Cl-]}{[NH3][HCl]}} \hspace{1cm} K_\mathrm{b}(\ce{NH3})=\ce{\frac{[NH4+][OH-]}{[NH3]}} \hspace{1cm} K_\mathrm{a}(\ce{HCl})=\ce{\frac{[H3O+][Cl-]}{[HCl]}} $$
$$K_\mathrm{a}(\ce{HCl}) \cdot K_\mathrm{b}(\ce{NH3}) = \ce{\frac{[H3O+][Cl-]}{[HCl]}} \cdot \ce{\frac{[NH4+][OH-]}{[NH3]}}=\ce{\frac{[NH4+][Cl-]}{[NH3][HCl]}} \cdot \ce{[H3O+][OH-]} = K_\mathrm{w}K_\mathrm{eq}$$
$$K_\mathrm{w}K_\mathrm{eq}=K_\mathrm{a}(\ce{HCl}) \cdot K_\mathrm{b}(\ce{NH3}) \Rightarrow$$
$$K_\mathrm{eq}=K_\mathrm{a}(\ce{HCl}) \cdot \frac{K_\mathrm{b}(\ce{NH3})}{K_\mathrm{w}}=\frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{NH4+})} \Rightarrow$$
$$\mathrm{p}K_\mathrm{eq} = -\log \left(\frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{NH4+})} \right) =  \mathrm{p}K_\mathrm{a}(\ce{HCl}) - \mathrm{p}K_\mathrm{a}(\ce{NH4+})=-8-9.24=-17.24$$
Just from the $\mathrm{p}K_\mathrm{a}$ difference of the acid and conjugate acid of our reaction, we figured out that at equilibrium the products are present in a $10^{17.24}:1$ ratio to the reactants, meaning the reaction has essentially gone to completion.

With a strong acid you should "level" the $K_a$ to $1$, that is the hydrochloric acid is turned into $\ce{H3O^+}$ before reacting with the base. You still get a favorable equilibrium constant to form ammonium chloride but not as large.

It is the strong acid that is pushing the weak base to full protonation.
The more formal explanation requires you to be familiar with the ionisation of water $(1)$ and a generic Brønsted acid-base reaction including the acidity constant $(2\text{ – acid})$ and $(3\text{ – base})$.
$$\begin{align}\ce{2 H2O &<<=> H3O+ + OH-}&K_\mathrm{w} &= [\ce{H3O+}][\ce{OH-}] = 10^{-14}\tag{1}\\[0.6em]
\ce{HA + H2O &<=> H3O+ + A-}&K_\mathrm{a} &= \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}\tag{2}\\[0.6em]
\ce{H2O + B &<=> HB+ + OH-}&K_\mathrm{b} &= \frac{[\ce{HB+}][\ce{OH-}]}{[\ce{B}]}\tag{3}\end{align}$$
The first equation is basically the water-version of the second but without the activity of water in the denominator. Because water is a very weak acid, the equilibrium of $(1)$ lies strongly on the reactant side. If we consider $\ce{B} = \ce{A-}$, i.e. $\ce{B}$ being the conjugate base of $\ce{HA}$ and then add together $(2)$ and $(3)$, we arrive at $(4)$:
$$\require{cancel}\ce{\cancel{\ce{HA}} + 2 H2O + \cancel{\ce{A-}} <=> H3O+ + OH- + \cancel{\ce{HA}} + \cancel{\ce{A-}}} \quad K' = K_\mathrm{a}\times K_\mathrm{b} = K_\mathrm{w}\tag{4}$$
Using the more commonly applied $\mathrm{p}K_\mathrm{a/b}$ values, we see that:
$$\mathrm{p}K_\mathrm{a}(\ce{HA}) + \mathrm{p}K_\mathrm{b}(\ce{A-}) = 14\tag{5}$$
Knowing that strong acids are defined as $\mathrm{p}K_\mathrm{a} < 0$ and weak bases as $\mathrm{p}K_\mathrm{b} >0$, we basically have access to all our parameters; most importantly, the conjugate acid of the weak base must have an acidity constant fulfilling $\mathrm{p}K_\mathrm{a} < 14$ — but at the same time it would still be larger than zero because otherwise the compound would not be notably basic.
Putting this together, we start off with three compounds:

a strong acid $\ce{HA}$ with a $\mathrm{p}K_\mathrm{a} < 0$.
water, with $\mathrm{p}K_\mathrm{b} = 14$, i.e. $\mathrm{p}K_\mathrm{a}(\ce{H3O+}) = 0$
the weak base $\ce{B}$ for which we can say $0 < \mathrm{p}K_\mathrm{b} < 14$ and therefore also $0 < \mathrm{p}K_\mathrm{a} (\ce{HB+}) < 14$

It is clear that the strongest acid — $\ce{HA}$ — will deprotonate practically completely. The question is who gets those protons? We have the choice of giving them to water or the weak base. Whether you now wish to argue with relative basicities (water is less basic than the weak base) or relative acidities of the conjugate acids (hydronium (or oxonium, $\ce{H3O+}$) is more acidic than $\ce{HB+}$) does not matter, in both cases the conclusion is that the weak base acquires all the protons available for thermodynamic reasons.

This did it for me after grappling with it for so long. So from what I understand, the strong acid will donate its protons to the weak base because the weak base is still stronger than water?

@khajiit That’s the gist, yes. Because water is such a terrible base.

The definition of a weak base as given by Wikipedia is:

In chemistry, a weak base is a chemical base that does not ionize
  fully in an aqueous solution. As Brønsted–Lowry bases are proton
  acceptors, a weak base may also be defined as a chemical base in which
  protonation is incomplete.

Where Brønsted–Lowry is they type of acid-base classification used for your question.
A simple qualitative way to envision what is happening when you add a strong acid to a weak base is that you are "flooding" the unprotonated fraction of the compound with protons, greatly increasing the probability that each of them will become protonated.  This is equivalent to saying the reaction of the weak base to it's protonated form essentially goes to completion.

Ugh... Let's not make this hard.
I start off with $n_1$ moles of the free weak base before the reaction and I'll end up with $n_2$ moles of the unprotonated base and $n_3$ moles of the protonated base after the reaction.
So "essential complete” is a significant figures argument. It means that:
$n_3 \approx n1$ and $n_2 \ce{<<} n_1$

Here are four answers that skip the math:

Strong acids are strong enough to protonate water (the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid $\ce{H3O+}$ is less than zero). The conjugate acid of a weak base is a weak acid, with a $\mathrm{p}K_\mathrm{a}$ bigger than zero. So if given a choice, the proton will react with the weak base rather than with water.

Imagine just adding the strong acid first. This will be very acidic solution. In very acidic solutions, weak bases get protonated. Granted, if the weak base is in excess, all the strong acid will be gone, so the pH will not be very acidic anymore. However, this shows that when the weak base is limiting, it will become fully protonated. When the strong acid is limiting, it will become fully deprotonated. So in either case, the reaction goes almost to completion.

There is a big difference between the $\mathrm{p}K_\mathrm{a}$ of strong acid and the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of the weak base. The species with the lower $\mathrm{p}K_\mathrm{a}$ become (or remain) deprotonated, and the species with the higher $\mathrm{p}K_\mathrm{a}$ ends up (or remains) protonated. The larger the difference of the $\mathrm{p}K_\mathrm{a}$ values, the more complete the transfer (no matter whether in aqueous solution or otherwise).

What are the products of a strong acid reacting with a weak base? We get a counter ion (from the strong acid) and a weak acid (from the weak base). If in doubt whether this reaction goes virtually to completion, consider the reverse reaction. Would a weak acid react with a spectator ion (or with water)? Yes, a little bit, but mostly it would hold on to its proton.

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

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


text:
Finding the mystery ester strucuture using NMR
I have an unknown ester with the chemical formula $\ce{C9H10O2}$ that is used as a flavoring agent in candies. It exhibits the following H-NMR and C-NMR (Attached file: unknown ester NMR)
Based on the NMR peaks I believe the unknown is benzyl acetate, as the C-NMR and the first two peaks of H-NMR are identical to the unknown NMR (Attached file: benzyl acetate NMR).
Knowing that aromatic protons are not equivalent, as shown in the unknown's H-NMR at 7.4 and 7.5 ppm, I do not know why the aromatic protons for benzyl acetate are integrated and shown as a singlet.
Is this just due to the sensitivity of the prediction algorithm (I am using Chembiodraw)? Or benzyl acetate is not the unknown?
Your help is greatly appreciated!

Benzyl acetate seems very plausible to me, although if this is supposed to be your homework, using ChemDraw to predict NMR is probably not encouraged. Better to use chemical reasoning to rationalise why benzyl acetate fits the spectrum you're given. As for ChemDraw's prediction, did you take a look at whatever was below the spectrum? As far as I know, the software should tell you how it calculates the chemical shifts. https://i.stack.imgur.com/WBVRA.png

Thank you for the quick response! Yes, I have looked at Chemdraw's numerical output, which was the same as the picture you've linked. And that is source of my confusion as in why the program is considering the aromatic protons to be equivalent :s

Prediction software always has its limitations, and there is always a degree of error in the calculation. For the ChemDraw predictions, you will see that for the 3 aromatic environments, it has done 3 independent calculations, and has happened to arrive at the same chemical shift. This simply means that these shifts are coincident, not equivalent.
Remember, prediction software is like any tool — only as good as the person using it, and should not replace a proper assessment of the data.
Your assessment here should first look at your proton environments; all 10 protons are accounted for. We have a $\ce{CH3}$, a $\ce{CH2}$, and a mono-substituted benzene. The $\ce{CH3}$ and $\ce{CH2}$ are not directly connected to any other group that is causing obvious splitting. Secondly, the $\ce{^13C}$ spectrum shows 9 peaks, consistent with our $\ce{CH3, CH2}$ and a mono-substituted benzene. We also have a peak at ~δ170, which is a $\ce{-C(O)-{}}$
There are only a few ways you could put these groups together. Rather than using ChemDraw to predict the spectrum and see which is identical to your question, you should rationalise why each possibility is or isn't the correct answer.

Trying to rationalise a bunch of peaks around δ7.4–7.5 is the of minimal interest, and remember that a real spectrum will almost always look different to a simulation. The peaks you should focus on willjustify the chemical shifts for the $\ce{-CH3}$ group and the $\ce{-CH2}$ group in the proton spectrum, and the $\ce{-CH3, -CH2}$ and $\ce{-C(O)-{}}$ in the carbon spectrum. There is only one possible correct answer.

LOL - As I have heard it put "A fool with a tool is still a fool."

I would like to argue that $170~\mathrm{ppm}$ clearly shows a carboxy group or amide group and a pure ketone would have something closer to $200~\mathrm{ppm}$. That would cut down the number of possibilities to two from four. Otherwise have my full approval and upvote.

@Jan - I think that is exactly Long's point. The OP shouldn't just rely on pattern matching but use some knowledge of spectral interpretation to solve the problem.

@Jan - precisely. Leaving this deduction to OP. Similar arguments can be made for the methyl group - it is clearly not oxygen-bound. And so on...

I am just learning about NMR specteoscopy. But one thing I don't realize is why the measurementss are in unit ppm. Isn't it should be a unit of magnetic field?

@Mockingbird - The whole point of NMR is that you want to build you a reference catalog of spectra taken under the same experimental conditions. So you could either keep the magnetic field strength constant and vary the frequency, or keep the frequency constant and vary the magnetic field strength. Instrumentally it is easier to keep the magnetic field strength constant. So the shift is in terms of frequency.

@MaxW is ppm a unit of frequency?

@Mockingbird - No parts per million related to the base frequency. Tetramethylsilane is the typical reference standard.

@Mockingbird - if you have a different question you should consider asking a new question, rather than trying to engage in a conversation in the comments. But, yes, ppm is a dimensionless, field independent unit of frequency.

@long I am sure if I ask that you would close it as a 'homework question'.

@Mockingbird not if the question is constructed fairly, demonstrating that you have made your own attempts to understand the concepts


text:
Calculate optimal surroundings for given reaction mechanism to lower activation energy?
Let's take a reaction that's fairly well understood, which in this case means that the reaction mechanism is known, maybe something like this from here:

If we know the reaction mechanism, we also know or can infer with some reasonable degree of accuracy the paths of all the atoms between reaction start and finish.
Is there a way to calculate surroundings, perhaps just electron densities to begin with, in which the activation energy is minimal for this reaction? (Yes, I realize that another way to put this is "can we begin designing a catalyst for this reaction mechanism?")

The only generally applicable way is brute force.  Example: https://peerj.com/articles/111/


text:
Is there a way to find the mixture type with just the molecular formulas and masses of the solute and solvent?
If you had the molecular formula and molar mass of a solvent and a solute and no other specific information about the two, could one deduce the type of mixture (suspension, colloid, solution) they would create?

Only in rare circumstances. For instance I know $\ce{NaCl}$ in $\ce{H2O}$. But what is $\ce{C8H18}$?

Formula alone does not give you the structure. But if you have the structure of the molecules, there are molecular/thermodynamic models which take into account group contributions for every section of a molecule. These can help calculate and predict phase behaviour and mixture type. See PC-SAFT as an example.

Thanks for the response! Could molar volume suggest behavior? For example, if the molar volume of the solute is 231.9±5.0 cm3 and the molar volume of the solvent is 59.1±3.0 cm3, is it likely the solute would settle rather than form a colloid? I had read that the size of the particles (regarding generally very tiny particles) is a greater determinant of mixability than mass, due to the interactions between the molecules being  much more impactful than the gravitational force. Thanks so much again for your insight!

Size plays a role but it is not the only indicator. Functional groups (hydrophobic vs hydrophilic interactions etc) and geometry are also crucial. Take hexane and decanol for example. hexane is smaller than decanol in size, while decanol is more than 300 times more soluble in water than hexane. For very large molecules such as proteins, the folding and total geometry also play a role in forming colloid vs settling. See globular proteins and their solubility properties.

Aah I see, thanks so much again for shedding more light on the topic, it is certainly far more complex than one might assume!


text:
Finding the concentration of ferrous thiocyanate in calibration solutions
Concentration of $\ce{Fe(SCN)^{2+}}$ in calibration solutions
Recall that the concentration of $\ce{Fe(SCN)^{2+}}$ in the calibration solutions is essentially the concentration of the $\ce{SCN-}$ initially in the solution. So, determine the concentration of $\ce{SCN-}$ initially in each of the $\pu{10.0 mL}$ calibration solutions. This is a simple dilution!

The lab checker is saying that this is wrong. Am I even going about doing this problem correctly?

So, to find the concentration of $Fe(SCN)^{2+}$, the problem says "the calibration solutions is essentially the concentration of the $[SCN^−]$ in the solution." So, you simply take the volume of $K(SCN)$ solution divided by the total volume, then multiply by $[K(SCN)]$:
$[Fe(SCN)^{2+}]=[SCN^−]=[K(SCN)]=0.002M*\frac{0.5mL}{10.0mL}=0.0001M$ (repeat for 2, 3, 4)
The reason we do not involve $K_c$ is because the large amount of $[Fe^{3+}]$ drives the reaction to the right. The absorbance measurements are going to be useful for determine the amount of $[Fe(SCN)^{2+}]$ in a system where $[Fe^{3+}]$ is not high. In these cases, the concentrations $[Fe^{3+}]$ and $[SCN^-]$ are known, and you use the absorbances you collected to draw a line relating concentration $[Fe(SCN)^{2+}]$ to absorbance. When you know the concentration of your three parts of $K_c$, you can simply use arithmetic to find $K_c$.


text:
Why don't decomposition reactions form pure elements as products?
For example, if I had a reaction like this:
$$
\ce{NaHCO3 -> Na2CO3 + H2O + CO2}
$$

Why does it not break down all the way down to its elements?
What makes it form such "intermediate" products, stopping at water, carbon dioxide, etc.?

If you heat sodium bicarbonate strongly you can convert it to $\ce{Na2O}$. In general the reaction favors the most thermodynamically stable chemical.

Using your example, you simply have not given the system enough energy to do so. Water and $\ce{CO2}$ are very stable compounds under ordinary conditions of temperature and pressure, and thus are the preferred products under standard conditions for that reaction.
If you were to heat the system further, further decomposition would occur. For example, at higher temperatures you will convert $\ce{Na2CO3}$ into $\ce{NaO}$ and more $\ce{CO2}$.  Under extreme conditions, i.e. thousands of degrees K, you will ultimately have enough energy to break all of the chemical bonds and then you will be left with the elements.

Quibble - I'd explicitly note that "extreme conditions" mean that the whole sample will be atomized to the gaseous elements.

Just one disagreement with this answer. Not elements, rather atom of those elements.

Or a plasma I guess.

@CupC_56, my plasma comment was a response to Max. Another good point though. And for that matter we could go nuclear with our energy source to get to subatomic particles, but I guess at that point we have to move the discussion to Physics.SE ;)


text:
Osmotic pressure of a solution of a protein?

A small protein molecule, code-named "sloth", has a MM of $\mathrm{1.50 x 10^4}$ g/mol. What is the osmotic pressure exerted at $\mathrm{24.0^oC}$ by 25.0 mL of an aqueous solution that contains $\mathrm{3.75 x 10^{10}}$ nanograms of "sloth"? R = 0.08206 (atm L)/(mol K).

The equation my professor gave me is: $\mathrm{O.P. = T*R*M}$. So I did:
$\mathrm{O.P. = (24.0 + 273) K * 0.08206 (atm*L)/(mol*K) * ([(3.75 x 10^{10})x 10^{-9} g] / 1.50 x 10^4 g/mol) / 0.025 L}$
I keep on trying and I get 2.44 atm, but my professor marked the correct answer as $\mathrm{2.44 x 10^{-3}}$ atm. I think he may have divided the moles of sloth by 25 instead of .025 L, which would make his answer $10^3$ smaller than mine. Did I do something wrong?

Hello and welcome to Chemistry.SE! If you have any questions about how this site works, a good starting point is taking the short [tour]. If you have any questions about homework-type of questions in particular, you should read through this discussion. Good luck!

Ok let's start with $\text{OP} = \text{T}\times\text{R}\times\text{M}$
$\text{T} = (24.0 + 273) \text{K} = 297 \text{K}$
$\text{R} = 0.08206 \dfrac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}}$
$\text{M} = \dfrac{\text{moles}}{\text{L}} = \dfrac{\frac{37.5 \text{ g}}{1.50 \times 10^4 \text{ g/mol}}}{0.025 \text {L}} = \dfrac{2.50\times10^{-3}\text{ mol}}{0.025\text{ L}} = 0.100 \dfrac{\text{mol}}{\text{L}} $
$\text{OP} = 297\text{ K}\times0.08206 \dfrac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}}\times0.100 \dfrac{\text{mol}}{\text{L}} = 2.44 \text{ atm} $
So I get your answer too...


text:
Find Equilibrium Constant with only balanced equation?

Calculate the value of $K_p$ for the following equation:
$$\ce{PCl3 (g) + Cl2 (g) <=> PCl5 (g)} ,\qquad \mathrm{R = 0.0821 \frac{L\times atm}{mol\times K}}$$

This question appeared on my test. I know how to find $K_p$ if I was given $K_c$ $(K_{\mathrm{p}} = K_\mathrm{c} \times (\mathrm{R} \times T)^{n})$ or if I was given the partial pressures of all of the parts at equilibrium, but none of those values were given; only the equation. My professor marked the answer as $0.041$ (no units), but I have no idea how to calculate that if I wasn't given any values.

Notice that I replaced \mathrm with \ce for your chemical equation. Just like that, it properly rendered the subscripts and reaction arrows!

Interesting that the answer is 1/2 of R, and that there 1/2 as many moles of gas on the right side of the equation.

Thanks for the help! Interesting observation. I wonder if my professor accidentally took the moles of each side and divided them (products/reactants) instead of using the partial pressures.

Were you given any other information at all for the problem? I think the key is that the total pressure of the products is half of the total pressure of the reactants if the volume is constant.

I have the test in front of me, and there's absolutely nothing else in the question. If my professor was thinking about the total pressure of each side, I don't see how R would be involved in the equation to begin with.

Were you allowed to use tables of free energies of formation or enthalpies and  entropies of formation?

The only attached information that we were given was the periodic table

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


text:
Typography of atomic orbital subscripts p_x
It the notation $\mathrm{p_x}$, $\mathrm{p_y}$ and $\mathrm{d_{x^2{-}y^2}}$, are the subscripts variables (and therefore should be in italics) or are they labels (labels for directions, I suppose, and therefore should be upright)?

The notation is for the subshells of electrons in atoms. The uses of x,y, and z correspond to the x,y and z axis in 3D space, and are therefore labels not variables.  See: https://en.wikipedia.org/wiki/Atomic_orbital#Orbitals_table for the visualized shapes.

@Max you sure they should be upright? I thought they're used slanted-ly everywhere.

I'll defer to the ACS style guide -- italic.

The subscripts that specify orbital axes are written in italic. This is mentioned in ACS style guide.$^{[1]}$ So for instance
$$\mathrm{d}_{x^2 - y^2}.$$
Be careful though. Other subscripts for orbitals are generally upright, e.g.,
$$\mathrm{t_{2g}}.$$
$[1]$ Anne M. Coghill, Lorrin R. Garson. ($2006$). The ACS Style Guide. Effective Communication of Scientific Information. American Chemical Society. DOI: 10.1021/bk-2006-STYG, ISBN: 9780841239999 (print), 9780841228306 (online). (p 256)

I would say this contradicts the basic rule of labels being upright (and I consider an axis name a label), but there we go.

@mhchem It shouldn't contradict. Consider $\mathrm{p}_x$ for example. It specifies in which coordinate axis the $\mathrm{p}$-orbital is "directed". So if it is pointed along the $x$-axis, here $x$ varies and is not a constant.

I know this is nitpicking, but I would say x does not stand for a number, it does stand for a (fixed) direction. Or, putting it differently, x is a shorthand label for $x$-axis. It a label. It's a label of direction, just as "south" is. I don't see any 'placeholder of a value' property here.

@mhchem No, I don't think that is correct. It does not refer to the axis. It refers to a mathematical function which appears in the wavefunction corresponding to the orbital. There is no "x2-y2" axis. It just means that that d orbital has a x2-y2 term in its wavefunction.

The p orbitals just happen to be "(a whole bunch of other stuff) multiplied by one of (x, y, z)". The p_z orbital does indeed point along the z-axis but it is not named so because it points along that axis.

Yes, I believe @orthocresol is correct. The terms are still called orbital axes though AFAIK, but the analogy indeed only works for $(x, y, z)$. (Correct me if I am wrong.)

See also: Why do the d orbitals have these notations?

My interpretation came from IUPAC's Green Book: "The overall rule is that ... labels, are roman. Sometimes there may seem to be doubt as to whether a symbol represents a quantity or has some other meaning (such as a label): a good role is that quantities, or variables, may have a range of numerical values, but labels cannot." However, they continue themselves with electric field strength in x direction as $E_x$. I would have said $x$ here (!) is a label, because it cannot take multiple values, but is just a label/name for a direction.

Another interesting find: ACS Style Guide reads "Use italic type for: ... axes: the y axis" (p. 212). (I understood that we have no axis labels here.)


text:
Why is zinc oxide reduced at a temperature of 1400 degrees Celsius?
I am attaching a graph of the Ellingham diagram on which my question is based.

If we interpret the graph, $\ce{ZnO}$ is reduced by $\ce{C}$ at a temperature of $\pu{800 ^\circ C}$ to $\pu{1000 ^\circ C}$. However, the actual temperature required is about $\pu{1400 ^\circ C}$. Why is this the case?

Kinetics? What is the source of the 1400 C value?

Which Carbon are you talking about, The carbon which is oxidizing into CO or CO2..?

If the answer is $1400\ \mathrm{^\circ C}$ it must imply that the user does not want $\ce{CO}$ in their product. Based on the diagram, the intersection of $\ce{ZnO + C -> Zn + CO}$ occurs around $1000\ \mathrm{^\circ C}$. Full conversion of $\ce{CO}$ to $\ce{CO2}$ requires a temperature of around $1400\ \mathrm{^\circ C}$ as shown in the diagram.
As to why the reaction does not follow the $\ce{C -> CO2}$ curve (no $\ce{CO}$ involved), this is a kinetic reason. Subtract the $\Delta G$ values of each line at a specific temperature (where the lines intersect, the value is 0, meaning the reductions of the two species are in equilibrium). You will find that at the temperature of intersection between $\ce{2ZnO -> 2Zn + O2}$ and $\ce{C + O2 -> CO2}$ $(\Delta G = 0)$, there is sizeable energy driving force for $\ce{ZnO + C -> Zn + CO}$ $(G = -300\ \mathrm{kJ/mol}\ \ce{ZnO})$. The difference of the Gibbs energies is related to the ratios of the $\ce{CO}$ and $\ce{CO2}$ products by
$$\Delta G_\mathrm r(\ce{ZnO + C -> Zn + CO})-\Delta G_\mathrm r(\ce{2ZnO + C -> 2Zn + CO2}) =-RT\ln\frac{[\ce{CO}]}{[\ce{CO2}]}$$
Through plugging in the number it is a very high proportion that goes through the $\ce{CO}$ pathway. Again, the only way any of this is important is if it mentions that $\ce{CO}$ must be eliminated, else the reaction proceeds at low temperature as you say.


text:
Why isn't this resonance possible in an imidazole ring?

My question is why isn't the resonance structure drawn in pencil on the right a valid one? My review book states that after initial protonation on the indicated nitrogen no resonance is possible.

What's the valency of N?

Well N is 5 valence electrons

The top C=C is less stable than N so it can definitly rotate its pi electrons to meet N's valence deficiency .. right?

It has 5e, so it makes 3 bonds, getting 1 election from each to complete octet.

The amine is partially positive while the imine has electron pair. In order to satisfy the partial positive, the the double bonded alkene can donate its pair to the N and have its positive deficiency satisfied by the electron pair of the Nitrogen on the bottom?

How many bonds are on your protonated N in 'your' resonance structure? (it's more than 4)

Ah thank you for explaining it, for others: N though has 5 valence, only can form 3 bonds and not 5 bonds because it only needs 3 more electrons  to reach the 8 electron octect for maximum stability.

You need to understand the basic reason for the aromatic nature of materials. A molecule is said to be aromatic only if:

A ring in it is planar,
Continuous delocalisation is possible throughout the concerned cyclic ring, and
This delocalisation stabilises the ring.

Now, for continuous delocalisation to be possible, the $\mathrm{\pi}$ orbitals of all the elements in the ring need to be parallel to each other ( They need to overlap sideways for formation of  $\mathrm{\pi}$ bonds).
In your example, consider the carbons in the ring. All of their $\mathrm{\pi}$ orbitals are parallel to each other and PERPENDICULAR to the plane of paper. On the other hand, the imine nitrogen is $\mathrm{sp^2}$ hybrid. The lone pair of electrons is in an  $\mathrm{sp^2}$ hybrid orbital, i.e, IN the plane of the paper. And mind you, it's fixed (can't rotate without distorting the geometrical shape of the ring, that is, which would be too much to expect from the poor guy.)
So the imine Nitrogen cannot give the lone pair anywhere. it can only play around with the  $\mathrm{\pi}$ bond already on it.
Moreover, as soon as the amine Nitrogen takes up a proton, it loses its ability to use the lone pair for delocalisation. The condition for planarity itself is not fulfilled ($\ce{(R)2-N+-(H)2}$ has tetrahedral geometry), so there's no use of discussing aromaticity. Before the proton uptake, the ring was aromatic because the amine Nitrogen was using the lone pair for continuous delocalisation, effectively attaining planar geometry. But now... It can no longer be done.
Conclusion: Resonance is still possible(between all atoms of the ring except the amine Nitrogen), but Continuous Delocalisation is not. So, the system loses its aromaticity.

The key to understanding why your proposed resonance structure is not possible is geometry. The five-membered ring is planar, or otherwise there wouldn’t be any aromaticity in the first place. The aromatic system is made up of a set of p orbitals perpendicular to the ring plane — so if you decided that your ring is lying in the $x,y$-plane then the aromatic system consists of $\mathrm{p}_z$ orbitals. For the atoms to have available $\mathrm{p}_z$ orbitals, they need to have $\mathrm{sp^2}$ or lower hybridisation; in this case, all atoms of the aromatic ring are $\mathrm{sp^2}$ hybridised, giving them three $\mathrm{sp^2}$ orbitals lying fully within the $x,y$ plane or ring plane.
One of these orbitals contains the lone pair of the non-protonated nitrogen. Thus, this lone pair is perpendicular to the aromatic system and cannot take part in resonance. The situation is similar to furane, in which oxygen has two lone pairs but only one of the two is in a $\mathrm{p}_z$ orbital and therefore able to participate in resonance.

Now that we have taken care of that, I need to point out a textbook-erratum — at least I would have to point it out from what I see, I don’t know what else the book writes on the page.
They drew side-chain protonated histidine as a nonaromatic system with one nitrogen having two protons. That is not what side-chain protonated histidine looks like. The lone pair of the other nitrogen (the one you wanted to use for resonance) is

pointing away from the ring
not participating in resonance
and therefore much more basic

So before the $\ce{R2N-H}$ nitrogen’s lone pair — which, as you should remember, is delocalised — is ever protonated, protonation will happen at the other nitrogen giving $\ce{{Ring}-\overset{+}{N}H={Ring}}$. This protonated histidine still features an aromatic imidazole ring.

I'm guessing that the question was something like, "Why is this nitrogen much less basic than the other one?" We are seeing the answer key that explains it. I don't think this is textbook erratum.

@jerepierre I think you're right, especially given what it says at the top of the picture. The diagram above will be the one with the correct nitrogen protonated.

@jerepierre I don’t really like guessing. I added it to guard OP from any misconceptions that may occur. If further textbook context is given that confirms your guesses, I’ll happily edit.

Above comment CC @bon

Very simple: The resonance structure in pencil on the right is not possible because you have five bonds to nitrogen, breaking the octet rule. This is the one and only reason why the structure drawn in pencil is not correct. Realizing this, the point that the review book is trying to make (about why the imine nitrogen is protonated) should become even more clear.
The other answers are not actually addressing the question you asked.


text:
Quantify Sweetness?
My textbook gives many examples of sweetening agents. For Alitame it says its 2000 times sweeter than cane sugar. Aspartame is 100 times sweeter than sugar.

I am wondering how they quantify sweetness?

Dilute the compound until you can't tell it from pure water, then check the concentration.

But that measures the intensity not sweetness right? Maybe Aspartame isnt really all that sweet but the intensity is very high?

What's the distinction between the two?


text:
Copper Sulfate Pentahydrate experiment, help please?
My science project specifies me to make seed crystals of copper sulfate. We must use the powder form to make seeds. There are two types of copper sulfate pentahydrate; one is granular form (small seeds), and the other one is powder form. We must produce crystal seeds from copper sulfate (powder form). We are not allowed to use the granular form, because it forms seeds on its own already.
Steps I have taken:

I saturated 50 ml of distilled water (room temperature) with copper sulfate pentahydrate (powder form).
I added about 15 ml of purified sulfuric acid to the mixture, so it prevents the hydrolysis with water.This purified sulfuric acid I got from boiling battery acid. I boiled about 50 ml of batter acid (sulfuric acid mixed with ionic water). I waited till it reached about 15 ml (water evaporation). The fumes were getting strong at that stage. I'm not sure if purified enough.
I then again mixed more copper sulfate to the mixture, and stirred it very well.
I put the china dish (mixture) on a hot plate, and waited for it to reach crystallization point, while stirring it very well.
5 I know the water has to evaporate for the mixture to produce seed crystals. I waited for the mixture to evaporate about to 20 ml.
I had no results.

I'm not sure what I'm doing wrong. I'm doing my project according to this video which I found on youtube. It is a very short video, and my main problem is the sulfuric acid part.
https://youtu.be/SAU-gptAFe0
Help.

See, "no results" leaves an awfully broad field to our imaginations. What exactly happened? Copper sulfate never crystallized? Or it crystallized back into powder form? Or it turned into a brown goo? Or it magically vanished from the dish? Or the dish itself got stolen?

It turned into white powder. As the water evaporated in low fire, it accumulated a white residue at the bottom. Yes, the copper sulfate never crystallized.

Aha! That's much better. So it did crystallize after all, only in a wrong form (not the pentahydrate we're after). Now try adding less acid and applying way less heat. The solution must not be nearly boiling; just moderately warm. Yes, it is going to take quite a while (a few hours, maybe). No, you don't have to stir it all the time.

Alright. I will try it soon (tomorrow). Please, stay put. Thank you for your help.

Yeah, I believe I have to lower the heat much more, and mix maybe 5 ml of sulfuric acid (from battery acid). Sources say battery acid is about 35% sulfuric acid and the rest is just ionized water. So, I think I'm doing the purification correctly. I stop the boiling of sulfuric acid, when I can smell the weird acidic fume. Im boiling 50 ml of battery acid, and turning the boiling off at 15 ml about. I think I'm on the right track, just measurement as you say. I will do it tomorrow.

Experiment was a success. Thank you very much for your input. I just have one more question: why is sulfuric acid added to the solution? what is its role exactly?

The acid serves to prevent hydrolysis. Actually, you need very little of it. You'd probably be fine without any acid at all.

So, if I was to carry out the same experiment all over again, same details (powder form to crystals form) without adding the sulfuric acid, would it still nucleate seed crystals (monocrystals)? The same crystals which I have nucleated in my experiment?

It definitely would. But there might or might not be some visible signs of hydrolysis. There is only one way to find out. Chemistry is an experimental science, they say.

Frankly you're going about this all wrong. First you want to exploit the difference in solubility of the salt in hot water versus cold water.
Copper sulfate's solubility is 32 g/100ml water at 20 C, and 114 g/100ml water at 100 C.
2-3 drops of sulfuric acid is all that is needed. You're just trying to prevent the formation of $\ce{Cu(OH)2}$ or $\ce{CuCO3}$.

I'm not sure about battery acid. I assume that you're using new battery acid, not actually from an old battery. So the sulfuric acid will be a little less concentrated. Use 5-6 drops instead of 2 or 3.

After you make a saturated solution of the copper sulfate (aim for 100g /100ml water) in hot water, let it cool slowly and don't stir at all. Stirring will promote nucleation of many small crystals.
To get a really large crystal you'd make a saturated solution at room temperature, filter it, and add one of these seed crystals. You'd then let the water evaporate over weeks very slowly.

Yes, I'm using a new battery acid (Autozone). It is for motorcycles. I'm assuming that all batter acid has the same make up; 35% (about) sulfuric acid, and 65% ionized water. I do not think batter acid has any more additives added to this mixture. I have researched quiet a lot about battery acid, I'm pretty sure it is only sulfuric acid mixed with ionized water. Do you think it could have any additives? I hope not. And if so, what could I use to purify the sulfuric acid completely? The 15 ml of the sulfuric acid, which I purify, changes color to transparent goldish a bit.

@ArchAlegre - You don't need to "purify" it. You're adding it to water, Just use a little bit more. Acid that concentrated is nasty stuff. Be sure to wear eye protection when using it.  // Acid from a used battery would have lead in it...

Yes, it is new battery acid. I turn off the heat plate at about 15 ml from 50 ml. The fumes are acidic strong, so, I imagine that it must be a signal that the sulfuric acid is concentrated. I will do as you say: I will add 2 - 3 drops of it when I redo the experiment.

Experiment was a success. Thank you very much for your input. I just have one more question: why is sulfuric acid added to the solution? what is its role exactly?

@ArchAlegre - Noted why in 3rd paragraph of my answer...

I retried the experiment, same details, but this time without adding sulfuric acid. The experiment reached crystallization point. It was crystallizing similar to sugar granules on the top, forming a thin layer of 'copper sulfate milk skin'. I think I must allowed too much water to evaporate. It never crystallized monocrystals seeds, but just tiny 'sugar' granules. The fire was kept very low, and it was kept as such for about 3 hours. Can this experiment be successful even if sulfuric acid is not added to the solution? Is sulfuric acid a must for this process?

Depending on you water source. Even distilled water which has been sitting around with have carbonate. Copper carbonate is very insoluble. So the solution gets a lot of nucleation points.

I used only distilled water. I will retry it again, but this time I will turn the heat plate off earlier.


text:
What is the oxidation state of sulphur in sulphur dioxide?
So for $\ce{SO2}$, here's the solution when finding for the oxidation number of sulfur.
$$x + 2(-2) = 0 \\
x - 4 = 0 \\
x = +4$$
$x$ here would be the oxidation number of sulfur, so the compound would now be
$\overset{+4}{\ce S} \overset{-2}{\ce{O2}}$.
Now if I tried using oxygen as the $x$ it would become:
$$1(-2) + 2x = 0 \\
-2 + 2x = 0 \\
2x = 2 \\
\frac{2x}{2} = \frac 22 \\
x=1$$
So, in this case the compound would actually be $\overset{-2}{\ce S} \overset{+1}{\ce{O2}}$.
I'm pretty confused about this since we just got into this today and haven't checked many examples. There is probably something I'm missing in the solution or a rule. Can someone explain to me the correct one? And also can't sulfur's and oxygen's oxidation number both be -2 as they're in group VI(A)?

Your second example seems to have a sign flipped somewhere. It assumes Sulphur exist as S(-2) which would lead to O(+1)*2, but you then describe SO2 as S(+2)O(-1)*2.

@MSalters Agreed; edited to reflect.

Your teacher used an interesting example that doesn't obey the rules cleanly. Sulfur can have many oxidation states in this method. In general, larger elements can have oxidation states different than suggested by the group they are in.
If you have learned the electronegativity rules yet, here is a good place to use. If you haven't, higher electronegativity basically means that the element has stronger driving force to reaching an octet; you can look up an electronegativity table if you like. The more electronegative element takes the electrons first. So, in your example, $\ce{SO2}$ has $\ce{2O^2-}$, because of the group oxygen is in. Then, because the $\ce{SO2}$ is neutral, there must be a $\ce{S^4+}$.
But, in a compound such as $\ce{Na2S}$, you can see that S is closer to reaching an octet than Na. So, this becomes $\ce{S^2-}$ while each sodium become $\ce{Na+}$.
Try finding the oxidation states (numbers) for $\ce{PF5}, \ce{WO3}, \ce{SOCl2}$ for understanding. Solutions in the spoiler below.

 $\ce{PF5~is~P^5+,5F-}$
 $\ce{WO3~is~W^6+,3O^2-}$
 $\ce{SOCl2~is~ S^4+, O^2-, 2Cl^-}$

This is a straight forward application of rules that should be in your text. These rules are based, in general, on preference given to more electronegative elements. From Zumdahl, and in order of application:

raw element = 0
monatomic ion = same as charge
fluorine = -1
oxygen = -2 except peroxides and when combined with fluorine.

So, oxygen is a bully until fluorine shows up.


text:
Recovering iodine from activated carbon
I have a sample of an iodine solution that has been dyed brown to mask the amount of iodine within.
I have added activated carbon to 100 mL of the sample and mixed for 3 hours to remove the iodine from the solution but now I'm having trouble recovering the iodine back into solution.

What is the solution?

An iodine teat dip, so i'm not entirely sure if its going to be PVP iodine or not

I was going to try addition of an organic solvent and giving it some gentle heat for 30 or so minutes

That would be my best guess too. I don't know what kind of equipment you have, but you could reflux or even distill it off if you were set up to do so.

I'll give it a go this afternoon and let you know how it goes :)

Don't think I heated the ethanol mix enough and it didn't work at all :( Try again tomorrow

Do you want $\ce{I2}$ back or just the iodine (whatever the chemical species...) back into solution?

I'm looking to do an iodine titration to test the amount of free iodine in the solution.

Based on Recovery of Iodine with Activated Carbon from Dilute Aqueous Solutions,Petrochemical Center of Excellence, Department of Chemical Engineering, Amirkabir University of Technology, No.424, Hafez Ave., Tehran, Iran , you should get good recovery of your iodine by heating in a solution of NaOH. According to the paper:

The optimal desorption conditions obtained by the software are as
  follows: temperature of NaOH solution: $\mathrm{50^oC}$, concentration of NaOH
  solution: 0.5 % and agitation speed: 300rpm. Desorption experiments
  under these optimal conditions showed 98.9% desorption. The Taguchi
  design software results also showed that in the desorption section,
  the contributions of each factor on desorption percentage are as
  follows: concentration of NaOH solution (55%), temperature of NaOH
  solution (39.4%) and agitation speed (4.6%).
  [...]
  the necessary time for equilibrium desorption is obtained as 6h.

The paper does not discuss the algorithms used to calculate the relative importance of the different parameters, but obviously temperature and NaOH concentration are the important factors to getting such good recovery.

I saw this paper and did try the NaOH method but I don't think I reacted for long enough. I'll give it another go and leave it spinning for 4-5 hours. Thank you :)

I hope it works, I've got my fingers crossed!


text:
Deriving an alternative expression for the Van der Waals equation using given parameters

The question is to find out an alternative expression for the Van der Waals equation of corresponding states for 1 mole of gas using the parameters given as $$A=\frac{P}{P_c},\quad B=\frac{V}{V_c},\quad C=\frac{T}{T_c} ,$$ where $P_c,T_c,V_c$ represent the critical pressure, temperature and volume respectively.

I am aware of the simple form of the Van der Waals equation which is $$\left(P+\frac{a}{V^2}\right)\left(V-b\right)=RT .$$ The values of $P_c , T_c \text{ and } V_c$ can also be written down in terms of the Van der Waals constants but I have no idea on how to proceed after this. I tried to substitute the known values but it made things complicated. Is there any other approach to this problem?

$\def\p{\partial} \require{enclose}$Start with the van der Waals equation $(1)$.
$$\left(P + a\frac{n^2}{V^2}\right)\left(V - bn\right) = nRT\tag1$$
Note that you have a slightly different form of the equation, namely you have used molar volume $V_m$. Divide both sides of $(1)$ with $n$, use $V_m = V/n$, and $n^2/V^2 = 1/V_m^2$ to reach your variant.
Values for critical parameters in terms of $a, b, n, R$
The critical point $(P_c, T_c, V_c)$ is a critical point and an inflection point of the isotherm by definition.$^{[a]}$ Therefore we arrive at two equations;
$$\left(\frac{\p P}{\p V}\right)_{T\ =\ T_c} = 0,\tag2$$
$$\left(\frac{\p^2 P}{\p V^2}\right)_{T\ =\ T_c} = 0.\tag{2'}$$
Equation $(2)$ is a sufficient condition for a critical point (mathematically, a derivative's non-existance works too). Equation $(2')$ gives a necessary condition for an inflection point.
So our first step should be to find the equation $P = P(T, V)$. Just use equation $(1)$.
$$P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\tag{3}.$$
Apply conditions $(2)$ and $(2')$ on $(3)$.
$$\left(\frac{\p P}{\p V}\right)_T = -\frac{nRT}{(V - bn)^2} + 2a\frac{n^2}{V^3} \overset{(2)}{=} 0\tag4$$
$$\left(\frac{\p^2 P}{\p V^2}\right)_T = \frac{2nRT}{(V-bn)^3} - 6a\frac{n^2}{V^4} \overset{(2')}{=} 0\tag{4'}$$
Both $(4)$ and $(4')$ are equal to zero. Thus we may multiply either equation by non-matching real numbers, and still maintain equality: $(4) = (4')$. We will use this to our advantage by multiplying $(4)$ with $2/(V-bn)$ and $(4')$ by $-1$.
$$-\frac{2nRT}{(V - nb)^3} + 4a\frac{n^2}{V^3(V- nb)} \overbrace{=}^{(4)\ =\ (4')} -\frac{2nRT}{(V - nb)^3} + 6a\frac{n^2}{V^4}\tag5$$
$$\frac{2}{V - nb} = \frac{3}{V} \implies 3V - 3nb = 2V \implies \enclose{box}[mathcolor="green"]{V_c = 3nb}\tag{5'}$$
Plug result $(5')$ into equation $(4')$.
$$-\frac{nRT}{(3nb - nb)^2)} + 2a\frac{n^2}{27n^3b^3} = 0 \implies -\frac{RT}{4} + \frac{2a}{27b} = 0 \implies \enclose{box}[mathcolor="red"]{T_c = \frac{8a}{27bR}}\tag6$$
The outcomes $(5')$ and $(6)$ are to be placed into equation $(3)$.
$$P_c = nR\frac{8a}{27bR} : 2nb - a\frac{n^2}{9n^2b^2} \implies \enclose{box}[mathcolor="blue"]{P_c = \frac{a}{27b^2}}\tag7$$
Values for $a, b, R$ in terms of critical parameters
$$b \overset{(5')}{=} \frac{1}{3n}V_c\tag{8a}$$
$$a \overset{(7)}{=} 27b^2P_c \overset{(8a)}{=} \frac{3P_cV_c^2}{n^2}\tag{8b}$$
$$R \overset{(6)}{=} \frac{8a}{27bT_c} \overset{(8a, 8b)}{=} \frac{8P_cV_c}{3nT_c}\tag{8c}$$
Reduced van der Waals equation
Replace the values of $a, b, R$ from equations $(8a)$$-$$(8c)$ into the original equation $(1)$. Immediately after this substitution multiply both sides by $3/(P_cV_c)$. You should have the result
$$\left[\frac{P}{P_c} + 3\left(\frac{V_c}{V}\right)^2\right]\left(3\frac{V}{V_c} - 1\right) = 8\frac{T}{T_c}.\tag9$$
Define $P/P_c = \pi, V/V_c = \varphi, T/T_c = \tau$. Finally, enjoy the reduced van der Waals equation
$$\enclose{box}[mathcolor="orange"]{\left(\pi + \frac{3}{\varphi^2}\right)(3\varphi - 1) = 8\tau}.\tag{10}$$

What does equation $(10)$ imply by there being only reduced quantities $\pi, \varphi, \tau$?

 Hint: theorem of corresponding states.

$[a]$ While I believe this to be technically true, I also think it deserves a question of its own. The fact that a phase critical point physically always has to (?) be a inflection point confuses me a bit, so is best left to another answerer.

The a,b,c in your question are the reduced values not the a, b of the vdw eqn and so its best to write them as $T_R=T/T_c,~ V_R=V/V_c, ~ P_R=P/P_c$.
The critical constants $T_c, P_c, V_c$ can be found from the vdw eqn. by writing it as a cubic in volume. The roots are real for temperatures below the critical temperature $T<T_C$, ($T_c$ itself is a point of inflexion)$^*$.
The result after some messing around with the algebra is
$$V_c=3b, ~ P_c=\frac{a}{27b^2}, ~T_c=\frac{8a}{27bR}$$
You can then substitute into the vdw equation and you should get an equation that depends only on reduced values. Have a go at this for yourself, if you get stuck add a comment :)
$*$ The solution to a cubic is a well known problem and several web pages describe this, but its very messy. It is easier in this particular case to use a 'trick' based on the physics which is that at the critical volume $(V-V_c)^3=0$ , expand this out as a cubic and compare it term for term with the expansion of the vdw equation. The critical values given above are then apparent.

Thanks for your answer.I tried to go in for direct substitution it made it cumbersome .Meanwhile I just got hold of the reduced form as $[A-\frac{3}{B}][3B-1]=8RC$.Is there any  method to arrive at this which requires ingenuity and not just plain substitution.

Its almost correct. Use the method given at the bottom of my text is by far the easiest.

I just noticed you changed A, B C which confused me. You should get in your notation $(A+3/B^2)(3B-1)=8C$

yes sorry for that .I just read your first line and edited the question to avoid confusion with van der wall constants

Can you please explain a bit more on your last para.I am aware of the 'trick' you mentioned and couldnot grasp it properly.

expand and re-write the vdw eqn as $V^3 + (...)V^2 +(...)V +(..)= 0$, then expand $(V-V_c)^3=0$ and compare coefficients of $V^3,V^2 $ etc. The first term is $3V_c=b+RT_c/P_c$ where you need to change T to $T_c$ etc after you have the coefficients.


text:
How to model Chemical Kinetics when both ligand and receptors are moving?
I am not a chemistry student but trying to understand something that requires Chemistry. My problem is that when we have Ligand and receptor reaction to form complex we normally assume as from wikipedia
"In protein-ligand binding, the ligand is usually a molecule which produces a signal by binding to a site on a target protein."
My issue is do we have chemical kinetics giving us some clue when both of Ligand and receptor is moving, of course this is something like a reaction in a solution but I am majorly concerned with probability that Ligand be able to bind with receptor when both of them are in random motion and any literature related to that.
I may not that much clear in my presentation, but it will be really helpful if any kind of support be given.

This is a non-issue. All molecules are always in random motion. Still, chemical reactions do occur sometimes.

This should be describable by chemical kinetics as $\ce{L + P <=> LP}$ where the ligand L, protein receptor P and when bound the complex LP are in equilibrium. The equilibrium constant determines what fraction of the ligand is bound.  The reaction is likely so slow that how quickly the molecules diffuse is irrelevant.

Also, the probability of the ligand to be able to bind with receptor is affected by the intermolecular interactions of both.


text:
How to perform molecular dynamics simulations of charged systems?
The GROMACS official documentation (see here) states that a system with non-zero total charge will yield an error:

System has non-zero total charge
Notifies you that counter-ions may be required for the system to neutralize the charge or there may be problems with the topology.
If the charge is a non-integer, then this indicates that there is a problem with the topology. If pdb2gmx has been used, then look at the right hand comment column of the atom listing, which lists the cumulative charge. This should be an integer after every residue (and/or charge group where applicable). This will assist in finding the residue where things start departing from integer values. Also check the capping groups that have been used.
If the charge is already close to an integer, then the difference is caused by rounding errors and not a major problem.
Note for PME users: It is possible to use a uniform neutralizing background charge in PME to compensate for a system with a net background charge. There is probably nothing wrong with this in principle, because the uniform charge will not perturb the dynamics. Nevertheless, it is standard practice to actually add counter-ions to make the system net neutral.

If I understand this correctly, it means that a uniform background charge will be added to a net-charged system. What if I wanted to study a charged system, e.g. a peptide with basic groups in the gas phase, where counter ions are absent. Is it still possible to perform a gromacs calculation on such a system and how will the introduction of the compensating background charge affect the molecular dynamics?
I then looked into other packages and noted that net-charged systems seem to be generally problematic in mm calculations. Can anyone point me into a direction like literature on this?

A system with net charge is not stable - electrostatic repulsion will make it fly apart. That would seem to be fairly problematic, no?

@JonCuster I am not sure if I understand correctly what you mean: there are tons of systems with a net charge that are incredibly stable. For example, ammonium is a stable ion, a "molecule" with a net charge. In mass spectrometry, you always observe all kinds of molecules with a net charge. So, molecules with a net charge don't fly apart.

@JonCuster Then, when I do MD simulations with semi empirical or ab initio methods, handling of the charge is no problem. The system does not "fly apart",
so I guess you meant to say that molecules with a net charge can not be handled by MM methods, as in MM methods, they will fly apart due to the electrostatic repulsion term. Right?

A single charged molecule will not fly apart, most likely. Trying to simulate two protons with no electrons will cause you a few problems. So, you need to clarify what your 'system' is since the program is not sentient (yet). If I were writing the documentation, I would probably say something similar - if the user doesn't balance charge on some arbitrary input, the program will do so since it cannot understand what the user really wants to get at.

@JonCuster Then let's say the system is lysin in charge state  +1. It is stable, otherwise biology would be very unhappy right now. Can gromacs (or any mm for that matter) handle it without adding an unwanted compensating background charge?

(Note - I'm just a simple materials scientist - I have no idea what a lysine molecule actually looks like!). I'll ask this question - in biology is the lysine(+1) isolated, or is it in solution? If it is in solution, isn't the system (molecule + surroundings) net neutral? I would think there is an OH- hanging around somewhere? What I do know is various mod/sim techniques have real problems modeling charged defects in solids, and an even harder time modeling charge capture/emission from said defects. Convergence can be very hard in those situations.

@JonCuster I agree. But isolated charged systems really aren't that exotic. E.g. in mass spectrometry, you will observe them all the time. But most MM/MD systems are laid out for solution state ...

The target for most (all?) MD software is biochem and/or materials science, so calculating isolated single molecules charged or not is not their main focus. Your best chance for calculations with periodic boundary condition is using PME or an artificial counterion.

This is intrinsic to Ewald summation methods, not software implementations. The uniform charge arises from neglect of a reciprocal sum term. It does not directly affect the dynamics and may be a reasonable model of a spatially homogeneous system. See https://mailman-1.sys.kth.se/pipermail/gromacs.org_gmx-users/2015-October/101544.html for further details and https://www.mpibpc.mpg.de/14063977/Hub_2014_JCTC.pdf for a description and explanation of the likely artefacts.

I also suspected that it has something to do with the mm calculation, but that really made it clear. Thanks!

You can also correct for the net charge post simulation

How would you change the sampling after the fact?

@mabraham sorry, I didn't get notified you commented, and just saw this randomly. Check out https://sourceforge.net/p/towhee/mailman/message/24030424/. A good question would be... what kind of sampling are you setting yourself up for by putting in an explicit counter ion as is often done? how does that not skew sampling?

I could imagine such a correction working for e.g. chemical potential. The OP's problem suits using a charged system and they should just do that. However, many cases do not suit that treatment - the sampling of a water box with four chlorides will differ whether four explicit counter ions are present, or not, and so many observable properties will not be able to be corrected by a general correction term.


text:
How can I relate the conductivity (uS/cm) at various temperatures of an aqueous solution of NaCl with ionic strength?
Can someone give the relationship between conductivity (uS/cm) with ionic strength please ? In some sources they that there is a linear relationship with a constant, but with no explanation to come up with it.


text:
Why is the principle of separation in GFC opposite to electrophoresis despite using same material for the gel?
In GFC, larger proteins get eluted out first while in electrophoresis the larger proteins move slower. Why is this?

Gel electrophoresis and gel filtration chromatography both use gel materials (though often polyacrylamide gels are used for GE and cross-linked dextran gels for GFC), but their operating principles are different.
In GE, the gel is a solid block of gel which contains only small pores for proteins to pass through.
The proteins generally have to be denatured (and given a charge) to pass through the pores as native proteins are too bulky.
The denatured proteins are like long strands of noodles being drawn through the gel and shorter proteins pass through more quickly because they don't have to contort as much to fit through all the pores—the larger the protein, the more interactions with the gel to slow it down.
However, GFC is a subset of what is known as Size Exclusion Chromatography.
SEC techniques generally also use porous gels, but instead of using a solid block of gel, the gel is made into small beads.
A similar situation to GE exists where small things can enter the pores of the gel and larger molecules suffer more resistance to pass through the gel, only now, the larger molecules have an alternative route with lower resistance—going completely around the beads.
The effect of this is that small molecules pass into the pores of the beads and slow down, while larger molecules bypass the beads entirely and travel faster (what is considered small and large depends on the gel's pores).
GE has the benefit of much better size separation, but requires denatured proteins.
SEC can handle native proteins (that are still functional on the other side of your column), but the separation is more akin to dialysis, where small and large molecules are easily separated from each other across a cutoff molecular mass, but there is otherwise poor resolution between species.

From a brief inspection of the materials, it appears the agarose, and other media choices, used in GFC is prepared in such a way as to be "beaded". The beads have pores in them which are of a size that interacts with smaller molecules to slow their passage.
This link shows several media used in GFC, scroll down to Sepharose for a purely agarose product. Also read the intro paragraph for a description of the mechanism.http://www.sigmaaldrich.com/life-science/proteomics/protein-chromatography/gel-filtration-chromatography.html#sephacryl


text:
Acetic(vinegar) and citric acid as the rust remover
I tested, separately, acetic and citric acids for cleaning steel parts. After 24 hours I have moderate results for acetic acid, and good results for citric acid. But in both cases, rust doesn't disappear fully.
I have two question:

If I mix two these acid, do I get the synergy?
Is there another acid which easy to buy?

P.S. I'd like to get clean iron without gray protective coat, like rust remover from auto parts shop.

Long, long time ago I was employed to formulate rust removers for iron. We needed to be concerned with toxicity, corrosivity, cost, waste disposal, effectiveness, speed, damage to the (uncorroded) metal, and flash rusting (rapid rusting once cleaner is rinsed off).
Rust isn't just one chemical compound. It is generally hydrates of $\ce{FeO}$ and $\ce{Fe2O3}$ (and there are a bunch of them, even if you exclude hydroxides).
So, since different acids will have different effects with different chemicals, you won't necessarily be able to find the one magic bullet for any and all types of rust. The stronger the acid, the more metal will be attacked, unless passivation is used (there's two types chemical and electrical). Amines are also often used. Strong acids Hydrochloric, sulfuric, nitric, are all more aggressive than the weaker organic acids. Naval Jelly (which I was involved with) uses phosphoric acid, but as you noted, it coats the steel/iron. These mineral acids can be dangerous and can't just be thrown out when finished.
There are alternatives. One trick is to get the chloride, sulfate, or nitrate into solution by using a compound other than the acid. For instance, adding sodium chloride to a solution of citric acid should make the citric acid more aggressive, and yet, won't change the pH. (One reason for this is that $\ce{FeCl3}$ is more soluble than the iron citrate.)
Some de-icers use nitrate or calcium chloride which may also be effective, and of course some fertilizers use nitrates, too (even some for indoor plants). So, if you'd rather not work with the strong acids (unless you're a careful person, you shouldn't work with them), you could try that approach.
As far as I know, the only other acid commonly used for rust removal is oxalic acid. I think Zud cleanser uses it, if you can find it. It's a poison, so don't be licking your fingers until you've washed them (yes, even assuming you DID wear gloves!). I don't think it is much better than citric acid. In fact, I'd guess it's worse in terms of rust removal.
Nitric acid will probably be difficult to find/buy. Hydrochloric acid is available (last I looked) at Home Depot ect. (Look in cleaning products, drain cleaners, and cement cleaners). Sulfuric acid, too. That is also used as battery acid, but I'm not sure if you can buy that at an auto parts store without the lead in it. Do not buy anything with lead in it! Do not use anything (like old battery acid) with lead (Pb) in it. Do not touch...you get the idea, right? Finally, look at a table of pKa values for the acids.
The only other ions I might fool around with would be iodide and bromide, and maybe fluoride, just for grins and snickers. I've no idea where you'd get them. (some aluminum cleaners (Aluminum Jelly?) have fluorides). Hint: one reason citric acid works pretty well is because the salt it forms is more soluble than something like iron phosphate, so it leaves the surface allowing further attack.
So, effectiveness will depend not only on acid strength, but on iron salt solubility. Ideally, you want to avoid any oxidizing acid (since that will dissolve (oxidize) the metal) but in reality, a bit of metal loss helps get rid of that stubbornly clinging rust.

Thanks for your answer. I'll use electrolytic method with electrolite based on sodium carbonate.

Awesome answer! Is there a safe amount of chlorides which can be added to solubilize rust without contributing to flash corrosion as the part dries, or is it always a risk?

To answer your first question, no, you would mainly just get dilution of the stronger acid (citric) with the weaker (acetic). If you used a citrus fruit as the source of citric acid, it's important to note that there is a fair bit of variation in acidity from one citrus juice to another. The best bet would be lemon or lime juice, which have among the lowest pKa values (measurement of acidity, lower is more acidic) of the common citrus juices.  If you just used something like orange juice, you might notice significant improvement by switching to lemon or lime juice.
Your second question probably is the right way to go. Rust is fairly soluble in dilute solutions of strong acids like hydrochloric acid (the acids you have used so far are weak acids).  You can get consumer-grade hydrochloric acid from a number of sources, the best probably being an aquarium or pond shop. These products usually have names like "pH down" or something indicating that they are acidic. What you want to do is look at the ingredients for a product that is just dilute hydrochloric acid, without a bunch of other stuff. Be sure to read and follow all of the precautions on the container, particularly with respect to wearing eye protection.
Another alternative source for dilute hydrochloric acid would be rust-cleaning products from a hardware store. These typically have a detergent of some sort mixed in with the acid, which is OK, and this may actually be your best solution depending on availability where you live.

If you used a citrus fruit as the source of citric acid

No, I use dry citric acid from supermarket.

OK, I see.  Even though that would be very "concentrated" it still wouldn't be as strong as a solution of HCl like I discussed. That is still by far the best advice I have for you. Like I said it should be readily available from multiple possible sources, although I guess I don't know if it's available where you live. If it is, I would definitely try it. Good luck!

Thanks for your answer. I test mix of two these organic acid. Acetic is weak. After Citric acid iron coat  with dark gray. In the future I'll try to use electrolytic method with sodium carbonate electrolite.

Mix of two these organic acid no take effect, pure citric is better.

I'm not surprised, that's what I said would happen in my first sentence ;)  To get a stronger acid, you really need to get something like HCl from an aquarium / pond maintenance store or possible a hardware store (and don't mix it with anything else). Depending on where you live, this should be a simple and cheap solution. Anyway, that's still my recommendation...good luck!


text:
Colour change on exposure to singlet oxygen?
Is there a dye that is particularly well suited to radical colour change on interaction with singlet oxygen, preferably non-toxic?
Based on this answer: https://chemistry.stackexchange.com/a/27633/41616 some compounds, such as tetraphenylcyclopentadienone, seem to lose their colour intensity, but what about changing / inducing colours completely, and what is the proper term for this process (chromism?) called?

Lookup dye degradation. But the challenge is to find something that is only oxidized by singlet oxygen. The problem is that there are superoxide or hydroperoxy radicals that can do the job and you may not be able to figure out whether the singlet oxygen was involved.

Besides being oxidized by singlet oxygen, a dye could act as a quencher for it. The resulting excited triplet state of the dye could prove long-lived enough to be detected spectrophotometrically.

Am I correct in assuming that the singlet oxygen would react with the dye more rapidly than triplet oxygen?


text:
What is the symmetry of the cuboctahedron (FCC metal)?
The background to this is that I've recently given a tutorial wherein we had to go through the determination of point groups for atoms in various lattices (BCC, FCC/CCP and HCP). BCC and HCP, I have no problem with. However, I've now acquired a copy of the lecturer's "official" solutions and there's a bit of discrepancy with the FCC, that I can't fathom: I get an $O_\mathrm{h}$ point group, while he gives a $D_\mathrm{3d}$ point group.
Here goes:
The coordination geometry of an atom in the FCC structure (assuming one atom/lattice point and only one species of atom, as in an FCC metal) is a cuboctahedron:
,
By my reckoning we have $3$ $C_4$ axes, one through each pair of square faces, and a $C_3$ through each pair of triangular faces.
Following the classic flowchart:

The number of high-symmetry rotational axes is surely enough to exclude $D_\mathrm{3d}$ already?
The only thing that makes me think that the lecturer hasn't just made an error is this line from the Wikipedia page for the cuboctahedron:

With $O\mathrm{h}$ symmetry, order $48$, it is a rectified cube or rectified octahedron (Norman Johnson).
With $T_\mathrm{d}$ symmetry, order $24$, it is a cantellated tetrahedron or rhombitetratetrahedron.
With $D_\mathrm{3d}$ symmetry, order $12$, it is a triangular gyrobicupola.

This is no doubt due to deficiencies in my understanding of group theory, but I don't understand how the same shape can simultaneously have multiple point groups, nor how you would select which one is relevant to a particular analysis.

I will update the question to make this clearer, but the assumption when solving was that we had an FCC metal structure, ie. only one atom.
The Wikipedia quote was from a geometry article and, afaik, refers only to the polyhedron and doesn't consider any "atoms" or similar.

Oh is correct, so they went off the rails there somewhere.

@WanderingChemist You are right, the symmetry of a cubooctahedron is similar to that of a cube and/or octahedron, i.e. $O_h$.

@JonCuster, IvanNeretin, Thanks! In that case, do you know what the quote from the Wikipedia article is on about when it discusses an cuboctahedron "with Oh symmetry", "with Td symmetry"? Is that just a situation where something is added to certain locations on the shape to reduce its symmetry?

I might suggest that you seem to be blurring the line between point groups and space groups in your latest comment. The 'adding things to reduce symmetry' is, roughly speaking, what happens to turn the (fewer) point groups in to (the many) space groups.

About the multiple point groups: you have noticed that the classic flowchart focusses only on perpendicular/parallel planes or axes. In your example, however, the square faces and the triangular faces do not obey that rule.


text:
Is it correct to use the symbols (l) and (g) in every reaction?
I know that this question might sound stupid to most of the people, but for a chemist noob like me is not very clear. So, here's the main question:
Is it appropriate to use the symbols (l) for liquids and (g) for gas in any reactions, or do the reactions have to be specific for a certain application?
The reason why I'm asking this is that in most of the cases, you don't see this notation. Also, are there other notations, for example (v) for vapor, or (s) for solid??

I'd rarely, if ever, use (l) for liquid. I'd never use (v) for vapor. I'd only use (s) and (g) if they really added information to the equation. I'd be more inclined to use up-arrow for gas being given off and down-arrow for a precipitate. Some reactions do use a gas reacting with a solid.

In addition to @MaxW's suggestions, the (aq) identifier for aqueous species is commonly seen and useful at times, especially if there is ambiguity concerning whether the solvent is water or something else.

Take for example H + OH + M -> H2O + M*. I would not use any of these indicators here. M could be anything really, H and OH are highly reactive radicals and defining them as "being a gas" is certainly questionable.

The one place that I'd used additional symbols everywhere is when calculating Standard Enthalpy of Formation. With all the swirling around it is easy to get lost.

Is there really (v) for vapor? Is there any arguments from authority (IUPAC or similar)?

@mhchem I am always a bit scared to answer with a certain no but I believe no is the correct answer here. Vapor is the gaseous phase of a substance that is mostly solid or liquid at normal conditions. Whether it's a gas or a vapor too should not have any difference for chemical reactions. But who knows (see first sentence of this comment) :p


text:
Boiling point comparison between cyclohexane derivatives
I am trying to order these according to boiling point. I think that B is the highest because it has hydrogen bonding. I am not sure how to differentiate A and C though. They both have about the same molar mass. The only thing I can think is that C has a double bond. Does this double bond make the boiling point of C higher than A because it is stronger than a single bond? Or perhaps the double bond is shorter which decreases branching in C making the boiling point lower than A?

It's not the double bond in C that is significant here, it's the oxygen. The carbon-oxygen bond is polar and will make C more "electrically sticky" with its neighbors compared to A. Since the $\ce{O-H}$ bond in B is even more polar than the ketone in C, its boiling point will be even higher. Read more about it on this page.
(If you'd like to skip to the relevant section, on Google anyway, search "ketone vs alcohol boiling point".)


text:
Why does the reduction of NAD+ to NADH change the absorbance so much, and cause fluorescence?
This is an updated question:
I don't have a very strong background in biophysics, but I want to understand the theory behind $\ce{NAD+}$ and $\ce{NADH}$ absorbance and fluorescence.

Background: I understand that the additional hydrogen that $\ce{NAD+}$ gains to form the reduced $\ce{NADH}$ results in the molecule being able to absorb light at $\pu{340nm}$. And that only the reduced $\ce{NADH}$ is able to produce fluorescence. This is also the property which is explored when studying, e.g., enzyme kinetics - if the enzyme binds $\ce{NADH}$.
Question: The oxidized form, $\ce{NAD+}$, still has an aromatic ring - why is it that it is not able to fluoresce? Why, in a nutshell, is the transfer of the hydrogens causing fluorescent in $\ce{NADH}$ and such a big change in absorbance spectrum?
Also, perhaps a bit naive question, but why is it fluorescence better to measure than absorbance?
I have tried to look for articles (the old and original) that explores these properties, but I can't seem to find any that purely speak about the spectroscopic properties.

Related: Why does NADH have 2 peaks in its absorption spectrum but NAD+ has only one?

In a nutshell absorption is caused by photon interacting with an electron  in an occupied orbital and exciting the molecule by "transitioning" the electron to an unoccupied molecular orbital. This happens over a wide range of photon energies. When NAD is protonated the energies and bonding simultaneously change so different transitions occur. The different transitions may be different molecular orbitals entirely (new types of bonds), shifts in energies of the same molecular orbitals (same bonds), or changes in intensity of the transitions.

Essentially NADH has an additional pair of electrons. These additional electrons which are oftein drawn as a lone pair of electrons alters the molecule in a similar way to the way in which many pH indicators change upon protonation.
Here are diagrams of the reduced and oxidized forms of NADH / NAD+

A typical indicator when protonated appears to have a colour, however upon deprotonation the indicator loses a hydrogen and gains a lone pair of electrons now appearing to have a different colour. The addition of the lone pair of electrons actually lowers the transition energy between the pi bonding and the pi anti-bonding states.
For example for methyl red it can be predicted that the unprotonated form has a HOMO/LUMO gap of 2.175 eV (according to extended Huckel calculations). For the protonated form of methyl red has a HOMO/LUMO gap of 3.275 eV.
This lowered energy correlates with a decreased absorbance energy, ultimately reflecting a higher energy wave of light (Frequency of Blue > Red). NADH's absorbance wavelength his greater than NAD+'s absorbance wavelength, because NADH is absorbing lower energy light. This, like pH indicators, is due to the lone pair lowering pi and pi anti-bonding energy state differences.


text:
Is it possible for the remaining solute of a saturated solution to form a single even phase from top to bottom (similar to that of a colloid)?
If a solute that dissolves in a liquid solvent has reached high enough levels to cause saturation within the solution, is it possible for the remaining solute particles to form an evenly distributed phase from top to bottom similar to that of a colloid (where settling would not occur or take very long to occur)? I realize behavior varies depending on the solute and solvent in question, I am just curious if this is possible. Thanks so much!

In inorganic qualitative analysis this happens frequently. The precipitates are highly insoluble so they form many small particles instead of a few large ones. A centrifuge is then used to force the particles to settle since they wouldn't own their own.

The density of the solute relative to that of the solvent will primarily determine whether the excess solute will float, sink or tend to disperse throughout the solution.  Additionally, the particle size and porosity will play a role in the rate of settling or floating, as will the viscosity of the solvent, but it is the density that determines whether the particles will float or sink.
But, the answer to your question is no. Even if the density of the undissolved solute is essentially the same as that of the solvent, and the undissolved solute is very fine, then you could have a uniform looking solution, but it would still be a two-phase solution, with a solid phase suspended in a liquid phase.

Okay I see, sorry I was communicating a bit vaguely, I meant to say could the solid particles of the remaining solute form their own roughly evenly distributed phase from top to bottom, suspended by the separate liquid solvent phase that's also spread from top to bottom. It seems that if the density and other factors are aligned, it is possible for the remaining solute to distribute roughly evenly from top to bottom within their own phase and that it is possible that this solute phase wouldn't settle easily?

Speaking of the densities, is there a general minimum in difference of density of solute and solvent for settling to occur, assuming in this situation that density is the primary determinant of that? Thanks so much again for your insight, it has been a huge help!

Regarding your first comment, yes you are correct, and I kind of thought that was what you really meant. Technically there are still two phases, but it can appear as a homogeneous suspension. Regarding your second comment, there is no general difference in densities, they just have to be "close" and it helps if the undissolved particles are small and the solvent is viscous. I hope that helps!

Thank you so much again for continuing to check back and sharing your insight, it really means a lot and it's helping a ton! Would you say something like 338 daltons (solute) vs 46 daltons (solvent) be close enough? If the solute molecules are very small in terms of volume, perhaps that would change things given their seemingly disparate densities? Unfortunately, the solvent isn't viscous :(


text:
What causes the smell of something being "musty"?
You know the smell that arises when you leave a wet sponge out for too long? Does anybody know the chemical structure of the molecule that causes that smell? Similarly, what chemical process might be underpinnig that smell from occuring in the first place?

The "musty smell" is from mold, sometimes bacteria. Different compounds are responsible depending on the species. Some examples: trichloroanisole, tribromoanisole, dimethylmethoxypyrazine. Search "mold smell molecules" for many links.
Anisole is also called methoxybenzene, structure here: https://en.wikipedia.org/wiki/Anisole


text:
Electrons split in Stern-Gerlach experiment but not in Thompson's experiment?
So when electrons are fired through a magnetic field in Stern-Gerlach experiment they split into two beams depending on their spin but where was this phenomena when Thompson was calculating charge to mass ratio of electron? Shouldn't his cathode ray beam have split up too?

Hello and welcome to Chemistry.SE! If you have any questions about how the site works I suggest that you take the short [tour]. Good luck!

Have you considered how spin interact with magnetic field vs how it interacts with electric field?

Of course it was split; it's just a matter of precision. Where was the Einstein's relativity when Newton was discovering his laws?

Besides, Stern–Gerlach is normally done with neutral atoms and not with free electrons.

As has already been suggested by some of the comments to your question, there is a fundamental difference between a Stern-Gerlach-type experiment and a Thompson-type experiment, and this difference lies in the type of interaction of the particles.
First of all, as Ivan Neretin points out, a Stern-Gerlach experiment is performed with neutral species (the original experiment involved silver atoms), while Thompson's experiment employed a beam of charged electrons.
The Coulombic or Lorentz interaction between charged particles (or charged particles and fields) is rougly four orders of magnitude larger than the interaction between neutral particles and fields. In the latter case, one can have interaction between a magnetic dipole moment and a magnetic field and/or interaction with a electric dipole moment and an electric field. These interactions are referred to as Zeeman and Stark effects. The important thing to realize is that the sign and magnitude of the interaction depends on the quantum state of the particle (or in other words on the relative orientation of its angular momenta). It is therefore a quantum-state specific interaction. As a consequence particles in one state may be attracted to a region with high fields, while particles in another state may be attracted to regions with low fields, as if you have two different compasses; one that points to the North pole and one that points to the South pole.
Electromagnetic interactions between charged particles and fields are so strong that the quantum-state specific effects are much harder to see (and the Coulomb and Lorentz forces dominate).
Let's go back to the Stern-Gerlach experiment. In their experimental setup, Stern and Gerlach used an inhomogenous magnetic field with a well-defined gradient. The silver atoms they employed have one unpaired electron that could be oriented with respect to the external magnetic field "up" or "down", these different quantum states experience different forces and are therefore separated.
In Thompson's experiment charged electrons are send through a magnetic field where the force they experience only depends on their charge and velocity vector with respect to the magnetic field. This force is identical for all electrons, so no splitting is observed.

Hosein Majlesi's paper at https://arxiv.org/abs/1504.07963 appears to show that the Stern-Gerlach experiment works with electrons.

It might be better if you discuss the literature you cite rather than just giving a reference.