text:
Calculating the dissociation constant of one acid in a mixture

A solution contains $\pu{0.09 M}~\ce{HCl}, \pu{0.09 M}~\ce{CHCl2COOH}$ and $\pu{0.1 M}~\ce{CH3COOH}$. The $\mathrm{pH}$ of the solution is $1$. Given that the $K_\mathrm{a}$ of acetic acid is $10^{-5}$, calculate the $K_\mathrm{a}$ of $\ce{CHCl2COOH}$.

My attempt is as follows. We have $[\ce{H+}] \text{ from } \ce{HCl} = [\ce{HCl}] = \pu{0.09 M}$.
Then, for the other two acids:
$$\begin{align}
[\ce{H+}] \text{ from } \ce{CH3COOH} &= x \\
\frac {10^{-2}\cdot x^{2}}{10^{-1}-x} &= K_\mathrm{a} \\
[\ce{H+}] \text{ from } \ce{CHCl2COOH} &= y \\
\mathrm{pH_\text{total}} &= 1 \\
-\log_{10}(0.09+x+y) &= 1 \\
0.09+x+y &= 10^{-1} \\
\end{align}$$
But the answer key says the answer is $\pu{1.25*10^{-2}}$.

I just wanted to add an answer based on "no" assumptions, except (and that is occasionally a big assumption) that concentration equals activity. At first it does not seem very practical, as of its size, but I couldn't stop me from adding it here as I like to see the whole picture. In the end, it is actually not that hard.
The exact equation for the proton concentration is:
$$[\ce{H+}] = [\ce{OH-}] + [\ce{Cl-}] + [\ce{CHCl2COO-}] + [\ce{CH3COO-}]$$
which is the sum of the hydroxide ions from the autodissociation of water and the sum of the resulting acid residue ions from the three acids. As is shown here [1], this system expands to the following (and can be reduced again, if appropriate simplifications are taken into account for each acid, as is shown here [2]):
$$
[\ce{H+}] = [\ce{OH-}]
+ \frac{[\ce{HCl}]_0 K_{\mathrm a1}}{[\ce{H+}] + K_{\mathrm a1}}
+ \frac{[\ce{CHCl2COOH}]_0 K_{\mathrm a2}}{[\ce{H+}] + K_{\mathrm a2}}
+ \frac{[\ce{CH3COOH}]_0 K_{\mathrm a3}}{[\ce{H+}] + K_{\mathrm a3}}
$$
where $K_{\mathrm a1}$, $K_{\mathrm a2}$, and $K_{\mathrm a3}$ are the $K_{\mathrm a}$'s of $\ce{HCl}$, $\ce{CHCl2COOH}$, and $\ce{CH3COOH}$ respectively.
Now what you need to find is $K_{\mathrm a2}$. Solving it at this stage for $K_{\mathrm a2}$ is probably not a good idea aka nothing nice to handle, as the system is quite large. But as you know all values except $K_{\mathrm a2}$, it's actually not that bad.
$$
\underbrace{[\ce{H+}]}_\pu{0.1 mol/l} = \underbrace{[\ce{OH-}]}_\pu{10^{-13} mol/l}
+ \underbrace{\frac{[\ce{HCl}]_0 K_{\mathrm a1}}{[\ce{H+}] + K_{\mathrm a1}}}_\pu{0.09 mol/l}
+ \frac{[\ce{CHCl2COOH}]_0 K_{\mathrm a2}}{[\ce{H+}] + K_{\mathrm a2}}
+ \underbrace{\frac{[\ce{CH3COOH}]_0 K_{\mathrm a3}}{[\ce{H+}] + K_{\mathrm a3}}}_\pu{10^{-5} mol/l}
$$
$\ldots$ I'll leave units from here on (everything is in $\pu{mol/l}$). This simplifies quite much into:
$$\begin{align}
0.1 &= 10^{-13} + 0.09 + \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}} + 10^{-5}\\
0.1 - 10^{-13} - 0.09 - 10^{-5} &= \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}}\\
0.01^* &= \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}}\\
0.01\,(0.1 + K_{\mathrm a2}) &= 0.09\,K_{\mathrm a2}\\
0.001 + 0.01 K_{\mathrm a2} &= 0.09 K_{\mathrm a2}\\
0.001 &= 0.09 K_{\mathrm a2} - 0.01 K_{\mathrm a2}\\
0.001 &= (0.09 - 0.01)\,K_{\mathrm a2}\\
0.001 &= 0.08\,K_{\mathrm a2}\\
K_{\mathrm a2} &= \frac{0.001}{0.08}\\
&= 0.0125
\end{align}$$

* I rounded to $0.01$; it was actually $0.00999\ldots$ which would lead to $\pu{0.0124861... mol/l}$
[1] Equations 2.1 and 2.2
[2] Assumption 1 for HCl, Assumption 3 for acetic acid

Let's use $K_1 = K_\mathrm{a}$ for $\ce{CH3COOH}$ and $K_2 = K_\mathrm{a}$ for $\ce{CHCl2COOH}$
\begin{align}
K_1 &= \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} \\[3pt]
K_2 &= \dfrac{[\ce{H+}][\ce{CHCl2COO-}]}{[\ce{CHCl2COOH}]}
\end{align}
Given that the $\mathrm{pH}$ is $1$, we'll assume that that can be written to at least four decimal places $1.0000$, and that concentrations can be used instead of activities.
Thus for acetic acid
$$ \frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{K_1}{[\ce{H+}]} = \frac{10^{-5}}{0.1} = 10^{-4}$$
so for all practical purposes acetic acid doesn't contribute any significant amount of the $\ce{H+}$.
Since we know that $\pu{0.09 M}$ (we'll assume $\pu{0.0900 M}$) of the $\ce{H+}$ comes from the $\ce{HCl}$, that leaves $\pu{0.0100 M}$ of $\ce{H+}$ from the $\ce{CHCl2COOH}$.
Thus
\begin{align}
[\ce{CHCl2COO-}] &= \pu{0.0100 M} \\
[\ce{CHCl2COOH}] &= \pu{0.0900 M} - [\ce{CHCl2COO-}] \\
&= \pu{0.0800 M} \\
K_2 &= \frac{[\ce{H+}][\ce{CHCl2COO-}]}{[\ce{CHCl2COOH}]} \\
&= \frac{(\pu{0.1 M})(\pu{0.01 M})}{\pu{0.0800 M}} \\
&= \pu{1.25\times10^{-2} M}
\end{align}


text:
Brown particles in electrolysis of brine
This is not a duplicate... pls read the whole question....
I am attempting electrolysis of brine  (tap water + table salt). I am using HB pencil lead (graphite) as my two electrodes. I am using 12 volt- 1.5 amp DC power supply (using an adapter from the wall).
The solution is getting slightly alkaline (~8 pH) and hydrogen is confirmed by pop sound.
However, the whole solution is getting filled with some brown particles (resembling Dirt). If by chance my alligator clips get submerged, the get a brown coating (which I am able to wash it off).
What are this brown particles?
I have a few theories:
1. Pencil lead also contains clay.
2. Something with the water (though it is filtered)
I had attempted this earlier (~3y) and had a similar but a more dramatic result.
Then I used iron/steel electrodes and they became completely brown. The plastic container I used also became completely brown coated. I was unable to wash it off.

Is the tap water coming straight from the ground? (groundwater)

@HugoWoesthuis, IDK, Ill try to check where my water is coming from. Your answer seems the likliest but any idea how to fix it!?!

@Zhe, I am getting bubble at both sides...

You're using a significant over voltage, so you might be oxidizing both to oxygen gas and to chlorine gas. Is the proportion of oxygen to hydrogen as expected or less?

Concerning your first theory: Typical HB pencils contain a mixture of about 2 parts of graphite and 1 part of clay.

@Loong Thanks but i always had doubts about that theory, seems that answer is the likeliest... But i hope someone as an easy way t avoid that complication.

If your tap water is groundwater, it will contain a lot of metals that will start rusting if reacting with O2 and Cl (your salt). The reaction causes the brown particles (rust/ Ferric Oxide). The only way to prevent this is to filter the water continuously. This reaction is also common with salt-electrolyzed pools. Here is a forum thread that has exactly the same issue you have.
This was the answer:

That looks like rust, Ferric Oxide, exactly!
  However, the cell should be Titanium, it contains no iron, but that doesn't mean it can't oxidize. Or "Rust", all metals do, Ti's oxidation would be white and powdery.
So either there is a non Ti component inside that cell that contains iron, such as steel that is breaking down, or there's something upstream of it, a Ferrous metal that's rusting, thus producing Ferric Oxide and dribbling down the pipe and into the cell. Is there any on the upstream side of the cell?
  If not, then it has to be localized to inside the cell.
Another though is that you have a very high iron content in your water, mixed with the electrolysis process is causing it to oxidize on the anode, where it combines /w oxygen molecules in the water to form rust, Fe2O3.
  This would keep it localized to the cell, and not be upstream, only at the cell, since that's the catalyst, or source and slightly downstream as the flow of generated Ferric Oxide tapers off.
  This shouldn't effect the Ti plates though, just simply clean them with MA and you're golden!!
  Ti is virtually indestructible, so they should be fine.
EDIT: Hayward says to submerge it just to the top of the cell, so the wiring harness, the electronics on the cell are not touching the acid, coiling the harness may be helpful prior to immersing. 4:1 dilute, 1 gallon of water to 1 quart of MA. And remember, always, always add the acid to the water, never, ever the other way around.
  The manuf. says to soak for a few, then rinse with a high pressure hose, then repeat if needed, until the plates are clean.
  BTW, Acid attacks Oxides, including Ferric Oxide.

Full thread here
Related Stackposts:
Salt product in water electrolysis

I was under the impression "rusting" is used to refer to the corrosion of Iron or iron-based alloys O.o. Personally, I doubt whether that's rust. I don't really see what the chlorine would do to promote rusting (if it is rust), perhaps the H+ would do the job?

@AaronAbraham It could be, but considering groundwater with so many metals that can react with each other, it will be unlikely to be completely sure if the brown particles are rust. I searched on Google and found out that this is a common thing in pools with salt-electrolyse. They say that if you don't filter the water correctly, and the salt (NaCl) is added too early, the metals in the water will react with the O2 and the salt together (I doubt salt plays a role in this, but the oxygen is for sure), making the water brown (rust).

Any idea how to fix this without continuously filtering., I only have one water source... All help appreciated..

@Srathi00, you can't. Use normal water/groundwater and filter or consider using demi-water (demineralized water).

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text:
If PQ, T2U2, RS2 and VW4 have the same Ksp, which has the lowest solubility in moles of solute per litre?
If the solubility products of the following ionic compounds all have the same numerical value, and all solubilities have values between zero and one, which one would have the lowest solubility in moles of solute per litre?
$\ce{PQ}$
$\ce{RS2}$
$\ce{T2U2}$
$\ce{VW_4}$

Their chemical equations and $\ce{Ksp}$ values would be:

$\ce{PQ <-> P +Q}$  and $\ce{Ksp=[P][Q]=[P][P]=[P]^2}$
$\ce{RS2  <-> R +2S}$ and  $\ce{Ksp=[R][2S]^2=[R][2R]^2=4[R]^3}$
$\ce{T2U2  <-> 2T +2U}$ and $\ce{Ksp=[2T]^2[2U]^2=[2T]^2[2T]^2=16[T]^4}$ (although the textbook says $\ce{[T]^4}$ which I don't understand)
$\ce{VW4  <-> V + 4W}$ and $\ce{Ksp=[V]^2[4W]^4=[W]^2[4W]^4=256[W]^5}$ (although the textbook says $\ce{256[V]^5}$ but I am pretty sure its equivalent whether it is written in terms of V or W)

FYI my thought process here is that:

Based on the chemical equation RS2⟷R+2S
The concentration of one product (S) is twice that of (R).
So if there is X moles of RS2, and there is X moles of R and 2X moles of S
Hence in the Ksp=[A]^n⋅[B]^m , A=X and B=2X

Now the textbook says

As $\ce{[P][P]=4[R]^3=[T]^4=256[V]^5}$
  [P] must be the lowest because $\ce{[P]^2}$  has the same value as  $\ce{4[R]^3 and [T]^4 and 256[V]^5}$

I can't get my head around this. I would have thought this proves
$\ce{[P]}$ has the highest value not the lowest. Why is [P] the LOWEST?
E.g. noting that all values are between $0$ and $1$ (as defined in the question)
$\ce{[P]^2=256[V]^5}$  (the other two options are in the middle so don't matter)
If both sides equaled e.g. $0.25$ (which is between $0$ and $1$)
$\ce{[P]^2=0.25 and 256[V]^5=0.25}$
${[P]=0.5}$ and ${[V]=(0.25/256)}^{1/5} = 0.25$
Showing [P]>[V], so [P] (moles/L) is the greatest solubility, not the lowest.
Is anyone able to spot my error?

Question and answer as shown in book below (I have typed up the relevant parts above but sometimes I interpret parts of the textbook question incorrectly (making my stack exchange question invalid or misleading):

You could help yourself out a lot if you wrote out the final concentrations in terms of the original solid, instead of arbitrarily (?) picking one of the ions to express it in. $\ce{PQ}$ in the aqueous phase is essentially the solubility. My guess is that book is missing that factor of 16 for $\ce{T2U2}$.

Thanks for the suggestion Zhe. You are right that I did arbitrarily pick one. I am not sure what the textbook's reasoning was in doing that. If I write them in terms of the original solid I get [PQ]^2=4[RS2]^3=16[T2U2]=256[VW4]^3  and so am stuck with the same problem I was having (although it does help give me a more systematic approach). The issue being that [PQ]^2=256[VW4]^3  to me shows that [PQ] > [VW4] whilst the solution says [PQ] is the lowest solubility.

Also thank you for confirming what I thought about the factor of 16 for T2U2. Do you have any idea as to the other part of my question regarding why the book says [P] < [VW4] or [V] ? (Because the 16*T2U2 answer is somewhat irrelevant as if falls between the other  two ([PQ] and [VW4]) options and can be excluded).

Here's the issue. The value $a^{x}$ gets bigger for bigger values of $x$ only when $a > 1$. For small values of $a$, it's the reverse. Notice the key phrase "all solubilities have values between zero and one." This means for a fixed small value of $K_{\mathrm{sp}}$, $a$ will be bigger if $x$ is bigger. This means that your assertion of $\ce{[PQ]} > \ce{[VW4]}$ is incorrect based on the previous equation.

I think you have spotted my mistake, but when I test it numerically I can't seem to prove it. E.g. if I assumed Ksp was 0.25 (i.e. a small value i.e. between 0 and 1 as specified).  $[PQ]^2=Ksp=0.25=256[VW4]^5$ ... This simplifies to $[PQ]=sqrt(0.25)$ = $0.5 & [VW4]$ = $(0.25/256)^(1/5)=0.25$. Hence $[PQ]$ (which is 0.5) > $[VW4]$ (which is 0.25) (please ignore the box, I can't seem to fix the formatting)

Pick a really small value of $K_{\mathrm{sp}}$. Like $10^{-10}$. To be fair, I think the textbook question is horrible.

Fantastic, that solves it. Using $Ksp=1E-10$ I get $[PQ]=1E-5$ which is smaller than $[VW4]=3E-3$. I really appreciate the guidance.

Easier just to migrate my comment into an answer. Your assumption
$$\ce{[PQ]} > \ce{[VW4]}$$ is not valid based on
$$\ce{[PQ]^{2}} = 256\ce{[VW4]}^{5}$$
For $y = a^{x}$, if $a>1$, larger $x$ leads to larger $y$. But for $a<1$, larger $x$ leads to smaller $y$, and the question indicates that solubilities are between zero and one, i.e., small.
If you plug in $K_{\mathrm{sp}}= 10^{-10}$, you get $\ce{[PQ]}=10^{-5}$ and $\ce{[VW4]} = 3.3\cdot 10^{-3}$.
As I said in my comment, I believe that this is a very poorly designed question: the constraints are too broad and there is a typo in the key. Someone was very sloppy...

Is anyone able to spot my error?

I think so. There is a systematic error in your equations when it comes to stoichiometric factors.
What is the solubility product constant $\mathrm{K_{sp}}$ for the reaction $$\ce{A_nB_m(s) <=> nA(aq) + mB(aq)}$$

 $$\mathrm{K_{sp}} = [\ce{A}]^n\cdot [\ce{B}]^m$$

Thank you for the answer Klaus. I agree with that formula. But based on the chemical equation RS2⟷R+2S ... doesn't this show the concentration of one product (S) is twice that of (R). So if there is X moles of RS2, doesn't that mean there is X moles of R and 2X moles of S? Hence in the Ksp=[A]^n⋅[B]^m , doesn't A=X and B=2X?
This seems to be what the solution shown at the bottom of the question has done.

As they have done here in step 2 http://i.imgur.com/uUGtdT8.png  (Source http://mhchem.org/223/223PPT/pdf223/III18b01Chapter18bCG.pdf)

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text:
Solubility product of HgS
The reported solubility product of $\ce{HgS}$ is $4 \times 10^{-54}$. How is this possible? The minimum concentration that is possible is 1/avogadro's constant which shall be about $10^{-23}\ \mathrm{M}$. Hence the minimum value of solubility product for a diatomic specie like $\ce{HgS}$ would be about $10^{-46}$. What am I missing?

True, homeopathic concentrations like that can't be measured directly. I guess the solubility product is calculated from thermodynamic data.

Can you provide a source for the "minimum concentration that is possible" being $1/N_A$? I was unaware of such a limitation.

It's just an implication. The minimum amount of a specie you can have in any solution is one molecule of it. Hence the inverse of Avogadro's number.

Can't you have ten times the Avogadro's number of water per one dissolved molecule?

Please read my comment to the answer below.

True, we'll run into weird statistical problems, hence my first comment.

Imagine you have one googolplex ($10^{10^{100}}$) liters of water. Now let's dissolve one mole of sugar in your water. Setting aside where would you find so much water, the concentration would be way lower than the inverse of Avogadro's constant.
Long story short, there's no lower limit to the concentration of a substance. You can always add more solvent to the mixture thus making the concentration smaller.

Aren't all the results and laws that we apply, statistical i.e., a collective picture and hence even if you keep adding the solvent there is only one molecule in there and hence won't the statistical compilation be pointless.

you say the concentration does not have a lower limit but isn't it a little illogical to consider one molecule to be distributed throughout the solvent uniformly.

Swaroop - it isn't necessarily illogical.  It depends what you are trying to describe about the system.  And no one (except you) said "one molecule".  Variax said "one mole".

I still think my reasoning is pretty solid. Dissolve one molecule of substance in one liter of water and you get a molarity of 1/NA . Add an extra liter of water and you halve that concentration. Granted, a concentration so enormously low may not be evenly distributed throughout the whole volume of the solution, but that's not the point at all. We're describing a macroscopic system, not its microscopic state.

The microscopic states statistically give the macroscopic system . Here the microscopic states being are so limited that taking a statistical conclusion is pointless.

Over short times, and short here could mean comparable to the life time of the Universe, the ergodic hypothesis (https://en.wikipedia.org/wiki/Ergodic_hypothesis) breaks down and hence statistical thermodynamics as usually formulated breaks down because, put simply, the one molecule won't have had time to visit every point in the solution. But wait long enough, and you will have to be very patient, and the standard statistical description will be regained.

Variax's answer addresses the minimum concentration misconception. Perhaps you are also wondering how we could conceivably measure concentrations that small.
If:
$$\ce{HgS(s) <=> Hg^2+(aq) + S^2-(aq)}\ \ \ \ K_{sp}=4\times 10^{-54}$$
then, in a saturated solution prepared by adding solid $\ce{HgS}$ to ultrapure deionized water:
$$[\ce{Hg^2+}]=[\ce{S^2-}]=\sqrt{4\times 10^{-54}}=2\times 10^{-27}$$
As an aside, actually $[\ce{S^2-}]\approx 0$ because $\ce{S^2-}$ is a strong base with a very large $K_B$. So in fact, the reaction is
$$\ce{HgS(s) +H2O(\ell) <=> Hg^2+ (aq) + SH^- (aq) + OH-(aq)}\ \ \ \ K_{sp}=4\times 10^{-54}$$
However, let's stick with the simpler case. If we can measure $[\ce{Hg^2+}]$ or $[\ce{S^2-}]$, we can compute the value of $K_{sp}$. According to this pdf presentation from Perkin Elmer, their IPC-MS system can detect mercury at 0.016 ppb and sulfur at 28 ppb ppb limit of quantitation will be a little higher). 0.016 ppb is approximately $0.016\ \mathrm{\mu g \ per\ L}$:
$$
0.016\ \mathrm{\mu g/L} = 1.6\times 10^{-8}\ \mathrm{g/L}\\
1.6\times 10^{-8}\ \mathrm{g/L\times \dfrac{mol\ Hg}{200.59\ g\ Hg} = 7.98\times 10^{-11}\ M} $$
Clearly, this instrument is not sensitive enough.
One way the might work is by using a concentration cell. Because of Le Chatelier's principle, an electrochemical cell containing the same ions at different concentrations will produce a voltage. The larger the concentration difference, the larger the potential difference:
$$E_{cell}= E^\circ -\dfrac{RT}{nF}\ln{\dfrac{[\ce{Hg^2+}]_{dilute}}{[\ce{Hg^2+}]_{concentrated}}}$$
If we take our saturated $\ce{HgS}$ solution and couple it to a $1\ \mathrm{M}\ \ce{Hg(NO3)2}$ solution, we get $(E^\circ = 0)$:
$$E_{cell}= \mathrm{-\dfrac{(8.314\ J/mol)(298\ K)}{2(9.65\times 10^4\ C/mol}\ln{\dfrac{2\times 10^{-27}\ M}{1\ M}}=0.781\ V}$$
As mentioned by Ivan in the comments, this $K_{sp}$ value could also be estimated from thermodynamic data using
$$\Delta G^\circ = -RT\ln K$$

Determining the solubility product is accepted. My doubt is at the validity of the assigning such a value to such extreme solutions.

This is interesting, but I do not see how it can work. Suppose you have just one liter of solution with the tiny mercuric ion concentration you used. Then it is probable that there are no mercuric ions present. Scaling up the solution volume to get a reasonable probability of having a few mercuric ions present then means larger electrodes or dealing with solution quantum effects, i.e., mercuric ions will not always be in the vicinity of a small electrode in a large vat of solution. My gut feeling is this is a classical answer that cannot be valid down in the lowest concentration regime. ???


text:
Calculating Heat Release from Bomb calorimeter

When $0.400\ \mathrm g$ $\ce{CH4}$ is burned in excess oxygen in a bomb calorimeter that has a heat capacity of $3245\ \mathrm{J/^\circ C}$ a temperature increase of $6.795\ \mathrm{^\circ C}$ is observed. What is the value of $Q_V$?

I thought it was the regular $Q=mc\Delta T$ problem but it isn't and not only that I did the equation but it didn't work.
I am not sure what to do right here at the moment

Using the data determine standard enthalpy change for the combustion of methane

Also I would like the definition of the standard enthalpy change and what it tells you and the purpose of it.

Heat of combustion is the amount of heat (enthalpy) liberated when 1 mole $\ce{CH4}$ reacts (combusts) with oxygen.
In other words, what is the heat of reaction in this case?
$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$
Then you must ask yourself, what happens after the heat is produced i.e. where does it go? It quickly turns into an accounting problem...
Take a look at: Constant Volume Calorimetry

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text:
Why in Hamiltonian for hydrogen atom we have $\nabla_e$ and $\nabla_N$ and one does not operates on other?
Laplacian is $$\frac{\partial ^2}{\partial x^2}+\frac{\partial ^2}{\partial y^2}+\frac{\partial ^2}{\partial z^2} $$
Then why in Hamiltonian for hydrogen atom we have $\nabla_e^2$ and $\nabla_N^2$ and one does not operates on other ?

Please refrain from using mathjax in your title.

First, $\nabla$ is the del operator, the Laplacian is $\nabla^2$. Second, the mere existence of derivatives with respect to different variables does not imply that there should necessarily be mixed terms in each and every expression. For instance, the Hamiltonian does not contain any such mixed terms by its very definition: it consists of kinetic energy term plus potential energy one, where the first is additive.

In quantum mechanics the Hamiltonian is commonly expressed as the sum of operators corresponding to the kinetic and potential energies of a system,
$$
\hat{H} = \hat{T} + \hat{V} \, ,
$$
where the kinetic part is additive, i.e., for an $n$-particle system
$$
\hat{T} = \sum\limits_{i=1}^{n} \hat{T}_{i} = \sum\limits_{i=1}^{n} -\frac{\hbar ^2}{2 m_i} \nabla_i^2 \, .
$$
Thus, by definition, the Hamiltonian does not contain any mixed derivatives of $\nabla_i \nabla_j$ form.


text:
Is HPLC the best way to detect residues from anti-coagulant rodenticides
I am a beginner chemistry student studying anticoagulant rodenticides. The literature shows that high-performance liquid chromatography (HPLC) with fluorescence or UV light is the most commonly used method for the detection of residues in tissue samples. I want to understand on a basic, conceptual level why this particular analytical method is the preferred method.
I understand that HPLC can separate complex organic compounds and enable detection at extremely low concentrations, but I do not know whether there are other analytical methods that can also do this.
Is this particular analytical method more effective in detecting the presence of residues from anticoagulant rodenticide in tissue samples? If so, why? Is the answer related to the chemical properties of the compounds in rodenticides (hydroxycoumarins and indandiones)?

Without a doubt there are other techniques that could be used. But chemistry employs the KISS principle. A HPLC setup is much cheaper to buy and easier to maintain than say a  GC/MS. // If the compound is fluorescence you get a better S/N ratio than measuring absorption, and also relatively few compounds are fluorescent so there is an additional specificity factor.

UV(/VIS) spectroscopy is the most common HPLC detection method, it's quite cheap and powerful but you can only detect substances which absorb light in the range you are using for detection. You also don't get any real structural information, most of the time identification is only possible using reference material. But most HPLC systems will be equipped with such a detector.
Fluorescence detectors can be used for fluorescent molecules and usually you got a much lower limit of detection. Like for UV/VIS you don't get structural information, but identification works quite well, since you can selectively only detect molecules with the used excitation and emission wavelengths.
Other detectors would be for example refractive index detector, which only really work well if HPLC is run isokratic or MS, which gives you structural information.

In this case the technique is also very adequate since the aromatic structures of hydroxycoumarins and indandiones will all absorb in the UV spectrum. I guess at least some of them might be fluorescent too. With a good collection of standards and a column with high resolution power there's no need for costly detectors such as MS.

Is this particular analytical method more effective in detecting the presence of residues from anticoagulant rodenticide in tissue samples?

Yes, it is :)

You might want more than just detection: quantitative determination.
You might want to determine a one or more compounds that are only available in traces.

Extraction of the relevant compounds from tissue samples will remove the biological background and allows for aggregation/enrichment prior to HPLC.
HPLC will allow to separate structurally similar rodenticides.
Fluorescence is extremely sensitive and will help to quantify trace amounts.

Update
In Multicomponent Determination of 4-Hydroxycoumarin Anticoagulant Rodenticides in Blood Serum by Liquid Chromatography with Fluorescence Detection, published in J. Anal. Toxicol., 1991, 15, 126-129 (DOI, free PDF), this has been demonstrated for the separation and fluorescence detection ($\lambda_\mathrm{ex} = 318~\mathrm{nm}, \lambda_\mathrm{em} = 390~\mathrm{nm}$) for a mixture of

warfarin
coumatetralyl
difenacoum
brodifacoum and
bromadiolone

Is this particular analytical method more effective in detecting the presence of residues from anticoagulant rodenticide in tissue samples?

It is not by any means more effective then other methods. The reason for popularity of HPLC is it's fairly universal nature, being adaptable to a variety of anayltical problems. This makes it very popular and widely available in analytical labs. Other separation techniques like capillary electrophoresis would also work for those analytes (also fluorescence and absorbance detector).

Is the answer related to the chemical properties of the compounds in rodenticides (hydroxycoumarins and indandiones)?

Some of those compunds show fluorescnce, which can be detected with high sensitivity but that doesnt make HPLC the obvious choice. Other separation techniques are possible too.


text:
Why does hydroxide attack the carbon atom of methyl iodide (in an Sn2 reaction)?
As the electronegativity difference between carbon and iodine is negligible, why does $\ce{OH-}$ attack the carbon atom (in an $\mathrm{S_N2}$ mechanism) even though carbon doesn't seem to be electrophilic? (Since the $\ce{C-I}$ bond is not polar enough to give a sufficiently positive partial charge on carbon)

Of course, just looking at methyl iodide and the partial charges, one would assume that there is no real reason for any nucleophile to attack the carbon atom for the reasons you stated. I’ll just point out and leave as an exercise to the reader that there is equally hardly any reason for the nucleophile to attack iodide, though.
However, the reaction also has a product side. Remember that what is happening is the following:
$$\ce{OH- + CH3-I -> HO-CH3 + I-}$$
While the attack onto the only very slightly polarised carbon in itself may not be particularly beneficial, iodide is a very good leaving group as it is able to distribute the resulting negative charge well. (Large size, many electrons, small charge per volume ratio.) Likewise, if we consider thermodynamics, the $\ce{C-I}$ bond is not a particularly strong bond while a $\ce{C-O}$ bond is in the region of $\ce{C-C}$ or $\ce{C-H}$ bonds — so overall we are gaining a much stronger bond for a rather weak one.
Consider what had happened had oxygen attacked the iodide side of iodomethane. While oxidising iodide is something rather common in chemistry, the leaving group would have to be $\ce{CH3-}$ — a terribly non-stabilised leaving group, the conjugate base of an extremely weak acid; something that simply does not leave. Thus, there is no way hydroxide would attack the iodide side in an $\mathrm{S_N2}$ reaction — even aside from the reasons above left for the reader to figure out.

I would rate this answer even more highly if we were to connect the stability of the leaving group to the energy of the transition state.


text:
Why is there no non-bonding molecular orbital in the overlap between H1s and Be2s in BeH2
I've only recently started learning about this and I'm stating to get confused. (The orbital in the middle belongs to Be) So you have a + (+) + which gives a sigma bonding orbital and a + (-) + which gives a sigma anti-bonding orbital. Why isn't there a + (-) - which would result in a non-bonding orbital?

If you look up the MOs of the allyl cation, it's exactly the same system here, except that you're dealing with s orbitals instead of p orbitals. The nonbonding orbital has no coefficient on beryllium because the antibonding linear combination of hydrogen orbitals ($A_{1u}$) does not have an appropriate symmetry match with the beryllium 2s ($A_{1g}$). Of course, this treatment is a bit simplistic because it neglects the beryllium 2p orbitals entirely; DSVA's answer covers those nicely.

Well, you need two sigma bonds here, and two anti-bonding ones. And there are actually two non-bonding orbitals here.
First of all: how you write it down doesn't really work, since there are p-orbitals in Be, not only s-orbitals. You need to look at them too, not only the s-orbitals. And I'm not quite sure if you did, but don't confuse Valence Bond and MO theory.
So for Valence Bond Theory:
We know this molecule is linear. Therefore we need two orbitals on the Be which can form bonds with the hydrogens and are in an 180° angle. So we do sp-hybridization and with one sp-orbital from Be and the s-orbital from hydrogen we can form a $\sigma$ and $\sigma^*$ orbital on each side. Two sigma bonds and two empty p-orbitals on Be which are non-bonding.
For MO Theory:
We construct a $\ce{H2}$ first, in this case we could write + + and + -, so we got a bonding and an antibonding orbital. Now we put our Be in the middle. And now it get's quite complicate to write down, so here on page 10 you can find the MO of $\ce{BeH2}$. You see that the bonding "$\ce{H2}$" orbital was combined with the 2s orbital of Be in an bonding and antibonding way, and the antibonding + - orbital of "$\ce{H2}$" was combined with a p-orbital in a binding and antibinding way. The two other p-orbitals aren't used and are non-bonding, since they would have equal constructive and deconstructive interaction with both "$\ce{H2}$" orbitals. In the end same result as with the valence bond theory.

I've been over what it actually looks like, but if I was to start a MO diagram is there no other way to do it other than from starting with H2.

Of course there are, you could also just "throw" all three atoms together, but it get's much harder to understand which orbitals will combine to form a new one. In case of molecules with one central atom and "ligands" around it the best way to do this is combine the ligands first

So if I was to do a MO for H2O, I'd take the same approach. Start with H2 then see how those overlap with the orbitals on the O atom.

Yes exactly. That's also described stepwise in the linked pdf starting on page 6, LGOs are the ligand group orbitals, the "H2" molecule.

@ShubhamR You'll find that this approach is pretty general. For example in one of my answers here you'll see that I constructed the "F6" ligand group orbitals (elsewhere you might also see the terminology symmetry-adapted linear combinations) first before deciding which sulfur AOs to overlap them with.


text:
Why does q/m ratio(specific charge) depend on nature of gas in anode ray experiment?
I can't understand why $q/m$ ratio changes in anode ray experiment (discharge tube experiment) (which was carried out by Goldstein) with change in nature of gas?
Since $q=1.6\times10^{-19}\ \mathrm C$ and $m=1.6726219\times10^{-27}\ \mathrm{kg}$, shouldn't the ratio be constant?

Because different ions can have different charge states, and different nuclei have different masses. So, q/m is different for H+, He+, and He++...

I really don't see how you got that from the comment. A hydrogen atom without the electron (thus H+) is a proton. To be more specific, a protium atom without an electron is a proton. A deuterium atom without the electron is D+. But, since you specified a gas, one would go with hydrogen, or H+, since protons don't generally form an independent gas.


text:
Creating an efficient synthetic route to this elastomeric block copolymer

This was a question my professor gave me in which I have a couple of ideas to create a synthetic route which include: Either an anionic or cationic polymerization? If so, I don't think it would be cationic because I can't seem to see if there would be some sort of initiator system and i don't think there could be carbocation formation either from the looks of it. The only feasible conclusion I have come to is that it would be an anionic polymerization where you create a living polymer and react this with a second monomer to create a block copolymer? I could maybe use an organolithium to start the initation with a polar solvent however I don't actually know which monomer to start off with nor how the reaction mechanism would work out. Can someone help me? Thank you very much!

Should the copolymer units have m,n type subscripts as opposed to being of a specific ratio?

sigh It's polyaddition. You don't need any catalyst.

How about starting with a difunctional isocyanate, adding hydrazine in step 1 and polyethylene glycol in step 2. I would call it an addition polymerization. Your structure does not show the number of repeating units, but that could be largely controlled by varying the monomer proportions.


text:
Saline removing rust from counter?
Background (pointform for readability):

I prepack oral medication for a hospital.
Our work area consists of a long stainless steel countertop.
We use plastic bins to transport the medication around the pharmacy. To clean the bins, we use cloths premoistened in a 0.3-0.5% solution of Hydrogen Peroxide (can't remember the exact concentration).
While the bins are drying, they're stacked on top of the counter. The bins at the bottom are often left facedown, which prevents then from drying properly.
We noticed bin-shaped stains appearing on the counter where we let the bins dry. I figured the Peroxide was oxidizing the counter.
70% alcohol, water and goo-b-gone had no effect on the stain.

Today, something was allowed to dry on the counter. It dried to a white powder. I'm almost certain it was a 4 mmol/ml solution of NaCl, but I don't know for sure. When I cleaned the counter off, the part underneath the power was completely clean! It looked brand new. We tried applying more NaCl to it, but we didn't have time to let it dry, and there was no noticeable change when we cleaned it off.
What I know/don't know:

I'm pretty sure the counter is stainless steel. We have signs up in the area with directions for cleaning stainless steel countertops.
The solution was almost definitely sodium chloride. If not, the only other thing I can think that it could have been is Sodium Phosphate; but I've never seen it dried, so I don't know for sure.

What could have happened here? I know rust can't be "undone". The only thing I could think of is somehow the top layer was removed; but I can't see saline doing that.
Can anyone think of what may have happened?

Most likely the abrasive action of the dried salt being rubbed off removed the minor rust stain.

@DrMoishePippik I soaked the area before I wiped. It was completely dissolved.

Actually, rust can be "undone"... was there oxalic acid or other organic acid, or perhaps phosphoric acid, in one of the solutions, which removes rust?

@DrMoishePippik No. 99% of the meds we use are either suspended in a sugary vehicle, or dissolved in one. The drug I cleaned up definitely wasn't sugary at all. The only non-sugary drugs are the ones intended for Injection, and the only injectable drugs we used were sodium chloride, sodium phosphate, and pyridoxine.

Unless somehow the phosphate in the sodium phosphate somehow was turned into phosphoric acid (?). But I'm not sure how possible that is.


text:
Show that decomposition of AsH3 follows first order kinetics
The decomposition of $\ce{AsH3}$ into solid $\ce{As}$ and $\ce{H2}$ is given in terms of pressure of the gas in a closed vessel from time to time.
$$\ce{2AsH3 -> 2As + 3H2} $$

Show that this reaction follows first order kinetics.
My attempt :
So to prove that the reaction follows first order kinetics, I tried finding the relation between pressure in the vessel and the time taken.
I assumed the initial concentration of $\ce{AsH3}$ (which is the pressure of the whole vessel at $t=0$) as '$2a$'. Assuming the dissociation factor to be $\alpha$, at any time $t$, the composition of the vessel can be :
$$\ce{AsH3} \rightarrow (2a - 2\alpha)$$
$$\ce{As} \rightarrow 2\alpha$$
$$\ce{H2} \rightarrow 3\alpha$$
(We are working with gases so the concentration of solid As can be assumed to be constant and have no effect on the pressure of the vessel)
Assuming all gases to be ideal.
So I was able to pull through till here. Assuming temperature to be constant, $a$ can found out:
$$2a = {\frac{P}{RT}} \Rightarrow a = {\frac{733.22} {2 \cdot 0.081 \cdot  T}} = {\frac{4465.4}{T}}$$
I am not able to move forward after this.

In a first order reaction $\ce{A \rightarrow B + C}$ with rate constant k the rate equation is
$$ {\frac{d\ce{[A]}}{dt} = - k\ce{[A]} } $$
which  if integrated to give the concentration vs time where the initial concentration is $\ce{[A_0]}$ produces
 $$\ln\left(\frac{\ce{[A]}}{\ce{[A_0]}}\right)=-kt$$ or $$ \ce{[A]= [A_0]}e^{-kt}$$
Thus a log of the ratio concentrations or pressures (or any measurement proportional to concentration) vs time should be a straight line if the reaction is first order.
In your case you are observing the product which appears as the $\ce{AsH3}$ disappears therefore
$$ {\frac{d\ce{[B]}}{dt} = + k\ce{[A]} } $$
You can substitute for $\ce{[A]}$ and integrate again to get $\ce{[B]}$.
$$\ce{[B]-[B_0]=[A_0]}(1-e^{-kt})$$
However the total pressure is measured which is the sum of $\ce{2[A] + 3[B]}$ and if $\ce{[B_0] =0}$ then
$$\ce{3[B] + 2[A] =3[A_0]}(1-e^{-kt}) + \ce{2[A_0]}e^{-kt}$$
if the measured pressure is proportional to sum of concentrations with $p_0$ as initial pressure then
$$p = p_0(3-e^{-kt})$$
At $t=0$ the initial pressure is $2p_0$ which is the value given in the table. At very long times the pressure becomes constant at $3p_0$ and calculating $3p_0 - p$, rearranging and taking logs produces
$$\ln\left(\frac{3p_0 - p}{p_0}\right) =  -kt $$
which when plotted is the 'normal' equation for a first order equation.


text:
Does computational chemistry include neutrons in ground state calculations?
I was under the impression that most particles in computational chemistry were treated as point charges. If this is true, wouldn't there be no need to include neutrons as a separate entity in ground state calculations? Or are particles not treated as point charges?

Depends on what you mean by 'computational chemistry'.
In the most common sense of the term, where the aim of a given calculation is determination of the electronic structure of a particular system, each entire nucleus is treated as a classical point charge/mass.  Thus, neutrons are not explicitly accounted for, aside from their effect on the various nuclear masses in, say, vibrational frequency analysis or in estimates of the kinetic isotope effect.
If you're referring to quantum models of the nucleus, though, then this statement of yours is incorrect:

... most particles in computational chemistry [are] treated as point charges.

In these nuclear computations, if the nucleons were being treated classically it would be correct to say that they are treated as point particles since, as you correctly note, neutrons carry no charge, and proper representation of both the neutrons and protons is necessary in order to obtain an accurate model.
However, as noted quite correctly by Mithoron in a comment, for quantum modeling of the nucleus, the protons and neutrons are not treated as particles but instead as contributors to the nuclear wavefunction.

Just curious: do you know software that simulates quantum models of the nucleus?

@QuantumCAPUCCINO Not off the top of my head, unfortunately. I know groups are working on it, but I think their codes are all specialty, in-house software.  I turned up one arXiv paper on a quick search; there appear to be several good hits on Google searching for 'atomic nuclei computation'.

I don't think they can be treated as points in nuclear calculation https://en.wikipedia.org/wiki/Atomic_nucleus#Composition_and_shape

You might be interested in the nuclear-electronic orbital method, which should be available in GAMESS.

@pentavalentcarbon Nice!!


text:
Difference between homonuclear and heteronuclear aromatic compounds
What is difference between homonuclear aromatic compounds and heteronuclear aromatic compounds?

Please have a look at the definition of aromaticity in the IUPAC Gold Book.

Benzene may be the prototypical aromatic molecule, but it is not the only aromatic ring system. According to the IUPAC gold book, any molecule that is cyclic and conjugated that displays resonance stabilization beyond that expected of a delocalized pi system is aromatic.
While experimental confirmation is necessary to verify a molecule's automaticity (e.g. certain NMR behavior, resonance stabilization energy from heats of combustion, or certain reactivity behavior), the Hückel rule gives a good way to predict aromaticity for monocyclic rings systems. In general a monocyclic ring system is aromatic if:

Planar
Conjugated
Has (4n+2) pi electrons, where n is an integer.

Heterocyclic compounds certainly qualify.
Larger and polycyclic systems may be aromatic while appearing to violate the Hückel rule.


text:
Relation of pressure and temperature for an ideal gas at constant volume
I have a tank of ideal gas with volume $3.1\ \mathrm L$; I have $R = 8.314$, and 11 moles. The tank is filled at $23\ \mathrm{^\circ C}$ (not sure why that's relevant) and the tank will rupture at $P = 100\ \mathrm{atm} = 1.013\cdot 10^7\ \mathrm{Pa}$. At what temperature does the tank rupture?
I divided $PV$ by $nR$. However, plugging those numbers in gives
$$\frac{3.1 \times 1.013\times 10^7}{8.314\times 11} = 342374$$
in kelvin. However, my book gives $70.2\ \mathrm{^\circ C}$ as the correct answer. I assume my units are somehow profoundly messed up. Or maybe that mysterious 23 has something to do with it? What am I doing wrong?

Our units of volume are $m^3$, not $L$.

When using the value of $R=8.314$, the units are $\mathrm{\frac{J}{K\cdot mol}}$. One joule is equal to:
$$\mathrm{J=\dfrac{kg\cdot m^2}{s^2}=N\cdot m = Pa\cdot m^3 = W\cdot s = C\cdot V}$$
If you want to use volume in liters, you need to use a different value for R.

The tank is filled at 23 °C (not sure why that's relevant).

This information is crucial :)
For the initial situation, the number of moles, the volume and the temperature are given.
$$P\,V = n\,\mathrm{R}\,T$$
Plugging these into the ideal gas equation will give you the initial pressure.
Given that the number of moles and the volume don't change, increasing the temperature will raise the pressure until the tank eventually bursts.
Just remember Gay-Lussac's law:
$\Large{\frac{P_1}{T_1} = \frac{P_2}{T_2}}$

Hmm... Why can't I just plug into the ideal gas law at 100 atm? There would only be 1 unknown.

@RenéG You certainly can. I thought that mentioning the underlying, older law might help.


text:
Understanding mzML, specifically precursor mass
I'm a Biotechnologist and computer scientist, migrating into the field of Mass Spec. for my new position. Some things I know well, and others not at all. Please be patient :)
I'm working with OpenMS, Python, MGF and mzML files.
Q1: Is there a better resource for understanding the format of mzML files than PeptideAtlas specifications?
Q2: In the MGF specification you get a listing of the precursor protein mass (PEPMASS) followed by a list of ions in m/z format.
Is the precursor (protein) mass listed similarly in the mzML file? Is it the BPI in (for example)
      <cvParam cvRef="MS" accession="MS:1000796" name="spectrum title" value="runID:160812_29,index:2120,ID:controllerType=0 controllerNumber=1 scan=2121,path:160812_29.raw,scan:2121,activation:HCD,isolation:745.306640625,selectedIon:745.306640625,z:2,precursorScan:controllerType=0 controllerNumber=1 scan=2118,spectrum:MSn spectrum,MSlevel:2,scanstartmin:7.4891189999999996,scanstartsec:449.34713999999997,BPmz:745.4205487,BPI:15730.942999999999,TIC:54362.472999999998"/>

EDIT:
This is what I have so far:
In the  tag, we have...
<cvParam cvRef="MS" accession="MS:1000796" name="spectrum title" value="runID:160812_29,index:2060,ID:controllerType=0 controllerNumber=1 scan=2061,path:160812_29.raw,scan:2061,activation:HCD,isolation:555.730041503906,selectedIon:555.730059736802,z:2,precursorScan:controllerType=0 controllerNumber=1 scan=2060,spectrum:MSn spectrum,MSlevel:2,scanstartmin:7.3208402000000001,scanstartsec:439.25041199999998,BPmz:129.1023507,BPI:17504.615000000002,TIC:1.8637816e05"/>

We have ...
selectedIon:555.730059736802,z:2,precursorScan:controllerType=0 controllerNumber=1 scan=2060,

...followed by...
<cvParam cvRef="MS" accession="MS:1000500" name="scan window upper limit" value="1155.0" unitCvRef="MS" unitAccession="MS:1000040" unitName="m/z"/>

...implying the upper limit is roughly the Precurser mass.
Later we find, in the (of course) precursorList count="1" tag, subtag selectedIonList count="1"  the double the above mass, with a charge count of 2...
<cvParam cvRef="MS" accession="MS:1000744" name="selected ion m/z" value="555.730059736802" unitCvRef="MS" unitAccession="MS:1000040" unitName="m/z"/>
<cvParam cvRef="MS" accession="MS:1000041" name="charge state" value="2"/>

All of this together indicates that precursor mass is ~550 x 2 = 1110.
I can't find any more direct route to this discovery than the deduction.
Is this correct and is it the best route?

You are calculating the precursor ion mass, which is not the same as the peptide neutral mass.  Most peptides probably ionize as $\ce{(M + nH)^{n+}}$ but that is an assumption.  mzML has to handle small-molecule spectra as well, where things like $\ce{(M + Na)+}$ happen a lot.

Also unfortunately most vendors do not fully adhere to formal mzML specifications in their converters.  I'm suspicious that the scan window upper limit tag is just an upper limit and not related to any detected ions in any way.


text:
What gas expands most in a vacuum?
Layman here.  I need a gas that occupies the most volume under the least pressure.
I am trying to design a water-conservation toilet. i.e- using grey-water from the tub to fill a tank above the toilet.
I want it to be VACUUM driven, so no pumps.
If the storage tank is (say) 1/2 gas and 1/2 grey-water.  I need a gas that will expand to the FULL tank volume, creating a vacuum, as the water is used.  Then when the tub is used, the vacuum will refill the tank.  I have the valve system worked out, but the physics is over my head.
Basically the straw full of water with your finger over one end;
IF there is a gas bubble, can I get a gas to expand enough to let water out and create a vacuum?  Which gas is best?  Assume temp is constant (ish)
Any discussion of weight of the water and opening sizes is welcome.

That question doesn't really make sense for several reasons.
First of all: All gases will expand to the whole volume of the vacuum chamber, so "which expands the most" doesn't make sense.
I think your question is which gas will produce the best vacuum if the volme doubles and for that: all gases will pretty much behave the same here. Double the volume, half the pressure.
Other than that: this setup won't ever work for a ton of reasons, the most obvious one: If the build up vacuum isn't strong enough to hold the water back from flowing out it won't be strong enough to suck water in. It also would be a perpetual motion machine, wich cannot exist.
Take a straw, put it in water and close the top hole with your finger. Then pull it out of the water. You will see that some of it comes out at the bottom, but a little bit stays in there. Why? Because there's a vacuum obove the water in the straw. So now put the bottom tip of the straw onto the water surface, it won't suck in additional water.


text:
Why are sodium and potassium displaced from soap by calcium and magnesium ions in hard water?
Sodium and potassium are more reactive than calcium and magnesium still they get displaced. Why?

Metallic sodium and potassium are reactive not their ions.

I have no external justification to point to on this, but I've always thought one aspect contributing to this behavior is the fact that the divalent $\ce{Ca^2+}$ and $\ce{Mg^2+}$ bind two soap fatty acid molecules, whereas $\ce{Na+}$ and $\ce{K+}$ as monovalent ions can only strongly bind one molecule.
As a result, the alkaline earth cations have two fatty acid "tails" that can 'glom' around them, shielding them better from the water molecules and making the resulting complex more stable, as well as less soluble.
On the other hand, since the alkali cations only have one fatty acid "tail" they're strongly associated with, that one tail can't wrap around nearly as completely, and so there is significantly greater accessibility of the solvent to the cation.


text:
What exactly is the d-orbital splitting and how does this affect the colors for transition metal compounds?
How does d-orbital splitting form colors for transition metals?
Why can't the electrons in the d-orbital split for the element Zinc?

Electrons don't split. Orbital energies do. They do it in zinc all the same, but it has completely filled d-orbital, hence no electronic transitions, hence no color.

In any form of spectroscopy (nmr, IR, microwave, uv visible etc)  a photon is absorbed and the molecule/atom goes to a new state. In your case an electron is promoted from one level into an empty or partly empty level. The colour appears as the photons absorbed can no longer be observed in the light source. If the levels are maximally filled then no transitions are possible

Let’s first address what leads to the colour of a given species. Colour is the absorption of one or more wavelenghts of visible light. Considering that visible light is rather energetic (but not as energetic as UV light and way less than X raays), this energy difference typically corresponds to the difference between occupied and unoccupied (spin-) orbitals in compounds. Thus, if a photon of the correct wavelenght interacts with an electron present in orbital $\Psi_1$, that electron can be excited to orbital $\Psi_2$ with the photon’s energy. Later relaxation processes are often complex, but sometimes visible light can again be liberated upon relaxation; typically as phosphorenscence or fluorescence. I have been glossing over a lot of important issues here, but realise that ‘what is colour’ is too broad a question for this margin.

d-orbital splitting does not form colour in itself nor does it happen in itself. In an uncoordinated metal ion (or atom) in vacuum, the five d-orbitals are degenerate and thus have the same energy level. It is only upon bonds being formed to other species that any type of splitting can occur. This is typically discussed in the context of coordination complexes.
The simplest model (but also the weakest) to explain d-orbital splitting on coordination complexes is the crystal field model. It basically assumes negative point charges approaching the central metal from certain axes which leads to destabilisation (increase in energy) of any orbitals that point towards that axis. In an octahedral complex, the axes are $x,y$ and $z$ so the orbitals whose energies are increased are $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$ — the ones with lobes pointing in the direction of the $x,y$ and $z$ axes.
A more sophisticated model to describe complexes and d-orbital splitting is the use of molecular orbital theory which comes in multiple levels (inclusion or ignorance of π-symmetric ligand orbitals). Scheme 1 shows the typical molecular orbital scheme of an octahedral complex.

Figure 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.
On the left-hand side of scheme 1, you can see the metal’s 3d, 4s and 4p orbitals, on the right-hand side six σ-symmetric ligand orbitals; in the middle their interactions are given. As we can see, a principal stabilisation derives from the $\mathrm{e_g}$ orbitals (corresponding to $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$) forming bonding-antibonding interactions with the corresponding ligand group orbitals. This leads to the metal-centred orbitals becoming antibonding ($\mathrm{e_g^*}$) and thus rising in energy. The $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}, \mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$) are nonbonding with respect to σ interactions, only.
The energy difference, marked in the scheme as $10~\mathrm{Dq}$, typically corresponds to the visible wavelength a complex will absorb and thus to the complementary colour of the complex. Chromium(III) complexes (three d-electrons; all populating $\mathrm{t_{2g}}$ orbitals) are typically green, the complementary colour of which being red, meaning they absorb red light. The wavelength can be calculated.
And finally, scheme 1 above also tells you what happens if zinc forms complexes. For a start, there is d-orbital splitting to be observed in zinc complexes. However, the overall stabilisation is very low since zinc is a $\mathrm{d^{10}}$ ion meaning that all five d-orbitals ($\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$) are fully populated — since two of those are antibonding, all bonding interactions derive solely from zinc’s 4s and 4p orbitals. Finally, since $\mathrm{e_g^*}$ is fully populated, the lowest energy difference with which a photon can be absorbed is no longer that marked $\mathrm{10~Dq}$ but that which corresponds to the $\mathrm{e_g^* \rightarrow a_{1g}^*}$ transition. This very high energy difference is firmly within the ultraviolet range and invisible to the human eye.

Hey @Jan you seem to be a quantum guy, can you answer my question here on colors  https://chemistry.stackexchange.com/questions/172258/colored-inorganic-liquids-why-is-scl2-red-and-s2cl2-orange


text:
Stability of cyclic radicals
How do we arrange the following in increasing order of stability?

The answer given is: $5<3<1<2<4$ (numbering according to their order in the image).
My Approach:
I can figure that in the second and the fourth structure, resonance stabilization will be applicable and the fourth structure will be more stable than the second.
But I am not able to figure out how do I arrange the remaining structures.

Well, 1 isn't stabilised and 3 and 5 are less table because of stronger s character of orbitals with unpaired electron.

Did this question just get asked again today? https://chemistry.stackexchange.com/questions/83360/need-an-explanation-for-stability-of-some-free-radicals @Jan

@Zhe Yes it did, so I decided to give the dupe target a proper answer.

Radicals are electron-deficient compounds. Therefore, according to Bent’s rule a radical should always populate an atomic orbital with a maximised $\mathrm p$ contribution. Ignoring 2 and 4 for a second, the radical occupies an $\mathrm{sp^2}$ orbital in 3 and 5 while it occupies an $\mathrm{sp^3}$ orbital in 1.[1] This already gives us $3\approx5<1$.
In a next step, let’s consider 2 and 4. These, as you noticed, are adjacent to a π bond, so the radical can occupy a $\mathrm p$ type orbital and participate in π resonance. This is the main way to differentiate 1, 2 and 4, since all three could essentially adopt a planar configuration with the radical confined to a $\mathrm p$ orbital — yet in 2 it can also resonance with the allyl system and in 4 it can resonate across two double bonds. Thus, we now intermediately have $3\approx5<1<2<4$.
All that remains is to determine the difference between 3 and 5, which is not trivial. In 3, you have a single double bond. This structure can distort a little bit, increasing the bond angle on the radical-bearing carbon towards something closer to $\mathrm{sp}$. The freely rotatable single bonds making up most of the ring allow this distortion. The phenyl radical can distort much less, since its geometry is basically already defined from the six-membered aromatic ring. Thus, we may assume 3 to be more stable than 5. Therefore: $5<3<1<2<4$.

Note:
[1]: In fact, rehybridisation is also possible in 1, giving the radical a pure $\mathrm p$ orbital and making the arrangement of substituents planar around the radical centre.

I hope with rehybridisation you mean distorting towards a trigonal planar coordination.

@Martin-マーチン Yes.

How do you explain radicals are electron-deficient? And apply Bent's rule to them in this way? The formal charge of a radical is 0. Could you cite some sources to justify your claims?

@TanYongBoon Very simple. Atoms desire to achieve an octet. A radical is a septet. Thus, an electron is missing; therefore it is electron-deficient.

Electron deficiency of the radical clearly does not suggest that the lone electron should be held in orbitals with higher p character, by Bent's rule. Bent's rule looks at the electropositivity/electronegativity of the substituents, not the overall electron deficiency of the central atom (in this case, carbon).

In that case, are you then saying that the lone electron of the radical can be seen as a rather electronegative substituent (i.e. not very charge-dense) and would preferentially be held in a p orbital?

@TanYongBoon When I wrote this, I thought I understood Bent’s rule but Martin later told me where I was wrong. Since then, I’m not sure whether I understand it (out ever did). But we don’t really need it: it suffices to say that the unpaired electrons is an electron hole (or half-hole) which should always populate an orbital as energetic as possible as anything else would mean an electron could relax down. Thus, ideally a pure p orbital since all hybrid orbitals are lower in energy. Maybe I should rewrite the answer without Bent’s rule … soon™

While checking stability of Carbon free radical it is treated as positive charge but we don't see resonance or mesomeric effect because for these effects we have to break bonds Homolatically, which is possible only at high temperature. So we see only H effect or I effect.
Now 5 and 3 are sp2 which are least stable and + I of sp3 carbon chain is more so 3 is more stable than 5.
In 4,2,1 we see H effect so 6,5,4 conjugated hydrogen atoms are there respectively in 4,2,1 so stability order is
4 > 2 > 1 > 3 > 5

We should definitely not treat a radical like a positive charge — even though I will admit that they are both electron-deficient compounds and similar rules apply to both of them. The main difference is that positive charges should always be confined to p-type orbitals while radicals can also occupy $\mathrm{sp}^n$ hybrid orbitals. Also, radicals do benefit from mesomeric stabilisation where applicable; we don’t have to ‘break bonds’.

@Jan Call me kleinlich, but you cannot put a positive charge into an orbital, you cannot confine a positive charge either. However, if an orbital is empty, it will be the one with the least s character.


text:
How to get anhydrous triflic acid from triflic anhydride?
I require a small amount of triflic acid, and while other research groups do not have any, I did locate some trflic anhydride. Does anyone have a published protocol for obtaining the anhydrous acid from the anhydride?
I tried to distill under a slow stream of nitrogen about 10 mL of triflic anhydride (yellow/brown solution) with about 5 mL of DI H2O. I got a clear colorless fraction coming off with B.P. 66 C, and afterwards a second fraction with a slightly lower boiling point. The boiling point of the free acid is 162C, the anhydride is 82 C, and there is a monohydrate of the free acid with a melting point of 34 C, but I couldn't find the boiling point.
I did not find an entry in Perrin & Armarego Purification 3rd edition, or in my Fieser and Fieser.
I'll run 19F NMR on the 2 distilled fractions I got, but as the BP was so low, I expect not to get the free acid.

...google ....trifluoromethanesulfonic acid CF3SO3H

If I had done that and found the answer, I wouldnt be asking here, would I?

My comment was merely to identify the substance. Anyway, what did you expect to get from that distillation procedure? As your supervisor, I would be frowning at you, badly.

Normally anyhydrides will hydrolyze to the free acids in the presence of water. That's basic undergrad organic chemistry. If you dont know that you really shouldnt be attempting to comment.

Well, "basic undergrad chemistry" seems to have failed you. Hydrolysation of anhydrides if often slow, from my own undergrad and graduate OC experience. And how do you expect to isolate the highest boiling substance in that hydrolysation by distillation? At ambient pressure no less, and nitrogen stream? That's simply bad laboratory practice, sorry. Not talking of a second fraction that comes at lower temperature.

I would try to cleave the anhydride in an excess of water, and afterwards remove H2O in the rotational evaporator. Recrystallise the acid from isopropanol perhaps.


text:
Electronic configuration of chromium(II)

Which is the electron configuration for $\ce{Cr^2+}$
(A) $\mathrm{[Ar]4s^23d^4}$
  (B) $\mathrm{[Ar]4s^23d^2}$
  (C) $\mathrm{[Ar]4s^03d^4}$
  (D) $\mathrm{[Ar]4s^43d^2}$

I referred to the periodic table and got the orbital notation for $\ce{Cr}$ as $$\mathrm{[Ar]4s^23d^4}\tag1$$
So $\ce{Cr^2+}$ would be $\mathrm{[Ar]4s^23d^2}$, or (B). But the teacher counted that wrong and said it was actually (C).

Question: Shouldn't the answer be (B)? Why does $\ce{Cr}$ lose the $\mathrm{s}$ shell first?

Chromium’s standard electron configuration as given by the aufbau principle is actually more of an exception than you may think initially. See for example this question to catch a small glimpse of the truth behind the simplifications.
The aufbau principle tells you to fill in the order $\ce{1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, }\dots$ That order, however, is only valid for neutral atoms (and not always as the linked question implies). As soon as you add charge to an atom, the relative orbital energies shift. The $\ce{3d}$ orbitals, belonging to the M shell, are stabilised more than the $\ce{4s}$ orbital which belongs to the N shell. This is a general rule: charges typically stabilise orbitals belonging to lower shells more than those in higher shells.
Thus, while in neutral chromium the $\ce{4s}$ and the $\ce{3d}$ shell are very close in energy, this balance completely tips towards $\ce{3d}$ (which then has a lower energy) upon oxidation. Therefore, one should always assume an empty $\ce{4s}$ orbital when discussing the electronic configuration of charged d-block elements.

Yup. I agree with this much more. Exchange energy, being a relatively small factor compared to orbital energies, should really always be the last-resort explanation when everything else doesn't work. And I once wrote a bit about the 3d vs 4s matter, so I'll dump this here: http://chemistry.stackexchange.com/a/33310/16683


text:
What makes a good antioxidant?
I am doing a school project on antioxidants and I would like to find out what makes a good antioxidant. I know examples of antioxidants such as Vitamin A, C and E but I was wondering what commonality exists between them in terms of chemical structure and bonding.
I am a novice in chemistry so a simple but informative answer would be appreciated (little to know nomenclature).

Antioxidants are molecules that help pair up electrons in the cell.
The way I see it, antioxidation in the body is all about getting electrons to pair up. dioxygen, $\ce{O2}$ is a molecule with an overall even number of electrons (12), but the two highest energy electrons are in different molecular orbitals, so it is actually a diradical. Species with unpaired electrons, like ground state oxygen, cannot react with the body's even electron count organic molecules. This is why the body must employ metals, which have unpaired electrons in order to reduce oxygen to water to form cellular energy in the mitochondria. On its way to being reduced to water, oxygen can form reactive species which can react with organic biomolecules, like $\ce{O2-}$, $\ce{HO-OH}$, and $\ce{HO.}$ These are known as Reactive Oxygen Species (ROS).
ROS cause oxidative stress, which is a process that damages important biomolecules like lipids and nucleic acids through oxidation. Unstable ROS like peroxides can initiate radical chain reactions, leading to oxidation of important molecules in the cell. Two good examples of ROS damage are lipid peroxidation, which damages the cellular membrane and nucleobase oxidation, which leads to DNA damage. I will show mechanistic examples for both:
Lipid peroxidation:
$\hskip1.8in$ (1)
formation of 8-oxoguanine:
$\hskip0.9in$ (2)
This creates a hydrogen bonding situation where guanine can match with adenine instead of cysteine:
$\hskip1.4in$ (3)
This is an example of base pair mismatch, a common form of DNA damage, potentially leading to genetic mutations.
Because of these possibilities, it is very important for the body to have safety mechanisms to protect itself against these (and other) forms of oxidative damage. There will always be some concentration of ROS in the body's cells, but the body's anti-oxidation systems should help minimize this.
"Antioxidant" is a broad class of chemicals, including both small molecules and peptides. The general idea is that they are molecules that prevent or reverse oxidation of important biomolecules, or that terminate radical chain reactions. In the lipid peroxidation reaction shown above, the final step is propagation of the radical chain reaction. Since radicals always form new radicals when reacting with non-radicals, the chain reaction will only terminate in one of two ways: either the ROS concentration gets soo high that two radicals bind (very bad), or an antioxidant like vitamin E or Vitamin C will "quench" the chain reaction. Vitamin E will "scavenge" ROS in a process coupled with reduction by ascorbate (vitamin C anion):
$\hskip1.6in$ (4)
Since vitamin E is fat soluble, it can help to protect the lipid bilayer. The semidehydroascorbyl radical that's produced is fairly stable, with a half-life of $10^-5$ s. Now, we have a less reactive radical. Processes like this are known as radical "quenching". two molecules of semidehydroascorbyl radical will quickly disproportionate to dehydroascorbate and ascorbate. This is a process where two odd numbers of electrons turn into two even numbers. 1 + 1 -> 0 + 2
Another sort of antioxidant is the sort that dimerizes. Because one electron needs to be donated to each radical to form a non-radical, one way of removing them is to convert two radicals and two non-radicals into three non-radicals. This is a process where two odd and two even numbers of electrons turn into three even numbers. 1 + 1 + 2 + 2 -> 2 + 2 + 2
Glutathione is probably the most important antioxidant in the body's anti-ROS arsenal. Glutathione can convert a molecule of of $\ce{HO-OH}$ to water in a processes catalyzed by the enzyme glutathione peroxidase. A peroxide is not a radical, but it's a high energy ROS which can easily form two $\ce{HO.}$. The overall equation for the reaction is as follows:
$$\ce{2GSH + H2O2 → GS–SG + 2H2O}$$
Glutathione can also turn the dehydroascorbate we mentioned previously back into ascorbate (vitamin C) through catalysis by the enzyme dehydroascorbate reductase. All oxidized (dimerized to GS-SG) glutathione is finally converted back to reduced glutathione by glutathione reductase.
Depending on your level of knowledge of chemistry and biochemistry, some of the specifics here may be difficult to understand, but the key is simple: The commonality between all antioxidants is that they are able to convert odd numbers of electrons into even numbers of electrons in processes like these:
1 + 1 -> 0 + 2 or 1 + 1 + 2 + 2 -> 2 + 2 + 2 and probably many others.
Antioxidants are just molecules that help pair up electrons.
ROS comes into play mostly as exhaust from the mitochondria, the location where oxygen is turned into water. Antioxidants take care of it, and many of them are regulated through the action of glutathione, which is dimerized on oxidation, then split apart again during enzymatic reduction (along with the cofactor NADPH). Through a system of enzymes and coupled reactions, the body tries to maintain a low ROS concentration.
(1) Wikipedia: https://en.wikipedia.org/wiki/Antioxidant#/media/File:Lipid_peroxidation.svg
(2) Jena, N. R.; Mishra, P. C. "Mechanisms of Formation of 8-Oxoguanine Due to Reactions of One and Two OH. Radicals and the H2O2 molecule with Guanine: A Quantum Computational Study" J. Phys. Chem. B, 2005, 109, 14205-14218
(3) Wikipedia: https://en.wikipedia.org/wiki/8-Oxoguanine#/media/File:8-oxoG_forming_Hoogsten_base_pair_with_dA.svg
(4)Valko, M.; Izakovic, M.; Mazure, M.; Rhodes, C. J.; Telser, J. "Role of oxygen radicals in DNA damage and cancer incidence" Faculty of food technology, Slovak Technical University, 2004


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Why does NaCl act as a catalyst for the reaction of CuSO4 + H2O + Al?
I am analyzing a reaction between water sulfate copper and aluminum. A chunk of $\ce{Al}$, e.g., aluminum foil. I realize that a layer of $\ce{Al2O3}$ forms on the surface of the $\ce{Al}$, passivizing it, and I realize that the $\ce{NaCl}$ (or rather just the $\ce{Cl^-}$) cleans the aluminum oxide off the surface of the chunk of aluminum, but I don't understand why. I'd appreciate greatly if somebody could offer me an explanation.
Furthermore, (I didn't exactly include this in the topic of my question,) why is it that the reaction looks like this:
$$\ce{6H2O + 3CuSO4 + 5Al \rightarrow 3H2 + Al2(SO4)3 + 3Al(OH)2 + 3Cu}$$
(or something along the lines of that) rather than:
$$\ce{3CuSO4 + 2Al -> Al2(SO4)3 + 3Cu}$$ ?
Something along these lines has to be the case, because the reaction yields gas..

You should split your first equation into two, one where Al reduces water, one where Al reduces Cu. Both reactions happen more or less independently, there is no absolute stochiometric relation between them.

Would that imply that clean Al would react with H2O to produce H2 gas equally as vigorously as when it's reacting with CuSO4 + H2O solution and forming H2 gas, if at the same temperature?
Also, thanks for the suggested edit.

No, that's what I meant by "more or less". I would not remove the first equation, just add a third that describes the evolution of H2 from water + Al. There is surely some interesting interaction between the two parts. Btw. Al-sulfate is soluble in water.

Aluminum resists corrosion in neutral or slightly off-neutral water because of the very insoluble Al2O3 film on the metal. If you break this film, it will corrode the bare metal and reform.
But if you scratch the Al and attach a cathode (a less active metal), you have a galvanic cell, and H2 can be evolved from this cathode as Al dissolves. The simplest example of this is when you put Al in H2O in contact with a drop of mercury. Scratch the Al thru the Hg; the Hg amalgamates with the Al (attaching itself as cathode), allowing H2 to be evolved as Al dissolves elsewhere.
The experiment with CuSO4 is similar: scratch the Al; some Cu will be deposited and act as a cathode and the rest of the Al will eventually dissolve in the H2O. If you are too lazy to scratch the Al (I'm being facetious!), you can add a little Cl- ion to the H2O, which will begin corrosion of the Al, deposition of Cu, galvanic cell production, dissolution of all the Al.
There is an electrochemical example posed as a test question: would it be better to make a ship out of aluminum with copper rivets or out of copper with aluminum rivets? An aluminum boat? Nah! It would corrode in seawater. Copper would survive longer. But NO! The copper boat would have a huge cathode and tiny aluminum anodes (the rivets), which would corrode rapidly and the copper plates would fall apart. On the other hand, an aluminum boat would not last forever in seawater, but the tiny cathodes would limit corrosion current, and the large aluminum anodes would have corrosion distributed over the whole ship, so it would survive longer than the other way. Of course, any intelligent person would use copper rivets on copper plates, and aluminum rivets on aluminum, but it's just to make a point.

Chlorides are known to be an aggressive ion for metal corrosion, this is thought to be because of the ability of chlorides to destabilise the passivation film on metals which protect the metals from corrosion.  What is known as pitting corrosion can occur, where a breakdown of the passivation layer at specific points of the metal surface leads to corrosion there by other oxidising agents in the aqueous medium, leading to the formation of corrosion pits.  Exposed metal surfaces due to the breakdown of the passivation layer  releases metal ions, which complex with the chloride anions; the metal-chloride complex reacts with water to form hydrochloric acid and a metal-hydroxide complex, reducing the pH of the medium close to the metal surface, further accelerating corrosion. (A reference can be found here: http://sassda.co.za/stainless-
steel-and-corrosion/).
Aluminium is an "amphoteric" metal which reacts with alkalis; water first oxidises Al to Al (III), producing hydrogen gas (probably the effervescence that you observe) due to the reduction of water.  Al (III) ions then complex with hydroxide ions to form a soluble complex in solution.  In fact, when an alloy of Cu and Al is placed in NaOH, the Al is seen to dissolve away, leaving the copper behind in a nanoporous structure, a material that has potential for catalytic applications.
As for your question as to whether Al can react as vigorously with water as copper (II) sulfate solution, I think yes.  But if your Al isn't as "clean" as you think and still has a layer of passivation on the surface, sulfate ions are known to also destabilise passivation layers of metals so that might enhance the corrosion (in fact some sources claim sulfate to be a better corrosive agent than chloride).
Sorry that I don't have any references, most of this is based on literature that I have read quite a while ago.  Also, I am just a beginner learner in chemistry and corrosion science, so please help to improve my answer and correct any mistakes!

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


text:
Can sonochemistry be used as a catalyst for basic reactions?
Recently, I have been interested in the field of sonochemistry, using ultrasonic irradiation to induce chemical changes in a system via the phenomena of acoustic cavitation.
From my rather brief research, I have come to the realization that this phenomena can be used as a catalyst for many reactions. However, I want to know if the most basic form of sonochemistry (using an ultrasonic bath) can be used to speed up very simple inorganic reactions involving dissolved common chemicals.
I am very interested in the implications of this emerging field in chemistry, but am curious as to whether I can perform a few basic experiments using common chemicals and an ultrasonic bath and compare reaction times for a high school science project.

You migth find the  Sonocatalysis  chapter  from the Handbook of Heterogeneous Catalysis interesting.


text:
Hydrolysis of starch by means of amylase to unclog bathtub drain
After my son decided to pour a substantial amount of uncooked rice down the bathtub drain, I find myself in the situation of removing it.
After trying some common methods - to no avail - I thought I could resort to chemistry.
Specifically, I thought about dissolving (hydrolysis) the main component of rice (starch) by means of amylase (possibly bought in a shop, not salivating over the drain).
Would this be a good solution to unclog the drain?
Under what conditions should I operate to achieve the result hoped?
Would there be a better solution?
As you might guess, I'm not a chemist.

The problem you'd face using any chemical to dissolve the rice is that rice forms a thick gelatinous surface coating as it dissolves, so that it would take an extremely long time for the amylase to penetrate... perhaps months or years. You'd be better off using some of the standard mechanical drain-cleaning techniques:

Disassembly of the sink trap (and drain, if needed) and cleaning or replacing it
Drain auger
Drain snake
Water-pressure drain cleaner

Note that augers and snakes can scratch sinks or toilet bowls, and that water-pressure cleaners can cause pipes to come apart.
N.B. If you've already poured a potentially dangerous cleaner into the drain, remove it safely before trying anything else!

I'll emphasize the point about safety. The various liquid drain cleaners usually contain strong lye compounds and they are extremely caustic. Getting them on your skin or in you eyes would be very very bad.

Thanks for listing other techniques: no 1 is not really feasible unfortunately (my preferred if we were talking about a basin sink) and I'm not super keen on option 4 for the reason you have already described. I guess I'll go with option 2/3 and give up the amylase plan for the time being!


text:
What is a virtual state?
In talking about Raman spectroscopy, one finds the Stokes line is simply the difference between the energy of an incoming photon and an emitted photon. This energy corresponds to a vibrational transition in terms of energy. Yet, the light sent at the molecule is usually much higher energy than that vibrational transition (or else we would just be doing IR).
The explanation I've heard is that the state which holds all that energy in between the ground vibrational state and decaying back to the first vibrationally excited state is a virtual state.
From doing some reading, I have gathered that a virtual state is called virtual not because it's imaginary, but because it's not an eigenfunction of the hamiltonian of the system.
Where does this state come from? I've heard that it drops out of some perturbation theory. Because Raman is intricately connected with polarizabilities, I imagine this might be perturbations to the time-dependent Schroedinger equation.
So, to put some actual questions out there:
What is meant by "virtual state"? In what sense is it virtual?
Can anyone show me the perturbation theory where this comes from? Even if it is just a reference, that would be greatly appreciated.
I have also seen this is somehow connected with Feshbach Resonance which I know nothing about, so an answer which addresses that point would be a good one.

Also take a look at this almost identical question on Physics.SE.

In most Raman experiments, the incident radiation is not near (or at) an absorbing wavelength, and so you will never access a real, honest-to-God excited state (stationary state). (If it was close to an absorbing wavelength, we would be taking about a resonance Raman experiment.)
Now obviously something is going on between the molecule and the incident light. What you are really doing is preparing a superposition state. A superposition of real excited states, or stationary states, of your Hamiltonian. This is what we call a "virtual state," and it is not an eigenstate of your Hamiltonian, so its energy is undefined. (You are free to compute an expectation value, though.) This virtual state doesn't last forever, and, when it decays, scatters radiation away in some direction (it doesn't matter where, unless you are doing an angle-resolved Raman experiment).
(By the way, if you look closely at the equation porphyrin gave, the sum over $v$ virtual states is actually a sum over real $v$ excited stationary states, such that $E_v$ is defined. Otherwise that expression is nonsense. This equation essentially is the sum of overlaps of the initial state into the states that make up your virtual state, and then multiplied by the sum of overlaps of the excited states that make up the virtual state into the final state. This is the Kramers-Heisenberg-Dirac expression for Raman amplitudes.)
If the superposition collapses back to the ground state, the scattered radiation will have the same wavelength as the incident light, and we have Rayleigh scattering. If the superposition collapses to an excited vibrational states, we have Stokes Raman. If your initial state was vibrationally excited and you collapse to a lower vibrational state, you get anti-Stokes Raman. In the actual experiment you will observe all types and can see a Rayleigh line in the spectra. Most spectrometers will filter this wavelength out, however, leaving just the Raman peaks. Usually the virtual state is dominated by the ground state and most of the scattered light is Rayleigh. This is why Raman spectroscopy was too weak to be useful until we had better optical components.
Obviously this isn't the only way to think about Raman. You can use path integrals and/or QED if you want, which is much closer to Wildcat's answer. In fact, I think you must use QED if you want to work out virtual state lifetimes. Here is a good reference if you want to dig a bit deeper:

Williams, Stewart O., and Dan G. Imre. "Raman spectroscopy:
  time-dependent pictures." The Journal of Physical Chemistry 92, no. 12
  (1988): 3363-3374.

Ah. This helps very much to clarify some of the other answers. Just out of curiosity, are there an infinite number of possible virtual states which can be formed as a superposition of two eigenstates? I would assume for a single frequency of incident light the system will enter the same superposition state each time, but if I change the frequency of light, does the system enter a different virtual state or are the virtual states also quantized and thus only possible for certain frequencies of light?

Also that article was extremely helpful and more or less answered the question in my comment above. Thanks for the reference.

Ordinary absorption and emission processes depend on the matrix element of the electric dipole moment operator. In Raman transitions it is the matrix elements of the molecular polarisability between the initial and final states that are important. This polarisability difference can be considered to be the 'virtual state'. The Raman scattering probability involve terms from a second order perturbation expansion of the interaction between the molecule and the radiation field which have the form $$\sum_v\frac{\langle f|\mu\cdot \epsilon|v \rangle  \langle v|\mu\cdot \epsilon|i \rangle f}{h\nu_{ex}-(E_v-E_f)}$$
where $\epsilon$ is the electric field of the lightwave and $v$ a 'virtual' state. However, transitions to and from these do not occur but are the means by which the quantum formalism takes place. This is somewhat unsatisfactory because to evaluate these integrals something must be known about the wavefunctions $\nu$ which is that the wavefunction will be proportional to the symmetry species of the vibration involved in the point group of the molecules.  In a Raman process only the initial and final states have any population.
There are several different types of Raman processes; spontaneous, stimulated, coherent, resonant, non-resonant and methods such as CARS & CSRS. see Mukamel 'Nonlinear Optical Spectroscopy' ; Yariv 'Quantum Electronics'; Marcuse 'Engineering Quantum Electrodynamics'; Herzberg 'Infrared & Raman Spectra'

I think this makes a decent deal of sense, especially keeping in mind the fact this is a scattering process when I'm more used to thinking about absorption and emission processes. What, however, does the $E_v$ term mean? I get that it's the energy of the virtual state, but I don't understand where this would come from? That is, is this energy the eigenvalue of any particular operator? The perturbed Hamiltonian I suppose. If that's true, is there not a corresponding eigenstate? That makes the state seem more real than virtual, but simply with a low probability of being occupied.

In exact state theory, $\hat{H}\left|v\right> = E_v\left|v\right>$. The Hamiltonian is the unperturbed one, the trick in response theory is to express perturbed states as combinations of unperturbed excited states.

I could well be totally wrong here, but I think that the virtual state is in fact the tail of a real excited state which has a much higher energy than that of the photon (in non-resonant scattering). The excited state is (a) never sensibly populated and (b) it is not a stationary state so has a finite lifetime. Thus the width of this state (in energy) gives the scattering some very , very small probability of occurring when the photon is non-resonant.

Could one postulate the virtual state to be related to the Heisenberg Uncertainty principle? That as long as the duration of the excitation is short enough, the large energy uncertainty could accommodate a range of virtual states? Not sure if this makes sense though because then scattering could occur for any arbitrary incident frequency.

It looks simply as mind games: in principle, we are free to explain any observed phenomenon any way we like unless our explanation do not contradict well-established foundations of physics. For instance, in the case of Raman spectroscopy we observe changes in rovibrational states of molecules, but the problem is that the incident radiation is not resonant with any real state, so the corresponding direct transitions are not possible.
To sidestep the problem, we could explain Raman scattering as follows:

Virtual excitation: first, a photon is thought as of being absorbed by a molecule exciting it from a rovibrational state of some electronic state to a shortly-lived intermediate state called the virtual state.
Virtual de-excitation: then, a photon is thought as of being emitted by a molecule causing de-excitation from the virtual state back to a rovibrational state of the same electronic state.
Both these processes are not real physical processes, it is all just a mental construct. And if the wavelength of the emitted photon is different from the incident one (Raman scattering), then, in order to conserve the energy, some change in rovibrational state of the molecule is though of as taking place in the course of these virtual processes.

The above described mental scheme relies on the so-called energy-time uncertainty principle, due to which a photon can be (though of as being) absorbed without conservation of energy as long as it is (though of as being) emitted a short time after.
Now, the only problem is where do we get these virtual states mathematically to do some predictions? And this is where perturbation theory come into the picture: virtual states are constructed by perturbation theory treating the incident electromagnetic wave as time-dependent perturbation.

you are not correct when you write that  a virtual excitation causes 'absorption'  & there is a 'short-lived intermediate' state as this also implies  absorption leading to formation of a state. Raman is scattering; the excitation photon is destroyed and a vibrational quantum and a new photon from the vacuum state is produced (not emitted by the molecule). No time is involved. The change in electronic polarisability is the origin of the scattering.

@porphyrin, but I did not say that this virtual excitation actually happens, just that things can be understood in that way. All this is virtual, as I mentioned few times in my answer. Scattering is real, the mathematical machinery involving virtual state to describe it is not.

yes I understand what you are aiming at  but I thought that saying ' is thought of a being absorbed' etc. could be misleading. There are physical processes involved via the polarisability. The virtual state is in effect the polarisability of some upper electronic state in the molecule whose energy is far removed from the Raman excitation frequency.


text:
What classifies as a solvent/base in SN1, SN2, E1, and E2 reactions?
I need to clarify some terminology for an intro to organic chem class. We are studying substitution and elimination reactions and it has not been made clear what exactly these terms describe. For instance, in the bottom example of this image I have seen CH3OH on both the top and bottom of the arrow. Does this imply that it is the nucleophile/base when it is on top, and a solvent when it is on the bottom?

Your assignment (top/bottom) seems to work in the first and the case, but breaks in the others.

The "main" reagent is usually put on top of the arrow. However, this doesnt mean that you should disregard solvents as these may favor some reagents, create/prevent side products. If it is written, you should consider it.

Okay in the question i am only seeing CH30H that simply means that methanol is acting both as a base and a solvent.

Since methanol is acting both as a solvent and a reagent, it will undergo E1 reaction as you will know that it is a polar protic solvent and if the solvent was like **water(polar) or DMSO **  then it would follow SN1 and SN2 respectively and *CH30 (-) * would act as the nucleophile.  Thus solvents play an important role in understanding the type of reaction it would undergo.  So it is really important to understand the solvent. As per as i know there are no such conventions..( That  solvent is written below and reagent/base at the top).  If only one reagent  is given it is acting as a both solvent and a base. I would like to tell another thing that sometime we are mistaken a reagent to be a solvent :
DMSO is often used as a oxidizing agent in presence of phosphoric acid.  We mostly confuse it and hold phosphoric acid as the oxidizing agent.

Please take a second to read our meta post on formatting.

@ringo Thank you so much, its really nice that you gave me such a useful link.


text:
What is the difference between these wave functions?
My chemistry textbook introduces the wave function as

$$\psi(x)= A \sin\left(\frac{2\pi x}{\lambda}\right)$$ Therefore, the Schrödinger Equation is:
$$\frac{d^2\psi(x)}{dx^2} = -~\left(\frac{2\pi}{\lambda}\right)^{2}\psi(x)$$
and by multiplying each side by $\frac{-h^2}{8\pi^2m}$ you get:
$$-~\frac{h^2}{8\pi^2m} \frac{d^2\psi(x)}{dx^2} = \frac{p^2}{2m}\psi(x) = (T)(\psi(x))$$ where $T$ is kinetic energy.

Now, I don't fully understand that last bit, and I am willing to see an explanation on it, but my main question is that, when the book transitions from discussing the Schrödinger Equation in the context of a "particle-in-a-box" to the context of a one-electron atom, it seemingly changes the wave function to
$$\psi_{(n,\ell,m)}(r,\theta,\phi) = R_{n,\ell}(r) Y_{\ell,m}(\theta,\phi)$$
where n = principle quantum number, $\ell$ = angular momentum number, m = magnetic quantum number, r = radius (distance between nucleus and electron), R = the "radial part" of the equation, Y = the "angular part" of the equation, and $\theta$ and $\phi$ are some angles that I still don't know how to get.
I guess my question is did the wave function $\psi$ change? Why is suddenly so different and why do r,$\theta$ and $\psi$ suddenly come into the picture? Is it because you are going from the one-dimensional, Cartesian "x" version of the wave function to the three-dimensional, polar coordinate version of it?

Don't worry too much about the wave function at this point. I think what they did in the third equation is the application of the momentum operator. Operators are a fundamental concept of quantum mechanics. So be sure that you are familiar with them.

The third equation is, as you mention, the wave function of an electron in an atom.

This wave function, is different to the particle in the box wave function since the potentials of the two systems are different. They are two different systems! As such, the variables that need to be considered are different (hence why the form of the wave functions are different)

I think your textbook is trying to prove to you that some waves are eigenfunctions of the momentum operator in the first equations.

Wave function is still just a function. Yes, it is just that we switched to a 3D version, and also to polar coordinates, and then turned on the Coulomb potential, so the function looks pretty different now. Other than that, everything is the same.

Your book seems to approach things backwards. The Schrodinger eqn says that kinetic energy + potential energy = total energy E, as one might expect. The eqn is  $-(\hbar^2/(2m))d^2\psi/dx^2 +V(x)\psi = E\psi$, where wavefunction is $\psi$.  Different systems, particle in box, H atom, vibrations etc have different potential energy V(x) . (The momentum operator is $p=-i\hbar d/dx$, so kinetic energy is $p^2/2m$). To solve any differential eqn. one must specify boundary conditions and in this case also the potential energy.

@IvanNeretin ahh, ok so, kinda like miguel said below, the wave function $\psi$ is just a general function like $f(x)$ and the actual equation for it can change between one and three dimensions, coordinates, and the fact that the potential energy must be considered when considering an electron in an atom?

Yeah, like that.

The first function you have there
$\psi(x)=A\sin(\frac{2\pi x}{\lambda})$,
is very similar to the function of a particle in a monodimensional box.
This function is a very helpful example in order to understand how does the quantum mechanics works.
The other function is
$\psi_{(n,m,l)}(r,\theta,\phi)=R_{n,l}(r) Y_{m,l}(\theta,\phi)$.
This function represents how an electron moves in a hydrogen atom and the full expression is quite different to the first function. Just see the representations:

Ok. The functions do not mean the same concept. So... Why do we use "$\psi$" in both?
In math we use $f(x)$ to speak about any function. In quantum mechanics we use $\psi(x)$ for the same: represent a function easily.
Yeah, but... Why do i have to study the first function if it's just a lie and not the full story?
Remember it's only a very good example. However, the particle in a box function can be used to determine the energy of an electron in a conjugated system as beta-carotene and it's energy for the first excited level. If you subtract the second energy to the first one it gives you the energy for a photon in the visible region: the carrot's orange color.
This means that the particle in a box is a good example if you start to study quantum chemistry and it even works as a good approximation for some systems.
Ok. Too much information but I want to know more: How can I get the second function from the first?
Well, that's... Impossible...
To obtain the functions we use the Hamiltonian operator, which contains the kinetic energy operator and the potential energy operator.
For the particle in a box, we only use the kinetic energy operator because we don't have any potential into the box. This leads to the first function you have.
For the hydrogen atom we must use both operators (all the Hamiltonian) which means to include a function for the potential and we must work in 3D. The potential function can be obtained using the angular momentum. The spherical coordinates makes easier to work in 3D. This leads to radium and angular functions.
Summarizing, we have a different wave function for every different situation. This situation is what determines the function and it's conditions:

The particle in a box → Quantization.
The free particle → No quantization.
The electron in a hydrogen atom → Quantization + quantum numbers + rotation +... → Atomic orbitals.

Ok, seems legit, but it's hard to understand
I recommend you the books "Physical Chemistry" from Atkins, Levine or Engel to understand better the ideas and see the complete development.
In case you hate textbooks you can visit hyperphysics web page for a general looking.

thank you! so the first wave function (the one for the particle in the box) is basically considering other things than the second wave function (the one for the electron in the hydrogen atom) but because the potential energy is 0 inside the box it doesn't matter but in an atom it does, so you have to include that, and then convert to 3 dimensions and then convert to polar coordinates, making the whole thing look very different, right?