text:
Ionic compounds and van der Waals forces
I'm having trouble understanding exactly what intermolecular forces occur in ionic compounds.
What I understand is that ionic compounds are held together by electrostatic interactions between opposite charged ions and that they aren't "finite" in the way covalent molecules are, and therefore they do not have intermolecular forces.
However, while my textbook says that only covalent molecules have van der Waals forces, my professor said that all compounds, even ionic compounds have London dispersion interactions.
Isn't London dispersion a van der Waals force? How does that work?

Generally while studying you tend to talk in absolutes: crystalline solids are fully ionic and molecules are fully covalent. However, often reality is a middle-ground; covalent bonds don't always mean 100% electron sharing (which is why we have the concept of polarized bonds), and ionic bonds don't always mean 100% electron abstraction (which is why we know the concept of polarizable anions and polarizing cations). The interaction between polarized ions is indeed the London force.

London dispersion forces are instantaneous dipole-induced dipole interactions. All compounds can have an instantaneous dipole moment, including ions, so London dispersion forces exist for all compounds.
Unlike molecules, ions are charged. The electrostatic interactions between charges are far stronger than the London dispersion forces between ions, so normally dispersion forces are ignored when considering ions.
Therefore, both your textbook and your professor are correct.

Just to help OP even more: while a covalent crystal relies only on dispersion forces, an ionic one would be hold even if the latter forces wouldn't exist.


text:
Acid catalyzed enolization and alpha hydrogen
In acidic catalyzed enolization of butan-2-one:

We have 2 alpha hydrogens available. From the mechanism of acid catalyzed enolization there is a certain carbocation character at the carbonyl carbon because of resonance, and thus I thought that the elimination would be an E1 styled one where the alkene formed is the more substituted one.
But if we look at from an 'acidic hydrogen' perspective the hydrogen on the least substituted end is the most acidic whose extraction will lead to a less substituted alkene in the enol form.
So which enol would be major in the enolization of butan-2-one under acidic conditions? Would the same enol be major in basic medium?

In the acidic conditions, first the oxygen atom of $\ce{C=O}$ gets protonated by taking  $\ce{H^+}$ from  $\ce{H3O^+}$ and after that water molecule abstracts proton from the $\alpha$ - carbon to from the enol in which there is  $\ce{C=C}$ double bond. So, definitely more substituted enol will be the major one, as the proton abstraction factor doesn't play a key role in acidic conditions, but in basic conditions it does.

@SoumikDas But as per http://research.cm.utexas.edu/nbauld/teach/ch610bnotes/ch19.htm in both acidic and basic medium proton abstraction is RDS.

I agree that the proton abstraction step is the rds, so the less substituted enol will be kinetically favoured, as it's easier, whereas the more substituted enol will be the thermodynamically controlled product. If you let your reaction run for much longer time, you will mostly get the more substituted one.

@SoumikDas Why is the kinetic product favoured over the thermodynamic product in  basic medium?

@YUSUFHASAN It is not E2. You need a considerably strong base incase of E2. Even aqueous KOH fails to do E2. But some sources write the proton abstraction in a concerted fashion(but they do show carbocation resonance structure).

@YUSUFHASAN I feel that it is E1, but the carbocation never rearranges because of backbonding with oxygen.

@AkshatJoshi Yes,my bad. You are correct about both the mechanism and the explanation for the acidic medium. In basic it is still E1CB, and that's why you end up with the less stable alkene

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text:
Is there a stable and non-toxic hydro-nitrogen-oxygen compound that's liquid in room temperature?
Is there a stable (i.e. does not turn into other substance when stored in a bottle exposed to the elements  or explode/combust spontaneously or under mechanical impact) compound that contains just $\mathrm{N}$, $\mathrm{H}$ and $\mathrm{O}$ but no $\mathrm{C}$ and is (a) either liquid or can exist as a high concentration ($80\%+$) water solution at the range of $-50\,{}^\circ\mathrm{C}\sim+70\,{}^\circ\mathrm{C}$ and (b) preferably non-toxic (c) combustible, but this seems to be usually implicit from its composition?
(Aside from $\mathrm{N}\mathrm{H}_3$, $\mathrm{N}_2\mathrm{H}_4$, $\mathrm{H}\mathrm{N}\mathrm{O}_3$, or $\mathrm{N}\mathrm{H}_4\mathrm{N}\mathrm{O}_3$ that is.)

Why there is an [organic-chemistry] tag if you don't want to look into C-containing compounds?! Also, since there is [fuel] tag, does it mean it also should be a high-energy material?

@andselisk the high-energy part is not a must.

How about ammonium nitrate water solution or just nitric acid?

@andselisk That seems to be explosive? The other one is unstable stored too.

Hydroxylamine in water - not explosive but there toxicity issues

Ammonium nitrate is not an explosive on its own, it's a component of explosives. Nitric acid is unstable in a sense that concentrated solution is light-sensitive and is a strong oxidant. I'm afraid that's the best you can get. Probably it would be beneficial if you share what do you need the stuff for.

@andselisk I'm just curious if there could exist a non-lh2, non-hydrazine jet/rocket fuel that produces absolutely no soot whatsoever. Energy density doesn't matter as long as it is not orders of magnitudes lower than kerosene.

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text:
Oil + UV radiation = cholecalciferol?
According to a web site about health (Antranik):

The oil that naturally comes from our skin is a type of cholesterol.
  Whenever we stand under direct sunlight, the ultraviolet rays react
  with this oil.  The ultraviolet radiation breaks some chemical bonds
  in the oil so that it forms a different molecule.  This molecule is
  known as the infamous Vitamin D3 which is why it’s also called the
  “sunshine vitamin.”

According to the web site, cats lick this converted oil off their fur to get Vitamin D (cholecalciferol).
What exactly is the reaction here? How could UV radiation be converting oil into a hormone? I thought hormones were much more complex molecules than oils, not simpler molecules.

The wikipedia page on Cholecalciferol gives a more scientific and accurate  account of this process.

@porphyrin The article does not completely describe the reaction, but only outlines it in general terms. Also, the article says it happens within the skin, but for a cat, according to the article quoted, it does not happen on the skin but on the fur of the cat.

Estherols such as cholesterol aren't simple oils. They are immersed in an oil matrix, yes, but the molecules are completely different; check out the structures of triglycerids vs. cholesterol in Google Images or similar. However the difference between cholesterol and cholecalciferol is smaller; cholesterol is transformed to 7-dehydrocholesterol in the body, and then the sun breaks this intermediate in a single step to produce vitamin D3. Note that this isn't the only way to get vitamin D; you can also eat vitamin D2, manufactured industrially from vegetables.


text:
Methane T2 SALCs
So I have a question on the form of the $T_2$ SALCs of methane. Below I show the $T_d$ character table and the accompanying reducible representation for the sigma framework (using, in the case of methane, the $4$ $1s$ H orbitals).

It is easy to show, and many people have, that the irreducible representations contained therein are $A_1$ and $T_2$.
If we then perform the projection operation for all of the elements of $T_2$, I get the following:

(The X's are intended to be "Chi's" here, signifying the $4$ $1s$ H orbitals that go into the procedure).
Meaning that, if we take the first line for instance, and perform the sum in the projection operator, we get:
$$
3\chi_1-\chi_2-\chi_3-\chi_4
$$
As our first un-normalized $\phi_t$ SALC. I can easily see by intuition how we can obtain the (again, not normalized) known SALCs for a $\sigma$ framework tetrahedron from this:
$$
3\chi_1-\chi_2-\chi_3-\chi_4 =
$$
$$
\chi_1-\chi_2+\chi_3-\chi_4
$$
$$
\chi_1+\chi_2-\chi_3-\chi_4
$$
$$
\chi_1-\chi_2-\chi_3+\chi_4
$$
What I do not understand, is how we can derive this result without intuition. If I perform any number of Schmidt orthoganalizations using the other 3 rows of the projection operator table, I always end up with an exceedingly strange SALC, regardless of normalization and inclusion of overlap integrals $\chi_i{\times}\chi_j = S_{ij}$.
Is there any mathematical tool that can be used to generate these known SALCs from the one above?
Again, let me stress, I can clearly see the intuitive path, but I always rest easier when I know there is a reproducible, rigorous theory that leads there.

This is worked out in David Bishop's book "Group Theory and Chemistry" on p. 238 and 239, which you might be able to view in Google Books. It's a bit tedious to write out. The gist is to set up each of the combinations you have (eg 3x1-x2-x3-x4) equal to a linear combination of the three p orbitals (since they are the T2 AO's). Then use the linearity of the symmetry operators to write equations to solve for px, py and pz in terms of x1, x2, x3 and x4. The solutions will be the three combinations you figured out intuitively. If you can't find it, I can type it out at some point.

Thank you Andrew! I've got that book nearby. I've noticed that accepting my own answers when others have provided citations is frowned upon, would you like to write either this, or the actual solution as an answer so I can accept it? Or would you rather I do so?

I can write something up. In case the page numbers in my version are different, look for equation 11-5.13 in the section "hybrid orbitals". That's the final answer you're looking for, and its derivation is the preceding few pages.

I don't think there's any stigma against self-answering based on resources suggested by others, as long as you give appropriate credit (which isn't so much of an intellectual property thing, but more of a courtesy, I think).

UPDATE: added more detail for those without access to the referenced text.
As noted in the comments, the details of this process can be found in "Group Theory and Chemistry" by David M. Bishop (Courier Corporation, 1993) p. 238-9. Here's a summary:
Following the same process that you started, you can get two other projections as (using your notation):
$-\chi_1+3\chi_2−\chi_3−\chi_4$
and
$-\chi_1−\chi_2+3\chi_3−\chi_4$
All three of these normalize with a factor of $\frac{1}{\sqrt{12}}$.
The challenge with degenerate orbitals is that there is an infinite number of linearly independent combinations that will combine to give the desired $T_2$ symmetry properties. Schmidt orthogonalization can be used to generate an orthogonal set, but there is an infinite number of those as well.
To prove that the canonical set is valid, one need only show that all three are linear combinations of the above projections and that they are orthogonal to each other. But to derive the canonical set without prior knowledge, we use the procedure below.
We want a set that is orthonormal and matches up with the central atomic orbitals. That is, each has symmetry that matches a member of the $p$ orbital set of the central atom. First, we set the $t_2$ central atom orbitals as $p_x$, $p_y$ and $p_z$. Then we consider that we must be able to make the hybrid orbital that coincides with each ligand bond from a linear combination of the $s$ and $p$ atomic orbitals. This is essentially a statement of the principle underlying the concept of hybridization. Furthermore, since the $s$ orbital only adds magnitude, not direction, an unnormalized hyrbid orbital can be made from a linear combination of $p$ orbitals only. Finally, any member of the $t_2$ ligand group orbitals must similarly be able to be constructed from a linear combination of the central atom $p$ orbitals.
Thus, $(3\chi_1-\chi_2−\chi_3−\chi_4)/\sqrt{12} = (a_1^2+b_1^2+c_1^2)^{-1/2}(a_1p_x+b_1p_y+c_1p_z)$ for some $a_1, b_1, c_1$ and likewise for the other two projections (adding a minus sign where necessary for directionality with respect to the positive lobes of the $p$ orbitals).
We can then take advantage of the linearity of the symmetry operators to set up equations and solve for $p_x$, $p_y$, and $p_z$ in terms of $\chi_1$, $\chi_2$, $\chi_3$ and $\chi_4$, and the coefficients of those solutions are the coefficients for the H1s orbitals in each of the three $t_2$ ligand group orbitals.
For example, let $\chi_1$ be the hybrid orbital oriented in the positive $x,y$ and $z$ octant of Cartesian space, and let the $C_{3a}$ axis run through it. Thus,
$O_{C3a}(\chi_1)=\chi_1$
$O_{C3a}(\chi_2)=\chi_3$
$O_{C3a}(\chi_3)=\chi_4$
$O_{C3a}(\chi_4)=\chi_2$
Because the operator is linear, we now have that
$O_{C3a}[(3\chi_1-\chi_2−\chi_3−\chi_4)/\sqrt{12}]=(3\chi_1-\chi_2−\chi_3−\chi_4)/\sqrt{12}$,
i.e. this projection is unchanged by that transformation because $\chi_2$, $\chi_3$, and $\chi_4$ are equivalent in the expression.
Therefore, we can further state that
$O_{C3a}[(a_1^2+b_1^2+c_1^2)^{-1/2}(a_1p_x+b_1p_y+c_1p_z)]=(a_1^2+b_1^2+c_1^2)^{-1/2}(a_1p_x+b_1p_y+c_1p_z)$
We can also determine by inspection that $O_{C3a}(p_x)=p_z$,$O_{C3a}(p_y)=p_x$, and $O_{C3a}(p_z)=p_y$. Therefore, we also have that
$O_{C3a}[(a_1^2+b_1^2+c_1^2)^{-1/2}(a_1p_x+b_1p_y+c_1p_z)]=(a_1^2+b_1^2+c_1^2)^{-1/2}(a_1p_z+b_1p_x+c_1p_x)$
and we can conclude that $a_1=b_1=c_1$. Thus,
$(3\chi_1-\chi_2−\chi_3−\chi_4)/\sqrt{12}=(p_x+p_y+p_z)/\sqrt{3}$.
Repeating this for the other symmetry transformations (for example $C_{3b}$ and $C_{3c}$) gives a system of equations that can be solved to yield:
$p_x=\frac12 (\chi_1−\chi_2+\chi_3−\chi_4)$
$p_y=\frac12 (\chi_1−\chi_2-\chi_3+\chi_4)$
$p_z=\frac12 (\chi_1+\chi_2-\chi_3−\chi_4)$
The coefficients of $\chi_1$, $\chi_2$, $\chi_3$ and $\chi_4$ in these three expressions are the coefficients of the H1s orbitals to each canonical $t_2$ ligand group orbital.

A clarifying question: If the ligand orbitals are offset from the central atom in a tetrahedron, would it not be impossible to combine the central atomic p orbitals in such a way as to be equivalent the SALCs?
That seems to be the content of the equation with $a_1$, $b_1$, and $c_1$ coefficients.
Put another way, because there is some non-perfect overlap between the ligand SALCs and the atomic orbitals, does this not imply that they cannot be made exactly equivalent?
Why does the requirement that the SALCs and the AOs coincide imply an algebraic equality instead of just equal symmetry?

I believe it works because the hybrid bonding orbital and the ligand orbital are both treated as a unit length position vector extending from the origin (ie the central atom) in the direction of the ligand. Conceptually, the ligand is a point at the end of the vector and the bond is the vector itself, but mathematically they are the same object. Does that make any sense?

Not entirely, but I'll keep digging. Thanks so much for your response!

I fixed some obvious typos, hope you don't mind. I think the equality symbols in the last $4$ equations might better be replaced by a symbol that means 'transforms like'. Also it's not clear to me from the discussion where the $\chi_2$, $\chi_3$, and $\chi_4$ orbitals lie, although from the last $3$ equations they seem to be in the $\langle-1,-1,1\rangle$, $\langle1,-1,-1\rangle$, and $\langle-1,1,-1\rangle$ directions respectively. Also I think $O_{C3a}(p_x)=p_y$ and so on: you seem to have written the operator's inverse.

This is an addendum to @Andrew 's answer which I will still keep as the correct one.
I went messing about in Mathematica, and because these SALCs are generated from 1s orbitals with no imaginary component, we can actually check the projection method for the correct production of these SALCs above.
And in fact, if you utilize the Schmidt orthogonalization process as I allude to above on the 3 projections $3\chi_i-\chi_j−\chi_k−\chi_l$, you do get the qualitative form of the expected SALCs (the ones with form $\chi_1−\chi_2+\chi_3−\chi_4$ instead)
The actual form of the Schmidt orthogonalized SALCs is incredibly ugly (especially since I used cartesian coordinates), but they reproduce the following density slices:

This is exactly what Andrew's answer from David Bishop's book leads to as well! The two methods appear to be one and the same!
I short, this is done by inputting
$$
R_{1s} = 2Z^{3/2}*e^{\frac{-2Z \frac{\sqrt{(x+c_x)^2+(y+c_y)^2+(z+c_z)^2})}{n}}{2}}
$$
and
$$
Y_{1s} =\frac{1}{4\pi}^{1/2}
$$
Such that
$$
\chi_1 = R_{1s}*Y_{1s}
$$
Four of these AO's are then placed at the corners of an tetrahedron (using the constant offsets for x, y, and z in the expansion of r), and the combinations thereof which the projection operator indicates are tested.
The first combination is $3\chi_1-\chi_2−\chi_3−\chi_4$ with normalization (divide by $\sqrt{12}$), but the next two require Schmidt orthogonalization w.r.t the first function, ie:
$$
|\chi_2> = |\chi_2> - <\chi_2||\chi_1>|\chi_1>
$$
Do this for $\chi_2$ and $\chi_3$ and then plot the density surfaces. They will be the ones shown here, and completely in keeping with the expected $t_2$ SALCs.

I feel like the hard approach to this problem is the easy one here. First consider the partners of an irreducible representation $\mu$ of a group $G$ of degree $n_{\mu}$ consisting of the $n_{\mu}\times n_{\mu}$ matrices $D^{(\mu)}(R)$ for all $R\in G$:
$$R\phi_j^{\mu}(\vec r)=\phi_j^{(\mu)}(R^{-1}\vec r)=\sum_{k=1}^{n_{\mu}}\phi_k^{(\mu)}(\vec r)D_{kj}^{(\mu)}(R)$$
Where the $\phi_j^{(\mu)}(\vec r)$, $1\le j\le n_{\mu}$ are the partners that engender the irreducible representation $D^{(\mu)}$. Notice that we have to rotate coordinates backwards to rotate functions forwards! This is a tricky point that must always be kept in mind. For example, consider what happens if $\vec r=\langle2.3,1.7,-2.9\rangle$, $\phi_1^{(T_2)}(\vec r)=x$, and $R=C_{3(1\bar1\bar1)}$, which means that if we set our viewpoint at $\langle1,-1,-1\rangle$ and looked towards the origin, functions would be rotated $120°$ counterclockwise about a line containing our viewpoint and the origin. Rotating coordinates backwards, we have $R^{-1}\langle2.3,1.7,-2.9\rangle=\langle-1.7,-2.9,-2.3\rangle$ and picking out the $x$-component we get $R\phi_1^{T_{\mu}}(\vec r)=-1.7$, that is, $Rx=-y$ as it should be.
Now let's make sure that group multiplication works. For any $R,S\in G$
$$\begin{align}RS\phi_j^{(\mu)}(\vec r)&=R\phi_j^{(\mu)}(S^{-1}\vec r)=R\xi(\vec r)=\xi(R^{-1}\vec r)=\phi_j^{(\mu)}\left(S^{-1}(R^{-1}\vec r)\right)=\phi_j^{(\mu)}\left((RS)^{-1}\vec r\right)\\
&=R\sum_{k=1}^{n_{\mu}}\phi_k^{(\mu)}(\vec r)D_{kj}^{(\mu)}(S)=\sum_{\ell=1}^{n_{\mu}}\sum_{k=1}^{n_{\mu}}\phi_{\ell}^{\mu}(\vec r)D_{\ell k}^{(\mu)}(R)D_{\ell j}^{(\mu)}(S)\\
&=\sum_{\ell=1}^{n_{\mu}}\phi_{\ell}^{(\mu)}(\vec r)\left(\sum_{k=1}^{n_{\mu}}D_{lk}^{\mu}(R)D_{kj}^{(\mu)}(S)\right)=\sum_{\ell=1}^{n_{\mu}}\phi_{\ell}^{(\mu)}(\vec r)D_{\ell j}^{(\mu)}(RS)\end{align}$$
And on comparing coefficients of $\phi_{\ell}^{(\mu)}(\vec r)$ we see that the representation matrices $D^{(\mu)}(R)$ satisfy the group homomorphism. In the above we made the substitution $\xi(\vec r)=\phi_j^{(\mu)}(S^{-1}\vec r)$ to make clear where $R^{-1}$ should act and noted that
$$(S^{-1}R^{-1})(RS)=S^{-1}(R^{-1}R)S=S^{-1}ES=S^{-1}S=E$$
so $(RS)^{-1}=S^{-1}R^{-1}$. Now we are ready to show that for any function $\phi(\vec r)$ and $j,k\in(1,n_{\mu})$,
$$\sum_{S\in G}S\phi(\vec r)D_{jk}^{(\mu)}(S^{-1})$$
transforms as the $k^{th}$ partner of the $\mu^{th}$ irreducible representation of $G$: note that $S^{-1}=S^{-1}E=S^{-1}R^{-1}R=(RS)^{-1}R$ so for any $R\in G$
$$\begin{align}R\sum_{S\in G}S\phi(\vec r)D_{jk}^{(\mu)}(S^{-1})&=\sum_{S\in G}RS\phi(\vec r)D_{jk}^{(\mu)}\left((RS)^{-1}R\right)\\
&=\sum_{T\in G}T\phi(\vec r)D_{jk}^{(\mu)}(T^{-1}R)\\
&=\sum_{T\in G}T\phi(\vec r)\sum_{\ell=1}^{n_{\mu}}D_{j\ell}^{(\mu)}(T^{-1})D_{\ell k}^{(\mu)}(R)\\
&=\sum_{\ell=1}^{n_{\mu}}\left(\sum_{S\in G}S\phi(\vec r)D_{j\ell}^{(\mu)}(S^{-1}\right)D_{\ell k}^{(\mu)}(R)\end{align}$$
Where we have used the fact that as $S$ varies over all elements of $G$ so do all of its products $RS$ with and fixed element $R\in G$. The other partners have the same $j$ as above, but different $k$.
Now, let's do this for the $s$-orbital of the $4$ hydrogen atoms, $\psi_1$, $\psi_2$, $\psi_3$, and $\psi_4$ located in the $\langle1,1,1\rangle$, $\langle1,-1,-1\rangle$, $\langle-1,1,-1\rangle$, and $\langle-1,-1,1\rangle$ directions respectively. Let's start with a table of the operators of $T_d$, the representation matrices of $D^{(T_2)}$ engendered by $\{x,y,z\}$, and the transformed $s$-orbitals:
$$\begin{array}{cccccc}E&C_{3(111)}&C_{3(111)}^2&C_{3(1\bar1\bar1)}&C_{3(1\bar1\bar1)}^2&C_{3(\bar11\bar1)}\\
\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}&
\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}&
\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}&
\begin{bmatrix}0&0&-1\\-1&0&0\\0&1&0\end{bmatrix}&
\begin{bmatrix}0&-1&0\\0&0&1\\-1&0&0\end{bmatrix}&
\begin{bmatrix}0&0&1\\-1&0&0\\0&-1&0\end{bmatrix}\\
\psi_1&\psi_1&\psi_1&\psi_4&\psi_3&\psi_2\\
\psi_2&\psi_3&\psi_4&\psi_2&\psi_2&\psi_4\\
\psi_3&\psi_4&\psi_2&\psi_1&\psi_4&\psi_3\\
\psi_4&\psi_2&\psi_3&\psi_3&\psi_1&\psi_1\\
\hline
\end{array}$$
$$\begin{array}{cccccc}C_{3(\bar11\bar1)}^2&C_{3(\bar1\bar11)}&C_{3(\bar1\bar11)}^2&C_{2(100)}&C_{2(010)}&C_{2(001)}\\
\begin{bmatrix}0&-1&0\\0&0&-1\\1&0&0\end{bmatrix}&
\begin{bmatrix}0&0&-1\\1&0&0\\0&-1&0\end{bmatrix}&
\begin{bmatrix}0&1&0\\0&0&-1\\-1&0&0\end{bmatrix}&
\begin{bmatrix}1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}&
\begin{bmatrix}-1&0&0\\0&1&0\\0&0&-1\end{bmatrix}&
\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&1\end{bmatrix}\\
\psi_4&\psi_3&\psi_2&\psi_2&\psi_3&\psi_4\\
\psi_1&\psi_1&\psi_3&\psi_1&\psi_4&\psi_3\\
\psi_3&\psi_2&\psi_1&\psi_4&\psi_1&\psi_2\\
\psi_2&\psi_4&\psi_4&\psi_3&\psi_2&\psi_1\\
\hline
\end{array}$$
$$\begin{array}{cccccc}S_{4(100)}&S_{4(100)}^3&S_{4(010)}&S_{4(010)}^3&S_{4(001)}&S_{4(001)}^3\\
\begin{bmatrix}-1&0&0\\0&0&-1\\0&1&0\end{bmatrix}&
\begin{bmatrix}-1&0&0\\0&0&1\\0&-1&0\end{bmatrix}&
\begin{bmatrix}0&0&1\\0&-1&0\\-1&0&0\end{bmatrix}&
\begin{bmatrix}0&0&-1\\0&-1&0\\1&0&0\end{bmatrix}&
\begin{bmatrix}0&-1&0\\1&0&0\\0&0&-1\end{bmatrix}&
\begin{bmatrix}0&1&0\\-1&0&0\\0&0&-1\end{bmatrix}\\
\psi_4&\psi_3&\psi_2&\psi_4&\psi_3&\psi_2\\
\psi_3&\psi_4&\psi_3&\psi_1&\psi_1&\psi_4\\
\psi_1&\psi_2&\psi_4&\psi_2&\psi_4&\psi_1\\
\psi_2&\psi_1&\psi_1&\psi_3&\psi_2&\psi_3\\
\hline
\end{array}$$
$$\begin{array}{cccccc}\sigma_{110}&\sigma_{101}&\sigma_{1\bar10}&\sigma_{10\bar1}&\sigma_{011}&\sigma_{01\bar1}\\
\begin{bmatrix}0&-1&0\\-1&0&0\\0&0&1\end{bmatrix}&
\begin{bmatrix}0&0&-1\\0&1&0\\-1&0&0\end{bmatrix}&
\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}&
\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}&
\begin{bmatrix}1&0&0\\0&0&-1\\0&-1&0\end{bmatrix}&
\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}\\
\psi_4&\psi_3&\psi_1&\psi_1&\psi_2&\psi_1\\
\psi_2&\psi_2&\psi_3&\psi_4&\psi_1&\psi_2\\
\psi_3&\psi_1&\psi_2&\psi_3&\psi_3&\psi_4\\
\psi_1&\psi_4&\psi_4&\psi_2&\psi_4&\psi_3\\
\end{array}$$
Starting with any of the $4$ $s$-orbitals and any of the $9$ matrix elements we could do $36$ projections if we were so disposed, but here $1$ will suffice. I will choose the row $j=1$ and the column $k=2$ and orbital $\psi_3$. The first $2$ columns of the table have $0$ in the $1,2$ position, but the third has a $1$ there. We have to be careful, though, because the formula requires the inverse operation to the matrix, so we select from the $C_{3(111)}$ column of the $\psi_3$ row because $\left(C_{3(111)}\right)^{-1}=C_{3(111)}^2$. Going through the table in this fashion we get
$$\psi_4-\psi_1-\psi_3+\psi_2-\psi_1+\psi_4-\psi_3+\psi_2$$
Normalize to $\phi_2^{(T_2)}=\frac12\left(-\psi_1+\psi_2-\psi_3+\psi_4\right)$ transforms as $y$.
So you can see that you can directly project to partners using the matrices of the irreducible representation rather than projecting through traces and subsequently resolving the partners. Just use the matrix elements rather than the traces in the similar formula.

Thanks for the corrections above. A question on your method, though: by choosing j=1, k=2 and phi3, aren't you choosing the projections that give the desired result? If we know beforehand what we want, we can also manipulate the projections resulting from the traditional projection method (which gives the +3, -1,-1,-1 combinations) to show that the canonical orbitals are linear combinations thereof, but if we imagine not knowing what the answer is ahead of time, how would you know in your method which projection would give the orbital that transforms as $y$?

@Andrew, I chose $j=1$, $k=2$, and $\phi_3$ more or less at random. Choosing $k=1$ would produce the partner that transforms as $x$ or $k=3$ as $z$. Choosing a different $j\ne1$ might produce a result with a different phase or normalization or even a partner from a different set or zero (although not in this case). For a more interesting example of what this method can achieve, try $$\phi(\vec r)=x+xz+2z^3-3x^2z-3y^2z$$ with all $9$ possible projections. You should find that it picks of any $j^{th}$ partner and multiplies by $|G|/n_{\mu}$ and transfers to the $k^{th}$ partner.

So after reading your answer it seems you are just cutting to the chase and using what Cotton would call the "full" projection operator, instead of what I did (using the incomplete projection operator, as in summing over the trace).Thank you for calling attention to the distinction.

I was thinking more in terms of Melvin Lax, Symmetry Principles in Solid State and Molecular Physics, John Wiley and Sons, New York 1974, pp. 99-101. This seems to go into more detail than F. A. Cotton, Chemical Applications of Group Theory, John Wiley and Sons, New York 1963. When you want a partner it seems to make sense to project directly to a partner. Of course given one partner here it's easy to get to the others because $C_{3(111)}y=z$ and $C_{3(111)}z=x$.


text:
Confusion in the name of this organic compound

I think the name of the above compound should be 3-Ethyl-2-methylcyclohex-1-ene. However, according to my book (Organic Chemistry Resonance DLPD, page #43, question #5), this organic compound is 6-Ethyl-1-methylcyclohex-1-ene.
A quick Google search shows relevant results for the later name but not for the former.
As far as I know as a high school student, substituents should be given the lowest number and double bonds should be given the number 1 whenever possible.
The name I thought of first satisfies these conditions better than the name the book gives as the answer.
What did I do wrong? Which rule did I not follow?

we prefer rule for substituted cyclohexene then lowest set must be occupied by substituted alkene so answer will be 6 ethyl 1 methyl cyclohexene

From here:
"Number the carbon atoms of the double bond as “1” and “2” in the direction that
gives the substituent encountered first the smaller number."
The methyl group right on the double bond is deemed first, so the carbon at that point on the ring gets 1 and the other double-bonded carbon gets 2, forcing 6 for the ethyl substituted carbon.

The rules for naming state that if all groups have the same priority (which all alkyl groups and alkenes are), then you should assign numbering such that the lowest numbers are given. This isn't a sum of numbers; rather, it's the maximum possible groups with the lowest numbers. That might be hard to understand, but the idea is that if there's a chance to make 2 groups have the lowest possible number (1), then it would be preferred to anything that doesn't have that feature. The textbook answer has two groups with a "1" location (both the double bond and the methyl group attached to it), whereas your proposed answer doesn't. Therefore, it is preferred over your answer.


text:
How much drinking water can I get back
If I was to boil a gallon of sea water and collect the condensation, how much drinking water would I get back?

It's sort of a loaded question (actually, not clear enough) since it doesn't specify temperature or location.

@TryHard I guess the "location" is sea level, and that pretty much also defines the temperature, about 100°C. ;-)

I am going to interpret this as a question asking for the maximum theoretical water recoverable, and not a reasonable expectation for a distillation apparatus (which would vary from apparatus from to apparatus).
According to Encyclopedia Britannica, seawater is 96.5% water by mass. In addition, the density of seawater is around 1.030g/mL.
There are around 3785mL in a gallon, so the mass of a gallon of seawater is 3898g. Multiply that by 0.965, and the mass of water in it is 3762g. Since the density of water is 1.00, you could expect to recover 3.762L of water with a perfect distillation apparatus and 100% recovery.


text:
Inactivation of enzyme: Why inactivation constants do not follow the Arrhenius equation?
The Arrhenius equation may not account for the temperature effect on thermal inactivation rate constants. Why can the thermal inactivation kinetics not be expected to follow Arrhenius? The reason that the inactivation constants do not follow Arrhenius is it because of other chemical processes occuring like diffusion?

Are you asking how it is possible in general to observe this or is the question result of having encountered a specific example (in the literature say) of this being violated or ?

Possible in general

Also denaturization might occur

Without attempting to elaborate (this is not my area of expertise) I rely on the abstract to an article [1] reviewing the subject for an answer:

The Arrhenius equation has been widely used as a model of the
  temperature effect on the rate of chemical reactions and biological
  processes in foods. Since the model requires that the rate increase
  monotonically with temperature, its applicability to enzymatic
  reactions and microbial growth, which have optimal temperature, is
  obviously limited. This is also true for microbial inactivation and
  chemical reactions that only start at an elevated temperature, and for
  complex processes and reactions that do not follow fixed order
  kinetics, that is, where the isothermal rate constant, however
  defined, is a function of both temperature and time.

Reference
[1] Peleg et al Critical reviews in food science and nutrition 52(9):830-51 (2012)


text:
What would be the name of this compound?

Although I assume this compound is likely unable to be formed, my chemistry teacher put it up on the whiteboard to show the extent of what you can name using IUPAC nomenclature, out of interest I tried to name it and got hexcyclo[2.2.0.0.0]cyclohexane, is this correct or am I mistaken?

What's up with the central atom? If it's carbon, then you missed something.

I think the idea my teacher had was that there wasn't a central carbon atom and so each bridge had 0 carbons. I think it was more just a quick drawing so I'm unsure if it's even possible to form but even theoretically I thought it was interesting.

Well, you either need to fix the drawing so that it makes sense, or sort out the correct structure with your teacher. Neither an incorrectly drawn compound, nor a compound that even you don't know the structure of, can be given a meaningful name.

@andselisk It is usually drawn that way (albeit incorrectly) https://en.wikipedia.org/wiki/Claus%27_benzene

@Mithoron Cyclohexane ring with three bridges #1; cyclohexane ring with three bridges #2. If it is supposed to be a cycloxexane, then it should've been drawn correctly (as #1, with the blank space denoting there is no central atom). The way OP drew it implies C7 skeleton.

@andselisk Well, I said it's incorrect, your #1 is better but not more simple. If you checked out the link I put, it's drawn exactly the same and OP provided the context. He all but named it as Clar's benzene.

Let us continue this discussion in chat.

"hexacyclo[2.2.0.0.0]cyclohexane" is an interesting attempt; however "hexa…" is obviously wrong (you should count again), one "cyclo" is enough, and the locants are missing.

I think it is Cubane ($\ce{C8H8}$), a synthetic hydrocarbon molecule (Wikipedia), which has synthesized using following scheme:

I am confident about that because your teacher had said that there wasn't a central carbon atom (assumingly). See following illstration:

Well, if someone drawn cubane in this way, it would be downright misleading. Did you read comments?

@Mithoron: I have read all comments and they were correct about their impressions. I got the idea about cubane with a touch of mechanical drawing. :-)

Well, while in theory it could be a weirdest pic of cubane, the question is still about nomenclature, not synthesis and I still think the compound is Clar's benzene.

@Mithoron No one denies the correctness of your opinion: if you would've posted this as a comment, before the question got closed, I'd happily upvoted it.


text:
Molarity and equilibrium reactions Kc

Samples of $\ce{A}$ ($\pu{1.40 mol}$) and $\ce{B}$ ($\pu{0.80 mol}$) are placed in a $\pu{2.0 L}$ container and the following reaction takes place:
$$\ce{2 A (g) + B (g) <=> C (g) + D (g)}$$
At equilibrium, the concentration of $\ce{A}$ is $\pu{0.40 M}$. What is the value of $K_c$?

I thought you had to convert Sample B into molar concentrations using its mols and the 2.0 L volume, but i still do not know anything about C or D, and I can't use an ICE table because I am not given a Kc value. Do I assume molar concentrations are all the same because it is in equilibrium? I also tried using mole ratios to determine moles of C and D but I wasn't sure which mole value to use (A or B) and in the end it didn't work.
I have this question in a past paper sample exam. I know the answer already, but I require the solution as to know how to solve future questions similar to this.

You started with 1.40 moles of A and at equilibrium you have 0.8 moles of A. How many moles of A reacted? Then how many moles of B reacted? Finally, how many moles of C and D would you get from that reaction? @q123

First, start from a few basic definitions applicable in the case of this specific problem:
$$\ce{2 A (g) + B (g) <=> C (g) + DA (g)}$$
a) Relationship between number of moles, molar concentration, and volume
$$ [i] = \frac{n_i}{V} $$
b) Equilibrium constant in concentration units (when activity coefficients are ~1)
and in terms of moles of each substance at equilibrium:
$$ K_c = \frac{[C][DA]}{[A]^2[B]} = V\frac{n_{C,eq}n_{DA,eq}}{n_{A,eq}^2n_{B,eq}}$$
Strategy:
i) Compute final (equilibrium) moles of A, and moles of A that reacted:
$$ n_{A,eq} = [A]_{eq}V = 0.40 M \times 2.0 L =0.80 mole$$
$$ n_{A,reacted} = n_{A,init} -n_{A,eq} =(1.40 - 0.80) mole = 0.60 mole$$
ii) Compute moles of B that reacted and final moles of B:
$$ n_{B,reacted} = n_{A,reacted}/2 = 0.30 mole$$
$$ n_{B,eq} = n_{B,init}-n_{B,reacted} = (0.80 - 0.30) mole = 0.50 mole$$
iii) Compute moles of C and D that formed:
$$ n_{C/D,eq} = n_{C/D,formed} = n_{B,reacted}= 0.30 mole$$
iv) Put it all together into the expression for $K_c$
$$ K_c =  2.0  \frac{0.30^2}{0.80^20.50}M=0.56 M$$
Note in another common problem one sets up an equation starting from the definition of $K_c$, something like this:
$$ K_c =  V  \frac{x^2}{(n_A-2x)^2x}$$
and solves for an extent (or moles) of reaction x. In the case of this problem $K_c$ is unknown but x can be determined.

:"nC/D,eq=nC/D,formed=nB,reacted=0.50mole".I think nC/D,eq=0.30mol

I thought you had to convert Sample B into molar concentrations using its mols and the 2.0 L volume,

Yes,you can calculate the concentration of  $B$ at equilibrium by the following expression:
$$[\ce{B}]_\mathrm{e}= \frac{(n_{(B)})_e}{V}$$

Do I assume molar concentrations are all the same because it is in equilibrium?

NO ,you must calculate the molar concentrations of each species at equilibrium.

but i still do not know anything about C or D,

The initial amount of C =The initial amount of D =zero
The amount of C at equilibrium  =The amount of C formed from the reaction
The amount of D at equilibrium  =The amount of D formed from the reaction

I also tried using mole ratios to determine moles of C and D

$$\begin{align}\text{ The amount of C at equilibrium}
  &=\text{The amount of D at equilibrium} \\&= \text{The amount of B reacted} \\
&= \frac{\text{ The amount of A reacted}}{2}
\end{align}$$

but I wasn't sure which mole value to use (A or B) and in the end it didn't work.

You have the initial amount of A equal $\pu{1.40 mol}$ ,the initial amount of B equal $\pu{0.8 mol}$ and the concentration of A at equilibrium  equal $\pu{0.40 M}=\pu{0.40 \frac{mol}{L}}$.

So start with A ;

Calculate the amount of A at equilibrium :

$$\begin{align}\text{Amount of A at equilibrium}
&=\text{concentration of A at equilibrium}\times{\text{Volume of the container}}\\
(n_\mathrm{A})_e &= C_\mathrm{A}\times{V}\\
 &=\pu{0.40\frac{mol}{L}}\times{\pu{2L}}\\&=\pu{0.8mol}
\end{align}$$

Calculate the amount of A reacted   :
$$\begin{align}\text{Amount of A reacted}&=\text{Initial amount of A }-\text{Amount of A at equilibrium}\\
(n_{(A)})_\text{reacted} &=(n_{(A)})_I -(n_{(A)})_e\\&=1.4-0.8\\&=\pu{0.6mol}\end{align}$$
Calculate the amount of B reacted   :
$$\begin{align}
\text{The amount of B reacted} &= \frac{\text{ The amount of A reacted}}{2}\\ &= \frac{\pu{0.6mol}}{2}\\&=\pu{0.3mol}\\&=\text{The amount of C at equilibrium}\\  &=\text{The amount of D at equilibrium}
\end{align}$$

Calculate the amount of B  at equilibrium   :
$$\begin{align}\text{Amount of B at equilibrium}&=\text{Initial amount of B }-\text{Amount of B reacted}\\
(n_{(B)})_e &=(n_{(B)})_I -(n_{(B)})_\text{reacted}\\&=0.8-0.3\\&=\pu{0.5mol}
\end{align}$$

and I can't use an ICE table because I am not given a Kc value.

You can  use the following table to determine the amount of each species at equilibrium
in order to  calculate $K_\mathrm{C}$ :
$$\begin{align}
&&\ce{2A_{(g)} &&+ &&B_{(g)}&<=>&C_{(g)} &+&&D_{(g)}\\
I &&(1.4\pu{mol})  &&&&(0.8\pu{mol})  &&(0\pu{mol}) &&&(0\pu{mol})\\
C &&(-0.6\pu{mol})  &&&&(-0.3\pu{mol}) &&(0.3\pu{mol})  &&&(0.3\pu{mol})\\
E &&(0.8\pu{mol})  &&&&(0.5\pu{mol})  &&(0.3\pu{mol})  &&&(0.3\pu{mol})}
\end{align}$$
Use the expression $[i]_\mathrm{e}= \frac{(n_{(i)})_e}{V}$
to calculate the concentration of each species at equilibrium:
$$\begin{align}[\mathrm{A}]_\mathrm{e}= \frac{\pu{0.8mol}}{\pu{2L}}=\pu{0.4M}\\
[\mathrm{B}]_\mathrm{e}= \frac{\pu{0.5mol}}{\pu{2L}}=\pu{0.25M}\\
[\mathrm{C}]_\mathrm{e}= \frac{\pu{0.3mol}}{\pu{2L}}=\pu{0.15M}\\
[\mathrm{D}]_\mathrm{e}= \frac{\pu{0.3mol}}{\pu{2L}}=\pu{0.15M}\\
\end{align}$$
Substitute the concentration of each species at equilibrium in the following formula to calculate $K_\mathrm{C}$:
$$\begin{align}
K_\mathrm{C}&=\frac{[\mathrm{C}]_\mathrm{e}[\mathrm{D}]_\mathrm{e}      }{[\mathrm{A}]^2_\mathrm{e}[\mathrm{B}]_\mathrm{e}     }\\
&=\frac{(\pu{0.15M})(\pu{0.15M})      }{(\pu{0.4M)^2}(\pu{0.25M})     }\\
&=0.5625\pu{\frac{L}{mol}}
\end{align}$$


text:
Do self-growing hydrogels grow and strengthen infinitely?
Recently, there was a publication [1] (/., YouTube) on self-growing materials using mechanoradically-generated double-network (DN) hydrogels, closely resembling self-strengthening of muscles.

We used a series of DN gels consisting of poly(2-acrylamido-2-methylpropanesulfonic acid) sodium salt (PNaAMPS) as the brittle network
  and poly(acrylamide) (PAAm) as the stretchable network. Each network is covalently cross-linked.

Fig. 1. Conceptual scheme of the self-growth of materials induced by mechanical training. (A) A skeletal muscle becomes bigger and stronger
  under the application of repetitive mechanical stress and a continuous nutrient supply through metabolic cycles. (B) As a result, such a tissue is
  initially weakened (purple arrows) but then recovers or even shows enhanced performance (red arrows) due to mechanical stimuli. The green outline indicates one mechanochemical cycle. (C) Our strategy to develop self-growing materials based on mechanical training of DN gels. Mechanical stress leads to breakage of the brittle network (blue), whereas the highly stretchable
  network (pink) maintains the integrity of the gel. The mechanoradicals generated at the broken ends of the brittle network strands trigger polymerization of monomers supplied from the external environment to form a new network
  (orange). The chemistry of the newly formed network can be the same or differ from that of the original brittle network, as required.

The process is reported to be repeatable and reproducible:

With proper conditions, our DN gel exhibits persistent growth in strength and size under repetitive mechanical training. Two requirements must be satisfied: Enough monomers should be continuously externally supplied, and the newly formed
  network should serve the role of a brittle network that then breaks during the next deformation to trigger subsequent network formation.

However, with muscle growth there are limiting factors such as hormones and bones structure, and even supplying enough protein one cannot grow the mass infinitely*. Considering constant feeding with monomer, is there a theoretical/practical limit for strength and growth for these DN-hydrogels?
*not even Tetsuo Shima — sorry, couldn't resist considering the origin of this work and the anime.
References

Matsuda, T.; Kawakami, R.; Namba, R.; Nakajima, T.; Gong, J. P. Mechanoresponsive Self-Growing Hydrogels Inspired by Muscle Training. Science 2019, 363 (6426), 504–508. https://doi.org/10.1126/science.aau9533.

The comparison to muscles is a rather frivolous one. Scar tissue would probably be more apt. ;-)

"Strength" is a dubious quantity. The E-modulus will rise, and as the material swells with more and more monomer diffusing in, it will at some point rupture when the brittle network cracks. Like a fruit gum that was put in water overnight.
I have no idea if the two networks actually stay separated, or if they interconnect more and more. I suspect the latter, with radicals at work. That would make the brittle part softer -> no further hardening, and the soft network more brittle -> it snaps under increasing load.
But the point is moot imho: "Self-healing" makes sense if you had a smaller damage due to light overload. Then this material here can become twice as hard, and you're fine. I don't think it makes sense to "train" sth to become fifty times as strong.


text:
How do I explain these bond angles in SF4 and SOF4?
From here and here I found that the bond angle between the equatorial $\ce{F}$'s  in $\ce{SF4}$ is less than that in $\ce{SOF4}$. I understand that is because of greater repulsion by lone pair than by double bond.
But the angle between the axial $\ce{F}$'s  in $\ce{SF4}$ is more than that in $\ce{SOF4}$. I am not able to understand why is this so. Shouldn't the lone pair repel more than double bond here too?

I'd say the effects are too subtle to be explained that easily.

You are trying to compare 2 different geometries: one with 4 vertices and the other with 5! Of course angles won't match

@SteffX Try counting again and don't forget about lone pair.

I suspect that some subtle molecular orbital effects are going on here.  Oxygen has two pairs of pi-symmetry electrons to donate and is tending to pull the fluorines towards a symmetric structure to optimize that interaction.


text:
Can a half-reaction be oxidation if it has the higher reduction potential?
I am working on a problem that provides these two half-reactions:
$$
\begin{align}
\ce{Fe^3+ (aq) + e- &-> Fe^2+ (aq)} &\quad E^\circ_\mathrm{red} &= \pu{+0.77 V}\\
\ce{S2O6^2- (aq) + 4H^+ + 2e- &-> 2H2SO3 (aq)} &\quad E^\circ_\mathrm{red} &= \pu{+0.60 V}
\end{align}
$$
What is confusing me is the way the question is worded:

Write the balanced chemical equation for the oxidation of $\ce{Fe^2+ (aq)}$ by $\ce{S2O6^2- (aq)}$.

Isn't this backwards though? Because the $\ce{Fe}$ half reaction has the higher reduction potential, shouldn't it be reduced while the $\ce{S2O6^2- (aq)}$ half reaction reverses to become oxidation?

What you propose is what's expected in terms of thermodynamics, and the standard reduction potential is the way to realise this. However, you have the freedom to write the equation backwards if you like; it won't happen, but you can certainly write it and calculate E° (which will be negative).

Furthermore, just because it is not spontaneous under standard conditions doesn't mean it can't be done. This is what electrolysis does: it reverses the direction of spontaneous change. So in theory, what your question is asking could be achieved by applying an EMF in the direction of non-spontaneous change and thereby force it to happen. This is similar to how in thermodynamics you can have a change that is non-spontaneous, but can become spontaneous at higher temperature. In electrolysis the energy for that change is supplied by an EMF instead of by heat.

It's also concentration-dependent (which include pH-dependence because one of the half-reactions includes hydrogen ions). If you start out in the absence of products, it will go forward (if just a tiny bit).

Redox reactions are reversible. Despite the forward reaction is not spontaneous (as indicated by the unfavourable potential), it can still proceed.


text:
How can I increase the viscosity of dilute hydrochloric acid
I'm cleaning the mortar off a new clinker brick wall. Walls actually. I hate this job. The builder told me to wash down the wall with weak HCl, leave it for a minute or two, and then spray it off with a pressure hose (eg Kärcher or Gerni). There are lots of problems with this. The main one is that the HCl gets painted on with a coarse bristled brush and splashes everywhere you don't want it to be. The protective clothing is a huge impediment to being able to do the actual job. The safety eye protection fogs up and you can't see how well you've cleaned. Once the bricks are wet, the mortar residue becomes transparent, so it looks clean, only to dry out and turn white again. Then there's the problem with blasting weak acid all over the place. I really hate this job.
So I was thinking that in order get some control over the application of the acid, it would be great if it was the consistency of paint. Then I could apply to only those areas that it needs to be. Question is, how do I make my dilute HCl the consistency of paint? Please feel free to correct me if viscosity is not the right word.

If the solution is not too concentrated and you dont need it to be stable for a prolonged period, many viscosity modifiers can be used. I would start trying something simple, like CMC based wallpaper paste.

So, for the uninitiated, CMC is carboxymethyl cellulose. Will adding the acid to a CMC solution cause a reaction that will neutralise the effectiveness of either or both constituents?

I wouldn´t expect any chemical reactions right away. Over time the cellulose polymer will probably hydrolyse in an acid solution but it will be a slow process. The effect of polymers like CMC on viscosity  is however significantly influenced by pH, so if the solution is too acid, you will lose some effect. Give it a try, manipulating the viscosity of liquids is an art you have to learn by experience :-)


text:
Is Fe(OH)2 soluble in alkalis?
I have just started inorganic chemistry in school and I have just learnt about metal aqua ions and how they react in water. I learnt that $\ce{[Fe(H2O)6]^2+}$ reacts in water by losing a $\ce{H+}$ acting as a Bronsted-Lowry acid producing $\ce{H+}$ ions. Due to this reaction we can test for metals such as $\ce{Fe^2+}$ by reacting it with $\ce{NaOH}$. This is because the $\ce{OH-}$ ions react with the $\ce{H+}$ ions produced pushing the equilibria to the right forming a insoluble precipitate. My questions is, why does the reaction stop there? The reaction stops once $\ce{[Fe(OH)2(H2O)4]}$ forms which is the insoluble precipitate but why doesn't it continue to form $\ce{[Fe(H2O)3(OH)3]-}$ and become soluble again?

Actually precipitate isn't  [Fe(OH)2(H2O)4]. Precipitation starts with it, but it's rather complex condensation/polymerisation process.

Since you said you have just started inorganic chemistry in school, I give you some simpler insight:
Simply, a complex metal ion has a metal ion at its center with a number of other molecules (e.g., $\ce{H2O}$) or ions (e.g., $\ce{CN-}$) surrounding it. When you start learning complex ions (or molecules) you must learn why complex ions of the type $\ce{[M(H2O)6]^n+}$ (hexaaqua ions) are acidic. Examples are, $\ce{[Fe(H2O)6]^2+}$, $\ce{[Fe(H2O)6]^3+}$, $\ce{[Cu(H2O)6]^2+}$,  $\ce{[Co(H2O)6]^2+}$, $\ce{[Ni(H2O)6]^2+}$, $\ce{[V(H2O)6]^2+}$, etc. The first thing you need to know is the solutions of hexaaqua ions with equal concentrations have varying range of acidity (pH of solution), which depend a lot on type of metal's charge, ionic radius, etc. See the following graph extracted from the web page, Complex Metal Ions - The Acidity of the Hexaaqua Ions.

Let's take the hexaaquairon(III), $\ce{[Fe(H2O)6]^3+}$ as typical complex ion. The structure of the ion (octahedral) is given in the diagram A below and an explanation for its acidity is given in the diagram B (From Complex Metal Ions - The Acidity of the Hexaaqua Ions):

This explanation is given by visualizing the 3+ charge of the ion is located entirely on $\ce{Fe}$ in the center. When one of lone pair on the $\ce{O}$ of each $\ce{H2O}$ form a coordinate bond with the iron, it can be visualize that these lone pairs move closer towards $\ce{Fe}$. That has an effect on the electrons in all $\ce{O-H}$ bonds within the complex (imagine all $\ce{O}$ atoms have partial positive charges due to this electron movement). The electrons of $\ce{O-H}$ bonds, in turn, get pulled towards the oxygen even more than usual because of this. That leaves each $\ce{H}$ nucleus more exposed than normal (compared to that in $\ce{H2O}$ molecules in the solution). The overall effect is that each of $\ce{H}$ atoms is more positive than it is in ordinary water molecules. Thus, the 3+ charge is no longer located entirely on $\ce{Fe}$ but spread out over the whole ion--much of it on the hydrogen atoms of coordinated water.
The $\ce{H}$ atoms attached to $\ce{H2O}$ ligands are sufficiently positive that they can act as Brønsted acid and can be pulled off in a acid-base reaction involving $\ce{H2O}$ molecules in the solution according to the following sequence of equations illustrating the first three stages of the process:
$$\ce{[Fe(H2O)6]^3+ + H2O <=> [Fe(H2O)5(OH)]^2+ + H3O+} \\\ce{Fe(H2O)5(OH)]^2+ + H2O <=> [Fe(H2O)4(OH)4]^+ + H3O+} \\\ce{Fe(H2O)4(OH)2]^+ + H2O <=> [Fe(H2O)3(OH)3] + H3O+}$$
As you see, the process end up with a complex with no charge on it after third stage. This complex is described as a Neutral Complex. Because it has no charge, thus, it doesn't dissolve in water to any extent, and will precipitate as it is formed. See below diagram for color illustration of the reaction of $\ce{[NaOH}$ with $\ce{[Fe(H2O)6]^2+}$ and $\ce{[Fe(H2O)6]^3+}$, respectively.

This is a pretty interesting question. First of all, you are right, if you ever happened to see a yellow $\ce{Fe^3+}$ solution that was no pure aqua-complex but more likely $\ce{[Fe(H2O)5(OH)]^2-}$. The real $\ce{[Fe(H2O)6]^3+}$ for example, like it is formed in conc. $\ce{HNO3}$ or $\ce{HClO4}$ is actually somewhat purple. We call these compounds cationic acids. So the aqua-complexes will act as acids and become deprotonated in water. For your $\ce{Fe^2+}$ example it's pretty much the same. I often read in student reports that, if for example a hexaqua complex is treated with a base the $\ce{OH^-}$ would exchange for one water ligand. Many then concluded that $\ce{OH^-}$ had to be the stronger ligand in the spectrochemical series. This is, of course, wrong and not related to each other here. What is far more logical, at least to me is, that by coordination the water gets closer towards the highly charged metal center. But that means there is a repulsion from the $\ce{H}$ in $\ce{H2O}$ and the metal center since both have a positive (formal) charge. This makes it much easier to become deprotonated by a base.
But why does this stop at some points? I asked myself the same question for the example of $\ce{BeO}$ reacting with $\ce{OH^-}$ to form $\ce{[Be(OH)4]^2-}$. Why is Beryllium oxide amphoteric forming a berrylate while $\ce{MgO}$ to $\ce{BaO}$ do not show this ability? It took me quite a while to find an answer to this. There is an equation by  Cartledge that tries to describe this phenomenon:
$$\ce{Φ = z/r}$$
Where Φ is the ionic potential, z the charge and r the radius. If $\ce{Φ^{1/2}}$ is smaller than 2.2 the ion is considered basic, above 3.2 it's considered to be acidic. In the middle, it's considered amphoteric. What does that mean? It seems like there is a ratio of radius to charge. Small, highly charged ions seem to polarize the surrounding water molecules so much that they quickly lose a proton.
In your case, we are not specifically talking about oxides but I guess that something similar should be happening. The high charge on the metal and the fact that $\ce{Fe}$ is already in the middle of the d-row where atomic radii decrease along the row should be enough to actually cause even more deprotonation, this time perhaps not of surrounding water molecules like in the case of $\ce{BeO}$ but the ones that are already ligated to the metal center. At some point it will require additional help which happens when the pH is increased. And in fact for $\ce{Fe^2+}$ a  $\ce{Na4[Fe(OH)6]}$ seems to exist.
I guess the comparison with the $\ce{BeO}$ may perhaps be a bit far fetched but it would be an actual example where you can see how the charge to radius ratio acts on the way soluble hydroxo-complexes are formed.

Can you give a reference, other than textbook problems that often make things up, for $\ce{Na4Fe(OH)6}$?

Good question. It is mentioned in the Riedel - Anorganische Chemie and the new edition of the former Holleman Wiberg to exist. But I will check the Gmelin on iron later.


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Does 1 mole have the same molar mass?
I have tried to look it up, and find answers, it really intrigues me!

One mole of nitrogen does not have the same mass as one mole of oxygen...

The question doesn't make much sense. It's like asking "Does 1 metre have the same colour?"

A mole of any substance is defined as Avogadro's number of molecules of that substance. So every mole has the same number of molecules. Since the masses of molecules differ, the mass of a mole of any substance will be dependent on the mass of one molecule of that substance. That is,
(Mass of one mole) = (Avogadro's number) x (the mass of one molecule).


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Aqueous copper and how it reacts with NH3
I have learnt that you can test whether an unknown metal is copper by doing a simple test tube reaction. The reaction involves adding aqueous $\ce{NH3}$ to the test tube which contains the copper.
The $\ce{NH3}$ as it is dissolved in water will form $\ce{NH4+}$ ions and $\ce{OH-}$ ions. The $\ce{OH-}$ ions react with the copper forming a precipitate $\ce{([Cu(H2O)4(OH)2])}$ but many other metals form a precipitate with $\ce{NH3}$ as well such as $\ce{Fe^2+}$, $\ce{Fe^3+}$ etc. So to make sure that it is copper present in the test tube you add an excess of $\ce{NH3}$ as this causes a ligand exchange reaction to occur producing $\ce{[Cu(H2O)2(NH3)4]2+}$.
Why doesn't this occur with the other metals? Why doesn't this occur for say $\ce{Fe^2+}$ to form $\ce{[Fe(H2O)2(NH3)4]2+}$ which would also be soluble?

I think the formation of a blue precipitate is already an evidence that the metal is copper. Further addition of ammonia should be unnecessary.

[Fe(H2O)6]2+ is a high-spin d6 complex, while [Fe(H2O)2(NH3)4]2+ is a low-spin d6 complex. The ligand substitution reaction is not favourable in terms of spin, so aqueous Fe2+ does not react with excess ammonia.
However, both [Cu(H2O)6]2+ and [Cu(H2O)2(NH3)4]2+ are d9 complexes. Their spins are the same, and being a stronger field ligand, NH3 increases the crystal field stabilization energy, so ligand substitution is favourable.


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Volume of air required to oxidize glucose

In the body, the oxidation of glucose is according to the reaction:
$$\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O}$$
Knowing that glucose consumption is $\pu{2.47 mol/d}$, What is the volume of air that must be breathed at $\pu{25 °C}$ to accomplish this oxidation for $24\ \mathrm h$ at the pressure of $1\ \mathrm{atm}$?

We are given:

The atmosphere is composed of $21\,\%$ $\ce{O2}$ and $79\,\%$ $\ce{N2}$ (% volume).
$R = \pu{8.314 J mol-1 K-1}$; $\pu{1 atm} = \pu{760 mmHg} = \pu{1.014e5 Pa}$.
The molar volume of a gas at $\pu{298 K}$ is $\pu{24.45 l mol-1}$.

So, I have been a little stuck on this question. I have been trying to use the formula:
$$pV = nRT,$$
but it seems that I am missing something and I am not getting the right answer. It's possible that it might include molar fraction or partial pressure. I am confused.
The answer is $1\,724\ \mathrm l$. I would like to know how to get to this point.

Following the advice of @matt_black:
Here are my calculations that doesn't give me the right answer of $1\,724$ liters.
$$V_\mathrm m = 24.45\ \mathrm{l/mol}$$
$$n(\ce{C6H12O6}) = 2.47\ \mathrm{mol}$$
$$n(\ce{O2})= 6\times2.47\ \mathrm{mol} = 14.82\ \mathrm{mol}$$
$$V =\ ?$$
By the ideal gas law:
$$V= n\times24.45\ \mathrm{l/mol}$$
$$= 14.82\ \mathrm{mol} \times 24.45\ \mathrm{l/mol}$$
$$= 362.349\ \mathrm l \neq 1\,724\ \mathrm l$$
Tried using $pV=nRT$
$$V= 14.82\ \mathrm{mol} \times 8.314\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 293\ \mathrm K / 760\ \mathrm{mmHg}$$
$$= 47.5\ \mathrm l \neq 1\,724\ \mathrm l $$
I don't know how to implicate the fact that the atmosphere is composed of $21\,\%$ of $\ce{O2}$ and $79\,\%$ of $\ce{N2}$. Any ideas?

Why do you need PV=nRT? The question already gives you the molar volume of a gas.

@matt_black I have tried calculating without using that formula, by using V = n*24.45 l/mol, however I don't get the right answer, the reason why I thought I should use the PV=nRT...

You might want to tell us your calculations steps in detail as your error may be nothing to do with this.

In one of your answer, you have used $R = \pu{8.314 J \cdot K^{−1} \cdot mol^{−1}}$. But answer is in L. You may need to convert units appropriately as well.

As matt_black commented, there is no need to use ideal gas law as you are given the molar volume already. Volume of the air $V_\mathrm{air}$ can be found from the volume fraction $\varphi(\ce{O2})$:
$$V_\mathrm{air} = \frac{V(\ce{O2})}{\varphi(\ce{O2})}$$
Volume of the oxygen can be found from the molar volume $V_\mathrm{m}(\ce{O2})$ and the amount of oxygen $n(\ce{O2})$ required to oxidize given amount of glucose:
$$V(\ce{O2}) = n(\ce{O2})\cdot V_\mathrm{m}(\ce{O2})$$
$$n(\ce{O2}) = 6n(\ce{C6H12O6})$$
Getting things together:
$$V_\mathrm{air} = \frac{6n(\ce{C6H12O6})\cdot V_\mathrm{m}(\ce{O2})}{\varphi(\ce{O2})} = \frac{6\cdot\pu{2.47 mol}\cdot\pu{24.45 L mol-1}}{0.21} = \pu{1725 L}$$
Note that you have to use appropriate units, e.g. you cannot use all base SI units in the deal gas law and put the pressure in mm Hg! Also, never drop any units — they do participate in calculations as well.

thank you! So that's how you implicate the 0.21l of O2.   It was so much simpler than I thought. In what cases will the pressure be in Pa or atm? Until now, I never seen the use of the other units besides mmHg.

No, it's not 0.21 L of oxygen, it's volume fraction and is dimensionless! As for pressure, mm Hg is a worthless unit, especially in physical chemistry. If you want to use ideal gas law with SI units, use Pa, as it's very convenient: $$\pu{1 Pa} = \pu{1 J m-3}$$ and goes along with volume in $\pu{m-3}$ and gas constant $R=\pu{8.314 J mol-1 K-1}$.


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Why does acetylene form a triple bond?
I'm going through basic tutorials about hybridisation, but I am having trouble understanding the reason for drawing a triple bond for C2H2. It seems to me that a double bond is enough for the octet rule. What dictates the triple bond between the two carbons?

Could you show the Lewis structure with a double bond that satisfies the octet rule?

Maybe it does not have to satisfy the octet rule.  Cyclopropenyl cation gets along well without it, and molecular oxygen is just weird.  Acetylene forms a triple bond because it has the right molecular orbitals to make that bonding most stable.

Firstly, nothing "dictates" the triple bond between 2 carbon atoms. The hybridisation of the carbon atom determines the number of pi bonds it has, and a carbon with 3 pi bonds and 1 sigma bond will have 3 unhybridised p orbitals and hence 2 regions of electron density, thus it is sp hybridised. For example, an alkyne can be formed from the two-fold elimination of either a geminal or vicinal dihalide. I think u may be having a misconception here, hybridisation of carbon atoms do not simply change and alkenes do not interconvert to form alkynes; strictly speaking, it is an oxidation from alkane to alkene to alkyne since oxidation state of carbon atom increases.


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Hydrogen vs neon in terms of entropy
I know that both molecular size and complexity increase overall entropy, but between the two I'm not sure which increases it more. Would 1 mole of hydrogen have the greater entropy because it is two atoms instead of one, or would 1 mole of neon have the greater entropy because it is a larger atom?

Entropy inherent to atoms itself is treated as zero, because it can't be lowered with cooling.

Assuming you have equal moles of two different perfect gases, according to the Sackur–Tetrode equation the difference in their molar translational entropy is:
$$\frac{\Delta S_\mathrm{trans}}{R} = \frac{3}{2}\ln\left(\frac{M_2}{M_1}\right)$$
where the $M_i$ are the molecular masses. For instance, since $\ce{D_2}$ and $\ce{He}$ have identical masses this term would vanish, i.e. $ \Delta S_\mathrm{trans} = 0$.
If the atoms are in their electronic ground states then the electronic state makes no contribution to the entropy. On the other hand, a diatomic molecule has rotational and vibrational degrees of freedom. Assuming it's a rigid rotor, then in the high temperature limit (in which many rotational quantum levels are significantly occupied),
$$\frac{S_\mathrm{rot}}{R} = 1 + \ln\left(\frac{8\pi^2IkT}{h^2\sigma}\right)$$
where $I$ is the moment of inertia, $\sigma$ is a symmetry number.
Assuming the diatomic molecule is a harmonic oscillator, then
$$\frac{S_\mathrm{vib}}{R} =  x(e^x - 1)^{-1} - \ln(1 - e^{-x})$$
where $x = hν/kT$ and $T$ is around room temperature. However, the contribution of the vibrational entropy is tiny compared to the rotational component, so it can be ignored.
That would wrap up the story, in any case, if nuclear spin didn't have something to say about it. Nuclear spin contributes additionally to the entropy of $\ce{H2}$ and $\ce{D2}$ as explained in this post, resulting in an altered rotational entropy term. This rotational/spin contribution lifts the entropy of these molecules above those of Ne and He.
Therefore at room temperature and $\pu{1 atm}$ pressure, the molar entropy of $\ce{D2}$ is expected to be larger than that of helium due to the rotational contribution of $\pu{31.3 J mol-1 K-1}$.
Similarly, the translational entropy of neon is $\pu{3.5 J mol-1 K-1}$ higher than that of hydrogen gas, but $\ce{H2}$ has a rotational entropy of $\pu{18.7 J mol-1 K-1}$ so the entropy of neon is expected to be smaller by about $\pu{15.2 J mol-1 K-1}$.

The entropy of deuterium is a little larger than you quote (about 35 rather than 22 ) due to nuclear spin statistics, see the answer https://chemistry.stackexchange.com/questions/70620/why-is-the-entropy-of-deuterium-greater-than-that-of-hydrogen-gas

@porphyrin well done catching that- I'll have to edit...

Assuming the same conditions, there is a difference in translational entropy proportional to log of the mass, see the Sakur-Tetrode eqn.  A diatomic molecule has the ability to vibrate and rotate and this adds to the entropy but an atom has neither of these motions.


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Strength of a nucleophile
A nucleophile should be stronger if it can donate a pair of electrons more easily. We say that a more electronegative atom should be less nucleophilic , generally. I understand this is because the more EN an atom gets, the more strongly it pulls electrons towards itself. But why does the fact that such an atom would ,generally, have more electron density not override this?

As a donor atom gets more and more electronegative, the outer orbitals are more closer to the nucleus and electrons in those orbitals become lower in energy, i.e the energy of the HOMO decreases, so it becomes progressively difficult for them to donate as now they need to give away the electrons which are more stable in a sense.

Yes. But does the partial negative charge make them less stable?

Electronegativity and nucleophilicity are different concepts. Electronegativity refers to how strongly an atom pulls bonding electrons towards itself. Nucleophilicity refers to ability of atom to donate a lone pair of electrons. Different factors affect nucleophilicity, such as orbital size, charge of ion, size of molecule, hybridisation etc etc.

Yes. But does the fact that an atom has a partial negative charge help its nucleophilicity?

Yes. Read that I wrote "charge of ion" as a factor affecting nucleophilicity.

And that is where I am confused. More electronegatvity - - - > more partial negative (ceteris paribus) - - - - > more nucleophilicity. But more electro negative - - - > less able to give up electrons because of stability - - - > less nucleophilicity. Which one dominates?

@Sal_99 stop thinking about electronegativity affecting nucleophilicity so much. EN is just a minor factor affecting nucleophilicity. Bulkiness of the base and ionic charge are larger factors at play. Anyways, I'm treading on thin waters, a generalisation is that the more electronegative the atom, the worse nucleophile it is.

Can we say that higher EN will mean greater electron density surrounding the atom?


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Reasoning about a separation problem

Describe exactly how you would go about separating a mixture of the three water-insoluble liquids, aniline (b.p. 184 °C), n-butylbenzene (b.p. 183 °C) and n-valeric acid (b.p. 187 °C), recovering each compound pure and in essentially quantitative yield.

This is a question from Organic Chemistry  by Morrison and Boyd. My problem is I know everything that is going to apply in this problem but I'm unable to execute the solution.
Things that I know:
1. Aniline is an amine attached to benzene group, n-butylbenzene is a benzene with a butyl group attached to one of its carbon atom, Valerie acid is pentatonic acid.
2. Aniline are polar and basic, Valeric acid are acidic, benzene is an electron withdrawing group.
3. Aniline are insoluble in water due to lack of hydrogen bonding.
How can solve this problem? Is my knowledge about the problem is insufficient? I have given this problem only to illustrate my problem with reasoning in chemistry, so please illustrate your answer by giving the solution of this problem or the way you find better.

Knowing too many useless things can obscure your vision. True, aniline is basic; what does that really mean?

Simple extraction by treatment with aq. acid followed by by aq. NaOH will seperate all the three compounds.

@SoumikDas Of course it will, but the question was not so much about the separation per se as it was about the reasoning involved, and that's what I tried to stress in the first place.

The question is a little strange as both aniline and valeric acid are soluble in water to the extent of a few grams/100ml which offers an immediate path to separation from the butylbenzene.

Several methods to separate substances exist in chemistry. A non-exhaustive list:

Distillation
Extraction
Chromatography
(Fractional) crystallization

which may be supplemented by chemical modification.
Let's consider some of the methods mentioned. Distillation is a classic, but the boiling points given in the problem are very close to each other and would probably prevent effective separation. No data is given on the freezing points of the substances, so direct selective crystallization may be hard. Chromatography (as in silica column chromatography) would probably work in some fashion, but is probably not the intended answer here. Other, more exotic methods such as the creation of complexes using cryptands and the like are probably possible as well.
What about extraction using water? The compounds are not soluble in water, but we can modify some of them to be so. If we add a sufficiently strong acid in a sufficient amount of water and mix vigorously, we can probably get most of the aniline into the aqueous phase. After separating in an appropriate funnel, we would make the aqueous phase slightly basic and extract the aniline using some organic solvent that does not mix with water. Likewise, a strong base (and subsequent "inverted" procedure) can be used to separate out the n-valeric acid.


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Effective nuclear charge and repulsion of electrons
Effective nuclear charge increases across a period. This is justified because number of protons increase in nucleus across a period, and there's no change in the number of "shielding" electrons. However, wouldn't valence electrons repulse each other? Why does this repulsion not seem to matter for effective nuclear charge?

for those who haven't seen it before (I hadn't, at least not recently) https://i.stack.imgur.com/8pKaH.png

They do, it's a reasonable approximation because valence electrons occupy a common shell further from the nucleus than the core electrons.


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Rationalising the reactivity of phosphoryl chloride
$\ce {POCl3}$ is a reagent well-known for enacting various functions in organic synthesis. For example, together with pyridine, it is able to dehydrate alcohols to give alkenes. Also, reacting it with a substituted amide gives chloromethyleneiminium ions. Salts containing these ions are also known as Vilsmeier reagents and they have great practical use in synthesis. The above two reactivity patterns seem to suggest that $\ce {POCl3}$ reacts with oxygen atoms containing rather high electron density, such that such reactions with other carboxylic acid derivatives at the carbonyl oxygen is not really observed. Firstly, I would like to ask if my rationalisation of its reactivity here is correct.
Also, as suggested by the same site, $\ce {POCl3}$ is also able to perform the halogenation of alcohol, in a manner similar to how $\ce {PCl5}$ performs the reaction, but only in the presence of dimethylacetamide (DMA) or dimethylformamide (DMF). Why is that the case? What particular properties of the solvent allow the transformation to be effected?

POCl3 reacts with DMAc or DMF to make the Vilsmeier reagent. Presumably this is the actual chlorinating agent.

@orthocresol Do you care to post an answer detailing the mechanism for better illustration of how it works?

@Tan Yong Boon https://en.wikipedia.org/wiki/Vilsmeier_reagent

AFAIK it's not a particularly common choice, but Vilsmeier reagents are sometimes used to convert alcohols to chlorides (see e.g. Tetrahedron Lett. 1992, 33 (34), 4901–4904). This explains the requirement for DMF/DMAc as solvent.
A reasonable mechanism is as follows:

(see also Org. Lett. 2002, 4 (4), 553–555 for a similar mechanism)


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How does di-isobutylaluminium hydride ( DiBAl-H ) reduces carboxylic acid
$\text{DiBAl-H}$ reduces by electrophilic attack and provides hydrogen by $\text{H}^-$ transfer. Unlike $\text{LiAlH}_4$ it does not directly provide $\text{H}^-$ ion. So it should not undergo acid base reaction with carboxylic acid. But my friend argued that it undergoes acid base reaction. I can't find any reaction mechanism for reduction of carboxylic acid with $\text{DiBAl-H}$, can someone propose a mechanism for it?

Reduction of esters of carboxylic acids directly to aldehydes is of great synthetic interest. Zakharkin and Khorlina in 1962 (Ref.1) has shown that, diisobutylaluminum hydride (DIBAL-H) could be very useful in reducing esters to aldehyde in high yields at low temperatures (about $\pu{-70 ^oC}$). They also found that solvents affect reduction significantly. For example, the reduction reaction in toluene or hexane proceeds to give corresponding aldehydes in yields l0-15% more than those in ether as the solvent. Previously, in 1959, Miller et al. (Ref.2) found that DIBAL-H is a more selective reagent than lithium aluminum hydride in the reduction of nitriles to aldehydes, and it can also be used to reduce benzoic acid to give 72% yields of benzyl alcohol (they also showed esters can be reduced to corresponding alcohols). The conditions and yields of these references are summarized in this page. In Ref.2, during acid reduction, Miller et al. have used 3 equivalents of DIBAL-H and observed that a gas (hydrogen) evaluation upon the addition of the hydride to benzoic acid until one mole of the reducing agent had reacted. Thereafter, the addition of more DIBAL-H gave no gas. Thus, the first reaction is, presumably an acid-base reaction. However, DIBAL-H here should be acting as a Lewis-acid instead of a base, as illustrated in scheme 1:

The ist mono-adduct, then further reacted with second DIBAL-H to give di-adduct, which upon acid hydrolysis would give expected aldehyde (Scheme 2):

References:

L. I. Zakharkin, I. M. Khorlina, “Reduction of esters of carboxylic acids directly into aldehydes with diisobutylaluminum hydride,” Tetrahedron Letters 1962, (14), 619-620 (https://doi.org/10.1016/S0040-4039(00)70918-X).
A. E. G. Miller, J. W. Biss, and L. H. Schwartzman, “Reductions with Dialkylaluminum Hydrides,” J. Org. Chem. 1959, 24(5), 627–630 (DOI: 10.1021/jo01087a013).

At lower temperatures (around -78oC), carboxylic acids (with 2 equiv. of DIBAL-H) is reduced by DIBAL-H

What does aqueous base have to do with the question? Two equivalents of DIBAL-H may be required. One as a base; one as a reductant. The bis-aluminum complex (lower right corner) must be stable prior to work up at low temperature.

@Sarahjane this does not answer the initial part of my question. Further this mechanism does not seem to be correct. 1 eq. of DiBAl-H can reduce ester, here's the mechanism (
https://i.stack.imgur.com/1Dwzk.jpg ). So why does carboxylate Ion require 2 eq. of DiBAl-H inspite being more electron rich?


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Is electrolysis of molten silver nitrate possible?
Since the salt is molten, the only ions are $\ce{Ag+}$ and $\ce{NO3-}$, and both of them can only do a reduction reaction. Who goes to the anode? Also, there is no acid in the solution — only the salt — therefore the following reaction cannot occur, right?
$$\ce{ NO3- + 4H+ + 3e- -> NO + H2O}$$
So, what happens? The only possible reaction is the reduction of the $\ce{Ag+}$ to $\ce{Ag (s)}$. There is literally nothing else.

You can't have reduction without oxidation.  Then again, who said they can only do a reduction reaction?

Why oxygen shouldn't evolve?

@IvanNeretin the nitrogen and the silver already have their highest oxidation numbers.

True, but are they the only elements here?

Then you might want to read the formula again: AgNO3. Still no other elements besides Ag and N?

but what's the potential: O(2-) = 1/2 O2 + 2e(-)? How do I know this is the non-spontaneous reaction?

Well, if it were spontaneous, then AgNO3 would not exist in the first place, would it?

$\ce{NO3- + Ag+ -> NO + O2 + Ag(0)}$ is a balanced reaction. Based on published half-reaction potentials, you should be able to work out the thermodynamics of the overall reaction.

@Andrew I'd opt for NO2 (not that it matters much). More importantly, the published potentials are for water solutions, and hence useless.

@Ivan Neretin that's pretty obvious i just want to know if the electrolysis can actually happen. Im now convinced it can but we can't calculate the potential to apply to the cell, and the reaction forms O2 at the anode and Ag at the cathode

That's right. Molten AgNO3 is a pretty exotic medium, so I wouldn't expect the potentials to be readily available in the reference books. Chemistry is an experimental science, people say.

In principle, a nitrate reduction in the presence or silver ions only would have to give silver oxide; but this decomposes between 200 and 300°C.  So, perforce, you would ultimately get elemental silver from electrolyzing the nitrate melt.
However, that does not mean it's best.  With a little additional heating, to 440°C, you get decomposition to elemental silver just with the heating.


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tolerable error, CASSCF simulation of UV/Vis spectrum
Please forgive my ignorance of experimental spectroscopy. I am a computational chemist, not an experimental one. My question is: what would be considered a tolerable or reasonable difference between an experimental lambda max from that taken from a CASSCF simulation? I am currently trying to design my active space, etc. My target is 4.56 eV but I can only get as close as 5.26 eV (+0.7 eV error) so far. Is it bad or should I proceed to my next step?

0.7 eV is around 16 kcal/mol, i.e., quite a large error for many applications. ("Chemical accuracy" is normally defined as errors below 1 kcal/mol, i.e., <0.05 eV, but that may be too much for many applications too.)

If your active space is indeed complete, then there would be nothing else you could do with CASSCF. It's a qualitative approach either way, so the numbers should point in the right direction, but you won't get them to match the experiment. I guess a thorough literature review would give you an idea what others do and consider working, and you might find out what is necessary to compare your results. Without knowing more of the context, it is hard to judge this error.


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Organic Layer/Aqueous Layer Question
I could not answer this question, it would be appreciated if someone could help.

Lost in translation: hydroxyde de sodium pur ou solution aqueuse?

You could improve your question by adding what part of the question you understand, and where exactly you need help. Or you could give an example of a simpler question you already understand, or a worked example you are aware of.

@orthocresol Note that the OP did not ask for the answer to the question, but for help. I agree that some pointers about what parts of the question are clear to the OP and where the hurdles are would make the question much better. Also, this could lead to an actual question.

@KarstenTheis, (1) the interpretation of "help" is subjective: I read it as being "help me answer it". (2) that is part of the point of closing questions; it is not a final statement, if questions are edited, they will be reopened. It does not have to be OP who edits it, you can do it too, if you think that you can make it into a question that satisfies the community guidelines. (3) Fundamentally, all homework questions can lead to an actual question. That doesn't stop us from closing them.

In my experience, right now this question is a very clear-cut case where the question should be closed. If you disagree feel free to post on Meta where we can discuss in greater detail and solicit more opinions! Again, closure is not the end of a question; there is always the edit + reopen route.

it would be appreciated if someone could help

To answer this question, you need to know about liquid:liquid extraction, acid-base chemistry, and intermolecular interactions. These are the type of questions you would have to consider to answer your question:
Extraction: What is meant by sodium hydroxide as a solvent? What are the two layers? Which one is the aqueous layer?
Acid-base chemistry: Are the solutions neutral, acidic, basic, or something else? What form (as is, protonated, deprotonated) will species A-F be converted to in an acidic, neutral or basic solution? In other words, what acid-base reactions could occur between A-F and the extraction layers?
Intermolecular interactions: What intermolecular interactions can species A-F, in their various protonation states, have with the aqueous layer or with the other layer?
Once you are comfortable with these concepts and have answered or considered all these questions, you should be able to come to a conclusion.


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Identifying orbital given electron probability density plot

Being a high-school student, I only have a qualitative understanding of the wave mechanical model of atom.
Below are my inferences regarding the plot shown in the above picture:
The concerned orbital has $3$ radial nodes (I'm counting the places where the value of the y-coordinate is zero excluding the ones at the extreme left and right).
Hence, the orbital should satisfy $n-l-1=3$. The orbitals satisfying the condition are $4s, 5p, 6d, 7f,\ldots$.
Are my inferences correct?

I think your inference is correct, though the plot looks very awkward. There should be an exponential decay at large values of $r$, but in the plot the curve plunges sharply onto the $x$-axis. Also the shapes of each peak should not be so symmetric, but instead skewed towards the left.

Exactly... Which probably means that the right end is also a node, and the plot is not complete, or it's plotted incorrectly


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Are risks (hexavalent chromium) during Chrome plating temporary, or remain with the product?
Since some chrome plating production processes generate hexavalent_chromium compounds, which are reportedly highly toxic/carcinogenic, I wonder:
Do the hexavalent chromium substances only occur (temporarily) during the production process (i.e. the chrome plating) or will they remain with the product?
My naïve impression has been that after the chrome plating process is over, the product will have an additionally a more or less thick layer of native Cr. Is this correct?
Should I expect impurities or residuals of toxic forms (e.g. hexvalent chromium) to be with the chrome plated this is possible do primitive test exist to detect product?
If it is possible that the chrome plated object could have
worrisome concentrations/amounts of Cr(IV), is there a primitive test to detect this?
Update
Some internet resources, such as tests/detection procedures for Cr(IV) described here, mention

The screening method can be used to determine chromium(VI) in
  passivation layers on metal pieces.

Also test kits for Cr(IV) sold online mention in their description to be

test[s] to detect leachable Cr (VI) ions on virtually any solid surface.

both giving reason to thing that the Cr (VI) might remain in worrisome amounts after the chrome plating. True.
update 2
one of the answers mentioned that "chrome plating", being understood as a general term, does not necessarily imply which material is being plated. If permissible I would like to add that I am most interested in those risks of the base material being copper.
update3
The most upvoted answer has stated little risk if chrome plating procedure is carried out well.
As new information the CHROME PLATINGANDCHROMIC ACID ANODIZING SURVEY strongly suggests that the term chrome plating, can be divided regarding the categories a) Tank process (Hard plating; decorative; chromic acid anodizing) and b) *Tank constituent(s) (Hexavalent chrome; trivalent chrome).

This information to raises the aspect, if the outer passivated layer should be expected to have only Cr(III), irrespective of the used tank constituents.

Many production processes involve toxic intermediates that don't make the final product. PVC doesn't contain vinyl chloride and polystyrene doesn't contain benzene but these are big in the manufacturing routes for the polymers.

Multiple questions should be broken in multiple items, but anyway,
If your plating process is well done, and everything is thoroughly cleaned after, the level of Cr(III) or Cr(VI) remaining should be low enough to be inconsequential to health and the environment. You're correct stating that regardin chrome, thre will be only a passivated layer of metallic chromium and not the trivalent or hexavalent.
Please do remember that the main reason that chrome plating makes things resistant to corrosion is that as soon as chromium is exposed to air, it forms a very thin layer of chromium oxide attached to the surface of the metal, preventing further corrosion. So, in fact, there is some Cr(III) on a finished piece of chrome plated material.
We aren't considering the environment were the finished part will be exposed/used. If aggressive enough conditions are present, the chromium layer will be corroded away and will enter the environment as Cr(III)/Cr(VI).
Regarding Cr(VI) detection, there is a "simple" method involving a solution of diphenylcarbazide described at https://www2.mst.dk/udgiv/publications/2009/978-87-7052-987-7/html/kap04_eng.htm
However, one can easily buy test kits for Cr(VI) on Amazon, ready for use, starting at about (February 2019) USD 110.

Thank you. Since you mention the concept "further oxidation is prevented by a tiny outer fine of chrome (III) oxide", this is also true for many so called stainless steel alloys, correct? I mean with regard to Cr(III)-oxide, that is the outer layer of most kitchen pots, right?

In the link you posted (which describes Cr(IV) detection), it lists quote: "The screening method can be used to determine chromium(VI) in passivation layers on metal pieces." which would suggest, that passivation layers (what I understand to be the Crome plating, correct) are indeed something at risk to contain "leachable Cr(IV)".

Might be worth noting that in the past there have been plenty of occurrences of companies mishandling Cr(VI), leading to permanent disabilities and illness like leukaemia. In other words, it's quite easy to screw it up.

@Mast Does this extend beyond mishandling in Chrome plating? For instance most stainless steel (especially US 304, US 316 food grade) contain considerable Cr amounts, could those be misshanded or are alloys impossible to contain CrIV in considerable amounts?

If chrome-holding paint is not considered plating, then yes, it goes beyond that. With regards to stainless steel, I honestly don't know. I wouldn't exclude the risk without further study.

If we assume that the plating process was done correctly onto tin then according to a XPS study on a chromate conversion layer on tin plate (J. Sun et. al., SURFACE AND INTERFACE ANALYSIS, 2009, volume 41, issue 6, pages 449-452) the conversion layer is a combination of mainly chromium(III) oxide, chromium(III) hydroxide, tin and tin oxides.
However in a study on a chromate conversion layer on zinc, it was found that much of the chromium was still present in the form of chromium(VI). The paper in question was X. Zhang et. al., SURFACE & COATINGS TECHNOLOGY, 2005, volume 197, issue 2-3, page 168-176.
Now this suggests that chromium(VI) is present in the films, I know that the chromium in soil can change from one oxidation state to another. If the soil is alkaline and MnO2 is present then it is possible for chromium(III) to be converted into chromium(VI). I think that the question suffers from the problem that we do not know what the metal is under the chromate conversion coating , also we do not know what environment the object is kept in. I think we know too little to be able to give a perfect answer.
In a study on chromate conversion layers on aluminium, it was found that soaking a film in sodium chloride caused chromate to be leached from the layer. This was published by J. Wan, Physica B, 1995, volume 208, issue 1-4, pages 511-512.
So I think it is possible for chromate to be released.

I thank you for bringing the aspect of the base material. I did not specify because I thought that different processes (i.e. different chemical composition of solutions used in the plating process) have a higher impact on how much Cr(IV) will be contained in the outcome. However if there was something specific to Cu as the base material being chrome plated, then it would be great to know about that.

The answer given above by Flyingfenix is on the spot. However, I felt reader may need to know the possible interference of the test mentioned to check $\ce{Cr(VI)}$ in finished products. This method is called Method 7196 by EPA (USA). They have published that Method 7196 may be used to analyze samples containing from $0.5$ to $\pu{50 mg}$ of $\ce{Cr(VI)}$ per liter ($\pu{0.5-50 ppm}$ levels). In Method 7196A, it was mentioned that:

3.1 The chromium reaction with diphenylcarbazide is usually free from interferences. However, certain substances may interfere if the chromium concentration is relatively low. Hexavalent molybdenum and mercury salts also react to form color with the reagent; however, the red-violet intensities produced are much lower than those for chromium at the specified pH. Concentrations of up to $\pu{200 mg/L}$ of molybdenum and mercury can be tolerated. Vanadium interferes strongly, but concentrations up to $10$ times that of chromium will not cause trouble.
3.2 Iron in concentrations greater than $\pu{1 mg/L}$ may produce a yellow color, but the ferric iron color is not strong and difficulty is not normally encountered if the absorbance is measured photometrically at the appropriate wavelength.

The other issue I have with the test is its use of ethanol as a solvent to prepare diphenylcarbazide solution. However, major theory of the test is $\ce{Cr(VI)}$ would oxidize diphenylcarbazide to diphenylcarbazone, which would produce a red-violet color. But, it's a known fact that ethanol itself would get oxidized by $\ce{Cr(VI)}$. For example, $\ce{K2Cr2O7}$ would oxidize ethanol at $\pu{25 ^oC}$ with rate $K_{\mathrm{obs}} = \pu{0.193 min^{-1}}$ (pseudo-first-order kinetics; Ref.1). Don't you think it would (negatively) interfere as well? Other than that, this is a good test for quick determination of the presence of $\ce{Cr(VI)}$.
Reference:

M. Hassan, A. N. Al-Hakimi, M. D. Alahmadi, “Kinetics of Oxidation of Aliphatic Alcohols by Potassium Dichromate in Aqueous and Micellar Media,” S. Afr. J. Chem. 2011, 64, 237–240 (https://www.ajol.info/index.php/sajc/article/view/123748).

In your answer you mention that "[t]he answer given above by Flyingfenix is on the spot.", which suggested that "If your plating process is well done, and everything is thoroughly cleaned after, the level of Cr(III) or Cr(VI) remaining should be low enough to be inconsequential to health and the environment.". I would be interested in how you understand this sentence, in particular what is meant by "cleaned after" part.

Your concern should be answered by @Flyingfenix, because you are directing to his comments. My article is purely based on test for the presence of $\ce{Cr(VI)}$.


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Is it possible to make polyethylene in a high school laboratory?
We learned how to make small amounts of ethene from ethanol using sulfuric acid.
Is there a way to make a visible amount of polyethylene from that using materials I could get for a school lab? We can't do pressures of 1000+ bars obviously, but we don't need high efficiency either -- just enough of a solid for the students to see that something happened.

Suggestion: Perhaps the aim is to show a / any polymerization?  The RSC put it's LearnChemistry section teaching ressources together, e.g., about an addition polymerization (http://www.rsc.org/learn-chemistry/resource/res00000479/addition-polymerisation?cmpid=CMP00004755) about polystyrene or polymerization at an interface to form Nylon (http://www.rsc.org/learn-chemistry/resource/res00000755/making-nylon-the-nylon-rope-trick).  Even after years I recall the later one ...

Not polyethylene, but I recall doing an experiment in school where hexamethylenediamine and adipic acid were poured carefully so one was layered on top of the other.  They did not mix in bulk, but reacted at the contact surface to form a Nylon-type polymer barrier.  We poked a rod in to pull the polymer out (enabling more reaction) and wind it around the rod!

Yup! I know that one. It's cool, and a nice demonstration of polymerization, but not exactly what I was looking for.


text:
Inductive vs conjugation effects in ester hydrolysis
I just came across the following problem about alkaline ester hydrolysis:

Since all of the substituents are in the para position, I immediately thought about the ability of those substituents to contribute electron density (through conjugation) to the ring and, therefore, the carbonyl group. Thus, I ruled out the p-methoxy and the p-toluene compounds since their conjugation effects lowers the electrophilicty of the carbonyl group.
Since I knew that chloro groups are ortho-para directors in electrophilic aromatic substitution, I also concluded that the chloro group also donates electron density (as shown below) and should also be less reactive at the carbonyl group than the non-substituted benzene ester, which I concluded to be the most reactive compound.

However, the answer key says that the p-chloro compound is the most reactive due to its electron-withdrawing abilities (which should be through induction only).

However, wouldn't those effects be minimal in comparision to the conjugation effects, or does the large nature of chlorine's 3p orbital vs carbon's 2p orbital make conjugation mininmal?

Clearly, you have not read benzene halogenation before. In the electrophilic substitution of halogens in a benzene ring, the halogen, taking Cl for example, deactivates the ring precisely because its inductive effect of withdrawing electron density from the benzene ring outweighs the resonance effect of Cl donating its lone pair into the ring. It is experimentally proven that chlorobenzenes are less reactive, and you can understand the reason why inductive effect outweighs the resonance effect as the Cl atom, being highly electronegative, is "not willing" to readily donate its lone pair of electrons into the ring; High nuclear charge causes Cl atom to hold tightly to its valence electrons, thus they are not available for donation. Also explains why Cl- ion is a weak conjugate base and weak nucleophile.

Actually, I am familiar with electrophilic aromatic substitution of halogenated benzenes. I just wasn't sure how the effects would change in the presence of an ester group. I would also like to point out that the main reason for weak conjugation is the weak overlap of the carbon 2p and chlorine 3p orbitals, not electronegativity; after all, oxygen is more electronegative than chlorine, and the methoxy group is a strong activating group.


text:
Oxidation products of propylene glycol
Working in a building that uses propylene glycol (PG) as antifreeze in a hydroponic heating system.  With complaints of chemical odors in offices, I conducted air sampling for glycols, aldehydes, and PAHs.  No airborne glycol or PAHs were detected, but low levels of aldehydes; acetone, acetaldehyde, butyraldehyde and formaldehyde are present building wide.  Not a chemist, so question is, if there is a leak in a closed hydroponic heating system, does the PG oxidize to aldehydes when it encounters occupied air?
A thick bioslime was present on the interior surface of the mechanical room glycol tank.  I was told that when glycol is exposed to air, it generates nutrients for microbes.  The bioslime analyzed as Yeast, and some PG is generated by Yeast and carbohydrates - all news to me.

By "bioslime" you probably mean "biofilm". You will find a lot more articles about "biofilms" on the web


text:
Why cant nail polish be used as an acid mask for making a PCB?
In the question: PCB etching with sodium persulfate, permanent markers seem the most used and experimented with. The thickness of nail polish would give more solid planes and traces.
Might nail polishes resist most common copper etchants mentioned such as:

Sodium Persulphate
Ferric Chloride
Peroxide + Vinegar + Salt

sufficiently for etching a PCB?

Chemistry is an experimental science. Part of the problem would be with removing the nail polish which takes a very reactive solvent. So try experimenting....

Nail polish works well with nitric acid and removal is done quite easily with acetone. Not sure it helps but the experiment is really easy.  Just try on some copper of no interest.