text:
Does a diatomic gas have one or two vibrational degrees of freedom?
Many sources claim that the vibrational degrees of freedom for a diatomic gas is one, but there are also a few which claim it to be two. Which is correct, and why is there even any ambiguity to begin with?

Because the few that claim it to be two are wrong.

How many do you think there are?

Related: How does the equipartition theorem apply to vibrational modes?

One vibrational degree of freedom but two rotational degrees of freedom.

I think this confusion arises from quite different definitions of degrees of freedom. Here are three different definitions that could reasonably arise:

Positional degrees of freedom. This leads to the common $3N$ degrees of freedom based on movement in three directions. Refer to Jan's answer. (Question: Bonds constrain atoms, so why don't they reduce the number of degrees of freedom? Answer: Bonds are not rigid constraints; the atoms don't necessarily remain a fixed distance apart.)
Phase space degrees of freedom. Here we have $6N$ degrees of freedom: $3N$ from the positions of all particles, $3N$ from the velocities of all particles. Positions and velocities are independent, so to specify a system of $N$ particles exactly requires $6$ degrees of freedom per particle. (We also presume no internal structure; as example, for rigid bodies, there are $12$ degrees of freedom: the $6$ mentioned above, $3$ more for specifiying the orientation of the body, $3$ more for describing the speed of rotation of the body.)
Quadratic degrees of freedom. This may seem like a strange distinction to make, but is a key requirement of the equipartition theorem, which is one of the areas where the concept of degrees of freedom is first introduced. These degrees of freedom are the subset of the phase space degrees of freedom whose elements all contribute quadratically to the system energy.

We will clarify these degrees of freedom with the example of a diatomic gas. In this case, there are $6$ positional degrees of freedom, $12$ phase space degrees of freedom, and $7$ quadratic degrees of freedom. Jan's answer describes the $6$ positional degrees of freedom well; specifying the $6$ analogous velocity degrees of freedom gives the $12$ phase space degrees of freedom. Let us enumerate the $7$ quadratic degrees of freedom: $$\dot{x}, \dot{y}, \dot{z}, \dot{\theta}, \dot{\phi}, q, \dot{q} \quad \Longleftrightarrow \quad \frac{1}{2}m\dot{x}^2, \frac{1}{2}m\dot{y}^2, \frac{1}{2}m\dot{z}^2, \frac{1}{2}I\dot{\theta}^2, \frac{1}{2}I\dot{\phi}^2, \frac{1}{2}kq^2, \frac{1}{2}\mu\dot{q}^2$$ Positional degrees of freedom do not play into this, because there is no quadratic energy contribution associated with them. Instead, we have the six velocity degrees of freedom from the two atoms, $\{\dot{x}_1, \dot{y}_1, \dot{z}_1, \dot{x}_2, \dot{y}_2, \dot{z}_2\}$, which we replace by the conventional center-of-mass coordinates $x, y, z$ and the two angles $\theta, \phi$ that describe the orientation of the molecule, $\{\dot{x}, \dot{y}, \dot{z}, \dot{\theta}, \dot{\phi}\}$. One degree of freedom has "gone missing", because by symmetry there is no (quadratic) energy associated with rotation about the bond axis. The bond also produces two more quadratic degrees of freedom, represented by the vibrational coordinate along the bond, $q$, corresponding to the potential and kinetic energies of vibration.
Hence in one sense there exists only one vibrational degree of freedom; in another sense, two.

Remark. One might argue that the two vibrational degrees of freedom $q, \dot{q}$ are not independent, as they should be related by the equation of motion for a harmonic oscillator; this is important for the statement of the equipartition theorem. This is true for an isolated system, but presumably collisions decorrelate $q$ and $\dot{q}$---I don't have a better answer for this.
At any rate, experimental evidence (heat capacities) does corroborate seven quadratic degrees of freedom for diatomic molecules at high enough temperatures, when vibrational modes are active.

Excellent, an answer clarifying this was much overdue! There's too much confusion over why one vibrational mode leads to two quadratic DOF and neither of the other answers really got to the crux of the matter, imo.

Awesome,  your answer clarifies everything, Thank you cyclohexane

Couldn't the confusion also be between vibrational and rotational degrees of freedom?

I had never heard about all the other degrees of freedom you mention. Thanks for educating me, too!

@md2perpe, why do you say that?

Because that's my guess. Doesn't a diatomic gas have one vibrational degree of freedom and two rotational degrees of freedom?

@md2perpe, oh, I see what you mean.

@a-cyclohexane-molecule In the case we have a real gas, how we take into account for intermolecular interactions? Assuming that the intermolecular potential follows an inverse relationship (between the positions of the particles e.g. $\frac{1}{r^2}$), there is no degree of freedom we can assign to it (based on the above definitions). I am asking this, because some consider the intermolecular potential as an extra dof (another way we can "put" energy into the system).

You can think of the degrees of freedom with respect to rotation, translation and vibration in the following way:
Each atom can, in theory, move in all three orthogonal directions of your standard coordinate system ($x, y, z$). Depending on which coordinate system you use (e.g. polar coordinates would be a valid alternative) you may have different ‘directions’ but the core principle remains: each atom has three ‘ways to go’. Therefore, the entire molecule has $3N = 6$ degrees of freedom.
Next we can try to analyse how this molecule can exercise its freedom. For example, there are three different ways in which both atoms move in the same direction at the same speed. These are the degrees of translation, and there are always three ($x, y, z$ or whichever coordinates you chose).
We might also assume that one molecule travels in $x$ and $-z$ direction while the other one travels in $-x$ and $z$ direction; this would be a rotation around the $y$ axis. The same with reverse coordinates can be constructed around the $x$ axis (assuming the $z$ axis coincides with the bond direction). These are the only two rotational modes, any other attempted rotational mode is actually non-movement (around the $z$ axis) or a combination of the former two. Thus, there are two rotational degrees of freedom ($R_x$ and $R_y$).
Finally, you can think of a degree of freedom in which one atom moves in $z$ direction while the other moves in $-z$ direction; this is the vibrational mode, characterised by the bond length being streched or shortened. There is only one possible vibrational mode $V_z$.
These degrees of freedom add up to $6$, so we have them all. Any other movement you can construct can be broken up into a linear combination of the previous six modes.

Now let’s assume for a second the statement of two vibrational modes were true. Let’s attempt to find the second vibrational mode by assuming it to be a flip-flop vibration (one atom moves in $x$ direction, the other remains still). Careful analysis will reveal that this hypothetical $V_x$ mode is actually a linear combination of $R_y + V_z + T_x$ — rotation around the $y$ axis (moving the $x$-displaced atom one way and the other one another way while moving them both closer to the centre in $z$ direction), $z$-directional vibration (countering the movement in $z$ direction giving an apparant movement in $+x$ for the $x$-displaced atom and $-x$ for the other) and $x$-directional translation (putting the non-displaced atom back on its original position).
If the reader wants to take up the exercise, they can attempt to find any other, potentially more complex ‘second vibrations’. They will find that all can be written as a linear combination of the six previously mentioned modes. Without formal proof, we can therefore reasonably assume that the statement ‘diatomic molecules have two degrees of vibrational freedom’ is false.
On the other hand, the reader may attempt the exercise of trying to describe any of the six modes presented before the horizontal line as a linear combination of any of the other five. I hope I am not spoiling the fun if I tell them they will fail, since these six modes are indeed orthogonal to each other.

There are two ways to think about the degrees of freedom for a vibration. One of them is to think only about the number of vibrational modes. The number of vibrational modes for a linear molecule is given by the formula
$$3N-5$$
Where N is the number of atoms. For a diatomic molecule, $N=2$. Hence, the number of vibrational modes is considered to be 1. This is why the degrees of freedom for a molecule would be considered to be 1.
Another way to think about the degrees of freedom Or they may be talking about the energy  degrees of freedom associated with each of the vibrational modes. Vibrating atoms will have two forms of energy: potential and kinetic. The kinetic energy comes from the fact that vibrating atoms are moving closer and farther from each other. The potential term comes from the fact that the atoms are "bound" to each other; they will experience a force that depends on the positions of the atoms. If we treat the bond as a harmonic oscillator, the potential energy will be equal to
$$k(x-x_0)^2$$
Where $x_0$ is the equilibrium length of the bond, $k$ is force constant, and $x$ is the position. Hence, each vibrational mode will contribute two degrees of freedom.
Therefore a diatomic molecule would have 2 energy degrees of freedom since it has one vibrational mode. A linear triatomic molecule would have 4 normal modes.
$$3N-5=3(3)-5=4$$
And it would have 8 energy degrees of freedom associated with it
However, the first interpretation is more common.
TLDR: Both of them are correct. It depends on what type of degree of freedom you are looking at (as answered by a-cyclohexane-molecule).

Helpful link: https://physics.stackexchange.com/questions/206303/the-potential-energy-degree-of-freedom

You explanation is self-contradictory. The sentence "Another way to think about the degrees of freedom is to consider the kinetic energy of the atoms in addition to of the vibrational mode" is discussing degrees of freedom as opposed to vibrational degrees of freedom. Also if you want to throw in kinetic energy there are  rotational motions also.

@MaxW Sorry, I should have written the answer more carefully. I edited the answer. Hopefully that addresses your concern.

Degrees of freedom are related to the quadratic terms in final energy expressions, but also are related to the number of variables needed to describe the movement of a system. To specify the movement you need 3N variables and to specify vibration (movement) you need 3N-5=1 (only one possible movement). But talking in equipartition language, it has 2 degrees of freedom. Why? Because the energy of a spring has the following expression:
$$\frac{1}{2}mv^2 + kx^2$$
so we have 2 degrees of freedom. It answers your question.
I will add something related to quantum chemistry.
The previous statements are derived from classical statistical mechanics (which result on equipartition principle). However, if the system is at low temperatures (below $T_{amb}$ for some diatomic molecules as $\ce{N2}$) vibrational modes aren't excited. In that condition the molecule only store energy in rotation and translation, but not in vibration. You can check it here. And classical mechanics fails to describe molecules.
The previous phenomena is explained by quantum chemistry. When can we safely neglect quantum effects? When $kT>>\Delta E_n$ being $\Delta E_n= E_n -E_{n-1}$ and being $E_i$ the energy of the $n^{th}$ state. A proof of this can be done by measuring heat capacity of gases.


text:
How to properly store acids at home?
I bought different kinds of acid for experiments and home usage that I stored in secured containers into an IKEA cabinet.
I just realized that despite the fact that all the containers are properly closed, the metal parts of the cabinets are completely rotten. I am quite surprised because there shouldn't be much vapor in there.
What is the correct way to store acids at home? Should I purchase some specific container? How can I prevent vapors from escaping? Perhaps the problem is somewhere else and my containers are not good enough for my acids.
The acids I have: $\ce{H2SO4}$ (30%), $\ce{HCl}$ (32%), $\ce{H3PO4}$ (85%), $\ce{HNO3}$ (10%), $\ce{CH3COOH}$ (98%).

Note: the chemicals are actually stored in my electric/electronic workshop which is a separate room at home and always locked when I have visitors.

The correct way is not to store acids at home.

Sorry about that, but you see yourself that an IKEA cabinet is not good enough.

@IvanNeretin well, so the solution is to buy something else like a chemical cabinet. Unfortunately these cabinets are also made of steel. So either I buy a hermetic cabinet or something else. I mean the problem would be the same if I want to store them in a workshop...

An proper acid cabinet isn't hermetically sealed. It is vented using an outside vent.

@MaxW I was afraid of this answer, but I can't afford this yet so I am looking for something else. What I am wondering is why do I have acid vapors since the bottles are all properly closed and made of thick plastic.

A "sealed" bottle contains liquid well. It does not contain gases well. There is also typically a bit of liquid left on the bottle threads after pouring some acid.

Unless your experiments require conc. HCl, you could store something about 10 ‰ and you should not see corrosion anymore, or at least just slow down the process. Among your list, problem is HCl 37 % with its fumes. You could also store the bottle you have in an almost sealed plastic box on the bottom of which you make a bed of baking soda, to be periodically renewed.

@dotancohen, the door wasn't closed

How long did it take to rust like that? Several years or just a few months or even weeks?

I would say several months.

@nowox several months sounds quite fast, but if it was a few years, than meh. In all 6 places storing acids I worked in, acid cabinet did rust somewhat heavily. Nobody really cared except to ensure reasonable ventilation. It takes too much effort to ensure sealing tight enouhg for it to be practical.

Salt water is enough to corrode some metals that bad. I once used metal tweezers to pick salt crystals from salt water solution and after a couple of weeks of just lying there they totally corroded.

First I'd locate the bottle which causes the problem. Usually HCl is #1 suspect, but to be sure you can put a vial with smelling salts (aqueous solution of $\ce{(NH4)2CO3}$) or ammonia in the box with acids; white coating of $\ce{NH4Cl}$ on the bottle signifies the leak.
It's also a good practice to store acids in glass bottles with a proper joint (teflon ring) and a screw cap (e.g. Merck's SafetyCap). I would strongly recommend to get the proper bottles as soon as possible. Plastic bottles are only used to reduce the production and transportation costs, they are a poor choice for a long-term storing of chemicals. Even thick plastic is prone to diffusion, whereas glass is not; also plastic is, well, plastic, and is prone to mechanical deformations, so it's also tricky to maintain an impermeability of a bottleneck-cap joint over time.
In the meantime I'd wrap the necks of the bottles with parafilm, and/or pour some baking soda (sodium bicarbonate) on the bottom of the container where you store the bottles. This should help to neutralize the vapors before they reach the furniture and your nose, but it's a temporary solution.

Very good answer. I will try your smelling salts trick and look for ammonium chloride on the bottles. I am also suspecting the HCl. I will buy glass bottles with PTFE rings as well.

I'd check for chemical vendors in your vicinity, or maybe eBay or LabX (sometimes labs do sell equipment they don't need, so you might also get other stuff for the home lab for a reduced price, or even for free), and search specifically for the bottles to store acids with.

I am working in a research facility, I will ask some chemist there if they can help me with the purchase.

@nowox Oh, I forgot to mention that it's a good idea to get an amber glass bottle for $\ce{HNO3}$ as it's considerably light-sensitive. Other acids you listed are going to be fine in a regular glass bottles.

Glass works fine , except for HF.

Do try to store these Acids in Glass containers..i have seen this in my school/college where they keep it inside glass cylinders.
Also glass containers being strong they hold the acids quite strongly and prevent any leakage to outside air.

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Ellingham Diagram
My textbook says the following about the Ellingham diagram:

Each plot is a straight line except when some change in phase ($\ce{s -> liq}$ or $\ce{liq -> g}$) takes place. The temperature at which such change occurs is indicated by an increase in the slope on the positive side (e.g., in the $\ce{Zn}$, $\ce{ZnO}$ plot, the melting is indicated by an abrupt change in the curve).

However, I do not understand this. If a solid metal oxide changes into a liquid or a liquid into a gas, shouldn't the entropy increase? And as a result, the change in entropy should become less negative with regards to the equation:

Ellingham diagram normally consists of plots of $\Delta_\mathrm fG^\circ$ vs $T$ for formation of oxides of elements i.e., for the reaction,
$$\ce{2x M (s) + O2 (g) -> 2 M_xO (s)}$$
In this reaction, the gaseous amount (hence molecular randomness) is decreasing from left to right due to consumption of gases leading to a negative value of $\Delta S$ which changes the sign of the second term in equation (6.14). Subsequently $\Delta G$ shifts towards higher side despite rising $T$ (normally, $\Delta G$ decreases i.e. goes to lower side with increasing temperature). The result is positive slope in the curve for most of the reactions shown above for formation of $\ce{M_xO (s)}$.

If this is the case, the slope should become less positive according to the equation:
$$\Delta G = \Delta H - T\,\Delta S$$
where $-\Delta S$ is the slope. Am I missing something?

Qualitative explanation
We can all agree that
$$S_{solid}<S_{liquid}<S_{gas}$$
Metal oxides are generally ionic and have high melting points. Let us say that they do not change phase. So, let us assume $S_{oxide}$ doesn't change as much.
Metals, however, may change their phase.
$$(S_{oxide}-S_{solid})>(S_{oxide}-S_{liquid})>(S_{oxide}-S_{gas})$$
Note that $(S_{oxide}-S_{metal})$ is your $\Delta S$. As $T$ increases, at the phase-change temperatures, $\Delta S$ decreases, $-\Delta S$ increases $\Rightarrow$ the slope increases.

We can explain this with the example of $\ce{CO + O2 <=> CO2}$.
Here we have to focus that there are two reactants on one side and there is only one reactant on the other.
The sum of the entropy of the reactants is subtracted from the entropy of the products, which would result in a negative entropy change in the present example of ideal gases, and hence a positive slope.

In the reaction of solid metal with oxygen the product is also solid. So CHANGE in entropy in the reaction is comparatively lower. Because solid has lower entropy and the product as well as one of the reactants are solid.
But when a molten metal and a gas (oxygen) react the initial entropy is much higher (liquids have more entropy than solids) and they react to form a solid, so change in entropy is much more negative (higher entropy to lower entropy).
So in the equation since $-\Delta S$ is the slope if $\Delta S$ is more negative slope becomes more positive

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How to calculate the percentage dissociation of phosgene using the pressure equilibrium constant?

In phosgene gas reaction at $\pu{400 ^\circ C}$, the initial pressure $p_{\ce{CO}}= \pu{342 mmHg}$ and $p_{\ce{Cl2}}= \pu{352 mmHg}$ and the the total pressure at equlibirum is $\pu{440 mmHg}$.
  $$\ce{CO + Cl_2 = COCl_2}$$
  Calculate the percentage dissociation of phosgene at $\pu{400^\circ C}$ at $\pu{1 atm}$.

Let $x$ be the decrease in pressure of both $\ce{CO}$ and $\ce{Cl_2}$
\begin{array}{lcccc}
 & \ce{CO &+& Cl_2 &<=> &COCl_2}\\\hline
\text{Initial pressure of substance} & 342 && 352 && 0\\
\text{Final pressure of substance}   & 342-x && 352-x && x\\\hline
\end{array}
$$342 - x + 352 - x + x = 440 \implies x = 254$$
Using the formula for $K_p$ I found it to be $\approx 13$.
For the dissociation at $\pu{760 mmHg}$,
$$\ce{COCl2 <=> CO + Cl2} \implies K_p= \frac{1}{13}$$
Let $y$  be the decrease in pressure of phosgene which is proportional to the amount of substance of it lost or it's degree of dissociation.
$$\frac{1}{13} = \frac{y^2}{760-y} \implies y \approx 7.6$$
The value of $y$, that I have obtained, must be proportional to percentage dissociation, so percentage dissociation should be $7.6 \%$. But the answer given is $20.6 \%$.
I am not looking for the entire solution but would like to know where have I misapplied the concepts. I am doubtful about my second step, but can't figure out how it's wrong.

I don't understand how you computed $K_{\mathrm{p}} = 13$. I got $34.0$ Double check your work.

Here goes a possible answer, I tried to describe it step by step so you can see your (and my) possible errors. The first part of your calculation was conceptually good, but you make the calculation wrong.
As one of the comments say, $$K_p=\frac{(342-254)*(352-254)}{254}=33.95$$
So you have $K_p$. Then
\begin{array}{lcccc}
 & \ce{CO &+& Cl_2 &<=> &COCl_2}\\
\text{Initial pressure of substance} &0 && 0 && A\\
\text{Final pressure of substance}   & x && x && A-x\\~\\
\end{array}
With the condition that $x + x + A-x=760$, we have $A+x=760$.
Using our first calculation:
$$K_p=\frac{(x)*(x)}{A-x}=33.95$$
(Kp is the same at same $T$)
$$K_p=\frac{x^2}{760-2x}=33.95$$
here we get x=130.229, then:
dissociation of $A= 1-\frac{(A-x)}{A}$
we get dissociation of A=1-0793=0.207
or 20.7% (0.01% flew away)

@Abcd Neither 13 or 1/13 are correct... Also, I think you'd need to assume $K_{\mathrm{p}}$ was relatively constant over a range of pressures to solve this problem.

@Zhe Isn't $K_p$ independant of pressure? I mean is this not true generally?

We generally assume it to be independent, and it likely is fairly constant for a range of pressures, but there's no reason why it should be constant for all pressures. $\Delta G^{\varnothing}$ isn't even constant at all temperatures.


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Is associative ligand exchange possible with octahedral complexes?
Can associative ligand exchange produce an octahedral intermediate as long as the initial complex has 16 electrons (unsaturated)?
Not sure. all I ever see is a square planar turning into a trigonal bipyramidal intermediate?

It could — and there isn’t even a general restriction on the electron count.
Any pentacoordinated species can easily be expanded to a hexacoordinated one. It is just a question of the ligand approaching from the correct angle. In fact, I would even want to prefer associative ligand exchange for pentacoordinated species over dissociative since there is less reorganisation involved (only two ligands need to shift positions compared to at least three in dissociative).
Also, it should be mentioned that the number of electrons in each structure does not have to be 18 or less. Many (pseudo)octahedral complexes such as $\ce{[Co(H2O)6]^2+}$, $\ce{[Cu(NH3)4(H2O)2]^2+}$ or $\ce{[Ni(H2O)6]^2+}$ have more than 18 electrons and are fully comfortable with it.


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Could the loss of mass in the kilogram be due to oxidation?
I was wondering about this. I have heard, and known for a while, that the famous and celebrated Prototype Kilogram - the lump of metal whose mass is used to define the standard mass scale worldwide - is slowly changing, degrading in some fashion, and this has prompted the search for a new and better definition of the kilogram mass unit on something more stable (the lead is now Planck's constant, but the change has not yet been adopted by the BIPM and looks like it won't be until at least next year.).
However what I'm curious about is the reason for the mass change. Apparently it's said it's not clear as to what exactly could be causing it, but I was wondering: is it chemically possible that this mass change could be due to oxidation - tarnishing - of the kilogram and removal of the oxide? I would imagine that the scientists who have studied this must have thought of this, so I'd be more interested in why it's dismissed as an explanation. In particular, it's said that the material it's made of - mostly platinum - has a very low oxidation rate, but "very low" need not equal "zero", and the mass change is reported as up to (unclear, because it's the measuring standard itself!) 50 micrograms over 100 years - which is nonetheless a very slow rate of loss - about $10^{-16}$ mol Pt/s, so that it would take $10^{16}$ s (10 Ps) to oxidize just one mole, or about 300 million years. Could this be within range of the very-slow-but-not-zero range of oxidization of platinum in air? In particular I find this:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0ahUKEwjTsN2bleLWAhWB5YMKHf7_A14QFggyMAI&url=http%3A%2F%2Fwww.technology.matthey.com%2Fpdf%2Fpmr-v19-i4-135-140.pdf&usg=AOvVaw2D87XQhqSGonkmrjf0kD_a
talking about platinum oxidizing in air upon mild heating, and even losing mass. Now it seems that a rule of physics is "whatever isn't forbidden is compulsory", that is, whatever isn't absolutely impossible will eventually happen given a suitably long time period, and so if the oxidation is possible could it still be going on at very reduced rates at room temperature that nonetheless might still be sufficient to degrade the kilogram? Or is the rate then so low as to not contribute at all? (I'd think about Arrhenius rate law, but no idea what the relevant constants would be. Nonetheless exponentials are pretty steep, so this might rule it out.)

What says they're storing it under inert gas?

I would think that no one here has enough data to answer this question.

@Zhe Not enough data? https://en.wikipedia.org/wiki/Kilogram#Stability_of_the_international_prototype_kilogram

@Mithoron "The reason for this drift has eluded physicists who have dedicated their careers to the SI unit of mass. No plausible mechanism has been proposed to explain either a steady decrease in the mass of the IPK, or an increase in that of its replicas dispersed throughout the world." I think they would say they need more data...


text:
Enantiomers of 1,2-dichlorocyclohexane
The other day in class we were discussing the enantiomers of 1,2-dichlorocyclohexane and my teacher told us that (R,R) and (S,S) are different isomers while (R,S) and (S,R) are the same.
I understand how (R,S) and (S,R) are the same (Because they can be rotated) but I don't understand why (S,S) and (R,R) are not the same, even though they are the same if you rotate them too.

Classical example - you can also rotate your hands, but that wouldn't make them identical.

The point is that you cannot rotate them. Rotation doesn't change the identify of the stereocenter. That's why rotation of (R,S) becomes (S,R). But rotation of (S,S) is still (S,S).

The one on the left (RR) and the right (SS) are not superimposable mirror images of one another and are thus enantiomers. In your head, try flipping one of them. For instance, if you flip the RR molecule on the left as I drew it, it will look something like this:

The two pictures I posted are depicting two same molecules, except the one on the left was flipped around an axis in plane of the screen. With the second picture, it's easier to see that these two molecules are non-superimposable mirror images. Also notice how neither of these molecules have a plane (or point) of symmetry.
I recommend trying to build this with a model kit if you can't visualize it in your head easily.

I actually want to know how two o,p dichloro cyclohexane with two chlorides projecting out are different from two projecting inwards. I can visualise this thanks

RS and SR are the same molecules. There's a plane of symmetry going through the molecule. Since they also have chiral centers, you could characterize them as meso.

The same goes for the two molecules given in the answer you could view one from under the plane of the screen and they're the same

No, that is false. There is no plane of symmetry for the molecules I drew pictures of, and their mirror images aren't the same. If you view the left molecule on the top picture from the other side of the plane of the screen, you'll see the left molecule in the lower picture. That is not the same molecule as the molecule on the right. I think you're having trouble visualizing the molecules in 3D.

Yes, you're right. Sorry, my bad. I didn't visualise it properly


text:
Rate equation for an elementary reaction
Yesterday, I had a debate with a friend of mine over the determination of the rate of elementary reactions. He said:

Rate equation of elementary reaction can be directly given by just raising the reactants concentration terms to the respective coefficients of the balanced chemical equation whose rate is to be determined.

and he also gave the example of this reaction
$$\ce{2NO (g) + O2 (g) -> 2NO2 (g)}$$
whose rate is
$$
\ce{[NO]^2[O2]}.
$$
Questions:

Can I write the rate equation of an elementary reaction by just simply looking at the stoichiometric coefficients in the balanced equation and raising them over the concentration of reactants?
Are there any examples of elementary reaction where the exponential powers on the concentration terms in the rate equation comes different than the stoichiometric coefficients of that reaction?

Yes you can write the rate equation of an elementary reaction like that. The problem is, you often don't know if the reaction is really elementary.

There's a reaction of decomposition of ammonia on Pt at high temperature. It goes like this 2NH3 --> N2 + 3H2. The order of this reaction is zero. Does that mean the reaction is not elementary?

Yes, pretty much so.

Your NO oxidation is blatantly multistep.

@Mithoron Yep! I just realised this

Even if it is an elementary step, the order might not be given by the stoichiometry. The observed order of a component $C$ will correspond to the amount of $C$ that is participating in the rate-determining step. E.g.
SN1 has only one component taking part in the rate determining step, thus first order, SN2 has two components and is thus second order (and first order with respect to each single component).
So here is the problem. Say you are looking at an E2 reaction. The reaction is second order, as there is a base and a starting material that gets deprotonated.
However, just by looking at the stoichiometry:
$$
 \text{Propylchlorid} + \text{Base} \rightarrow \text{Propene} + \text{Base}\cdot\text{HCl}
$$
you will not know if it is an E1 (first order), if the chlorid dissociates first, or E2 where the base directly participates in the rate determining step.
To make it even worse, two equivalents of base could participate in the rate determining step.
Therefore, the stoichiometry has nothing to do with the rate order.
You can imagine it like this:
If I take a look at your shoes, I see two shoe laces and you have two hands. But given this information, it would be wrong to consider that to bind you shoes, you take only one hand, as there is one hand per shoe lace. Instead, you probably took both hands to bind your shoes, as it is much easier this way. Binding your shoes is kinetics, shoe lace/hand is stoichiometry, and they are completely orthogonal.

The reaction order with respect to a reactant species cannot be determined from the chemical equation alone. You would have to be given more information. In real life, orders of chemical reactions are found experimentally.


text:
Lack of colour in Nitric Oxide
Usually compounds' colours are explained by the presence of unpaired electrons which is free to excite and de-excite giving the compound a charectistic colour. NO has atleast one unpaired electron but even then it is colourless. Can anyone tell me what the reason is?

I am not a physics major, this post will be rather free of equations and spherical chickens. This is a chemist's view of the situation ;)
You need more than an unpaired electron to create color in a substance. You need two places for it to be, with a given energy difference in the visible color range. Looking at the molecular orbital diagram of NO we see that no such organisation is possible (within reasons). The only choice is to go from one $\pi^*$ to the other, but those are at the same energy level. There is no way for this diatomic molecule to deform in any meaningful way so as to create a difference in energy levels between them. Metal complexes, especially iron- and cobalt-based ones, can do this, but not with the $\pi^*$ in any case. There is only $\sigma^*_2$ left, but the distance from $\pi^*$ to $\sigma^*_2$ is in huge, larger than visible spectrum (an event that knocks an electron up there is likely to unravel the entire molecule).
All in all, there is no possibility for the electron to move anywhere that requires the emission of a photon in visible range. It cannot produce the color you seek.

What do you mean by HOMO, and when you said 'it needs to change it's spin direction to "there" '. What exactly did you mean? I understand that you mean other places it can go to but what are the other places and why does it need to change it's spin to go there?

@Beast You would probably benefit from reading on the molecular orbital theory, you need that to understand fully. From MO theory: HOMO is the highest occupied molecular orbital. The uppermost arrow in the diagram. For 2 electrons to co-habitate in a single orbital, they need to have opposing spin. (Pauli exclusion principle)

I know the basics of MO theory and Pauli's Exclusion Principle, I just wanted to know where else the electron could go where it had to change it's spin, did you mean into N or O's orbitals?

@Beast Well, at second glance, I see I was a bit fast writing that. There is no open singly occupied orbital in the MO. So nowhere to go even if the spin flips. Sorry for the confusion there. I'll update the answer.


text:
i need help for a IUPAC name for a branched alkane

Help me finding the IUPAC name for this compound please!

Perhaps 3,4-diethyl-2-methylhexane.


text:
Reverse reaction of acid dissociation
Consider an acid HA with Ka 10^-10 dissociating into its ions as
HA⇌H+ +A-
My question is-if we introduce solution of pH=3(let's say any weak acid or a buffer) and put ions of A- of 0.1M into it (using strong electrolyte),using expression for Ka(in this case we are operating reverse reaction so k will be reciprocal of Ka) we get concentration of HA to be 10^6 M.Although such a high concentration is theoretically not obtained (assume for my doubt to get clarified) my focus is on the following question (value of 10^6 from FORMULA only):-
How is it CHEMICALLY CORRECT that concentration of products has become more than reactants?While dissociating we can make the argument that some acid is left undissociated so concentration decreases.But how does concentration INCREASE in reverse reaction and what is its chemical significance?
Please help.
(I came across this doubt when I saw a question asking to calculate solubility of AgCN in buffer of pH=3 ksp=1.2×10^-15 and ka of hcn 4.8×10^-10.Solution implied concentration of hcn was much greater than cn from AgCN which confused me as explained above.)

Sure enough, $10^6\rm M$ screams nonsense, no matter how it came about.

Can you please write down what exactly you want to add to what? I recommend you also translate that into specific amounts of real chemicals. It will become very clear where you are going wrong then.

The concentration of an acid will never be greater than the concentration of A-. Just because there is not enough of A's to make HA. You are doing something completely wrong in your calcs.
In your problem of AgCN, you are not given the amount of AgCN, so how can you actually think of possible concentration of cyanides?

Can you explain that last sentence "it's still a buffer"

That means the concentration of H+ wouldn't change no matter what happens, as long as we assume it's a good buffer solution, and no "external" equilibria influence it.


text:
Calculating dipole moments from atomic partial charges
How can I calculate dipole moment of a complex molecule where the partial charges are available from density functional theory? I have the coordinates of each atom from atomistic simulations of an isolated molecule in a box along with the Bader charges. Will a simple charge times distance work in this case? The charges are not equal and opposite in many cases. What should the approach be to calculate the dipole in such cases?

We used to do this as a student exercise, albeit using Mulliken charges - it does not work well. The DFT calculation can provide the dipole moment (and higher ones) by using the appropriate operator.

@TAR86 Indeed. Mulliken charges can be very different. In case of stacked molecules, it's even worse. Sometimes totally opposite. DFT does provide dipole moment, but If I want contribution of just one part of a complex structure, then? Something like a local dipole.

Atoms and electrons are not point charges, therefore this approach will likely fail in many cases.

@user3840530 My point was not about Mulliken vs. some other charges - but pentavalentcarbon's answer partially covers that.

@TAR86 Ok. Anyway, sometimes I get really weird result using mulliken charges right from DFT output without any additional post processing, say a positive partial charge on an expected negatively charged species while the whole structure seems fully relaxed. That's why I pointed  it out.

From a theoretical standpoint, that approach is correct:
$$
\mu_i = \sum_{a}^{N_{\text{atoms}}} \sum_{i\in\{x,y,z\}} r_{ia}q_{a},
$$
where the set of atomic partial charges $\{q\}$ could come from partitioning the density in AO space (Mulliken, Lowdin), partitioning the density in real space (Bader, Voronoi, Hirshfeld), partitioning the electrostatic potential (ESP; Merz-Singh-Kollmann or ChElPG), or other methods I'm certainly forgetting. Another interesting method is based on atomic polar tensors [1], called GAPT or IR charges, which are a natural byproduct of harmonic vibrational frequency calculations.
This doesn't mean the results will be of good quality. Those based on partitioning the AO density seem to be the worst overall. Anecdotally, those based on ESP calculations are better, however if your system has a lot of "buried" atoms that aren't solvent surface-accessible, results may be questionable. Real-space partitioning is an attractive method since it is independent of direct basis set effects and surface accessibility. GAPT results also seem to be good, but they require a harmonic frequency calculation, which can be expensive.

Milani, A.; Castiglioni, C. Atomic charges from atomic polar tensors: A comparison of methods. Journal of Molecular Structure: THEOCHEM 2010, 995 (1-3), 158-164. DOI: 10.1016/j.theochem.2010.06.011

I am using the above mentioned formula for calculating the dipole. But it is also dependent on r. if you change your reference point for r, the simple multiplication will change the summation as well. How, then, we should calculate the dipole moment, let's say for a H2O molecule in a box (in the context of DFT) ? What should be the reference for the origin of distances (r) ? Can we  obtain the value of ~1.8D for H2O then?

For a neutral (sub)system, the dipole moment is origin-invariant. Otherwise, the center of nuclear charge is a reasonable choice. There is no guarantee you'll reproduce an experimental or analytic result by the partial charge approach.

I cannot test GAPT at the moment. I think the rest answers my question within the scope of the theory. Though the real problem of getting the dipole from partial charges remains, I will mark it as accepted. Thanks for the explanation and the link. Should be useful in the future.

If you want to try APT (there is a difference from GAPT that I don't remember), it should be available in DALTON under "Dipole-gradient based population analysis".

Another alternative that requires more manual labor is between the point charge approach and the fully analytic approach. Take your subsystem of interest as fully quantum mechanical, treat the rest of your system as point charges (calculated by your method of choice), and calculate the analytic dipole moment. This is something like an electrostatically-embedded QM/MM calculation.

Can you, please, provide a reference to your last suggestion? If possible, I will give it a shot.

@pentavalentcarbon - one quick note is that you should remember to convert to Debye. Your formula will normally give e * Å if you don't convert: http://cccbdb.nist.gov/debye.asp


text:
Does water self ionize to protons or hydronium ions, or both?
The most common equation for the self ionization of water I've seen is with protons and hydroxyl ions as products.
H2O ⇌ H+ + OH-
I've also seen the following equation with hydronium ions as a product instead of protons
2H2O ⇌ H3O+ + OH-
Does the first equation actually happen, or is it just an abbreviation for the second? Or do both species (hydronium and protons) exist in solution, or something more complicated?

Bare protons do not exist in chemistry. First form is a shorthand for the second.

Ivan is right but technically even the second equation is a short hand for how acidic protons are held by a whole network of H2O.


text:
What is the pH of HCl solution at $10^{-8}$M?
After using the negative logarithmic value of $\ce{H^+}$ ion concentration,  I get a value of pH that results in base. Can you please help me?
Thank you in advance

Do you know what is the pH of neutral water (with zero HCl)?

Well this is not pretty difficult, let's go !

Because chloride ions don't react with water we have,
$$c_0=\ce{[HCl]=[Cl^-]}\tag1$$
Now for hydrogen ions, $$\ce{[H^+]=[HO^-]}\ce{+[Cl^-]}\tag2$$
We all, must, know that $$K_e=\ce{[H^+][HO^-]}\tag3$$
Then we have $$\ce{[H^+]}^2-c_0\ce{[H^+]}-K_e=0\tag4$$
You'll find your pH solving $(4)$.

Now between theory and practice you may not find a big difference...

Why chloride ions do not react with water ? @ParaH2

When you want to find $\mathrm{pH}$ of acid whose concentration is less than $10^{-6}$ you have to consider water's dissociation as $\ce{H+}$  and  $\ce{OH-}$ too. Which is $10^{-7}$ at $\pu{25 ^\circ C}$. So concentration of  $\ce{H+}$ into this case is actually, $10^{-6} + 10^{-7}$, giving $1.1 \times 10^{-6}$. Which will give you a $\mathrm{pH}$ of about $6.9$. We usually ignore water's dissociation because it is negligible but in cases like this it becomes necessary to consider it too.

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


text:
Converting plastic back to bio-degradable substances
I have read that every plastic ever made still exists. Plastic must be made of some substances that are biodegradable. Even if they aren't, why can't we change their form? Why can't we change plastic into something else?

You're contradicting yourself and apparently never heard of recycling.

Some plastics decompose, albeit slowly, e.g. urethanes from photodegradation. Some decompose rapidly,  e.g. celluloid. Polyvinyl alcohol is a water-soluble, biodegradable plastic. Question your source of information!

Some definitions: plastic is a synthetic (man-made) organic polymer usually composed of a large number of simple monomer units chemically united (generally chemically) and usually used for its structural properties. For instance, the polypropylene fibers which make up a polypropylene rope would commonly be called a plastic but the polyvinyl alcohol used as a water thickener would normally not.
Biodegradation is any sort of chemical decomposition by environmental causes. These causes may include sunlight, moisture (water), and microbes. Heat, erosion, and oxidation are generally not included.

If you have read that every plastic ever made still exists, then you need to be aware that your gullibility is extreme and the source for that claim should not be relied upon. Surely, you've seen plastic burn. Surely, you don't believe that the burned plastic magically reassembles itself. Hence it is obvious that not all plastic ever made still exists.
We can change plastic.  Because most plastic items contain many other chemical compounds, (often colorants, antioxidants, plasticizers, & fillers) it can be quite expensive to decompose them into useful materials.
Chemistry isn't magic. Any industrial chemistry involves people, equipment, transportation, and energy. That is, it takes money. Most plastic is made from natural gas (the polymer, that is), which is cheap and fairly pure (consisting of mostly a half-dozen chemical compounds). There are tens of thousands, perhaps hundreds of thousands, maybe millions, of different plastics that have been made. Because of all the other stuff mixed into the polymers, and because the polymers can be very different, chemically, it is usually cheaper and easier to make a plastic starting from natural gas than starting from pre-existing plastic. Just the mix of materials separated from the polymers would be terribly expensive to dispose of.
The forces of nature will eventually bury some plastic, break up some plastic, and decompose some plastic. Nothing lasts forever. Very little of the plastic existing today will be present in 10,000 years. Erosion, aging, and biodegradation are the 3 categories of this degradation, along with mechanical forces. Much plastic exists in the oceans as microscopic bits. Some bacteria are able to eat some plastics. All plastics burn. Eventually, all of the polymers present in the plastic here on Earth will either be converted to CO2 (either in the air or captured in the ground (eventually as carbonate) OR subducted with the tectonic plates and thermally decomposed.

Plastics are usually organic polymers. During the process of polymerization covalent bonds are formed. This reaction is exothermic, meaning it sets free energy. So, in order to reverse the process of polymerization and break these bonds again, it would take a lot of energy.
It is fair to assume the energy needed will also break other bonds in the polymer, forming multiple products at once. I guess some kind of inseparable tar will be formed if oxygen is excluded. When oxygen is included the polymers will probably just burn, also forming multiple products at once - and CO2.

That being said, some polymers can be depolymerised. Indeed various byproducts are created, but you can get big part of monomer back.

I wasn't really aware of that. I guess my answer is not fully correct then. Hmmm.

The initial statement is incorrect. Plastics contain carbon-carbon bonds and everything with carbon-carbon bonds can be burnt to $\ce{CO2}$ given sufficient burning temperature. Some plastics are indeed burnt so it is impossible for all plastics still to be around.
However, most plastics can be reasonably approximated by a long, saturated carbon chain much like crude oil except with less other organics mixed in. As is the case with crude oil, there are very few organisms capable of digesting plastic in any form so its bio-lifetime is very long. However, given enough plastics and evolutionary pressure, microorganisms that can digest it will come along. I believe there have been some found already but I cannot point to any sources right now.

I am aware that this answer is basically a shorter rehash of alphonse’s.


text:
Fabrication and/or Selection of a Single Crystal Ionic Conductor
I want to fabricate or buy single crystal ionic conductors for usage as solid state electrolytes in batteries testing. However, I am having trouble finding anyone that can sell appropriate materials.
From my understanding, lanthanium series elements typically have good ionic conductivities. Thus, it seems like Strontium Lanthanum aluminum oxide (SrLaAlO4) or lanthanum aluminum oxide-strontium aluminum tantalum oxide (LSAT) which are both sold in single crystal form, may be suitable. However, I'm having trouble determining if they are compatible with the lithium ion.
Meanwhile, I know that I have seen SiO2 and Al2O3 used as barriers in battery electrodes, and it's possible that single crystals of either of these may be possible to use. However, just because it is a barrier doesn't necessarily render it a solid state electrolyte. Meanwhile, it appears alpha-Al2O3 has previously been used as a separator, I'm unsure if this is truly because the alpha-Al2O3 can transport the lithium versus the pores of the material transporting the lithium. When trying to discover which was the case, I found this paper discusses the transport mechanism of lithium in Al2O3, but I am unsure if this applied to alpha-Al2O3.
Lastly, I can also find MgAl2O4 single crystals, which appear to be good ionic conductors of lithium but only if doped with lithium. However, I do not know a way to create single crystal LiMgAl2O4, starting with MgAl2O4 single crystal.
Does there exist a suitable single crystal material which can be easily fabricated or bought for my purposes?

I'm confused about what you are trying to do. The electrodes in batteries are solids in general. Are you trying to make a completely solid state battery?

At what temperature would this operate? Beta-alumina, for example, provides appreciable current only above 200°C. See http://www.ionotec.com/conductive-ceramics.html for supplier,  https://en.wikipedia.org/wiki/Beta-alumina_solid_electrolyte for info.

To be clear, I'm wanting to use this purely as a solid state electrolyte. That means that it separates the two electrodes, providing a path for lithium ions but having little electrical conductivity.


text:
Why would this be 5-ethyl-6,6-dimethyl-3-heptene when in aplhabetical order it should bebe: 6,6-dimethyl-5-ethyl-3-heptene?

Name the following compound:

Why would this be 5-ethyl-6,6-dimethyl-3-heptene when in alphabetical order it should be 6,6-dimethyl-5-ethyl-3-heptene?

Actually, since 1993, the correct name according to IUPAC recommendations is 5-ethyl-6,6-dimethylhept-3-ene.

The prefixes di-, tri-, etc. are not considered when determining the alphabetical order of the substituents. So the order is actually determined like this:

5-ethyl-6,6-dimethyl-3-heptene

You can read up on this matter when following this Wikipedia link.


text:
Naming of asymmetric organic molecules with two nitroxides
When writing my diploma thesis I had to name my molecules following IUPAC nomenclature rules. There were several molecules like the following one, that contained two TEMPO groups, but were not symmetrical. The following is one example:

While I was able to find IUPAC recommendations on how to name radical organic molecules I could not find sufficient information for this case and ACD/Labs was no help either. I went ahead and replaced the two TEMPO groups with COOH, as this is a high ranking moiety, got a name for it and replaced the corresponding parts of the name with what I found (carboxy to ylooxy; carboxylic acid to oxidanyl) in the recommendations.
This was the name I came up with:

(4-(4'-((adamantan-1-yl)((1-ylooxy-2,2,6,6-tetramethylpiperidin-4-yl)carbamoyl)carbamoyl)-N-((adamantan-1-yl)carbamoyl)-[1,1'-biphenyl]-4-carboxamido)-2,2,6,6-tetramethylpiperidin-1-yl)oxidanyl

In the meantime, the results have been published but I always wondered, if this was correct. I checked on SciFinder (CAS No. 1947339-77-6) and an index name has still not yet been assigned.
Is there a more elegant way of finding a name for a molecule like this?

The structure can be redrawn somewhat more clearly:

Let's call the basic structure, where R1 = R2 = H, ‘the skeleton’.
If your name was, after fixing some enclosing parentheses to square or curly brackets, even the ‘preferred IUPAC name’, it's not a nice one, because it does not reflect the symmetry of the skeleton, and repeats the TEMPO and adamantane building block names.
(From now on, I'm gonna use the blockquote for IUPAC organic nomenclature rules 2013 generated names, or the book citations, if section name included.)
"Almost preferred" IUPAC name, substituted carbamoyl biphenyl dicarboxamide
The skeleton is named

N4,N4′-dicarbamoyl[1,1′-biphenyl]-4,4′-dicarboxamide.

This preferred name is obviously based on the biphenyl dicarboxylic acid diamide

  [1,1′-biphenyl]-4,4′-dicarboxamide

substituted with carbamoyl ($\ce{NH2-CO\bond{-}}$) groups at each nitrogen.
Simple monosubstituted derivatives of the skeleton (there are only two kinds of them initially) are named

N4,N4′-dicarbamoyl-N4-methyl[1,1′-biphenyl]-4,4′-dicarboxamide

and

N4-carbamoyl-N4′-(methylcarbamoyl)[1,1′-biphenyl]-4,4′-dicarboxamide.

R2 (“TEMPO-4-yl”) can be named,

  2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl (group)

according to the rules of making the radical to be prefix (e.g. when there is yet more prioritized radical):

P-71.5 Prefixes denoting radicals
(…)
Examples:
(…)
$\ce{-O^.}$
oxyl (preselected prefix)
  ylooxidanyl
    (not ylohydroxy)
(…)
P-71.7.d.1 (…)

(3-oxylbenzoyl)oxyl (PIN)
  [3-(ylooxidanyl)benzoyl]oxidanyl

however the rules do not make a good distinction between residues (prefixes) and radicals.
So the resulting name would be
N4-(adamantan-1-yl)-N4′-[(adamantan-1-yl)carbamoyl]-N4′-(2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl)-N4-[(2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl)carbamoyl][1,1′-biphenyl]-4,4′-dicarboxamide. But it still redundantly repeats some building block names.
Unfinished IUPAC name based on substituted "diallophanylated biphenyl"

P-66.1.6.1.1.4 Carboxylic acid derivatives of urea
Two carboxylic acids are related to urea; have been known as ‘allophanic acid’ for $\ce{H2N-CO—NH-COOH}$ and ‘hydantoic acid’ for $\ce{H2N-CO-NH-CH2-COOH}$. These names are no longer recommended. Preferred IUPAC names for these two acids and their derivatives are formed systematically.
Examples:
$\ce{H2N-CO—NH-COOH}$
  carbamoylcarbamic acid (PIN)
  (aminocarbonyl)carbamic acid
$\ce{H2N-CO—NH-CO\bond{-}}$
  carbamoylcarbamoyl (preferred prefix)
  [(aminocarbonyl)amino]carbonyl

The allophanoyl- prefix can be found in the older IUPAC 1979 nomenclature, rule C-971.
The skeleton could be named

4,4′-diallophanoyl[1,1′-biphenyl], or
4,4′-bis(carbamoylcarbamoyl)[1,1′-biphenyl],

if you want.
But now we would have to repeat the two differently substituted allophanoyls in the name. Or derive a name from the symmetrical one, but how to number it?
Urea based IUPAC name
Closer look at your structure reveals that it in fact is two N,N′ asymmetrically substituted urea

P-66.1.6.1.1 Urea and its substitutive derivatives
P-66.1.6.1.1.1 The compound $\ce{H2N-CO-NH2}$ has the retained named ‘urea’, Which is the preferred IUPAC name, With locants N and N′, as shown above the structure below. The systematic name is ‘carbonic diamide’. The locants 1, 2, and 3 have been used in the past and may be used in general nomenclature.
$$\begin{array}{ccc}
\ce{H2\overset{$N$}{\underset{1}{N}}-\underset{2}{C}O-\overset{$N'$}{\underset{3}{N}}H2} & \hskip{3em} &
\ce{H2\overset{$N$}{N}-CO-\overset{$N'$}{N}H2} \\
\text{urea (PIN)} & &
\text{carbonic diamide}
\end{array}
$$
$$\bbox[0.5em,border:1px solid black]{\text{Numerical locants for urea are no longer used in the IUPAC preferred names.}}$$

molecules, N,the-other-one's-N′-interconnected with

  [1,1′-biphenyl]-4,4′-diyldicarbonyl (group).

And that's a correct name, used e.g. in

  2,2′-([1,1′-biphenyl]-4,4′-diyldicarbonyl)dipropanoic acid.

(If these propanoic acids were interconnected by different carbons, 2,3′ (there's no other option), such multiplicative name would be no longer a preferred one; see e.g. P-15.3.2.1.1. But never mind, we're making a nice one.)
The skeleton now can be named ([1,1′-biphenyl]-4,4′-diyldicarbonyl)diurea. (More precisely N,N′-([1,1′-biphenyl]-4,4′-diyldicarbonyl)diurea. Or N,N′′-…? And what about such multiplicative names of asymmetrical derivatives? There's no definitive answer, but this would be for another question.)
So we'll use the other urea numbering, the numerical, non-preferred one.
Now we get a nice name:
1,3′-([1,1′-biphenyl]-4,4′-diyldicarbonyl)bis[1-(adamantan-1-yl)-3-(2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl)urea]
(But I would like similarly short and non-redundant name that would end like “…bis[(2,2,6,6-tetramethylpiperidin-1-yl)oxidanyl]” yet more.)

This is a very comprehensive answer. I really like the concise urea name, as we described the products as urea derivatives. I'd just like to point out that the correct name of the oxygen-radicals (according to IUPAC RC-81.3.3.1) is supposed to be ylooxy. Generally, the prefix ylo should be used for radical bearing groups. See the following document for further reading: http://www.sbcs.qmul.ac.uk/iupac/ions/RC813.html#p331

@basseur Please note that I was mentioning that we are making the radical subtituent to be named by prefix. I have updated the 2013 rules citation with the example containing the "oxyl" preselected prefix (P-71,5). (I did not track consistency with previous version 1993 rules you linked.)


text:
Why is fermentation of cellulose to produce biofuel and nutrients so difficult?
The formula for glucose is $\ce{C6H12O6}$ and that of cellulose is very similar $\ce{C6H10O5}$.  Glucose can be readily fermented by yeast and other micro-organisms to produce carbon dioxide and ethanol. Whereas cellulose is not ferment-able by yeast and only very specific anaerobic Archae bacteria can digest it, but the end product being methane, water, sugars (primarily lactose), and volatile fatty acids (acetate, proprionate, butyrate); not ethanol.  These bacteria reside in the guts of termites that allow them to digest wood, and in the intestines of cattle and other ruminants.  The termites and ruminants use the sugars and fatty acids as food.
The enzymes used by the host anaerobic bacteria in the gut of termites are of particular interest because:

termites produce very little solid waste as feces
termites can digest wood, leaves, and broader variety of plant fodder than ruminants, which mainly feed on grasses

We are not really interested in the termites, nor the anaerobic bacteria, but rather specifically the enzymes and chemical processes used by the bacteria that break down the cellulose into the lactose and fatty acids.  Can this be accomplished synthetically in an anaerobic lab setting outside of said bacteria?
This article describes a company that tried to use cellulose raw material to produce ethanol on a commercial level, but recently went bankrupt.  This company apparently was not getting ethanol directly from fermentation of cellulose but rather the result of a multiple step process of generating syngas from the cellulose, then creating alcohol from the syngas; a much less efficient process than direct fermentation of cellulose into alcohol.
Obviously, such a process if achievable would relieve the need to use food crops to generate ethanol and put to use otherwise inedible cellulose plant waste. What is the main challenge here.  Cellulose is just one water molecule different from glucose.

Cellulose was designed so as not to be easily digested. No one likes to be someone else's food.

Cellulose is polyglucose not one water molecule different from glucose. the acetal linkages are beta essentially equatorial to the pyranose rings and therefore of lower energy than the Alpha axial linkages in starch. This means starch is more easily hydrolyzed than cellulose and more enzymes evolved to digest starch. Cellulose is a linear, unbranched, molecule that crystalizes, and it is not crosslinked. Morrison and Boyd explain this nicely in the chapters on carbohydrates. Wikipedia also.

Your question is a little bit all over the place, but I believe I can answer it anyway. First, though, allow me to point out that your sum formula for cellulose is wrong. While glucose is indeed a single $\ce{C6}$ compound with the exhaustive sum formula of $\ce{C6H12O6}$, cellulose is, in fact, a poorly defined polymer consisting of multiple $\ce{C6H10O5}$ subunits, linked together both linearly to form long chains but also cross-linked to form extensive networks. Therefore, the better sum formula would be $\ce{(C6H10O5)_nH2O}$ or $\ce{C_{6n}H_{10n+2}O_{5n+1}}$ for very large values of $n$.
There are two reasons why glucose, its dimer maltose and its polymers amylose, amylopectin and glycogen are much more easily digested by all organisms than cellbiose (the glucose dimer that can be found in cellulose) and cellulose itself.
The first one is the type of linkage: the first set of compounds all contain α$\ce{{1}\bond{->}4}$ connections while cellbiose and cellulose contain β$\ce{{1}\bond{->}4}$ connections. This seemingly minor difference means that while the starch subunits form spirals, the cellulose subunits actually form fibres and all the enzymes designed to break up spiral subunits cannot work with fibres.
The second is the number and type of cross-linkages. Each organism has a set of enzymes that can break up the cross-links present in amylopectin or glycogen (or both) but these again are different in shape from those in cellulose owing to the subunits’ β configuration in the latter.
Enzymes to break down amylose, amylopectin and glycogen are pretty much present in every multicellular organism since these compounds are used to store energy in an available form — which would not make sense if the organism could not break the storage form back down to monomers. However, cellulose was not ‘intended’ to be energy storage and thus there was no evolutionary pressure to evolve enzymes that could digest it. Therefore, only very few species that found and populated this ecologic niche are able to digest cellulose and even less are able to digest the additional compounds present in wood that makes it hard.

Now on to the process design part of your question. In general, if we have the enzymes we should be able to degrade wood to useable monomers much like whichever organisms do that in whichever environment they live. Not only the symbionts of termites are of interest but also wood-digesting fungi, naturally. However, the key difficulty here is acquiring the enzymes required in sufficient amounts to make the process economic. I am only aware of very few enzymes that can be fermented and isolated in sufficient quantities and purity to make for a possible economic use; among them a generic lipase. That is one significant hurdle process chemists must overcome.
However, there’s another bit. It may work if we have access to the enzyme but it doesn’t have to. A number of times the most significant issue for chemical engineers is the scale-up problem: reactions work fine on a small scale but once you try to scale them up to higher quantities for more throughput they break down in one way or another. I am sure without knowing the exact titles that there are shelves of books concerning the scale-up problem. To the best of my knowledge (caveat: I am not an industrial chemist or chemical engineer), it is often down to trial and error which processes work well and which don’t; and what works well for one type of product does not work at all for others.
Needless to say that one has to have access to a significant amount of funds to bridge the research gap until one actually is able to make money so it doesn’t surprise me (although I am unhappy) that the company you mention went bankrupt.

The ultimate reason is that cellulose is designed to be hard to digest
The specific chemical reason is well covered in Jan's answer, but there is an explanation that is simpler and more fundamental: cellulose is designed that way. Nature has created some organisms that need a structural component that isn't easy for other organisms to break down (for the avoidance of doubt evolution sometimes produces outcomes we call designed we don't need an external designer).
Trees want to last for a long time and, if other organisms could digest their structural material easily, they wouldn't achieve that goal. Cellulose is one of the key structural compounds enabling tree trunks to be strong and long-lasting. If they were built from a different sugar polymer like starch, they would be easy to digest (starch is designed to be a storage medium for sugars and is relatively easy to convert back to digestible forms which is why a lot of our foods are mostly made from it).
So, ultimately, the indigestibility of cellulose is a product of an evolutionary history where some organisms want to be big, strong and long-lasting. Very few other creatures have evolved good ways to beat this property (the symbiotic bugs in termite stomachs are a rare example).

The main challenge is the macroscopic structure of cellulose. It's a solid, enzymes obviously can only start working on the surface, and hydrolases will work only on a few random loose chains.
This is too slow, even if you have a lot of time and are not trying to make any money. So you need to break apart your cellulose first using a process that's cheap and results in a product clean enough for the enzymes (you can use acids, but not too much or you'll prevent the enzymes from working). This is not too easy.
After you break the crystalline structure you need to use an enzyme mix, as you need still need some more "starting points". You need enzymes to make some breaks in the middle of the chain, some enzymes too loosen the chain from the bulk, etc. before you can use an enzyme that digests cellulose from the end (like happens easily with starch). The hydrolases are fast enough, but you need quite a few more activities.
Then all of this needs to fit together, you need a source that's really really cheap, a process that breaks it down enough for the enzymes to do their work, and then this has to result in a mix of glucose and other compounds that are good food sources for the fermentation step, and this has to contain no competing micro-organisms or other inhibitory stuff. And remember, all this time you're competing against dirt-cheap fossil fuels.
Creating some ethanol is trivially easy at the moment, you can order some enzymes, boil your cellulose in acid, wash it, add the enzymes, add some yeast and distill off your ethanol. But doing this in an economically feasible way is not that easy.
One company failing is also not an indication that it doesn't work / isn't feasible. Dupont for example is producing a lot of ethanol using an (partially) enzymatic process. You can browse around and get an idea of the challenges, bottlenecks and the scale of the operation. Other companies are also doing a lot of stuff, for example POET-DSM and Abengoa.

Just for amusement, connected only by the crystalline structure and lack of surface to attack: once in my youth I bathed well formed calcite crystals in sulfuric acid, expecting some spectacular, violent reaction but nothing happened. Probably the small and even surface was converted to an inert layer of gypsum that sealed the calcite even more.

There are quite a number of cellulases in nature (enzymes to split cellulose into shorter oligosugars). See e.g. https://en.wikipedia.org/wiki/Cellulase
They are needed by any plant that grows, to be able to do some restructuring. They however only work on amorphous chains, and only one type (endocellulases) is actually able to cut into the middle of a long chain or network segment. The rest just bites pieces off (the resulting) dangling ends.
Crystalline segments (they're what makes trees so hard and elastic) are virtually immune to enzymes. They only degrade after previous e.g. oxidative or UV damage. Wood under water does not decay on human timescales, unless you go to higher temperatures and/or acidic conditions.
Your termites et al. have a lot of tiny, sharp teeth (figuratively speaking) and mechanically split the wood into very small particles, the huge surface of which allows the enzymes to get to work again.


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What is the chemical reaction between Ferric Chloride and Baking Soda
Awhile ago I tried to neutralize ferric chloride with baking soda once I had finished using it for PCB etching. Once I added the baking soda it started bubbling up a lot and I assumed it was releasing a gas.
Does anyone have any idea what gas it was releasing? I read different things on different websites and I wasn't sure which one was right.

Baking soda is basically $\ce{NaHCO3}$. The only way it can typically react is by proton transfer; either deprotonation of the hydrogencarbonate in basic solution to give the carbonate anion or protonation to give $\ce{H2CO3}$. The latter is unstable and quickly breaks down into $\ce{H2O}$ and $\ce{CO2}$ — carbon dioxide and water. It is the carbon dioxide bubbles you see when adding baking soda to neutralise.

FeCl3 or ferric chloride get ionised in water to Fe+3 and Cl-. Fe+3 then reacts with OH- of water forming Fe(OH)3 a brown ppt and H+ of water reacts with the remaining Cl- to make the aqueous solution acidic by forming HCl. If you put baking soda (NaHCO3) i.e. sodium bicarbonate in this ferric chloride solution it will react with hydrochloric acid and will give out CO2 gas as effervescence. That is why gas bubbling is seen. Hope this piece of information will help you.

Where does the water come from?

@NilayGhosh It is quite obvious that in the problem stated by Matt121, he has used aqueous solutions, and that is the source.

He has not mentioned aqueous solution. You are assuming it.

PCB Etching
Currently used for print circuit board etching of solvents are iron (Ferric Chloride), cupric chloride (Cupric Chloride), alkaline ammonia (Alkaline Ammonia), sulfuric acid hydrogen peroxide (Sulfuric Acid + Hydrogen Peroxide) etching fluid, ammonium persulfate, sulfuric acid – chromic acid etching solution. Like the PCB manufacturing in Fusion PCB of Seeed.
Etching liquid are mainly copper chloride solution, ferric chloride liquid, alkaline etching, acid/hydrogen peroxide (hydrogen peroxide) etching.
Chloride copper etched liquid by chloride copper (CuCl2 Castle, 2H2O) + hydroch

Hope this will remove ambiguity.


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Why is this PH3 considered to be an unhybridized molecule?
Also, don't hybrid orbitals appear in all molecules?

The answer by "ron" says "the molecule can be viewed as being unhybridized":
How does lone pair of a central atom affect the dipole moment?

It is considered unhybridized because of its geometry. Also, hybridized orbitals are a mathematical abstraction, so the words "appear", "observe", or even "exist" do not apply to them.

Read: Hybridization in PH3  (maybe dupelicate) Related: Why is the bond angle H-P-H smaller than H-N-H? In general: Bond angles for the hydrides

I'll try to explain this by comparing this with the case of NH3.
N is a better electron withdrawing group than P. So in the case of NH3 the electron cloud is more closer to N. And therefore the repulsion in the bonds causes the 3 H atoms to be farther away from each other resulting in a bond angle of about 109°. But in the case of PH3 the electron cloud is farther away from the central atom(P) than in the case of NH3. So the bond pair - bond pair repulsion is comparatively lesser, causing the 3 H atoms to move closer together to an angle of almost 90°, resembling the px, py, and pz orbitals, as a consequence it's lone pairs are present in almost pure s - orbital. So it is is said that PH3 can be viewed as unhybridized.

It is due to greater difference in energy of atomic orbitals of phosphorous such as 3s and 3p.


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Hybridisation in phosphorus allotropes
The white phosphorus has a tetrahedral structure, like this:
 [Image source: Wikipedia]
The red phosphorus, on the other hand, has a polymeric structure:
 [Image source: google images]
Then what is the hybridisation state of P atoms in these allotropes?
For tetrahedral $\ce{P_4}$ in white phophorus, we can take $\ce{sp^3}$ hydbridisation, but some sites give information that P atom uses atomic 3p orbitals, as then the the angle strain is reduced from $\ce{109^\circ\rightarrow 60^\circ}$ to $\ce{90^\circ\rightarrow 60^\circ}$.
I can't find any website which gives information about the hybridisation state of red phosphorus. My chemistry textbook also does not give any information about this.
If you can give a reference to back up your answer, it would be useful to me.

Hybridisation is in our heads, not out there in the P allotropes.

Related: Hybridization in PH3; Why is the bond angle H-P-H smaller than H-N-H?

@IvanNeretin Here, I am considering that hybridization actually occurs in atoms, and trying to deduce whether the phosphorus atoms are hybridized or not.  I don't understand why you say that hybridisation is in our heads; because atomic orbitals undergo hybridisation if it produces lower energy state, and it does explain many properties of compounds.


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Bonding in protonated molecule
We have $\ce{H2O}$ with two lone pairs on the oxygen atom and $\ce{H3O+}$ with only one lone pair on the oxygen atom.
One electron from the four non-bonding electrons has made a bond with the third hydrogen atom, so we should still have three electrons left. Why do we have only two?

What is the charge of $\ce{H3O+}$...

The answer is quite simple: When you have $\ce{H2O}$, it means you have two hydrogen and one oxygen atoms. Each hydrogen brings 1 electron, while the oxygen brings 6.
Your assumption now is, that you have another hydrogen atom bringing in another electron. In this case, you'd be right to assume that there are three electrons from hydrogen, totaling up to 9 electrons in the molecule. But this would violate the octet rule and consequently Paulis exclusion principle.
The solution to your question is, that you do not have $\ce{H3O}$, but $\ce{H3O+}$! The plus means, there is one electron less in the molecule, because you don't bring another H (with one electron) into the molecule, but rather just a proton ($\ce{H+}$ with the electron removed).
So, you have 6 electrons from the oxygen atom and only two from the hydrogen atoms, totaling up to 8 electrons. Three covalent bonds take up 2 electrons each, leaving you with two electrons in the remaining sp3 hybrid orbital as the lone pair.

If we brought a proton, how did it connect to the O, it doesn't even have electrons to share?! and when drowing  Lewis structure the O may have a positive formal charge, and the H may have no formal charge, why, if its a proton?

Well, it only formally is a proton. Actually, there are no protons in aqueous solutions. $\ce{H+}$ is usually just short for $\ce{H3O+}$. But that is not important to answer the question. It is important, however, to count the available electrons and deduct 1 for the positive charge!

Another thing: the electrons for a covalent bond do not have to come from both atoms, but can also be completely donated by only one (here the oxygen atom). The only important thing is that there are two electrons forming the bond In the end.


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Real Gases vs Ideal Gases
Real gases behave less ideally at low temperatures because intermolecular forces have a greater effect.
But wouldn't they still be non-ideal at high temperature? Even though they overcome the intermolecular forces, they aren't moving with their normal energy. Doesn't that make them non-ideal, because the pressure is greater than it would be in an ideal gas?

Well, that's right: at higher T they are closer to ideal, but not quite there yet.

"because the pressure is greater than it would be in an ideal gas? " What do you believe is the pressure of an ideal gas?

Low pressure and high temperature are the necessary condition for ideal behavior of real gases. So if temperature is high but pressure is not very low then the gas will show non-ideal behavior.


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My question is about stoichiometry, limiting reagent problems
So I'm really struggling with this particular concept of these problems. I know how to convert to moler mass and all that, but I don't know how to do the ratio part, I can do them when the ratio is 1/2 or 1/3 but I don't know what to do with it when the ratio is 5/3 or others that don't include a 1. Any help would be greatly appreciated, thanks

What's the difference? Presumably, you are multiplying with the ratio. Why does it matter if there is a 1?

Let's look at the reaction between permanganate and oxalate in acid as an example.
$\ce{5C2O4^2- + 2MnO4^- + 16H^+(aq) -> 2Mn^2+ + 10CO2(g) + 8H2O}$
So if I have 13 moles of oxalate and 6 moles of permanganate what is the limiting reagent?
An easy way to solve is to divide the chemical equation by 2 to get
$\dfrac{5}{2}\ce{C2O4^2- + MnO4^- + 8H^+(aq) -> Mn^2+ + 5CO2(g) + 4H2O}$
So $\dfrac{\text{moles oxalate}}{\text{moles permanganate}} = \dfrac{5}{2} = 2.5$
Given that we started with
$\dfrac{\text{moles oxalate}}{\text{moles permanganate}} = \dfrac{13}{6} = 2.167$
there is too little oxalate to react with all the permanganate. For a complete reaction we'd need 6 moles of permanganate and $\dfrac{5}{2}(6) = 15$ moles of oxalate.
Also note as a different problem we can look at oxalate and $\ce{H+}$.
$\dfrac{\text{moles } \ce{H+} }{\text{moles oxalate}} = \dfrac{16}{5} = 3.2$
So if we started with 13 moles of oxalate, we'd need (3.2)(13) = 41.6 moles of H+.


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Have we found stable, super-heavy elements?
Have we found a stable super heavy element or created one in a lab, or are they still hypothetical?

More sci-fi then hypothetical. There's no reason to think any superheavy nuclide wouldn't be highly radioactive (half-life measured in minutes, at most).

It depends on how you define "stable." The notion is that there is a region of super heavy isotopes that would have a longer half-life than isotopes which are lighter or heavier. But the "stable superheavies" won't be as stable as say helium-4.

There already is an element, heavier  (higher atomic and mass numbers) than all naturally occurring ones, that is stable enough for everyday use.  You are probably using it, well, everyday, in your smoke detector.
It's americium, element 95, with three isotopes having 100+ year half-lives.  Americium-241 has a half-life of over 400 years, for americium-243 it's over 7000 years and there is even an odd-odd isomer, with mass number 242, having a half-life over 100 years.  (https://en.wikipedia.org/wiki/Americium).
This is not perfect (and whether it's called "superheavy" depends on definition), but the americium is out-lasting other components of smoke detectors, or industrial gauges where americium is also used.  In that sense we can indeed call the most long-lasting americium isotopes stable.

This is just wrong. For some background on stable super heavy elements see https://www.scientificamerican.com/article/superheavy-element-117-island-of-stability/

I thought superheavies have $Z \geqslant 104$, so actinides are supposed to be out of scope.

So, will we hit the jackpot again like with americium?  No one knows.

Am isn't "hitting jackpot", it's just after plutonium, which has most stable isotope with half-life over 80 mln years...


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O-H infrared band dependence of alcohol concentration in water
With regard to IR spectroscopy, how would the O-H band change as an alcohol solution was diluted with water? I am referring to any alcohol, but for the sake of example we can use ethanol. I think first that the $\ce{O-H}$ band shown would increase in intensity. I believe this is the case because the water is becoming a greater proportion of the mixture and water is explicitly made from $\ce{O-H}$ bonds whereas the $\ce{O-H}$ bond in ethanol only represents a fraction of the bonds in the molecule. Is this correct? Is there also a wave number shift that would occur too that I am missing?

The easiest thing to do is look at some example spectra. All of them are taken from the Spectral Database for Organic Compounds.
This is 1-propanol, first as a liquid film and then in $\ce{CCl4}$.

In the liquid, the broad peak at $\pu{3333 cm^{-1}}$ is due to intermolecular hydrogen bonding, just as you would see in an IR spectrum of bulk pure water. The $\ce{CCl4}$ spectrum shows that as you reduce the probability of 1-propanol (or any alcohol) molecules contacting each other, the free stretch appears at $\pu{3637 cm^{-1}}$. This is similar to water in a gas-phase spectrum.

water is explicitly made from $\ce{O-H}$ bonds whereas the $\ce{O-H}$ bond in ethanol only represents a fraction of the bonds in the molecule

Close. It isn't about the fraction of total bonds, in this case it's simply that water has two hydroxyl stretches and ethanol only has one. As water is added, the intensity of the hydrogen bonding $\ce{O-H}$ band should increase over the pure alcohol hydrogen bonding band, but we cannot know quantitatively what the increase will be without performing an experiment.

Is there also a wave number shift that would occur too that I am missing?

Yes. In the gas phase, the water symmetric and asymmetric $\ce{O-H}$ stretches are at $\pu{3657 cm^{-1}}$ and $\pu{3756 cm^{-1}}$, which then red-shifts to $\pu{3404 cm^{-1}}$ in the bulk due to hydrogen bonding. Compare this to 1-propanol, where the "free" and hydrogen bonding values are $\pu{3637 cm^{-1}}$ and $\pu{3340 cm^{-1}}$, respectively. The water stretches are blue-shifted (higher in energy) compared to the alcohol.
Water vibrational frequency values taken from http://www1.lsbu.ac.uk/water/water_vibrational_spectrum.html.

So just so I understand. It "blue shifts" or shifts towards a higher frequency because the O-H bond in water is more polar than the O-H bond in alcohol? is that correct? thanks so much for your help also.

I'm not sure about the general case, but I know that charge transfer (additional electron density) into the $\ce{O-H}$ bond causes a red-shift due to populating previously empty antibonding orbitals. If you assume that a decrease in frequency is purely due to an increase in the bond's electron density, it's consistent with a little more electron donation from a $\ce{CH3}$ group than a hydrogen atom, like https://en.wikipedia.org/wiki/Electrophilic_aromatic_directing_groups.

It doesn't look like there are large changes going from 1-propanol to 1-butanol, but the broad band in methanol (liquid) is at 3347, compared to 3333 in 1-propanol, presumably because the electronic structure of $\ce{CH3CH2CH2\bond{-}}$ isn't very different from $\ce{CH3\bond{-}}$.

The water and ethanol -OH stretch bands are too close to be distinguished in a system with hydrogen bonding (liquid systems). You could expect a broadening in the signal as you add water since you will have the two compounds in your system, then the ditribution of frequencies will be broadened by both water and ethanol absorptions. You are right on the assumption that the intensity will be larger since water has 2 OH groups while an alcohol only has 1. As your system becomes very diluted, it will eventually show the IR spectrum of pure water, with a small wavenumber shift towards the water -OH stretch (3700-3100 cm-1).
The situation stated above is for alcohols such as ethanol that are capable of doing hydrogen bonding. Some alcohols don't have such capability and the -OH stretch band in a sample of this pure alcohol would show a sharp and intense -OH band, such as the remarkable -C=O band. If you would add water to such substance, then you would observe the disappearance of this band and the appearance of the "upside down bell" band characteristic of water. (Both assumptions are for liquid systems, not for gases).

In pure alcohol the band is narrower than in alcohol/water however indeed is not sharp. Broadening occurs because not only there are two similar vibrations within alcohol and water, but because an ensemble of new vibrations occurs, those due to alcohol - water hydrogen bonds.


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Is the surface of glass a fine layer of silanol?
Today we learned in class through a vivid demonstration that water sticks to glass. We were then taught that water sticks to glass because while glass is mainly composed of silicon dioxide, the surface of glass is actually silanol. Hydrogen bonds form between molecules of water and silanol, which is why water sticks to glass.
Is this correct?

Yes, it is correct that the surface of glass has silanol ($\ce{Si-O-H}$) at the surface. This is because water will react at the surface to relieve bond strain of a $\ce{Si-O-Si}$ bond to form $\ce{Si-O-H H-O-Si}$ pairs. additionally there is sodium siloxide at the surface ($\ce{Si-O-Na}$) which will release sodium hydroxide on contact with water $(\ce{Si-O-Na + H2O -> Si-O-H + NaOH (aq)})$ and is major reason why normal glass cannot be used for biomedical implants.
Additionally these surface silanols can react with chlorosilanes to produce hydrophobic surfaces $$\ce{Si-O-H + Cl-SiR3 -> Si-O-SiR3 + HCl}$$
where the R is a non-polar organic pendant group that is hydrophobic. This coating causes water to bead off of treated glass that would priory be wetting.


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Can a bond be electrophilic or nucleophilic?

The indicated bond is:

A) Nucleophilic because it is electron-deficient.
B) Electrophilic because it is electron-rich.
C) Nucleophilic because it is electron-rich.
D) Electrophilic because it is electron-deficient.

The answer is D. I'm arguing with my friend because of this particular question.
I argued that there may be an error in this question. The arrow is pointing toward a carbon atom, so the question should be "The indicated carbon atom is:". But, if you take the question as it is, I answered C since I think that the "indicated bond" is the double bond, and is therefore nucleophilic and electron-rich. If it is not the double bond, then what is the indicated bond in this question? And is it possible for a single bond to be electrophilic?

How does the arrow even indicate a bond?  It seems to be pointing to a C atom that is involved in three bonds...

Bonds can indeed be considered to be nucleophilic. For example, in the electrophilic aromatic substitution mechanism an electrophile interacts with benzene’s π system. By analogy, we can also see this as a nucleophilic electron pair (situated in the HOMO) interacting with the electrophile’s LUMO.
However, since bonds are considered to be the bonding electrons which by definition carry a negative charge, bonds can only be nucleophiles or non-nucleophilic; never electrophiles.
Bonds also cannot be electron-deficient. A bond in a Lewis structure is made up of two electrons, full stop. In MO theory, we may no longer consider the bond localised but we still have an electron pair in a certain bonding orbital and thus never electron-deficient.
I suspect the book made an error here and wanted to point at the carbon atom so the text should be as you suggest.


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Can someone explain me the difference in boiling point of water when I add: KBr; CaCl2; AlCl3?
I have noticed (during an experiment) that the boiling point of the solutions created by H2O+KBr; H2O+CaCl2 and H2O+AlCl3 see their boiling point increase.
During the experiment, I thought that, since every time the number of Ions dissociated in each solution increases, the boiling point of each solution should increase with the number of Ions formed: BP(H2O+AlCl3)>BP(H2O+CaCl2)>BP(H2O+KBr)
But apparently this is true only when the concentration of the salt in the water is small: in fact, with high concentrations, the solution formed by CaCl2 reaches much higher boiling points than the one formed by AlCl3.
Is there someone who can help me to understand?

At high concentrations you'll run into all sorts of problems: non-linear effects (which can't be predicted quite as easily), solubility limit, and hydrolysis of AlCl3.

I believe that Raoults law (as all properties of solutions) is only correct at finite concentrations if you take don't take the concentration, but the activity of each ion.


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Dissolving collagen in water: Why does the weighing scale reading fall?
Every day I take 10g of hydrolysed collagen dissolved in about 200ml of water.
First I put the cup of water on my digital scale (which measures in 0.1g increments) then I add exactly 10g of collagen and stir it with a spoon until it dissolves.
As the collagen dissolves, the total weight always gradually decreases by about 1.5g - 2g in total. If the total weight of the cup, spoon and water is 300g, when I add the collagen, it becomes 310g, then as it dissolves over the next 15 seconds it decreases to about 308g. All I've done is stir it.
I can't see how this is due to evaporation since I can leave the pure water on the scale for 5 minutes beforehand and the weight doesn't change; but as soon as I stir the collagen in it always decreases.
What's happening?

It is likely that the scale does not keep the tare. What happens if you stir the water only? Or does your container at start and after you have prepared your solution and emptied it weight the same? Check theses.

The change in density might make up for a few micrograms (buoyancy in air). Do you use a large spoon, that you first lay on the cup, but afterward leave in the cup, so it creates a torque to one side?

@Karl: With our scales, torque was never an issue. The maintenance guy explicitly showed me after gauging how it was irrelevant where you put your object (for our lab scale at least). You could even place very heavy and very light object right on the edge of the scaling plate.

This is quite not like scale should be used - that's the problem. No adding anything to container on scale, no mixing on scale, otherwise you see what happens.

True, this only leads to contamination and even corrosion of the scale. Seen it in other labs in our institute.

Because collagen does not react with water and you say, no (measurable) evaporation of water is evident, there are only two options as to what is happening:

Matter is converted into energy:
If virtually no evaporation or spillage took place, and your scale is working properly, this is what must have happened. Using the formula of mass-energy equivalence $E=mc^2$ (E = energy, m = mass, c = speed of light in a vacuum) we can calculate the amount of energy that should result from a "loss" of 2 grams:
$$ E = 0.002~kg\cdot(2.998\cdot10^8~m/s)^2 $$
$$ E = 0.002~kg\cdot8.988\cdot10^{16}~m^2/s^2 $$
$$ E = 1.798\cdot10^{14} (kg~m^2)/s^2 = 1.798\cdot10^{14}~J$$
so... that would be a LOT of energy.
Your scale is not working as intended.

I'd go with option 2.
Now on a serious note: We do have a scale (precision of 0.0001 g) in our lab that occasionally has the same problems. Sometimes I would measure an empty reaction flask and it would be fine but some other time I could watch the measured weight drop by the second, "losing" up to 100 mg. The scale was checked by a professional but he could not find anything wrong with it.
My advice is: try another scale, see if the problem persists.

Option 3 OP isn't using his scale properly.

I tried stirring some plain water and the weight decreased by about one gram over 20 seconds, so the machine is the reason.

The apparent weight displayed on the will depend how much of the spoon is below the water surface. You should be able to demonstrate that effect easily in a controlled, repeatable way. (This is just Archimedes' principle, and buoyancy).

"weighing accuracy of 0.0001 g" - perhaps you mean "precision", especially since you seem to indicate it is nowhere near as accurate as that?

lol, of course. English is not my native language. ;) Thanks for the hint.

@alephzero Sorry, that is just plain nonsense.


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Effect on solubility by dissolution of other substance (no common ion)
In one of our recent classes on mass and momentum transfer, we studied qualitatively how blood drug concentrations vary after consuming a tablet containing that drug.
We considered the following:

The region 2 is what I was concerned with. Greater the difference between the actual drug concentration and the threshold concentration, higher are the chances of side effects. The drug will show effect at any concentration above the threshold.
We said that the tablet-stomach acid interface were in equilibrium. So, the maximum achievable concentration of the drug at any point corresponds to the saturated solution of the drug in stomach acid. The concentration of the drug in blood will go on increasing depending on the value that corresponds to the saturated solution concentration. So, lower the saturation concentration, lower would the height of region 2 be.
My question is: Is it possible to decrease the solubility of the drug by adding another substance in the tablet, which is soluble in water? The thought process is - The soluble additive and the drug, both, will dissolve in water. The solution at the tablet-stomach acid interface will be saturated. The soluble additive will "engage" some water molecules. These water molecules will then not be available for the drug molecule to be solvated, reducing the saturation solubility of the drug and, sort of, solving the problem of the region 2.
Is this a feasible idea? What calculations does one do to find this? Please consider this approach for solving the region 2 problem, though there are other solutions possible.

I think your question should be cleared of the biological/pharmaceutical part first. Is "can the presence of a solute influence the solubility of a second compound? (no common ion). As such the answer is yes. This is however of limited general relevance, unless one" solvent" in a solvents mixture is named as "solute"

Pharmaceutically speaking I would go with little doses and multiple takings, but I am not a specialist. I have trouble on see how long one can keep a tablet in his her stomach. Better you isolate a just chemistry question...

"corresponds to the saturated solution of the drug" no it absolutely does not. You don't understand how pills work.

@Mithoron We were working on an 'over'simplified model...

The actual question bolded in your post must be answered with yes. There will most certainly be some compound that reduces the solubility of your drug in aquaeous solution.
However, it is exceedingly unlikely that one of these compounds can be used in the way you intend. The most important reason is that drugs are typically consumed in tablets containing only some hundred micrograms of the active compound (or less). This is so far away from the solubility limit that you would need tens of grams of the other compound to cause some drug precipitation at all. These tens of grams will be far from harmless and would make pills non-consumable due to the much larger size.


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In a substance do atoms actually touch each other?
In a substance do atoms actually touch each other?

The problem here is that you're using your macroscopic intuition of the concept of "touch" and applying that to a microscopic system.

That is terrible reasoning and I wonder how it got even single upvote. There's no fixed orbits and "touching electrons" are even worse.

Quantum mechanics fuzzes out the electrons into clouds of probability distribution (the motion of little electrons orbiting the nucleus is from an old outdated model).  In molecules these electronic clouds do not so much touch, rather they overlap.  See, for instance, https://en.wikipedia.org/wiki/Covalent_bond.

Your question contains a number of conceptual errors that make it clear you have not covered quantum mechanics in any form in your classes yet.
Traditionally, because that was their experience in the real world, scientists would indeed think of atoms as compact spheres (Dalton), hollow spheres with a nucleus and electrons revolving (Rutherford) or the same type of hollow sphere with a nucleus but different orbits on which electrons can be located (Bohr). Using real-world examples, the Dalton model is basically a marble, the Rutherford model a type of cherry and the Bohr model like the solar system.
In all of these models, an atom would have a defined outer boundary which would mark the end and electrons were thought not to go past this boundary except in a reaction. However, these models pose a number of problems that could not be explained using traditional mechanics. For example, an electron revolving around a nucleus would constantly be accelerated towards the nucleus and therefore would emit electromagnetic radiation according to classical physics. This is not observed for atoms. Since there was no way to reconcile the atomic models of Rutherford and Bohr with classical mechanics, an entirely new way to think of atoms needed to be discovered.
This is where quantum mechanics come in. Using the wave-particle dualism and a set of axioms, wave functions for electrons can be calculated. These wave functions give rise to a certain energy (and also to some other observables) that can be measured and that explains the energy levels Bohr had in mind when devising his atomic model very well. Another consequence of this wave function model is that it no longer makes sense to imagine a fixed atom boundary. Theoretically, an electron of an atom can be infinitely far away from the nucleus while still belonging to it. (In mathematical terms: even for very large distances $r$ the absolute value of the wave function $\Phi\ne0$.)
This means that your notion of atoms ‘touching’ does not make sense at microscopic levels, since there is no boundary that could be touching the other side. Instead, the nuclei will come close enough to each other so that the electrons can interact in a favourable way while just far enough apart so that their electrostatic interaction does not dominate.