text:
What is the most stable conformation (Newman Projection) of CF3CH3?
Is a totally eclipsed Newman projection the most stable conformation for a 2 carbon compound, with one carbon bonded to more electronegative atoms (making them partially negative), and the other carbon with something less electronegative (making it partially positive)?
For example, let's imagine $\ce{CF3CH3}$. If they are eclipsed, the fluorine side is partially negative and the hydrogen side is partially positive, so wouldn't they attract?
Or do the electron clouds always repel each other, no matter what atom is used (making a staggered conformation the most stable, not eclipsed)?

Electron clouds are always negatively charged and repel if they come too close, no matter if the atoms they belong to have an overall attraction to each other.

Hmmm, how do hydrogen bonds (or intermolecular forces) exist in that case? Wouldn't the atoms repel there too? I'm getting confused haha.

All bonds have the lenght where long range attraction and short range repulsion cancel each other.

For $\ce{H3C-CF3}$, the stable conformation is staggered. I would expect this to be case for any similar, freely rotating molecule (the exception being conformations imposed by rings etc.), with the reason being the repulsion between the $\ce{C-F}$ and $\ce{C-H}$ bonds.
To confirm, I have optimized the molecular geometry on different levels of quantum chemistry, starting from a conformer halfway between staggered and eclipsed. In all cases, I have observed convergence to the staggered conformer. The levels are

HF/def2-TZVP
PBE-D3/def2-TZVP
B3LYP-D3/def2-TZVP
PW6B95-D3/def2-TZVP
PW6B95-D3/def2-QZVP

In response to comment: I have performed PW6B95-D3/def2-TZVP optimizations for propene, acetaldehyde, and propanaldehyde (the latter starting from two different conformers). The result is that in all cases, the double bond is eclipsed by a single bond (and two non-equivalent conformers exist for propanaldehyde).

So, the attraction between the fluorine and hydrogen is dwarfed by the repulsion by the C-F and C-H bonds?

@F16Falcon That's my observation, yes.

@TAR86 AFAIK eclipsed confirmation of propene, acetaldehyde (ethanal) and propanaldehyde is more stable than staggered and gauche form. Can you check for them too, if I stand correct or corrected...

@TAR86 I am sorry but I didn't get what you mean by .... and two non-equivalent conformers exist for propanaldehyde. And that means eclipsed confirmation of propene and acetaldehyde is the most stable.

@Zenix Propanadehyde has three stable conformers. Two of them are mirror images of each other and thus equivalent.

For 1,1,1-trifluoroethane the two extremes are the staggered and the eclipsed conformations. I have done some molecular mechanics calculations which indicate that the staggered conformation has an energy of 5.7 kCal per mole while the eclipsed it is 1.31 kCal per mole.
This is the result for the molecular mechanics, if we consider the charges calculated in extended Huckel theroy then the flourines and hydrogens do not have large partial charges.
The carbon bearing the three flourines has a charge of +0.865378, the flourines have a charge of -0.261741, the other carbon has a charge of -0.209224 and the hydrogens have a charge of + 0.0430
I have not done the calculations for the electrostatic attraction, but I think in this molecule that if we were to treat the atoms as point charges that the main effect would the flourine atoms and the carbon which they are attrached to. I think that a change of the conformation will make very little difference to the electrostatic attraction energy.

AFAIK eclipsed confirmation of propene, acetaldehyde (ethanal) and propanaldehyde is more stable than staggered and gauche form. Can you check for them too, if I stand correct or corrected...


text:
What is the best hydrogen bond acceptor in 2-amino-N-(3-hydroxyphenyl)acetamide?
I was preparing for my first exam in organic chemistry and came across this question.

Draw a water molecule that is hydrogen bonded to the best anticipated hydrogen bond acceptor.

  2-amino-N-(3-hydroxyphenyl)acetamide, PubChem CID: 20238693

I was thinking that it would be one of the oxygens, because of its greater electronegativity compared to nitrogen. However, I'm having a little trouble distinguishing which oxygen.
I'm leaning towards the alcohol group. My reasoning is that there is less steric hindrance preventing solvation, but I am not sure.

But also, the ketone oxygen is sp2 hybridized. Would this have an effect to increase the polarity of the bond?

There is no ketone in this compound the functional group is an amide.

As a general rule of thumb, a nitrogen is usually a better hydrogen bond acceptor than an oxygen. There is a brief overview on Wikipedia with a couple of values and where they were derived from.
This makes sense intuitively, too. Ammonia is a stronger base than water as it is a better proton acceptor. An amine is a stronger base than an alcohol.
In your example, 2-amino-N-(3-hydroxyphenyl)acetamide, there are three functional groups (or four if you want to count the aromatic ring, too): a primary amine, a secondary amide, and a phenolic alcohol.
The phenolic alcohol is rather acidic and not such a good hydrogen bond acceptor, as some of the lone pairs are in conjugation with the aromatic ring.
The amide is also a delocalised system, which makes the nitrogen pretty unavailable for hydrogen bonding. For this functional group, if it were separate, the π-orbital would be the HOMO, and this is where the protonation/ proton-accepting will happen; due to the delocalisation this is less so the case. An amide usually is not a good hydrogen-bond acceptor.
The amine is primary, which means there is much space around it and it can directly interact with the solvent. This is a good hydrogen-bond acceptor.
Therefore neither oxygen will be the best hydrogen-bond acceptor.

Thank you! That makes sense. I was a little confused about the relative strength of hydrogen bond acceptors/donors so that helped a lot.

As per the answer, it of course correct since this was an exercise from a book. But in real live the matter is more complex. In a polar solvent, the molecule will probably be ionized with a $\ce{-NH3^+}$ group and a $\ce{-O^-}$ group. Then the best hydrogen bonding group will be the phenoxide. Still, the ionization will not be complete, etc. It is a pity that the books for teaching new chemists do not pay any attention to the details.

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text:
N2 symmetry representation in higher symmetry groups (D3h...)
I'd like to describe the symmetry of the ground state of $N_2$ in the higher symmetry groups like $D_{3h}$ and $D_{\infty h}$.
In this answer the topic is covered very good, but only for $D_{2h}$ point group, where $N_2$ orbitals are described by the irreducible representations like this:
$$\begin{array}{cccc}
\hline
\mathrm{s}     & \mathrm{p}_x    & \mathrm{p}_y    & \mathrm{p}_z    \\
\hline
\mathrm{A_{g}} & \mathrm{B_{3u}} & \mathrm{B_{2u}} & \mathrm{B_{1u}} \\
\hline
\end{array}$$
But, when I have a look at character tables for $D_{3h}$ and $D_{\infty h}$, there are no irreducible representations with basis functions $x$ and $y$ as can be seen below.
So, how am I supposed to describe the symmetry of the $p_x$ and $p_y$ orbitals to be able to represent the symmetry of the whole system subsequently?

My attempt
I've tried to describe the symmetry of $p_x$ in $D_{3h}$ like this:
$$
\begin{array}{cccccc}
\hline
E & 2C_3(z) & 3C^{'}_{2} & \sigma_h(xy) & 2S_3 & 3\sigma_v\\ \hline
1 & 0 & -1 & 1 & 0 & 0\\
\hline
\end{array}
$$
Having this reducible representation, I've tried to identify needed irreducible representation using the reduction formula described here:
$$
a_i = \frac{1}{h}\sum_Q N\cdot \chi(R)_Q \cdot \chi_i(R)_Q
$$
For the first IR I'm getting the following result
$$
a_{A^{'}_1} = \frac{1}{12}\left[ 1\cdot1\cdot1 + 0 + 3\cdot(-1)\cdot1 + 1\cdot1\cdot1 + 0 + 0 \right] = -\frac{1}{12},
$$
which doesn't maky any sense, as it should be an integer, if I understand it well.
What am I doing wrong here? Is my reducible representation incorrect or does the formula not apply in this situation?

Will $p_x$ not transform as $x$ in the point group?  In $D_{3h}$ this would be $E'$.

@porphyrin $E'$ is two dimensional, $x$ is one-dimensional only. Otherwise, $E$ operation couldn't be +2.

Yes, that’s why you need to look at p_x and p_y together. If you sum up the characters for both orbitals under each symmetry operation, it should exactly match the E’ row. You can’t look at one part without the other, which is why you’re running into issues with the reduction formula.

@orthocresol Thank you! I'm trying that, but $C_3(x)$ is 0 for both $p_x$ and $p_y$. Then it doesn't correspond with -1 in $E'$.

@orthocresol Typo in my comment - I meant $C_3(z)$.

The angle rotated is $120$  and $2\cos(120) = -1$.

@porphyrin Ok, now it's probably a really silly question, but why is there a cosine?

Not at all; its the projection of the unit vector representing the p orbital on to the axis.

@porphyrin Oh, now I get it! Would you mind to write the full answer for $D_{3h}$ and $D_{\infty h}$, so I could accept it?

@porphyrin And, when is there supposed to be 0 in the representation? Only, when the vector is rotated by $\pi / 2$? Initially, I thought, that 0 is used always when the vector doesn't stay the same nor it's reversed.

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text:
Number of octahedral and tetrahedral voids present in a HCP structure
The given image depicts the tetrahedral and octahedral voids in HCP structure.

For the tetrahedral part
I figured out that there would be 6 tetrahedral voids in total.

My visualisation: In the image, 3 inverted and 3 upright tetrahedron can be formed. Is this correct?

For the octahedral part
It is hard to visualise the octahedrons for an atom. And so, I can't figure out the number of octahedral voids in HCP.

Also, I want to calculate how much part of the octahedron (see figure) is in each HCP structure.

Finally, the image shows only one octahedron. Can you help me in visualising the other octahedrons?

Think of it in terms of hexagonal layers. Any two layers are positioned the same way as in fcc, hence there must be exactly as much voids of each type per atom.

This picture is from How are the number of tetrahedral voids twice the number of octahedral voids in a CCP structure?:

Let's say the red circles represent a layer of atoms above, and the pink circles a layer of atoms below. The T's and O's represent voids with tetrahedral and octahedral coordination located between layers.

My visualisation: In the image, 3 inverted and 3 upright tetrahedron can be formed. Is this correct?

It is correct that there are inverted and upright tetrahedra. In the picture above, the T's squeezed between red circles are inverted, and the T's between pink circles are upright.
I can see eight tetrahedra with their center in the unit cell reposted below. Six have one of the orange atoms (in the image posted by the OP) as their tip, and two have all orange atoms as their base.

Complicating things, there are also tetrahedra with their center at the edge of the unit cell, making counting them difficult (see below).

It is hard to visualise the octahedrons for an atom. And so, I can't figure out the number of octahedral voids in HCP.

The atoms forming the octahedral coordination sphers are hard to visualize in the picture of the question because the unit cell does not fully contain a full set of six. Here is the view from the top again, with the base of the unit cell shaded (red circles correspond to orange atoms in the unit cell shown form the side):

Because we are only showing two layers of atoms, this reflects only half of the unit cell. In this half of the unit cell, there are three voids with octahedral coordination. To visualize them, you need three "pink" atoms at the base of the unit cell (the center one and two vertices of the hexagon) and three "red" atoms (located at c = 1/2, i.e. the second hexagonal layer). Of these three, two are in the unit cell and one is in a neighboring one. If you complete the unit cell by adding another layer, you get a total of six octahedrally coordinated voids in the cell. Because none are on the edge of the cell, this is the total number.

Also, I want to calculate how much part of the octahedron (see figure) is in each HCP structure.

I'm not sure how to do that. I can say that two of the coordinating atoms are in our unit cell, one is in the neighboring cell, and three are on the border of the unit cell. So the larger part of the volume is probably within the unit cell.

Number of octahedral and tetrahedral voids present in a HCP structure

We already know that there are 6 octahedrally coordinated voids per unit cell. Let's get the ratio of the two types of voids. If you look at the "red/pink" pictures, you notice that T's and O's come in horizontal rows, with one row of O's next to two rows of T's. So there are twice as many T's than O's, or 12 tetrahedrally coordinated voids per unit cell.
You can also figure this out in a more complicated way. There are 8 T-voids with center inside the unit cell, and 12 T-voids with center on the edge. These are shared among three unit cells touching at these centers (3-fold axis). So the math is 8 plus (12 divided by 3) is a total of 12, as we already figured out.
Or you can make an argument how many tetrahedral coordination spheres an atom partakes in. Each atom forms the tip of two (up and down) and the base of six (three directions, up and down), that is eight. A tetrahedron has four vertices, so two tetrahedra per atom. The unit cell contains six atoms, so twelve tetrahedra per unit cell.
Guide to related questions
Same question for cubic-close packing: How are the number of tetrahedral voids twice the number of octahedral voids in a CCP structure?
Counting atoms and voids is much easier in that case, but the answer is the same.
Shape of the voids and visualization: Distinguishing between octahedral and tetrahedral holes


text:
Why does lavender oil dissolve plastic?
A few weeks ago I decided to harvest the lavender in my garden and put it through a still to extract the essential oils. The still in question is essentially just an air cooled condenser.
I collected the first fraction in a clear plastic vial. Within a day the vial had become cloudy looking (like frosted glass - the lavender oil itself was not cloudy - just the container). After a week this had happened:

The whole top of the container has been eaten away and is now flexible where before it was rigid.
What's caused this, is it expected behaviour?
Is there anything I should consider before using the lavender oil in fragrances and soaps?

You do realize that "lavender oil" isn't a single chemical but a mixture.

I'm not convinced this question is too broad.

Related https://chemistry.stackexchange.com/questions/28430/how-do-cloves-break-down-plastic/28433

Based on appearance and extent of deformation, your bottle is likely made of PVC plastic which is not compatible with oils. Plastic bottles are made from blow molding. This process leaves residual stress in the polymer chains of the materials but  creates a smooth and transparent surface. Lavender oil can diffuse into PVC and make it softer. Once it is softer, the polymer chains can relax some of the residual stress which makes the surface rough and as a result appear cloudy due to increased scattering. Given Long enough more lavender oil can penetrate the plastic and soften it to the point that the plastic is allowed to flow which causes the observed deformation of the top.
Now I cannot say with certainty it was PVC that your bottle was made from, but if it was, then plasticizer would have seeped into the oil which isn't great for people. If you want to use your oil for air freshener or candles that is fine but I would not recommend it for skin contact applications.

"Isn't great for people" is a euphemism for "toxic, probably carcinogenic, but hopefully there's not too much of it", I take it?

@wizzwizz4 more like "has effects including but not limited to toxic, probably carcinogenic, endocrine disruptor but hopefully not at the current level"

@wizzwizz4 Plasticiser is used to lower PVCs Tg (80°C) so it becomes rubbery at RT. A solid vial, even if made of PVC, would not contain plasticiser. Otoh PVC typically contains a lot of processing aids and other stuff. Anyway all plastic bottles are made via blow molding, and all plastics are likely to plasticise and yield if exposed to unpolar solvents.

Are we sure that the plastic is PVC, and not a polystyrene?

A quick google search yields a number of public scientific pdf papers that report on the chemical composition of lavender oil.   Most simply, the wikipedia entry tells us the primary constituents are various esters, phenols and ketones.
These are organic solvents. It is well known that organic solvents soften or dissolve many plastics.  Try, for example, putting acetone or methy-ethyl-ketone (MEK) in a plastic container.  Put it in an outer glass container to avoid a mess.  Happy experimenting!
By the way, thanks for bringing this to my attention.  I'd never have imagined such potent solvents could be obtained so simply.

Acetone can be obtained quite easily.


text:
How is the light from a fire emitted?
When you see a fire burning, you have wood (fuel) interacting with oxygen which (through a chemical reaction) gives off heat - and light.
But the light appears to emit at some point well-away from the actual fuel - i.e. the flames appear to rise inches (or more) above the actual fuel source where (I assume) the actual combustion/reaction is taking place.
I assume this obviously has something to do with the fact that heat rises - and there is oxygen up there, but there is no fuel up there.
I might conclude that the light is artifact of the elevated temperature of the oxygen, alone up there - but this wouldn't explain the fact that different elements burn different colors - i.e. color is related to the fuel.
So it would appear that the only logical explanations would be either:
a. Fuel becomes "detached" from it's (solid) fuel-source, rises with the heat and emits light after rising.
b. Light is emanated near the fuel-oxygen reaction but appears higher due to some sort of bending it it's travel path due to high heating of nearby air - like how light squiggles over a hot BBQ grill.
There is also the fact that in a very hot "coal" fire the light does emit from the fuel - but in a "roaring" fire the flames appear higher. (Is this because a "roaring" file has higher heat - which correlates to the difference in flame color?) Why would these flames appear "above" the fuel?
Okay - I hope I've illustrated my line of questioning sufficiently. So how does it work?

With wood, it is purely option 'a'.

The final reaction steps of combustion (formation of water and CO2) have the major energy yield, and they largely occur in the gas phase. Wood fire additionally produces gaseous pyrolysis products, which obviously rise before burning up.

Light is emitted from flames by two primary mechanisms: one is small particles glowing incandescently because they are hot (the same mechanism that drives an incandescent light bulb); the other is from electronic transitions from specific energy levels in excited atoms in the flame produces as a by product of the combustion process (this is partly why flames can coloured specifically by elements present in the burning material).
But why do flames and their emissions appear so far from the primary materials that are burning? Clearly solids themselves can burn (eg charcoal in a barbecue is glowing from surface combustion of the carbon with atmospheric oxygen and the emission you see is because this makes the carbon hot enough to glow with heat).
But when things are that hot, plenty of other reactions can occur and this is particularly noticeable if the thing that is burning is less pure than charcoal (which is mostly carbon). Raw coal, for example, contains a lot of volatile impurities (smokeless coal is deliberately treated to reduce the volatile impurities). When the primary reaction of burning carbon happens in raw coal, those volatiles are often turned to gases and driven off the coal to combust as gases far from the solid coal. Of course this is happening at the same time as solid carbon in the coal is burning so the two effects are mixed up to give both glowing coal and gases burning some distance from the glowing coal. Old home chemistry sets used to contain experiments that allowed these effects to be separated. Raw coal is placed in a vessel connected to a tube allowing any emerging gas to be directed far away from the coal. The coal is heated with an external source (eg a Bunsen burner). The emerging volatile gases can then be lit without setting the coal on fire giving a flame a long distance away from the heated coal.
The same can be done with wood (which contains even more volatile and flammable components than coal).
So, when wood or coal burn in an uncontrolled way both the burning of the solid and the burning of the volatiles contained with the solid occur at the same time but the volatiles (as gases) can move far from the solids before they burn. This is why the position of the fame can be far from the burning solid.
Also worth noting
In addition to the volatiles being driven from the burning solid, partial combustion can also create gases like carbon monoxide which can then travel some distance away from the solids while burning (though CO tends to have not very bright blue flame). In fact a controlled version of this reaction (which sometimes also used steam to create hydrogen) was once the primary way of creating town gas (which was widely distributed to support gas lighting and cookers in cities.
If the primary goal is to create flames designed to illuminate rather than heat then burning can be designed to create a lot of small hydrocarbon particles in the flame. This is intentional for things like candles. Here the heat may come from burning the was but the light comes because that burning creates small incandescent soot particles which are heated enough to glow turning a lot of the energy into light.

Thanks for the explanation! So would it be correct to say that when fires "burn down to coals" and are still emmitting heat and "red glow" - that most of the gassious volatiles have been released and burned - and we're mostly down to the rawer carbon/charcoal?

@matt_black So would it be correct to say that the heat released from the reaction temporarily excites the electrons in the products to higher energy levels, which then restabilize by releasing light and returning to their ground state?

@DavidReed basically, yes.


text:
Total Structural Isomers of diphenyl chloromethane?

The maximum option is 8 available but while Drawing I found there are many more . Are there any other condition of structural formula except the fact that the compounds will have the same molecular formula but different structural formula?
Is there any other condition that I have to consider I think I am missing a point .

You were asked just to replace one hydrogen with chlorine, and not to rearrange the carbon skeleton.

Starting from methane, $\ce{CH4}$, as point of departure,

diphenyl methane (letter "d" as in Delta) has two phenyl groups that independently from each other are attached directly to this core (cf.).  Which is the situation described on the questionary, and considered in your option in line A) which answers correctly the question.
Alternatively, as a constitutional isomer, you may join one phenyl group to the next one (which then is called biphenyl, letter "b" as in Bravo), which in turn as one group is bound to the methyl group.  Which is what the is iterated in your options B), C), and D), departing from three different isomers of methyl-biphenyl, like 4-phenyltoluene.

As a side note -- even if options B), C), and D) do not correctly answer the questionary -- it is laudable that you started considering different positional isomers (i.e., the pattern of ortho-, meta-, para-substitution).


text:
Do polyfluorosulfuric acids like H2SO2F4 exist?
I was thinking about a compound resembling $\ce{H2SO4}$ but more acidic. I thought to replace the 2 oxygens with 4 fluorine atoms getting $\ce{H2SO2F4}$, which might be more acidic. Does such a compound even exist?
I also tried searching another compound i.e. $\ce{HOSF5}$ thinking of it to be more acidic than the previous but no results.

https://en.m.wikipedia.org/wiki/Fluorosulfuric_acid.

No.$\mathstrut$

As the comments imply, no dice.  You can, of course, make a six-coordinate sulfur compound with the formula $\ce{SF6}$.  But trying it with a mixture of oxide, hydroxide and fluoride ligands instead of just fluoride presents the opportunity to evolve $\ce{HF}$ leaving the sulfur with a lower coordination number.  There is a five-coordinate compound $\ce{SOF4}$, but otherwise you should expect the sulfur to get down to four-coordination.  So the only stable protic acids you can get with one sulfur atom and oxide, hydroxide and fluoride ligands are plain old $\ce{H2SO4}$ and the more strongly acidic $\ce{HSO3F}$.

It's a different thing with heaver chalcogens. Teflic acid is fine with hexa-coordinated Te.


text:
Diffusion equation I can't seem to find anywhere
So I was looking through some old lecture notes from a few years back when I stumbled upon a useful and simple equation that should, according to my former professor describe the diffusion in solid state compounds quite easily. As I forgot however what the single variables mean I tried to find the equation anywhere. I can only find Fick's laws and differential equations with their solution but I haven't found this equation yet.
$$D=\frac{f⋅σ⋅k⋅T}{n⋅(z⋅e)^2}$$
I remember that we used to describe a lot of things with the help of that equation like how diffusion is slower when higher charged ions are used or that warmer temperatures have to be used for reactions. Or how you can use fluxes to lower the concentration and therefore increase the mobility.
Does anyone either know what the other variables describe or can anyone tell me where to find this equation?
I could only guess here, k and T come from Boltzman's part, n should be something related to the number of ions, z and e have the charges and  f  is probably some specific diffusion constant. So σ may, for example, be something like the specific surface (hence describing why you have to grind your reactants to a fine powder) but I just don't know.

In the diffusion of charged particles $D=\mu k_BT$ where the mobility $\mu=\sigma/(nZ^2e^2)$ and $\sigma $ is the conductivity (siemen/m$^2$) for $n$ particles with charge $Ze$. $D$ has units $\mathrm{m^2s^{-1}}$.


text:
Electrophilic addition of IBr to alkene
I would assume options A and B are possible products due to the $\ce{I+}$ electrophile attacking the alkene, and hence are ruled out.
However, I am torn between options C and D. Both products are formed via the thermodynamically unfavored generation of $\ce{Br+}$, followed by an attack of either iodide or $\ce{H2O}$. While iodide might be a better nucleophile than water by virtue of the former being a charged ion, the concentration of water is likely to exceed that of the iodide ion manyfold. I am of the opinion that both C and D are possible products, and if one cannot be formed, I am unable to decide which one.

You need to think about the mechanism and in particular whether you form a bromonium ion or an iodonium ion as an intermediate. If you form an iodonium ion, then one of the two things that adds has to be an iodine. Unlike the addition of HX, this isn't an electrophilic addition where you form a carbocation.

@orthocresol if we're forming an iodonium ion, for the I to end up on the middle carbon then the attack by the nucleophile must be on the less substituted carbon, which I thought isn't the case for cationic 3-membered rings? Please correct me if I'm wrong.

Yes, you’re right. But I suppose there’s possibly a minor product arising from attack at the less substituted carbon.

But is the probability of an attack at the less substituted carbon greater than the probability of the Br+ electrophile generated?

Good question - not one that I can definitively answer with rigorous proof, but based on experience, I would say yes. By the way, there is no discrete (i.e. free) Br+ or I+ being generated - the alkene reacts directly with I-Br and the selectivity in that step is determined by the inherent polarisation of the I-Br bond.

I would say C is unlikely to form because this requires the formation of a primary carbocation like carbon with no opportunity for a hydride shift.

i think C this more or less requires a cation or attack by water on a less stable complex with an iodine passively sitting there.   I would thing that a small number might react in reverse. [I despise this type of question. If the reaction was run they knew not the results before so why expect a student to know. If the prof found it give the data and discuss then ask the question.]


text:
Finding the error in attempting the equilibrium question through isotherm equations

Consider the partial decomposition of A as:
$$\ce{2A(g) <=> 2B(g) + C(g)}$$
At equilibrium $\pu{700 ml}$ of gaseous mixture contains $\pu{100 ml}$
  of gas C at $\pu{10 atm}$ and $\pu{300 K}$. What is the value of
  $\mathrm{K_p}$ for this reaction?

Attempt:
$\ce{2A<=> 2B + C}$
Let $\ce{n_A}= 2x - 2y$
$\implies \ce{n_B = 2y}, \ce{n_C = y}$
$\dfrac{n_x}{n_{total}}= \dfrac{P_x}{P_{total}}= \dfrac{V_x}{V_{total}}$
For C,
$\implies \dfrac{y}{2x+y}= 100/700  = 1/7$
$\implies x = 3y$
Again for see, from pressure relation
$y/ (7y) = \dfrac {10}{P_t}$
$\implies P_t = \pu{ 70 atm}$
Now, $\mathrm{K_p} = \dfrac{(p_C)(p_B)^2}{p_A^2}$
$\implies K_p = \dfrac{10 \times (2y/7y \times 70)^2}{(4y/7y \times 70)^2} = 10/ 4 = 5/2 $
But answer given is $10/ 28$.
Please let me know my error.

While Solving this, you have assumed that the partial pressure is $10$ atm., which is not the case here. The question actually says that the total pressure ($P_t$) is  $10$  atm.

It will be very much easier if you do the whole calculation by considering partial pressures instead of no. of moles. Suppose the initial pressure was $P^0$, which is nothing but the initial pressure of $A$. Now, after attaining equilibrium let the new partial pressure of $A$ becomes $P^0 -P$. So, partial pressure of $B$ will be $P$, and that of $C$ will be $\frac{P}{2}$.

 Thus ,total pressure ($P_t $)= $P^0 + \frac{P}{2} = 10 $ and by the given volume condition, $\frac{\frac{P}{2}}{P^0 + \frac{P}{2}}  = \frac{1}{7}$.
Solving these two equations will give, $P = \frac{20}{7} $atm. and $P^0 = \frac{60}{7}$ atm. Thus, new partial pressure of $A$ will be = $P^0 -P = \frac{40}{7}$ atm. and partial pressure of $B$= $P = \frac{20}{7} $atm. and that of $C$ ($\frac{P}{2}$) = $\frac{10}{7} $atm.

So, $$K_p = \frac{p_C \times p_B^2}{p_A^2} = \frac{\frac{P}{2} \times P^2}{(P^0-P)^2}  = \frac{10}{28}$$ Thus, the assumption of total pressure as the partial pressure created the error.

Well the question doesn't clearly say that the total pressure is 10 atm, does it?


text:
Acids of noble gases
Are there any other acids of noble gasses besides xenic acid? Why other ones seem to be very unstable (like perxenic), or completely unknown (acids of any other noble gases). Particularly are there fluoroxenic acids, for example $\ce{HXeF5}$?

I rewrote your question. Check out if the changes are OK for you.

You don't make clear if Lewis acids are acceptable - if so XeF6 is an example in its reactions with CsF.

I want them to be h+ donor not lewis acids

Yes there is and it has a very simple structure and is the strongest acid known: HeH$^+$. You might find it interesting to know that HeH$^+$ is also most likely the first molecule to have formed after the Big Bang.

Is it stable i don't think so and acc. to me strongest acid is fluoriantimonic acid

Can it protonate fluoriantimonic acid

It's stable in vacuum and it is not really an acid in the classical sense as you cannot really make a solution of HeH$^+$. See also the discussion here.

Has it ever been created in pure has phase in space or vaccum chamber

It has been created in the laboratory in He plasma's around 1925. Though many people think it should have played an important role in the early Universe, it has no been detected there yet. A possible reason might be its high reactivity so that it may only be detected at very high redshift when there were no neutral particles  to accept its proton (He was the first element to become neutral in the post-Big-Bang plasma).

An acid of krypron is also known and its barium salt is known to be stable. From an abstract of a 1964 paper:

An acid of krypton is formed when krypton tetrafluoride is slowly
  hydrolyzed by ice at -30 degrees to -60 degrees C. The yield is 2 to 3
  percent (mole). A barium salt of this acid, thermally stable at room
  temperature, is formed by the hydrolysis of krypton tetrafluoride with
  a 0.35N solution of barium hydroxide at O degrees to 5 degrees C in a
  yield of approximately 7 percent by weight.

Also see argon fluorohydride, $\ce{HArF}$

Hmmmmm ... According to wikipedia at least the "discovery" of KrF4 was shown to be mistaken (https://en.wikipedia.org/wiki/Krypton_difluoride), thus throwing doubt on the formation of Kryptic acid. Can't find a better reference before I rush out, anybody?

@IanBush http://science.sciencemag.org/content/139/3559/1047


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Acid-base extraction lab exercise
I have a solid mixture of trans-cinnamic acid and menthol. I’m new to this organic chemistry business and I don’t know if I can react the acid directly out of the solid mixture with a basic solution or if I need to dissolve them both into a nonpolar solvent first. Probably a dumb question but thanks anyway

Sodium cinnamate is moderately soluble in water and menthol only slightly soluble so I think you could separate them without a second solvent. Much easier to use a second solvent such as ethyl acetate or diethyl ether though

If the grains of the two were mixed you might be able to dissolve one and leave the other. However if the two compounds coat each other then you might need to dissolve both to get a good separation.


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NMR of phenylethane amine
I did an undergraduate experiment and could use some help in interpreting the $\ce{^1H}$ NMR spectra.
This was the task:
About $\pu{5-10mg}$ (1-2 drops) distilled amine is weighed in a NMR tube. Put the top on the tube immediately to prevent reaction between the amine and $\ce{CO2}$  from the air. Weigh exactly 1.1 equivalent of (S)-alpha-acetoxyphenylacetic acid. Dissolve the acid in about $\pu{0.4 ml }\ce{CDCl3}$ and transfer the solution to the NMR tube.
This here is the $\ce{^1H}$ NMR:

Could I get affirmation of the choices drawn in red and what does the shift at $\pu{4.2ppm}$ correspond to? Could it be because $\ce{CO2}$ reacted with the amine?
(ChemdDraw was used to get the molecular figure and chemical shifts on the figure.)

Looks like your NH3+ group

but it says 7.0 for the chemical shift at that one.

I doubt it - the signal at ~7 looks to sharp to contain NH3+ . Shame that signal is not integrated. To be sure you need to do a D2O shake with the sample and rerun. Note that several of your other signals are not where they are predicted to be. Heteroatom-H signals are notoriously difficult to predict.

The integration came at 12.0 when not including the little top beside it, with the little top it came at 14

and you have 10 aromatic protons to account for. The rest of the integrations of your phenethylamine molecule are off so I would not be too concerned. If that hump at 4.2 is not your NH3+ then where is it?

Is your amine racemic or (S) as shown? Are the amine and acid stoichiometric (1:1)?

racemic. It said to weigh 1.1 equivalent acetoxyphen. acid so it should be 1:1, correct?

So you are likely looking at the spectrum of a mixture of two diastereomeric salts: (R)-amine-(S)-acid and (S)-amine-(S)-acid. The integrations do not seem realistic. You might try a D2O exchange in an attempt to identify NH3+.

I think some of your assignments are not correct. The doublet at $\pu{1.3 ppm}$ is correctly assigned. But close inspect of that signal suggests $\pu{12 Hz}$ coupling constant. Therefore, the signal given by the single proton causing that splitting should be a visual quartet with $\pu{12 Hz}$ coupling constant. But, your assignment for that proton is broad singlet at $\pu{4.75 ppm}$ with 2.7 integration (keep in mind that ChemDraw suggestion is not always true). I think that signal at $\pu{4.75 ppm}$ belongs to $\ce{NH3+}$. The small signal at $\pu{4.1 ppm}$, which is seemingly a quartet with large coupling constant is your $\ce{H-C(Ph)(CH3)NH3+}$ resonance. My best bet for the broad singlet next to it at $\pu{4.2 ppm}$ is crystalline water from  (S)-alpha-acetoxyphenylacetic acid ((+)-O-Acetyl-L-mandelic acid), which is crystalline solid.


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Stability of 3d metal fluorides and iodides in different oxidation states
I read that the fluorides of 3d metals in lower oxidation states, e.g. $\ce{VF2}$, $\ce{TiF2}$, and $\ce{CuF}$ are thermodynamically unstable. For example, $\ce{CuF}$ disproportionates to $\ce{Cu + CuF2}$, as described on Wikipedia.
On the other hand, their iodides tend to be more stable (so, for example, $\ce{CuI}$ can be easily obtained from a redox reaction between $\ce{Cu^2+}$ and $\ce{I-}$). Why is this the case?

In general, iodides stabilise lower oxidation states and fluorides stabilise higher oxidation states, e.g. $\ce{CuF2}$ versus $\ce{CuI}$. This can be explained with some thermodynamics. Consider
$$\ce{M(s) + $\frac{n}{2}\,$X2 (g/l) -> MX_n (s)}$$
In order to form a halide salt with a higher oxidation state (i.e. larger $n$), you need to pay some energetic costs to get the metal to the higher oxidation state, namely:

more electrons are removed from the metal, so you need more ionisation energies
you need to generate more $\ce{X-}$ anions, so you need to break more $\ce{X-X}$ bonds (bond dissociation energies) and add more electrons to them (electron affinities).

This seems bad, so why would any higher oxidation state ever be formed? The answer is that when you pay these costs, you recoup some of the energy by forming more bonds between $\ce{M^n+}$ and $\ce{X-}$. If this bonding is strong enough, this can outweigh the costs described previously, and that makes the higher oxidation state more stable than the lower one.
For fluorine, the $\ce{M-F}$ bond is strong (regardless of whether you consider it to be ionic or covalent) and hence this energetic gain is large. On the other hand, the formation of fluoride ion is not particularly costly: the $\ce{F-F}$ bond is easy to break and fluorine loves picking up electrons, so the electron affinity works in your favour.
Overall, this means that there is a thermodynamic force for formation of fluorides in high oxidation states. Conversely, if you consider iodides instead, the formation of iodide ions is not as easy as the formation of fluoride. And the $\ce{M-I}$ bonding is weaker, too, so there isn't even much incentive to go to higher oxidation states.

Why is the M-F bond stronger, though? I mean, even the 3d series metals are quite big, so the lattice will probably not stabilize there...?

But why does CuF disproportionate ..? In CuF, F already is present as F-, right ? It already has the extra electron that it wants ... what is the driving force behind it ..?

Let's look specifically at copper. Arguably the "outlier" here is not the nonexistence of $\ce{CuF}$, but the existence of the other halides with copper in the +1 oxidation state. Earlier transition metals in tge fourth period generally don't do that, even with iodine.
As explained here, copper has an increased second ionization energy. It's not enough to fully stabilize copper(I), but it does bring copper(I) into the game. Combine that with a soft base that takes advantage of the greater polarizabiluty of $\ce{Cu^+}$ versus $\ce{Cu^{2+}}$, and the former can be stabilized. For instance, copper(I) becomes stable in water solution when the soft base thiourea is added to the solution to provide a soft-base ligand. This does not occur with proposed metal(I) using an earlier transition metal in the period.
Among halide ions, chloride and its heavier congeners are sufficiently soft and polarizable to stabilize copper(I) and thus form $\ce{CuCl,CuBr,CuI}$; but the more compact and tightly bound fluoride ion is not. So we get specifically $\ce{CuCl,CuBr,CuI}$ as simple salts. This is still more than we see in the +1 oxidation state with earlier transition metals whose second ionization energies are lower than copper's.
As with the aqueous environment (and again with copper instead of an earlier fourth-period transition metal), $\ce{CuF}$ can become stabilized by providing a softer base that can exploit the greater polarizability of the less-positive metal ion. Wikipedia cites such complexes with triphenylphosphine and with molecular nitrogen.


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Oxidation state in transition metals
Why does scandium has +2 oxidation state unstable? It only has +3 oxidation state stable.
Scandium has outermost electronic configuration as 3d¹4s². Why it does not lose two 4s e- only instead of all three?

Because then it only has one single electon in all its outer orbitals.  Doesn't sound very desireable to me. Also https://en.m.wikipedia.org/wiki/Scandium#Uncommon_oxidation_states

Scandium actually does have oxidation states lower than +3.  Wikipedia reports a double chloride $\ce{CsScCl3}$ where scandium is clearly in the oxidation state +2.
The stability of oxidation states in transition metals depends on the balance between ionization energy on the one hand, and binding energy due to either ionic or covalent bonds on the other.   In the case of scandium the third ionization energy is low because all three valence electrons are held rather loosely, being in diffuse orbitals that are shielded from most of the nuclear charge by the argon core.  Thus it's easy to get from the +2 to the +3 oxidation state because of the greater binding energy enabled by the latter.  Later first-row transition elements like iron have more tightly bound $3d$ electrons and thus higher third ionization energies, making the +2 oxidation state more favorable.


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Chemical Reaction of aluminum foil in CuCl2 solution
In class we did a lab in which we first dissolved copper(II) chloride powder  in water to form a blue solution.  We then placed a crumbled up piece of aluminum foil in that solution.  It turned brown and the (originally blue) solution that   came in contact with the aluminum turned colorless. The aluminum foil turned brown.
My question here is, what is the brown substance on the aluminum foil, and how do you know it's not rust?
Thanks in advance.

I show the reaction, and copper metal produced, right here.

Rust is iron oxide; you have no iron in the system so it is not rust.
The brown substance is copper dust produced by the reaction $$\ce{2Al + 3CuCl2 -> 3Cu + 2AlCl3}$$
The aluminium chloride then hydrolyses to give $\ce{AlCl3}$ hexahydrate which is colourless and acidic explanation here.

The aluminum foil is a thin sheet of solid aluminum. You can see that it is white colored (or silver colored, if you prefer). So solid aluminum is white in color. Aluminum salts are colorless (or you see them as white). When aluminum salts are dissolved in water, the solutions are colorless.
Now, what is the color of solid copper? Remove the plastic insulation from a piece of electric wire to see the copper wire inside, or look at the copper bottom of a frying pan. You can see that copper is a brown solid.
Copper(II) salts are blue in color. So when you dissolved Cu(II)chloride in water, you got a blue solution. The blue color indicates the presence of Cu(II) ions in the solution.
Aluminum is more reactive than copper. Therefore when aluminum foil is put into the copper salt solution, aluminum atoms on the surface of the foil (in contact with the solution) reacts and takes the place of copper(II) ions in the solution (it now becomes aluminum chloride solution). That is why the blue solution becomes colorless.
The copper(II) ions that were originally in the solution are pushed out to form solid copper powder (brown) and sticks on the surface of the aluminum foil (in chemistry, we say that the aluminum "displaced" copper from the salt). So the brown coating which appears on the aluminum foil is solid copper.
Now how do you know that it is not rust? You must know that iron objects slowly rust as time passes. This is because the iron reacts with oxygen in the atmosphere and slowly changes to iron oxide in the presence of moisture. Therefore rust is iron oxide, which is also brown in color.
Now you can infer that the brown coating on the aluminum foil is not rust, because you did not use iron anywhere in your experiment. Aluminum and copper do not rust.

I agree with my colleagues to the extent that it is most probably copper metal.  However, given that we are starting with an acidic copper chloride solution, and likely there is some oxygen exposure either from the starting water or from air contact, a little more chemistry may be taking place, more precisely, electrochemical (or displaying a surface anode/cathode mechanism). The following reaction is known to occur, for example, with cuprous, ferrous and likely other transition metals (Mn, Co, Ce, Cr,..) in acidic solutions in the presence of oxygen:
Cu(l)/Fe(ll) + 1/4 O2 + H+ --> Cu(ll)/Fe(lll) + ½ H2O
Source for the equation above for ferrous (which is fully available as an html file at https://wwwbrr.cr.usgs.gov/projects/GWC_coupled/phreeqc/html/final-78.html and relatedly also, comments at https://pubs.acs.org/doi/10.1021/es0501058 ) and for copper, see, for example, Eq (7) at https://en.wikipedia.org/wiki/Dicopper_chloride_trihydroxide.
I would rewrite the above reaction given the propensity of some transition metals (like copper) to form basic salts as follows:
4 Cu(l) +  O2 + 2 H+  --> 4 Cu(ll) + 2 OH-
For college students and researchers, I have interestingly derived an underlying radical chemistry path to this reaction employing the supplement, "Impacts  of  aerosols  on  the  chemistry  of atmospheric  trace  gases: a  case  study  of peroxides radicals"', by H. Liang1,  Z. M. Chen1,  D. Huang1,  Y. Zhao1  and  Z. Y. Li, link: https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.atmos-chem-phys.net/13/11259/2013/acp-13-11259-2013-supplement.pdf&ved=0ahUKEwj64JTH4ejMAhVCHR4KHegXCu8QFggcMAE&usg=AFQjCNGZWxTNxFPmgaT1bARYjO08w2_wIg&sig2=Gbhj5orSHmeDIV9uy-moYg :
R24 O2(aq) + Cu+ → Cu2+ + O2−             (   k = 4.6xE05 )
R27 O2− + Cu+ + 2 H+ → Cu2+ + H2O2   (   k = 9.4xE09  )
R25 H2O2 + Cu+ → Cu2+ + OH + OH−    (   k= 7.0 xE03 )
R23 OH + Cu+ → Cu2+ + OH−              (  k = 3.0×E09 )
Net reaction again:  4 Cu+ + O2 + 2 H+ → 4 Cu2+  + 2 OH-
Now, for the experiment discussed with Al/CuCl2, I suspect elemental copper metal, especially freshly precipitated, will similarly react forming a coating of cuprous oxide per the normally slow observed oxidation of iron and copper metal in an acidic air/CO2 mix per the so called metal auto-oxidation reaction (Reaction R24 above):
O2(aq) + 4 Cu/Fe = Cu(l)/Fe(ll) + •O2−
which is also a reversible reaction. Upon acidification, the superoxide radical anion (written as O2−, •O2− or O2•− , but while the latter form is in accord convention, it can present significant issues  on readability and possible misinterpretation of the nature of the active specie) can lead to H2O2 (Reaction R27) and feed a Fenton (or Fenton-type with copper) reaction (see R25 above) creating a ferric or cupric salt. As a consequence, the seemingly observed darker coloration described may actually be due to  the presence of Cu2O, and at low pH and elevated oxygen levels, some much darker spots of CuO could develop.
At temperatures over 65 C, one study (see the fully available article as a pdf at https://www.sciencedirect.com/science/article/pii/0001616069900467 ) refers to “the arrival of electrons at the oxide-gas interface by thermionic emission as the rate controlling step”, where I would expect this leads to the superoxide radical anion via:
e- (aq) + O2 (aq) = •O2−
And per another study also the reaction:
e- (aq) + H+ = •H
where the hydrogen proton (derived from water vapor) apparently occupies surface Cu vacancies, formed from major defects associated with Cu2O (see discussion by Wang and Cho at https://www.jstage.jst.go.jp/article/isijinternational/49/12/49_12_1926/_pdf ).
I would also note that atmospheric superoxide plus water vapor, due to the change in the dielectric of the medium, •O2− becomes largely present as •HO2 (the proton being apparently readily donated by water). Further, the latter hydroperoxide radical is not only a slow precursor to H2O2 (and a possible fenton-type reaction noted above per R25) via the reaction:
•HO2 + •HO2 = H2O2 + O2
but also, recently recognized as an acidic radical in itself (see ‘Radical-Enhanced Acidity: Why Bicarbonate, Carboxyl, Hydroperoxyl, and Related Radicals Are So Acidic’ at https://pubs.acs.org/doi/abs/10.1021/acs.jpca.7b08081?src=recsys&journalCode=jpcafh ).

One equation cited by AJKOER is impossible. I rewrite it here :  $\ce{4 Cu(l) + O2 + 2 H+ -> 4 Cu(ll) + 2 OH-}$, because $\ce{OH-}$ ions cannot be created by acidic ions like $\ce{H+}$ ions

No Maurice, the cited equation is correct. In spite of all the cited chemistry, I should have also noted that for Copper, one needs a chelate (here chloride, as in for example the salt, NaCuCl2, as Cuprous salts are generally insoluble. However, no upgrades, as I did not clearly answer the question! It is not likely darker brown Cu metal, but a basic Cuprous/Cupric oxide layer from simply the cited chemistry. More simply the formation of Cu metal which 'rusts' per O2 exposure per the cited electrochemical reaction to Cupric in acidic conditions (aqueous AlCl3). Verify by removing all O2!

It is not a question of copper chemistry. It is a more fundamental question. The point is that it is impossible to have an equation written with $\ce{ABC + H+ ⟶ XYZ + OH−}$  Because if this would happen, the $\ce{OH−}$ produced at the right hand side will be immediately destroyed by some more $\ce{ H+}$  ions, so that the preceding equation would give $\ce{ABC + 2H+ ⟶ XYZ + H2O}$

Actually, this equation is the pathway to the creation of so-called basic salts, for example, basic copper chloride. See Wikipedia article on a commercial process for its preparation at  https://en.wikipedia.org/wiki/Dicopper_chloride_trihydroxide and in particular relevance here, the oxidation pathway outlined in Equations (6) and (7), namely: 6 NaCuCl2 + 3/2 O2 + H2O → 2 Cu2(OH)3Cl + 2 CuCl2 + 6 NaCl (eq.7).  In current context, replace NaCl with AlCl3. Note cuprous goes to cupric here.

Now, in the case of the discussed experiment with created Cu on Al, in the presence of any O2 and a source of H+ (via hydrolysis of AlCl3), I am unclear as to whether, due to likely limited oxygen presence, we just go from Cu>Cu+. One should think of basic copper salts here as uninvited guests that frequently appear in experiments with transition metals in the presence of oxygen and a source (even H2O) of H+.

For advanced students, note here with have Cu (cathode) and Al (anode), in an electrolyte, can result in a current flow. With O2 and any created solvated electrons (e- (aq) ) could lead to the superoxide radical anion (.O2-) which notoriously (as in health-related issues with humans and Al3+) may create a transient semi-valent species with superoxide. In acidic conditions, this leads to the radical .HO2 and H2O2.  This is a possible further oxidation pathway to convert Cu > Cu2O.


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Calculation of average bond length from force constant and IR absorption?
I need some help beginning a problem set problem (just want hints, I'd like to figure it out myself).
The problem gives me the IR absorption for a diatomic, as well as the force constant for the bond. It asks me to identify the the average separation (bond length) of the two atoms.
I am stuck because the typical Schrodinger equation for harmonic oscillation takes "displacement from equilibrium length" as its variable: $\hat{H}\psi_{v}(x)=E_{v}\psi_{v}(x)$. A change of variables such as $x = l-l_{0}$ would require I know the average separation, which I am trying to solve for.
Any help or direction to relevant resources is greatly appreciated.

You can get the bond lenght via rotational transitions, using the rotational inertia $I=µr^2$, where µ is the reduced mass. From pure vibrational data i don't think it works ... ?

As @Karl mentions, vibrational spectroscopy does not provide any information about bond lengths. Only if a vibrational-rotational (in the infra-red) or pure rotational (microwave) spectrum is measured can bond lengths be obtained.

As pointed out in the comments above, one usually needs information on the rotational structure to extract rotational constants and bond lengths. However, if you are willing to make some crude approximations you can estimate the order of magnitude of the bond length by calculating the classical turning points for the harmonic oscillator using the force constant and the reduced mass of your molecule.

See this analysis for HCl.

Of course if you have the well resolved IR spectrum of a diatomic gas, you have the necessary information about both vibrational and rotational transitions to calculate force constant and bond lenght. If the substance has an IR spectrum, that is, i.e. it's not a diatomic element.

Interesting! Perhaps it's my misinterpretation/misunderstanding of the problem  $-$ we are given a single IR absorption, which suggested to me that it is discussing the fundamental frequency for oscillation. We are also given the force constant of the bond in $Nm^{-1}$ and enough information to find the reduced mass (it is a diatomic @Karl). They then ask for the average separation of the two atoms. Am I just misunderstanding the question?

@Paul, an interesting idea but surely this type of calculation will only give the displacement relative to the average internuclear separation not the separation itself?

What is the actual wavenumber given? Is it an actual IR transition (e.g. $2900cm^{-1}$ for HCl) or a pure rotational microwave band, $~20cm^{-1}$? Otherwise I think there is a mixup in the question, not in your understanding. ;-)

@porphyrin, you're absolutely right of course, but I think one should get a rough idea of the order of magnitude of the bond length. After a bit more searching I came across a few articles who found empirical relations between the bond length and force constant (e.g. R. M. Badger, J. Chem. Phys. 2, 128 (1934)).  A more theoretical approach has been adopted in J. Chem. Phys. 86, 4801 (1998).

Been meaning to get back to y'all -- the professor admitted he should have given the average bond length, and that he intended the exercise to get us to conclude that the average deviation from equilibrium length is $0$.... Thank you guys for an opportunity to really grok the concept, though! I appreciate it.

@porphyrin https://mattermodeling.stackexchange.com/a/1720/5

The method involves fitting the whole ir spectrum (or at least as much as can be measured)  to a guessed potential such as Morse so the value of the bond length is a parameters  and so must depend on the assumed potential, however, this can still be a good result for a diatomic. I would imagine that a polyatomic would be much more problematic as normal modes are involved.


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How to determine the proton concentration in the following solution in mol/L?

A $\pu{0.8L}$ solution contains $\pu{0.60 g }\ce{ NaOH}$ in water. Then the above question follows.

It's been a long time that I made some chemistry exercises. Now I'm stuck with a couple of hesitations.
I know one needs to calculate the number of protons of the molecule sodium hydroxide. But do you need to take water into account or not? I'm thinking not but then the question asks for the solution. So I probably should do that which gets me to $\pu{29 g mol-1}$. Otherwise it is $\pu{20 g mol-1}$.
So that I would divide by $\pu{0.60g}$ which gives me for the first one $\pu{48 mol}$ and the second one $\pu{33 mol}$. Then I'd divide it by $\pu{8.0L}$.
I probably should have a larger answer than what I've done so it is not correct but I don't know exactly how I should work this out.
I'd appreciate a hint.

@Karl: I meant number of protons in the molecule natriumhydroxide. I changed it.

You need more than a hint. You need a hard shove...
The overall notion here is that water dissociates
$$\ce{H2O <=> H+ + OH-}$$
and that the equilibrium is given by
$$K_w = 1.00\times10^{-14} = \ce{[H+][OH-]}$$
where $\ce{[H+]}$ is the molar concentration of $\ce{H+}$ ions and $\ce{[OH-]}$ is the molar concentration of $\ce{OH-}$ ions.
NOTE: The pH of a solution is defined as the negative log of the $\ce{H+}$ ion concentration, and pOH of a solution is defined as the negative log of the $\ce{OH-}$ ion concentration. So
$$14 = \text{pH + pOH}$$
Sodium hydroxide is a strong base, so the assumption is that it completely dissociates.
$$\ce{NaOH(solid) <=> Na+(aq) + OH-(aq)}$$
The molecular mass of sodium hydroxide is 40.00 grams/mole.
$$\ce{[OH-]} = \dfrac{0.60\text{ g}}{40.00\text{ g/mole}\times0.80\text{ liter}} = 0.01875\text{ mole/liter}$$
Rearranging the $K_w$ equation
$$\ce{[H+]} = \dfrac{K_w}{\ce{[OH-]}} = \dfrac{1.00\times10^{-14}}{1.875\times10^{-2}} = 5.3\times10^{-13}\text{ mole/liter}$$

Ok, I didn't see that coming. I didn't yet cover that part of chemistry. I'm familiar with the formulas but that's just because I read it from Wikipedia. I'll cover that part now. Thx for the answer though. I think it would be enough if you'd say use pH and the solubility product. My book I used a while ago has a chapter about it.

Hmmm... water doesn't have a "solubility product." There is no precipitate. $K_w$ is known as the (a) ionization constant, (b) dissociation constant, (c) self-ionization constant, or (d) ionic product of water, but never as the solubility product.

I mean my book discusses that term in that chapter of acids and bases.


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Number of enolizable hydrogens in 2-methylcyclohex-2-en-1-one

The number of enolizable hydrogen atoms according to me is four, but the answer given for the question says it has seven enolizable hydrogens. Where are the rest of them?

Also this book counts also allylic H atoms in methyl group as "enolizable", even if "weakly acidic" would be correct. I have a feeling this site would be better off without all such "book erratums".

Eureka!! , the methyl group hydrogen are also active  , the double bond shifts towards methyl group when hydrogen is removed

@Mithoron I don't see any viable way to produce an enol with the methyl group. So I guess this again boils down to some semantics of what was actually asked. It would be nice to at least have a citation from where the exercise is from...


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Why does a colour change in neutralisation occur?
An indicator shows whether an substance is a base or an acid but during a neutralisation reaction the colour change occurs due to the solution be neutral (i.e pH is 7). But since the indicators only show colour change in an acidic or basic condition doesn't this mean the solution is either acidic or basic?

Different indicators change colour at different pHs, not necessarily at neutral pH

And change for different reasons. but as an example see: Phenolphthalein

You asked two different questions.
(1) Why does a colour change in neutralisation occur?
In general for acid-base indicators, the indicator molecule either gives up or acquires a $\ce{H^+}$ which changes the bonding levels in the molecule. Electrons traveling between the bonding energies thus have different energies which show up as different colors for the two different molecular configurations.
(2) But since the indicators only show colour change in an acidic or basic condition doesn't this mean the solution is either acidic or basic?
Essentially yes, but it is foremost a significant figures problem. So if 50 ml of the acid is 0.02 molar and the equivalence point is at pH = 7.00, the different in moles of base (used to neutralize the acid) between pH = 7.00 and say pH = 8.2 for the phenolphthalein endpoint is inconsequential.
Also not all "equivalence points" are at pH 7.00 The equivalence point for the titration of an acid with a base is the pH at which you have added as many moles of base as there are moles of acid. Since the pKa's of acids vary widely, so do the pH's of the equivalence points.

For a strong acid (ie assuming complete ionization and no protonation
of the conjugate base) then the "equivalence point" would be pH = 7.00.
For a weak acid, some of the conjugate base is protonated (ie $\ce{A- + H2O -> HA + OH-}$ ) so $\ce{pH_{eq} > 7.00}$. Thus the $pK_{eq}$ would be the same pH as if the sodium salt of the conjugate base ($\ce{Na+A-}$) was dissolved in the final volume of the solution.

So a colour change does not neccesarily refer to a neutral solution being formed.

Correct. The pH of the color change depends on the indicator. Link to table of acid-base indicators. Note the various transition pHs.

I'll add that the easiest transitions to detect visually are for indicators that go from colorless to colored.


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What exactly is a potential energy surface?
I first learned about potential energy surface (PES) when I was studying Born-Oppenheimer approximation. In the approximation, the PES is essentially electronic PES which is defined as follows:
Given the Hamiltonian of a molecule
$$\hat H = \hat T_{\text{nucleus}} + \hat T_{\text{electrons}}+\hat U_{\text{electron-nucleus}}+\hat U_{\text{electron-electron}}+\hat U_{\text{nucleus-nucleus}}$$
Under Born-Oppenheimer approximation, $\hat T_{\text{nucleus}} = 0$. So the equation about to be solved is:
$$(\hat T_{\text{electrons}}+\hat U_{\text{electron-nucleus}}+\hat U_{\text{electron-electron}}+\hat U_{\text{nucleus-nucleus}}) \psi_{\text{electron}} = E(\{ R_{\text{nucleus}}\}) \psi_{\text{electron}}$$
Clearly, $E(\{ R_{\text{nucleus}}\})$ gives a PES, and this PES is the relation:
$$\{ R_{\text{nucleus}} \} \rightarrow E_{\text{ground state electronic energy}}$$
However, when I learned about molecular dynamics (also in wikipedia), a PES is defined to be the relation
$$\{ R_{\text{nucleus}} \} \rightarrow E_{\text{energy of the system}}$$
How do we find this map in practice? The only way I can think of is:

We solve the electronic eigenequation for several $\{ R_{\text{nucleus}} \}$.
For each calculated $E(R_{\text{nucleus}})$ of electrons, we solve the equation
$$(\hat T_{\text{nucleus}} + E(R_{\text{nucleus}})) \phi_{\text{nucleus}} = E_{\text{nucleus}}\phi_{\text{nucleus}}$$
to find the energy of the nucleus. (I have never learned how to solve an eigenequation of a nucleus in quantum chemistry)

Is this the way that we find the 'generalized' PES? When we say PES, which is the PES that we are referring to?

$\vec T_{nucleus}$ is the kinetic energy of the nucleus.  Normally under BO approx. the equations are solved at a series of fixed nuclear $R$; this  is what you describe initially and a PES results. If you want to describe motion over a PES during a reaction (dynamics) a common method is to use a LEPS (London-Eyring-Polyani-Sato) potential, which although empirical is based on known potentials obtained from spectroscopic experiments. See Steinfeld, Franscico, & Hase ' Chemical Kinetics & Dynamics' publ Prentice-Hall, or Hirst 'Molecular Structure & Reaction Dynamics'.

The PES $V({R})$ (where $R$ are the nuclear coordinates, electronic coordinates are usually denoted by $r$) is the electronic energy of any electronic state. So there is one PES for the ground state and many more all the excited states.

Clearly, $E(\{ R_{\text{nucleus}}\})$ gives a PES, and this PES is the relation:
  $$\{ R_{\text{nucleus}} \} \rightarrow E_{\text{ground state electronic energy}}$$
  However, when I learned about molecular dynamics (also in [wikipedia][1]), a PES is defined to be the relation
  $$\{ R_{\text{nucleus}} \} \rightarrow E_{\text{energy of the system}}$$

"Energy of the system" is here referring to the "electronic system", so those two relations are the same.

How do we find this map in practice?

Your first step is correct, we need to solve for the eigenvalues of the electronic Hamiltonian, for many different parameters $R$. Each electronic eigenvalue is one point on the PES. So in the end we get a discretized representation of the (actually continuous) PES $V(R)$.
The second step is to solve the nuclear Hamiltonian
\begin{equation}
  \hat H_{\mathrm{nuc}} = \hat T_{\mathrm{nuc}} + V(R)
\end{equation}
There is only one such nuclear Hamiltonian for each PES.

So we need to do many times step 1, but only once step 2 (for each electronic state of interest).
How to solve the nuclear Hamiltonian?
This depends on what $V(R)$ looks like. In the simple case of a diatomic molecule, where $R$ becomes just the 1-dimensional internuclear distance, we may approximate it with a Morse potential or harmonic potential. We can fit the parameters of such analytic expressions to our discretized points from step 1 and then use the existing general solutions for those quantum systems. This will give us the vibrational eigenvalues and eigenfunctions.
Another option is to solve the nuclear Hamiltonian on a numeric grid over $R$. The grid is either defined by the points chosen in step 1, or we can interpolate to a new grid. How to solve this is a topic on its own, so will just give one literature reference here: The book "Introduction to Quantum Mechanics - A time-dependent Perspective" by David Tannor covers many things related to your questions (although it can be challenging to read it).

So altought it is called "potential energy surface" it also includes kinetic energy, that of the electrons but not of the nuclei or I am missing something?

@adosar Yes, this is correct.


text:
Is the mole fraction equal to the weight fraction?
Whenever there is a reaction is excess of air it is customary to assume mole ratio of $\ce{N2}$ to $\ce{O2}$ as $79/21$. However, in another reaction which states that it will contain $\pu{1 wt\%}$ of $\ce{C}$ in iron, I am absolutely confused whether weight ratio will suffice just like I did it for air. Is it absolutely necessary to convert mass to amount of substance and then take ratio or are they equal?

Yes $\vphantom{ghhh}$

Ratios can be quoted either as mass ratios or molar ratios
The trouble with the example you use is that, for air, the weight and mass ratios are close (oxygen and nitrogen molecule masses are only 14% different) so the ratio will similar either way. And you quote the ratio as a molar ratio but later describe it as a "weight ratio". The molar and mass ratios will not be remotely as close for carbon in iron. So you need to know which is which.
If you are trying to interpret a ratio don't overthink it, look at what the ratio claims to be. 1 wt% C in Fe claims to be a mass ratio (that's what wt% or weight percent means, if the ratio has been mol% it would have meant a molar ratio).
If you are given a wt% you don't need to work out the molar ratios, you can simply directly measure the relative weights of the components and be done. If you do convert to a molar ratio you will (usually) find that that is very different to the mass ratio.

Is the mole fraction equal to the weight fraction?

If you had a mixture of two compounds then typically the mole fraction and weight fraction would be different quantities.
The odd exception would be mixtures of isomers. So a mixture of n-hexane and 2-methylpentane would have the same weight fraction and mole fraction since the two different molecules have the same molecular weight.

Is it absolutely necessary to convert mass to amount of substance and then take ratio...?

Typically a problem is posed so that you need to use either the weight fraction or the mole fraction. However some problems could require you to use both.


text:
Ionisation energy is lower for higher energy shell?
Why does an electron in a higher energy orbital require less energy to remove it? Wouldn't it be harder to remove an electron from a higher energy orbital when compared to a lower energy orbital?

It is difficult to remove an electron from lower energy shells because of very large electrostatic forces of attraction between the electrons and the positively charges nucleus.
As the distance between the nucleus and the electron increases with the number of shells, the electrostatic forces of attraction between the electrons of higher energy level and the nucleus decreases, and it becomes easier to remove an electron from that shell.
That is why it requires a large amount of energy to remove an electron from lower shell with respect to the electrons at higher shells.


text:
Why $\psi(r_e, r_n) = \sum \psi_e^i(r_e|r_n) \chi^i(r_n)$ in Born-Huang expansion?
Given a molecular system, the time-independent Schrodinger Equation (SE) that we want to solve is
$$\hat H \psi(r_e, r_n) = E \psi(r_e, r_n)$$
$\hat H$ is hermitian, so the set of solutions $\psi(r_e, r_n)$ are orthonormal.
Under Born-Oppenheimer approximation, $\{r_e \}$ decouples with $\{ r_n \}$, thus each $\psi(r_e, r_n)$ can be represented as
$$\psi(r_e, r_n) = \psi_e(r_e|r_n) \chi(r_n)$$
in which $\hat H_e(r_n) \psi_e(r_e|r_n) = E_e(r_n) \psi_e(r_e|r_n)$.
My confusions:
In every reference I read, the eigenfunction of $\hat H$ is represented as
$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty}\psi_e^i (r_e|r_n) \chi^i(r_n)$$
in which {$\psi_e^i(r_e|r_n)$} is the set of eigenfunctions of $\hat H_e$
Why an eigenfunction of $\hat H$ can be represented in such a way? For me, this expression is just
$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty}\psi^i (r_e, r_n)$$
it makes no sense that an eigenfunction can be expanded in some bases.

Your last equation is just some basis set expansion, something you can always do and not an approximation. But it does not decouple electrons and nuclei, so it is not very useful on its own. As an approximation you can decouple them in each basis function, as is done in your second last equation. This is called Born-Huang-Expansion, which is closely related to the BO Approximation.

@ Feodoran Thank you for the clarification. But I still don't understand why the eigenvectors of $\hat H$ can be expanded by eigenvectors of $\hat H_e$. My understanding of basis expansion is limited to group representation theory, in which we can find the basis of the irreducible subspaces, then any functions can be expanded by the basis. I cannot see the relation between this and the expansion used in BO approximation.

As $\hat{H}_e$ is an hermitian operator it's eigenfunctions represent a complete basis. This in turns means that any arbitrary other function can be expressed as a linear combination of those eigenfunctions.

The eigenvectors $\psi$ of $\hat H$ are not expanded in the eigenvectors $\psi_e^i$ of $\hat H_e$, but in a product containing $\psi_e^i$ (and $\chi^i$). Also as @Raven already pointed out: From a mathematical point of view it does not matter in what basis we expand something, as long as it is a valid basis. We can choose a physically motivated basis, which makes interpretation of the results easier, or we can choose something more abstract. In this case, the choice of $\psi_e^i$ and $\chi^i$ is physically motivated as we need to include both $r_e$ and $r_n$ in $\hat H$.

We prefer to not use MathJax in the title field, see here for details.

After reading this lecture note I have a better understanding.
The reason I'm so confused is that the form
$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty} \psi_e^i(r_e|r_n) \chi^i(r_N)$$
seems to be a linear combination with coefficients that are all $1$. And this is very importand for the derivation of the set of entangled nucleus time-independent SEs since there is no arbitrary coefficients in the set of equations.
After I read the lecture note, here is my understanding:
For a fixed atomic configuration $R^{\alpha}$, the electronic state can be represented by a linear combination of the eigenstates of $\hat H_e(R^{\alpha})$
$$\psi(r_e| R^{\alpha}) = \sum_{i = 1}^{\infty} c_i(R^{\alpha})\psi^{i}_e(r_e|R^{\alpha})$$
although $\psi(r_e| R^{\alpha})$ is an electronic state, this is essentially a solution to $\hat H \psi = E \psi$ at a fixed atomic configuration.
Since by varying $R^{\alpha}$, the set of linear coefficients {$c_i$} will change. So the general $\psi(r_e, r_n)$ can be represented as
$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty} \chi^i(r_N) \psi_e^i(r_e|r_n)$$
in which the set of {$\chi^i(r_n)$} is just the set {$c_i({R^{\alpha}})$}.
Essentially, the $\psi(r_e, r_n)$ that we plug back into $\hat H \psi = E \psi$ is still an electron wavefunction.


text:
why is NaOH used instead of KOH for preparation of K2Cr2O7?
I have been told by my teacher that we use NaOH for the first step of preparation of potassium dichromate to form sodium chromate, which then reacts with conc H2SO4 to form sodium dichromate. Then it undergoes a double displacement reaction with KCl to form K2Cr2O7..My question is that why cant we directly use KOH in the first step so that we dont have to use the third step?

Manufacturers usually prefer sodium hydroxide over potassium hydroxide because it is cheaper. Sometimes, NaOH is preferred because a smaller quantity (as compared to KOH) serves the purpose. And also the product from by KOH is much more soluble than NaOH, make it harder to extract than using NaOH.


text:
Hygroscopic Substances and Drying Agents
My book says:

Most of the hygroscopic substances are drying agents

Based on what I read on several websites, I think dehydrating agents also act as drying agents. If so, all dehydrating agents would also be drying agents and that would contradict my book's statement. So, is there another type of hygroscopic substances or is the book wrong, or is my information or logic wrong?
P.S. I don't think the tags I've used are suitable. I just put the following tags as Wikipedia said that hygroscopy is caused due to either absorption or adsorption. I would appreciate if anyone could help me out.

Things aren't "drying agents" if they aren't used as such...

This question is about the language and logic in regards to chemistry terms that requires analysis of what those terms are. I don't think closing it and sending it off to the English and philosophy scholars would yield an answer other than ask a chemist.

Read closer:

Most of the hygroscopic substances are drying agents

not

all drying agents are hygroscopic substances.

First  Terminology:
Hygroscopic: Absorbs water from air
Drying agent: Removes water from organic substance
Your book did not exclude other materials from being considered drying agents. It just said many of the substances that are hygroscopic are also useful as drying agents.
Take for example anhydrous calcium chloride calcium sulfate hemihydrate (also called plaster or DrieriteTM) and phosphorous pentoxide. All of these are hygroscopic (sequester water from air)
$$\begin{align}
\ce{CaCl2 (s) + H2O (fl) & -> CaCl2 (aq)}\\
\ce{CaSO4.\tfrac12 H2O (s) + H2O (fl)&  -> CaSO4.2H2O (s)}\\
\ce{P2O5 (s) + H2O (fl) &-> HPO3 (cd)}\
\end{align}$$
Calcium chloride is deliquescent (hygriscopic to the point that it forms a liquid) and calcium sulfate hemihydrate is hygroscopic (stays solid but still absorbs water) and both can be used as drying agents for most solvents. Phosphorous pentoxide on the other hand cannot be used as a drying agent for some materials such as alcohols and styrene.
$$\ce{P2O5 (s) + EtOH -> EtH2PO4 (cd) + HEt2PO4 (cd)}$$
$$\ce{PhCHCH2 ->[P2O5] [PhCHCH2]_n \tag{explosive}}$$


text:
What is the rate-determining step in free radical halogenation?
Recently, we were learning about free radical halogenation in school. A set of notes given to us wrote that the rate-determining step is the propagation step. This seems to go against my understanding of the chemistry of this reaction. The step that likely has the highest activation energy should be the initiation step as it is the only step that involves bond-breaking solely. I have looked up various sources and they also seem to suggest that the rate-determining step ought to be the first step.
Advanced Organic Chemistry (5th ed.) by Carey and Sundberg states that:

Free radicals are usually highly reactive and the individual steps in a chain reaction typically have high absolute rate constants. However, the concentrations of the intermediates are low. The overall rates of reaction depend on the balance between the initiation and termination phases of the reaction, which start and end the chain sequence.

This will depend heavily on the concentrations involved. If the concentration is very low, the propagation step could be a lot slower because it's bimolecular.

It depends on halogen.

It does seem odd as a blanket statement. However, the bond dissociation energies (BDEs) for Cl2 and Br2 are quite low, 243 and 193 kJ/mol, respectively, while BDEs for the C-H bond in alkanes are much higher (eg. 435 kJ/mol for methane). So, it's not necessarily the case that events involving only bond breaking will be the rate-determining step, particularly since, as Zhe points out, the propagation steps are bimolecular.
Bromine radicals are significantly less reactive than chlorine radicals, leading to interesting selectivity differences between chlorination and bromination reactions - https://youtu.be/c5enwHzAwmA
Edit: BDEs are taken from Klein, "Organic Chemistry", 3rd ed., Chapter 10.

But the difference here is that the propagation steps are overall still more energetically favourable because formation of bonds like the H-Cl bond or C-Cl bonds are quite exothermic, compensating for the energy used to break the C-H bonds.

The initiation step, breaking of the X2 bond, is not rate limiting nor does it participate in the thermochemical calculations. The initiation step need only supply a catalytic amount of halogen atom for the two propagation steps to proceed. Heed @Tan Yong Boon


text:
How to determine the volume of water necessary to produce a solution with 4.9 g/l sulfuric acid?

The molar mass of sulfuric acid is $M(\ce{H2SO4}) = \pu{98.1 g/mol}$.
  We have $\pu{100 mL}$ of a $0.1$ molar solution of sulfuric acid. What volume $V$ in  $\mathrm{mL}$ of water, $\ce{H2O}$, do we have to add in order to obtain a solution that has $\pu{4.9 g/L}$ sulfuric acid?

I've done the following:
\begin{align}
M(\ce{H2SO4}) &= \pu{98.1 g/mol}\\
M(\ce{H2O})   &= \pu{18.0 g/mol}\\
\text{mass concentration of }  H_2SO_4 &= \pu{98.1 g/L}\\
c(\ce{H2SO4}) &= \pu{1 mol/L}\\
c_{\mathrm{contr}} \times V_{\mathrm{contr}} &=
c_{\mathrm{dil}}\times V_{\mathrm{dil}}\\
\Longrightarrow V_{\mathrm{dil}} &=
\frac{c_{\mathrm{conctr}}\times V_{\mathrm{conctr}}}{c_{\mathrm{dil}}}
\end{align}
The molarity of the diluted water I can't calculate with elementary formulas. So it leaves me with 2 unknowns.
I've tried using so many formulas to no avail.
What formula am I not thinking of? Or what variable am I missing?

The body says "0.1 molar solution", while later you write c(H2SO4) = 1 mol/L; shouldn't the latter be 0.1 mol/L? 2) Where did you get the value for the density (rho) from? Did you just re-use the value of molar mass and change the the units? Don't do that :) You have to look up the density in tables, if it isn't given. You will then notice that the density changes in dependence of the concentration - so you had to note down, which H2SO4 concentration you are referring to when writing down the density. Here is an example tabular

@Marzipanherz  I'll change that. Yeah I shouldn't use the same notations as in physics. I'll change that too.

I see. Are you aware that this mass concentration is also referring to a 1mol/L solution, not to a 0.1molar solution the question talks about?

We have talked about it in chat and maybe the question is a bit ambiguous. I would interpret the last line as a mass concentration. So what you are actually looking to create is a $\gamma(\ce{H2SO4}) = \pu{4.9 g/l}$ solution. Then the calculation becomes straight forward.
Keep in mind that water is your solvent, so you can't actually dilute it; also writing down the concentration is a bit pointless here.
You can easily convert the mass concentration to an amount concentration. The relative molar mass $M$ is given by the mass $m$ of the substance divided by the amount of substance $n$:
$$M = \frac{m}{n}\tag1\label{molarmass}$$
The mass concentration is defined as
$$\gamma = \frac{m}{V}.\tag2\label{massconc}$$
The amount concentration is defined as
$$c = \frac{n}{V}.\tag3\label{amountconc}$$
You can use \eqref{molarmass} in \eqref{amountconc} and obtain
$$c = \frac{m}{M} \times \frac{1}{V}
=\frac{m}{V} \times \frac{1}{M}.\tag4$$
Substitute \eqref{massconc} back in and you have
$$c = \frac{\gamma}{M}.\tag5\label{amounttomass}$$
Now in the question you have given the following:
\begin{align}
M(\ce{H2SO4}) &= \pu{98.1 g/mol}\\
c_\mathrm{i}(\ce{H2SO4}) &= \pu{0.1 mol/l}\\
V_\mathrm{i}(\ce{H2SO4}) &= \pu{100 ml}\\
\gamma_\mathrm{f}(\ce{H2SO4}) &= \pu{4.9 g/l}
\end{align}
I have given the initial values the subscript $\mathrm{i}$ and the final values the subscript $\mathrm{f}$ (instead of concentrated and diluted).
You are now looking for the volume of solvent you have to add, which is
$$\Delta V = V_\mathrm{f} - V_\mathrm{i}.\tag5\label{volume}$$
Then you are right, you need the following relation
$$c_\mathrm{i} \times V_\mathrm{i} = c_\mathrm{f} \times V_\mathrm{f},
\tag6$$
rearranged and using \eqref{amounttomass}
\begin{align}
c_\mathrm{i} \times V_\mathrm{i} &=
\frac{\gamma_\mathrm{f}}{M} \times V_\mathrm{f},\\
V_\mathrm{f} &=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i},
\end{align}
and using \eqref{volume}
\begin{align}
V_\mathrm{i} + \Delta V &=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i},\\
\Delta V &=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i} - V_\mathrm{i}\\
&= \left(\frac{M \times c_\mathrm{i}}{\gamma_\mathrm{f}} - 1\right)V_\mathrm{i}.
\tag7\label{dilute}
\end{align}
Now simply plug in your values and get the solution (hover over box to reveal):

 $\displaystyle \Delta V = \left(\frac{\pu{98.1 g//mol} \times \pu{0.1 mol//l}}{\pu{4.9 g//l}} - 1\right)\times \pu{0.100 l} = \pu{0.100 l}.$


text:
Why does a chiral carbon affect some hydrogen's hydrogen environment?

In the screenshot attached taken from Khan Academy's video (https://www.youtube.com/watch?v=p9B4s0N5yk8&list=PL0pJUwkI0YCZRYrgKqHXZ0XCIu11afbJ8&index=3) the thick pink dot refers to a chiral carbon atom. Why is it that the two hydrogens circled red and pink that are bonded to a carbon which is then bonded to a chiral carbon atom are in different hydrogen environments? However, the three green circled hydrogens that are bonded to a carbon which are then bonded to a chiral carbon atom are in the same hydrogen environment?
From my understanding of hydrogen environment, a hydrogen environment is characterised by the functional group(s) that is attached to that hydrogen. Therefore, I think that the two hydrogens circled red and pink are in the same hydrogen environment, just like the three green circled hydrogens are in the same chemical environment.
Therefore, my question is why does a chiral carbon atom affect the hydrogen environment of a hydrogen? And especially from Khan Academy's example, only some hydrogens are affected by the chiral carbon and some aren't affected by it. Is there some loop hole in my understanding of hydrogen environment?

Try this: https://chemistry.stackexchange.com/questions/86205/ch2-group-in-alkyl-glucoside-gives-two-separate-hnmr-signals/86451#86451. Occasionally Khan's Academy is weak on organic chemistry.

1
These two hydrogens are in different environments. Search SE for "diastereotopic protons/hydrogens". Substitute one hydrogen with deuterium and then the other one. Now you have two diastereomeric stereoisomers.

There is a big difference between the two hydrogens and the hydrogens in the two methyl groups.
The issue is that the methyl hydrogens are part of a group that can rotate quickly and so they see an averaged environment which is the same for all the hydrogens. If the bonds were "stuck" and the methyl groups could not rotate your argument might be correct, but the is not the case.
The two hydrogens on the carbon next to the chiral centre are also connected to bonds that can rotate. But they never see the same environment even on average because they will always have a different relationship to the hydrogen and hydroxyl on the next carbon. Whatever way you rotate the bond connecting their carbon, the two hydrogens will be in a different environment, so they appear with distinct shifts.


text:
Boiling point of water
I was doing an experiment on boiling-point of water and I was supposed to measure the temperature of the water when it boils, but I saw the there were droplets of water on the lid of the pot, and I was confused if I should take the temperature of the droplets too ,then record the average temperature.

No, the droplets on the lid are irrelevant. They can be pretty far from boiling, too.

I'm kinda curious why someone downvoted this question; yeah, it's pretty ignorant, but then, as someone once said, everyone's ignorant, just on different subjects. The OP asked a question because he/she didn't have the information to know better, but to downvote just because of that seems mean.


text:
Memorizing chemical equations
Is there's an easy or at least a practical solution for memorizing chemical equations? I'm currently in grade 12 and we're studying a whole chapter about Iron and how to prepare it, starting from the Iron ore till converting it to different Iron compounds used in the industry, and we have questions in the exams like How can you obtain Iron(II) sulfide from Iron(II) oxide, we have about 25 equations that should be memorized in order to solve such question!

We don't memorize equations. We memorize the general rules.

@IvanNeretin So what are those rules? where should I start from?

Why should you memorize them when you can write them yourself. In this case  of obtaining Iron(II) sulfide from Iron(II) oxide there are various ways like reacting the Iron(II) oxide with Hydrogen Sulfide (H2S) etc.

Your teacher should have taught you those rules. How oxidation numbers work, stochiometry, Budouard reaction, etc. Then you just memorise the general picture of the blast furnace process, and the equations fall into place.

This is a good question. Memorization is difficult. I'm a chemist, and I try not to memorize, but just stuff stuff into my memory. As Ivan said, you memorize the rules.
But I'm going to go one step further: in your imagination (because that's the only lab and production space available to you for this exercise), get a ton of iron ore (in whatever compounds are listed) and convert them into the various products that are desired. In other words, immerse yourself into the chemistry.
There are not 25 equations; there are about 10 compounds and about 50 ways to combine them, only some of which are reasonable. The unreasonable combinations are as interesting as the reasonable ones. Why don't some things happen? Why do some things go neatly? When you stack the 10 compounds up in columns or rows and draw lines between them, you have a road map - that you have traversed backwards and forwards! You have drawn an image that is engraved in long-term memory rather than memorize a list in short-term memory.
I think this enables you to become a chemist more easily. If you can't do it this way, hard work and memorizing the equations will probably get you thru the course with a good grade, but I bet you'll forget most of it within a year.


text:
Any consumable liquid that would dissolve ground coffee?
Figured I would try here rather than a food-oriented forum.
Is finely ground coffee (Espresso/Turkish) soluble in any consumable liquid?
Otherwise asked: is there any liquid available to the average consumer that will allow him/her to make instant coffee out of grounds?

No way to dissolve the grounds, and that would no doubt given the coffee a much different taste. // I have seen coffee marketed in tea bags.

This may seem ridiculously obvious, but hot water will do it. But it won't dissolve everything, just the stuff that makes coffee. Even if you could dissolve the remaining material it would not be coffee as the insoluble parts won't taste like the soluble parts (which is after all what makes the drink).

It's a fair comment. Yes, it's easy to simply add hot water to very fine coffee grounds and make coffee. But that requires boiling water and knowing when to stop drinking to avoid the grounds at the bottom. I was simply interested in knowing whether there was any liquid into which the whole coffee grinds would dissolve, at room temperature, for convenience.
Re: taste. Chewing roasted coffee beans (or consuming the inside of chocolate-coated coffee beans) tastes roughly the same as drinking a cup of coffee to me.

Why don't you just buy instant coffee then? That's what they make it for.

Dislike the taste, believe it contains less caffeine/weight than regular coffee, and am uncomfortable with the amount of processing required. So I guess the answer is no!

uncomfortable with the amount of processing in instant coffee powder, but you would consider drinking some strange liquid chemical that dissolves coffee beans? ;-)

I was hoping that there were some relatively un-exotic liquid that would dissolve ground coffee, in which case I would indeed be comfortable with ingesting it. Sure, instant coffee and pouring hot water over finely ground coffee will work, but that wasn't why I opened this question.

The water-soluble material in coffee beans (1-2.5% caffeine) and water-extractable oils (10-15%) represent a small percentage of the total bean. There is a lot of carbohydrate (~50%), not all of which is soluble, and some volatile oils (not so tasty).
The bean is not totally soluble in anything, but when ground very fine, it could be dispersed in a thick suspension (or a thin suspension, if you keep shaking it up). I wonder what coffee material is used in coffee-flavored ice cream: ground beans, powdered instant? The ground-up bean would probably be a good source of dietary fiber.

Thanks for the answer, James! Neither here nor there, but I believed that whether those coffee oils make it into the brew or not depends on the method used (and I believe that unfiltered coffee methods like pour-over and Turkish, including those oils, have been implicated in raised cholestrol levels). I presume that that is referring specifically to the water-soluble oils which you mention?
So hot water over finely ground beans will provide a suspension, but there is truly no liquid into which the bean will be soluble?

My interest in this, by the way, is in finding a way of benefiting from the residual fibre, antixodants (probable), and whatever else doesn't make it into an aqueous solution while still enjoying the benefits of caffeine (namely, caffeination and the ingestion of xanthines).

It was informative to address your question, so I looked up coffee beans in Wikipedia. Otherwise I just drink it. I must not be very discriminating; I use cream and sugar to taste. The grounds could be a good source of fiber - just look at what you discard!


text:
Addition of NaBH4 and LiAlH4

NaBH4 and LiAlH4 will be a source for the hydride ion and reaction will begin with addition of hydride by displacing the carbonyl group.
But I am not sure will it attack both the carbonyl or only one.
Moreover, Will it displace the chlorine at the bottom carbon?

There are three carbonyls here,  not two. Both of those reagents will actually react with all three of the carbonyls in this particular case.

In the book, when the compound reacted with NaBH4, it reduce only the first carbonyl and the bottom one to -OH. Whereas when it reacted with LiAlH4, it reduce all the carbonyls to -OH

This is a β-ketoester, though, and that changes things. I don't have time to draw out the reaction right now, but I wanted to make sure that if you get an answer, the person answering takes this into account.

What solvent is this in?

consider it as any acid.

LiAlH4/ether or THF will reduce the three functional groups forming a tetraol. NaBH4 in hydroxylic solvent will behave the same way. Normally, esters reduce slowly with sodium borohydride such that the acyl chloride and ketone can be reduced selectively. I have found beta-ketoesters to reduce not only the ketone rapidly, but also the ester group. Presumably, by initial reduction of the ketone and then the ester via a neighboring group participation.


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Is flour or baking soda really suitable for extinguishing grease fires?
I think most have heard the advise that you should put flour (or baking soda) on a grease fire to absorb the oil before. While this is certainly better than using water, I'm skeptical.
For starters it is know that flour is excellent at demolishing flour mills,  but in these instances it is a dust explosion (actually a deflagration) in a confined space versus a splash of mostly unaerated powder onto a fire. Even the fine particles that fall off should only cause mild flames compared to the on going grease fire. The main cause of skepticism is the potential charring of the flour releasing water and causing splattering (i.e. bigger fire).
Alternatively baking soda has been suggested, but this also seems dodgy as it starts to decompose at 50 °C releasing carbon dioxide and water, and how many people have enough baking soda to fill a pan in their kitchen ready to go? It seems that a lid would be the most appropriate response and of course would not recommend someone to do their own fire fighting if not prepared to do so, but when seconds count help is only minutes away.
I suppose that decomposition would be less of an issue if sufficiently slow kinetically. But debating physical sciences is a fools errand so I ask: Is flour or baking soda really suitable for extinguishing grease fires?

Not sure about flour, but baking soda definitely sounds good. Its decomposition is precisely why is it good.

If the flour is coarse, it might be OK. But if it's very fine and powdery, you might just be making a very flammable aerosol, depending on how you deliver it.

Putting on a lid is the best solution. Soda decomposes not at 50, but 1600°C. No chance of that happening, but if it did, it would, as you say, release CO2, which is also fine. You might have a larger supply of table salt in your kitchen, an excellent fire extinguisher. If none of the above is at hand, a kg of wheat flour  serves the same purpose, at the cost of a bit more of a mess to clean up afterwards.

Id have to look up the composition of flour, but unless the stipulation of grease fire has a caveat im unaware of... the decomposition of baking soda is good for two reasons off the top of my head. 1. It's a heat sink. Energy absorbed from the fire is energy that doesnt go into sustaining the chain reaction of combustion (ie. the fire) after ignition. NaCl salt is used similarly (against class D metal fires i believe). 2. Evolving CO$_2$ gas displaces the air around the fire and insulates it from oxidation by O$_2$ in the air. This use of CO$_2$ is a common mechanism for fighting fires.

@Karl Baking soda definitely does not decompose itself at 1600 Deg C. :-) "Carbon dioxide production from exposure to heat starts at temperatures above 80 °C " But I agree the lid is the best.

@Poutnik Dry soda melts at 850 or sth., and decomposes only much later. In a dough, thats different. Would be quite pointless to use a leavening agent that only works when your bread has long turned into charcoal. ;-)

@Karl But the topic is about the bicarbonate, not carbonate. I originally thought you just missed that.

@Poutnik baking soda == bicarbonate, got it. ;-)

Do not use flour, it may explode!

Use the pot lid to smother the fire.
Use a fire blanket - read the instructions, watch a you tube demo. People who think they know what they're doing often spread it or end up burned.
Salt, will put out small fires, such as a pan fire, you won't have enough salt to smother a pot or absorb the heat.
Good for putting out small fat flare ups on a barbie plate.

Baking powder and powder will also work on small fires. Baking soda ( or powder) fire extinguishers used to be common  but they made such a mess that generally they have been replaced with CO2 extinguishers.

@blacksmith37 Pure CO2 extinguishers are far less effective if used by untrained personnell. Powder extinguishers make a big mess, but they are not getting replaced. Who told you that?

The reason these substances are advocated for is to quickly remove the oxygen source from the burning oil. The lids to the pan is a great option, but if the flames are too high, someone might struggle to get the lid on.
The idea with a powder substance is to dump a bunch on, beating down the fire, then continue to fight it. Now put the lid on. If all the flames are out, you can add water to cool the mixture, perhaps, but be careful.
The main point is to have something other than water to reduce the flames. Yes, it will make a mess, but it will buy a little time by cutting off the oxygen supply.

Don't put water on it until everything has cooled well below 100°C! Pull the pan off the hot plate, extinguish the flames with lid, salt, soda, or whatever you have at hand, and wait.


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How to quantify the adsorption affinity of gases?
Is there a term/quantity which shows how 'sticky/adsorptive' a molecule is? I am interested in gas adsorption on steel surfaces in our mass spectrometer and would like to estimate which gases have a higher propensity to adsorb.

The term you are looking for is "adsorption enthalpy". However calculating that value will depend on a variety of factors mainly the absorbate and substrate, but it is not that simple. The surface structure of substrates matters and a $[111]$ oriented crystallographic face will have a different adsorption enthalpy with a given absorbate that say a $[001]$ crystallographic face.
I suggest you study this more, namely the Langmuir adsorption model and Brunauer–Emmett–Teller Theory (BET). Especially since BET is a popular technique to measure adsorption.

Okay, so there is no simple way of estimating this gas is stickier than that gas provided we talk about the same absorbent,or?

@Zorg
Critical point is quite useful in that sense. A least you can have some idea form the boiling point of the molecule.