Why should the scalar product of the components along two axes remain unchanged after transformation? A couple of proofs from Chapter 2 of "Space Groups for Solid State Scientists" are giving me a hard time (see attached image).

So what I understand is:
"r" was the original lattice point. "r'" is the lattice point r after the mirror operation, which should also be a point of the original lattice. I also have no trouble with how the transformation r->r' works, and how dot products being zero imply a right angle between the vectors.
So why should the two dot products be equal? And how should one interpret this dot product intuitively (component of one axis onto another?) ? (I can kind of visualize how the mirror image of a triclinic lattice wouldn't coincide with the original if one of the axes is not perpendicular to the other two, but how should one prove it rigorously, and how does the aforementioned proof work)

Your question
The scalar product depends on the length of the vectors and the angle between them. Rotations, translations and mirror operations do not change distances or angles (i.e. don't distort objects), so the scalar product is unchanged as well. There are other operations that do (stretching, warping, etc).
The proof
Without knowing the context, the proof seems wrong. Equation [2.18] is not general and only works for orthogonal coordinate systems. To give a simple example: if the b axis is not perpendicular to the c axis, and we rotate on a twofold along c, the b axis will not be transformed into -b, but rather into a linear combination of b and c. What this proof shows is that if axes are orthogonal, they are orthogonal (tautology).
What is true is that if there is a two-fold axis along c, you can always choose the two other axes such that they are orthogonal to c. This is a choice, however, and other choices are possible.

In the figure, b and c are not perpendicular. Points at the origin, at the tip of b and the tip of c are equivalent by translation. If there is a crystallographic twofold rotation along c, there is another equivalent point at the tip of the rotated b. The rotated b is not the same as negative b, as the figure illustrates by counter-example. However, if you want perpendicular axes (and that is the choice recommended by the international tables), you can switch from the axis b to the thin line (shown as dot-dash-dot-dash), giving you a new axis perpendicular to c. In the case shown, the lattice is now face centered.


What type of Organic Substance is used in LCD I know that the liquid crystal displays contain some class of organic compound, which is somehow used to polarise light to make contrast variations(In calculator displays).

My Question is What type of Organic Compound is used in this case and How does it polarise light only when stimulated by external agents such as current in LCD's

Liquid crystals are usually rigid rod-like molecules that tend to show some degree of intermolecular order even when notionally liquids (there is a good summary on Wikipedia). This happens because the rigid polarisable parts of the molecules can have relatively strong directional interactions even in the liquid phase.

Some of the structures they form can rotate the plane of polarised light You can get some intuition about why knowing that polarising lenses consist of highly ordered molecules whose interaction with light is highly direction dependent. In normal liquids the molecular orientations are effectively random so, even in each molecule interacts with light depending on its orientation, this effect washes out due to the randomness of the overall structure. Liquid crystals form structures where there is the possibility or large-scale order allowing macro-effects on polarisation (just like the rigid structure in a polaroid film).

The other thing that makes liquid crystals useful is that the details of the ordering in the liquid can be affected by external electric fields, especially in thin films of the compounds. In principle this is possible when any molecule is polar (that is it has a net electrical dipole). Of course, making an effective display requires a lot of fine tuning of the electrical, mechanical and chemical structure, but the basic details are simply that the molecules are affected by electrical fields and this allows some control of the macro-structure which, in turn, alters the way the liquid crystal interacts with light. the effect is not "turned on" by the field, just altered and that is enough to make a viable display technology.

Liquid crystals have a long history. The first compounds known to show liquid crystal behaviour is cholesterol benzoate, the first molecule shown below, which was dicoveredn in 1888. The other structures show some actual molecules from the early days of LCD displays to modern developments. For more examples of the history there is a great summary from Merck (one of the major manufacturers) in this slideshow.


Does acetone eat through Thermoplastic polyurethane and Polycarbonate I used nail polish remover on an expensive phone case to remove sticker residue (and it worked, no longer sticky). I didn't know it didn't mix well with plastic materials. Did the acetone damage it?

If there is no visual damage, e.g. cloudiness or crazing, the case should be OK. If there's some surface cloudiness, it might be possible to buff it out with mild abrasive, such as toothpaste or milk of magnesia.

In the future, try oil (olive, safflower, mineral etc., but not motor oil) to remove stickers. A bit slower, requiring subsequent cleanup, but safer.


Products for the hydrogenation of the cycloalkene in (3R,4R,5S)-4,5-diamino-3-(pentan-3-yloxy)cyclohex-1-ene-1-carboxylic acid Write the products (more than one) for the hydrogenation of the cycloalkene ring in (3R,4R,5S)-4,5-diamino-3-(pentan-3-yloxy)cyclohex-1-ene-1-carboxylic acid. Reagents are molecular hydrogen with Pt catalyst.

My problem here is that the question implies there are several products for hydrogenation of the cycloalkene. I only see one product, where each end of the alkene obtains a hydrogen, and the pi bond is gone. How can a hydrogenation of this alkene result in more than one product?

Hydrogenation of the alkene introduces a stereo centre, thus the products are:


Nucleophilic substitution with chloroethane: hydrogen cyanide vs. potassium cyanide Chloroethane can be used to make sodium propanoate.
$$\ce{chloroethane -> Q -> sodium propanoate}$$
The Intermediate, $\ce{Q}$, is hydrolysed with boiling aqueous sodium hydroxide, to give sodium propanoate.
Which reagent would produce the Intermediate, $\ce{Q}$, from chloroethane?
A concentrated ammonia solution
B dilute sulfuric acid
C hydrogen cyanide
D potassium cyanide

What is the difference between using hydrogen cyanide or potassium cyanide? Why would the answer be D?

Because hydrogen cyanide is not a very strong acid.

Potassium cyanide exists as potassium and cyanide ions, so here the cyanide ions can act as strong nucleophiles. Hydrogen cyanide is only a weak acid and so it will not dissociate to produce a significant amount of the cyanide ion nucleophiles. So it is a better idea to use potassium cyanide.


What does "EM grade" mean I'm looking at purchase orders and I see purchases for chemicals like the following.

Glutaraldehyde, EM Grade, 25%

What does the "EM Grade" mean in this text? It says that it's recommended for histological or immunological techniques, but how?

EM grade is a label for chemicals used in electron microscopy.


Volume of different gas in a container I know that technically two or more gases in one container should have different volumes. But when you imagine the gas molecules being mixed together, it look sorta like the two gases have the exact same volume, that is the volume of the container.

Also, in one of the questions I did on gas laws, wet hydrogen and hydrogen are shown to have the same volume in a container. Given the barometric pressure, I subs tract the vapor pressure at the certain temperature from it. Then I sub the pressure of dry hydrogen gas into the ideal gas law along with the volume of the container to find the number of mole of hydrogen gas.

This is the part in my textbook that I am confused about. It shows how partial pressure fraction is equal to the corresponding mole fraction. But this is all under the assumption that the volume and other things are constant. I am lost right here, because Avogadro's theory states the volume and mole number of a gas are directly proportional.
( wait...wouldn't this mean that pressure, mole and volume are all directly proportional to one another???)

Can anyone give a answer, along with a proof/ a source?

Your second idea, that they have the same volume, is correct.  Each gas in a mixture of gases stored in the same container has the volume of that container.  Remember that gases are mostly empty space, so it is not a problem that their volumes overlap.

As is implied by your comment about subtracting the vapor pressure of water from the pressure of wet hydrogen, the gases in a mixture can have different partial pressures. Perhaps that is your point of confusion.

The only case where they would have different volumes would be in a more accurate gas model than ideal gases, something like the Van der Waals model, where gas volume is corrected by a small amount to account for the portion of space that is taken up by the gas particles.

In short, each component of a mixture of gases stored in a given container shares the same volume and temperature and may (or may not) have different partial pressures and number of moles.


Relationship between pH and volume I had prepared two different volume of solution: $\pu{500 mL}$ and $\pu{100 mL}$ with same concentration ($1\%$), but I obtain different $\mathrm{pH}$. Will volume affect $\mathrm{pH}$?

If correctly prepared, the pH should have been the same. Logically, consider if the pH would be different if measured in 5 separate 100 ml samples poured from the 500 ml preparation.

Why might the pH differ in your tests? Some possibilities:

One container was not completely rinsed after cleaning and had some (often alkaline) detergent clinging to it.
The preparation absorbs $\ce{CO2}$ from the air, so the layer on top is more acidic.
Error in measurement.

Unless you have a buffered system, pH may vary abruptly with a slight change in ingredients.


Why does the chlorination of indene occur with syn selectivity in heptane? In the chlorination of indene to form 1,2-dichloro-2,3-dihydro-1H-indene in heptane as solvent, the syn dichloride is formed as the major product, as reported in J. Org. Chem. 1980, 45 (25), 5150–5155):

The explanation I was given for this was that "in a non-polar solvent, the ion pair collapses quickly". What does this mean, and why is the stereoselectivity reversed, as compared to typical halogenations which proceed via a halonium ion intermediate to give anti dihalides?

Key

The reaction is run in heptane, a non-polar solvent.  A non-polar solvent will not stabilize or support ions as well as a polar solvent.

Background

When the $\ce{Cl2}$ begins to interact with the indene an initial ion pair is formed.  In the case of chlorine, it is not always clear whether

a cyclic halonium ion (the 3-membered structure on the bottom left)
an open ion (the structure on the bottom right)
or an equilibrium between the two ions occurs

[Side Note: In electrophilic bromination reactions involving alkenes, bromonium ions are very common intermediates usually leading to trans addition.  However, chlorine isn't as polarizable as bromine and in this particular case with indene, the open ion is extremely stable due to its benzylic nature. Hence the involvement of only a chloronium ion is questionable in this case.]

Analysis

As mentioned above, ions are not very well stabilized in the non-polar solvent. As a result

the two ions (one positive, one negative) will tend to stay close together  (tight ion pair) and stabilize each other
and further due to this lack of stabilization they will have a short lifetime leading to quick collapse to products.

Both of these factors (chlorines located close to one another, short lifetime doesn't allow much movement of the second chlorine) favor syn collapse - addition of the second chlorine from the same side that the first chlorine is on.

If we were to repeat the reaction in a polar solvent that would allow the ions to separate and have a longer lifetime, more trans product from anti addition would be expected.


Charge-transfer absorption complex - solvent sensitivity This is a quote from Anslyn's Physical Organic Chemistry:

  [...] One way to think of these systems is that the ground state is a weak,
  intermolecular complex, (D • A). Often such molecules are polar and /
  or quite polarizable, facilitating complex formation. Because one
  partner is intrinsically a good electron donor (low ionization
  potential) and the other is a good acceptor (high electron affinity),
  the excited state will have a large contribution from states such as
  (D+• A-) that involve electron transfer, a configuration not possible
  for either isolated molecule. This preferentially stabilizes the
  excited state, leading to a lower energy (longer wavelength)
  absorption.  Because of this highly polar character, charge-transfer
  absorptions are extremely sensitive to solvent polarity, moving to
  longer wavelengths as the solvent polarity increases.

Earlier I read about Franck-Condon principle but I lack QM-mathematical background behind, so that's why this principle still leave me at unease, and hence my question. How is it possible that absorption wavelength is so sensitive to solvent polarity, when significant charge develops not until the complex is excited? I mean how the excited state of the molecule "knows" there is solvent ready to solvate it, since according to Frank-Condon principle solvent rearrangement is much slower than absorption.

In a charge transfer DA complex the lowest excited state is the charge separated species $D^+ A^-$. The 'normal' lowest excited singlet states of D and A are each at higher energy. As the molecules D and A come together the interaction between them, although small, (the absorption spectrum is only marginally altered) is nevertheless enough to cause electron transfer upon photon absorption.

The dipolar excited state possesses its own electrostatic self-energy E, as does any dipole. This is the sum of the Born charges $\pm q$ at infinity less the Coulomb energy of bringing the charges together to from the dipole. This works out as $E=  q^2/(8\pi\epsilon_0 \epsilon a)$ where $\epsilon_0$ is the permitivitty of free space and $\epsilon$ the dielectric constant of the solvent. The charges have a combined radius of $2a$. The equation can be re-written as $E=\mu^2/(4\pi\epsilon_0\; \epsilon\; L^3)$ if the dipole is defined as $\mu=qL$. Thus you can understand why the energy is lowered, and hence CT transition red shifted in a polar solvent, e.g acetoninitrile $\epsilon =37$ vs hexane $\epsilon \approx 2$.

Notes: Electron transfer upon absorption is in contrast to photoexcited electron transfer where the excited state of the donor or acceptor, whichever is excited, undergoes a thermal reaction in the excited state to form $D^+ A^-$. In the CT complex the charge is transferred on absorption between one molecule and the other, rather than from one part of the molecule to another part as in a normal absorption process. The absorption process is so rapid that, there is plenty of time for this to occur whilst the D and A molecules are close to one another and in the correct relative orientation because molecular diffusion is a relatively slow process.


What citrate is a common anticoagulant? When I underwent plateletpheresis, something that the staff called "citrate" was added to my blood as an anticoagulant. Everything I can find about this product online refers to it as just "citrate". But from what little I know about chemical names, citrate is an ion, normally found not by itself but rather as magnesium citrate, potassium citrate, calcium citrate, vel sim.

What's the other half of this anticoagulant: which citrate is it?

Per the NIH:

  ANTICOAGULANT CITRATE DEXTROSE A...
  ...
  Citric Acid, anhydrous, USP      0.073 g
  Sodium Citrate, dihydrate, USP   0.220 g
  Dextrose, monohydrate, USP       0.245 g
  Water for Injection, USP         q.s.
  pH: 4.5 – 5.5

They used the citrate ion in the form of citric acid and sodium citrate. The citric acid is for chleating calcium while the sodium citrate does this and acts as a buffer to prevent the citric acid from raising your blood pH too much when injected as this would be painful and hazardous. Sodium is naturally in your blood and will not interfere with action potentials and is a abundant and cheap ion to use in manufacturing so it makes a logical counter ion for the citrate buffer.


How do we weigh one C atom? As far as I get it, we did weigh one carbon (12,6-isotope) atom. Then we said that electrons have nearly zero weight and protons weight nearly the same as neutrons. Hence we took the weight of that one atom and divided it by twelve and this way we got the mass of a proton/neutron:

$w_\mathrm {n,p} = 1.6726 \cdot 10^{-24}\ \mathrm g$

Because that isn't easy to calculate with, we introduce a number which multiplied with the weight of a proton/neutron gives us one gram:

$N_\mathrm A = w_\mathrm {n,p}^{-1} = 1\ \mathrm{mol}$

So far so good. Now hydrogen weights 1 g/mol and carbon weights 12 g/mol. Two questions to that:

How did we actually measure the weight of one carbon atom in the first place?
Why is the molar mass not precisely 1 and 12?

This is a question worth debating and discussion. First of let us realize the fact that we are used to looking at relative atomic weights or masses (if someone wants to be too accurate). No Table lists the absolute atomic masses of atoms in the periodic table, i.e., through out the history we have to assumed either the weight of hydrogen atom is exactly 1.0000, or oxygen-16 had a weight of 16.0000 or lately the weight of carbon-12 atom is exactly 12.0000 units.  So let us forget the picture that someone actually made a measurement, took a single atom of hydrogen or carbon and weighed by some fancy means. These are all man made references and numbers. We could have taken the weight of C-12 as 100, and all the atomic masses would change proportionally.

  So far so good. Now hydrogen weights 1 g/mol and carbon weights 12 g/mol. Two questions to that: How did we actually measure the weight of one carbon atom in the first place?
  Why is the molar mass not precisely 1 and 12?

See above. You have to choose a reference. Fix hydrogen to 1 or carbon as 12 exactly but not both. Now you may ask how these numbers came up? You have to look at the question historically, when Avogadro's number was not used. If you understand German, have a look at Handbuch der Anorganischen Chemie by Abegg. There is long history how the number 12  got associated with carbon.

https://archive.org/details/a2handbuchderano03abeguoft/page/12;


Why do 1s and 2s orbitals intersect? So I was going through the Schrödinger wave equation to determine the shapes of orbitals. We know that an orbital is nothing more than a region where probability of finding an electron is maximum.
Further, we know that the probability of finding a 2s electron in 1s orbital is just too low! Why do 1s orbital and 2s orbital intersect? I mean both the orbitals have the same shape and the radial function of 2s orbital is more than that of 1s which clearly means that 1s orbital lies inside 2s orbital!
But as it's impossible to find a 2s orbital electron in 1s orbital, shouldn't the shape of 2s orbital be something like a hollow sphere with thick walls instead of a complete sphere?

I read your question a few times, and I think you're confusing occupancy with spatial probability distribution.
Orbitals have a lot of spatial overlap. But if the orbital has two electrons “in it,” then it cannot describe the probability behavior of any other electrons in that atom, as other electrons must be in another orbital.
A 2s electron absolutely can go into a 1s orbital provided there is space. This is exactly what happens with atomic emission spectra when an electron in energy level $n=2$ transitions to $n=1$ and releases energy.


What is the highest possible effective nuclear charge as a percentage of nuclear charge an electron in any neutral atom can experience? I was wondering about the following statement "The higher the principal quantum number of the orbital the electron is in, the lower the nuclear charge it experiences." I then argued that this statement is false because when an atom has more electrons, it also has more protons and as the orbitals become more diffuse, the attraction of the protons would significantly outweigh inter-electronic repulsion at higher atomic numbers.

This led me to think about the following question: What is the highest possible (nuclear charge an electron can experience divided by the nuclear charge of the atom)? My guess is that this electron would certainly belong to one of the more diffuse subshells. (i.e. d or f)

Please note that the quantity I am looking at is not effective nuclear charge but it is the effective nuclear charge divided by the nuclear charge.

A high principal quantum number does not necessarily mean more electrons and/or protons. For example Hydrogen can be in an electronic excited state with $n=4$ and higher, as can be observed for example in the Balmer series. So for a specific element your cited statement remains true.

But you want to compare between different elements now. What we are looking for is minimal screening of the nuclear charge by other electrons:

According to  Slater's rules the second $1s$ electron screens with $0.3$ (all other screening constants are larger). Therefore the ratio you are looking for is $\frac{Z_{eff}}{Z}=\frac{Z-0.3}{Z}$ and converges to $1$ for large atomic numbers $Z$.

The only way to get even less screening would be to have no further electrons. As you restricted the question to neutral atoms we are left with Hydrogen. The charge divided by the nuclear charge of the atom, an electron can experience here is $\frac{Z_{eff}}{Z}=1$ for any excited state (principle quantum number) of the Hydrogen atom.


Why are the absorption and emission spectra mirror images of each other? In one of my chemistry courses we learned that the absorption and emission spectra are mirror images of each other, but I can't rationalize why.

I know that the electronic transitions in emission spectra is not very different from that of absorption spectra, however the spectra is shifted slightly to the right due to the presence of non-radiative decays releasing some of the energy absorbed. I can see how similar energy transitions can account for a similar shape, but why is it a mirror image in most cases?

The sketch below explains how the mirror image spectrum comes about. (the figure is not at all to scale, its just illustrative, but does show the salient features.)

There are some conditions attached.

(a) mirror image spectrum is only observed in the condensed phase. This is because solvent collisions cause vibrational relaxation from the vibrational level excited and removes$^*$ energy to the solvent much faster than fluorescence can occur, thus all emission occurs from v=0. This vibrational relaxation occurs with a rate constant of $\approx 10^{13}$ s$^{-1}$.(This is not true in the vapour phase where if the pressure is low enough no collisions may occur within the excited state lifetime and fluorescence occurs from the level excited) The intensity of each transition (not wavelength) in absorption or emission is governed by the Frack-Condon Principle. The strength of a transition is proportional to the squared modulus of the two wavefunctions. i.e. $|\psi _{s0} \psi _{s1}  |^2$. In practice this means that overlap between vibrational levels in the two states is only large at the turning points of the wavefunctions, so transitions appear to be vertical. Classically this makes sense as momentum has to be conserved and this can only be achieved from vertical transitions starting and ending at the potential curve where kinetic energy is zero. In the quantum world the uncertainty principle limits change in position and momentum and so Franck -Condon Factors.

$^*$(The collisions preferentially remove energy as generally the excited state is at a higher 'temperature' than the surrounding solvent, i.e. it has more vibrational energy that the average $3k_BT/2$ of the solvent)

(b)The geometry of the excited state has to be similar to that of the ground state, thus mirror image spectra are observed for 'rigid'$^{**}$ molecules such as anthracene (3 benzene rings attached side by side) and not for flexible molecules such as cis or trans stilbene (phenyl=phenyl, where = is a double bond and this molecule isomerizes in the excited singlet state).

Relative displacement of potential energy surfaces (for the normal modes mode involved in the transitions) is not the only consideration, for the shape of the potential has to be similar in both states, thus the vibrational frequencies are similar and so gaps between vibrational peaks in absorption and fluorescence are similar. (Gaps between peaks in absorption give excited state vibrational frequencies and vice versa)

If the lowest triplet state has a similar potential to the singlet, which is usual if it derives from that singlet state (and not another higher singlet state) the the phosphorescence spectrum is very similar to the fluorescence, but shifter to longer wavelengths

$^{**}$ By 'rigid' is meant that the molecule has only its normal vibrational modes and no large torsional motion as is the case in stilbenes.


Perkin reaction for aliphatic aldehydes Why isn't Perkin reaction possible with aliphatic aldehydes? Since the carbon atom of carbonyl group of aliphatic aldehydes is more electrophilic than that of aromatic ones, the reaction must be more feasible in case of aliphatic compounds. Please explain.

In Perkins reaction, condensation occurs between aromatic aldehydes (which cannot undergo self condensation) with an acid anhydirde in presence of sodium or potassium salt of corresponding acid.

Consider Perkins reaction of benzaldehye with acetic anhydride to form cinnamic acid. The first two steps occur in weakly basic medium (sodium acetate). The last two occur in acidic medium (protonation and subsequent dehydration). Notice that the last step is particularly favoured owing to formation of a conjugated product.

This wouldn't be the case for aliphatic aldehydes (with no alpha Hydrogen , else self condensation may occur). Also, if the anhydride used has only one alpha Hydrogen (here acetic anhydride has 3 alpha Hydrogen), then the reaction barely proceeds to completion.Instead an aldol type product is obtained. This shows the role conjugation plays in making the last step fairly irreversible.

Furthermore, this reaction occurs with vinylogs, hetrocyclic aldehydes and even pthalic anhydrides. Notice that in all cases , final product is conjugated.

It would be worthwhile to mention another alternative mechanism, which has been mentioned in Peter Stykes, Wikipedia as well.

Some Support for this mechanism comes from fact that anhydrides with only 1 alpha hydrogen donot give this reaction. Again, here the base promoted dehydration step would be largely irreversible incase of aromatic aldehydes due to conjugation.

Hope this helps.

References:

Perkin, W. H. J. Chem. Soc. 1868, 21, 181
XXIII.—On the hydride of aceto-salicyl.DOI: 10.1039/js8682100181
Peter Stykes. Organic mechanism. Sixth Edition

3.https://en.wikipedia.org/wiki/Perkin_reaction

SN Sanyal: Reactions,Rearrangements and Reagents.

5.http://www.name-reaction.com/perkin-reaction


Why are oxides always of the form AyOx with A representing any atom except oxygen and fluorine I notice that for quite a few elements, oxygen comes before it in alphabetical order. Here is one of the rules in inorganic chemistry for molecular formulas:

Put all the elements in alphabetical order.

So why do oxides always have what is being oxidized followed by oxygen?

That makes some molecular formulas like PO4 for phosphate violate the rule in inorganic chemistry that the elements are arranged in alphabetical order.

I guess all oxides have some element(except for oxygen or fluorine) followed by oxygen because:

1) it clarifies that what you are talking about is an oxide of that element.

and

2) the structure of an oxide is always some element with oxygen atoms around it.

Generally the least electronegative element is written first. And while naming also, +ide is attached as prefix to the more electronegative element.
For Ex- $ICl$ is written as Iodine Chloride not Chlorine Iodide. Similarly $OF_2$ is written as Oxygen Difluoride.
Coming back to your question, Oxygen is the second most electronegative element in the periodic table after Fluorine. Therefore it is generally written after all the other elements (except fluorine).


How does NaCl (or any inorganic salt) increase surface tension? Does the compound "prefer" to stay in the solution?
If yes, wouldn't the surface tension of the solution be equal to that of pure water, since only water is in the surface?

See: Does NaCl reduce the surface tension of water?

Or is the compound distributed evenly and because the ionic bonds are stronger than the attractions between water molecules, the surface tension increases?

Read here a reddit answer to why do ionic compounds such as NaCl increase the surface tension of water?

Explain like I'm 5 please.

I hope this picture helps.

The interaction between a molecule of water and an ion is stronger than the hydrogen bonding that occurs between two water molecules.


Incorrect Ring Flip? Is the flip of the cis-isomer wrong? Specifically the image on the bottom right.

What I see is that on the cis-isomer in the first chair, we have an axial upward methyl group and an equatorial upward methyl group. When the ring flips we now have an equatorial upward methyl group (good) and a downward equatorial methyl group (bad). Shouldn't the second one be axial upward? From what I understand, ring flips shouldn't change the orientations of groups; just their axial/equatorial configurations.

Also, in general, would these two rings be equivalent?

Yes, the cis isomer on the right should have the methyl group on C2 in the axial (up) position.  The two rings you've drawn at the bottom of your post are equivalent, they can be interconverted by a 180 degree rotation about an axis that bisects the C1-C2 bond and the C4-C5 bond.


Why are real orbitals used to define the orbital of hydrogenic atoms instead of complex orbitals? On the physics site, I asked the difference between real & complex orbital. Real orbitals are the superposition of complex orbitals having definite magnetic quantum number states.

\begin{align}
p_z &= p_0 \\
p_x &= \frac{1}{\sqrt{2}} \left(p_1 + p_{-1} \right) \\
p_y &= \frac{1}{i\sqrt{2}} \left( p_1 - p_{-1} \right)\\
\end{align}
where $p_0 = R_{n1} Y_{10}$, $p_1 = R_{n1} Y_{11}$, and $p_{−1} = R_{n1} Y_{1−1}$, are the complex orbitals corresponding to $\ell = 1$.

My question is:

Why aren't the orbitals of hydrogenic atoms taken as complex orbitals instead of the real ones? They are valid states for an electron, aren't they? So, what is the problem in having complex atomic orbitals?

The main disadvantage of complex atomic orbitals (AOs) is that many of them are hard to visualize.  For example, try making a simple sketch explaining how $p_{-1}$ differs from $p_{1}$. Conversely the complex AOs have very few  advantages over real MOs - they are all eigenfunctions of $\hat{L}_z$ but that's not really important for molecular  and closed shell systems.

So with many key advantages and very few disadvantages real orbitals are understandably the most popular choice.


Why is the gradient of the curve of a strong base titrated with strong acid small up until equivalence? I understand the cause of the sharp decrease in pH upon reaching the equivalence point, but I don't understand why addition of relatively large amounts of the strong acid would be accompanied with small changes in pH before that.
When looking at how pH changes when adding a strong acid to water there is a very fast drop in pH, as to be expected due to the logarithmic nature of the pH scale. The pH scale (I believe) also explains the flat region at the start of the curve for titrating a strong acid with a strong base.

But I don't understand it for the titration in question. I presume it must be something to do with the neutralisation but I'm not sure.

Let's calculate the $p$H by changing the volume of $\ce{HCl}$ added. I will use the unit millimole ($1 mmol = 0.001 mol$), which is more practical than the mole.
In the very beginning the solution contains $50 mL~ \ce{NaOH}~ 0.1 M$, or $5 mmol$ $\ce{NaOH}$. Let's take, for example, the point when you have added $30  mL~ \ce{HCl} ~0.1 M$ (or $3.0 mmol$$ ˙\ce{HCl}$). After this addition, the solution contains $5 - 3.0 = 2.0 mmol$ NaOH in a volume $50 mL + 30 mL = 80 mL$. The concentration of $\ce{NaOH}$ is $\frac{2.0 mmol}{80 mL} = 0.025 M$. The concentration $\ce{[H+] = 10^{-14}/0.025 = 4 10^{-13}M}$. So $p$H = $12.4$. You see that the pH has fallen from $13.0$ to $12.4$ for more than half the trip to the equivalencce point. A change from $13.0$ to $12.4$ is rather small. The point of the curve is still nearly at its initial value. The gradient of pH between the beginning and here is small.
If you take the same calculation with an addition of $45 mL~ \ce{ NaOH}~ 0.1 M$, or $4.5 mmol$, the solution contains still $ 5 mmol - 4.5 mmol = 0.5 mmol ~\ce{NaOH}$ in $95 mL$. The concentration in $\ce{NaOH}$ is $\frac{0.5 mmol}{95 mL} = 5.26·10^{-3} M$. $\ce{[H^+] =  1,9·10^{-12}}$, and $p$H = $11.7$ We are near the equivalence point but the pH has not decreased a lot. It is still $11.7$. As you see the gradient (or the slope) of the curve is not important in the part of the curve between $0 mL$ and $45 mL$.
And suddenly between $45 mL$ and $50 mL$, the $p$H drops from $11.7$ to $7$. This is a huge change in just $5 mL$.
If you want you can calculate one or two other points along this titration. Try for example $49 mL$. You will be surprised ! It still gives a rather high $p$H value.


How to prevent petrol from being flammable? Is there any way to prevent petrol from being flammable without losing it's solvent properties? Preferably some affordable way that is applicable for DIY.
In my case, petrol is used for bearing cleaning, and there is a risk of inflammation due to accidental spark.

If no affordable way exists, what octane number is safer (less flammable)?

Nearly every solvent is flammable, some exceptions are halogenated hydrocarbons. But plain hydrocarbons are pretty much all flammable, though some are more dangerous than others like e.g. diethyl ether.

Chloroform is non-flammable, but due to the toxicity I would not recommend to use it. There are some other non-flammable solvents for cleaning mentioned in this presentation from NASA on the search for non-flammable solvents. But it seems that many candidates are being phased out because they are a hazard to the ozone layer.


Self tan mousse cannot contain UV, right? So, I bought this self tanner specificly to avoid UV radiation. Back home I noticed that the bottle says it contains ultra violet: "The inclusion of Ultra Violet fluorescence in the base colour" . I guess they meant it contains something fluorescent but what could it be and is it dangerous? (sorry about my bad English)

The product is Sunkissed Ultra Dark Selfie Ready Tan.

That's right, it can't contain UV, because UV light is not a thing you can put in a bottle. The rest I'd rather dismiss as some marketing gibberish akin to this:

(sold back in the days when radium was all the hype, this thing is probably less awful than it seems, because it could not realistically have contained more than a few atoms of radium.)

Then again, chances are that the gibberish bears some resemblance to actual facts; so what? Fluorescence is just the ability to absorb and re-emit light. Whatever its composition may be, this thing would not shine UV on you all by itself.

As to the possible dangers, you either trust the regulatory body acting in your jurisdiction, or you don't. At least these guys supposedly did test the product, which I didn't.


Effect of steric effect on acidic and basic strength I found a sentence given in my textbook:

More is steric effect more is acidic strength of carboxylic acids and
more is steric effect less is basic strength of aniline and other
amines.

Could someone please explain with an example?
I guess it might seem I am asking about ortho effect but I am not. I have gone through the post suggested but I am basically asking about subcases of ortho effect as in if we are asked to compare between ortho tertbutyl benzoic acid and ortho methyl benzoic acid then how would we go about it?  Or suppose the questions asks comparing the basic strength between orthotert butyl amine and ortho methyl amine how would we do it ?

Ortho tert butyl benzoic acid will be more acidic than the other one .because former is more bulkier so it will rotate the COOH group out of the benzene ring plane to a more extent .
For NH2 group Ortho to CH3 or C(CH3)3 .Ortho effect never applies NH2 can never go out of the plane .rather this time CH3 and C(CH3)3 goes out of plane .so hyperconjugation effect is lost only inductive effect is present .still tert one will be comparatively less basic because of STERIC INHIBITION OF PROTONATION (SIP) . Due to hindrance nitrogen can't donate it's electron pair easily to a proton


Lewis Structures - Peripheral Octets 1) Which Lewis structure is preferential of the two below? EN considerations suggest the first structure. Size/charge density suggests the second one; the second structure would likely be more stable because if we think of the halide series of binary acids - the bromide anion is more stable than the chloride anion - and hence hydrobromic acid is a stronger acid than hydrochloric acid - the sheer size of the bromide ion is able to effectively dilute its charge and overcome its lack in electronegativity.

2) I am told that peripheral atoms should have octets. Why is this so? What is the physical reasoning behind this? Does this have anything to do with minimizing the overall potential energy of the molecule - i.e. the peripheral atoms are perhaps more subject to attack than internal atoms (due to steric factors) - so molecules try to minimize such a possibility by completing the octet for peripheral atoms? On the other hand, the octet rule is more of a guideline rather than a rule.

3) What is the bond-angle of the Cl-Br-Cl bond? I am told that the bond angle is a perfect 180 degrees. What about the first structure? If we consider the first structure a resonance contributor, wouldn't that lead to a distorted/non-linear bond angle? Or would that still lead to a perfect 180 degree bond angle, because to me, it seems that all the lone pairs on the bromine can participate in resonance (if not all at once).

Quoting Greenwood & Earnshaw's Chemistry of the Elements, discussing the triatomic polyhalide anions:

  [I]nteratomic distances [...] are individually always substantially greater than for the corresponding diatomic interhalogen.

Unfortunately, the following data table does not contain an actual value for the bond length, though it does indicate that the two $\ce{Br-Cl}$ bonds are of equal length (which we probably could've safely assumed from symmetry considerations and the energetic degeneracy of the orbitals of the two chlorine atoms, if nothing else), and that the $\ce{Cl-Br-Cl}$ bond angle is (approximately) $180^{\circ}$. Bond length can often be correlated to both bond strength and bond multiplicity, especially when comparing compounds of the same atoms. Since the bonds in $\ce{BrCl2-}$ are longer than those in $\ce{BrCl}$, it's reasonable to infer that they are also weaker, and do not have greater multiplicity (i.e., there is no additional $\pi$-bonding). I would further reason that any $\pi$-orbital overlap would have to be very poor (judging by the mismatch in both size and energy of the $3p$ and $4p$ orbitals of chlorine and bromine, respectively), which makes $\pi$-bonding unlikely. (For the sake of completeness, I would note that I could be mistaken here, since $4p$ orbitals are somewhat smaller and lower in energy than expected, given the poor shielding of the nucleus by $3d$ electrons. That said, I don't think the magnitude of this effect is large enough to make a major difference here.)

A full quantitative molecular orbital theory treatment of the molecule would require computation, and I wasn't able to find many useful literature references. However, as the molecule is relatively simple, we can probably make reasonable progress with even a very naive analysis. On the valence level(s), we have a total of $22$ electrons in $12$ twelve atomic orbitals, all $s$ or $p$. These will combine to form twelve molecular orbitals, eleven of which are going to be doubly occupied. Note that for any bonding MO generated, an anti-bonding MO will also be generated. Hence, even if we had six bonding MOs (and, in reality, some would almost certainly be non-bonding), we would have five anti-bonding MOs occupied as well. This would yield an overall bond order of $1$, indicating a bond order of $\frac{1}{2}$ for each individual bond. By analogy to similar (but more common and hence extensively studied) polyhalogen anions (e.g., $\ce{I3-}$), this conclusion seems justified. $\ce{I3-}$, and numerous similar isoelectronic molecules, are often described using the three-center four-electron (3c4e) bond, and I think this model is probably applicable to $\ce{BrCl2-}$ as well.

The upshot of all this is that, while it's true that you can draw certain valid resonance contributors, they may not accurately reflect the structure and electron density of the actual molecule. The structure with the negative formal charge on the bromine is best, which agrees with your intuition regarding its size and polarizability being able to stabilize the charge. Due to the disparity in electronegativity, however, the bonds are nevertheless polarized towards the chlorine atoms, so the formal charge alone doesn't fully describe the electron density correctly.


Total oxidation of toluene free enthalpy I have trouble finding whether this reaction is spontaneous or not :

$$\ce{C7H8 + 9O2 -> 7CO2 + 4 H2O}$$

I consider (from the CRC Handbook of Physics and Chemistry) :

And so I find :

$$\Delta_r H° = 7\cdot (-393.5)+4\cdot(-241.5) - 9\cdot0 - 1\cdot 50.5 = -3670 \;\text{kJ/mol}  $$

and

$$\Delta_rS°= 7\cdot213.8+4\cdot188.8-9\cdot205-1\cdot0= 406.8\;\text{J/mol/K}$$

So at T=298 K I have

$$\Delta_rG°=\Delta_rH° - T\Delta_rS° = -3670\cdot 10^3-298\cdot406.8= -3.8\; \text{MJ/mol}$$

This would mean the reaction is very spontaneous, but according to experimental data, it should not. (The reaction only occurs when heating and with a catalyst).

What did I get wrong ?

Your problem is that you are trying to calculate a kinetic property (spontaneous versus nonspontaneous; on a compound level that’s inert versus reactive) with thermodynamic data (exergonic versus endergonic; on a compound level that’s stable versus unstable).

Thermodynamically, we should all be carbon dioxide and water, as should any other organic compound. However, the activation energy of combustion of organic compounds is rather large. Thus, while the reaction is thermodynamically favourable, it is not spontaneous in the absence of activation energy.


Stereo-isomerism in 7-methylbicyclo[2.2.1]hept-2-ene The above are two compounds.

The objective is to find if the compounds are identical/enantiomers/diastereomers

I can see a plane of symmetry if we keep the $\ce{H-C-CH3}$ in the plane. So, concluding it as identical looks great.
but, if so, how can one transform the first compound to the second?

Applying, CIP rules, we might invalidate the P.O.S. The solution given for this question is "identical".
Update:(attempt)
I have somehow manipulated the bonds such that I can prove the identity. Even though it is extremely-sterically hindered as indicated in the comments section (Now moved to chat). The forceful manipulations are done keeping in mind, not to break any bonds and only .

However this again proves that the molecules are not identical.
Now, since it is proposed that these molecules are different, can I get a lab or computer synthesized version of both molecules? Upon many search attempts, I could get this document which discusses about the same cyclohexene ring but Deuterium instead of the $\ce{CH3}$ at the bridge head and substituted oxygen ions.

7-Methylbicyclo[2.2.1]hex-2-ene exists as two diastereomers, anti-1 and syn-1, which differ by the relationship of the methyl group and the double bond. Both compounds are meso with the bridgehead carbons stereogenic and chirotopic (R/S) while C7 is stereogenic but achirotopic (r/s). The plane of symmetry  in each one does not produce an enantiomer of itself. Stereoisomers have the same molecular formula and the same atom connectivity. You probably know that the stereoisomers (R,R)- and (S,S)-tartaric acid are enantiomers and bear a diastereomeric relationship with (R,S)-tartaric acid, which is a meso-compound. A pair of meso-compounds with the same molecular formula and the same atom connectivity may also be diastereomers. Just like structures 1, the central carbons (C3) of ribitol and xylitol are stereogenic and achirotopic. The properties of these two pentaols are different as are the properties of anti-1 and syn-1.


How will the equilibrium partial pressure of carbon dioxide be affected? \begin{equation}\ce{CaCO3(s) -> CaO(s) + CO2(g)}\quad\Delta H^\circ = +556\ \mathrm{kJ/mol} \tag{endothermic}\end{equation}

How will the equilibrium partial pressure of  $\ce{CO2}$ be affected if the volume is decreased?

So, I thought of course the moles will increase when you decrease the volume hence $pV = nRT$, thus $p = \frac{nRT}{V}$. Therefore, $p$ is inversely proportional to $V$. But, the right answer is NO CHANGE, can anyone help me to understand why?

The equilibrium constant for the reaction is

$$K=\frac{P_{\ce{CO2}}}{P^\circ}=e^{-\Delta G^\circ/RT}$$

Changing the volume changes neither $\Delta G^\circ$ or $T$ (at least this is implicitly implied), so $K$ and, hence, $P_{\ce{CO2}}$ does not change.

For an ideal gas

$$P_{\ce{CO2}}=\frac{n_{\ce{CO2}}RT}{V}$$

So if $V$ decreases then $n_{\ce{CO2}}$ must also decrease in order for $K$ to remain constant. So, when the equilibrium is perturbed by the volume-decrease the equilibrium shifts to the left to re-establish equilibrium, in accord with Le Chatelier's principle.


Constant Pressure and Temperature during Phase Change One of the solution guides to a question I was working on said that pressure and temperature is constant for a phase change. I understand why temperature is always constant for a phase change, but don't understand why pressure is. If the heat coming in and out of the system and the volume is changing, shouldn't the pressure change too, or does the solution approximate that pressure is constant because it changes by a marginal amount?
The question is:

1.00 mol of steam is reversibly compressed to form water(l) at
100 degrees celcius ($\Delta H_\text{vap}(\text{water}) = \pu{2.259 kJ/g}$). Calculate the entropy of the phase change.

I attached the solutions on the answer key:

Constant pressure and isothermal. Only a change of state occurs. As steam condenses, heat is lost from the system.
$\begin{align}&\Delta H_{\text{condensation}} = \pu{-2.259 kJ/g}=q_{\text{rev}}\\
&\Delta S_{\text{condensation}} =q_{\text{rev}}/T = q_p/T = \Delta H_{\text{condensation}}/T_{\text{condensation}} \\ &m_{\text{steam}}=\pu{1.00 mol}\times \pu{18.0 g/mol} = \pu{18.0 g} \\ &\Delta H_{\text{condensation}} = q_p=(\pu{-2259 J/g})(\pu{18.0 g})=\pu{-40700 J}\\&\Delta S_{\text{condensation}} = q_p/T=\pu{-40700 J}/\pu{373 K}=\pu{-109 J/K}\end{align}$

For a pure substance, the phase rule implies that at coexistence of two phases only one intensive variable (pressure or temperature) can be altered independently, the other being dependent on the choice of the independent variable. So the answer is yes and no, you can choose to alter one of the two intensive variables and still retain two-phase equilibrium, but the choice of one intensive variable determines the value of the other. This fundamental principle derives from the conditions of thermodynamic equilibrium necessary for coexistence of the two phases. The set of $(p,T)$ points on the coexistence line can be described in terms of the Clapeyron equation $dp/dT = \Delta S/\Delta V$ and a single reference point $(p_{ref},T_{ref})$ on the line.

Consider a pure substance in a diathermal container whose volume we can alter at will, and kept at a fixed temperature. Assume we begin with the container completely filled with a condensed phase (say liquid) of the substance, that is, the volume is too small (or alternately, the pressure too high) to allow vapor to form. We have however chosen a temperature that corresponds to a point on the coexistence line at a lower pressure (or greater volume) than that currently in the vessel. You can bring the substance to an equilibrium between two phases by changing the volume until the pressure in the container equals the equilibrium vapour pressure - a fixed pressure at that temperature. You can then continue to change the volume (all the while keeping the temperature constant). The proportion of the two phases (their volumes) will readjust to accommodate the new total volume, with heat exchanging with the surroundings to keep the temperature constant, and the pressure returning to the value at equilibrium - the same vapour pressure - provided the volume change is not too big. Naturally if you increase or decrease the volume too much you will end up with a single phase and a different pressure below or above the vapour pressure.

Remember: you don't have to have two phases. But if you want to observe coexistence at a chosen temperature, you have no choice wrt the pressure that will ensure coexistence. What's cool is that the proportions of the two phases can be altered within a certain limited range of total volume so that p and T remain constant. I recommend consulting a p-V-T phase diagram and inspecting it carefully to see this graphically. The following is a slice in the p-V dimensions which I borrowed from this site:


Effect of His-tag on enzyme activity In my biochemistry laboratory class, I designed an experiment to study the effect of a His-tag on enzyme activity. First, I measured the activity of the His-tagged enzyme. Then, I cut off the His-tag (using enterokinase) and measured the activity again.
After data analysis, I was able to calculate the $K_\mathrm{M}$ values and $k_\mathrm{cat}$ values for both enzymes. How should I interpret the data? I hypothesize that the His-tag could interfere with kinetics. But, how do I use the data to show this? Should I compare $K_\mathrm{M}$ values or $k_\mathrm{cat}$ values?

I assume the excised-off His-tag as well as the enterokinase protease are removed from the system.
I would use the catalytic efficiency, i.e. the ratio of kcat/KM as the metric to compare the results of two conditions.


Is there a food-safe hydrophobic liquid with a substanially higher density than water? Requirements

In short, I am looking for a fluid that meets these criteria:

Hydrophobic
Denser than water
Low viscosity
Food safe

Background

We are working on adding a pool and designing a backyard recreational area.  One design element that we would like to include in multiple places is waterfalls of varying sizes and styles; I love flowing water.  As an extension of that I started wondering if it would be possible to create an underwater waterfall.

Initially I was thinking this could be accomplished by creating a regular waterfall in the pool with normal water, which could then be capped with a clear glass or acrylic watertight cover.  This certainly works as an underwater waterfall, but the cover will be visible, and will likely cloud over time, not to mention the condensation issues.

So I'm wandering now if some sort of fluid exists that could actually flow underwater.  If it's dense enough it will flow down a rockface even under the water, and if hydrophobic will not mix with the surrounding water.  As it will be within a pool that humans will be in for extended periods it must be externally safe, and I'd like something food-safe as kids swallow pool water and it's possible this stuff might get kicked up.  And to add to that tall order, the wrong viscosity will ruin the visual effect.

Does such a fluid exist in a form that a normal person can purchase for a reasonable cost?  I live in the US (Pennsylvania).  Another question, Is there a readily available liquid that's more dense than water and insoluble?, seemed to be looking for something similar but the answer doesn't meet my safety requirements.

Hello I know I went late to the party, but I have a solution to this. You all already have the right idea, but you need to put your ideas together.

Double layer your pool

Whether it's gallium, water, oil, or some other liquid it will all work and be safe under glass.

Their are hydrophobic resins that you can treat the glass with, and if you double paned the area that you're waterfall would be under, it would not cause a health problem because there would be no contact.

For example if you had made window panes in your pool (let's just say in a t-shaped stretching halfway under your pool), and you coated the window panes with a hydrophobic resin, and you kept it under let's just say a temperature controlled environment, that would one solve your pool problem of being boring and two, keep your waterfall from fogging up.

So in short
1 Removable hydrophobic panels for cleaning
2 Heated/ chilled pool or temperature controlled environment
3 Heated / cooling pump for waterfall
4 Well designed lighting ( you'd be surprised what a little bit of light in the right spot can do)

I do not know if it's available to the public yet but there is a coolant used for computers that is one consumable and two low maintenance called Coriant. It has a lighter density than water, if you're  panes are hydrophobate this would not present a problem.

Colored mineral oil or antifreeze would also work here.

If you use antifreeze and we're creative with its use, it may assist in some other mechanical uses as well as being a display piece.

Please note that any hydrophobic material or substance is going to be toxic to breathe or have extended contact with.

This should not be a problem if it is under a air / water tight encasing.

I apologize that I'm so late in responding to this but I hope someone reads this and finds it useful thank you for your time.


In water, will sucrose be hydrolysed by water to form glucose and fructose molecules? If I dissolve sucrose sugar ($\ce{C_12H_22O_11}$) in water, would the water readily hydrolyse the sucrose sugar to form glucose and fructose sugars?

Not without a catalyst present. The process is usually performed using invertase, but some other enzymes with similar activity exist and addition of acid also can catalyse the reaction.


Why are highly branched alcohols solid? Usually branching (or decrease in surface area) leads to increased volatility. But in case of alcohols, my book states that:

  The higher alcohols (butanols to decanols) are somewhat viscous, and some of the highly branched isomers are solid at room temperature.

Here the decerease in surface area did not increase the volatility; on the contrary, volatility has considerably reduced.

What is the reason of this anomaly? Or is this even an anomaly?

Well, if you're talking about melting points, the simple argument that more compact molecules should have higher mp because of increased dispersion forces can easily fail.  The problem is that melting is a much more complicated phase transition than boiling, with significant entropic effects coming from considerations of packing and entanglement -- especially relevant for floppy large molecules -- so that arguments based purely on the enthalpic changes due to intermolecular forces can easily be insufficient.


does aluminum oxide float on molten aluminum and by what mechanism? Not really a chemistry question since no reactions happening, but couldn't find a more appropriate stack exchange site.

Considering recycling aluminum cans for scrap metal using induction heating. I've seen people do it with very unsophisticated equipment, but what I don't get is Al2O3 is denser (3.95g/mL) than molten aluminum (2.38 g/mL).

I suppose surface tension can stop any oxide that forms at the surface of the melt from sinking, but how strong is that? Is it resistant to mild stirring (e.g. when you skim off the dross)? Are there other examples of denser materials floating?

A chunk of $\ce{Al2O3}$ would sink in molten aluminum. However, $\ce{Al2O3}$ usually appears in smaller pieces (down to microns), with irregular surfaces, probably filled with air ($\ce{N2}$) from when it was cool. A mix of metallic aluminum and $\ce{Al2O3}$ could allow aluminum to sink thru the solid particles after melting, allowing the particles to rise above the surface of the liquid. $\ce{Al2O3}$ has a very high melting point.

The mass of solid particles would have some very weak structure, and some buoyancy by trapped air. The surface tension of aluminum is very high (~900 dyne/cm) , so it might very well account for the covering of the metal by a heavier substance and the non-wetting of the oxide.

This is reminiscent of floating a needle on water.

In the electrolytic production of aluminum from $\ce{Al2O3}$:

  The electrolyte is a molten bath of cryolite ($\ce{Na3AlF6}$) and
  dissolved $\ce{Al2O3}$. Cryolite is a good solvent for $\ce{Al2O3}$
  with low melting point, satisfactory viscosity, low vapour pressure,
  and density lower than that of liquid aluminum (2 vs 2.3 g/cm3), which
  allows natural separation of the product from the salt at the bottom
  of the cell.

References

https://link.springer.com/article/10.1007%2Fs11663-999-0108-4
https://en.wikipedia.org/wiki/Aluminium_smelting


Boiling and Exploding Beaker? A few days ago, I was creating a solution which involved boiling water. I had two 4L beakers on identical hot plates. After around 10 minutes, the first one boiled. Waiting another 30 minutes, the second one had still not started boiling, although small bubbles had formed at the bottom. Shortly afterwards, I heard a ping and the second beaker cracked and leaked water.

Is there any possible correlation between the long boiling time and the beaker breakage? Both beakers should have been heat-safe, although they were from different brands. I'm not convinced of the cleanliness of the beakers, so if this could've been caused by a contaminant, that's a definite possibility.

I'm inclined to suspect that your identical hotplates were less identical than you think.  They can have substantial differences in heating rate as well as the actual temperature they reach, particularly if they have been around for a while.  You could try boiling another pair of similar beakers and check the time they take to boil to confirm.

Usually lab glass breaks because of thermal shock - either hot glass with cold water or cold glass with hot water.  However, if you heated the beaker steadily for 40 minutes, it definitely wasn't shocked into breaking.  An exception might be if the hotplate is malfunctioning and get extremely hot extremely quickly, but that doesn't seem likely.  It is definitely possible that there was already a small crack in the beaker and that heating caused it to open further.

A contaminant is very unlikely to have been the source of the problem.  You should be able to boil dirty beakers all day long without them popping on you.

Unfortunately it's pretty much impossible to tell what happened for sure unless it winds up being reproducible.


How to determine the volume of a substance produced when only given the equation at STP? At standard temperature and pressure, what volume in $\pu{mL}$ of $\ce{H2O}$ is produced in the following reaction?
  $$\ce{2 KOH(l)  +  H3PO4(l)  ->  K2HPO4(l)  + 2 H2O(l)}$$

Most questions I've found have more information than this so I'm confused as to how to do this problem.

I've tried multiplying $\pu{2 mol}$ of $\ce{H2O}$ by $\pu{22.4 L}$ and then converting it to $\pu{mL}$, but I've a feeling that isn't right.

I think you erred in multiplying 2 moles of water by 22.4 L. There are 55.5 moles of water per liter of water. Keeping this relationship in mind, use dimensional analysis to find the moles of water produced in your reaction.

You're confusing the number 22.4 with the fact that 1 mol of gas at STP occupies approximately 22.4 L.


How much heat (in J) is required to produce 2.0L of N2(g) at 77.36 K and 1.0 atm? The only thing that is given is:

The enthalpy of vaporization for $N_2(l)$ is 5.56 kJ/mol.

There aren't any examples similar to this problem in the textbook. I only know how to get energy using:

$q=mcΔT$ and $w=PΔV$

No clue what to do with the enthalpy of vaporization value.

UPDATE: Solved it by first getting number of moles ($n$) via $PV=nRT$ then multiplying moles by 5.56 kJ/mol. Moles cancel out, leaving only kJ, 1.75kJ to be precise.

Since nitrogen's boiling point is 77.36 K , it appears that the question is asking how much heat is required to boil nitrogen to produce 2.0 L at 1 atm.


Ionisation Energy vs First Ionisation Energy It was stated in a textbook that the first ionisation energy is

  the minimum energy needed to remove one mole of electrons from one mole of gaseous atoms in their ground state

Is there a difference between ionisation energy and first ionisation energy?

There is no difference between "ionization energy" and "first ionization energy". Another definition from Wikipedia:

  The ionization energy (IE) is qualitatively defined as the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation.

Here it's more clear that they are the same, since "the most loosely bound electron" is a single electron, and each successive ionization only removes a single electron at a time.

Additionally, if one were to remove all electrons from an atom, it would be explicitly stated as "complete ionization".


Why is fluorine a gas, but lithium isn't? $\ce{F2}$ is a gas, but pure lithium is a solid metal. Why is this? Lithium is much lighter, just 3 protons and 3.9 neutrons (average). Fluorine is much heaver, 9 protons and 10 neutrons.

A lithium atom has one valence electron, easily lost (shared), so it is connected to other atoms by a metallic bond. This is a  bit like the shell game where a pea (electron) is hidden under a walnut shell... the uncertainty of where it is at any instant implies, in a quantum sense, that all the atoms share it, and are held together. This bond is so strong that the next element, beryllium, has a boiling point of $\pu{3243 K}$.

Fluorine, on the other hand, has an almost complete shell and each atom forms a tight covalent bond with just one other fluorine atom. These molecules of $\ce{F2}$ keep their electrons to themselves and do not associate with each other.

The difference in bonding is like the difference between bricks mortared together in pairs, or mortared together in a three-dimensional structure, and is responsible for the difference in mp and bp.


Determining stoichiometry in ICP-OES data analysis I have questions regarding data analysis for an inductively coupled plasma (ICP) experiment from a paper by Wang et al. [1].

The generic formula for the chemical compound is $\ce{Na_{2−x}Fe[Fe(CN)6]}$.

In the paper, it was mentioned that chemical formula and ICP data are

\begin{array}{lrr}
  \text{Formula} & \ce{Na} / \mathrm{ppm} & \ce{Fe} / \mathrm{ppm} \\\hline
  \ce{Na_{1.53}Fe[Fe(CN)6].{4.2}H2O} & 112300 & 357200 \\
  \ce{Na_{1.67}Fe[Fe(CN)6].{3.9}H2O} & 116100 & 339400 \\
  \ce{Na_{1.73}Fe[Fe(CN)6].{3.8}H2O} & 118400 & 332400 \\
  \ce{Na_{1.68}Fe[Fe(CN)6].{3.9}H2O} & 112000 & 332700 \\
\end{array}

My question is how they determined the $\ce{Na}$ number ($1.53$, $1.67$, and so on) on each cases based from the $\mathrm{ppm}$? I tried to make a $\mathrm{ppm}$ ratio of $1:2$ for $\ce{Na}$ and $\ce{Fe}$ however the number does not match with the paper.

Reference

Wang, W.; Gang, Y.; Hu, Z.; Yan, Z.; Li, W.; Li, Y.; Gu, Q.-F.; Wang, Z.; Chou, S.-L.; Liu, H.-K.; Dou, S.-X. Reversible Structural Evolution of Sodium-Rich Rhombohedral Prussian Blue for Sodium-Ion Batteries. Nature Communications 2020, 11 (1), 980. DOI: 10.1038/s41467-020-14444-4

Let R be the ratio of number of Na atoms to number of Fe atoms. Then, for the first compound, we have

$$\pu{R = }\frac{\pu{112300 ppm } /\ \pu{22.98977 g mol^{-1} } }{\pu{357200 ppm } / \ \pu{55.845 g mol^{-1} } } = \pu{0.7637  }$$

Since there are two Fe atoms in the compound, the Na number for the first compound is simply 1.527, i.e., twice the value of R calculated above.

Following the same procedure for the remaining three compounds, I get Na numbers of 1.662, 1.731 and 1.635. The last one is furthest from the value of 1.68 the authors published, but maybe 1.63 was read as 1.68.


Decide major product for first order elimination reaction This is my own created question.

The favourable thing for (c) to be a major product is  conjugation of double bonds and highly acidic nature of hydrogen (resulting in elimination of hydrogen at a higher rate) adjacent to carbonyl group (credit $-$ Rover). According to information I have the above reaction is most probably reversible in nature.
(b) should be major product as per mechanism I have given. So which will be the major (or only) product. You may take high temperature conditions if temperature is required as we are doing elimination reaction.

The product that you have drawn is very unlikely to be the major product of a mixed aldol condensation. The actual major product in said reaction would be (assuming I get the substrates right) would be as follows:

However, your question is a valid one if we take this out of consideration. Now, If we consider just the elimination reaction and your reaction conditions, your mechanism is correct. However, if this reaction were to take place via E2, the rate determining step would be that of the removal of the hydrogen and this would mean that the product formed would be (c). In case of E1, (b) would be the major product since the rate determining step is that of the formation of the carbocation.


Isolating salt from sea water using filters Could a filter be used to isolate salt from sea water? Could we say that the water molecules are smaller than the molecules of salt dissolved in the water. I have no relation to chemistry or physics, so pardon the question if it's too naive.

Yes, this is possible and is a process called reverse osmosis. It is based on a membrane which lets water pass but not the salt ions (and other water contaminants). From Wikipedia:

  The membranes used for reverse osmosis have a dense layer in the polymer matrix -- either the skin of an asymmetric membrane or an interfacially polymerized layer within a thin-film-composite membrane -- where the separation occurs. In most cases, the membrane is designed to allow only water to pass through this dense layer, while preventing the passage of solutes (such as salt ions). This process requires that a high pressure be exerted on the high concentration side of the membrane, usually 2–17 bar (30–250 psi) for fresh and brackish water, and 40–82 bar (600–1200 psi) for seawater

Water desalination features prominently among applications of reverse osmosis:

  The largest and most important application of reverse osmosis is to the separation of pure water from seawater and brackish waters; seawater or brackish water is pressurized against one surface of the membrane, causing transport of salt-depleted water across the membrane and emergence of potable drinking water from the low-pressure side.


Acid-Base Titration of ammonium dihydrogenphosphate? How can I determine the concentration of ammonium dihydrogenphosphate solution?
In this compound, the ammonium cation and the dihydrogen anion are both weak acids.
I have a weak solution of this salt, $C_A\approx 0.05\mathbf{M}$, that I need to quantitatively determine.
My first thought was acid-base titration because it's cheap and easy but the presence of two weak acids complicates things.
Would it be possible to determine the concentration of ammonium and dihydrogen in one equivalence point, as $2\times C_A$?

The pH can be measured vs. the amount of $0.1$ M NaOH added to your solution with a pHmeter. Then two titrations would be visible : first the titration of dihydrogenphosphate ($p\ce{K_a = 7.21}$) and then the titration of ammonium ($p\ce{K_a = 9.22}$). If you don't have a pHmeter, you would probably use a colored indicator. But the end of the first titration will not be determined with precision, as the two $p\ce{K_a}$'s are not different enough. You will only be able to determine the end of the titration of the two acids together. To do this, a rather basic indicator is needed, like Alizarine yellow whose $p\ce{K_a = 10.5}$.


In which of the following compounds does Cl have the most positive partial charge? The problem is :

In which of the following compounds does Cl have the most positive partial charge?

SCl2 ,, HCl  ,, BrCl  ,, OCl2 ,,

before I go through how I will solve this problem .. There is a question : Can Cl have a partial positive charge?

Well, all of them are polar, so there is a partial charge on each one of them

but how to compare or specify the partial charges?

The only thing I can think about is electronegativity

is it like .. the greater the difference in electronegativity the more the partial charges ?

Or .. ?

The greater the difference in electronegativity the more the partial charges is true. Compare the electronegativities of the atoms bonded with chlorine and you can easily determine in which compound chlorine has the biggest partial charge. Chlorine can have a partial positive charge when it's bonded with a more electronegative element like oxygen or fluorine.


Mechanism of reduction by elemental sodium How exactly does sodium form an alkoxide from an alcohol? I understand the general idea - sodium has a low first ionization energy. Sodium can thus give up its electron to something that'll take it. Sodium can therefore somehow cause hydrogen gas to leave.

How exactly does this mechanism work? I'm imagining that because alcohols are weak acids, that some hydrated protons exist in solution, and that sodium can donate an electron to these hydrated protons, which then can pair up with each other and form hydrogen gas. Is this accurate?

When alcohol molecules adsorb on the surface of the sodium metal, an electron can be transferred from sodium to the alcohol O-H bond with simultaneous disproportionation  (homolytic cleavage) of the O-H bond thereby generating the alkoxide anion and a hydrogen radical.  The hydrogen radical (probably still on or near the metal surface) can then go on to dimerize and produce hydrogen gas.

EDIT: Support of Mechanism

The mechanism proposed above involves a single electron transfer (SET) from sodium.  This is based on the fact that:

All dissolving (sodium) metal reactions, such as the Birch reduction of alkynes and aromatics, proceed by the SET mechanism
Other reactions involving sodium metal such as the acyloin condensation, the pinacol coupling, the Bouveault–Blanc reduction, etc., proceed through the SET mechanism
The reaction of sodium and water still proceeds in the gas phase where ionization seems less likely, but electron transfer still viable.

Searching on-line I found that water can solvate electrons just like ammonia can.  The chief difference being one of stability; in ammonia solvated electrons are stable for days, whereas in water they are stable for less than 1 msec.  Given the preceding mechanistic evidence and the fact that solvated electrons can exist in water, it seems reasonable to suggest that the mechanism of alkoxide formation, when sodium is reacted with an alcohol, also involves the SET mechanism.


Phenyl or benzene when naming compounds These two molecules are both aromatic:

However, the first molecule has a halogen substituent and is named "bromobenzene". The second molecule has a chlorine and a hydroxyl group and is named "4-chlorophenol". Assuming there was no alcohol group, how would we know when to use benzene or phenyl?

I thought the following:

therefore, shouldnt the first molecule have been written as "bromophenyl" and not as "bromobenzene"?

You are confused about several overlapping but mostly unrelated issues.

First, you are right that "benzyl" is the $\ce{C7H8\bond{~}}$ radical and that "phenyl" is the $\ce{C6H5\bond{~}}$ radical.

Second, although "phenol" and "phenyl" sound almost the same, you would probably do well to think of them as entirely separate naming roots.  "Phenol" is the simplest aromatic alcohol.  Simple aromatic alcohols with other substitutions (e.g. 4-chlorophenol) can thus be named as substituted phenols.

Third, "benzyl" and "benzene" similarly sound the same but should be treated as distinct concepts.  Substituted aromatic rings can be named as benzene derivatives, such as chlorobenzene.

Fourth, for shorthand and convenience, "phenyl" and "benzyl" are sometimes used to name substituents.  Note that IUPAC has detailed priority rules for what is a substituent and what is the base molecule, but chemists don't always follow these.  "chlorobenzene" is a correct IUPAC name, and it views the molecule as a substituted benzene.  That is, benzene is the base molecule and "chloro" is a substitution.  Chemists sometimes find it convenient to reverse their perspective, even though the resulting names would not be IUPAC-approved.  From a reverse perspective, the chlorine atom is the "base molecule", and it is substituted by a phenyl group.  Thus you could call the chlorobenzene "phenyl chloride" if you wanted to take this reverse perspective.

Lastly, "bromophenyl" ends in "-yl" and so is the name of a "radical", i.e. a substituent or fragment, not of a complete molecule.  Without any other context, I would think that "bromophenyl" was a $\ce{C6H4Br\bond{~}}$ substituent.  In contrast, "bromobenzene" is a complete molecule, derived from benzene by substitution.


Thermal properties of aluminium alloy 6063 I am looking for the operating temperature range both in a cold and hot environments (minimum and maximum operating temperatures) for aluminium alloy 6063.
I'm new to finding this kind of information efficiently, and I would appreciate a reference document as well.

American Society for Metals has many publications. Temperatures below ambient are no problem for any aluminum alloys. They are commonly used for natural gas  separation liquefaction exchangers and piping down to −100 °C (guess at the temp.).
In the T6 condition (hardened) room temperature yield is 31000 psi; at 300 °F the short term yield is 20000 psi; 6500 psi at 400 °F, and falling fast as the temperature raises.
ASME Boiler Code will have allowable stresses which will be lower than these values. I looked in ASM Materials Handbook vol 1, 1961, it is probably not online. If this is long term, you will want creep or stress/rupture data which is not commonly available for aluminum; try in the aircraft industry.


Chemical kinetics with the reaction of tert-butyl bromide with azide ion The reaction of tert-butyl bromide with azide ion in aqueous solution
  is proposed to proceed by the following mechanism:

  $\ce{(CH3)3CBr(aq) <=>(CH3)3C+(aq) + Br–(aq)}$

  $\ce{(CH3)3C+(aq) + N3–(aq) \to (CH3)3CN3(aq)}$

  Assuming that $\ce{(CH3)3C+ (aq)}$ achieves a steady-state concentration, but making no further assumptions about the relative magnitudes of the three rate constants,
  what is the rate law for this reaction, where $k_1$ is the forward
  reaction rate of the first equation, $k_{-1}$ is the reverse reaction
  rate of the first equation, and $k_2$ is the forward reaction rate of
  the second equation?

  (A) $\ce{k_1[(CH3)3CBr]}$

  (B) $\ce{k_2[(CH3)3CBr][N3-]}$

  (C) $\ce{\displaystyle\frac{k_1k_2[(CH3)3CBr][N3-]}{k_{-1}[Br-]}}$

  (D) $\ce{\displaystyle\frac{k_1k_2[(CH3)3CBr][N3-]}{k_{-1}[Br-] + k2[N3-]}}$

My thoughts: First of all, I read that $\ce{(CH3)3C+ (aq)}$ is a steady state concentration, which meant that $\frac{d\ce{[(CH3)3C+ (aq)}]}{dt} = 0,$ which prompted me to write out differential equations for each reactant. Thus, I got a system of differential equations:
$$\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} =-\frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}]$$
$$0=\frac{d\ce{[(CH3)3C+ (aq)}]}{dt} = -\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} - \frac{d\ce{[(CH3)3CN3 (aq)}]}{dt} = \frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}] -k_2 \frac{d\ce{[(CH3)3CN3 (aq)]}}{dt}$$
$$\frac{d\ce{[Br- (aq)}]}{dt} = -\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} = \frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}]$$
$$\frac{d\ce{[N3- (aq)}]}{dt} = -k_2\ce{[N3-(aq)]}$$

But I was unsure of how to use these differential equations to actually find the rate law. Could anyone provide any suggestions?

You really don't have to write all that differential equations.

Note: I'll write all my steps on a paper, since MathJax formatting is a little time-taking at the moment.

Consider the rate laws for the two reactions:

Since the $\ce{[(CH3)3C+]}$ isn't changing with time, we get

From which, you get

Plugging these in the Rate2 which I mentioned, you will get,

Which is the same as what's given in option (D).

I hope I have satisfied your queries.


Why are carbocations with sp3 hybridisation highly unstable? Why is it that sp3 hybridized carbonium ions do not exist? Or why are they very difficult to form?

For instance, nucleophilic substitution in bicyclo compounds at the bridge head carbon is not very easy. I don't understand the explanation given in the book. Thanks for any help.

When you hybridize towards $sp^3$, you use one $p$ orbital more compared to the $sp^2$ hybridization.

The additional $p$ orbital raises the energy of the 4 $sp^3$ orbitals higher with respect to the energy level of $sp^2$-hybridized orbitals ($p$ orbitals are higher in energy than $s$ orbitals), which in the energy-minimizing world of chemistry is not favoured by the system.

The energy is minimized by hybridizing in a $sp^2$ fashion (in which all the $\sigma$-bonding electrons go) and adopting a planar geometry.


Why is the first ionization enthalpy of molecular O2 less than atomic O? A cousin of mine asked for my help to answer one of his homework questions. It stated-

Explain why- The first ionization energy of molecular oxygen $\ce{O2}$ ($\pu{1175 kJ mol-1}$), is lesser than the first ionization energy of $\ce{O}$ ($\pu{1314 kJ mol-1}$)?

It took me some time trying to find out perfect reasons for it, because generally, bonded electrons are harder to remove than valence electrons in atoms. But indeed, not just with $\ce{O2}$, but $\ce{F2}$ ($\pu{1515 kJ mol-1}$) also has a first ionization enthalpy lesser than atomic $\ce{F}$ which is $\pu{1681 kJ mol-1}$. (I wonder why there aren't any more talked of!). So I ended up taking a "class" on the basics of MOT, trying to reason that "the energy required to remove an HOMO electron from an ABMO in $\ce{O2}$ is lesser than removing an electron from a bonding orbital. That could probably lead to this anomaly."
But this wasn't convincing enough for me. I could sense a lack of conviction, a whole list of factors that could go wrong, also with the possibility of a much simpler explanation. But what made me post it up here, despite its dubious look, is the lack of convincing information addressing this issue even on the internet. Hence, a proper answer would be helpful for the community at large.

A cousin of mine asked me to share her answer here:

In the molecular orbital diagrams shown in textbooks, the highest occupied molecular orbitals of $\ce{O2}$ and of $\ce{F2}$ are shown as antibonding. If the molecules and their cations had the same bond length, and the electrons in a species didn't "talk to each other", and the orbital energies of an isolated atom were not influenced by the approach of another atom, you could predict that the ionization energy of the molecular species is lower than that of the atomic species. The argument would be that these HOMOs have higher energy than the corresponding atomic orbitals (see comments to question).
As these assumptions are not quite true, it is more complicated, but luckily the quick and dirty analysis yields the correct answer.
According to the MO diagrams, the bond order of $\ce{O2}$ is 2 and that of the cation $\ce{O2+}$ is 2.5 (after all, we are removing an antibonding electron). Indeed, the bond length of the cation is shown as smaller in this source. I did not find a bond length for the $\ce{F2+}$ species. This shows that you have to at least adjust the MO diagram for the change in bond length.
To get a rationale for the ionization energies in a quantitative manner, you could calculate the energies of the four relevant species ($\ce{O, O+, O2, O2+}$) and look at the differences.


How soluble is uranium in molten iron? As I know, uranium and iron simply aren't mixable, which means there are no U-Fe alloys. The Google says, uranium is insoluble in molten iron.

But, if something is "insoluble", it actually means only that there is a very low maximal concentration.

What if we want to mix only 1ppm of U in molten Fe? What if 0.01 ppm? What is the maximal concentration?

According to Uranium partitioning between liquid iron and silicate melt at
high pressures: implications for uranium solubility in
planetary cores

  The solubility of U in
  liquid Fe is in the range of 0.6 to 800 ppm and increases with temperature (T) and
  pressure (P).

See also U solubility in Earth’s core which finds a solubility of 2 ppm at zero pressure (run number 198).


Why are enthalpy of combustion and molar mass of organic compounds proportional? I noticed that when you graph heat of combustion against molar mass, you get a completely straight line with a correlation of 1. I know that the longer the chain is, the stronger the LDF get, so it is expected to see an increase in heat of combustion if molar mass increases. Though, I don't know why the curve is completely straight.
You can see heat of combustion against molar mass of alkanes in the graph below.

Data source: Organic Compounds: Physical and Thermochemical Data

The biggest contributor to the heat of combustion is the difference in covalent bond energy of reactants and products. The intermolecular forces that have to be overcome for combustion of liquids (with products all gases) are an order of magnitude lower (adapted from https://chemistry.stackexchange.com/a/125013):
$$
\begin{array}{lrr}
\hline
n\text{-Alkane} & T_\mathrm{b}/\pu{K} & \Delta_\mathrm{vap}H(T_\mathrm{b})/\pu{J mol-1}\\\hline
\text{Methane} & 112 & 8176\\
\text{Ethane} & 184 & 14640\\
\text{Propane} & 231 & 18832\\
n\text{-Butane} & 273 & 22390\\
n\text{-Pentane} & 309 & 26352\\
n\text{-Hexane} & 342 & 28850\\
n\text{-Heptane} & 372 & 31800\\
n\text{-Octane} & 399 & 33972\\
n\text{-Nonane} & 424 & 36910\\
n\text{-Decane} & 447 & 38750\\
\hline
\end{array}
$$

[OP] I know that the longer the chain is, the stronger the LDF get, so it is expected to see an increase in heat of combustion if molar mass increases.

Some of these alkanes are gases, so there are no dispersion forces to overcome. Also, you got the sign of the contribution wrong. If a reactant has higher LDF than the products (which have none), this would make the reaction less exothermic. So you would expect lower magnitudes for the larger alkanes, opposite of what is observed.
Instead, the heat of combustion scales with the number of carbons (see e.g. this table and graph) because the number of $\ce{C=O}$ and $\ce{O-H}$ bonds made (and the number of $\ce{C-C}$ and $\ce{C-H}$ bonds broken) also scales with the number of carbons.


Why is acetic acid more acidic than phenol? Acetic acid (ethanoic acid, $\mathrm pK_\mathrm a \approx 5$) is more acidic than phenol ($\mathrm pK_\mathrm a \approx 10$), which is reflected in their reactivity with a weak base such as sodium carbonate (adding acetic acid leads to effervescence of carbon dioxide).

However, if we look at the structures of acetate and phenoxide ions, the delocalization of negative charge should be greater in the phenoxide ion as the charge moves inside the benzene ring. On the other hand, in the acetate ion, the negative charge is delocalized between two oxygen atoms only.

How is it possible that acetic acid is still a stronger acid, even when its conjugate base has fewer resonance structures?

There are actually several factors that affect the stabilization of negative charge. The first is what kind of atom centers the charge can be delocalized on and the second factor is how many atom centers the charge can be delocalized on. The first factor is more important than the second. Just think it like quality is more important than quantity. Symmetry is another factor but not very important in your case.
Let's look at the ethanoic acid vs phenol case. In the acetate ion, the negative charge is delocalized on two oxygen atom centers, while in the phenoxide ion, the charge is delocalized on one oxygen and three carbon atom centers. Because oxygen is much more electronegative than carbon, the delocalization of negative charge over two oxygens is better than delocalization over one oxygen and three carbons. Quality beats quantity here.

If you have 2,4,6-trichlorophenol, because the strong election withdrawing chlorine atoms is helping carbon center to stabilize negative charge, it is a much stronger acid than phenol itself. This is an example of the inductive effect at work, as opposed to resonance.


To heat or not to heat; distillation? One of the questions in a recent past paper I completed was:

  One way to make water from the reservoir suitable for drinking is by distillation.

  Describe how water is distilled.

And I answered (perhaps too succinctly) that "The water is vaporised and the condensation is collected."

I would have thought my answer correct, in that it fully addresses the question, however, the mark scheme mentions two points I did not make:

I should have mentioned heat.
I should have mentioned cool (the mark scheme's words, not mine).

But I was wondering, are these two (common) aspects of distillation actually necessary to distill water?

Surely heating and cooling are not really necessary to distill water; couldn't pressure be reduced to evaporate the water, and increased again to condense it?

Or, in the context of the question, is the heating perhaps necessary to sterilise the water (for killing microbes)?

You obviously already have a general idea of what is going on in a distillation, so I will spare you the picture and everything.

Your question can be broken into two parts:

Do I need heat for vapourisation?
Do I need cooling for condensation?

I am going to address them one after the other.

Vapourisation

As you probably also know, reducing pressure is one means of evaporating liquids. This is used in rotavaps which use a $40~\mathrm{^\circ C}$ bath and reduce the pressure to make sure the solvent evaporates. You could think that just reducing the pressure should be enough to vapourise any liquid.

But if you vapourise a liquid only by reducing pressure, you will reduce the temperature of the liquid. This is the evaporation cooling that your body uses when sweating: If the water in sweat evaporates it cools the surface it evaporates from. Thus, after a certain time and especially if you attempt to distill large quantities of water, it will freeze and further evaporation will become exponentially harder. Therefore, a practical setup will always require heating.

Condensation

Effectively, this is the same discussion but in reverse. Note that if you boil water in a closed pot, the lid of the pot will get successively hotter due to condensing water releasing heat. After a certain time, the temperature of the surface you want to condense on will be so hot that the liquid will no longer condense.

This is especially a problem if you decided to reduce pressure for evaporation: If you have to reduce the pressure so far that the liquid would evaporate at $10~\mathrm{^\circ C}$, then there is no way that it will condense at a cooling trap of $25~\mathrm{^\circ C}$. Therefore, the condensation area must be cooled which is typically achieved with Liebig coolers or similar.

What about vapourisation followed by condensation in the same vessel with changing pressure?

Well, that is possible. It is not practical, though, since the inside of a single vessel can only have a certain pressure (within slight margins). And vapour spreads out uniformly. Your vessel would have to have an initial ‘bowl’ and a collecting one. You reduce pressure to evaporate out of the initial bowl and then re-raise pressure. The liquid will condense uniformly and you will lose a significant amount of it because it will drip back into the initial bowl. Your yield with a typical common setup is much better.


Chirality of spirocyclic compound I am facing trouble with the following problem.

Is (R)-1,7-dioxaspiro[5.5]undecane chiral?

My idea is that the compound possess a plane of symmetry and thus should be achiral. However if we take its mirror image then the bond pointing below of the plane will become above the plane and thus it will not form a nonsuperimposable mirror image indicating that it is chiral. Certainly one of the approach is incorrect. I couldn't understand which one.

The only place where a plane of symmetry may make sense would be vertically through the centre, through the spiro carbon. However, as the image by Tetrahydrocannabinol shows, these rings are not in the same plane (implied by the wedge and dash notation). Therefore, using that plane as a plane of symmetry would map each of the two rings into the void.

The only element of symmetry I can see in the compound is a $C_2$ axis (rotation by $180^\circ$). This axis would also bisect the spiro carbon, but it would map the oxygens onto each other. Having only a $C_2$ axis allows a compound to be chiral (point group $C_2$) and thus your compound is chiral.


How is the acidic dissociation of acetic acid thermodynamically feasible? For a reaction between an acid and a base

$$\ce{A1 + B1 <=> A2 + B2}$$

to be spontaneous $\ce{A1}$ must be a stronger acid than $\ce{A2}$.
With this in mind, how can this reaction  be understood in terms of spontaneity
    $$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
Also sources on the internet say that $\ce{pK_a}$ of $\ce{H3O+}$ is $-1.74$ while that of acetic acid is $4.74$.

To expand on my comment above, I will reiterate that all chemical reactions are equilibria and this is due to the interplay of enthalpy and entropy. More specifically, it is due to enthalpy and entropy of the individual compounds of the reaction as well as an additional entropy component due to mixing.

Consider the general reaction $$\ce{A<=>B}$$ and let $\chi_\ce{i}$ denote the mole fraction of compound $\ce{i}$ in the solution or gas mixture. The enthalpy and the entropy of the compounds, combined together to form $\Delta G_{compounds}=\Delta H_{compounds}-T\Delta S_{compounds}$ or alternatively $$G=(\chi_\ce{B})\cdot H_{\ce{B}}+(1-\chi_\ce{B})\cdot H_{\ce{A}}-(\chi_\ce{B})\cdot T\Delta S_{\ce{B}}-(1-\chi_\ce{B})\cdot T\Delta S_{\ce{A}}$$ has a particular side of the reaction it favors. This component of the free energy leads to a complete preference; if this were the only factor, the solution would be all $\ce{A}$$(\chi_\ce{A}=1)$ or all $\ce{B}(\chi_\ce{B}=1)$. If we were to plot $G$ as a function of $\chi_\ce{B}$, it would either be linear increasing (reactants favored), linear decreasing (products favored), or flat (reactants and products equally favored).

However, there is an additional entropy component due to mixing, which leads to an additional free energy component. We can write the entropy of mixing as $$\Delta S_{mixing}=-\mathrm{R}[\chi_\ce{A}\ln(\chi_\ce{A})+\chi_\ce{B}\ln(\chi_\ce{B})]=-\mathrm{R}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$
where we have used the fact that $\chi_\ce{B}+\chi_\ce{A}=1$. We can then write the free energy of mixing as $$\Delta G_{mixing}=\mathrm{RT}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$ We can see graphically what these two mixing equations would look like

Combining all this together, we can write $$G_{total}=(\chi_\ce{B})\cdot H_{\ce{B}}+(1-\chi_\ce{B})\cdot H_{\ce{A}}-(\chi_\ce{B})\cdot T\Delta S_{\ce{B}}-(1-\chi_\ce{B})\cdot T\Delta S_{\ce{A}}+\mathrm{RT}[(1-\chi_\ce{B})\ln(1-\chi_\ce{B})+(\chi_\ce{B})\ln(\chi_\ce{B})]$$ We can take the derivative of this function with respect to $\chi_\ce{B}$ and set it equal to zero to determine where the miniumum free energy is $$\frac{\delta G_{total}}{\delta\chi_\ce{B}}=0= H_{\ce{B}} - H_{\ce{A}}- T\Delta S_{\ce{B}}- T\Delta S_{\ce{A}}+\mathrm{RT}[-\ln(1-\chi_\ce{B})+\ln(\chi_\ce{B})]$$ $$H_{\ce{B}} - H_{\ce{A}}- T\Delta S_{\ce{B}}- T\Delta S_{\ce{A}}=-\mathrm{RT}[-\ln(1-\chi_\ce{B})+\ln(\chi_\ce{B})]$$

This second equation is the crux of all this. While we cannot solve this for an exact value without knowing the enthalpy and entropy of the compounds, we can definitively say that $\chi_\ce{B}\ne1$ and $\chi_\ce{B}\ne0$. Why? Plugging either of the values into the equation will result in the right side being $\infty$, meaning the differences in enthalpy and entropy of the components would also have to be $\infty$ for the equation to be satisfied. Thus, the minimum free energy will never be attained by letting the reaction go to completion; it will always be more favorable to allow at least some mixing even if one side of the reaction has a much lower enthalpy and higher entropy of its compounds.

Thus all reactions are technically equilibria. I say technically because the mixing free energy might be negligible compared to the overall free energy and so the reaction could appear to have gone to completion to the extent that we can observe experimentally.

So for your reaction, the reason it doesn't all remain as $\ce{CH3COOH}$ and water is that the entropy of mixing makes it favorable for some of the $\ce{CH3COOH}$ to dissociate.

(For a brief version of this argument and where I got the graph, see http://rkt.chem.ox.ac.uk/tutorials/equilibrium/entropy_mixGas.html)


Anti-bonding molecular orbitals The anti-bonding molecular orbitals are higher in energy than the constituent atoms in a diatomic molecule. That means it is less stable. Why does it form? Why can't all the molecular orbitals be bonding molecular orbitals?

Molecular orbitals form when you apply the "Linear Combination of atomic orbitals" (LCAO) model to atoms bonding. Where you take a linear combination of the relevant atomic orbitals to create an MO. In each case of combination a bonding and an anti bonding (ab) combination forms. The ab MO form as a result of applying the Huckel model which results in a square root in the expression for the energy, thus giving a negative and a positive contribution to the energy. One of the expressions E1 will be lower in energy and will form the bonding and the second E2 will form the higher energy MO - anti bonding.
Sometimes a non-bonding orbital forms, which doesn't contribute to the total bond energy, where the squared root expression equals zero.

To understand it more thoroughly one needs to understand the hydrogen atom quantum model and it's solution which give the orbitals we know as: s,p,d,f etc. In the case of MO we apply say that each atom "brings" it's set of occupied orbitals from it's valence electronic shell, combining these gives MO diagrams.
It's an approximation but the results it gives describe to some extent what can be measured in the lab.