Is the Lowdin Orthogonalization used in diagonalizing the atomic orbitals really a basis change? This is kind of more on the mathematics side of quantum chemistry, but I can't quite figure out why the Lowdin Orthogonalization is called a basis set change. I get how it works from the perspective of matrices, and that it creates an orthonormal set as long as you have linearly independent input, but I don't get why it works as a change of basis.

Typically to change the basis set of a matrix $M$ you use a non-singular linear transformation $A$, which need not be unitary, as such: $$M'=A^{-1}MA.$$ Quantum mechanics, we are dealing with Hilbert Spaces, which produce an isometry between the Hilbert space and the dual (I think?), and then for post HF methods most methods often use unitary transformations, so that $A^\dagger=A^{-1}$, and so it becomes trivial.

The Lowdin transformation between nonorthogonal AO $|\phi\rangle$ and orthogonal AO $|\phi\rangle_\perp$ involves the overlap matrix $S$ as:
$$ |\phi \rangle_\perp = S^{-\frac{1}{2}}|\phi\rangle  $$
Then, the corresponding bra just has the adjoint, which is also $S^{\frac{-1}{2}}$ because the matrix is self-adjoint: $$ \langle \phi|_\perp = \langle \phi|(S^{-\frac{1}{2}})^\dagger = \langle \phi|S^{-\frac{1}{2}}$$

But, considering the matrix analog, why don't we use the inverse?

For instance, we have the identity: $$ D^{AO} = C  D^{MO}C^T $$ where $D$ is the 1-electron reduced density matrix. In my thinking this has to be a transformation from the MO to the AO on the right of $D^{MO}$, and from the AO to the MO on the left, which gives you $D^{AO}$. However, $C$ is not unitary obviously, because it comes from the Lowdin procedure, and so while $C$ on the left is from the AO to the MO, $C^{-1}$ is not equivalent to $C^T$.

Furthermore to get $D^{MO}$ matrix in terms of the $D^{AO}$ you DO have to apply the inverse relationship with $C^{-1}$. $$D^{MO} = C^{-1}D^{AO}(C^T)^{-1},$$ because $C^{-1}$ and $C$ and related non-trivially through $S$.

Hopefully this is appropriate to ask here. I think the answer involves Hilbert spaces but I can't find a lot on non-orthogonal transformations. Thanks!

Consider a nonsingular, linear transformation $\mathbf A$ of a set of vectors arranged into a matrix $\mathbf V$ $$\mathbf V'=\mathbf {VA}$$

We can say the matrix $\mathbf V'$ is orthogonal (and thus that we have linearly transformed the vectors to form an orthogonal basis) if $$\mathbf {V'^{\dagger}V'}=\mathbf {A^{\dagger}V^{\dagger}VA}=\mathbf {A^{\dagger}SA}=\mathbf{1}$$

In general, this occurs when $\mathbf{A}=\mathbf{S}^{-1/2}\mathbf{B}$, where $\mathbf{B}$ can be any unitary matrix. The Lowdin orthogonalization commonly used is just the case where $\mathbf{B=1}$, so we can see that we are properly changing the basis.

But your issue is specifically with transformations of the density matrix, where the transformation matrix is not unitary in general. The standard change of basis formalism would suggest that to go to AO basis from MO, we would do a transformation something like this:
$$\mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{-1}=\mathbf{D}^{AO}\mathbf{S}$$
Ensuring orthogonality of the MOs requires that $\mathbf{C^{\dagger}\mathbf{S}\mathbf{C}}=\mathbf{1}$, which we can sub in to the equation above to give: $$\mathbf{C}\mathbf{D}^{MO}\mathbf{C^{\dagger}\mathbf{S}\mathbf{C}}\mathbf{C}^{-1}=\mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{\dagger}\mathbf{S} =\mathbf{D}^{AO}\mathbf{S} \to \mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{\dagger}=\mathbf{D}^{AO}$$

So the form in terms $\mathbf C$ and its adjoint is a direct result of the conventional similarity transform combined with the orthonormality condition on the MOs.

This latter derivation was adapted from$^1$ which can be found here.

As to why we avoid using transformations involving the inverse of the transformation matrix, I suspect Zhe is correct that it is a matter of cost, since obtaining the transpose/adjoint of a matrix requires almost no work at all, while an inverse is relatively difficult to obtain.

T. Helgaker et. al Chemical Physics Letters 327 (2000). 397–403


What factors are important for quantitative analysis of a proton 1D-NMR spectrum? Besides elucidating or verifying a chemical structure, NMR can also be used e.g. for quantifying a mixture of different chemicals.

Depending on the quality of the spectrum and the specific substance, integrating the NMR signals for the same molecule can result in significant variations from the actual ratio of nuclei, which is known from the chemical structure.

What factors do affect the quantification of NMR signals in a proton 1D spectrum? What precautions should one take during acquisition and processing of NMR spectra to make sure that the spectra can be analyzed quantitatively?

The signals you get in a simple 1D proton spectrum are in most cases roughly quantitative, though there are some exceptions and some aspects you need to consider if you need high accuracy.
Acquisition
There are several things you need to keep in mind while setting up the experiment to ensure that the resulting NMR spectrum will be quantitative.
The spectrum should be of high quality, any artifacts and noise will affect the quantification.
Relaxation delay
The relaxation delay is the time between experiments, if it is too short some protons won't relax completely back to the equilibrium state. This affects the intensity of the NMR signal and will cause the integrals of those signals to be off.
The relaxation delay (including the acquisition time) should be around $5 \cdot \mathrm{T_1}$. You can either estimate the $\mathrm{T_1}$ or measure it.
13C satellites
The 13C satellites are around 1% of the total signal, if you need high accuracy you should use decoupling to get rid of those.
Other aspects
You also need to ensure that the pulses used in the experiment have a reasonably flat excitation profile for the area you're looking at. This is usually no problem for standard 1D proton experiments, but might be a problem for other kinds of experiments or if you have an unusually long pulse and/or and unusually wide distribution of your chemical shifts.
Processing
Baseline
The baseline should be absolutely flat and at exactly zero, any error there will lead to large errors in the quantification. Using a digital filter during acquisition is a good idea for that (baseopt option for Bruker spectrometers), this will ensure a very flat baseline at zero, as long as no other problems distort the baseline.
Phasing
The phase-correction should be exact, it might be necessary to perform a manual phase-correction.


What are the bond angles of propane? In an organic chemistry book, I find out that the observed C-C-C bond angle is approximately $111.66^\circ$. What about the C-C-H and H-C-H bonds shown in the picture below?

According to “Structure of Free Molecules in the Gas Phase”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL., the bond angles in propane are

$\angle\ce{CCC}=112^\circ$

and

$\angle\ce{HCH}=107^\circ$

Therefore, the remaining bond angles should each be about

$\angle\ce{CCH}\approx109^\circ$


Why is $\Delta G = -T \Delta S_{\mathrm{total}}$ valid only at constant pressure? According to my book(Elements of Physical Chemistry by Atkins and de Paula, 5th ed.), $\Delta G = - T \Delta S_{\mathrm{total}}$ is valid only for constant pressure and temperature.

Enthalpy is defined as $H = U + PV$. So, when pressure and volume both are not constant, $\Delta H = \Delta U + \Delta (PV)$. So, enthalpy can be defined even if the pressure is not constant.

Now the criterion for spontaneity is that $\Delta S_{\mathrm{total}} > 0$. But $$\Delta S_{\mathrm{total}} = \Delta S_{\mathrm{sys}} - \frac{\Delta H}{T}$$ where $T$ is the temperature of the surroundings. Now, if any heat enters the surroundings, the temperature of the surroundings does not change provided that the surroundings are large. I think that due to this the temperature of the system has essentially to be the same or infinitesimally greater/lesser than the surrounding temperature as the heat transfer has to take place reversibly.

But now, as $G = H - TS_{\mathrm{sys}}$, $$\Delta G = \Delta H - T\Delta S_{\mathrm{sys}}.$$

Thus, from comparison of equations, $\Delta G = -T \Delta S_{\mathrm{total}}$. So, I did not need to specify that pressure is constant and why does my book do so?

You have actually assumed in your equations that both pressure and temperature are constant. Firstly, we have that

$$\mathrm{d}S_\text{total} = \mathrm{d}S_\text{system} + \mathrm{d}S_\text{surroundings}.$$

By definition, differential of entropy is

$$\mathrm{d}S = \frac{\delta q_\text{rev}}{T}.$$

From the first law of thermodynamics, and the definition of entalphy, for a reversible process

$$\mathrm{d}H = \mathrm{d}U + P\mathrm{d}V + V\mathrm{d}P= \delta q_\text{rev} + \delta w + P\mathrm{d}V + V\mathrm{d}P.$$

When all work done is to expand, then $\delta w = -PdV$, and thus

$$\mathrm{d}H = \delta q_\text{rev} -P\mathrm{d}V+ P\mathrm{d}V + V\mathrm{d}P = \delta q_\text{rev} + V\mathrm{d}P.$$

Only if pressure is constant, do we get the relationship

$$\mathrm{d}H \overset{\mathrm{d}P=0}{=} \delta q_\text{rev} \to \mathrm{d}S = \frac{\mathrm{d} H}{T}.$$

Therefore,

$$\mathrm{d}S_\text{total} = \mathrm{d}S_\text{system} - \frac{\mathrm{d} H}{T}\tag{constant pres., exp. work}.$$

Additionally for constant temperature, or $\mathrm{d}T = 0$,

$$\mathrm{d}G = \mathrm{d}H - S_\text{system}\mathrm{d}T - T\mathrm{d}S_\text{system} \overset{\mathrm{d}T = 0}{=} \mathrm{d}H - T\mathrm{d}S_\text{system}.\tag{constant temp.}$$

Hence the result $$\mathrm{d}G = -T\mathrm{d}S_\text{total}$$ is true when pressure and temperature are constant (in addition to possible other constraints).

It is probably not the best practice on my part to write things like

$$\mathrm{d}H = \delta q_\text{rev} $$

which has the impression of sloppy (but probably not wrong) notation. Interpret this as that, given our assumptions, the number $1$ becomes an integrating factor for $\delta q$, equivalent to the inexact differential becoming exact under special conditions.

A better option might be to write this expression with definite integrals, i.e.,

$$\int_{\text{state 1}}^{\text{state 2}} \delta q_\text{rev} = \int_{H_1}^{H_2}\mathrm{d}H.$$


What is occuring on the quantum level when a molecule rotates plane polarized light? What is occuring on the quantum level when a molecule rotates plane polarized light?

Also, why do enantiomers then rotate light in opposite directions? I would think that the electromagnetic waves would interact with the transient electric and magnetic fields within the molecule itself as it propagates and therefore the light changes direction as it passes through the molecule. However with this idea there seems to be a problem with the fact that all the molecules wouldn't be oriented in the same directions and thus scatter the light in all directions rather than a net rotation in one direction of the light. I am looking for an explanation using electromagnetism if it can be explained at that level or if it cannot, the quantum mechanical answer would be fine as well.

As mentioned in the comments, the answer to your $2^{nd}$ question regarding why optical rotation isn't simply cancelled out in solution due to random orientations is given in ManishEarth's answer to Molecular chirality and optical rotation

For your first question, I will try to explain what actually causes rotation when light impinges on an optically active molecule. For chiral molecules, it is simplifies the picture to think of each enantiomer as a screw of negative charge that is either right or left handed. Plane or linearly polarized light can always be decomposed into circularly polarized clockwise and counterclockwise components. When the light collides with the molecule, the clockwise component and counterclockwise components interact differently with the electric field of the molecule, leading one of these components to travel faster than the other, which alters the angle of the plane polarized light. A good visualization of this effect (for a whole solution rather than an individual molecule) can be found at http://cddemo.szialab.org/.

As an interesting aside, you might imagine that because we have a reasonable understanding of why optical rotation occurs, we should be able to predict, either heuristically or from a first principles calculation, the magnitude and direction of optical rotation for a molecule just based on its structure. This, thus far has not been the case. Not only are there very few ways of heuristically guessing the magnitude/direction, but quantum mechanical calculations at high levels of theory have proven unable to even consistently obtain the correct direction of the optical rotation. Given a particular chiral molecule, we still can't predict which direction, let alone magnitude, that a particular enantiomer will rotate light.

My main point in this little tangent is that even something as seemingly simple and in such widespread use as optical rotation is still not very well understood and is a difficult area of active research.


How does this ring expansion on a steroid system occur? The correct answer is supposedly B, as shown by Gravestock et al.[1] In this reaction after ozonolysis of the C=C double bond, two ketones are formed. How is a rearrangement or expansion of the ring possible, since carbocations are not formed as intermediates?

Gravestock, M. B.; Johnson, W. S.; Mccarry, B. E.; Parry, R. J.; Ratcliffe, B. E. Acetylenic bond participation in biomimetic polyene cyclizations. Model studies directed toward the synthesis of 20-keto steroids. Synthesis of dl-progesterone and dl-Δ4-androstene-3,17-dione. J. Am. Chem. Soc. 1978, 100 (13), 4274–4282 DOI: 10.1021/ja00481a044.

It is clear that the first step is ozonolysis with a slightly reductive workup, yielding two ketones. You seem to have already established that. The second step features a base and an alcohol as a solvent. The hydroxide can react with the ketones in two ways: 1) nucleophilic attack at the electrophilic carbonyl carbon 2) abstraction of the mildly acidic $\alpha$ hydrogen. With the methanol in large excess as the solvent, it could also potentially act as a nucleophile to attack the carbonyl group to eventually form hemiacetals. However, what seems to be the most preferred reaction pathway, based on thermodynamics, here is the aldol condensation to form the $\alpha$,$\beta$-unsaturated ketone. I will now go through the considerations that a problem solver could have made.

Why is the aldol condensation thermodynamically favoured over the other alternative reation pathways?

The key reason is that the formation of the conjugated enone $\pi$ system is quite energetically favoured. On the mechanism of the aldol condensation, Carey & Sundberg (2007) wrote the following, on  p. 684:

  In general, the reactions in the addition phase of both the base- and acid-catalyzed
  mechanisms are reversible. The equilibrium constant for addition is usually unfavorable
  for ketones. The equilibrium constant for the dehydration phase is usually favorable
  because of the conjugated $\alpha$,$\beta$-unsaturated carbonyl system that is formed. When the reaction conditions are sufficiently vigorous to cause dehydration, the overall
  reaction can go to completion, even if the equilibrium constant for the addition step is
  unfavorable.

What about the ketone-hemiacetal equilibrium?

The addition products (i.e. hemiacetals) formed between methanol and the ketones in the basic medium can easily be converted back to the reactants (i.e. reaction is highly reversible). The depletion of the ketone by it reacting in the aldol condensation pathway thus pushes the equilibrium towards the ketone form. However, this consideration is not even important in this case since the equilibrium constant for the ketone-hemiacetal equilibrium is much less than one.

There are numerous $\alpha$ hydrogens but which one would be removed?

To answer this question, usually we would have to know if the conditions provide thermodynamic or kinetic control. Not much information has been provided in this case so I believe it is reasonable to assume thermodynamic control since only one product would clearly dominate while there are various possible kinetic products.

In this case, the hydrogen is first abstracted from the terminal methyl group of the newly-formed ketone, forming the enolate ion. This then acts as a nucleophile attacking the carbonyl carbon of the other newly-formed ketone. Upon protonation, the $\beta$-hydroxyketone is formed. Finally, there is E1cB elimination to form the product, as shown in B.

Other addition products may result either from 1) formation of other enolates from abstracting other acidic $\alpha$ hydrogens or 2) formation of other possible addition products from the enolate attacking other carbonyls (possibly even those on other molecules). Firstly, many of the alternatives would result in ring sizes that would possess too much strain. In this case, a 4-membered ring could form if deprotonation took place at the other $\alpha$ carbon of the same ketone. Intermolecular reaction would not be as favoured here since the proximity of the groups simply make the intramolecular reaction "hard to resist".

Conclusion

I have just gone through some important considerations behind arriving at the structure B, as well as the mechanism of the aldol reaction. Many details provided may seem extraneous though... Hopefully, you are able to understand how the "ring expansion" took place now.

References

Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.


General rules for deciding volatility Given an organic compound, is there any way to decide if it is volatile (or compare volatility--everything is volatile in the end)?

Volatility is due to the tendency to evaporate. The conclusion I can draw from this is that a more volatile compound will be lighter and/or will have less intermolecular forces. But this doesn't seem to work--I recall that methanol is less volatile than ethanol ([citation needed]).

So are there any general guidelines for this or is it another experimental property?

This answer to this question applies to your query. Many physical properties depend on the intermolecular forces experienced in the liquid or solid. These intermolecular forces are hydrogen bonding (strongest), dipole-dipole attractions (in polar molecules), and London dispersion forces (weakest of the forces when comparing molecules of the same size, but dependent on the number of electrons in the substance). Stronger intermolecular forces would make the substance less volatile.

As noted in the first response, methanol is more volatile than ethanol. Methanol and ethanol would both have both hydrogen bonding (a relatively strong type of dipole-dipole attraction) and London dispersion forces. Because ethanol has more electrons (because it is a bigger molecule, but not necessarily because it is a heavier molecule) it would have more London dispersion forces in comparison to methanol, so, with stronger intermolecular forces, it would be less volatile. So, this example follows the general guidelines--figure out the intermolecular forces present in the molecules you want to compare and the molecule with the weakest cumulative intermolecular forces will be the most volatile.

Intramolecular bonding (covalent bonds, when you're talking about molecules) does not affect physical properties.


In an closed adiabatic container, if we carry out an exothermic process, what will happen to the temperature of the system I know that a closed adiabatic container acts as an isolated system, so I expect the internal energy of the system to be a constant as in an isolated system. But since it is an exothermic reaction, temperature will increase, and since internal energy is a function of temperature it should increase. This is a direct contradiction. Can anyone please help?

Temperature will increase. If system is adiabatic, it doesn't neccesarily mean that temperature of the system can't change because even though system can't exchange energy with surroundings, processes inside can generate heat. In your example, exothermic reaction. Note that, internal energy in your case stays the same even though temperature increased. This is because internal energy is the sum of kinetic energy of all molecules and atoms and all potential enegy of interaction between atoms chemically bonded and between all molecules (intermolecular bonds). When exothermic reaction happens, potential energy of interaction between atoms is partially converted into heat because if reaction is exothermic, chemical bonds formed (of products)are stronger than of reactants. Potential energy is only converted to heat which is kinetic energy of atomic and molecular motion (energy conservation - 1st law of thermodynamics). Since internal energy is sum of potential and kinetic energy of atoms and molecules it stays the same, only potential enegy is smaller and kinetic is bigger after exothermic reaction. If you have any misunderstandings, fire them to me.


What atom would be protonated first in a strongly acidic solution? The answer is A. I thought it would be D since the ester group has the highest pka and the more basic functional group would be more prone to protonation. Also, I don't really get how to rank C in terms of atoms that would be protonated first as I don't know the pka of the ether group.

In short, none. As Mithoron said, the oxygen atoms can all act as Lewis bases and will all be protonated to an extent. If you rephrase the question to "Which of the atoms has the highest Kb?", well that's another story. Atom D has the lowest Kb, because the carbon in a carbonyl has a slight positive charge. Atom C would be difficult to protonate, the oxygen is already donating a pair to the furan ring to make it aromatic. So the debate is between a carbonyl and a primary alcohol. After consulting some pKa tables I found that protonated ketones have a pKa of around -7.3 while protonated primary alcohols of around -2.4 thus the alcohol is a stronger base (higher pKa means weaker acid thus stronger conjugate base). Finally we can conclude that the primary alcohol -OH group will be the first to accept a proton.

P.S. Fun fact, the quickest to be protonated was none of the ones marked with letter, it would have been the ester's carbonyl group!


Why does a Lower Concentration of SO2 call for an increase of equipment dimensions in a Contact Process Plant? I'm working on a presentation based on an article about the Contact Process (production of sulfuric acid), and it says that using a low concentration of SO2 at the entrance of the converter (in which the oxidation of SO2 to SO3 takes place as an equilibrium reaction) is better in terms of conversion, but disadvantageous in that it would require using bigger plant equipment dimensions.

Why does using a lower concentration of SO2 require a higher volume of such equipment?

Wouldn't using more gas require more size?

You need higher quality of equipment and larger "chambers" to deal with low concentrations. Besides, in order to meet needs of plant (I mean, each plant has production output) you will need stated amount of raw materials, which you can find in material balance sheet of the applied technological processes. So, when using low-concentrated things as raw materials, even if yield is higher (it's higher usually by less than 10%, not really much) you will need larger volume of raw materials (since needed amount is usually given in mass of pure $SO_2$ in entering gas).


Why is nitroglycerin so unstable? I know that nitroglycerin is an incredibly powerful explosive due to its three nitro groups, but why is it so unstable?

What in its chemical structure makes it so sensitive to shock and temperature? Why is TNT, in contrast, so stable?

An explosion is a type of spontaneous chemical reaction that, once initiated, is driven by both a large exothermic change (great release of heat) and a large positive entropy change (great quantities of gases are released)   $\ce{^1}$ from  reactants to products (a thermodynamically favorable process).

The energetic stability of the gaseous products and hence their generation comes from the formation of strongly bonded species like carbon monoxide, carbon dioxide, and (di)nitrogen, which contain strong double and triple bonds having bond strengths of nearly 1 MJ/mole.

Consequently, most commercial explosives are organic compounds containing $\ce{-NO2, -ONO2}$ and $\ce{-NHNO2}$ groups that, when detonated, release gases like the aforementioned (e.g., nitroglycerin, TNT, HMX, PETN, nitrocellulose).

In general, Detonation is used to describe an explosive phenomenon whereby the decomposition is propagated by an explosive shock wave traversing the explosive material at speeds greater than the speed of sound within the substance.

The following factors affect the stability of an explosive: (1) Chemical constitution.
(2) Temperature of storage. (3) Exposure to sunlight (4) Electrical discharge.

Certain groups like nitro  (–NO2), nitrate (–ONO2), and azide (–N3), are intrinsically labile or Kinetically there exists a low activation barrier to the decomposition reaction which makes compounds exhibit high sensitivity to flame or mechanical shock.

Nitroglycerin $\ce{^2}$ is a primary explosive  that is extremely sensitive to stimuli such as impact, friction, heat, static electricity, or electromagnetic radiation ultimately leading to detonation.

You can see that 4 moles of nitroglycerin produces 35 moles of hot gases.The concentrated gases are under very high pressure, so they expand rapidly. The heat speeds up the individual gas particles, boosting the pressure even higher. The gas expands faster than the speed of sound, it generates a powerful shock wave.

As a rule of thumb, explosives with an activation energy of decomposition
higher than 170 kJ/mol are stable for thousands of years at room temperature, whereas for values below 155 kJ/mol, chemical stability is limited.$\ce{^3}$

As can be seen from Table below $\ce{^4}$, the commonly used explosives from the classes of aromatic and aliphatic nitro compounds, secondary nitramines, and organic azides are very stable, whereas aliphatic nitrate esters suffer from much lower chemical stability.

TNT is a secondary explosive less sensitive than a primary explosive and requires substantially more energy to be initiated.

2,4,6-trinitrotoluene (TNT) is a molecular explosive that exhibits chemical stability due to its high performance, comparatively low shock sensitivity, low melting temperature of TM = 353–354 K (80–81 °C)$\ce{^5}$ $\ce{^6}$

References

https://en.wikipedia.org/wiki/Explosive#Stability
http://www.ch.ic.ac.uk/rzepa/mim/environmental/html/nitroglyc_text.htm
Chimia 58 (2004) 401 -408
https://www.ingentaconnect.com/content/scs/chimia/2004/00000058/00000006/art00010?crawler=true
M. L. Hobbs and M. J. Kaneshige, in Proceedings of the Society for Experimental Mechanics Annual Conference, Albuquerque, NM, 2009
APPLIED PHYSICS LETTERS 104, 021911 (2014) ,
" Chemical stability of molten 2,4,6-trinitrotoluene at high pressure "
Dana M. Dattelbaum, Raja S. Chellappa, Patrick R. Bowden,Joshua D. Coe,
and Madeline A. Margevicius


Naming of an organic compound How would I determine the systematic name of $\ce{(CH3)2CHCH2CHClCH3}$?

First of all, how do we determine which side we number the carbon atoms from? It has an alkyl group ($\ce{CH3}$) and a functional group ($\ce{Cl}$). I got lucky by going from right to left so I wrote 2-chloro-4-methylpentane but why couldn't we also label this as 4-chloro-2-methylpentane?

Do we take functional groups in preference to alkyl groups when naming them?

The substituents go alphabetically, and as 'chloro' is before 'methyl' in the alphabet, the correct name is 2-chloro-4-methylpentane.


How to explain the increase in current with the increase in the concentration of responsible ions? I have conducted an experiment wherein I have used different concentrations of $\ce{H2SO4}$, $\ce{HNO3}$ and $\ce{H3PO4}$ as the electrolytes. I filled the acid in glass tube and used a 9V battery, wires and two graphite electrodes to complete the circuit. I also attached an ammeter to check the current variation during a time period of 1 minute. I observed that the current was higher with more concentrated solutions, although my query is that when I add more of these acids, the concentration of the $\ce{OH-}$ ions reduces; it is the $\ce{OH-}$ ions and the $\ce{H+}$ ions that dissociate. How exactly do I explain the increase in current due to increase in concentration?

The electrical current is carried by all ions in the solution, including the $\ce{SO_4^{2-}}$/ $\ce{HSO_4^-}$, $\ce{NO_3^-}$, and $\ce{PO_4^{3-}}$/ $\ce{HPO_4^{2-}}$/ $\ce{H_2PO_4^-}$ ions, and not just by $\ce{H+}$ and $\ce{OH-}$.  So, even though the $\ce{OH-}$ concentration is decreasing with increasing acid concentration, like you say, it's much more than made up for by the increased concentration of the anions from the acids.

EDIT (in response to a comment): The contribution of ions that don't participate in the electrochemistry differs depending on the system.  For example, in $\ce{NaOH}$ the $\ce{Na+}$ effectively just distributes to balance the charge from the concentration gradient in $\ce{OH-}$:

$\ce{H2O}$ is reduced to $\ce{H2}$ and $\ce{OH-}$ at the cathode, increasing the $\ce{OH-}$ concentration there
$\ce{OH-}$ is oxidized to $\ce{O2}$ and $\ce{H+}$ at the anode, decreasing the $\ce{OH-}$ concentration there
Thus, $[\ce{OH-}]$ changes from high at the cathode to low at the anode
In order for the solution to maintain electroneutrality, something has to balance the higher concentration of negative charge at the cathode, and in this system $\ce{Na+}$ is the major contributor.  (Water self-ionization can't provide enough $\ce{H+}$ to maintain the charge balance.)

In other systems such as $\ce{H3PO4}$, the acid-base equilibration behavior of an electrochemically inert species can contribute to the charge carrying function:

Adjusted to pH 7 with, say, NaOH, phosphoric acid exists mainly as $\ce{H2PO4-}$ and $\ce{HPO4^{2-}}$
Water is oxidized at the anode, producing $\ce{H+}$, which reacts with $\ce{HPO4^{2-}}$ to form $\ce{H2PO4-}$
Water is reduced at the cathode, producing $\ce{OH-}$, which reacts with $\ce{H2PO4^{-}}$ to form $\ce{HPO4^{2-}}$ and $\ce{H2O}$
So, $\ce{HPO4^{2-}}$ has a higher concentration around the cathode than the anode, and $\ce{H2PO4^{-}}$ has a higher concentration around the anode than the cathode.  These concentration differences result in a net transfer of charge between the two electrodes.
All of the negatively charged phosphate ions are balanced out by a corresponding distribution of positively-charged $\ce{Na+}$, similar to the $\ce{NaOH}$ case.


Mechanism of combustion of hydrocarbons While this question comes close, it doesn't quite answer my question.
Adding heat to oxygen & a hydrocarbon breaks the atoms apart and they reform into carbon dioxide and water. I get that.
I want to know specifically at what point the water and carbon dioxide are formed?
As I understand it the heat vibrates the atoms, breaking bonds.  Those atoms reform with oxygen forming stronger molecules ($\ce{H2O}$ and $\ce{CO2}$) which can withstand the combustion heat.  I reached this far with help from a Reddit thread.
Where that thread seemed to disagree is in my main question, on where/when the formation actually happens, it also raised another question of how long atoms are broken apart for.
In this article, it would suggest that free atoms are created which find oxygen:

$\ce{CH4 + 2O2 → CO2 + 2H2O + energy}$
[...] the arrow represents the moment the reaction happens. But there are a lot of interesting things happening that are hidden behind that arrow. A more detailed equation would look something like this:
$\ce{CH4 + 2O2 + a little energy →C + 4H + 4O →CO2 + 2H2O + lots of energy}$
The first arrow represents the breaking of the bonds, which requires energy. On the middle line are the atoms, now broken out of molecules and free to react. [...]

Which at least one of the Reddit thread contributors disagreed with.
So! (TL;DR)

When are the new molecules formed (cylindar, afterwards or both)?
How long are atoms "free", how long do they take to form new molecules, is it instantaneous or ... what?

Combustion reactions are complicated even when they result in simple products
Part of the problem with answering the question "at what point are water and carbon dioxide formed" is that the reactions occuring in combustion are very complicated and not at all well represented by the simple versions shown in the question.
We can say for sure that the picture where all the components are atomised and then recombine is not the way the reaction works. This simplistic idea is part of the reason for disagreement on the reddit thread.
It is an incorrect view for many reasons but mostly because the reaction doesn't all happen at once. Take petrol (that's gasoline for americans) engines. The initiation of combustion is started by a spark and then (relatively slowly in chemical terms) progresses through the rest of the fuel/air mixture (if it were very fast the engine would explode rather than smoothly driving the cylinder, if it were only a little too fast the engine would "knock", damaging the cylinder). So we need to think–even before considering the detailed chemistry–about the (slow) steps involved in the reaction.
The major steps involved in most combustion reactions involve radical chains. Though the full detail is not understood (though this answer gives some more detail), it is thought that the initiating step involves a hydrocarbon reacting with an oxygen molecule to give a hydroperoxide radical and a hydrocarbon radical. Very simplistically this looks like:
$\ce{O2 + CH3(CH2)nCH2-H -> HO2. +CH3(CH2)nCH2.} $
But neither of those radicals is stable or long-lived. In an environment flooded with more hydrocarbons and oxygen, they can both react further to give more products and more radicals. For example, the hydroperoxide may also fall apart to hydroxide radicals which can react with parts of other hydrocarbon molecules to give water and yet more radicals. In the early stages, most reactions create more radicals; later in the process some radicals can combine into stable products (like water and carbon dioxide). Once the reaction is started it can be self-sustaining as the formation of those stable products release a lot of heat, enough to keep creating new radicals and new chains until the source hydrocarbons or the oxygen is depleted.
So, even thinking about the detailed, step by step, chemical reactions suggests that the process is messy. Even early on in the process some final products are being created, though this continues until the fuel/air is mostly used up. if you plotted the amount of final product over time you would see some very early and a rising amount until the reaction had finished. It is pretty meaningless to ask "at which point" the products are formed.
In short, the initiation does not "atomise" the components: it creates radicals (this doesn't need as much energy input as "atomisation"). They can continue to react creating chains of yet more radicals and some chains can stop when stable products are formed, even early in the reaction. The net reaction releases enough heat to sustain combustion while fuel remains. Final products start to form early in the process and continue to be formed until the air and fuel have gone.


Filling a buret tip before an experiment I have a question that asks if the buret tip is not filled before the experiment what would happen (effect on molarity for acid/base titration). The way I'm thinking about it, it's not like you're putting a pre measured solution into it. You fill it as high as you fill it and you can't put anything in without getting the tip filled up too, so I don't see where that will change anything. Am I thinking about that right?

You fill it as high as you fill it and you can't put anything in without getting the tip filled up too,

No I don't think it will work like that. Usually you have the stock or valve closed at the bottom. Consequently, the tip will be empty until you bleed the buret. Further, you will often bleed (allow the titrant to flow) a buret while thumping the tip to remove any air bubbles that may form directly under the stop-cock when your titrant flows initially through the tip. Once you have titrant in the tip and the tip is devoid of air bubbles you will then close the valve and fill the buret to the desired level.

At this point you can better determine just how much you have used.

The measure marks on the buret itself take into account what is in the tip below the stop cock. But if you do not bleed the tip initially, filling it with your titrant then your first addition will be a bit off because the tip does not fill entirely with liquid before you allow some to drip through.

Here is a good write-up on titrations with burets.

  A bubble in the nozzle of a buret will produce an inaccurate volume reading if the bubble escapes during a titration. Bubbles may be large and visible as shown above left or so small as not to be seen, above center. During a titration such small bubbles begin to move in the direction of the nozzle but may remain in place even though there is a moderate flow of titrant (above right). Even when the buret valve is wide open some bubbles remain in place until you take your eyes off them. Then they sneak through the nozzle and ruin your titration. Also if you let a an air bubble stay because you think it will remain and then later it leave to the tip you will have to start over because you have no idea what


How can we calculate enthalpy change in a closed tank containing methanol adsorbed by activated carbon? Here's the problem in the simplest way:

I have a closed box containing methanol adsorbed by activated carbon. The box is made from copper. A pipe is connected to that box, which has condenser on the other end to return the methanol back to the liquid state.

The box is painted black and put in the sun to absorb heat.

The data I have collected are:

initial and final pressure
initial and final temperature
0.26 kg of methanol was condensed in the other end
0.3744 kg of methanol remained in the tank

How can we calculate enthalpy change in the box (heat absorbed by the box from the sun)?

Enthalpy ($H$) is defined as, $H=U+PV$ where $U$ is the internal energy and $P$ is pressure and $V$ is volume. The change in enthalpy is given by

$$\Delta H = \Delta U + \Delta(PV)$$

And $\Delta U$ is equal to the sum of the heat transferred $q$ and the work done $w=-P\Delta V$.

At constant $V$ (which is what you have in a closed system), $\Delta V =0$ (and thus $w=0$) and $\Delta(PV)=V\Delta P$.

Thus...

$$\Delta H = q+V\Delta P$$

You can determine $q$ based on the amount of methanol that vaporized using the heat of vaporization of methanol. We can assume that the majority of heat absorbed was used to distil methanol.

$$q_{vap}=m\cdot c_{vap}$$

How can we determine the volume of the system? Or... if the change in pressure is very small, perhaps that term can be ignored.


Feasibility of another product in peroxide effect mechanism of propene In the Chain Propagation step of the reaction of propene  ($\ce{CH3CH=CH2}$) and hydrogen bromide ($\ce{HBr}$), why is $\ce{CH3-\dot{\ce C}H-CH2-Br}$ free radical formed when a more stable, resonance stabilized allyl free radical ($\ce{\dot{\ce C}H2-CH=CH2}$) is possible?

Source  : https://www.chemguide.co.uk/mechanisms/freerad/alkenehbr.html

The energy cost of bromine radical abstracting hydrogen from propene is 89 kcal/mole. The gain of forming $\ce{HBr}$ is 18-28 kcal/mol (for a $\ce{H-Br}$ bond from $\ce{H2}$  and $\ce{Br2}$; variation is due to solvent effects) + 23 kcal/mol (from the $\ce{Br2}$ which was fractured into 2 Br radicals by 46 kcal/mol bond energy); so the gain is 41-51 kcal/mol vs a cost of 89, or a net cost (net energy requirement) of 38-48 kcal/mol. This would be a slow reaction with an activation energy even higher than 41-51 kcal/mol. (All data from CRC Handbook)

The alternate reaction, of adding a Br radical (with 23 kcal/mol energy above ground state) to propene, loses the pi bond and its energy (84 kcal/mol) and gains a $\ce{C-Br}$ bond (70 kcal/mol). The energy cost of forming the bromoradical is 84 but the gain is 70 kcal/mol for $\ce{C-Br}$ plus the 23 kcal/mol from the Br radical, or 93 kcal/mol. Net gain is 9 kcal/mol. This reaction will go spontaneously while the first will not (rate determined by activation energies, of course).

It turns out that allylic stabilization is only part of the story, and is, in fact, rather small: the bond strength of $\ce{H - allyl}$ is 89 kcal/mol vs the bond strength of $\ce{H - n-propyl}$ which is 98 kcal/mol.


Monosodium Diphosphate (NaH3P2O7) Preparation I looked on wikipedia for an answer but I couldn't even find a page for this salt, a subsequent google search didn't further enlighten me.
Does any one know how monosodium diphosphate can be prepared?

$\ce{NaH3P2O7}$ is an intermediate acid-salt formed during the neutralization of pyrophosphoric acid ($\ce{H4P2O7}$) which becomes unstable as pH rises and leads to formation of trisodium diphosphate, $\ce{Na3HP2O7}$:
$$\ce{H4P2O7 + NaOH ->[pH < 3] NaH3P2O7 + H2O}$$
$$\ce{NaH3P2O7 + NaOH ->[pH = 3-9] Na2H2P2O7 + H2O}$$
$$\ce{Na2H2P2O7 + NaOH ->[pH > 9] Na3HP2O7 + H2O}$$
This acid-salt can also be produced by the reaction of trisodium phosphate and pyrophosphoric acid:
$$\ce{Na3PO4 + H4P2O7 -> NaH3P2O7 + Na2HPO4}$$
Notes and references:

Study of crystallization processes in some rare earth vanadate, tungstate and phosphate systems under hydrothermal conditions, K Byrappa and B Nirmala, Indian J. Phyx. 73A (5), 621-632 (1999) (PDF)
Its CAS number is 13847-74-0
The first dissociation of pyrophosphoric acid gives the anion, $\ce{H3P2O7^-}$. The dissociation constant is pretty large compared to 2nd, 3rd, and 4th dissociation of the acid.

$$\ce{H4P2O7 + H2O <=> H3O+ + H3P2O7¯ ~~~~ K_{a1} \sim 10^{–1}}$$


Deciding which α-Η gets abstracted in aldol condensation Consider the base catalysed aldol condensation of methyl ethyl ketone. The first step involves the abstraction of a proton by the base. However I am having trouble deciding whether the proton will be abstracted from the $\ce{-CH2-{}}$ group or the $\ce{-CH3}$ group.

I think it should be from the latter as the $\ce{CH^-}$ ion so formed in the former experiences undesired $+I$ effect by the two groups neighbouring it; but I'm not sure.

There have been tons of articles, review articles and even books written on the subject. The tl;dr version is that it depends on the conditions you use.

In general, the less-hindered enolisation (in your case: enolising towards the methyl) will be more rapid. This leads to the possibility of kinetic control to selectively generate the terminal enolate by very strong bases and low temperatures such as LDA or even BuLi if you’re daring (both at $\pu{-78^\circ C}$).

On the other hand, the more substituted enolate is more stable thermodynamically as you noted. Thus, if you use a weak base that binds and deprotonates reversibly and if you use relatively high temperatures or longer reaction times, the more stable enolate will be formed. An example for these conditions would be an alcoholate base at room temperature.

Unfortunately, I can give you no simple rules. You will have to check the literature or your lecture notes to see which combinations of reagents give which selectivity.


Mechanism for reaction between hydrochloric acid and magnesium Balanced equation:

$$\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}$$

However, Cl- is a spectator ion so the actual reaction equation is:

$$\ce{Mg + 2H+ -> Mg^2+ + H2}$$

Here is my proposed reaction mechanism:

A magnesium atom reacts with a $\ce{H+}$ ion: $\ce{Mg + H+ -> Mg+ + H}$
The $\ce{Mg+}$ intermediate produced reacts with another $\ce{H+}$ ion: $\ce{Mg+ + H+ -> Mg^2+ + H}$
The two hydrogen atoms (from step 1 & 2) react together: $\ce{H + H -> H2}$

However, my problem arises when I compare this to my rate law that I found experimentally ($\ce{Mg}$ is not in the rate law as it is a solid and has no concentration):

$$\mathrm{rate} = k[\ce{H+}]^2$$

None of these mechanism steps above conform to my observed rate law (or do they?), could any of you enlighten me to any other mechanisms?

There are only two mechanisms I can think of: a one-step mechanism where it is literally $\ce{Mg + 2H+ -> Mg^2+ + H2}$ (though I've read that a one-step reaction is unlikely) and the other way is via an intermediate (as described above).

Based on the three step mechanism you have provided, what if the slow step is your step 3 which is H + H --> H2  so rate law is now k3[H]^2. Since H is an intermediate, you can substitute for this intermediate from earlier steps and you will end up with
Rate = (k3.k1.k2.[Mg][H+]^2)/[Mg2+]   and if [Mg]/[Mg2+] is constant as the reaction of a solid always happens at the surface, you can get a rate law = k [HCl]^2


Intrahydrogen bonding Is it okay to conclude that in 2-nitrophenol where intra hydogen bonding is experienced, two similar molecules are only held by the weak van der waal's forces and is the reason for it to have a relatively lower boiling point than other nitro phenols

From the abstract to ref 1:

  We investigate both experimentally and theoretically the structure and photodynamics of nitrophenol molecules and clusters, addressing the question how the molecular photodynamics can be controlled by specific inter- and intramolecular interactions. Using quantum chemical calculations, we demonstrate the structural and energetic differences between clusters of 2-nitrophenol and 4-nitrophenol, using phenol as a reference system. The calculated structures are supported by mass spectrometry. The mass spectra of 2-nitrophenol clusters provide an evidence for a stacked structure compared to a strong O–H···O hydrogen bonding for 4-nitrophenol aggregates.

The key words in the context of 2-nitrophenol are "stacked structure". This suggests that Mithoron's comment might be correct. However "stacking" is not evidence of a particular attractive interaction mechanism, it doesn't tell us of the relative importance of classical pi-stacking versus dispersion interactions between 2-nitrophenol molecules.

I'll throw up my hands and claim ignorance, deferring explanation to the Wikipedia:

  In chemistry, pi stacking (also called π–π stacking) refers to attractive, noncovalent interactions between aromatic rings, since they contain pi bonds. These interactions are important in nucleobase stacking within DNA and RNA molecules, protein folding, template-directed synthesis, materials science, and molecular recognition, although new research suggests that pi stacking may not be operative in some of these applications. Despite intense experimental and theoretical interest, there is no unified description of the factors that contribute to pi stacking interactions.

References

Photochemistry of Nitrophenol Molecules and Clusters: Intra- vs Intermolecular Hydrogen Bond Dynamics. Kateryna Grygoryeva, Jakub Kubečka, Andriy Pysanenko, Jozef Lengyel, Petr Slavíček, and Michal Fárník. The Journal of Physical Chemistry A 2016 120 (24), 4139-4146. DOI: 10.1021/acs.jpca.6b04459


How is sulphur trioxide formed and why is it stable? Structure of $\ce{SO3}$ (sulfur trioxide):

In the molecule, if each oxygen atom shares two electrons with sulfur atom then how does the sulfur atom remain stable? It already has 6 valence electrons and needs only 2 more to become stable but the oxygen atoms share total of 6 electrons with sulfur atom. So how is the molecule stable?

Synthesis

Synthesis of $\ce{SO3}$ can be done by heating $\ce{CaSO4}$ with $\ce{SiO2}$, which produces $\ce{CaSiO3}$ and $\ce{SO3}$. Another way of synthesis is the heating of $\ce{NaHSO4}$ to $\ce{Na2SO4}$, $\ce{SO3}$, and water. (source)

As to why $\ce{SO3}$ is stable...

Sulfur forms an expanded octet. That means that it doesn't really obey the octet rule, allowing it to take on extra electrons. Sulfur is a 3rd-period element; hence it can use its 3d orbitals to make more than 4 bonds. Wikipedia has a page on hypervalency, and from an answer here:

  Regarding the octet concept, it's only a guideline at best, and nothing more. I wouldn't put too much stock into the concept. If you insist on using it, then only apply it to the following elements: C,N,O, and F.

EDIT:

Users such as @Mithoron have pointed out that this analysis is not correct. Papers seem to show that there is no d-orbital utilization. According to this comment:

  ...the doubly bonded picture is a very crude oversimplification, though. -@Martin

This enlightening image

shows that sulfur trioxide is actually a resonance of several different structures, and should be more accurately expressed as several single bonds with a dashed line to show that the bond order is not actually two. This also solves the question as to why it is stable with apparently six bonds, as in reality it only has four bonds.


Represent bonds spanning a unit cell boundary plane Is there a chemical input format that is-

Supported by Open Babel (and hence supports visualization in avogadro??)
Can represent a bond connecting an atom to its partner in a neighbouring unit cell, for crystals/systems with periodic boundary conditions

Follow up (added later):
I could not understand how cif does it, or cml. But it would be nice if I could get rid of the symmetry folds in the markup given below.
Here is the cif file. The cml output using babel is --
<?xml version="1.0"?>
<molecule id="Calcium titanate" xmlns="http://www.xml-cml.org/schema">
 <crystal>
  <scalar title="a" units="units:angstrom">5.380000</scalar>
  <scalar title="b" units="units:angstrom">5.440000</scalar>
  <scalar title="c" units="units:angstrom">7.639000</scalar>
  <scalar title="alpha" units="units:degree">90.000003</scalar>
  <scalar title="beta" units="units:degree">90.000003</scalar>
  <scalar title="gamma" units="units:degree">90.000003</scalar>
  <symmetry spaceGroup="-P 2c 2ab">
   <transform3>1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1</transform3>
   <transform3>1 0 0 0.5 0 -1 0 0.5 0 0 -1 0 0 0 0 1</transform3>
   <transform3>-1 0 0 0 0 -1 0 0 0 0 1 0.5 0 0 0 1</transform3>
   <transform3>-1 0 0 0.5 0 1 0 0.5 0 0 -1 0.5 0 0 0 1</transform3>
   <transform3>-1 0 0 0 0 -1 0 0 0 0 -1 0 0 0 0 1</transform3>
   <transform3>-1 0 0 0.5 0 1 0 0.5 0 0 1 0 0 0 0 1</transform3>
   <transform3>1 0 0 0 0 1 0 0 0 0 -1 0.5 0 0 0 1</transform3>
   <transform3>1 0 0 0.5 0 -1 0 0.5 0 0 1 0.5 0 0 0 1</transform3>
  </symmetry>
 </crystal>
 <atomArray>
  <atom id="a1" elementType="Ti" formalCharge="4" xFract="0.000000" yFract="0.500000" zFract="0.000000"/>
  <atom id="a2" elementType="Ca" formalCharge="2" xFract="0.006480" yFract="0.035600" zFract="0.250000"/>
  <atom id="a3" elementType="O" formalCharge="-2" xFract="0.571100" yFract="-0.016100" zFract="0.250000"/>
  <atom id="a4" elementType="O" formalCharge="-2" xFract="0.289700" yFract="0.288800" zFract="0.037300"/>
 </atomArray>
 <bondArray>
  <bond atomRefs2="a1 a4" order="1"/>
  <bond atomRefs2="a4 a2" order="1"/>
 </bondArray>
</molecule>

As the developer of both Open Babel and Avogadro I can say the answer is "not yet."

The main thing isn't the file format. It's that Avogadro at the moment doesn't have support at the moment to show bonds across a unit cell.

I believe there was a patch for that, but it was too slow for typical use:
https://github.com/dlonie/avogadro/commits/ENH_intercell_bonds


Why is acetone so cold? I have a summer job as a custodian and one cleaner I use is called Goof Off, which is pretty much just acetone. Another is a disinfectant which is mostly water.

Why is it that whenever I put the acetone on my rag, it is always colder than when I use another cleaner or plain water?

The acetone also seems to stay cold longer than water does. Both cleaners are left in the open so they are both the same temperature.

Evaporation is an endothermic phenomenon, i.e. it absorbs heat in order to proceed. Acetone is a volatile solvent (it evaporates easily) so it absorbs much heat when evaporating, and your skin gets colder because of that. That is what you are feeling. So acetone is not "colder", it just cools down your skin more easily when evaporating.

Ether (aka diethyl ether) is even more powerful at this.

On another note, while acetone is a pretty safe solvent, it can damage your skin because it can take away skin's lipids and you get "dry hands". In France, your employer would be forced to provide you with protecting gloves, e.g. latex gloves.


Is scopolamine same as scopolamine hydrobromide? Is scopolamine extracted with methanol or acetone from Datura or Brugmansia scopolamine free base or scopolamine hydrobromide?

If you take a look at the Wikipedia page for hyoscine, the Synonyms part lists scopolamine hydrobromide as a synonym for hyoscine even though neither molecular structure nor chemical formula contains bromine.

If the extraction procedure does not involve HBr, then I would doubt it is the hydrobromide because:

hydrobromide salts don't dissolve well in acetone;
hydrobromide salts are rare in nature.

According to the work by Bracci et al. [1], the major component of MeOH extraction of Brugmansia is atropine.

Śramska et al. [2] used ethyl acetate extraction and isolated scopolamine and atropine.

No mention of hydrobromide salts in either.

References

Bracci, A.; Daza-Losada, M.; Aguilar, M.; De Feo, V.; Miñarro, J.; Rodríguez-Arias, M. A Methanol Extract of Brugmansia Arborea Affects the Reinforcing and Motor Effects of Morphine and Cocaine in Mice. Evidence-Based Complementary and Alternative Medicine 2013, 2013, 1–7. https://doi.org/10/f96zpc.
Śramska, P.; Maciejka, A.; Topolewska, A.; Stepnowski, P.; Haliński, Ł. P. Isolation of Atropine and Scopolamine from Plant Material Using Liquid-Liquid Extraction and EXtrelut® Columns. J. Chromatogr. B Analyt. Technol. Biomed. Life Sci. 2017, 1043, 202–208. https://doi.org/10/f9wvv8.


Calculate the radius of an Ag atom The question I have been given is:

  Silver atoms in a metallic lattice only fill up $88\,\%$ of the space ($12\,\%$ is empty). The density of silver is $10.5\ \mathrm{g\cdot cm^{-3}}$. Assuming that silver atoms are hard spheres ($V=\tfrac43\cdot\pi\cdot r^3$, when $r$ is atomic radius), what is the radius of a silver atom? Give the answer in units of $10^{-12}$ meters.

The atomic mass of $\ce{Ag}$ is 107.8682.

My solution:

$$V=0.88\times V$$

$$V=\frac{0.88\times10.5\times6.022\times10^{23}}{107.8682}=5.158\times10^{22}\ \mathrm{cm^3}$$

$$V=\frac43\cdot\pi\cdot r^3 \Rightarrow r=\left(\frac34\cdot\frac V\pi\right)^{1/3}$$
Then I switched to the $10^{12}$ meters, the result was $4.953\times10^{17}$ and it is not correct. What am I doing wrong?

If you had included the units in your calculation, you would have noticed why your equation is not correct.

Molar mass $M$ is defined as
$$M=\frac mn\tag1$$
where $m$ is mass and $n$ is amount of substance.
Since the Avogadro constant $N_\mathrm A$ is
$$N_\mathrm A=\frac Nn\tag2$$
where $N$ is the number of particles, the mass $m$ of one atom $(N=1)$ is
$$m=\frac M{N_\mathrm A}\tag3$$

Density $\rho$ is defined as
$$\rho=\frac mV\tag4$$
where $V$ is volume.
Thus, the volume of a sample is
$$V=\frac m\rho\tag5$$
Using Equation $\text{(3)}$, the volume $V$ can be calculated for a single atom:
$$V=\frac M{N_\mathrm A\cdot\rho}\tag6$$

Assuming that a fraction of $88\,\%$ of the volume $V$ is filled with a hard sphere, the volume $V_\text{sphere}$ of the sphere is
$$\begin{align}
V_\text{sphere}&=0.88\times V\tag7\\[6pt]
&=0.88\times\frac M{N_\mathrm A\cdot\rho}\tag8
\end{align}$$

Since the volume of a sphere is
$$V_\text{sphere}=\frac43\pi r^3\tag9$$
where $r$ is the radius of the sphere, the radius $r$ is
$$\begin{align}
r&=\sqrt[3]{\frac{3V_\text{sphere}}{4\pi}}\tag{10}\\[6pt]
&=\sqrt[3]{\frac{3\times0.88\times M}{4\pi\cdot N_\mathrm A\cdot\rho}}\tag{11}\\[6pt]
&=\sqrt[3]{\frac{3\times0.88\times 107.86820\ \mathrm{g\ mol^{-1}}}{4\pi\times
6.02214076\times10^{23}\ \mathrm{mol^{-1}} \times 10.5\ \mathrm{g\ cm^{-3}}}}\\[6pt]
&=1.53\times10^{-8}\ \mathrm{cm}\\[6pt]
&=1.53\times10^{-10}\ \mathrm m\\[6pt]
&=153\times10^{-12}\ \mathrm m\\
\end{align}$$


What program could I use to create a protein model from scratch? I would like to create protein models such as the ones in the Protein Data Bank. I have got GROMACS but I don't know if it can be used to model macromolecules.

You have several options to generate smaller peptides using the peptide builders in

Avogadro or
PyMol


How can I find out the concentration of chlorine in an acid bath? A professor I have gave the class this question, can anyone help? This is the first part of the question. (Also, my professor is from Bangladesh...so yea...)

It's the year 2395, and the world is in chaos. Two nations have conquered the world, USA and Bangladesh. You have been hired by Bangladesh's government to monitor acid baths used to remove oxidation on aerospace metallic material desperately needed for the new   F-95 fighter jet engine.

The acid bath is made up of three components:

25-30% HNO3
4-5% HF
Balance H2O

Unfortunately, a spy has broken into your facility and added large amounts of Chlorine to the acid bath. As the lead chemist, it is your job to determine how much Chlorine was added to the acid bath.

During sample collection, it is uncertain what any of the concentrations are, however once a sample is obtained you may manipulate it with any method you wish. Devise a method to determine the concentration of chlorine in the solution.

The addition of chlorine to the bath will produce hydrochloric acid.  If the chloride ion concentration in the bath sample could be determined, then you could back-calculate to the amount of hydrochloric acid added.  The standard method to determine the amount of chloride ion in an aqueous sample is to precipitate it by adding silver nitrate solution followed by collecting the precipitate by filtration and then drying and weighing it.  You could use that approach here because silver nitrate and silver fluoride (which would be created from silver nitrate and the hydrofluoric acid) are soluble* in water, so only the silver chloride would precipitate if silver nitrate were added to your acid bath sample.  Here's a link to an actual gravimetric procedure for determining chloride ion content.

Note: silver fluoride is soluble to the extent of 1.8 kg/L, therefor silver fluoride should not precipitate upon addition of silver nitrate to your chlorine-contaminated sample because the hydrofluoric acid content of the bath was only 4-5%.


Thermodynamics: Is Isolated system possible? I have just started learning thermodynamics. I have learned that The system which is not able to exchange energy and matter on its own with surrounding is called isolated system.

I wanted to know that practically is it possible to make 100% isolated system?

There are no 100% isolated systems - even the universe itself may not be truly isolated (depending on how you define universe).

However, in practice, many systems are "isolated enough" that we can treat them as conserving mass and energy, especially over short time frames or small distances.

For example, we often treat a well-insulated and closed reaction vessel as isolated when doing thermodynamic calculations, and this approximation works very well for fast reactions.

Another example where this works is an adiabatic process. Many processes can be successfully modeled as adiabatic even though they happen in poorly-insulated containers - for example, the rapid expansion or compression of a gas.

Another example is the use of small sub-domains of a non-isolated system, which over short time scales behave as isolated systems themselves. For example, in finite element methods (FEM), a system is broken up into very small sub-domains. Each one acts as a small system in which mass and energy can transfer across the boundaries. Locally, an element is an open system, and over long time scales the entire system is open, but on short time scales, the sets of domains that share boundaries are treated as isolated. If energy leaves one domain, it enters the neighboring domain - it doesn't completely leave the system. In effect, you can model an open system as a large collection of isolated systems, with the outermost having open boundaries.

As you can see, even though the assumption of an isolated system is not really true, you can find many, many places where it is "true enough" that we can use it practically. In fact, it would be very hard to do anything in thermodynamics without it!


Half-Reactions of Hydroxide The following is the equation I have:

$$\ce{Ca(s) + 2H2O(l) -> Ca(OH)2 v  + H2 ^}$$

I am tasked to write the half-reaction equations for both the oxidation and/or reduction process. So, currently I have:
$$\ce{ Ca(s) -> Ca^{2+}(aq) + 2e-}\quad\text{(oxidation)}$$

but I'm not sure how do you write the reduction equation for the water and hydroxide.

Any help is appreciated :)

P.S.: I have no idea how to superscript or subscript in the forum.

Alright, so I've worked out my answer (thanks to @user55119). The Reduction step was:

  $$\ce{ 2H2O_{(l)} + 2e^- -> 2OH^-_{(aq)} + H_2_{(g)}}\quad\text{(Reduction)}$$


Residual stress in polymer coatings A paper talks about the residual stress when coating polymers.
It says that residual stress is the sum of the "heat stress" (due to the difference in thermal expansion coefficients) and the "shrinkage stress" (due to the solvent evaporation and changes in volume). It then states that heat stress usually increases with temperature while shrinkage stress decreases with temperature. Therefore the shape of the graph of residual stress vs temperature would be a parabola with the minimum changing depending on the polymer/system system.
It then goes to find the starting point of the shrinkage stress and states that it is dependant on the glass transition temperature.
From what I understand, that would happen when the glass transition temperature is between room temperature and the drying temperature.
However, if the glass transition temperature is below both room temperature and drying temperature, what would happen? How this would affect the shrinkage stress? Or did I get the whole connection between glass transition temperature and shrinkage stress wrong?
Thank you

EDIT:

I read the paper linked in your comment, and see that my initial guess was incorrect (see below) - for this system, it is as you say - for temperatures above the bulk pure polymer glass transition temperature, you would not expect to see shrinkage stress due to evaporation.

END EDIT

I don't have access to that paper - so this is just an educated guess, but usually physicists use the glass transition temperature to characterize polymer behavior as a function of temperature because things roughly (very roughly) scale with $T_g$. In other words, if I know how the behavior of one linear homopolymer depends on temperature, and I know a certain response curve scales with $T_g$, then I can scale the temperature curve of another linear homopolymer using the $T_g$ and sometimes get accurate results.

They could be linking it to shrinkage stress because $T_g$ also characterizes the dynamic behavior of polymers (how they "move" in response to thermal fluctuations and external stress). This is essentially the principle of time-temperature superpostion - or the idea that as temperature increases, polymers move the same way, only faster. So heating up is like fast-forwarding time, and cooling down is like slowing time down.

The glass transition temperature is a discontinuity in this behavior - at a certain low temperature, the polymers freeze in place (on large length scales). This phenomenon is not well understood, but we have found empirically that the temperature at which it occurs can be used to scale response curves in certain situations, and since it is fairly easy to measure, it gets used a lot in these types of models.

So in summary, and again this is an educated guess, I would imagine that the glass transition temperature would be used in this model regardless of the temperature you were actually operating at - it is probably a fundamental parameter of the curve that is used to put the "reality" (the type of polymer) into the model.


Why, in titration, is universal indicator not commonly used? My teacher said that we always used indicators such as methyl red, methyl orange, litmus, phenolphthalein, etc. rather than universal indicator. I don't understand why, though, would not universal indicator serve the same function, but for essentially all titrations rather than select few?

The thing with universal indicators is that you have 6 different colors or so differentiating certain pH-ranges.

Now if you want to titrate to a certain pH-Value, it is easier to have an indicator like phenolphtalein that changes from colorless to a color at a certain point (phenolphtalein being colorless from 0 to 8 and red for $< 0$).

So you can titrate more accuratly to a certain pH-Value with specialized indicators. On the other hand you can't see where on the pH scale you are.

With an universal indicator you can get a feeling for where you are approximatly, but titrating to a certain value is very difficult.


Is LiOH a weaker base than NaOH? Is $\ce{LiOH}$ a weaker base than $\ce{NaOH}$?

Note: I'm not interested in "why," but rather what the "real" $\mathrm{p}K_
\mathrm{b}$ values are.

$$
\begin{array}{lcc}
\text{Data for Alkali Metal Hydroxides} \\
\hline
\text{Cation} & \mathrm{p}K_\mathrm{b} & \text{Solubility} & \text{Solubility} \\
\text{} & \text{} & \pu{g/100 ml} \text{ at } \pu{20 ^\circ C} & \pu{moles/L} \text{ at } \pu{20 ^\circ C} \\
\hline
\ce{Li} & -0.36 & 12.8 & 5.34\\
\ce{Na} & 0.2 & 111 & 27.75\\
\ce{K} & 0.5 & 112 & 20.0  \\
\ce{Rb} & ? & 180 & 17.6\\
\ce{Cs} & -1.76 & 200? & 13.3?\\
\hline
\end{array}
$$

The $\mathrm{p}K_\mathrm{b}$ data is from Wikipedia ($\ce{Li}$, $\ce{Na}$, $\ce{K}$, and $\ce{Cs}$). Wikipedia doesn't have a value for $\ce{Rb}$ and I couldn't find one.

The solubility data is from the Wikipedia Solubility Table. Wikipedia doesn't have a solubility value for $\ce{CsOH}$ at $\pu{20 ^\circ C}$  and I couldn't find one, About $\pu{200 g}$ would be a reasonable guess, which would be $\pu{13.3M}$.

One point is that the Wikipedia article for lithium hydroxide says "It ... is the weakest base among the alkali metal hydroxides."

Another point is an earlier question "Why is $\ce{LiOH}$ a weaker base than $\ce{NaOH}$". User Lighthart references a paper "Ions in water: The microscopic structure of concentrated hydroxide solutions" by Imberti et al. but the abstract at least doesn't seem to address the strength of the alkali hydroxides directly (I don't have access to the paper itself since it is behind a paywall).

Is this one of the Bertrand Chord paradox problems where it depends on how we define "stronger"?

With a $\mathrm{p}K_\mathrm{b}$ of -0.36, $\ce{LiOH}$ would seem to be a stronger base than either $\ce{NaOH}$ or $\ce{KOH}$.
Certainly more moles of $\ce{NaOH}$ dissolve in a liter of solution at $\pu{20 ^\circ C}$ than any other alkali hydroxide.

EDIT 3/22/2016 11:31 pm

Found a different table of $\mathrm{pK}_\mathrm{b}$ data for the alkali hydroxides: Source-2

EDIT 3/23/2016 12:12 am
Source-3 CRC Handbook of Chemistry and Physics, 92nd edition, "Dissociation Constants of Inorganic Acids and Bases. Only $\ce{Li}$ $(\mathrm{p}K_\mathrm{a}= 13.8)$ and $\ce{Na}$ $(\mathrm{p}K_\mathrm{a}=14.8)$ are listed as $\mathrm{p}K_\mathrm{a}$ of the cations. It is the same data as in this pdf file which is from another edition.

$$
\begin{array}{lcc}
\hline
\text{Cation} & \text{Wikipedia} & \text{Source-2} & \text{CRC}\\
\hline
\ce{Li} & -0.36 & 0.18 & 0.2 \\
\ce{Na} & 0.2 & -0.56 & -0.8 \\
\ce{K} & 0.5 & -1.1 & \\
\ce{Rb} & ? & -1.4 & \\
\ce{Cs} & -1.76 & -1.76 & \\
\hline
\end{array}
$$

In Proton Affinities of the Alkali Hydroxides J. Am. Chem. Soc., 1969, 91, pp 2810–2811, it is shown that gas phase affinities in $\pu{kcal/mol}$ are:

  $\ce{LiOH}$: 240.7
  $\ce{NaOH}$: 247.6
  $\ce{KOH}$: 262.6
  $\ce{CsOH}$: 269.2

So, yes, $\ce{LiOH}$ is weaker, at least in the gas phase.

In aqueous solution, ACIDITY FUNCTIONS FOR STRONGLY BASIC SOLUTIONS Chemical Reviews volume 66, pages 119-131:

  the order of basicity for the same molarity of aqueous solution, $\ce{LiOH < NaOH < KOH}$.

For example, for a $\pu{1M}$ solution, the acidity function (H) is:

  $\ce{LiOH}$: 13.48
  $\ce{NaOH}$: 14.01, 14.16 (two different measurement techniques)
  $\ce{KOH}$: 14.17

So in both the gas and aqueous phases, $\ce{LiOH}$ is a weaker base than $\ce{NaOH}$.


Chemicals for electrochromic reaction ? I would like to detect the electric current going through the water solution using the colour reaction in this water solution. What chemicals can be used for this solution to make it electrochromic.

Easiest way is with an iodide, e.g. $\ce{NaI}$ or $\ce{KI}$, and starch.

An old trick was to sprinkle a bit of $\ce{KI}$ on a (damp) slice of potato, stick a couple of wires into the potato and attach a small cell to the wires. Iodine is liberated at the positive terminal and it reacts with starch in the potato to make a deep blue color.

There are more suggestions from Flinn Scientific, such as using an indicator to show acid or alkali liberated at electrodes.


What does the Clausius Inequality mean in thermodynamics? (entropy) $$\delta S\ge\frac{\delta q}T$$

I don't understand how a quantity (the change in entropy in this case) can be greater than the very thing that defines it ($\mathrm{\frac{\delta q}{T}}$).

Also what is the significance of this inequality?

Your definition of entropy is incorrect. The significance of the Clausius inequality is that it shows that the definition of entropy, i.e. $\mathrm{\delta S=\cfrac{\delta q_{rev}}{T}}$ (note that entropy change is defined for a reversible process) is consistent with observed reality: the entropy of an isolated system does not decrease spontaneously.

We know that $$\mathrm{\delta w \geq \delta w_{rev} \Rightarrow \delta w-\delta w_{rev}\geq 0}$$

and according to the First Law of Thermodynamics

$$\mathrm{\delta U = \delta q + \delta w=\delta q_{rev}+\delta w_{rev}}$$

Therefore $$\mathrm{\delta q_{rev}-\delta q=\delta w-\delta w_{rev} \geq 0}$$

Now using the definition of entropy yields

$$\mathrm{\delta S=\cfrac{\delta q_{rev}}{T} \geq \cfrac{\delta q}{T}}$$

Taking $\mathrm{\delta q=0}$ (since we have an isolated system no heat can flow into it), produces

$$\mathrm{\cfrac{\delta q_{rev}} {T} \geq 0 \Rightarrow \delta S \geq 0}$$


What is the significance of Sphingolipids in human? I want to know the significance of sphingolipids in human. I have learnt that sphingomyelin is the most significant type of sphingolipid in human. Also that the sphingomyelin serves as a structural constituent of myelin sheath of nerve fibres.

Is that all? I need to clarify this. Please help me. Thank you

There are multiple known functions of sphingolipids and their metabolites in humans, although the roll played in the structure of myelin might be the most important and well understood.

However, according to this Wikipedia article, sphingolipids also play a roll in cell recognition and signalling, among other specialized functions. Additionally, some of it's metabolites have been shown to perform numerous functions. From the above article:

  Recently, simple sphingolipid metabolites, such as ceramide and sphingosine-1-phosphate, have been shown to be important mediators in the signaling cascades involved in apoptosis, proliferation, stress responses, necrosis, inflammation, autophagy, senescence, and differentiation.

The bottom line answer to your question is that there are many other functions of sphingolipids in humans, particularly if you include it's metabolites.  The article cited above is well referenced and should serve as a good starting point to investigate (or just enumerate) the other functions of sphingolipids in humans.


Complete transfer of Heat into Work During an isothermal reversible expansion of a gas, all heat absorbed by the gas from the reserviour is being used up to push the piston up. But doesn't this contradict something like the second law?
When I push a brick on a rough table, the motion will eventually stop and some heat will be generated due to friction, but If I get a bunsen burner and supply the same amount of heat, it brick won't move back to where it started.

In an isothermal reversible process the total entropy change of piston and energy reservoirs (surroundings) must be zero, $\Delta S_T = 0$. Thus the entropy of the expanding gas may increase but only if it is compensated for by a decrease in entropy of the reservoirs. So no contradiction of second law. (The heat absorbed from the surroundings is $T\Delta S $ and if this is positive more work can be done by a chemical reaction than the heat of reaction.)

In the second question you will realise that you can turn all the work into heat but not all the heat into work. Its entropy again.


Negative charges in acid-catalyzed reactions Whenever my professor draws a mechanism in an acid-catalyzed reaction, he always says you can't form negative charges. Why is that?

The acidic environment would not allow creation of anions but anions of strong acids, that are too weak bases to be significantly protonated.


How to determine the amount of barium in a sample of contaminated water? Here is the assignment I have:

  You are working for Matrix Pro - a company involved in analytical assessment and method development for the clean-up of contaminated waters.

  You received water samples from a potential client. The samples may contain the following ions: $\ce{H+, Ba^2+, OH^─}$

  The potential client requested a qualitative analysis of the sample as well as the development of a treatment technique that will bring the pH value to neutral and eliminate the barium ions. For comparison another water sample (considered to be non-contaminated) was also sent for similar qualitative analysis.

  Assignment

  Develop an experimental procedure to comply with the request of the potential   client.
  Draw a flow chart.
  Tabulate all your observations and data.

What I'm not sure about is finding the amount of barium present in the unknown sample so I can determine the amount of precipitating reagent to add.  I just want to know what type of analysis to do.  Chromatography? Spectrophotometry? Something else?

Because the client is ultimately interested in removing the barium by precipitation, gravimetric analysis is likely a suitable solution. This can be done in a few hours in a lab with basic equipment like a filter flask, a drying oven, an analytical balance and the appropriate reagents.

You can take advantage of the fact that barium sulfate is very insoluble in water by adding some saturated solution of sodium sulfate to a known volume of a sample of the water. Then just filter out the barium sulfate onto some pre-weighed filter paper, dry, weigh, and calculate the mass of barium (from the measured mass and the molecular weight of barium sulfate) per volume of water. You should repeat the analysis to be sure you added enough sodium sulfate solution to precipitate all of the barium.

If there is little or no precipitate found, you have a couple options. Firstly, you can optimize the gravimetric analysis by using large volumes of sample with a small filter and a good analytical balance. Also be sure to pre-filter the sample prior to adding the sodium sulfate in case there are suspended particulates, which would get measured just as if they were barium.

If a gravimetric analysis does not provide the detection limit required by the client, the next best options are atomic absorption spectroscopy (AAS) and inductively coupled plasma - atomic emission spectroscopy (ICP-AES). As these analyses can be done very cheaply (~\$10-$30 per sample) you may actually want to do this to begin with. The down side of outsourcing is that the turnaround time could be weeks, or the price could quadruple if you need results fast.


Etching around corners In etching of mono silicon, why does the etching process around the corners differ from the etching process around the flat edges? What brings about this difference and is there some way that MEMS engineers could quantify the process around the corners?

What happens is that convex (outside) corners get severely undercut in wet etching. On a <100> cut wafer, the edges leading up to the corner reveal <111> planes which are resistant to etching, but right at the corner, where the two planes intersect, etching reveals a <110> plane, which is etched much more quickly. The difference in etch rates has to do with the number of "dangling" bonds—faces with fewer Si-Si bonds at the surface etch faster. The reason why concave corners don't undercut, despite having the same <111> plane intersection is that there are no dangling bonds in an inside corner so it doesn't etch any faster than the <111> planes.

It should also be noted that adding a small amount of surfactant to the etchant can dramatically decrease undercutting by adsorption to the corners.

This (fortunately open-access) article explains the phenomenon quite nicely.


What defines a discharged battery? In the case of an AA Alkaline battery, what actually happens inside the battery once it is fully discharged? What components of the battery change as we use it?

what actually happens inside the battery once it is fully discharged?

Alkaline batteries use the exchange of electrons from zinc to manganese dioxide to produce electricity

$$\begin{array}{rcl}\\
\ce{Zn (s) +2 OH- (aq)}& \ce{->}& \ce{ZnO (s) + H2O (l) + 2 e-}\qquad \text{(anode)}\\
\text{(high potential anode)}\quad \ce{2e-}&\ce{->[load][\ ]}&\ce{2e-}\quad\text{(low potential cathode)}\\
\ce{2MnO2 (s)+ 2e- + H2O (l)} &\ce{->}&\ce{Mn2O3 (s) + OH- (aq)}\qquad \quad\text{(cathode)}\\
\end{array}$$

typically a battery voltage would decline according to:

$$\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF}\ln Q$$

but in alkaline batteries all of the products and reactants are solids, which without going into heavy detail of the thermochemistry means that $Q = 1$ thus $\frac{RT}{nF}\ln Q = 0$ which means cell potential remains fairly constant ($\mathscr{E} = \mathscr{E}^{\circ}$). But this is only in theory as we can see alkaline batteries do lose voltage:

This is partly because the internal resistance of the cell will increase with time meaning that for a given external load the total load will increase with time reducing current and apparent voltage. But even with no load voltage drops, the the zinc oxide will dissolve in alkaline medium meaning the reactions depart from ideal equations:

$$\ce{ZnO (s)  ->[+ H2O (l)] Zn(OH)2 ->[+2OH- (aq)] [Zn(OH)4]^2-}$$

this is not to say that $\mathscr{E} = \mathscr{E}^{\circ} - \frac{RT}{nF}\ln Q$ is an unreliable equation, but it does not account well for the complexity of the system and does not apply well to alkaline batteries. Additionally it cannot explain the vertical slope of the discharge curve. What can explain the vertical slope is that once all of either zinc or manganese dioxide is consumed, there is no more matterial to react and the battery is considered fully discharged.


Combining Bond Enthalpies and Enthalpies of Formation Can I mix formation enthalpies and bond enthalpies in the same calculation?

I think the answer is no, since they are relative measurements, and possibly relative to different things, but I haven't seen it stated explicitly anywhere.

You can combine this data. You just need to find a sensible reaction cycle, and you have to make sure the temperatures (usually 298 K) match.

To give a simple example, consider $\ce{CO2(g)}$.

What is the $\ce{CO}$ bond enthalpy $\mathrm{\epsilon_{C=O}}$in $\ce{CO2(g)}$?

It is one half of the enthalpy for the reaction

$$\ce{ CO2(g)->C(g) + 2O(g)}~~~~~\mathrm{\Delta  H^\circ = 2 \epsilon_{C=O}}$$

The enthalpy of formation of $\ce{CO2(g)}$ is equal to the enthalpy of combustion of graphite:

$$\ce{C(s) + O2(g) -> CO2(g)} ~~~~~~\mathrm{\Delta _f H^\circ = -393.51}\pu{kJ mol^{-1}}$$

The heat of formation of $\ce{C(g)}$ is:

$$\ce{C(s) -> C(g) }~~~~~~\mathrm{\Delta _f H^\circ = 716.68}\pu{kJ mol^{-1}}$$

We also need a value of the $\ce{OO}$ bond enthalpy in $\ce{O2}$:

$$\ce{O_2(g) -> 2O(g) }~~~~\mathrm{\Delta  H^\circ = \epsilon_{O=O} =  498.3}\pu{kJ mol^{-1}}$$

Adding things up we obtain $\mathrm{\epsilon_{C=O}}$:

$$\mathrm{\epsilon_{C=O}} = \frac12(393.51+716.68+498.3)\pu{kJ mol^{-1}= 804.2 kJ mol^{-1}}$$


Disproportionation with aqueous NaOH Which of the following undergoes disproportionation with $\ce{NaOH}$?

  $\ce{SiO2}$ (silicon dioxide)
  $\ce{CO2}$ (carbon dioxide)
  $\ce{SO2}$ (sulfur dioxide)
  $\ce{SO3}$ (sulfur trioxide)
  $\ce{NO2}$ (nitrogen dioxide)

From the above we can conclude that $\ce{SiO2}$, $\ce{CO2}$ and $\ce{SO3}$ do not undergo disproportionation as they are all in the maximum oxidation states. $\ce{NO2}$ is in the +4 oxidation states so it disproportionates in to +3 and +5 upon addition of aqueous $\ce{NaOH}$. But what about $\ce{SO2}$? It can also disproportionates because it is in +4 oxidation no. But the answer states only $\ce{NO2}$. Can anyone please explain the reason?

Disproportionation is a redox reaction. Since
$$\Delta G = -zF\Delta E$$
you may compare the redox potentials for both half-reactions in order to determine whether disproportionation occurs spontaneously ($\Delta G < 0$).

Note that nitrogen dioxide ($\ce{NO2}$) exists in equilibrium with dinitrogen tetroxide ($\ce{N2O4}$)
$$\ce{2 NO2 <=> N2O4}$$
and literature values for redox potentials are often given for $\ce{N2O4}$ (values at $\mathrm{pH}=14$):

$$\begin{alignat}{2}
\ce{N2O4 + 2 e- \;&<=> 2 NO2- }\quad &&E_1^\circ = +0.867\ \mathrm{V}\\
\ce{2 NO3- + 2 H2O + 2 e- \;&<=> N2O4 + 4 OH-}\quad &&E_2^\circ = -0.86\ \mathrm{V}
\end{alignat}$$

The potential $E_1^\circ = +0.867\ \mathrm{V}$ for the reduction of $\ce{N2O4}$ is larger than the potential $E_2^\circ = -0.86\ \mathrm{V}$ for the oxidation of $\ce{N2O4}$; therefore, disproportionation should occur.

In equlibrium:

$\begin{align}
E_1^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},1}}{a_{\text{red},1}} &= E_2^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}}\\
E_1^\circ - E_2^\circ &= \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}} - \frac{0.059\ \mathrm{V}}{z}\log \frac{a_{\text{ox},1}}{a_{\text{red},1}} \\
\left(E_1^\circ - E_2^\circ\right) \cdot \frac{z}{0.059\ \mathrm{V}} &= \log \frac{a_{\text{ox},2} \cdot a_{\text{red},1}}{a_{\text{red},2} \cdot a_{\text{ox},1}} \\
\left(0.867\ \mathrm{V} + 0.86\ \mathrm{V}\right) \cdot \frac{2}{0.059\ \mathrm{V}} &= \log\frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{\left[\ce{N2O4}\right] \cdot \left[\ce{N2O4}\right]} \\
{10^{58.5}} &= \frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{{\left[\ce{N2O4}\right]}^2} \\
{10^{29}} &= \frac{\left[\ce{NO3-}\right] \cdot \left[\ce{NO2-}\right]}{{\left[\ce{N2O4}\right]}}
\end{align} $

Therefore, $\ce{N2O4}$ is unstable with respect to disproportionation:
$$\ce{N2O4 + 2 OH- -> NO2- + NO3- + H2O}$$


Is sodium "also known to turn purple and combust (like lithium)," when exposed to water? I understand that indicator solutions can change color depending on pH, but the following statements in the Phys.org article Super cheap earth element to advance new battery tech to the industry has me baffled.

Most of today's batteries are made up of rare lithium mined from the mountains of South America. If the world depletes this source, then battery production could stagnate.
Sodium is a very cheap and earth-abundant alternative to using lithium-ion batteries that is also known to turn purple and combust if exposed to water—even just water in the air.

Later it says:

They minimized sodium's exposure to the moisture that would make it combust by making the sodium powder in a glovebox filled with the gas argon. To make the powder, they used an ultrasound—the same tool used for monitoring the development a fetus — to melt sodium chunks into a milky purple liquid. The liquid then cooled into a powder, and was suspended in a hexane solution to evenly disperse the powder particles.

Phys.org news items are generally of very high quality, so I'm wondering if this is just a mistake (perhaps a reference to the use of a pH indicator) or if there is some reaction between sodium, water and perhaps air that involves a purple color.
The "purple" statement is echoed on Purdue University's web page Super cheap earth element to advance new battery tech to the industry as well.
So far I've only found the video Sodium gas... omg it's purple!.
The paper in question is Ultrasound-assisted synthesis of sodium powder as electrode additive to improve cycling performance of sodium-ion batteries. Tang,  Kye and Pol, Journal of Power Sources, 396, pp 476-482, https://doi.org/10.1016/j.jpowsour.2018.06.067

I found the paper in Researchgate.

I'm going to leave this tentative answer in place but hopefully I or someone else will be able to investigate further and post a better answer.

These are particles of molten sodium (>98C) dispersed in mineral oil. The linked references suggest that the ultrasonic dispersal technique can be used to make quantum dots, so if these particles are similar size, then the color could be some quantum mechanical effect for the conduction electrons in the liquid sodium nanoparticles.

Since that is a size-dependent QM effect, one would not see it in experiments on bulk sodium.

This doesn't completely rule out Rayleigh scattering effects, but since the color changes as the particles cool (and presumably solidify) then that suggests it's the properties of the particles themselves, rather than there dispersal in oil.

  Sodium Powder Preparation: Sodium powder was prepared via ul- trasonic heating, melting, and subsequent fragmentation of solid sodium chunks in an organic solvent. A similar technique was utilized in previous studies involving Sn nanoparticles 32 and dispersed sodium 33. Ultrasonication was generated from a Sonics VCX500 probe equipped with a stepped microtip. In a typical synthesis, 15 mL of mineral oil was first degassed by ultrasonication for 15min inside a 150 mL cone-shape sonochemical reactions vessel; then, about 100 mg of fresh metallic sodium chunks were added into the oil. A continuous argon flow was fed to the vessel to maintain an air/moisture-free environment. The Na/oil mixture was then exposed to ultrasonic irradiation at 40% amplitude. Pulsed ultrasonic irradiation was applied in 4 cycles of 59s-sonication and 30s-rest intervals. The formation of sodium powder took place in two stages. At the first stage, solid Na chunks were melted into liquid sodium when the mixture temperature rose above the melting temperature of sodium (mp = 98 °C). The temperature rise is a result of ultrasound induced heating to the oil. At the second stage, the molten Na particles were dispersed into mineral oil via ultrasound to form a homogenous mixture with purple coloration. Once cooled, the solution color changed to grey. A simplified synthesis method is illustrated in Fig. 1a (A video of the sodium powder synthesis can be found in Movie S1 in the supporting information). The mixture was then washed with anhydrous hexane and centrifuged three times to obtain a dispersion of Na powder in clean hexane. All washing procedure except the centrifugation were completed inside an Ar-filled glovebox to minimize air exposure.


Sulfuric acid fumes from cleaning shower drain This morning a workman came by to my newly rented property to fix a few things, among which a slightly blocked bathtub drain.

He let some mild water run continuously and he poured down a chemical, and it wasn't before I saw a huge vapor coming out and an intense boiling-like sound, I realized something was wrong. Basically he let down an entire bottle of >95% sulfuric acid down the drain just like that, without even wearing safety goggles or gloves. A pungent odor emanated through the whole flat and I had to open every window (he hadn't advised to open them beforehand).

Now, I don't have much experience with chemicals of this sort, so here are my questions:

Did I inhale toxic fumes?
What should I do to make the bathroom/shower usable again? I noticed that the mirror and several water tiles that are not in the close proximity of the drain but are on the way to the nearest window opening, show condensation streaks, such as those left by drops of water that run down. They were definitely not there before. Is that condensed sulfuric acid vapor? Do I need to clean with a sodium bicarbonate solution?
Needless to say, I also had towels, toothbrushes, and other items in the room, do I need to take any precaution before using them (cleaning, rinsing, etc?).

In short, how do I clean up this mess?

If the workman did use concentrated sulphuric acid (which is not the best choice for drain cleaning) then the best solution if to flush the drain with large quantities of water and wash the surfaces with water.

Sulphuric acid is not toxic but is very corrosive when concentrated. But it is highly water soluble and not harmful when it is dilute enough. So dilution with water will solve whatever problems it seems to cause. If it got splashed around, wash everything thoroughly and there will be no further problem. If you use enough water then it won't even be necessary to neutralise the result as any remaining acid will be swept through the drain and highly diluted by the flow of water.


How does 3dCarbon CLIP printer work? Recently there has been announced a new 3d printing technology which uses UV light, from lasers or a projector, to harden slices of epoxy resin so that the printed object rises out of a pool of epoxy.

There is some information about an oxygen boundary and UV light, can someone explain the chemistry of the process?
https://www.youtube.com/watch?v=UpH1zhUQY0c

Actually, the technology isn't that new. It seems however, that the setups are becoming more affordable in the hobby market.

The technology is know as stereolithography and is a form of light-induced polymerization (UV curing).

The liquid monomers used have a substituent that can be activated using UV light. Out of practical considerations, UV is more convenient: polymerization of the resin will not occur under natural light.

In principle, the procedure is comparable to UV curing in dental repair. There, a lamp that irradiates the whole area is used: you want a fast hardening all over the tooth.

In 3D printing, this is obviously different! The light source is a laser with a narrow beam: you only want to irradiate a very small volume element at a time.

You don't want a rapid polymerization either: only the volume element irradiated should react!

The laser beam is focussed at the boundary of the resin and different volume elements are addressed by moving the focus in the $xy$ plane.

Rather than moving the (heavy) laser itself, the beam is "moved" using a mirror optic. This allows for a very fast adressing of reaction points within the plane.

UPDATE
In the example linked, in the question and a comment, however, the setup is different from laser-based STL printers.

Irradiation from the bottom of of the resin tank through an oxygen-permeable membrane indicates indeed that the UV curing proceeds in the presence of oxygen. Anyway, if the light source would be just a lamp, one would nevertheless achieve homogenous curing of the whole layer.
But the manufacturer doesn't use the term lamp or light source. Instead, they write about a projector and that's probably what they do:

They are not illuminating the whole layer, but are projecting masks for every slice of the object to be printed.


How to calculate the heat capacity of a calorimeter? I need to find the heat lost of an unknown metal dropped into a calorimeter with $70~\mathrm{g}$ $\ce{H2O}$.
The initial temperature and final temperature of the $70~\mathrm{g}$ $\ce{H2O}$ and the calorimeter are $21~^\circ\mathrm{C}$ and $34~^\circ\mathrm{C}$. I already know that the heat gained by the water is $3807.44~\mathrm{J}$.

The metal's starting temperature and mass are $100~^\circ\mathrm{C}$ and $180.45~\mathrm{g}$, but that didn't help me much as I don't know the $C_{sp}$ of the unknown metal.

What I don't know is how to find the capacity of the calorimeter, any thoughts?

I already tried a number of solutions that didn't work:

$C_{sp} \times \ce{H2O} = q(\mathrm{calorimeter})$
$q(\ce{H2O})/m(\ce{H2O}) \times \Delta T = q(\mathrm{calorimeter})$
$q(\ce{H2O}) = q(\mathrm{calorimeter})$
I knew the unknown metal was one of three metals (lead, aluminum, copper),     so I tried finding $q(\mathrm{metal})$ using the 3 heat specific heat capacities, but it didn't work out since the data provided were only approximations.

This is impossible to answer. Usually you have to assume that when no calorimeter heat capacity is given, then it negligible (i.e. you only use the heat capacity of the 70g $\ce{H_2O}$). You know the temperature drop of the metal and the energy increase of the water, combine both to obtain the heat capacity of the metal.


At the eutectic point, what is the effect on the melting point of adding impurities? At the eutectic point, what is the effect on the melting point of adding impurities?

  Increase the melting point
  Decrease the melting point
  No effect on melting point

I know adding impurities decreases the melting point of a compound in comparison to the pure compound. But, the eutectic point is the lowest temperature where the new compound (the compound + impurity) melts. I'm indecisive between 1 and 3.

Lets state my assumption that we can represent the free energies of the mixtures in fairly regular thermodynamics of phases ways (as I will get to here quickly). I will also assume fairly limited solid solubilities (think something like the Au-Si system).

OK, just take a system of A and B. The Gibbs free energy of a binary mixture (either solid or liquid) is:

$G(T) = x_{A}G_{A}^{sol/liq} + x_{B}G_{B}^{sol/liq} + x_{A}x_{B}H_{AB}^{sol/liq} + RT(x_{A}\ln x_A + x_{B} \ln x_{B})$, with $x_{A} + x_{B} = 1$, $G_{A,B}^{sol/liq}$ being the Gibbs free energy of pure A (or B) in the solid (or liquid) at that temperature, and $H_{AB}^{sol/liq}$ being the enthalpy of mixing of A and B.

For a eutectic, the free energy of the liquid drops below the tie-line connecting the free energies of the solid phases, and it clearly does this in such a way that the eutectic point has a composition that is intermediate to the compositions defining the solid tie-line. In a perfectly symmetric and regular system, this puts the eutectic at $x_{A} = x_{B} = 0.5$, but at least keep in mind that liquid composition has reasonably large values of $x_{A}$ or $x_{B}$, certainly larger than the minority composition in either solid end-member.

Now, we add a third element. The Gibbs free energy formula becomes longer, since it now takes in to account the binary and ternary interactions.  So, one gets:

$G(T) = x_{A}G_{A}^{sol/liq} + x_{B}G_{B}^{sol/liq} + x_{C}G_{C}^{sol/liq} + x_{A}x_{B}H_{AB}^{sol/liq} + x_{A}x_{C}H_{AC}^{sol/liq} +  x_{B}x_{C}H_{BC}^{sol/liq} + x_{A}x_{B}x_{C}H_{ABC}^{sol/liq} +  RT(x_{A}\ln x_A + x_{B} \ln x_{B} + x_{C} \ln x_{C})$, with $x_{A}+ x_{B} + x_{C} = 1$. This is pretty ugly, but we just added a little bit of C, so $x_{C}$ is small.

Now, back to the assumptions on the system being considered. For the liquid, $x_{A}$ and $x_{B}$ are reasonably large ($\sim 0.5$). In solid A, $x_{A}$ is near 1, $x_{B}$ is small(ish), and $x_{C}$ is really small.  In solid B, $x_{B}$ is near 1, $x_A$ is small(ish), and $x_{C}$ is small. Really small.

What has really changed? Well, for $x_{C}$ being really small, you can throw out all terms linear in C. The only thing left (!) is the entropy of mixing terms, where adding a little C results in a change of  $RT x_{C} \ln x_{C}$. But, this gets applied to both solids (A and B), and also to the liquid. All the free energy curves shift down by the same amount on an absolute scale. By geometry this means that the tie line connecting the two solid phases and the liquid at the eutectic point does not shift to first order in $x_{C}$.

The eutectic point will not shift.


How is benzene prepared from acetylene? The mechanism of this polymerization reaction which finds its utility in the preparation of benzene in a red hot iron tube piques my curiosity.

I've read this Wikipedia article on cyclotrimerization, but it's too high a level for a high-schooler to comprehend. Can you please explain this to me in simpler way?

Walter Reppe developed a process for the "safe" trimerization of acetylene to benzene and cyclooctatetraene. Although completely superseded by more modern alternatives, this process was used before/during WW2 because starting from only coal, calcium carbonate and electricity one could synthetise calcium acetylide and therefore acetylene, having no need for oil derivatives, scarce during war.

The mechanism (in simplified terms - hold onto the stones!) is that on the surface of the catalyst (iron, copper or nickel depending on the variation of the process) there would be formation of transient "clusters" of three or four molecules of Fe/Cu/Ni-acetilyde. That close arrangement facilitates one of the electrons of the triple bond of the acetylide "attacking" one of the carbon atoms of one of the nearest acetylide neighbours (again, simplifications). That molecule with an excess of one electron would in turn atack its other in vicinity and finally this one would attack the first one (that became deficient in the first step).

Since the aromatic structure of benzene is rather stable compared to ethyne, and the product is 1/3 (or 1/4) the number of moles of reagents, the formation of product is favored under pressure.

Some sources:
Walter Reppe and the chemistry he developed: https://en.wikipedia.org/wiki/Walter_Reppe
Trimerization: https://en.wikipedia.org/wiki/Trimer_(chemistry)


Is there any substance that's a 0-0-0 on the NFPA diamond? Is there any substance that's a 0-0-0 on the NFPA diamond? Basically a perfectly safe chemical.

Water!

You can find plenty of others if you perform this search on Wikipedia (but then, go check the MSDS for the individual compounds before doing anything serious with them).