Which has a greater entropy? Which of the following has greater entropy?

a.  $\ce{CH3COOH~ (aq)}$

or

b.  $\ce{CH3COO^{-}~ (aq) + H+~(aq)}$

They are both in the same state, and have the same number of particles.. so how do I figure this out? My instinct says a, because the products will be more "random", but I'm not completely sure.. Any ideas?

In the gas phase I would suspect that the entropies would be equal as both systems have the same number of translational, rotational and vibrational modes (18).

However you specifically ask about the liquid phase system, since you show that both "a" and "b" are solvated in aqueous media. In this case, the ionized molecule (b) has charges on both ions and this will cause increased ordering of the solvent molecules surrounding the ions, more so than neutral molecule "a".  This increased ordering of the system in case "b" will reduce the entropy of the system.  Consequently system "a" will have greater entropy.


Is there any notion of pH out of solution? For example, could one define a $\mathrm{pH}$ for pure acetic acid? It's a weak acid in water, but if someone handed you $1~\mathrm L$ of pure acetic acid, what would its $\mathrm{pH}$ be?

The IUPAC definition of pH is:

The quantity pH is defined in terms of the activity of hydrogen(1+) ions (hydrogen ions) in solution:
$pH = − lg [a(H^+)] = − lg [m(H^+) γ_m (H^+) / m^⦵]$
where a$(\ce{H+})$ is the activity of hydrogen ion (hydrogen 1+) in aqueous solution, $(\ce{H+})$(aq), $γ_m(\ce{H+})$ is the activity coefficient of $\ce{H+}$(aq) (molality basis) at molality m$(\ce{H+})$, and m$^⦵ = 1$ mol kg $^{−1}$ is the standard molality.

So since the definition specifically refers to "aqueous solution", pH is undefined unless an aqueous solution is being considered.
There would be a p[H+] in pure acetic acid based upon the self dissociation of acetic acid, where [H+] is the concentration of hydrogen ions in acetic acid solution, but this is not pH according to the IUPAC definition.
According to Acid-Base Equilibria in Glacial Acetic Acid. III. Acidity Scale. Potentiometric Determination of Dissociation Constants of Acids, Bases and Salts J. Am. Chem. Soc., 1956, vol. 78, pages 2974–2979:

the autoprotolysis constant of acetic acid is calculated to be $3.5 \times 10^{-15}$ (pK = 14.45)

Therefore p[H+] = 7.2 for pure acetic acid.
This value is not a measure of acidity, but simply the concentration of solvated hydrogen ions in pure acetic acid.


Where to buy/How to build a Devanathan-Stachurski cell? I'm preparing to do some hydrogen permeation tests to calculate the diffusivity of some iron samples. I would like to do this using the Devanathan-Stachurski cell, which consists of two cells, one where the hydrogen is charged and the other where it is oxidized (each containing an electrolyte solution), with the sample in the middle. The current produced by the diffusing hydrogen can be used to calculate diffusivity.

Does anyone have experience using this cell and know where to purchase one? So far I've only been able to find one company that sells it, but I would like to know about more options. If anyone has built it from scratch, can you provide some specifications on dimensions and materials used and how it worked for you?

I would recommend you to do a research to find scientific articles on using DS cells. In this case if you have any specific questions you can try to contact authors directly, hopefully they will reply and be able to help you.

You can start from this article (Mike J. Danielson, Corrosion Science 2002 44 (4), pp. 829-840) http://www.sciencedirect.com/science/article/pii/S0010938X01001032.
The author made his own DS cells as described follow


Boiling point of alkynes from n = 2 to 10 From a video on organic chemistry, I found this table that shows the boiling point temperatures of alkynes $\ce{C_{n}H_{2n-2}}$ with $\,n=2...10$.

$$\begin{array}{|c|c|c|c|}
\hline n & t_\mathrm{boil}/^\circ\mathrm{C} & m & T_\mathrm{boil}/\mathrm{K}\\
\hline 2 & -83.6 & 26 & 189.6\\
\hline 3 & -23.2 & 40 & 250.0\\
\hline 4 & 8.1 & 54 & 281.3\\
\hline 5 & 39.7 & 68 & 312.9\\
\hline 6 & 71 & 82 & 344.2\\
\hline 7 & 99.7 & 96 & 372.9\\
\hline 8 & 125.2 & 110 & 398.4\\
\hline 9 & 150.8 & 124 & 424.0\\
\hline 10 & 174 & 138 & 447.2\\
\hline\end{array}$$

Here $\,m=14n-2\,$ is the molecular mass and $\,T_\mathrm{boil}\,$ is the boiling point temperature converted to kelvins. Surprisingly, when I plot $T_\mathrm{boil}$ v.s. $m$, I find a power law $T_\mathrm{boil}\propto \sqrt{m}$. The melting point temperatures were also shown in the video but were not so regular. Can someone explain why the boiling point satisfies such a simple power law?

I would like to try to give a more detailed answer, but this is all I can give right now. I am fairly confident this is what is going on.
The dominant intermolecular force which determines the boiling point is dispersion. Because dispersion is the result of instantaneous dipoles, the strength of the interaction is proportional to the "surface area" of the molecule. There is no well-defined measure of the surface area of a molecule, but the most common way to describe this would be by using a van der Waals' radius for the atoms/groups.
The important point, however, is that when we add another carbon with its hydrogens, to a hydrocarbon, the mass increases linearly, while the surface area increases proportional to the square of the van der Waals' radius.
So, if we believe that dispersion forces are truly proportional to surface area and hence truly proportional to the boiling temperature, then a change of mass proportional to $\sqrt m$ should make a linear change in the surface area of the molecule, which should make a linear change the boiling temperature.
Notice, however, that this argument should be exactly the same for a linear alkane, and not just an alkyne. I made this same plot for the first fifteen n-alkanes, and the trend is very linear.

As to why there isn't a similar trend for the melting points, melting temperature is much more sensitive to symmetry than the boiling point, so how the molecules stack will make quite a difference. I would expect that hydrocarbons do not form very good crystals, so the trends may be highly dependent on the specific molecule.
Also, notice that the more branched these molecules are the less this argument should apply to the boiling point because each groups added at branching points will add less surface area than for the straight alkane because some of the surface area will be "redundant" so to speak.
Great observation! I hope my answer is right. I'm sure this has been noticed before, so it would be interesting if someone could find references.

Another way of thinking about this which I like is this:
Imagine that we can add mass to the carbon chain continuously rather than in lumps of atoms. Because adding mass also adds length to the chain, adding mass adds surface area. From this perspective, the relationship between $\sqrt m$ and surface area is the same as the relationship between length and surface area. This follows because these molecules are shaped a lot like cylinders.
So, the only jump left is from surface area to boiling point, which is already understood to be related through dispersion forces.


Verifying data acquired from a molecular dynamics simulation This question is about verifying a system simulated with MD. I am currently building a model of a two phase gas-liquid system, where I initially have $\ce{N2}$, $\ce{CO2}$, $\ce{Ar}$, $\ce{He}$ and $\ce{CH4}$ in the gaseous phase and $\ce{Na+}$ with $\ce{Cl-}$ in an aqueous phase, which is modeled with TIP3P water. The most abundant component of the gases is $\ce{N2}$, which comprises over 95% of the gas.

I have a certain, non-ambient $p/T$ setting with a relatively high pressure (30-40 kbar). My idea is to study partition coefficients using molecular dynamics in multiple different $p/T$ settings. I'm currently using OPLS/AA force field which models intramolecular potentials with harmonic potentials and intermolecular potentials through Ewald-summation and LJ-612 -potential.

Currently I have just equilibrated the system, so that $p,T,V$ and $E$ do not fluctuate much. Before moving onwards with this system, I would like to verify that it produces reasonable data which is on par with other classical computational simulations.

I have noticed that there are not many papers which have focused on a system similar to mine, and I'm probably not going to be able to find an exact match in which to compare to. So, while I've been looking through the literature, I've looked at the following parameters to get a feeling of how well the system is behaving:

Radial Distribution Functions (RDF).
Partition Coefficients of the gases.
Long term behaviour of the system (no odd behaviour such as crystal forming etc.).
Densities of phases ($\rho_{\text{gas}} < \rho_{\text{liquid}}$)

So far these have all been quite reasonable. Are there any other standard methods and/or values that I should be looking at? I am obviously constantly reading through literature to find other results for comparison.

With the simple molecules in your system, using the Gibbs Ensemble Monte Carlo would be the way to go. Each phase gets its own simulation box, and particles are swapped back and forth until equilibrium is reached.

Cassandra, DL_MONTE and other programs can do this easily for you. Once equilibrium is reached you can sample each box in the NVT or NPT ensemble at that composition.


Why are there two contradicting formulas for Enthalpy (Delta H): 1) Delta H = Bonds broken - Bonds Formed

2) Delta H = $H_{products} - H_{reactants}$

In the first formula, bonds break in the reactants and form in the products, so its basically saying "bonds energy of reactants" minus the"bond energy of products", which is the complete opposite of what the second formula gives. How so? And when you know to use each formula?

This is my take on this

  1) Delta H = Bonds broken - Bonds Formed

  2) Delta H = $H_{products}−H_{reactants}$

Equation 2 is pretty straightforward, (change in enthalpy is the enthalpy of the products minus the enthalphy of the reactants).

Equation 1 is a bit less straightforward. Let's define the system as where you are having the reaction, and the surroundings as everything else. When you break bonds, you take energy from the surroundings into the system. Then, when bonds are formed, energy is released from the system into the surroundings. Therefore, if there is more energy released in the bonds formed than there is energy gained in the bonds broken, the system loses energy. How I see it is, you are looking from the surrounding's perspective at equation 1 and from the system's perspective from equation 2, that's why you got two opposite statements (from your POV).


How can there be an enthalpy of formation for gaseous water? Under standard conditions( 1 bar and 298.15 K), water will be in the liquid state. So how can you find the enthalpy of formation for gaseous water at standard conditions(as found in the back of all chemistry textbooks)?

The only explanation that comes into my mind is that,

$$\ce{H2 (g) + \frac{1}{2} O2 (g) -> H2O (g) <-> H2O(l)}$$
 And
$$\Delta H_f^\circ(\ce{H2O (g)}) = \Delta H_f^\circ (\ce{H2O (l)}) + \Delta H_\mathrm{vap} (\pu{25^\circ C}) = -285.8 + 44.99 = \pu{-241.8 kJ/mol}$$ almost as tabulated depending the sources, here the calculation is perform with tabulated values found in the 87th Edition of the Handbook of Chemistry and Physics.

This is a hypothetical state but actually, vapour water is in equilibrium with liquid water even under the standard state. It's just that this equilibrium is very unlikely to go in direction of formation of the gas water. But if you find an ingenious way to vaporize water under these conditions then the heat of formation of gaseous water will be the same as I just calculated.

Also in the 87th Edition of the Handbook of Chemistry and Physics, it is said (5-4):

  The standard state of a pure gaseous substance is that of
  the substance as a (hypothetical) ideal gas at the standard
  state pressure.

Although I've never figured that out and I can't really tell you about the need for this value, it exists. Probably you'd like to have this kind of values when performing a calculation considering a particular state but as far as I used to do, I've never used this one.


Chiral centers in albuterol-like compounds My question is apart from the C in the straight chain with $\ce{-OH}$ group, should $\ce{N}$ be counted as chiral center since it also has tetrahedral geometry and all different substituents?

There are more such examples.

You can surely count that nitrogen atom as a "chiral center" because it has four different groups attached to it.

However, the net optical activity due to that chiral center is zero because the two optical isomers from that nitrogen atom are rapidly interconverting with each other ("Amine inversion"). They always exist as a racemic mixture, and hence are optically inactive.


Why Frenkel defect does not impact the density of the solid? The following text is from Chemistry Part I - Textbook for Class XII (NCERT), chapter "The Solid State", page 23, topic "1.9 Imperfections in Solids", sub topic "1.9.1 Types of Point Defects - (iii) Frenkel Defect":

This defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location.
Frenkel defect is also called dislocation defect. It does not change the density of the solid.

(Emphasis mine)
Earlier I've learnt that a vacancy defect results in decrease in the density of the substance, and an interstitial defect increases the density of the substance. I could understand, due to the presence of vacancy or interstitial defects, the density varies locally or in other words density of the substance is not uniform throughout.
According to the quoted text, Frenkel defect creates a vacancy defect at one location and an interstitial defect at a different location (not at the same location), so I think the density will again have local variations and so the density changes. But it is given Frenkel defect doesn't alter the density. What is the reason for this fact? Why does the same logic fail here?
The Wikipedia article on "Frenkel defect" gives the following statement:

Frenkel defect does not have any impact on the density of the solid as it involves only the migration of the ions within the crystal, thus preserving both the volume as well as mass.[citation needed]
Source : Frenkel Defect - Effect on density

If density is plainly defined as the ratio of mass and volume it doesn't seem to change, but in reality I think there will be variations within the crystal having this defect. But why is this fact not specified in many sources?

The sources may not explicitly say this, but I suspect that they consider mass density to be a macroscopic property, to be measured from at least milligrams and cubic millimetres.

If the volume considered is small enough, then yes, one would see local variations of the density. However, if you make the volume even smaller, you would see huge variations in the density as additional nuclei are included. (Similar statements will be true for Frenkel defect concentrations.) On this level, the definition of mass density simply breaks down and is no longer very useful.


Understanding Beta Decay Reading Wikipedia on "Beta Decay" they list the equation of Carbon-14 decaying into N-14.
see equation https://en.wikipedia.org/wiki/Beta_decay

I presume the 14 refers to total protons and neutrons.  The 6 in C, I believe means six protons.  In N, I believe the 7 refers to protons. Where did the seventh proton come from the decay of  C-14 into N-14?  I understand neutrons can decay into protons, but the math doesn't work for me.

For all radioactive decay (or other nuclear reaction) of a nuclide into other nuclides, the atomic number $Z$ and mass number $A$ need to be conserved.

$$\ce{_{Z_1}^{A_1}X -> _{Z_2}^{A_2}Q + _{Z_3}^{A_3}R }$$
$$Z_1 = Z_2 + Z_3$$
$$A_1 = A_2 + A_3$$

Additionally, the charges must be conserved. If $\ce{_{Z_3}^{A_3}R }$ is an alpha particle $\ce{_2^2\alpha^{2+}}$ or $\ce{_2^2He^{2+}}$, then:

$$\ce{_{Z}^{A}X -> _{Z-2}^{A-2}Q^{2-} +_2^2He^{2+}}$$

Neutrons decay into protons and electrons (beta particles). Both particles are needed for the math to work. And in order for the math to work, you need to know that the atomic number $Z$ is $0$ for a neutron (no protons), and $1$ for a proton (1 proton, or the $\ce{H+}$ ion). The electron does not have an atomic number. In fact, it is not that $Z$ and charges must be conserved separately. The sum of atomic number and charge is conserved.

$$\ce{_0^1n -> _1^1p+ + _{0}^0e-}$$

For the decay of $\ce{^{14}_6C}$ into $\ce{_7^{14}N}$, we first have a neutron decay into a proton and an electron, which is captured by the newly formed $\ce{_7^{14}N^+}$. The nitrogen nuclide is initially a cation because carbon only had six electrons, and nitrogen needs seven.

$$\ce{_6^{14}C -> _7^{14}N+ +e- -> _7^{14}N}$$


N-N bond length in Dinitrogen Tetroxide The $\ce{N-N}$ bond length in $\ce{N2O4}$ is $\pu{1.64 Å}$ which is much longer than in the case of Hydrazine $\ce{N2H4}$ $\pu{(1.47 Å)}$. This means that the bond is quite weak. Another unusual case I see that $\ce{N-N}$ bond length in $\ce{N2O3}$ ( the blue asymmetric form ) is $\pu{1.86 Å}$ !!
That is very weak for a Nitrogen single bond. Is there a satisfactory explanation for the same? And amazingly molecules with such long bonds tend to have a nonplanar structure due to bond rotation but $\ce{N2O4}$ is planar.
I found a similar question on N2O4 but the answer to the planarity question wasn't so satisfactory. Also, I wanted the extend upon the weak bond question.

Images are from Wikipedia.
The first obvious thing that you can see is that N2O4 is planar but N2H4 is not. And the primary reason starts here.
N2O4 can be viewed as two nitrogen dioxide molecules joined together. If you draw a Lewis structure of NO2, you will recognize two things. Firstly, it has one unpaired electron. So imagine when two NO2 molecules come near together, they will share their two unpaired electrons (like how radicals react that you might be familiar with in organic chemistry) to form a single sigma bond. This is the very bond you are asking about.
And secondly, it is sp2 hybridized at the nitrogen atom. Thus, when you join the two NO2 molecules together, taking the relative orientation of the sp2 orbitals and how they should overlap to form a sigma bond in mind, you can see why N2O4 is planar. Because the N=O bonds (I write the "=" symbol for double bond for simplicity, it should be about 1.5 order due to resonance) heavily repulse each other when they are in the planar position, it weakens the N-N bond. Indeed this bond is so weak that it is homolytically dissociated at near below room temperature or above, creating basis for the popular equilibria between NO2 and N2O4.
The situation is even worse for N2O3. N2O3 can be viewed as formed from a NO (itself has an unpaired electron and a sp2 nitrogen) and a NO2 molecule, with the exact mechanism as above. And guess what? N2O3 has a lone electron pair on the nitrogen (in place of the missing oxygen compared to N2O4). Lone pairs repulse much, much more heavily than the bonding pairs do, and this further weakens the N2O3 N-N bond. This is illustrated by the easily disproportionation of N2O3 into NO and NO2.
Then why N2H4's bond is stronger? Because the nitrogen atoms are sp3 hybridized in this case (note that oxygen is divalent but hydrogen is only monovalent), thus they form a tetrahedral electronic structure. Thus they can comfortably adopt a skew conformation to minimize the repulsion. Either N2O3 or N2O4 could achieve this because the sp2 hybridization confines them to planar geometry.
Using Pauling's formula with the 1.14 A triple bond in N2 as a reference:
D(n) = D(m) - 0.6log(n/m)
With D(n, m) is the length of the bond with order n, m respectively.
The N-N bond order in N2O4, N2O3 and N2H4 can be estimated to be 0.44, 0.19 and 0.85 respectively.


What happens to the colour change of an equilibrium reaction when one of the reactants begins to form a complex? (specific problem included) I'm working on an equilibrium lab/quiz for my chemistry class and I came across this reaction:

$\ce{Fe^{+3}\ \text{(pale yellow)} + SCN- <=> FeSCN^{+2}}\ \mathrm{(red)}$

Then the lab said a stressor was added: $\ce{Na2HPO4}$ was added to the equilibrium reaction -- which formed a complex with some of the $\ce{Fe^{3+}}$ ions.

This stressor changed the solution's color from an orange to a pale-yellow as seen in the image below:

(left: before stressor ($\ce{Na2HPO4}$) was added; Right: after stressor ($\ce{Na2HPO4}$) was added)

I was asked to explain why that after $\ce{Na2HPO4}$ is added (which actively forms a complex with the iron ion ($\ce{Fe^{3+}}$)) the system/solution turns increasingly pale-yellow.

Basically I want to know what is happening chemically when $\ce{Na2HPO4}$ is added, and more specifically why their is the resultant color change from orange to pale-yellow.

My basic confusion: wouldn't the formation of a complex between $\ce{HPO4^{2-}}$ and $\ce{Na2HPO4}$ actively remove the $\ce{Fe^{3+}}$ ions responsible for the pale-yellow color of the solution? Making the solution less rather than more yellow?

Thanks!

Going by the right hand picture, what was formed there is the very poorly soluble and lightly yellow $\ce{FePO4.2H2O}$. A precipitate and clear supernatant liquid can clearly be seen.

If so, the reaction is:

$$\ce{FeSCN^{2+}(aq) + HPO4^{-}(aq) + 3 H2O(l) \to FePO4.2H2O(s) + SCN-(aq) + H3O+(aq)}$$

This explains why on adding the $\ce{Na2HPO4}$ the red colour of the $\ce{FeSCN^{2+}}$ complex disappears and a yellowish, flocculant precipitate appears. This is not however, as the OP suspected, a complex. It's merely an almost insoluble salt.

Some info on $\ce{FePO4}$.

A true complex would have been formed by the addition of $\ce{CN-}$, which would have led to the formation of $\ce{Fe(CN)6^{3-}}$, which has a much higher formation constant ($K_f$) than the thiocyanate complex. It's also weakly coloured and would not form a precipiate. The reaction would have been:

$$\ce{FeSCN^{2+}(aq) + 6 CN-(aq) \to Fe(CN)6^{3-}(aq) + SCN-(aq)}$$

The solution would have evolved from bordeaux to weakly amber/red.


Will a strong acid still dissociate "completely" even in the presence of a stronger acid? For example, nitric acid or sulfuric acid in an equal-molar solution of perchloric acid. Won't perchloric protonate a significant concentration of nitrate ions?

Suppose you have nitric acid (about Ka=20) and perchloric acid (Ka > 100,000).

$\frac{[A-][H+]}{[HA]} = K_a$

In 1M nitric acid alone, let x be [$\ce{NO3-}$], that is dissociated nitric acid

$\frac{x^2}{(1-x)} = 20$

x = 0.954M

the nitric acid would be about 95% dissociated.

If the solution is 1M nitric and 1M perchloric, with all the perchloric acid dissociating, the nitric acid equilibrium would be:

$\frac{x(1 + x)}{(1-x)} = 20$

x= 0.913

the nitric acid would be about 91% dissociated.

For 2M nitric acid alone:

$\frac{x^2}{(2-x)} = 20$

x= 1.83M

or 92% dissociated.

For 2M nitric and 2M perchloric:

$\frac{x(2 + x)}{(2-x)} = 20$

x = 1.69M

or 84% dissociated.

So yes it can make a difference and it is more apparent at higher concentration.


Which side of the C-O-C bond breaks in esters during hydrolysis in basic medium? When an ester undergoes hydrolysis which side of the $\ce{C-O-C}$ breaks for instance in the following example:

I believe the first is correct but is it a rule that the salt of a carboxylic acid is formed (and then of course in the presence of $\ce{-OH}$ an alcohol also forms)?
Basically which carbon does the original O of the $\ce{C-O-C}$ stay with?

To answer this,Think about how an ester is formed.

In the formation of an ester, wherin you react an alcohol with an acid in presence of conc.$\ce{H2SO4}$

$\ce{RCOOH + R'OH -> RCOOR' + H2O}$

Now what we have found by replacing the oxygen with an isotope of oxygen is that

$\ce{RCOO'H + R''OH -> RCOOR'' + H2O'}$

What this reveals is that the acid loses an $\ce{OH- group}$ and the alcohol loses an $\ce{H+}$

So summing up, O atom stays with the carbon which is not attached to the $C=O$ group.


Can gas be made to block radiation better? Can any gas block radiation? I ask this because I would like to know if the properties of any gas element would chemically react differently with radiation from adding an electrical current.

This picture shows an Earth magnet's field deflecting plasma naturally in a vacuum.

Does any gas extend the electromagnetic shielding properties of a magnetic field?

I am interested if the lack or composition of an atmosphere's chemical properties would inhibit or complement the deflection and absorbing of cosmic radiation when coupled with a magnetic field.

Can contained pure $\ce{O2 O1 O3}$ be cycled back with electricity to replenish $\ce{O3}$ broken down by radiation while giving the gas an electromagnetic field?

Your question seems to be about the ozone layer, but shows some misunderstanding.

First, ozone, $\ce{O3}$, absorbs some "radiation", specifically electromagnetic radiation, e.g. visible light or ultraviolet light (UV), as do many other gases. This absorption is not the same for all wavelength ("colors"), but in the UV region peaks about 250 nm, which is a good thing for us, because plants and animals have not evolved good protection from that light, and the small amount of $\ce{O3}$ in the upper atmosphere is enough to block most of that UV. Sulfur dioxide, $\ce{SO2}$, released in volcanic eruptions, is also able to absorb 250 nm UV. Molecular oxygen, $\ce{O2}$, absorbs "vacuum UV", 10-100 nm, which otherwise would be a problem.

Second, Earth is continually bombarded by another type of "radiation", particles such as cosmic rays and solar energetic particles. Rather than electromagnetic energy, these are made of particulate matter such as protons or nitrogen nuclei. All gases, in fact all matter, help block these particles. Magnetic fields such as that of the Earth help divert them around the planet, too, but there is some question whether a magnetic field is essential to blocking particulate radiation. Here at the bottom of the atmosphere there is a column of air weighing about a kilogram over each square centimeter, which is sufficient to block primary cosmic rays, though a showers of weaker secondary particles do reach the surface. There is no advantage to a conductive Faraday cage to protect the Earth.

Third, all gases conduct electricity when ionized. The majority of air is nitrogen, oxygen and argon, which produce various colors in an aurora. Though magnificent to watch, aurorae are simply evidence of energetic particles trapped in a magnetic field, ionizing the upper atmosphere. This layer of the atmosphere is called the ionosphere, and is responsible for long-distance radio communication. It's height varies from day to night and with solar storms.


Can activated charcoal be used to filter bacteria out of air? As the title says: Can activated charcoal be used to filter bacteria out of air and or gas?

Activated charcoal has been shown to be able adsorb certain species of bacteria, but the effectiveness varies between species [1] (which is potentially useful for medical applications), and has not been tested for many, so it is unlikely that activated charcoal could be used to adsorb all the bacteria from the air or a gas.

Activated charcoal general usage is in the adsorption of toxins.

[1] Naka, Katsumi, et al. "Adsorption effect of activated charcoal on enterohemorrhagic Escherichia coli." Journal of Veterinary Medical Science 63.3 (2001): 281-285.


What home products are reactive with glass? Glass seems to be one of the most stable solid substances - which chemicals, particularly home products that react with glass?

Some rust and stain removers (like Whink) contain hydrofluoric acid* at low concentrations (1-2.3%) Here is the MSDS for Whink: https://www.whink.com/cmssites/ws0811www.whink.com/uploads/Documents/Rust%20Stain%20Remover%20SDS.pdf

Now, at those concentrations it would probably take a while to dissolve silica, but it would be arguably quicker than a $\ce{NaOH}$ cleaning product.

*I do not recommend playing around with $\ce{HF}$ and its products unless you are experienced with them.


How does the electronic configuration for this acid oxide correspond to its structure? The following electronic configuration and drawing of the structure are given:

$\mathrm{(1s)^2 (2s)^2 (2p)^6 (3s)^2 (3p)^6 (3d)^{10} (4s)^2 (4p)^4}$

How does this image correspond with the electronic configuration?

My train of thoughts so far:

Since the configuration has 34 electrons, I assume the element Z is actually selenium (atomic number 34). It could also be gallium since Z has no charge and has five covalent bonds to look like krypton. This is mere speculation since I just don't know how to approach the problem.

I am unsure what kind of bonds are present here. The tetrahedral form, even though it consists of a total of five bonds, indicates hybridization, but I am completely unfamiliar with hybridization outside of carbon-like hydbridization (four bonds). I am guessing one double bond leads to sp$^2$ hybridization.

The compound is $\ce{SeO3}$, and here is my reasoning:

From the electronic configuration given in the question, the central atom is selenium (atomic weight 34).

As you can see from its electronic configuration, the valence shell has 6 electrons and it goes on to form 4 covalent bonds with oxygen, of which 3 are $\sigma$ bonds and 1 is a $\pi$ bond. Therefore, it shares 4 electrons and has a lone pair. Its steric number is 4 and hence its hybridisation is $sp^3$.

With the tetrahedral geometry, the lone pair will occupy the uppermost position (which is shown by a dash) and the three oxygen atoms will occupy the base of the distorted tetrahedron (due to the lone pair and arising from VSEPR theory). Further, there would be a negative charge on two of the oxygen atoms.

The central atom is not gallium since its electronic configuration is $(4s)^2(4p)^1$. And krypton doesn't react with oxygen as far as I know.


Oxidation of Carbon atom in Alkene, Alkyne and Alkanes of functional groups? Nomenclature priorities are in order of the degree of oxidation of the
  carbon atom.

I am not sure if it's advisable to just take it as it is and use it or try to understand the basis of the functional group prioritization. I guess the latter interest me. So looking at the priorities for functional groups,

Alkene has higher priority than Alkyne and followed by Alkane. I started reading some materials about Oxidative Cleavage of Double Bonds and rest of the reactions, which kinda went way over my head.

Can someone explain, how and why the carbon oxidation in Alkynes are lower than Alknenes? Or put in another word, why would Alkene carbon oxidate faster than Alknenes?

Nomenclature priorities are in order of the degree of oxidation of the carbon atom.

This may be true for the heteroatom functional groups, but not for alkenes, alkynes, and alkanes.

Consider that each $\ce{C-H}$ bond contributes $-1$ to the oxidation state and each $\ce{C-C}$ bon regardless of order contributes $0$ to the oxidation state of carbon.

Thus:

hydrocarbon   formula   functional group   oxidation number of C
ethane        CH3CH3    alkane             -3
ethene        CH2CH2    alkene             -2
ethyne        CHCH      alkyne             -1

  Alkynes have a higher oxidation number than alkenes.

Alkenes react better in oxidation reactions than alkynes because alkenes are less oxidized than alkynes.

The organic chemistry textbook that I teach from explains that alkenes have higher nomenclature priority than alkynes because alkenes are more common. I cannot find a more authoritative reason/source.


is dental acrylic dangerous when heated? I work in a research lab where we use small screws to attach onto hard objects and secure it with dental acrylic, to maintain durability. The screws are expensive, so when we are done with the particular project, we use a soldering iron to melt through the acrylic to get to the screws. I am concerned that when dental acrylic is heated, does it release toxins that can be harmful to my health?

What do you mean by acrylic, if it is PMMA. Then be aware that when it is heated that it cracks with unusual ease into the monomer (MMA) which is a strong irritant.

I have (in a fume hood) taken lumps of PMMA and heated them in a test tube with a bunsen, if you distill the product out and then redistill it you can get very pure methyl methacrylate. I have found that if you heat some of this MMA with some AIBN then it will then reform the PMMA.


Stability comparison of 2-methylpropene and trans-but-2-ene Actually, in a book I found that  2-methylpropene is more stable than trans-but-2-ene. But, my teacher and also I am convinced that trans-but-2-ene is more stable than 2-methylpropene. My reasoning is that 2-methylpropene is less stable due to the steric repulsion of the -CH3 on one sp2 hybridised carbon. Also, 2-methylpropene is more reactive for the electrophilic addition reactions involving carbocation intermediate as a 3° carbocation will form and in case of trans-but-2-ene, a 2° carbocation is formed. Please say if I am correct and if not, please explain the correct comparative analysis.

Since these two compounds are constitutional isomers, their heats of formation may be compared directly. 2-Methylpropene (isobutene) has a heat of formation = -4.0 kcal/mol. Trans 2-butene's heat of formation is -2.7 kcal/mol. Isobutene liberates 1.3 kcal/mol fewer kcal/mol than trans-butene (E-2-butene) upon combustion. The more negative the heat of formation, the more stable. The book gets an A.

Resource: http://webbook.nist.gov/chemistry/


Why are more electronegative atoms better leaving groups? My book says, "halide anions are electronegative and can easily stabilise
negative charge, hence are good leaving groups."
What exactly does "stabilising negative charge" mean?
I have read:  "It is crucial that the leaving group be able to stabilize the additional electron density that results from bond heterolysis"

Where heterolysis means "bond breaking in which the bonding electron pair is split unevenly between the products."
So is this stabilization referring to stabilizing the charge on the  $\ce{Cl-}$ ion in the above example? Does it just mean it is easier for something that can become stable to leave?
Does being more stable equate to being closer to neutral charge, or has it got something to do with steric hindrance ("steric hindrance occurs when the large size of groups within a molecule prevents chemical reactions that are observed") preventing reactions?
Another issue I have with this is why are more electronegative atoms better at stabilizing charge?

You can interpret the term "stabilizing" to mean that some effect leads to a  more energetically favorable state (i.e. lower in energy) relative to your original state. So to answer your first question, if atom A is better at stabilizing negative charge than atom B, then it means that it is more energetically favorable to place a negative charge on atom A and have B remain neutral than the other way around. This could be because placing additional electron density around A is more energetically favorable than placing it on B, or because it is less energetically unfavorable to place more electron density on A than it would be to place it on B.

To answer your second question, more electronegative atoms are generally better at stabilizing negative charge because, by definition, electronegative atoms have a greater tendency to attract electron density. However, it is not only the electronegativity of an atom which determines its leaving group ability, but also its size. If an atom is larger, then it means the electrons will be spread out over a larger volume, which is more energetically favorable. This is one reason why iodide is a better leaving group than bromide, even though bromine is more electronegative than iodine. In addition, iodine is larger and it forms weaker bonds than bromine with other atoms due to poorer overlap of atomic orbitals, meaning it is easier to break iodine bonds (e.g. the bond dissociation energy is 284 kJ/mol for Br$-$CH$_3$ versus 232 kJ/mol for I$-$CH$_3$ at 298 K). Because iodine bonds are easier to break, it is easier for iodine to "leave" the molecule and is thus a better leaving group.

Along this same line of reasoning, fluorine, being very electronegative, can form a very stable anion, but it forms very strong bonds which are difficult to break, and fluoride cannot leave if the fluorine bond doesn't break. That is why fluoride is a poorer leaving group than the other halogens.


Is the density of the solution significantly different from the density of the solvent? I am working on a mathematical model for the dissolution of coffee in water. I need to know some precise details about the effects of dissolution on the density of the mixture of the substances (that is, the density of the solution). The system is isothermal and the pressure is constant. I just want to focus on mass, volume, density, and concentration as the relevant variables.

In the drawing (below), there is a graphical representation of the system in question (in the bottom image, I should have mentioned concentration of water and coffee instead of density). We have a container of water with a single coffee solid and I've drawn in 20 molecules for both the water and the coffee. The volumes are obviously not to scale (it looks like approximately 50% of the coffee has dissolved in the second pic, not 20%!).

Start

So at the beginning, $t = 0$, no coffee has dissolved so far and the solution liquid is simply the water. We have the following information for the density and the volume of the coffee, water, and the solution, respectively

$\rho_c = 1500$ kg m$^{-3}$

$\rho_w = 1000$ kg m$^{-3}$

$\rho_s = \rho_w = 1000$ kg m$^{-3}$

$V_c = 0.1$ m$^{-3}$

$V_w = 0.5$ m$^{-3}$

$V_s = V_w = 0.5$ m$^{-3}$

The solution is just the water at the start.

Finish

At, say, $t = 100$, 20% of the coffee has dissolved and has diffused out evenly into the solution and the system is in equilibrium.

Now the solution contains both water and coffee. I realise that the chemical reaction dissolution and interactions at a molecular level between the coffee and the water have an impact on the volume and density of the solution. I need information on the density of the solution to decide whether the compressible Navier-Stokes or the incompressible Navier-Stokes govern the flow in the solution.

What is the density  and volume of the solution? If experimental measurement is required for this can you instead give me an approximate answer.

I need to know how much of a change in solution density occurs from $t = 0$ to $t = 100$.

This depends a great deal both on the solute and the concentration.

At the concentrations that we normally drink coffee at, you can almost certainly ignore the difference in density of the solution. But ccording to your question you are trying to dissolve 0.1m3 coffee (I assume the minus sign is an error) in 0.5m3 water. That's 100L of coffee in 500L of water, a 100L/600L=16.667% volume/volume solution! A lot of very strong coffee!

As a general rule, the density rho is given as:

rho = (mass component 1 + mass component 2) / (volume component 1 + volume component 2 )

This assumes (correctly) that masses can be added. Strictly speaking, the volume of a solution is not equal to the sum of the volumes of its components, as there is a slight volume change on dissolution, but this is normally insignificant.

The hard part is to know the solid density of coffee, because it cannot be readily measured, as coffee comes in powder form. If you are not using instant coffee there is the problem that the husks of the grains will not dissolve. But in your case, you are told the density of the coffee is 1500kg/m3

So you can easily work out the masses of water and coffee, and then the density of your enormous vat of extra strong coffee.

Hint: it's going to be more than water, but not different enough to make any difference regarding the correct navier stokes model to follow.

EDIT: on re-reading, it seems you are assuming that only 20% of your 0.1m3 of coffee will dissolve: that is 0.02m3 in 0.5m3 of water, an approximate concentration of 4%. That's not quite as absurdly strong as before. I was a little confused because 0.1m3 is 20% of 0.5m3. However I'm not sure how you decided that you would reach equilibrium with 20% of the available coffee dissolved.


Colloids dispersion classification I've been teaching chemistry for 25 years, but now I'm confused.  Is a colloidal dispersion a heterogeneous or homogeneous mixture? I've seen both answers on the web.

Words like "homogeneous" and "heterogeneous" do not have a well-defined meaning.  Different authors use them differently.  The fact is, all matter is made of atoms, and atoms are not homogeneous.  Thus, from a certain point of view, nothing is homogeneous.  Matter is lumpy!

But suppose we don't care about things on the scale of atoms, i.e. angstroms.  Then the question is "what scale do you care about?"  Nanometers?  Hundreds of nanometers?  Chemists are making materials that have lumpiness on more more less all of these scales.

Back before folks invented centrifuges, filtration, and microscopes, bacterial cultures, with cells as small as 1 μm in diameter, probably seemed homogeneous.  But van Leeuwenhoek peered through a microscope and dispelled that notion.


Why are arenes with conjoined benzene rings drawn as they are? Whilst we are taught to represent benzene as below (A) due to its delocalised electrons, however when two benzene rings share carbons (e.g. naphtalene) it seems to be more commonly represented in textbooks as two conjugated rings (B).  Whilst I realise that the result is the same, what is the advantage gained by drawing them like this rather than (C)?  Is it simply a matter of clarity?
A:

B:

C:

Whilst draing those in Marvin I wondered if it was as the delocalisation between the shared carbons isn't shown?  I.e. it looks like the electrons are delocalised across both rings separately rather than as one big ring?

Questions about why we represent things in one way or another are hard to answer in general, because they appeal to one’s sense of taste… de gustibus non est disputandum. Here, I want to state the position and the reasoning of IUPAC on this matter. The relevant publication on this are the IUPAC 2008 Recommandations, titled Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008) [1].

Section GR-6 of the report deals with “Aromatic rings and other types of electron delocalization”. I quote part of it (emphasis is mine):

  GR-6.5 Curves should only be used when delocalization is being represented

  Curves represent delocalization, yet lots of structures have some delocalized elements. Although all such elements could be depicted with curves, there is little gained by doing so. On the contrary, the arbitrary use of curves can draw the viewer’s attention to insignificant portions of a structural diagram, and away from areas that are chemically more important such as an active site or a reactive group. Accordingly, curves should only be used when the delocalization is specifically being highlighted as an important feature of the structure.

  When curves are not used, any alternating configuration of double bonds is acceptable within the further constraints discussed in GR-3.5.

The text then goes further to note:

  It is generally not acceptable to use curves in two adjacent fused rings, since such diagrams are at best ambiguous in terms of the character of the shared fusion bond between the two rings.

Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008), Pure Appl. Chem., 2008 DOI: 10.1351/pac200880020277

PS: hmm, the leader for the IUPAC report works at CambridgeSoft…


What happens to the NMR of an alcohol when D2O is added? What difference you expect in the 1H NMR spectrum of phenol in $\ce{CDCl3}$ after a drop of $\ce{D2O}$ is added to the test tube?

1) Since the $\ce{-OH}$ of the phenol is quite acidic, it undergoes rapid $\ce{H+/D+}$ exchange with the $\ce{D2O}$. My expectation is that the $\ce{-OH}$ signal from the original phenol will get reduced and broadened (maybe even disappear, but I am not sure) because it is replaced with an $\ce{-OD}$ group.

2) Aside from this, will the aromatic regions of the NMR spectrum change at all? I expect that the ortho and para $H$ can slowly turn into $D$ through EAS, but I am not sure if that is possible when only "one drop of $\ce{D2O}$ is added", especially without acid and base present.

My book does not contain an answer key, but I would like hear if my thought process is correct.

Thank you for the question. This is a very basic physical chemistry question but the way it has been asked is interesting.

In order to understand the probability of displacement of a H from any X-H bond (X is any atom), you need to know its pKa. The pKa values are as follows (ref.):

pKa of water = 14
pKa of OH (phenol) = 9.95
pKa of CH (benzene) = 43

If the pKa of a bond is $x$, it means that for one $\ce{X^- + H^+}$, there has to be $10^x$ no of $\ce{X-H}$ compound. Thus, it clear that for one dissociation of OH bond of phenol, ~$10^{10}$ phenol molecules are required, while for one dissociated CH bond of benzene, $10^{43}$ molecules of benzene is reuqired. To put it in perspective, in 1 mole of phenol, there will be ~$10^{13}$ dissociated OH bonds but only ~$10^{-20}$ dissociated CH bonds.

This hereby implies that dissociation of CH bond and exchanging with $\ce{D_2O}$ is highly improbable, but dissociation of OH bond and its exchange with $\ce{D_2O}$ is highly probable. That's why you get the OH peak to be lowered due to exchange but no similar phenomenon occurs for CH bonds in the system.

Please reply in comments if you have doubts.

Greetings from India. जय हिन्द. Jai Hind.


How can ionic bonding in lanthanides occur without valence orbitals available for overlap? I've recently been taught that the 4f orbitals in lanthanides are "core-like", supposedly meaning they have radius smaller than the 4d orbitals, therefore they are not available on the outside of the atom to form covalent bonds with ligands. The lecturer then went on to say that, therefore, the bonding of lanthanides to ligands is entirely ionic.

It is my understanding that a bond is deemed ionic when there is a large electronegativity difference between the two atoms, i.e. the bonding molecular orbital is near-identical in energy to the atomic orbital of the more electronegative atom, as shown below for sodium fluoride:

However, if this is the case, how can lanthanides form any bonds if their valence orbitals are not available? Surely orbital overlap with the ligand is still required to form a molecular orbital, even if the resulting orbital is ionic in nature?

Lanthanide atoms also have $5d$ and $6s$ valence orbitals available for bonding, and these can overlap effectively with ligands.  Whether this overlap leads to a bonding orbital encompassing both metal and ligand (covalent bond) or transfer if the electrons to a much low-energy ligand orbital (ionic bond), a bond is formed either way.  The bond may not be entirely ionic but is entirely (or nearly so) derived from the $5d$ and $6s$ orbitals of the lanthanide atom instead of the $4f$ orbitals.
Lanthanides do tend to form complexes with high coordination numbers, exceeding the coordination number of six commonly seen with $d$-block transition metal complexes.  From here:

Lanthanides share many similar characteristics. These characteristics include the following:
...
Adoption of coordination numbers greater than 6 (usually 8-9) in compounds
...
A preference for more electronegative elements (such as O or F) binding
...

The coordination number thereby exceeds the number of separate covalent bonds that can be formed from $5d$ and $6s$ valence orbitals alone, so a limited number of covalent bonds must be de-localized over the many linkages.  This would lead to the bonds defaulting towards a more predominantly ionic mode, requiring the ligand orbitals to be low-energy and associated with highly electronegative ligand atoms (also noted in the above excerpt).


Bond dissociation energy order for C−C, N−N, O−O and F−F single bonds Data from Chemistry LibreTexts — Strength of Covalent Bonds, Table 8.8.1: Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K:

$$
\begin{array}{cr}
\hline
\text{Bond} & D/\pu{kJ mol-1} \\
\hline
\ce{C-C} & 346 \\
\ce{N-N} & \approx 167 \\
\ce{O-O} & \approx 142 \\
\ce{F-F} & 155 \\
\hline
\end{array}
$$

I couldn't understand the trend of bond dissociation energy of various single bonds. Why

$$D(\ce{C-C}) > D(\ce{N-N}) > D(\ce{F-F}) > D(\ce{O-O})?$$

I had mugged up the trend the now, and according to me BDE should continuously increase while going from left to right in a period due to decrease in size and increase in ENC. Why this irregular trend?

In chemistry there is nothing like always. You have to see any element characteristics than decide its property. The case you are asking about:

Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., $\ce{N, O, F}$), which are unusually weak compared with single bonds between their larger congeners. It is likely that the $\ce{N–N, O–O,}$ and $\ce{F–F}$ single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms.
$\ce{N-N}gt \ce{F-F}$, in $\ce{F}$ there are greater number of lone pairs than in $\ce{N}$ so it experiences strong repulsion between its electrons that make it vulnerable to be broken.
$\ce{F-F} \gt \ce{O-O}$ it's so because $\ce{F}$ atom is very small that enables strong electrostatic attraction between nucleus and electrons
For further information on 3) you can follow this link.


Mass of salt required to make a 1.2 g/ml solution I would like to make an NaCl and water solution with a density of 1.2 g/ml. How would you calculate the grams of NaCl required to make a 1000 ml solution?

I have used density of mixture = (mass NaCl + mass H2O)/1000 ml and solved for the mass of NaCl. However, the resulting mass does not give a density of 1.2 g/ml in practice (and in fact does not dissolve) so is there another way of calculating this? (at room temperature (~21 °C))

To answer in detail Jan, I didn't make stranged calculus, now I answered more than a year ago I can't tell you if my calculus are the same, but here is the logic.

For any solution (aqueous) we have,
$$m_{solution}=\rho_{solution}\times V \tag1$$

We also have,
$$\rho_{solution}=d\times \rho_{water} \tag2$$

where d is the relative density of the solution and then, merging both $(1)$ and $(2)$,
$$m_{solution}=d \times \rho_{water}\times V \tag3$$

The mass of solute in the solution is a percentage of the masse of the whole solution then, because,

$$n_{solute}=\frac{m_{solute}}{M_{solute}} \tag4$$

We have,

$$n_{solute}=\frac{p\times d \times \rho_{water}\times V}{100\times M_{solute}}\tag5$$

Then find my previous result with $C\times V=n_{solute}$.

Now I think everyone is able to goes from molar to mass concentration then I can stop here. If not ask me.


How to make elemental silicon? I want to extract elemental silicon from pure silicon dioxide powder.

From what I've read, one way to do it is by adding carbon from graphite at very high temperatures, which is difficult.

A better way to do it is to mix with magnesium powder and heat it to about $800 ^\circ \rm C$.

Some places mention the use of a hydrogen atmosphere, is that necessary?

What is the best way to do it, the best temperature? If I use a torch, wouldn't the exhaust gases purge the oxygen and serve as an inert atmosphere?

How do I know the reaction is over?

Making elemental silicon is easy, but not for amateurs. Since it is cheap (~$2/kg) there is little reason to do this yourself.

Industrial processes do indeed mix silica with a source of carbon like coal. These react when very hot to give silicon. But the process usually uses an electric arc furnace and controlled conditions with carefully controlled use of gasses (including extra oxygen late in the process to enhance the amount of impurities that gather in the surface slag). It is unlikely that an amateur can easily reproduce this control.

You might get a result at lower temperatures using magnesium. But you will struggle to do this safely without some control of the atmosphere above the mix as magnesium will burn violently in air (or even in carbon dioxide). You might even need to use argon to blanket the mix as even nitrogen will react with magnesium.

So, basically, this is not something to contemplate as an amateur outside a well-equipped lab.


What proportion of a combustion engine's output energy comes from the air? When trying to compare the efficiency and range of electric engines vs gasoline engines, it seemed at first that gasoline had a higher energy density than batteries. But then I realiced that a significant proportion of the energy of combustion must come from the oxygen in the air, whereas in an electric engine all the energy comes from the batteries, so it's not a fair comparison.

So, assuming a typical air-fuel ratio of 14:1, what proportion of the energy produced by combustion comes from the air and what proportion from the fuel itself?

PS: Please assume usual values of any other variables if you need to.

In combustion the "energy" is really the heat that is released. This heat causes the expansion that moves the piston upwards.

However, the fuel and the oxygen did not enter the engine with energy, and then leave without it. The fuel and the oxygen did not leave at all, different molecules did. The heat that was released was due to the reaction that changed the molecules into different molecules, it was not "brought" and handed off by the molecules. I don't think you can divvy this energy up, since it would not have happened without either the fuel or the oxygen. 0% of the reaction would have happened without the oxygen, but that doesn't mean the oxygen was responsible for 100% of the heat generated.

If the oxygen had entered at one temperature, and left at another, you could calculate the heat the oxygen released or absorbed by its change in enthalpy. However, this does not happen in combustion, the oxygen that is responsible for the reaction does not leave as oxygen.

In fact, if anything, the uncombusted oxygen absorbs heat that could otherwise have been used in the expansion of the piston and this quantity would be measurable. So the only thing I think you can quantify is the amount of work not done because air is used (same goes for all the nitrogen that goes in cold and leaves hot). If pure oxygen was used rather than air, engines would get more expansion work since less heat would be absorbed by  nitrogen.


Is H+ (in an aqueous solution) = H3O+? I was learning about autoionization in science and my science teacher wrote $\ce{H3O+}$ as $\ce{H+}$ — now, my intuition would lead me to assume that:

$$\ce{H3O+} = \ce{H+}$$

But, $\ce{H+}$ is just an atom with a positive charge because its lack of electrons, while $\ce{H3O+}$ is the product of $\ce{H2O}$ nabbing that $\ce{H+}$, right?

So, is saying that $\ce{H3O+} = \ce{H+}$ just for the purpose of simplicity in the sense that we are assuming that in the solution (aqueous, is the context I am dealing with), the positive hydrogen atom (is it more specifically called an ion?) will bond with a water molecule?

P.S. I know that you would call $\ce{OH-}$ hydroxide, and $\ce{H3O+}$ hydronium, so  what would you call $\ce{H+}$ (if it even has a name)?

When dealing with an aqueous solution, you are correct that the $\ce{H+}$ ion is equivalent to $\ce{H3O+}$ for all intents and purposes. Due to the abundance of water in solution, molecules of $\ce{H2O}$ will readily pick up the hydrogen ions, meaning that most of the $\ce{H+}$ in an aqueous solution is actually of the form $\ce{H3O+}$. While this is the case, using $\ce{H+}$ over $\ce{H3O+}$ is more correct in general contexts. If you are using something like ethanol ($\ce{CH3CH2OH}$) as your solvent, then instead of having hydronium ions, you would have $\ce{CH3CH2OH2+}$ as your measure of concentration of acid.

As for your other questions, the $\ce{H+}$ would be considered an ion rather than an atom. This is because it is a charged species. Either $\ce{H+}$ can be referred to as a proton or as a hydrogen ion, since $\ce{H-}$ is too unstable a species to be formed. While some of your $\ce{H+}$ in solution might be deuterium or tritium, most people will understand what you are talking about if you just call the hydrogen ion a proton.


Why is 15,16-dihydropyrene aromatic? Why is 15,16 dihydropyrene aromatic?

Here the middle two carbons are $\ce{sp^3}$ hybridized. Then how is it aromatic? Do we only consider the conditions for aromaticity to apply only at periphery in case of fused rings?

While doing calculations for Huckel's rule, in polycyclic compound only peripheral pi electrons are considered for aromaticity.
So, 7 double bonds in the periphery i.e. 14 pi electrons.
14=4(3)+2
Huckel's rule satisfied. Aromaticity achieved.
Why are only peripheral pi electrons considered?
The reason is that only these elections are participating in delocalization for conjugation. If multiple bonds occur in the centre like in pyrene these central bonds will not be localized. Thus, only peripheral pi electrons are considered in conjugation.

Therefore, first 2 compounds in this image are aromatic and last one is weakly Anti aromatic or Non aromatic.
We can't commit compound to be aromatic on ONLY Huckels rule with this compound must be

Cyclic
Planar
Complete cyclic conjuction
Huckel's rule

Here in 3rd compound,
Due to crowding of Interior Hydrogen compound loses its planer nature. Thus, not called to be Aromatic.
(I have attached 3d structure of cyclotetradeca-1,3,5,7,9,11,13-heptaene (3rd compound)
Please refer it.)


What chemical that when cured chemically cross-links to already vulcanized rubber I'm looking into ways to repair rubber materials and I was wondering if there are any chemicals that when cured will chemically bond to the already vulcanized rubber material.

I know this might involve having to break the strong carbon-carbon or carbon-sulfur bonds to have it re-bond to the chemical, so said chemical itself might have to be volatile.

So, is there a chemical like that?

Various elastomers, such as ethylene propylene diene terpolymer (EPDM), can be cured (crosslinked) using ionizing radiation.

The general radiation effects on polymers are

formation of gas,
cross-linking of polymer chains, and
scission of polymer chains.

However, EPDM rubber contains side chains, ring systems and π electrons, which results in a delocalisation of excitation energy and causes a greater stability of the excited states. Therefore, radiation chemical yields for all dissociative radiation chemical reactions on EPDM are decreased.

The mechanism of cross-linking involves the cleavage of a $\ce{C-H}$ bond on one polymer chain to form a hydrogen atom. The free radical released in the initial radiolysis reaction has enough kinetic energy to abstract a second hydrogen atom from a neighbouring chain to produce molecular hydrogen. Different recombination pathways follow the irreversible loss of hydrogen. Double bond formation results from two events occurring on the same chain; the recombination of two separate macroradicals leads to cross-linking. Cross-linking can also result from the reaction of free radicals with double bonds (comparable to free radical addition polymerization). The radiochemical ageing of EPDM shows the formation of unsaturated bonds (trans-vinylene, vinylidene, and vinyl) as well as the consumption of double bonds by radical addition reactions. Since EPDM already contains double bonds which originate in the diene, the last-mentioned reaction predominates.

The overall effect of cross-linking is that the molecular mass of the polymer steadily increases with radiation dose, leading to branched chains until ultimately a three-dimensional polymer network is formed when each polymer chain is linked to another chain. In contrast, scission is the opposite process to cross-linking in which the rupture of $\ce{C-C}$ bonds occurs. Under irradiation in absence of oxygen, EPDM shows some increased cross-linking with a plateau at roughly $10^5\ \mathrm{Gy}$.

However, further exposure to ionizing radiation makes elastomers hard and brittle until disintegration occurs.


Which of these thermodynamic relationships is true for a system at equilibrium? Which of the following is true at equilibrium

$$T \Delta{S} = \Delta{H}$$

$$T \Delta{s} = \Delta {G}$$

I think it's the first one but I just wanted to be sure.

At equilibrium, $\Delta {G}$ is zero, so according to the equation $\Delta {G} = \Delta {H} - T \Delta {S}$ the first option is correct.


How concentrated can an acid be? How concentrated can an acid be without it being supersaturated? Is there a certain limit to how concentrated an acid or base can be?

When I mean concentration, I mean molarity; so how concentrated can an acid or base be?

NOTE: Can be a strong or weak acid/base. Because we are just looking for the amount of solvent needed to dissolve to start the dissociation process.

EDIT: No autoprotolysis please. Acids that need a solvent to dissociate.

I am talking about dissolving in water since an acid needs some sort of solvent to dissociate.

No, it doesn't. You neglect autoprotolysis where the water-free acid itself is the solvent.

\begin{align}
\ce{2H2SO4 &<=> H3SO4+ + HSO4-}\\
\ce{3HF &<=> H2F+ + HF2-}
\end{align}


Why does NCl3 use sp³ hybrid orbitals? I have a question regarding the hybridization of $\ce{NCl3}$.
From what I've understood, hybridization occurs when electrons belonging to different orbitals mix.
For example an electron belonging to the orbital 's' moves to one of the 'p' orbitals creating a certain number of hybrid orbitals. This occurs when more individually occupied orbitals are needed in order to complete the bonds and form the molecule.

Now, considering the fact that in $\ce{NCl3}$, nitrogen already has three individually occupied 'p' orbitals, it doesn't need to hybridise these orbitals in order to create three bonds with chlorine.
So why does $\ce{NCl3}$ utilise sp3 hybridization?

Let us suppose that your assumption is correct and nitrogen has already $\ce{2p^3}$ configuration. Let's further assume that it doesn't undergo hybridisation. Instead, the pure atomic orbitals undergo overlap with 3 $\ce{Cl}$ atoms and the filled $\ce{2s}$ shell electron pair acts as a lone pair and thus $\ce{NCl3}$ is formed. This consideration creates problems.

If pure atomic orbitals of nitrogen, i.e. $\ce{2p_x}$, $\ce{2p_y}$ and $\ce{2p_z}$ undergo σ-bond formation with the chlorine atoms, then the bonds should be $90^\circ$ to each other, as $\ce{p_x}$, $\ce{p_y}$ and $\ce{p_z}$ orbitals are mutually perpendicular. But $\ce{NCl3}$ has 3 σ-bonds with bond angles between $\ce{Cl-N-Cl}$ close to $107.1^\circ$, which is close to the tetrahedral bond angle of $109.5^\circ$. Therefore the pure orbital undergoing bond formation doesn't support this observation.

If we consider $\ce{sp^3}$ hybrid orbitals of $\ce{NCl3}$, this tetrahedral arrangement of three σ-bonds and a lone pair (disregarding  the lone pair, it's pyramidal) is nicely explained. That's why, the hybridisation is $\ce{sp^3}$ which supports every observed chemical property of $\ce{NCl3}$.


Why can one double bond exist in some resonance structures but two can't? I am doing some biology and have not done chemistry in ages. I have two concepts I am struggling with.

When drawing Lewis structures, am I right in assuming that the central atom will always require covalent bonds to fulfil its octet before the outer atom's octets are filled?
For example when drawing lewis structures, the central carbon atom should be 'given' enough covalent bonds or lone electrons to have a full octet before lone electrons are allocated to the outer atoms in the structure.
I do not get why, for example, $\ce{HCO2-}$ can have resonant structures with one double bond (but not two double bonds) between the carbon and oxygens. (please see picture attached)

In your drawn example, the red-boxed structure requires the carbon atom to have five bonds, which, to channel my secondary school teacher, says "yuck, I don't like that".

This is a case of breaching the octet rule which has to with how the orbitals hybridise. Five full bonds on a carbon is not "allowed". However three full bonds and two partial bonds is common in both resonance and transition structures, and is in fact a much more accurate representation of the molecule you're drawing.

Don't forget too, that the location of the negative charge matters, and I can't see it in any of your drawings.


What ratio of 73% calcium hypochlorite powder to water mimics the strength of sodium hypochlorite (bleach)? Calcium hypochlorite is sold as pool shock in varying strengths.  I've read that it can be dissolved in water to create a 'bleach' (typically sodium hydrochloric or sodium hydroxide) and is often used as such to treat drinking water in developing countries, but every source I've read prescribes a different ratio. Strictly looking at chlorine levels for water sanitation, cleaning, etc., how much powder (at 73% calcium hydrochlorite) would you need to dissolve into 1 gallon of water to match common household bleach?

To start, we must first understand that we can ignore the counterions to the $\ce{Ca(ClO)2}$ and the $\ce{NaClO}$. Typical household bleach claims to have a concentration of 6% (w/v). Now we can do some basic arithmetic to convert the percent weight by volume of sodium hypochlorite in water to a molar concentration:

$\mathrm{\frac{0.06\,g}{1\,mL}}$$\mathrm{\frac{1\,mol}{74.44\,g}}$$\mathrm{\frac{1000\,mL}{1\,L}}$$\mathrm{\,= 0.8060\,M \,OCl^-}$

The first fraction in the above equation is based on the definition of w/v, which is the number of grams of a substance dissolved in one milliliter of a solution times 100%. In the next fraction, we convert the grams $\ce{NaClO}$ into moles of $\ce{NaClO}$. Finally, in the last fraction, we convert milliliters into liters to give the number of moles per liter of $\ce{NaClO}$, or simply $\ce{ClO^-}$ without the counter ion.

Now that we know the molar concentration of $\ce{ClO^-}$ we are aiming for, we can convert this into the number of grams of pure $\ce{Ca(ClO)2}$ we would need to to achieve this concentration in one gallon of water

$\mathrm{\frac{0.8060\,mol}{1\,L}}$$\mathrm{\frac{1\,mol}{2\,mol}}$$\mathrm{\frac{142.98\,g}{1\,mol}}$$\mathrm{\frac{3.78541\,L}{1\,gal}}$$\mathrm{\,=\frac{218.1\,g\,OCl^-}{gal}}$

In the second fraction, we converted the moles of $\ce{ClO^-}$ into the number of moles of $\ce{Ca(ClO)2}$ (as there is only one unit of $\ce{Ca(ClO)2}$ per two units of $\ce{ClO^-}$). In the third fraction, we converted the number of moles of $\ce{Ca(ClO)2}$ into grams. In the last fraction, we converted the number of liters into the number of gallons.

As you stated, the powder only contains 73% $\ce{Ca(ClO)2}$ by mass. We know we need a total of 218.1 g of $\ce{Ca(ClO)2}$, so we can use the following conversion factor:

$\mathrm{\frac{218.1\,g\,OCl^-}{1\,gal}}$$\mathrm{\frac{1\,g}{0.73\,g}}$$\mathrm{\,=298.8\,g\,powder}$ (Rounded to 0.3 kg since we only have one significant figure)

In the second fraction, we know that per every gram of your powder, 73 percent (0.73 g) of the $\ce{Ca(ClO)2}$ will be present. Thus, the conversion factor is valid.

Final Answer: 0.3 kg of Powder


How can I predict what dissolves what? Regarding solvents, the only thing I remember from school is my teacher saying: "The same dissolves the same." regarding the polar/non-polar solvents.

From this I understand that water is polar solvent, because it has polar bonds and therefore it dissolves table salt, which also has a polar bond.

What I do not understand for example is, why does water perfectly mix with acetone? X

Another thing is, that non-polar organic solvents seem to have some very complicated rules. For example, acetone dissolves most paints and all ABS plastics. Ethanol or tetrachloroethylene do not.

Why does ethanol dissolve acetone-peroxide while acetone doesn't? X

Q: Are there some general rules to this, or is this big science that requires years of study?

X: these are just rhetoric questions. You can answer in comments but what I'm looking for are rules.

As you note, there are some general rules about solubility. For example, acetone is a highly polar organic molecule, so it's not surprising it has high solubility/mixibility in water. The "like dissolves like" rule is generally based on polarity and size/shape considerations, and it's a fairly good rule.

But solubility is complex because you have two big sources of driving force:

enthalpy: The molecular and intermolecular interactions between the solvent and solute are favorable. Consider methanol dissolving in water: it forms hydrogen bonds with the water, so the enthalpy is favorable.
entropy: This is often the larger driving force. You create a far more disordered result with a mixture than with two pure components.

While most chemists develop reasonably accurate intuitions of enthalpy, the entropic effects are harder to predict.

Many efforts have been made to predict solubility, both qualitatively and quantitatively. The comments above give some good links. Consider that this problem is important to chemical synthesis broadly, but also to drug delivery (i.e., can a drug be given orally and dissolve in the stomach, can a drug go through the blood-brain barrier?)

Perhaps the most successful is the Abraham general solubility model which attempts to predict solubility of organic compounds in 84 different solvents. Here’s the article.

Basically, it breaks down solubility into multiple components (grabbed from Andrew Lang and Jean-Claude Bradley):

E is the solute excess molar refractivity in units of (cubic cm per mol)/10. It represents the solute's polarizability and gives a measure of the ability of a solute to interact with a solvent through n- and $\pi$- electron pairs.
S is the solute dipolarity/polarizability. It gives a measure of the solute's ability to stabilize a charge or dipole
A is the overall (summation) hydrogen bond acidity. The hydrogen bond acidity descriptor measures the extent of hydrogen bonding by the solute in a basic solvent.
B is the overall (summation) hydrogen bond basicity. The hydrogen bond basicity descriptor measures of the extent of hydrogen bonding by the solute in an acidic solvent.
V is the McGowan characteristic volume in units of (cubic cm per mol)/100.

The nice thing about this model is that it makes sense -- you want a solute that matches lone pairs, hydrogen bond donating or accepting ability, polarizability, and volume for the solvent.

The catch is that surprisingly little data is available. The late Jean-Claude Bradley encouraged his students to make accurate solubility measurements, but also to publish them on the web. In this way, they gathered a large spreadsheet of "compound X dissolves Y amount in Z solvent."


Do the concepts of Newtonian and non-Newtonian fluids still hold at a nanoscopic level? Recently I've spent a great deal of time thinking about enzyme diffusion in cytoplasm. One thing that keeps me awake at night is the constant referral to cytoplasm as a non-Newtonian fluid (which it surely is when looking at it as a whole), and the implications this has for the diffusive properties of singular enzymes.

If I think about a single enzyme in the cytoplasm with its immediate surroundings, I would expect to find the protein with its solvation shells, and only a few non-water components. In this context I'm unable to understand how this single enzyme could be thought to be in a non-Newtonian medium.

Furthermore, when thinking about the diffusion of the enzyme, quite a few authors refer to the Scallop theorem, and define it in the context of non-Newtonian fluids. If the enzyme is mostly surrounded by water molecules, and only occasionally runs in to other proteins or peptides, shouldn't the protein be considered to reside in a Newtonian medium?

Shortly put: How far down can you go until the classification between Newtonian and non-Newtonian fluids starts to break down?

Generally Fick's laws of diffusion have been found to apply pretty well to molecular translational and rotational diffusion in solution. This means that we assume that the solvent is a continuous structureless fluid, which in practice means that the diffusing molecule should be far bigger than that of the solvent so that the solvent's molecular nature can be ignored. Surprisingly this works well with 'slip' and 'stick' boundary conditions down to situations where the solute is not that much bigger than the solvent, e.g.  rhodamine and similar dye molecules in solvents such as acetonitrile or ethanol. In your example of a protein in water then you might expect that it would behave 'normally' as just described.

However, in the cytoplasm the close packing of many similar sized and large molecules means that these can easily block the diffusion of any other (except for far, far smaller ones such as water) and so the situation is totally different and will be more like a percolation process where a slight reduction in packing can lead to a large, almost stepwise increase in diffusion coefficient. The idea of a single diffusion coefficient is therefore not helpful.


Are all monomers in a polymer covalently bonded? I'm approaching chemistry and I read on Wikipedia that

  The modern concept of polymers as covalently bonded macromolecular structures was proposed in 1920 by Hermann Staudinger"

But then the article goes on and it says:

  Historically, products arising from the linkage of repeating units by covalent chemical bonds have been the primary focus of polymer science; emerging important areas of the science now focus on non-covalent links.

So now I'm wondering, are there polymers whose monomers are bonded in other ways?

I think your second quote is referring to some polymers such as sodium polyacrylate, $\ce{[-CH2-CH(COONa)$-$]_n}$. This polymer is basically just the sodium salt of polyacrylic acid (or polymerized sodium acrylate). You can see below the ionic bond between the anionic carboxylate group and the sodium cation:

However, the quote could be referring to so-called 'chain-link' polymers, in which the monomers are intertwined and looped/knotted together, like links in a chain, instead of just consisting of a bunch of long strings of monomers, like a rope. However, chain-link polymers are highly uncommon and rather difficult to synthesize. Note that these are not the same as cross-linked polymers.


Finding equilibrium concentrations in solution from initial reagent concentrations and equilibrium constant We have the following reaction:
$$\ce{A + B<=>2C +D}$$
The initial concentration of $\ce{A}$ and $\ce{B}$ are $\pu{1M}$ each. $K_c$ is $\pu{1.0E8}$. Find the equilibrium concentration of $\ce{A}$.

This is what I have tried:
The initial concentration of A,B,C,D are 1,1,0,0 respectively
The final concentrations of A,B,C,D have to be ($1-\alpha$), ($1-\alpha$), $2\alpha$, $\alpha$ respectively. Therefore
$$K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$
$$10^8=\frac{(2\alpha)^2\alpha}{(1-\alpha)^2}$$
$$10^8(1-\alpha)^2=4\alpha^3$$
But even on solving this cubic equation on a calculator, the values are incorrect.

No need for the cubic equation. Just firstly assume that $\ce{A}$ and $\ce{B}$ reacted completely, then $\ce{[C]}=\pu{2M}$, $\ce{[D]}=\pu{1M}$, and from $K_c$ you can calculate $\ce{A}$ which is the same as $\ce{B}$.
That's the approximation usually used with this kind of problem.
The answer should be $\pu{2E-4M}$.


Tri-iodination of 2-methylpropane In the reaction of tri-iodination of 2-methyl propane only five isomers can be formed, from which only the branch methyl can be iodinated if both the lateral carbond are iodinated .. Is this because of the inductive effect or the steric effect is the responsible here...if it is the steric effect which hinder the iodination of the branch carbon (2-methyl) why is it possible to tri-iodinate the lateral carbons( the primary carbons other than the branch one)  ?

The reason only five isomers can be formed is the three methyl groups are equivalent by symmetry.

I believe your arbitrary distinction of branch carbon/methyl from lateral carbon/methyl is creating confusion


Is there any easy trick to know if a substance is a Brønsted–Lowry acid or base? I know that Brønsted–Lowry acids can donate a proton and a Brønsted–Lowry base can receive a proton. I have also memorized the strong acids and bases, which makes it easier to figure out. However, if I was given a substance that I didn't memorize, I would struggle to tell if it is an acid or base. Are there any quick methods to know if a substance is an acid or base? Or do I have to memorize as many as I can?

Not perfect, but here are a couple rules of thumb to start with for the most common cases.  The key with both rules is to look at the molecular structure.

A Bronsted acid typically has hydrogen attached to oxygen, sulfur or a halogen, all these being relatively electronegative elements.  The hydrogen could also be attached to another nonmetal, especially nitrogen or phosphorous, if the atom is a positive charge center like the nitrogen in ammonium ion.  If the atom is both a positive charge center and oxygen, sulfur or a halogen, you likely have a strong acid.

A Bronsted-Lowry base typically has a nonbonding electron pair that can combine with a proton, thus resembling a Lewis base.  In stable compounds the nonbonding pair us usually on the same kind of atom as one of the elements mentioned above for acids.  This is one reason Bronsted-Lowry acid-base reactions are typically fast and reversible.


How to obtain pure alcohol from mixture of alcohol & carbonyl? (Grignard experiment) If one discovers that after finishing up a Grignard reaction that their alcohol product has been contaminated with the carbonyl starting material, how could they separate the alcohol from the carbonyl compound to obtain a purified alcohol product?

I was thinking that they could recrystallize the contaminated product using an alcohol as a solvent, but is there a better option for separating these two compounds?

I rather doubt that there is a better option in general. Other possible methods are

distillation
To give an example, acetone has a boiling point of 56 °C, while tert-butanol boils at 82 °C.
column chromatography on silica

But in many cases, your suggestion to separate the compounds by crystallization is a good idea. This is in particular true when your product has a high boiling point and/or you have to purify large amounts of substance.


Angular momentum - Question on Zare's use of the Vector Model to Estimate Probability My question regards a specific example, although I get the feeling that the answer is simpler than I imagine. In Zare's Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics pgs 53-54, He discusses the vector model for coupled and uncoupled representations for the addition of two angular momenta.

I've scanned and included the picture I am going to refer to for clarity:

The image above is a modified representation of the uncoupled representation, although it appears very similar to the coupled diagram, the difference is that j1 has been arbitrarily fixed, and j2 is precessing in the dashed circle, around an axis parallel to z.

Zare expands the dot product of $\mathbf{j_1}$ and $\mathbf{j_2}$
(Eq 2.34):

$\mathbf{j_1 . j_2} = |\mathbf{j_1}||\mathbf{j_2}| cos(\gamma)$

$=|\mathbf{j_1}||\mathbf{j_2}| [ cos(\theta_1) cos(\theta_2)+  sin(\theta_1) sin(\theta_2) cos(\theta_1) cos(\phi_1 -\phi_2)]$

So far, so good, no problem. Then he instructs:

"Let us differentiate Eq (2.34) with respect to time to obtain the rate of change of the length of $\mathbf{j}$. From the law of cosines

$|\mathbf{j}|^2 = |\mathbf{j_1}|^2 + |\mathbf{j_2}|^2 - 2 |\mathbf{j_1}| |\mathbf{j_2}| cos(\gamma)$

$=|\mathbf{j_1}|^2 + |\mathbf{j_2}|^2 - 2 \mathbf{j_1} . \mathbf{j_2}$

So that

$\frac d{dt} (\mathbf{j_1} . \mathbf{j_2}) = -\frac 12 \frac d{dt} [ |\mathbf{j}|^2 + |\mathbf{j_1}|^2 - |\mathbf{j_2}|^2 ]$"

And THIS next step is where I get lost:

He IMMEDIATELY equates that expression with the following:

$$ = -|\mathbf{j}| \frac {d|\mathbf{j}|}{dt} $$

So that there's no confusion, what I do not understand is the statement:

$$\frac d{dt} (\mathbf{j_1} . \mathbf{j_2}) = -\frac 12 \frac d{dt} [ |\mathbf{j}|^2 + |\mathbf{j_1}|^2 - |\mathbf{j_2}|^2 ] = -|\mathbf{j}| \frac {d|\mathbf{j}|}{dt}$$

Now, if I explicitly expand all components of the vector norms: $\mathbf{j_1}$,$\mathbf{j_2}$,and $\mathbf{j}$, and try to solve for what's inside the brackets on the LHS, I just regenerate the expression:

$|\mathbf{j}|^2 =|\mathbf{j_1}|^2 + |\mathbf{j_2}|^2 - 2 \mathbf{j_1} . \mathbf{j_2}$, which is of no use in aiding me as far as I can see.

Furthermore, it is explicitly stated that the only time-dependent component of this system is the dihedral angle $\phi_1 - \phi_2$, and in any expansion of the vector $|\mathbf{j}|$ into its components, a time derivative would render the any other independent parts mute.

I am thoroughly confused. I have consulted the following textbooks and not found any sign of this exact derivation:

Cohen-Tannoudji (volumes I and II), Feynman's lectures on Physics Vol III, Shankar, Atkin's physical chemistry, Engel and Reid's physical chemistry, Gaziorowicz, and Eisberg and Resnick.

What I suspect is that I'm simply lacking in some fundamental piece of calculus or arithmetic. Whether there's some simple trick of Vector Calculus that I haven't employed to give me the identity placed here? I would love a direct answer, but will also accept a nudge in the right direction. Right now it just feels like I'm leafing through books without a compass.

Best,

and THANKS!

As mentioned in the comment by Weijun Zhou, the length of the vectors $\mathbf{j}_1$ and $\mathbf{j}_2$ does not change as $\mathbf{j}_2$ precesses about $\mathbf{j}_1$, only the total vector $\mathbf{j}$ changes its length. Knowing this just requires application of the sum and chain rule of derivations that is

$
-\frac{1}{2}\frac{d}{dt}|\mathbf{j}|^2=-2|\mathbf{j}|\frac{1}{2}\frac{d}{dt}|\mathbf{j}|
$

and the time derivatives of $\mathbf{j}_1$ and $\mathbf{j}_2$ are zero.


Does the oxidation for all the aliphatic monoaldehydes to carboxylic acids have the same value of electrode potential? I have been wondering on this for quite a while. Consider aliphatic monoaldehydes. If I oxidize them, we get corresponding monocarboxylic acids. We can use certain oxidizing agents to perform the reaction.

For all the possible aliphatic aldehydes, it involves breaking a $\ce{C-H}$, and the formation of the $\ce{C-O}$ and a $\ce{O-H}$. I did calculations and got the enthalpy change for such a process to be $\pu{412kJ/mol}$.

My main motive is to find the electrode potential for the process. Unfortunately, I cannot determine the entropy here, although I don't think the side chain would matter much.

Here's my question. Does the oxidation for all the aliphatic monoaldehydes to carboxylic acids have the same value of electrode potential? In all the cases, it's the same bonds broken and same bonds formed, so I cannot see why the electrode potential should drastically vary for different aldehydes.

My apologies, I was unable to find a $\Delta\text{G}$ value table for such processes.

My answer to your main question is no.

Firstly, you are simplifying the process of oxidation of the monoaldehyde. The mechanisms for oxidation depending on the oxidising agent used, although they are generally the same (Refer to link). By considering only these 3 bond formation/cleavage events, you are not looking at this process in totality.

Secondly, the side chain does matter as it would affect the bond energies of the bonds that are broken and formed. Different side chains may give the aldehyde different steric strains etc. Depending on whether it is more electron-withdrawing or electron-donating, it can severely influence the bonds on the carboxyl group.

Due to the variations in oxidising agent, medium of reaction and side group of aldehyde, the oxidation of different monoaldehydes to carboxylic acids would have a different value of electrode potential as these factors would affect the feasibility and ease of oxidation.

Oxidation using chromic (VI) species:

http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch17/ch17-3-5-1.html


Priority of carbons attached to chiral group, one with double bonded oxygen, one with two oxygens I have a chiral carbon. Two of the atoms attached to the carbon are other carbon atoms. One of those carbons has a double bonded oxygen and a methyl group attached. The other has two oxygens and a methyl group. Which carbon substituent has higher priority?

Given the same atom, those bond via triple bonds have higher precedence over double and double bonds have higher precedence over single. This is exemplified here and, a bit more formal covering all CIP-rules, here.


What is the difference between "vapour" and "gas"? I have read the definition of vapour on Wikipedia, but I don't seem to have a clear idea about the difference between a vapour and a gas. I know what it means when it says, "gas at a temperature lower than its critical temperature".

In this graph, the blue curve, means the part where gaseous and liquid phase coexist, so did the scientists coined the term Vapour for only this reason so, that we can have orderly fashion of different states of matter ?
My question is Are we naming vapour and gas to the same thing just at different temperature or they are distinct in some other way too?
Are there some other things which justify that vapour and gas are actually different things than just a gas below and above critical temperature ?

Vapor is a much older word alluding to dampness and it was not coined by scientists. It is in use since the 1300s. The actual meaning of meaning of vapor is "Matter in the form of a steamy or imperceptible exhalation; esp. the form into which liquids are naturally converted by the action of a sufficient degree of heat. This is the original 13th century meaning of vapor. The German equivalent of vapor is Dampf (which is relatively clearer in terms of feeling- imagine entering a basement which is damp from water vapor). Gas is relatively a new word as per the unabridged Oxford English Dictionary, and it entered English in the 1600s.
In fact, the OED mentions the distinction "Gas: A substance in a state in which it expands freely to fill the whole of a container, having no fixed shape (unlike a solid) and no fixed volume (unlike a liquid); spec. (distinguished from a vapour) such a substance above a critical temperature such that it cannot be liquefied by the application of pressure alone"
You can also think of the general meanings of vapor in scientific usage:
A substance which is a liquid at room temperature or under ambient conditions, and if it is heated enough so that the liquid evaporates (same root of word), you can call that as vapor of that substance.
Examples: Mercury is a liquid metal, when it is heated enough in a an electric bulb, it becomes a gas, hence the name mercury vapor lamp. It is not called a mercury gas lamp.
It is not necessary that heat be involved. You will find the usage "cold vapor method for Hg analysis" again for mercury which means that Hg was a liquid in a sample but by blowing a gas over it, it has been evaporated.
The recent meaning of vapor from OED, "Vapor: In modern scientific use, a fluid that fills a space like a gas but, being below its critical temperature, can be liquefied by pressure alone."
This is clear enough. Now apply this distinction to your phase diagram and notice the position of the critical point. The vapor "state" is below the critical point.


Cis and Trans Isomerism given three different groups around the reference plane In the following problem I am meant to deduce the molecules that are cis-trans isomers of each other.
This question does not make much sense in my head because I thought that cis/trans isomerism was only applied to configurational isomers if there were only two types of groups bonded to the carbon atoms in the carbon-carbon double bond i.e. the reference plane. Such that, if there are more than two groups bonded to the carbons in the reference plane then E/Z isomerism should be used. However, then I thought that cis/trans isomerism is a subcategory of E/Z isomerism and therefore I tried to work out the solution using the E/Z method of finding configurational isomers by finding the relative priorities based on the atomic numbers for the atoms attached to the carbon atoms on each side. Using this method I discovered, that the solution could only be 'A. X and Z' since X seems to be an 'E' isomer since its high and low priority groups are across from eachother and Z seems to be a 'Z' isomer since its high and low priority groups are on the same side.

Am I on the right track?

Am I on the right track?

Generally, yes. For this question, yes but you have over thought this question.

For starters, any answer with Y can be totally eliminated as Y is neither cis nor trans. Now you are down to answers 'A.' and 'D.'

Here you correctly determined the E and Z configurations of molecules with identical functional groups, but could have saved yourself some effort if you had recognized W and Z are identical molecules that are inversions/180° rotations of each other. Thus it can't be D. and the answer is in fact:

  A.


Why is the amount in a mole equal to the Avogadro's number? This question just popped in my head as I was working with electric force problems involving atoms. Why is the amount in a mole equal to the Avogadro's number? Do atoms/molecules/etc. tend to group in this exact amount? Why do we use this number as a unit of measurement?

Avogadro articulated the concept that is Avogadro's number, but didn't pick the number. From Wikipedia:

  Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules".

https://en.wikipedia.org/wiki/Avogadro%27s_law

The Number is a logical extension of this law: So what is that number in relation to the volume and what is the mass of that amount of material?  By agreement, a mole is defined as the mass, in grams, equivalent to the atomic weight of a substance. This was laid out before Avogadro's Number was actually determined.  Then it was simply a matter of measuring the mass of a single atom to determine how many atoms were required to have a Mole of a material.  Over time, Avogadro's Number has been refined to more and more accuracy.


Why doesn't cottage cheese melt? Most kinds of cheese melt but cottage cheese does not at usual cooking temperatures.

During the preparation of various kinds of cheese, I have heard of a coagulant named ‘rennett’ being added along with the acidifying agent. For cottage cheese it is just lemon juice mostly (acidifying agent). Does this have anything to do with the difference in properties?

You essentially make cottage cheese in the process of making cheddar cheese. But cottage cheese is about 80% water. The "cheese" solids are soluble in oil, but not in water. So cheddar cheese will "melt" but cottage cheese won't.


Database for experimental diatomic ionization energies I am looking for a database, in which the experimental ionization energies (and e.g. heat of formation) of atoms and diatomic molecules are listed as files.
If it comes with a Python package, that would even better.

I found the Pitt Quantum Repository, but there seem to be only theoretical values.

NIST Atomic Spectra Database has the atom part. Not sure about molecules.
https://physics.nist.gov/PhysRefData/ASD/ionEnergy.html


Trend in the boiling point of the hydrogen halides The boiling points of the hydrogen halides are as follows:

$$\begin{array}{cc}
\hline
\text{Species} & \text{Boiling point / }\mathrm{^\circ C} \\
\hline
\ce{HCl} & -85.1 \\
\ce{HBr} & -67.1 \\
\ce{HI} & -35.1 \\
\hline
\end{array}$$
(source: Greenwood & Earnshaw, Chemistry of the Elements 2nd ed., p 813)

My textbook writes that this is due to an increase in the number of electrons, which increases the van der Waals forces between molecules.

But the $\ce{HX}$ molecules are polar. Shouldn't the boiling point be determined instead by permanent dipole-dipole attractions instead? Since the dipole moment decreases going from $\ce{HCl}$ to $\ce{HI}$, the boiling points should actually decrease. Why isn't this the case?

There are formulae you can use to calculate the strength of interactions between two molecules, which are derived from first principles. You can find an overview here on Chemistry LibreTexts. The two relevant ones are:

$$U_{\mathrm{dipole}} = -\frac{2}{3kT} \frac{p^4}{(4\pi\varepsilon_0)^2r^6}$$

$$U_{\mathrm{dispersion}} = -\frac{3}{4} \frac{\alpha^2 I}{(4\pi\varepsilon_0)^2r^6}$$

$p$ is the molecular dipole moment
$r$ is the average separation between neighbouring molecules
$\alpha$ is the polarisability of the molecule, which in a nutshell, measures how easy it is to create an induced dipole in the molecule
$I$ is the first ionisation energy of the molecule

If you plug all the numbers in for the hydrogen halides $\ce{HX}$, you will find that the magnitude of the dispersion forces is much larger than the magnitude of the dipole-dipole attractions, i.e. $\left|U_{\mathrm{dispersion}}\right| \gg \left|U_{\mathrm{dipole}}\right|$.

Unfortunately, I don't have all the necessary data on hand, but I can quote some results that were given in first-year lecture notes at Oxford (which should be legitimate). For $\ce{HCl}$, dispersion forces contribute $86\%$ to the intermolecular attractions, and for $\ce{HI}$, they contribute $99\%$.

Considering this fact, it is not surprising that variations in the magnitude of dispersion forces affect the boiling point much more than variations in the magnitude of the dipole-dipole attractions. If you had the all the data required and evaluated the interactions using the two formulae above, you would reach the same conclusion quantitatively.

In fact, this trend is hardly unique to the hydrogen halides. The hydrides of the Group 15 and Group 16 elements behave exactly the same way:

$$\begin{array}{|cc|cc|}
\hline
\text{Species} & \text{Boiling point / }\mathrm{^\circ C} & \text{Species} & \text{Boiling point / }\mathrm{^\circ C} \\
\hline
\ce{PH3} & -87.5 & \ce{H2S} & -60.3 \\
\ce{AsH3} & -62.4 & \ce{H2Se} & -41.3 \\
\ce{SbH3} & -18.4 & \ce{H2Te} & -4 \\
\hline
\end{array}$$
(source: Greenwood & Earnshaw, Chemistry of the Elements 2nd ed., pp 557, 767)

Of course, the first-row hydrides are left out of the discussion because of hydrogen bonding, which makes their boiling points anomalously high.


The use of ice water in an experiment An experiment from Vogel that I want to conduct requires "a mixture of ice and salt of about -10 °C". Reading trough the text it appears that I need at least 1-2 liters of this. Obviously the salt is added to lower the freezing point of the water. I guess the interaction of water with the dissolved salt trough hydrogen bonding / lone pair donation is more favorable (less energetic) than ordinary water-water interaction.

With "salt", does Vogel mean just $\ce{NaCl}$?
Since I'm a hobby chemist I do not have the same fancy machinery as they have in the lab (those cool things that generate rough ice shards and in which everybody stuffs food against all safety regulations lol). All I can use is a small refrigerator with a tiny freezer. It's easy enough to make some ice cubes, but how do I go about making these ice shards that are used in the lab? I don't think grinding the cubes is very practical as it is a tedious job that takes a lot of time in which the shards have probably already melted (also from hitting them with a pestle continuously)...

1.With "salt", does Vogel mean just $\ce{NaCl}$?

Yep, it's cheap and available.

  How do I go about making these ice shards that are used in the lab?

Wrap the ice cubes in a towel and hit the bundle with a hammer until you get whatever size you want.  That's how we did it in the lab.


Why aryl halides are very less reactive towards nucleophilic substitution reactions? I've often herd that aryl halides are very less reactive towards nucleophilic substitution reactions. One of the reasons for this I found was that there is resonance effect in the compound, but how is  that responsible for it to less reactive towards S.N. reactions and in addition to resonance effect are there other factors also responsible ?

No, that is a very incorrect explanation for why (unsubstituted) aryl halides do not participate in nucleophilic substitution reaction. Conjugating effect between the halide substituent and the benzene ring is not responsible for this. The problem is that there is simply no available mechanism to perform the substitution:

$\ce{S_{N}2}$ is impossible since  inversion would give an incredibly weird strained ring (even more weird than a bridgehead carbon that is already forbidden by Bredt's rule).
$\ce{S_{N}1}$ would leave a very unstable empty $\ce{sp^2}$ orbital. This can happen, but only for the very best leaving groups, like $\ce{N2}$.
There is no electron stabilizing group that is conjugated with the halide group. This renders an addition-elimination type mechanism impossible.


How to avoid the appearance of extra hydrogens from lattice packing of PDB file? I'm struggling with a pdb file of a unit cell that has a $\ce{CH3}$ which becomes a $\ce{CH6}$ when some software performs replication transformations. I've tried pack it with Avogadro and Mercury.

There is a way to avoid this by only changing something in the PDB file?

This is a bug, which is clear.

My guess is that you don't get $\ce{CH6}$ but that there's some disorder, so the transformations put two near-identical $\ce{CH3}$ groups on top of each other. Thus, one C becomes two, and each gains hydrogens.

There's usually a "slop" factor to detect near-equivalent atoms, but it's clearly failing.