Thermal oxidation of gold(I) sulfide As far as I know the chemistry of $\ce{Ag}$ and $\ce{Au}$ are almost the same; because $\ce{Ag}$ belongs to the 4d series and $\ce{Au}$ belongs to the 5d series, we know the size of 4d and 5d are similar, so their chemistry remains similar.

Thermal study and mechanism of $\ce{Ag2S}$ oxidation in air

In this paper, it is given that $\ce{Ag2S}$ can be oxidised thermally in the presence of $\ce{O2}$ to give $\ce{Ag}$ and $\ce{SO2}$ over 510 °C. I am not able to find any references where a similar experiment is done for $\ce{Au2S}$. Can somebody please help me ?

Gold is generally extracted by pressure oxidizing a refractory ore such as pyrite or chalcopyrite. They are common sulfide ores where finely divided nanosized gold are present. They are roasted to break the sulfide matrix and recover the metallic gold(Here). These refractory ores are considered low grade ores (where gold% is less) and are therefore roasted to extract the gold (Here).

Critically, the size at which one can efficiently liberate gold from the host mineral of the particle will determine the process to use for optimal gold recovery(Here). Gold(I) sulfide is not considered a refractory ore as because gold is not present in microscopic level and hence are generally not roasted (Here and Here).

But it is possible to roast gold(I) sulfide at 200-500°C(Here) but the yield will be less as compared to the gold  yielded when it is leached.

$$\ce{Au2S + O2 ->[200-500°C] 2Au + SO2}$$


Is circularly polarized light ever used to isolate particular chiralities? Chiral molecules tend to absorb one circular polarization of light more than the other. Is this ever used to isolate a particular enantiomer?

Yes. For some chiral compounds, each enantiomer in a racemic mixture absorbs certain frequencies of light differently depending on the circular polarization type of the light.

That means that if you can pump in sufficiently intense and energetic (e.g. ultraviolet) circular polarized light pulses into the racemic mix of such compounds, you may in some cases increase the rate of breakdown of the enantiomer that absorbs that particular circular polarization (left or right), resulting in an increase in the relative concentration of the other enantiomer.

You can view the first page of a paper on asymmetric photolysis using circularly polarized ultraviolet light here.


How to get (2R,3R)-2,3-dibromobutane from but-2-yne? What I had in mind was to form but-2-ene with $\ce{H2}$ and Lindlar's catalyst and then simply add bromine ($\ce{Br2}$). But this addition, if I'm not mistaken, is a trans-addition. So I would end up with (2R,3S)-2,3-dibromobutane instead of (2R,3R).

Note that this is (2R,3S)-2,3-dibromobutane:

Have I made a mistake, or is there something wrong in my way of thinking? If my solution is wrong, how could I get (2R,3R)?

I think I figured it out. Lindlar is a cis-addition so one would not obtain a (2R,3S), but rather (2S,3S) and (2R,3R) which happen to be enantiomers.


Which d-orbital configurations form an octahedral complex? With reference to crystal field splitting in octahedral complexes, my textbook shows the following table of distribution of $d$-orbital electrons in octahedral complexes, based on the energy difference of $t_{2g}$ and $e_g$ orbitals:

Table

Do ALL the $d$ orbital configurations (i.e. pertaining to all examples given) form octahedral complexes? I doubt it because not all of them have two vacant $d$ orbitals, especially in the case of weak field ligands where Hund's rule is never violated.

A metal ion doesn't actually need vacant d orbitals to form a coordination complex according to Crystal field theory. All the ions mentioned in the table form complexes with water (which is a weak ligand) - see this page.

Crystal field theory assumes that all bonding between ligand and metal ion in a complex is electrostatic (ionic). That means there is no exchange of electrons between the metal ion and ligand. Instead, the negatively charged (or polar in case of neutral ligands like $NH_3$) ligand is attracted to the positively charged metal ion by purely electrostatic forces. This is why vacant d orbitals are not required for bonding.


How to interpret the MSDS warning? Just a quick question; I was looking up the MSDS for o-Toluic Acid for an experiment when I came across this in the "Hazards Identification" section.

Very hazardous in case of skin contact (irritant), ....
...
Slightly hazardous in case of skin contact (permeator).

How should this warning be interpreted?

An irritant is a chemical, which is not corrosive, but which causes a reversible inflammatory effect on living tissue by chemical action at the site of contact. There are specific lab tests to determine the ranking of an irritant, but something which is rated as very hazardous is likely to cause a very severe reaction upon exposure.

A permeator is a chemical which can pass through the outer protective dermal layers and into the body, and can either expose the body to toxic effects of that chemical OR act as a carrier for other toxic/hazardous chemicals.

Other common classes of hazardous chemicals include corrosive and sensitizer.

o-Toluic acid is a strong irritant, but moderate permeator. Depending on your level of experience in handling chemicals, simply getting the advice to 'wear gloves' may leave you a little underprepared. You should take this to mean that you should not allow any skin contact; full length sleeves (lab coat), eye protection and THE CORRECT TYPE of gloves. Using the incorrect type of gloves is worse than using no gloves at all, as it gives you a false sense of security. Select the correct type of gloves using a suitable selection tool or guide, such as this one from Ansell.

As a strong irritant, you should also avoid breathing any vapours or fumes, so it means handling this chemical in a suitably ventilated area.

AS with any risk assessment, it also means that before you handle the chemical you have an understanding of the response procedures in the event of things going wrong. That is, from your MSDS, you need to know Section 4:First Aid Measures prior to any procedure you undertake. It is no good having to find and read the MSDS after you have splashed yourself with this stuff.


How does High Frequency water electrolysis work? I have read that combining the DC current with a high frequency AC current, the electrolysis of water speeds up. Is this true? In that case, how is less energy wasted as heat? Or does it simply catalyze the process?

First of all, have a look at the wikipedia page on electrolysis of water.

I also like this review: Zoulias et al.: A Review on Water Electrolysis, TCJST, 4 (2) (2004) 41-71
Specifically they list a number of actually existing installations (context: renewable energy) and their actually achieved efficiency.
Speed up does not necessarily have anything to do with higher efficiency. In electrolysis it is often the other way round: if you want to squeeze out the maximum free energy, you need to do the reaction infinitely slowly (despite thermodynamics having a dynamic name, it looks at infinitively slow processes).
Thus, speeding up usually means that you find a way to put more power through your system. The big issue is to find a way of doing this without loosing (too much) efficiency.
Pulsed/modulated DC: looking through a few papers I liked this one: Shimizu et al.: A novel method of hydrogen generation by water electrolysis using
an ultra-short-pulse power supply, Journal of Applied Electrochemistry (2006) 36:419–423, DOI 10.1007/s10800-005-9090-

They aim at avoiding the diffusion controlled situation by having the pulses short enough so that no depletion zone occurs.
Look at these diagrams:

So they report one setting where the pulsed electrolysis is actually more efficient than DC in their cell.

There's more research going on on this, however the papers I found reported increased efficiency compared to pure DC electrolysis, but the absolute efficiencies are around 10%.   However, compare their numbers with the 80% efficiency cited by the review for an industrial alkaline electrolysis. Note that one big difference is the voltage that is applied: for the DC it is around 1.85 - 2.05 V, so much less overvoltage. Note also that when they say that higher voltage speeds up ion transport, then this overvoltage is converted to heat (ions face friction in the medium when they travel) and thus basically lost.
Another line that looks real is that if you go to higher temperatures, a (small) part of the energy can be supplied by heat. As heat is cheap, this may help. One point one has to be aware though, that the efficiency calculations may be done with respect to the electric energy only (neglecting the heat input) and thus look artificially nice (like efficiencies of condensing boilers calculated against the lower heating value).
I found a bunch of nonsense claims in the internet, about the resonance frequency of water helping to split bonds.

The first thing to realize here is that there is no one resonance frequency of water. With suitable energy, you can excite rotational, vibrational and electronic states (I left out translation - there transition energies minute). At room temperature you can say as a rule of thumb that most molecules will be in some excited rotational state, but in the vibrational and electronic ground states. Excitation energies for rotation are in the far infrared or microwave energy/frequency region. Widely used e.g. in the microwave oven at 2.45 GHz ($\approx$ 12 cm). Actually, the whole region is full of bands where water absorbs. Note that microwave heating of water does not cause electrolysis. Vibrational transitions are around 2.9 μm = 105 THz = 3500 cm⁻¹ and 6μm = 50 THz = 1635 cm⁻¹ with lots of combinations and overtones throughout the near infrared region. Quite exceptionally, the visible region is basically free of water absorption. Electronic transitions (breaking of bonds) need energies in the UV, and here we meet bands that lead to photodissociation, e.g. at 166nm (taken from Wikipedia). That corresponds to 1.8 PHz = $1.8 \cdot 10^{15}$ Hz. Compare this to the kHz and MHz where your link claims dissociation.

This doesn't mean that the pulsed DC cannot help, nor that impedance spectoscopy won't give important information. But resonance frequencies in the kHz range are electrical LC-circuit resonances depending on cell and electrode geometries and electrical double layers etc. But neiter on vibrations nor breaking of the bonds of the water molecule.
To give the "method" you ask about some real world numbers,

at the very end of the Wiki page the energy efficiency for industrial water electrolysis is cited as usually between 50 and 80 %.
The paper then proposes to burn the gas in an internal combustion machine. As such a stationary process could be adjusted so that the engine is at its maximum efficiency, we may assume 1/3 or 35% efficiency here.
we then need a generator to convert the mechanical energy into electric energy. Fortunately, that step is rather efficient. Say, 95 %.
A fuel cell would be more efficient than the combustion - generator combination: ca. 40 - 60 % according to Wikipedia.
Unfortunately, also battery charging is not 100% efficient. Let's assume 80–90% (taken from Wikipedia on Li-ion batteries) For batteries that are charged with higher current (or current density) efficiency is less. Example would be lead-acid batteries as used in cars. Wiki quotes efficiencies between 50 and 80 %.

Taking these numbers together, I conclude that after going once through the cycle of the proposed "perpetuum mobile", 8 - 24 % of the energy are retained in a "useful state" while 76 - 92 % became heat. With fuel cell, we may be able to "boost" the energy efficiency to 43%.

Useful general knowledge (in addition to the law of energy conservation)

The US patent system is different from e.g. the German patent system in that here in Germany the patent application has to have commercial/industrial usability. This includes a technical argument why it works (according to the physical laws). A perpetuum mobile would be rejected on these grounds (of course the inventor could prove his case with a prototype). US patents to not have this technical check.
generators (mechanic -> electric conversion) are sources of current, while batteries (galvanic cells) are sources of voltage.


How are the hybrid orbitals of sulfur hexafluoride shaped? On the fluorine end the fluoride atoms are simply completing a $3p$ orbital.

On the sulfur end one could posit a hybrid of one $3s$ orbital, two $3p$ orbitals and a three inner $2p$ orbitals. However, I think this would result in a geometry which sulfur hexafluoride does not exhibit.

The explanation I read online is that the orbitals are a hybrid of an $s$ orbital, three $p$ orbitals and two $d$ orbitals.

This confuses me because the $3d$ orbital is quite high energy and I don't think it is possible to have a $2d$ orbital even in a hybrid.

The ground state of a neutral sulfur atom is:

$$ [\ce{Ne}] \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow \downarrow \vert \uparrow \vert \uparrow]}$$

With this hybrid the state would be something like the following but with the electrons paired up with the orbitals they are bonded with (also the orbitals are hybridized):

$$ [\ce{Ne}] \underset{3s}{[\uparrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \uparrow]} \underset{3d}{[\uparrow \vert \uparrow \vert \; \; \vert \;\;\vert \;\;]}$$

One approach is that sulfur hexaflouride could be thought of as a less extreme case of $\ce{S^{+6} + 6F^-}$.

Another approach is that sulfur hexaflouride could be a hybrid of a bunch of $\ce{2F2 + SF2}$ arrangements. The $\ce{SF2}$ part has a tetrahedral geometry and so $sp^3$ character. The bonds are then $sp^3-p$ bonds. The total result would be a bunch of partial $sp^3-p$ bonds between the center sulfur and outer flouride atoms and some outer partial $p^3$ bonds between flouride atoms. But I'm not sure what the actual geometry this interpretation would predict or how it would differ in properties.

How are the hybrid orbitals of sulfur hexaflouride shaped?

If you limit consideration to hybridization of atomic orbitals, a good reference to see is On the role of d orbitals in sulfur hexafluoride J. Am. Chem. Soc., 1986, 108 (13), pp 3586–3593.

First, it is found that of the 6 valence electrons that atomic sulfur has (two 3s and four 3p), in SF6 a total of 3.1 electrons worth of electron probability density remain in sulfur atomic orbitals.

Of the 3.1 electrons:

0.97 are in the 3s orbital

1.82 are in 3p orbitals

0.24 are in 3d orbitals

and small fractions are in 4s, 4p and 4 d orbitals.

For a more recent study, see Theory of Hypervalency: Recoupled Pair Bonding in SFn (n = 1−6) J. Phys. Chem. A, 2009, 113 (27), pp 7915–7926


Writing half equation confusion The following is the answer for a question that I am doing:

  Reduction: $\ce{X^2+(aq) + 2 e- -> X(s)}$

Can I write $\ce{X + 2 e- -> X^2-}?$ Why can't I start with $\ce{X}$ instead of $\ce{X^2+}?$

EDIT:
The question:

Deduce the balanced equation for the voltaic cell constructed using metals B and C as electrodes.

Context:

Can I write $\ce{X + 2 e- -> X^2-}?$ Why can't I start with $\ce{X}$ instead of $\ce{X^2+}?$

The question says that the metals are $\ce{A(s), B(s), C(s) and D(s)}$. That means the letters stand for the uncharged element, not for something more arbitrary. In the table, the letters are used again, in the following way (shown for A):

$$\ce{A(NO3)2(aq)}$$

Because nitrate has a negative charge, the metal A forms an ion in water with a positive 2 charge ($\ce{A^2+}$).

$$\ce{A(NO3)2(s) -> A^2+(aq) + 2NO3-(aq)}$$

With the same argument, you can figure out that B also forms a divalent cation in solution, while C and D are monovalent. There is no flexibility of having $\ce{A}$ stand for an ion, or for the formation of a metal anion; you should stick to the species and symbols in the text and table of the question.

Examples where letters stand for something else

In acid/base chemistry, A and B are used in the following way:

$$\ce{AH <=> A- + H+}$$

$$\ce{B + H+ <=> BH+}$$

If you look closely, $\ce{B}$ stands for a base (with conjugate acid $\ce{BH+}$) and - which might cause confusion - $\ce{AH}$ stands for an acid (with conjugate base $\ce{A-}$). Neither A nor B have to be a single atom, and their charge could be anything. However, $\ce{B}$ and $\ce{A-}$ have one charge less than $\ce{BH+}$ and $\ce{AH}$, respectively.

In redox (bio)chemistry, $\ce{NADH and NAD+}$ are used. Again, the charges are to be interpreted as relative, and the relationship between the two is given by:

$$\ce{NAD+ + H+ + 2e- <=> NADH}$$


What is the citric acid concentration, [HCit], in the lemon juice? The pH of lemon juice is about 2.1.  What is the citric acid concentration, [HCit], in the lemon juice?

I am given that $HCit \leftarrow \rightarrow H^+ + Cit^-$ and that $K_a = 8.4 \times 10^{-4}$.

$K_a = \frac{[H+][Cit]}{[HCit]} = 8.4 \times 10^{-4}$, and $[H+]$ is $10^{-2.1}$. However, without knowing $[Cit]$, how can I get [HCit]? Any help would really be appreciated!

For any mono-protic acid, when the acid dissociates in water, every H+ created will create one CB-.

  H-Cit  <--> H+ + Cit-

For each 1 H-Cit. 1 H+ and 1 Cit- is produced upon dissociation.
Therefore, the [H+] = [Cit-]

For a Ka expression, set these two [] equal to each other, and the expression for [H+]*[Cit-] = x^2

Use this to determine the amount of [Cit-] created when you create x amount of [H+].


What makes polyimides good polymers for fibers? Wikipedia says polyimides are known for “thermal stability, good chemical resistance, excellent mechanical properties”. I guess that's why they're used industrially as fibers.

But where do they derive there good stability (thermal, mechanical and chemical) from? How do these properties differ from, e.g., polyamides?

Polyimides (PIs) are high-temperature engineering polymers originally developed by the DuPont™ Company. Polyimides exhibit an exceptional combination of thermal stability (>500 °C), mechanical toughness, and chemical resistance.It was based on pyromellitic dianhydride and 4,4’diaminodiphenyl ether.

image source

The presence of n-π conjugation between non-pair electron of nitrogen atom and π electrons of the carbonyl group makes them resistant to chemical agents and moisture.

image source

The linearity and stiffness of the cyclic backbone allow for molecular ordering. This phenomenon results in lower coefficients of thermal expansion (CTE) than those found for thermoplastic polymers having coiled, flexible chains.

Additionally, the morphology of long, linear ordered chains provides solvent resistance to the aromatic polyimides.

The comparatively rigid structure of polyimides provides high glass transition temperature (Tg > 300°C) and imparts good mechanical strength and high modulus.

The thermal stability of polymer can further be improved by incorporating aromatic rings on the backbone and/or side groups. In addition to such high thermal stability, the nature of the chemical structure consisting of rigid imide and aromatic rings always provides:

Excellent mechanical toughness
Excellent dielectric properties
High chemical resistance

These result in  thermoxidative stability, high mechanical strength, high modulus, excellent insulating properties, and superior chemical resistance.

Polyimide layers are suitable matrix for incorporation of metal, salts, chromophores as nanoscale particles to obtain of nanocomposite materials. It was discussed the possibility of use polyimides in materials chemistry and nanomaterials, one of these applications is the use for making biomedical implants for neurology, ophthalmology, biosensor device and chips which are a powerful tool in clinical diagnostics.

Another important trend is use in electronics and optoelectronics such as dielectric substrates and intermediate barrier layers, creating nanocomposite films with various nanosized particles such as dyes, metal, dielectric and other clusters.

Typical polyimide parts are not affected by commonly used solvents and oils — including hydrocarbons, esters, ethers, alcohols and freons. They also resist weak acids but are not recommended for use in environments that contain alkalis or inorganic acids.


Does intramolecular hydrogen bonding occur with carbonic acid and geminal diols? Is an intramolecular hydrogen bond possible in such compounds?
What are the requisites to have intramolecular hydrogen bonding?

Is this possible?

Probably not possible. An elementary explanation would be "too much ring strain."

https://en.wikipedia.org/wiki/Ring_strain

  I'm a bit confused: What are the requisites to occur a intramolecular hydrogen bonding?

5 and 6-membered rings are more favorable. As a matter of fact, beta-ketoacids, when warmed, readily decarboxylate. Why? The intermediate to decarboxylation is stabilized through hydrogen bonding.

Note the 6-membered ring in the picture below.


Finding (∂U/∂V )T = n^2a/V^2 for a van der Waals gas $\displaystyle dU = \left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV \to dU = C_v dT - C_v\left(\frac{\partial T}{\partial V}\right)_U dV$

Not sure what I do here. Thank you in advance.

The more general equation you are looking for is called the thermodynamic equation of state. You can start from the differential form of U according to the 1st + 2nd laws of thermodynamics for pV work only:

$$dU = TdS-pdV$$

Taking the derivative wrt V at constant T

$$\left(\frac{\partial U}{\partial V}\right)_T= T \left(\frac{\partial S}{\partial V}\right)_T-p$$

At this point you want to use the Maxwell relation

$$\left(\frac{\partial p}{\partial T}\right)_V= \left(\frac{\partial S}{\partial V}\right)_T$$

(which follows from the Helmholtz energy being a state function) so that

$$\left(\frac{\partial U}{\partial V}\right)_T= T \left(\frac{\partial p}{\partial T}\right)_V-p \tag{tes}$$

This last expression is the thermodynamic equation of state.

The final step is to evaluate $\left(\frac{\partial p}{\partial T}\right)_V$ for the van der Waals equation of state.


Why must basis sets used in computations be non-orthogonal? While constructing basis sets for calculations using Hartree Fock, the basis set must be non-orthogonal. And because of this, we go through many steps to put the overlap and density matrices in the 'right' basis. (See Modern Quantum Chemistry (Szabo and Ostlund) or David Sherill's notes online).

My question is: why must this basis be non-orthogonal?

It does not have to be non-orthogonal. The basis set can be arbitrary, and one usually uses the chemically motivated Ansatz with atomic orbitals (LCAO), which just happen to be non-orthogonal. An alternative would be to use a plane wave basis set, which are orthogonal.


How to explain the selectivity between methanol and methanethiol in an SN1 reaction with an halogenated hydrocarbon? Consider the SN1 reaction shown below:

Would different products be formed if $\ce{CH3SH}$ was an additional reagent?

Superimpose both energy diagrams on top of each other.

For the first question, yes, two products would be formed. I'm confused about the second question though.  I know The rate determining step would remain unchanged, as the rate depends on the concentration of the substrate not the nucleophile (SN1 is unimolecular). However, methanethiol is a better nucleophile than methanol. Would the activation energy for the second transition state be less for methanethiol?

Obviously saying "the rate is same but the activation barrier is different" is a contradiction, as both are directly related via the arrhenius equation. The rate is predominantly determined by the rate determining step. If there are several rate determining steps (aka several steps with similar activation energies), the overall rate will be a product of all steps.

So what does that mean for your question?

You stated that methanethiol is a better nucleophile than methanol. This would suggest that the activation barrier in the last step is slightly lower in the case of methanethiol. However, the rate determining first step is unaffected. Hence, the overall rate will be only slighter smaller for methanol.

The ratio of both products will be determined by the last step.


producing ozone through electrolysis of water I would like to produce low concentrations of ozone through electrolysis of water. The standard setup to do this uses a graphite electrode and a platinum electrode in an $\ce{H2SO4}$ solution.  However, I'd like to produce the ozone using only water and preferably using inexpensive electrodes.  I don't seem to understand the exact chemistry of this, though.  I'd like to know:

If I use different electrodes, what changes?  I'm currently using zinc and graphite and, other than the fact that the zinc degrades, is there any downside to using it?

The $\ce{H2SO4}$ solution, I expect, increases the conductivity of the water, is that right?  Why is it used?  What if I don't use it?

How can I go about detecting the ozone in the water?  When I run my system I seem to smell something ozone-y, but I'd like a much less subjective method than that.

First of all $\ce{H2SO4}$ produces $\ce{H2SO5}$ which is a peroxyacid during electrolysis  which most electrolytes will not do, so I suspect that $\ce{H2SO4}$ will be mandatory unless phosphoric acid ($\ce{H3PO4}$) can.  This peroxyacid give you a potential for a mechanism of ozone creation.
$$\ce{H2SO5 + O2 -> H2SO4 + O3}$$
Note: I do not claim this to be the mechanism but only an explaination for why the $\ce{H2SO4}$ may be important.

The carbon and platinum electrodes are not only chemically resistant, but may likely have a catalytic effect in producing ozone.  You can get platinum wire at a reasonable price if you look around.

In regard to testing if you have ozone, you can use paper strips that have potassium iodide impregnated in them.  The ozone will oxidize the iodide ion and turn the paper strip brown.


Relationship Atmospheric Gases and Ocean Gases If this question is better suited for the Earth Science StackExchange or another StackExchange please let me know.

Edit 1: Modified for clarity and brevity.

Something that's unclear to me is the exact relationship between gases within an atmosphere and the concentration of those gases within a large body of water under that atmosphere (like an ocean).

For example, the solubility of $\ce{CO2}$ at $298\ \mathrm K$ and $1\ \mathrm{bar}$ of atmospheric pressure is $\approx 1.496\ \mathrm{g/L}$ in water. My understanding is that number indicates how much $\ce{CO2}$ can be dissolved within water before the water is saturated with $\ce{CO2}$ and won't accept any more. However, this doesn't tell me what the concentration of $\ce{CO2}$ would be under given conditions, it just sets a maximum cap. For one thing it could be much less because there isn't enough $\ce{CO2}$ in the atmosphere to saturate the ocean.

(I hope I have that correct! If not please let me know where my understanding is going astray.)

Key Question: What are the calculations I need to go through to figure out the amount of gases in a body of water under an atmosphere of a given gaseous composition and pressure?

We could clarify this question through looking at an idealized situation (below). I'd like to know the steps to take to figure this out. If we need to ballpark it then that's ok, though I'd prefer to keep it as reasonably accurate as feasible.

Scenario

Assume uniform temperatures at water surface. Assume no minerals dissolved in water. Assume no gases have yet dissolved in the water from the atmosphere.

SURFACE

Total surface area: 510 million $\mathrm{km^2}$
Water surface area: 361 million $\mathrm{km^2}$
$15\ \mathrm{^\circ C}$ uniform surface temperature

ATMOSPHERE

By volume: $99.5\ \%$ nitrogen, $0.45\ \%$ carbon dioxide, $0.05\ \%$ methane
Surface pressure: $1\ \mathrm{bar}$

OCEAN

Pure water ocean (nothing yet dissolved in it)
Volume: $1.4$ billion $\mathrm{km^3}$

Q: What will be the equilibrium composition of gases in the atmosphere and in the water?

Q: How much of each gas will dissolve into the water?

… and of course how you got there.

What happens when you increase pressure and decrease temperature? How would that effect the answer?

What I really want to know is both general principles as well as establishing those principles in actual numbers. I'd also love any insight you may have on how real-world numbers are likely to differ and why.

Q1: The equilibrium concentration c of a gas in a solution (typically water) which is present in the atmosphere with partial pressure p dictated by Henry's Law:

$H_{cp} = c/p$

where $H_{cp}$ is Henry's coefficient for that specific gas.

This law states that the concentration of the gas in the solution is directly proportional to its pressure in the atmosphere above the liquid phase and it is a simplification generally applicable to inert, ideal gases.

Corrections must be taken into account when considering the following factors:

Temperature: The enthalpy of dissolution of the gas changes according to van't Hoff's equation. The general result is that the solubility of most gases in water decline with temperature (i.e. solubility is inversely proportional to temperature).
Chemical equilibria: Any transformations (ionization, deprotonation, etc.) that happens to the gas in solution must be taken into account by expressing its concentration by the according equilibrium equation.
Ionic strength: Generally the higher the ionic strength of the solution, the smaller the solubility of the gases in it (Sechenov equation).
Non-ideal solutions: Instead of concentrations, activities of the substances must be considered.

Q2: How much of each gas will dissolve into the water?

The calculation of the concentration for carbon dioxide, for example, is:
$$c = [(0.45\%)\times(\pu{0.986923 atm})]\times[\pu{3.4 \times 10^{-2} M/atm}] = \pu{0.0001509957 M}$$

This calculation is a direct application of Henry's Law, where the first term is the partial pressure and the second is Henry's constant. The same type of calculation goes for the others.

Observation:

It is important to notice that those volume percentages given should already be considering measures taken for those gases in our atmosphere, so they are already taking in account the solubilization. The concentration of each gas in the atmosphere can be calculated by the ideal gas equation, where the number of moles divided by the volume will be the concentration.


SCF procedure performed for some quantum-chemical methods, but not others. Why? The SCF procedure is used to find solutions to the Schrödinger equation, and obtain energy of the molecular system. It uses a "guess-and-check" approach, where an initial guess is needed to solve the equation, giving the charge density, which can be used again and again until the results become self-consistent.

So, why don't some quantum-chemical methods perform SCF, while others like HF do?

Most computation methods write the wavefunction($\Psi$) in terms of Slater determinants ($\Phi$)

$\Psi = \sum_i^L c_i\Phi_i$

which in turn are made from MOs ($\phi$), which are linear combinations of basis functions ($\chi$)

$\phi_i = \sum_\mu^K C_{\mu i}\chi_\mu$

The molecular energy depends on the MO coefficients ($C_{\mu i}$) in a non-linear fashion, so if they are variationally optimized then you need to iterate (i.e. SCF). This is the case for HF (where $L$ = 1 and $c_1$ = 1) and MCSCF (where both $\{c_i \}$ and $\{C_{\mu i} \}$ are optimized).

For methods like CI, perturbation theory, and coupled cluster, the MO coefficients are not changed, so you don't need an SCF.


Why are basis sets needed? I am not sure whether this question is even reasonable, but here it goes. We are taught about the different types of basis sets (extended, minimal, double-zeta, plane wave), but I do not think it is clear as to why they are needed. After all, it is possible to do computational chemistry without a basis set (see James R. Chelikowsky, N. Troullier, and Y. Saad, Phys. Rev. Lett. 1994, 72, 1240).

From what I understand basis sets are needed because we use LCAO. We find a set of functions (a basis set) that resembles atomic orbitals. Is this true? Or is the picture more complicated?

Spatial orbitals $\phi_i$ in modern electronic structure calculations are indeed typically expressed as a linear combination of a finite number of basis functions $\chi_k$,
\begin{equation}
    \phi_i(1) = \sum\limits_{k=1}^{m} c_{ki} \chi_k(1) \, .
\end{equation}
In the early days, atomic orbitals were built out of basis functions, while molecular orbitals were built out of atomic orbitals, which is where the name of the approach, linear combination of atomic orbitals (LCAO) originated.
Today both atomic and molecular orbitals are built out of basis functions, and while basis functions for molecular calculations are still typically centered on atoms, they are usually differ from the exact atomic orbitals due to approximations and simplifications.
Besides, basis functions centered on bonds or lone pairs of electrons, or even plane waves, are also used as basis functions.
Nevertheless, the approach is still commonly referred to as linear combination of atomic orbitals.

Now, to the very question of why do we do things this way?

On the one hand, the LCAO technique made its way into quantum chemistry as just another example of a widely used approach of reducing a complicated mathematical problem to the well-researched domain of linear algebra. To my knowledge, this was proposed first by Roothaan.1 However, and as it was mentioned by Roothaan from the beginning, the LCAO approach in present day quantum chemistry is also attractive from the general chemistry point of view: it is tempting to construct molecular orbitals in modern electronic structure theory from their atomic counterparts as it was done by Hund, Mulliken and others already in the early days of quantum theory2 and as it is though in high-school general chemistry courses today.

1) Roothaan, C.C.J. "New developments in molecular orbital theory." Reviews of modern physics 23.2 (1951): 69. DOI: 10.1103/RevModPhys.23.69

2) Pauling and Wilson in Introduction to Quantum Mechanics with Applications to Chemistry refer to the following works in that respect (p. 346):

  F. Hund, Z. f. Phys. 51, 759 (1928); 73, 1 (1931); etc.; R. S.
  Mulliken, Phys.  Rev. 32, 186, 761 (1928); 41, 49 (1932); etc.; M.
  Dunkel, Z. f. phys. Chem. B7, 81; 10, 434 (1930); E. Hückel ,Z.f.
  Phys. 60, 423 (1930); etc.

Hund's papers are unfortunately in German, but Mulliken's ones are quite an interesting read. Especially the second one,

Mulliken, Robert S. "Electronic structures of polyatomic molecules and valence. II. General considerations." Physical Review 41.1 (1932): 49. DOI: 10.1103/PhysRev.41.49

which (again, to my knowledge) introduced the very term "molecular orbital".


Iodine formation reaction $$\ce{IO3-(aq) + 5I-(aq) + 6H+(aq) -> 3I2(aq) + 3H2O(l)}$$
Why does the reaction between iodide and iodate ions have to occur in an acidic medium ($\ce{H+}$) to form the iodine? what do the $\ce{H+}$ ions react with?

A bit more complex per a source 'Kinetics and mechanism of the iodate-iodide reaction and other related reactions', to quote from an article in the journal Physical Chemistry Chemical Physics (November 1999):

The reduction of iodate by iodide, known as the Dushman reaction, is one of the components of the oscillating Bray-Liebhafsky reaction. The large body of work devoted to the kinetics of the Dushman reaction gives seemingly contradictory results. This work analyses and summarizes them and shows that they are special cases of a complicated kinetics and that they can be explained by a common mechanism. This mechanism can be generalized to all the XO3- + Y- reactions (X, Y = I, Br or Cl). They all involve an asymmetric intermediate YXO2.

See reaction system depicted in Equations (8) to (13). Here is a partial discussion of the system from the available PDF to quote:

The essential hypothesis of this mechanism is the existence of as a definite intermediate species and not as I2O2 an activated complex. Its existence was already proposed in  1930 by Bray [17] in order to explain the classical rate law. We make no distinction between I2O2 and H2I2O3 (I2O2 + H2O) as these two forms are kinetically equivalent. In reaction (12), B~ represents the anion of the buffer....

As to the question on the role of H+, note that the starting Equations (8) and (9) of the system both involve H+, which results in the cited intermediate $\ce{I2O2}$:

$\ce{IO3- + H+ ⇌ IO3H}$  (8)
$\ce{IO3H + I- + H+ ⇌ I2O2 + H2O}$  (9)


How to report pH values? I have measured the pH in the water-soluble fraction of a powder. I weighed in a certain amount of powder and a certain amount of water, and extracted the water-soluble fraction. I then measured the pH in this extract.

How do I report this? The pH depends on the amount of powder and the amount of water. Should I give the weighed-in amounts along with the pH-value? Or is there a way to "normalize" the pH-values to make them comparable for all? Just giving the pH values make no sense, as no one can compare anything to them.

Will it become more clear for a reader if you use an an abbreviated, "coded" form to express what you have actually done?

If not, why waste your time to encode it and waste the time of the reader to decode it (and maybe still leave space for ambiguity)?

I'd be fine with a report that precisely describes the experimental procedure and the observations.


When using LPCVD, is it possible to pattern a silicon wafer with poly-silicon using a mask? I know it's possible to etch (i.e. pattern) poly-silicon, but I'm curious if one can just use a mask instead. Are there masks that inhibit the growth of poly-silicon?

In my fab experience, the answer is no, you are not going to inhibit deposition of silicon with some other material. You might well change the final grain size, but it will deposit. Furthermore, LPCVD generally has good sidewall coverage, so you can't use a lift-off mask either.

The OP asked for references, so lets dig in to those. First, it is somewhat difficult to find specific references to deposition rates on various substrates that can be broadly compared - reactor specifics, gas flow, temperatures, etc., all make it hard to compare experiment-to-experiment. But, we can pin down a few things.

First, going back to Andy Grove's (yes, Intel's Andy Grove) book "Physics and Technology of Semiconductor Devices" (John Wiley and Sons, New York NY, 1967) one finds no mention of polysilicon deposition. At that early stage in technology it was not used - metal gates were the way to go.

Polysilicon became a thing with poly-Si gates and the LOCOS (local oxidation of silicon) process. Here one might end up depositing poly over a wafer that included single crystal silicon (to make source-drain contacts), oxide (gate and field isolation), and silicon nitride (for the LOCOS process). Looking at cross sections from LOCOS devices (in various more modern text books - one I have is Stephen A. Campbell, "The Science and Engineering of Microelectronic Fabrication" (Oxford University Press, New York NY, 1996)), you see no variation in poly thickness across those different materials.

There is much more literature available around questions like polysilicon grain size vs initial surface roughness, annealing temperature, etc., such as Voutsas and Hatalis, "Surface Treatment Effect on the Grain Size and Surface Roughness of as-Deposited LPCVD Polysilicon Films", J. Electrochem. Soc. 140(1) 282-288 (1993).

Finally, a number of people have done work on the kinetics of LPCVD poly-Si growth, with one early reference being M.L. Hitchman et al., "Polysilicon growth kinetics in a low pressure chemical vapour deposition reactor", Thin Solid Films 59 231-247 (1979).

There are two takeaways from papers like this. First, under LPCVD reactor conditions, the mean free path of a silane molecule (or any other reactant) is quite long. This is, of course, highly desired to have uniform deposition across many wafers in, e.g., a tube furnace. The flip side is that the probability of a silane molecule reacting upon hitting a surface is quite low - they bounce around a lot. And, once a surface has any poly-Si on it, it all looks the same.

Second, the only way to not have poly-Si deposition is to never allow nucleation in the first place. Yet, point one above implies that lots of silane is hitting the surface, and if any one molecule manages to stick you now have a place to grow more. Further, all that seems to be needed to get some silane reactivity is the presence of hydrogen on the surface, and there will be lots of hydrogen around as silane decomposes elsewhere (plus hydrogen on surfaces is pretty much a given - it is by far the easiest surface termination on just about anything, particularly after a chemical process in an aqueous solution, like an acid clean).


Understanding Equivalent Wyckoff positions I'm somewhat confused by the concept of Wyckoff positions in crystal structures. From Quantum Chemistry of Solids by Robert Evarestov, the definition of a Wyckoff position is "all the crystallographic orbits which have the same (not only isomorphic, but the same) site-symmetry group". I was okay with this, but he then went on to define equivalent Wyckoff positions as positions such that "equivalent sites have the same point-group symmetry and the same orientations of symmetry elements with respect to the lattice".

What I fail to understand is how the equivalent Wyckoff positions don't collapse to a single Wyckoff position, as they seem to meet the criteria of having the same site-symmetry. Is there a clear explanation of what distinguishes two Wyckoff positions enough to be equivalent, but not just a single position?

To clarify my question a little more, here is the specific example given in the text for the space group 136, also called P42/mnm or $D_{4h}^{14}$.

  Pairs of Wyckoff positions a-b, f-g, and i-j have isomorphic site-symmetry groups
  ($D_{2h},C_{2v},\text{ and } C_s$ respectively). As is seen from the table, the pairs of points a-b and f-g not only have isomorphic, but also equivalent Wyckoff positions. The equivalent sites have the same point group symmetry and the same orientations of symmetry elements with respect to the lattice.

It seems to me that care has to be taken with the words 'equivalent'  and 'same' in the definitions above. Equivalent positions do not mean the same position. In the case, say, of a 4 fold axis any point say in the first quadrant can be replicated to the other three quadrants and these are equivalent points but not the same point, clearly. In some cases such as with reflections in a plane, two sets of identical (or equivalent) points may be produced and so and the number of equivalent points is reduced. The number of equivalent sets is called the multiplicity. These have been called special positions when they lie on a symmetry element

In the Wycoff notation equivalent positions are labelled consecutively $a, b, c\cdots$ starting with $a$ for the highest symmetry. A position 2f, for example, indicates two equivalent atoms at position given by symmetry f. For each space group a table of general and special positions can be made.  This includes point symmetries, positions multiplicities and Wycoff symbols. The table below shows examples.

Table copied from  page 75 of 'Introduction to Crystallography' by  D. Sands, publ. Dover


Configuration of endocyclic double bond in Vitamin A I think the double bond inside the ring of vitamin A should have an (E) configuration. Am I wrong? I don't think I'm assigning the priority correctly.

The endocyclic double bond of vitamin A has a (Z) configuration. Here’s why:

First, let’s determine priorities of the atoms connected on the top side of it:

C

C (ring)
C (methyl 1)
C (methyl 2)

C

C
(C) (ghost atom; due to the double bond)
H

So the ‘ring side’ — the atom on the left in the way it is drawn at the moment — has a higher priority than the side chain (C at (1.3) beats H at (2.3)).

What about the bottom side?

C (ring)

C
H
H

C (methyl group)

H
H
H

Here again the ‘ring side’ has a higher priority than the methyl group: (1.1) carbon beats (2.1) hydrogen.

Both left-hand sides have a higher priority than their corresponding right-hand side. Thus, the double bond is (Z) configured.


Why not to consider hydrogen and oxygen moles to determine an empirical formula A question says:

Find the empirical formula of an organic compound from the following composition:

34.62% C, 3.88% H, 61.50% O.

The answer is $\ce{C3H4O4}$.
It was found by using the mass (percentages) divided by the molar mass of each element.

But they didn't consider the (2moles) of hydrogen and oxygen in calculations. Why?
in this case it will be $\ce{CH3O3}$

On the other hand, in a similar question in the text book, they consider the moles of hydrogen and oxygen in calculations by multiplying by 2 after getting the moles.

Could anybody explain?

The answer is $\ce{C3H4O4}$. It was found by using the mass (percentages) divided by the molar mass of each element.

Exactly! That's the way to do it.

  But they didn't consider the (2moles) of hydrogen and oxygen in calculations. Why? In this case it will be $\ce{CH3O3}$

No, they did not! As you have pointed out yourself, it is about the empirical formula of a compound in question, such as one of the following:

It is completely irrelevant whether an element usually exists as a diatomic molecule in nature. If that would matter, how would you then treat sulfur-containing compounds? Note that natural sulfur typically consists of $\ce{S8}$ molecules! How would you treat elements that only appear in the form of compounds in nature but never as pure elements.

In summary, the "natural appearance" of elements in nature does not matter here, it is only about the relative atomic abundance of an element in the compound in question.


Kinetics of a simultaneous parallel radioactive decay Question:

A radioactive isotope, A undergoes simultaneous decay to different nuclei as:
\begin{array}{cc}
\ce{A->P}&\,(t_{1/2}=9\ \mathrm h)\\
\ce{A->Q}&\,(t_{1/2}=4.5\ \mathrm h)
\end{array}
Assuming that initially neither P nor Q was present, after how many hours, amount of Q will be just double to the amount of A remaining?

My solving:
I got the $t_{1/2}$ for $A$ as $T=\dfrac{T_1T_2}{T_1+T_2}\ =\pu{ 3 h}$
Now from Rutherford-soddy law the number of atoms left after $n$ half lives are,
$$N=N_0\left(\frac{1}{2}\right)^n$$
and number of half lives $n=\dfrac{\text{Total time of Decay}}{\text{effective half life}}$
For isotope $A$, $\dfrac{N_A}{N_{0_A}}=\left(\dfrac{1}{2}\right)^{n}=\left(\dfrac{1}{2}\right)^{T/3}$ where $T$ is the common time of decay for $\mathrm A$ and $\mathrm Q$
For $\mathrm Q$ I had to assume that $\mathrm Q$ is also undergoing radioactive decay to form a relation between $\mathrm A$ and $\mathrm Q$
Now using the same formula as above $\dfrac{N_\mathrm Q}{N_{0_\mathrm Q}}=\left(\dfrac{1}{2}\right)^{T/4.5}=\left (\dfrac{1}{2}\right )^{2T/9}$
Now applying the condition in the question, $$\left (\frac{1}{2}\right)^{2T/9}=2\cdot \left (\dfrac{1}{2}\right)^{T/3} $$
Raising by power $9$ on both the sides,
$$\left (\dfrac{1}{2}\right )^{2T}=2^9\cdot \left (\frac{1}{2}\right )^{3T}$$
which implies,
$$ 2^{-2T}=2^{9-3T}$$
Therefore,
$$9-3T=-2T$$
$$
{\bbox[10px, border:2px solid red]{
 T=9\ \mathrm h.
}}
$$
Although I have got an answer and it satisfies the condition I still think it's wrong because of the assumption made for $\mathrm Q$ and something else needed to be done to make the relation between $\mathrm A$ and $\mathrm Q$ Moreover, it's given in the question that initially neither $\mathrm P$ nor $\mathrm Q$ was present and I guess the assumption violates that.

The question has already been solved by Yashwini and the answer given is correct.$^2$ A more intuitive and specific to question explanation would follow here.
Now, the two reactions given are:
\begin{array}{cc}
\require{cancel}
\ce{A -> P} &(t_{1/2} = 9\,\mathrm h) \\
\ce{A -> Q} &(t_{1/2} = 4.5\,\mathrm h) \\
\end{array}
Now using the rate law, we get,
\begin{align}
-\frac{\mathrm{d}[A]}{\mathrm{d}t}&=k_\mathrm P [A] \tag{1} \\
-\frac{\mathrm{d}[A]}{\mathrm{d}t}&=k_\mathrm Q [A] \tag{2} \\
\end{align}
The rate constant for a first order reaction having a half life of $t_{1/2}$ is defined as:
$$k=\frac{\ln 2}{t_{1/2}} \tag{3}$$
Now, substituting the given values of $t_{1/2}$ into the equations, we get $2k_\mathrm P = k_\mathrm Q$ (since $k\, \alpha \frac{1}{t_{1/2}})$
Now, intuitively since both reactions take place together, it would mean that for every one mole of P formed, two moles of Q forms. Therefore, for every mole of P formed, three moles of A react (since one mole is required for each mole of P and Q).
Now, we add the rate laws ($1$) and $(2)$, since the reactions take place simultaneously, to get:
$$-\frac{\mathrm{d}[A]}{\mathrm{d}t}=(k_\mathrm P +k_\mathrm Q) [A] \tag{4} $$
Now, since using the relation between $k_\mathrm{P}$ and $k_\mathrm{Q}$, we get $k_\mathrm{P} + k_\mathrm{Q} = 3k_\mathrm{P}$
Therefore using the integrated rate law for a first order reaction on equation $(4)$, we get:
$$A=A_0e^{-3k_\mathrm Pt} $$
Now, the amount of $A$ used here would be $A_0 -A$, and we get that value to be:
$$A_\text{used}=A_0\left(1-e^{-3k_\mathrm Pt}\right)$$
Now, as we have previously noted, for every three moles of A used, two moles  Q are formed. This means that the amount of Q now in the mixture would be two thirds of $A_\text{used}$. Therefore the amount of Q would be:
$$Q=\frac{2A_0\left(1-e^{-3k_\mathrm Pt}\right)}{3}$$
Now, we are given the condition, $Q = 2A$, substituting values of $Q$ and $A$ into the given relation we get:
$$\begin{align}
\frac{\cancel{2A_0}\left(1-e^{-3k_\mathrm Pt}\right)}{3} &= \cancel{2A_0}\left(e^{-3k_\mathrm Pt}\right) \\
\implies 1 -e^{-3k_\mathrm Pt} &= 3e^{-3k_\mathrm Pt} \\
\implies 4e^{-3k_\mathrm Pt} &= 1
\end{align}$$
Solving for $t$, we get:
\begin{align}
3k_\mathrm Pt&=2\ln 2 \\ \\
t&=\frac{2\ln 2}{3k_\mathrm P}\\
\end{align}
Now, using equation $(3)$, we get the rate constant $k_\mathrm P$ to be $\frac{\ln 2}{9}$. Substituting this value into the expression for time, we get:
$$t=\frac{2 \cancel{\ln 2}}{\cancel{3} \frac{\cancel{\ln 2}}{\cancelto{3}{9}}}$$
Therefore, time taken for this condition to happen is:
$$t=2\times 3 = 6\ \mathrm h$$


Does nitrogen pressure in tyres fluctuate to much a lower degree than air with changes in temperature? This is my first question in this site. I am referring to this question.The linked question what is the advantage of filling nitrogen in bike tubes over air and I got some answers such as:

The biggest advantage to filling your tires with Nitrogen is consistency.
Air pressure fluctuates based on temperature, whereas Nitrogen fluctuates to a much much lower degree

and

Nitrogen changes pressure with temperature slightly less than water does

But I want to know the reason. Please help me understand this scientifically.

There can be no scientific explanation, because the argument itself is incorrect.

For the most part air is nitrogen.  To be specific, about 80% of dry air is made up from nitrogen.  In fact both nitrogen and oxygen, the second major components of air that takes up almost all the other 20%, behave extremely like ideal gas.   Pressures of dry air and pure nitrogen gas at different temperatures, from far below freezing point to above boiling point, have indeed been measured to incredibly high accuracy.  The difference between these two gases is below 0.1%.   In fact, near room temperature, many types of gas are close to ideal gas and have very similar temperature profiles.

I am not sure why people are comparing nitrogen with water in that question thread you mentioned.  Water is a minor component of air and its concentration varies so dramatically across places.  However, even in a summer day, at the maximum humidity of 100%, the amount of water in the air is only 10% of that of oxygen.

Basically, pressure of nitrogen gas do not fluctuate less with change of temperature compared to normal air.


Which chemical reactions don't involve bond breaking? I recently came across a question in which an option is 'Breaking a chemical bond is the first step in any chemical reaction', which lead me to think of reactions which do not involve bond breaking. The only reaction I can think of is dimer formation, but a Google search reveals no more. Surely this cannot be the only example?

The reaction of a Lewis acid with a Lewis base results only in bond formation. Your question gets at the essence of chemistry, though. A chemical reaction is the rearrangement of electrons between atoms. So generally, most reactions will involve breaking bonds. Even for monomers (e.g. alkenes) to polymerize, $\pi$-bonds within the monomers have to be broken to form $\sigma$-bonds in the polymer. You seldom have one without the other.


Organic Nomenclature: But-1-en-3-yne or But-3-yne-1-en Just wondering whether "but-1-en-3-yne" or "but-3-yne-1-en" follows the correct IUPAC naming convention for organic molecules. I'm going through a text book that uses the former in an example, but in the answer to one of the questions it uses the latter, so is it interchangeable or is one just wrong?

The IUCN Blue Book says that:

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride [...] is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

Therefore the double bond gets priority whatever be its locants (numbering) i.e., '-ene' comes before '-yne'. And in this case it's "but-1-en-3-yne".


Which formula is correct for calculating the heat of dissolution? I want to calculate the energy change when a solute is dissolved in water. I know that I can achieve this by using the equation $q=mc \Delta T$. My question is does the mass ($m$) change with the amount of solute added to the solvent?

For example, if I had 25g of water ($c=4.18$) as my solvent and 5g of sodium chloride as my solute. Would the energy change equal:

$$q=25 \cdot 4.18 \cdot \Delta T$$
or
$$q=(25+5) \cdot 4.18 \cdot \Delta T$$

Your energy conservation equation is $$q_{rxn} + q_{soln} = 0$$ so $$q_{rxn} = -q_{soln} = -m_{soln}c_{soln}\Delta T_{soln}$$ Note the minus sign; if $\Delta T$ is negative, the energy will be positive, and vice versa. And yes, you need the mass of the solution here, and the specific heat capacity of the solution too.

If you only need a couple of significant figures and the solution isn't too concentrated, you can assume $c_{soln} \approx c_{\rm H_2O}$.

If you need more precision, though, you'll need to measure or look up $c_{soln}$. And you'll need to include the heat absorbed by the calorimeter in your energy conservation equation, too.


I'm looking for a substance that can be used as an adhesive, but that will burn without producing any ash I need to 'adhese' paper to plastic, but I need the adhesive to be flammable and quick burning, but without producing any ash or too many toxic fumes.

I know this is an absurd question, but does anything like this exist?

Thanks so much

Nitrocellulose(flash paper). Using acetone or something you get a gel. Caution recommended.


MacGyvered chlorine gas protection I was watching Periodic Video's chlorine video, which discusses chlorine's ferocity in stripping electrons whenever possible, and the professors mentioned its consequent use in WWI as a chemical agent. This got me thinking, if I were in that situation (say Syria if they decided to go old-school or a chemical leak), what could I do to protect myself against gaseous chlorine using the chemistry of everyday materials around me?

I imagine a wet towel won't help much to prevent the chlorine from getting in my lungs. Are there any easily obtainable substances I could safely breath through that would neutralize the chlorine? Or are there any (realistic) MacGyver solutions to quickly building a semi-functional gas mask? Just kinda curious what could be done.

Interestingly, the British had such a solution in WWI: the Hypo Helmet. This was simply a bag soaked in sodium thiosulfate solution and worn over the head.  Thiosulfate reduces chlorine (and halogens generally) to chloride (halide) ions:

$\ce{2 S2O3^{2−} + Cl2 → S4O6^{2−} + 2 Cl^{−}}$

As the article notes, this was not the best solution, but I think it fits your criteria for rough and ready protection.


Diluting 4 % NaOCl to get a 1 % solution Assuming that one has a 1 litre bottle of liquid bleach containing $4~\%$ sodium hypochlorite. If I wish to dilute it down to $1~\%$ sodium hypochlorite (to use as disinfectant), would adding 3 litres of water to it, be sufficient? This should yield 4 litres of $1~\%$ sodium hypochlorite.

And that should be same as taking $\pu{250ml}$ of the liquid bleach, and then adding $\pu{750ml}$ of water, to yield 1 litre of $1~\%$ sodium hypochlorite.

Now the liquid bleach also contains a stabiliser to reduce the rate of decomposition of $\ce{NaOCl}$. How will diluting the solution affect the action of the stabiliser?

There is a simple dilution formula: $\mathrm{C_iV_i=C_fV_f}$, which is valid for all concentration units. The "i" and "f" indicate initial and final concentrations or volumes.

Using this formula, you can see that you don't need to prepare buckets full of 1% bleach. All you need to decide is what should be your final volume and final concentration.

As you highlighted, the issue, you cannot prepare a 1% solution and expect it to last for a day. It is better to prepare on need basis. The stabilizer in bleach is sodium hydroxide itself. Upon dilution, of course the base is also diluted and the pH is reduced.

One can look at the distribution diagram of "bleach" as a function of pH (yes this is from a Handbook of Surfactant. It is a 4000 paged volume- books are still useful)


Why is ammonia less acidic than terminal alkynes? Here's a picture from Organic Chemistry by Solomons, Fryhle & Snyder :

Why is it that ammonia is less acidic than terminal alkynes even though water, and even alcohols, are more acidic than them?

Water and ammonia seem to be very similar to me:

Electronegativity difference of oxygen and nitrogen is only 0.4 (Pauling scale)
Both are very similar and differ by only number of hydrogen atoms.
Both are $\ce{sp^3}$ hybridized
Oxygen and nitrogen atoms have similar size

However, the difference in acidic strength is huge. If my understanding of $\mathrm{p}K_\mathrm{a}$ values is correct, water is nearly $10^{10}$ times more acidic than terminal alkynes, whereas ammonia is less acidic than terminal alkynes by a factor of $10^{13}$!

Shouldn't the list be like: water > alcohols > ammonia > terminal alkynes > terminal alkenes > alkanes or water > ammonia > alcohols > amines > terminal alkynes > terminal alkenes > alkanes ?

It is the alkyne that is not in line.
In the alkynes, the carbon is sp hybridized and this gives it much more s character.
Because of this, the electrons are closer to the carbon nucleus and the $\ce{C-H}$ bond is slightly polarized.
The carbon has some δ - and the hydrogen δ +.

Therefore the terminal alkynes are more acidic and have a lower $\mathrm{p}K_\mathrm{a}$ than ammonia.


Truncated CI Wavefunction In a calculation of for example a Li-Atom in a DZ basis we have 4 MO.

For the alpha string (containing two electrons), that means we have for a FCI calculation the determinants:
|12>, |13>, |14>, |23>, |24>, |34> (electrons in MO 1 and 2, 1 and 3 and so on).

If I wanted to do a truncated CI calculation (CIS for example) only singly excited wavefunctions would contribute.

My (maybe extremely dumb) question now is: Are the determinants |23> and |24> singly excited or doubly excited? Is it the excitation
$$a^\dagger_3a_1|12> = |32> = -|23>$$
 or is it
$$a^\dagger_2a_1 a^\dagger_3a_2|12> = |23>$$
 Or is it both? And the coupling (1,-1) so to say determines whether this in used as a single or doubly excited determinant?

EDIT: From the "Molecular Electronic Structure Theory" book by Helgaker, Jorgensen and Olsen, from the chapter 11.8.5 (Direct CI algorithms for RAS Calculations) I read "In CISD theory, there are three such classes (string classes): class 0, with no electrons in virtual space; class 1, with one electron in virtual space; and class 2, with two electrons in virtual space".

This for me reads as if |23> would count as a class 1 string and not a class 2.

In configuration interaction (and in coupled cluster) we typically think of excited determinant being caused by single particle excitations of the form $\hat{t}^a_i = \hat{a}^\dagger_a\hat{a}_i$ where $i$ labels a virtual orbital and $a$ labels an occupied orbital, causing an excitation of the Hartree Fock reference determinant $\Psi_0$.
The resulting CI expansion would go:
$$
\Phi_\text{CI} = \Psi_0 + \sum_{i\in occ\\ a \in vir} \hat{t}^a_i \Psi_0 + \sum_{i,j\in occ\\ a,b \in vir} \hat{t}^{ab}_{ij} \Psi_0 + \sum_{i,j,k\in occ\\ a,b,c \in vir} \hat{t}^{abc}_{ijk} \Psi_0 + \: ...
$$
Hence, in the calculation on Li mentioned above, we would not consider the excitation $\hat{t}^2_1 = \hat{a}^\dagger_2\hat{a}_1$ in the first place as it would annihilate the Hartree Fock determinant $|12>$ containing a particle in orbital 2.
By this logic the determinant $|23>$ is a singly excited determinant caused by $\hat{t}^3_1= \hat{a}^\dagger_3\hat{a}_1$, or class 1 as described by Helgaker.
Note:
As determinants $|23>$ and $|32>$ only differ by a multiplicative constant (-1) only one needs to be considered and the resulting coefficient/cluster amplitude will take care of the relative sign.


treatment of alkyl chloride with aqueous KOH and alcoholic KOH The treatment of alkyl chloride with aqueous KOH leads to the formation of alcohol but in the presence of alcoholic KOH alkenes are major product. Why?

I have added an image to answer your question, always look at the acid base reactions which are possible to take place, that helps in understanding the reactions and their respective mechanisms.


Is azide paramagnetic? Is the azide ion ($\ce{N3-}$) paramagnetic?

How do I use molecular orbital theory to show that it is or isn't?

I drew a sine wave diagram and filled up the $1\pi$ and $2\pi$ MOs, and since there are no unpaired electrons, can I conclude that the ion is not paramagnetic?

Deriving a proper MO digram for N$_3^-$ is not completely trivial a task (e.g. a random google hit, which seems to me well done: www.chem.uci.edu/~lawm/Azide%20MO%20Diagram.pdf). In the end you will have to show that the degeneracies of the orbitals together with the electron count will not require open shells. Beware of Lewis-structures and mesomery, think of O$_2$, it looks perfectly o.k. to draw O=O, only the MOs and their degeneracies will reveal that its paramagnetic.


Freezing point vs Intermolecular forces Water has a higher boiling point (100°C) than cyclohexane (81°C). This is probably because of stronger intermolecular forces between water molecules as compared to cyclohexane molecules. Then, why is the freezing point of cyclohexane higher than that of water? By the previous logic, water should be having a higher freezing point as well.

This is one among many examples. I am pretty sure than boiling point depends primarily on intermolecular forces while freezing point depends on both intermolecular forces and packing efficiency in the solid state. I am sure that both factors support a higher freezing point for water. So, what other factors govern the freezing point of a liquid substance?

This is a tough question to answer because the intermolecular distances are similar in the  solid to liquid transition unlike those in the liquid to gas phase transition.

In the case of the elements there is a correlation between the Debye temperature and the melting temperature. The Debye temperature is that temperature at which the atoms gain their full degrees of freedom, and the correlation, while not prefect  provides evidence for a connection between melting point and inter-atomic vibrations.

The idea (originally by Lindemann ) is that when the temperature reaches a certain value, the potential energy of the (harmonic) intermolecular vibration has now increased by a  sufficient amount that atoms become independent of one another and the solid melts. Thus the energy profile of a solid is like an egg box with many depressions in which the atoms reside and have some limited freedom, given a bit more energy they become free. An approximate calculation shows that only a 5% increase in intermolecular spacing is enough to cause melting, which to me seems very small. Clearly this is an over simple model but does give some insight into the problem.

In your example it is far more difficult to give any quantitative answer and many examples should be studied to understand the general picture. If we follow this simple model we must assume that the vibrations are of whole CX or water molecules moving in their intermolecular potential.  Hopefully someone who is an expert in this area can provide you with a more satisfying answer.


Gas law with molecules at the same time How many grams of oxygen are present at 25.0 C in a 5.00L tank of oxygen enriched air? The total pressure is 30.0 atm and the only other gas is nitrogen with a pressure of 17.0 atm.

After converting the temperature to Kalvin, I do not really know what to do.
How can I find the grams of oxygen with another molecule in the container?

We have 2 gases in the container, oxygen ($O_2$) and nitrogen ($N_2$). Now we are given that $N_2$ has a pressure of $17$ atm out of $30$ atm, the total volume of the container is $5.00$ L and that the temperature is $25^o$C or $298$ K. So somehow, we need to find out the volume that oxygen occupies and from there we will be able to find out the grams of oxygen present in the container.

First let's find out the mole fraction $N_2$.
You can find mole fraction by the following formula,
 $$Mole\, Fraction = \frac{Partial\,Pressure\,of\,Gas}{Total\,Pressure}$$

So,
Mole fraction of $N_2$= $\frac{17}{30}$=$0.6$

Therefore mole fraction of $O_2$=$1-0.6$=$0.4$

Now we need to find out the total no. of moles by the formula,
$$PV={nRT}$$

Putting the values given for the whole system,
$$30\times5=n\times0.0821\times298$$
$$\Rightarrow\,n=6.1$$

So moles of oxygen in container = $0.4\times6.1$ = $2.5$
Consequently, grams of $O_2$ present in container = $2.5\times32$ = $78.4$

So that is how you get the grams of oxygen. Hope you understood everything. If not, feel free to ask by commenting. Good Luck!


Why is Na+ reduced instead of H+ in the electrolysis of dilute NaCl(aq) with Mercury cathode? In a class of electrolysis, my instructor told me that Hg forms Na-Hg in the electrolysis of dilute NaCl aqueous solution. For this reason, sodium cations are reduced in the cathode instead of hydronium ions. However, he did not give me any reason for how creating an amalgam affects the reduction of the metal or, why Hg creates amalgam in the first place.
So, I would like to know why Hg creates amalgam whereas other metals like Pt don't form them ? And also how this affects the reduction of Na.

The mental conflict occurs because two processes could theoretically happen (electrodeposition of sodium, or electrodeposition of hydrogen), and the theoretically unfavorable one occurs - and not only occurs, but is so favorable that it was used commercially to produce NaOH as EdV noted.
So why is the theoretically favored process not favored in this instance? Hydrogen evolution is a multi-step process, and if you wave your hands a lot, you can call it overpotential and be off to other projects.
When (or if) sodium is deposited (on mercury), it can dissolve and be removed from contact with water. So this process is a slam-dunk if you push with a high enough voltage.
But the hydrogen evolution should take place at a lower voltage. Why not? If/when hydrogen is deposited, its first appearance is atomic: one H atom sticks somewhere on the mercury surface; the second step is deposition of another H atom somewhere else. The next step, for the two atoms to form a molecule, requires some mobility on the part of the hydrogen and the mercury surface. At this point, conventional wisdom suggests that the molecule could dissolve into the solution, but eventually the aqueous solution becomes saturated. Then the first molecule must meet another molecule, and another, to begin the process of forming a bubble which will eventually rise to the surface of the aqueous solution. Somewhere in there is an impediment to the formation of bubbles, and it may involve mobility of atoms or molecules, or attachment of hydrogen to mercury, but the process is difficult enough to allow the voltage to be raised high enough to allow deposition of sodium to proceed.
And then we say that mercury has a high overpotential for deposition of hydrogen.
Now suppose we do the same, but with gallium. Sodium should deposit and become dissolved as with mercury, if the voltage is high enough. (Sodium alloys with gallium; amalgamation is a chemical term that should be reserved for mercury alloys.) But the hydrogen overpotential on gallium is less than on mercury (Ref 1), and gallium and hydrogen are deposited together from a solution of $Ga^{+3}$ (Ref 2). So, in the case of liquid gallium, it seems that hydrogen would be more easily deposited than sodium (unless, perhaps you used a very high voltage, which would override the thermodynamic preference because diffusion couldn't keep up with the current demands).
Ref 1. https://pubs.rsc.org/-/content/articlelanding/1966/tf/tf9666203524/unauth#!divAbstract
Ref 2. https://link.springer.com/article/10.1007/BF01016217


Intuitive explanation for depression of freezing point Why does depression of freezing point takes place if a non-volatile solute is added to the solution?

For more details about the question checkout this link below

Depression of freezing point

Imagine a pure liquid. If part of it freezes, what happens to the rest of the solution? Well, it looks exactly the same since it's also pure, so nothing.

However, this is not true for the system you've described.

What happens if there's a solute mixed in liquid? When, now you're concentrating the solute in the remainder of the liquid, so you're creating an entropic barrier to freezing the liquid.

Freezing is a process with $\Delta H < 0$ and $\Delta S < 0$. This means that with $\Delta G = \Delta H - T\Delta S$, the process is spontaneous when $T$ balances the favorable enthalpy change against an unfavorable entropy change. This $T$ is the freezing point. However, notice that due to the above argument, $\Delta S$ is now more negative. This means that the same balance will be achieved with a smaller value of $T$, hence, a freezing point depression.


Buffers: what does the strong base react with? I am a bit confused on buffers:
If I have a solution with a weak acid (say $\ce{NH4}$) and its conjugate weak base ($\ce{NH3}$), I have a buffer solution (in $\ce{H2O}$).

  $$\ce{NH4 <--> NH3 + H+}$$

But now If I add a strong base to my solution ($\ce{OH-}$ for example) I get that my pH will go up but I don't understand the dynamics of it.

Will my added $\ce{OH-}$ react with the weak acid $\ce{NH4}$ (more $\ce{NH3}$ produced)?

$$\ce{NH4+OH -> NH3 + H2O}$$

OR will my added $\ce{OH-}$ react with the $\ce{H+}$ produced by my weak acid (Less $\ce{H+}$ produced)?

$$\ce{NH3 + H+ (products of $\ce{NH4}$) + OH- -> NH3 + H2O}$$

I understand that both cases up end up with the pH going up but but I want to be sure to understand with what exactly my $\ce{OH-}$ molecule reacts once added to my solution.

Is it the $\mathrm{1^{st}}$ reaction which means more $\ce{NH3}$ produced?

Or is it the $\mathrm{2^{nd}}$ reaction which means less $\ce{H+}$ produced?

The reaction NH$_4^+$ $\Leftrightarrow$ NH$_3$ + H$^+$ is in a dynamic equilibrium that means the reaction will happen in both directions all the time and the acid dissociation constant will tell you how much on average you have all components in solution so by that you can also calculate the pH value. So of course since you have NH$_4^+$ in solution as well as H+ both can in principle react with the OH$^-$ in providing the proton to form water. The question is, which of the two reactions happens more. This will be determined by the two equilibrium constants

$K_{1} = \frac{[NH3][H2O]}{[NH4+][OH-]}$
$K_2   = \frac{[NH3][H2O]}{[NH3][H+]}$

You see the only difference are the concentrations of the educts, so how large are these? Suppose you prepare a 0.05M NH$_4$Cl solution. After a short time there will be thermodynamic equilibrium and by using the $pK_\mathrm{a}=9.25$ value for NH$_4$Cl you can calculate the respective concentrations of NH$_4^+$ and NH$_3$ in equilibrium:

$K_a=\frac{[H+][NH_3]}{[NH_4^+]}$

By using $[H^+]=[NH_3]$ we get

$10^{-9.25}=\frac{[x][x]}{0.05-[x]}$
gives $[x]=5 \cdot 10^{-6} M$, so actually not much has dissociated. So you can be nearly completely sure that the reaction takes place with the NH$_4$ rather than protons or hydronium ions. And that is what a buffer is for! The pH should not change even when you add $OH^-$ until you reach the stability window of the buffer, that is when the NH$_4^+$ concentration gets too low compared to the $H^+$ concentration. Free protons will also never exist in solution, they can nearly always be associated to some water molecule.


Plotting a Langmuir adsorption isotherm We have,

$$\theta = \frac{Kp_\mathrm a}{1+Kp_\mathrm a}$$

where, the fractional coverage $\theta$ is defined as the number of sites occupied by adsorbate $A$ over the total number of sites. $p_\mathrm a$ is the partial pressure of the adsorbate and $K$ is the equilibrium constant.
$$\theta = \frac{[A_{\text{ad}}]}{[S_{\text{total}}]}$$

I need to plot the following data to verify the monolayer adsorption of $\ce{CO}$ on 1 g of $\ce{Pt}$ powder.

$$\begin{array}{cc}
\hline
\text{Equilibrium Pressure (P)/mmHg} & \text{Adsoprtion Amount(V)/ mmHg} \\ \hline
 2.50 & 0.507\\
5.00 & 0.92\\
10.0 & 1.51\\
15.0 & 1.97\\
20.00 & 2.32\\
30.00 & 2.75\\
40.00 & 3.13\\ \hline
\end{array}$$

The question gives me the following hint

  Because adsorption amount $V$ is proportional to $\theta$, $\theta$ can be expressed as $\frac{V}{V_\mathrm m}$ where $V_\mathrm m$ is the adsorption amount at $\theta = 1$

I generated the following graph in Mathematica, simply by plotting $V$ vs $P$

I don't quite understand what the hint wants me to do? Does that require that I normalise the data fashion, i.e by dividing by a particular $V_\mathrm m$, so instead of having $V$ on the y-axis, I have $\theta$ varying from 0 to 1.

Is $V_\mathrm m$ something I obtain from the data/graph by calculating, or is it a tabulated value that I can look up in table somewhere.

Self Answer

Additionally, I plotted $\frac{1}{V}$ and $\frac{1}{p}$, as suggested by @Chester Miller in the comments below, and indeed I get a straight line.

A linear model fit was performed and the equation $4.39274 x+0.214246$ fits the data with an $R^2 = 0.999929 $. From my limited knowledge on this topic (I haven't formally studied this subject), the intercept should give me $\frac{1}{\text{adsorption capacity}}$, and the slope should give some indication of the affinity of the adsorption sites?

So going back to the initial equation, $$\theta = \frac{V}{V_\mathrm m} = \frac{Kp_\mathrm a}{1+Kp_\mathrm a} $$

And now considering $$\frac{V_\mathrm m}{V} = \frac{1+Kp_\mathrm a}{Kp_\mathrm a} = \frac{1}{Kp_\mathrm a} + 1$$

and dividing throughout by $V_\mathrm m$ we get,

$$\frac{1}{V} = \frac{1}{V_\mathrm m} + \frac{1}{V_\mathrm mKp_\mathrm a}$$

And we get the appropriate values for $V_\mathrm m$ and $K$ from the slope and intercept of the $1/V$ vs $1/p$ plot, and then obtaining $\theta$ and plotting vs. $p$ is a trivial exercise.

Anyway, using the calculated parameters, I once again plotted $V$ vs $p$ and fit a curve of the form $\frac{V_\mathrm mKp_\mathrm a}{1+Kp_\mathrm a}$


What are high throw solutions? I am trying to make a copper sulfate electroplating solution. I came across a document (Schlesinger and Paunovic 2011) which mentions the amount of copper sulfate that should be added to water to create such a solution. In Table 2.1 on the bottom of page 35, it has two columns: one for conventional solutions and one for high throw solutions.

I read another paper (Nikolova and Watkowski) which talks about high throw copper electroplating process. All I could glean from the paper was that high throw solutions enable higher aspect ratios for copper PCBs.

Please reply if anyone knows what high throw solutions are.

References:

Nikolova, Maria, and Jim Watkowski. “INNOVATIVE HIGH THROW COPPER ELECTROPLATING PROCESS FOR METALLIZATION OF PCB.”

Schlesinger, M., and M. Paunovic. 2011. Modern Electroplating. The ECS Series of Texts and Monographs. Wiley. http://books.google.com/books?id=j3OSKTCuO00C.

This page suggests that throwing power is the ability to coat a surface evenly regardless of current density differences, and thus to produce an even plating thickness across the whole of a surface.

A solution with high throwing power would thus have the ability to produce a good, close-to-uniform coating on more difficult surface shapes.


Turning of a paddle wheel when exposed to light So, we observe that in the discharge tube experiment when a very light paddle wheel is placed in the path of cathode rays, it starts turning due to the particle nature of the rays.

My question is if the same paddle wheel is placed in light, why does it not start turning ?
Light is also of particulate nature (in some cases- made up of photons).
Is it due to the less mass of photons?
Thanks.

Edit:-

Thank you, particularly @Poutnik for giving me such great answers.

I just read an article while researching that the paddle wheel didn't turn because of the transfer of momentum by the cathode rays but due to changes in pressure of the gas around the paddles due to heating, which is much same as that to Crookes radiometer.
This fact was discovered later, much later after the experiment.
This solves my problem that even in the cathode ray experiment, it wasn't the trasfer of momentum that moved the paddles, because the actual pressure of the rays is quite weak (again as suggested by @Poutnik). And since the ray can't turn the paddles, the light no way can.
And yes, I calculated the momentum of sunlight and that transferred by cathode rays to see for myself that movement of the paddle by each of these is not possible.

Thanks everyone once again.

Not the complete answer, but a short analysis of expectable forces:
The energy and momentum of light are related by equation $E=p \cdot c $.
Then force caused by light $F=\frac{\mathrm{d}p}{\mathrm{d}t}=\frac{\mathrm{d}E}{c \cdot \mathrm{d}t} = \frac Pc$
Therefore the light power 30 W (some cca 50-60W fluorescent bulb or cca 35-40 W LED bulb ) would cause direct  force cca  100 nN, equivalent of cca 0.01 mg.
Or twice as much if reflected.
Crookes radiometer shows the wheel is turning in the opposite direction than it should be, if caused by the light pressure.
The motion is caused by uneven heating and expansion of gas around the wheel, causing pressure differences in the opposite sides of paddles.
When pressure is too low, the gas there is not enough and there remains just light pressure, that is too weak to turn the wheel.

Quoting Effect observations from article.
The effect begins to be observed at partial vacuum pressures of several hundred pascals (or a few torr), reaches a peak at around $\pu{1 Pa}$ ($\pu{7.5e−3 torr}$) and has disappeared by the time the vacuum reaches $\pu{1e−4 Pa}$ ( $\pu{7.5e−7 torr}$) (see explanations note 1). At these very high vacuums the effect of photon radiation pressure on the vanes can be observed in very sensitive apparatus (see Nichols radiometer) but this is insufficient to cause rotation.

Nichols radiometer can register the true light pressure by mirrors on torsion balance.

For cathode rays = electrons
If there is
potential difference $U$
current $I$
elementary charge $e$
electron mass $m_e$
charge $Q$
Area $A$
pressure $P$ (avoiding confusion)
momentum of 1 Coulomb  $p$
electron momentum $p_e$
then for kinematics:
$$eU=\frac 12 m_e.v^2$$
$$v=\sqrt{\frac{2eU}{m_e}}$$
for momentum:
$$p_e=m_e.v=\sqrt{2m_eeU}$$
$$p=\frac 1e \cdot p_e=Q \sqrt{\frac{2m_eU}{e}}$$
for force and pressure
$$F=dp/dt= I \sqrt{\frac{2m_eU}{e}}$$
$$P=F/A= \frac IA \sqrt{\frac{2m_eU}{e}}$$
for $\pu{380 V}$ and $\pu{1 A}$ it gives force about $\pu{66e-6 N}$, what is equivalent to weight of the mass about $\pu{6.6 mg}$.
This is a model case. For a particular scenario, the geometry and electrostatic forces would play significant role.


Molecular orbitals of heteronuclear diatomic molecules For drawing MO diagram for heteronuclear diatomic molecule, how do I know which is lower in energy than the other?

For example, given a heteronuclear diatomic molecule $\ce{NO}$, which one is lower in energy?

The rule as generally stated in textbooks is that “atomic orbitals of the more electronegative element are lower in energy than the corresponding orbitals of the less electronegative element”.

Now, MO diagrams are only simple for elements of the second row of the periodic table ($\ce{Li}$ through $\ce{Ne}$). Involving heavier atoms makes it harder to guess at molecular orbital diagrams, and there is need for quantum chemistry calculations. Thus, the rule becomes: the further to the right your element is, the lower its energy levels are. The reason for this is that the charge of the nucleus increases with its atomic number, and thus lowers the energy levels.


Synthesis Golf V: Denopamine A full FAQ post has been written on meta.chem.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

This fifth round of golf concerns the synthesis of (−)-α-(3,4-Dimethoxyphenethylaminomethyl)-4-hydroxybenzylalcohol (denopamine), a synthetic drug indicated for use in the treatment of angina.

The challenge is to propose a route to denopamine. You may start with anything that has less than 10 carbons, but it must be available in the Sigma-Aldrich catalogue. A method should also be included to set the oxymethine stereocentre (no buying it!).

Given the relative simplicity of the target and the need for large quantities of the drug, consideration to step count and scalability should also be taken.

OK, new route. Retrosynthesis:

Denopamine 1 can be made by a reductive amination between amine 2 and aldehyde 3 (either of the two reductive amination disconnections are fairly obvious choices).

The requisite amine 2 contains 10 carbons and so it becomes necessary to make it using some other means - in this case, reduction of a nitroolefin 4, which can itself be made via a nitroaldol reaction from 3,4-dimethoxybenzaldehyde 5 and nitromethane 6.

The chiral α-hydroxyketone 3 can probably be made in numerous ways. Earlier I proposed organoocatalytic oxygenation with PhNO, but another possibility is to have an addition of "formaldehyde anion" to 4-hydroxybenzaldehyde 7. Nitromethane 6 was chosen as the acyl anion equivalent, as the asymmetric Henry reaction is well-established methodology, e.g. by the Evans group (J. Am. Chem. Soc. 2003, 125 (42), 12692–12693).

The forward synthesis is fairly straightforward:

5 steps from cheap starting materials, with longest linear sequence being 3 steps.


Why is lemon mixed with water considered alkalizing water? The benefits and facts of alkaline water are often skewed due to the marketing of companies and uneducated hype for the innocent idea of unlimited potential. I wanted to know the specific reactions occurring when mixing water and fresh lemon slices that lead to water becoming alkalized (higher in pH).

The claim that lemon water is alkaline is simply wrong.  However, that hasn't stopped people from claiming that it's true:

  Lemons are an incredibly alkaline food, believe it or not. Yes, they are acidic on their own, but inside our bodies they're alkaline (the citric acid does not create acidity in the body once metabolized). As you wellness warriors know, an alkaline body is really the key to good health.

Every sentence above is wrong.  Lemons do not turn alkaline inside the body.  They are not an alkaline food.  Citric acid creates neither alkalinity nor acidity in the body when it is metabolized.  An alkaline body is not the key to good health.  In fact, if your body were more alkaline by only 0.5 pH units from its usual state, you would probably die from it.

  I wanted to know the specific reactions that occur when mixing water and fresh lemon slices [...]

There aren't really many reactions that occur.  Lemon juice is mostly water.  Mixing the lemon juice present in lemon slices with more water just dilutes it.  Lemon juice is acidic, due to the presence of citric acid and other acids.  Diluting will very slightly lower the acidity.  None of that has any relevance to the pH inside a mammalian body, which is exquisitely regulated by other mechanisms.

I don't know why people make up weird claims that lemons are an "alkaline" food, but those claims, sadly, are out there everywhere.


Identify a molecule by structure I have a drawing of a molecule, I want to know its name. Is there a software, or website that can help me identify the chemical? Im not a chemist, its been a decade since I had chemistry in college, and I cannot simply guess the name of the chemical.

The model goes like this:
C4H9-Benzine-N=CH-Benzine-CH=N-Benzine-C4H9

(also if there is a simpler form of this molecule, Im not sure how to simply it) I'm hoping It is a known chemical and that I can get details about it.

Joh,

Please let me know if you referred to this compound:

Also, if this is your compound, please find the name.


Differences between ¹³C-decoupled NMR experiment vs a ¹³C-coupled How does the acquisition time changes between $\ce{^{13}C}$–decoupled NMR experiment vs a $\ce{^{13}C}$–coupled?

Would the acquisition time be longer if a decoupling of the protons (H) is carried out?

The answer is a bit complicated because there are a number of ways to apply decoupling (see eg this site which describes different decoupler gating modes), depending on the goal. For simplicity I limit the discussion to acquisition of a 1D $\ce{^{13}C}$ spectrum:

$\ce{^1H}~$ decoupling during the entire experiment: this provides a signal enhancement due to the NOE from $\ce{^1H}$ to $\ce{^{13}C}$, and suppresses couplings, which reduces spectral complexity and also increases carbon s/n due to collapsed multiplets.
Decoupling during acquisition only (inverse gated decoupling): this suppresses couplings simplifying the spectrum and providing some enhancement by collapsing multiplets, but does not give NOE enhancement.
Gated decoupling: this provides the NOE enhancement but gives a coupled spectrum showing multiplicity.

Applying decoupling to obtain a signal enhancement from NOEs allows you to reduce the overall acquisition time (reduce the number of FIDs you need to acquire) relative to no NOE. This typically works for all but quaternary carbons. Decoupling during acquisition also can allow you to reduce the overall acquisition time, even if you do not obtain an NOE enhancement, by collapsing signals from multiplets into single peaks. The longest experiment would be a truly quantitative one, and particularly if you want full coupling information and s/n equivalent to that from a decoupled experiment. Then you would not apply decoupling at all. In that case you would have to wait on the order of $\mathrm{5 \times } \ce{^{13}C} ~\mathrm{T_1}$ and acquire enough scans to get the desired s/n.


Formation of ferrous chloride or ferric chloride Question
When $\ce{HCl}$ reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why?
Answer. Its reaction with iron produces $\ce{H2}$.
$\ce{Fe + 2HCl -> FeCl2 + H2}$
Liberation of hydrogen prevents the formation of ferric chloride.

Now my question is why couldn't it be: $\ce{2Fe + 6HCl -> 2FeCl3 + 3H2}$ ?

Acids can oxidize iron since the redox potentials $E$ for $\mathrm{pH} = 0$ show that $\ce{H+}$ can oxidize $\ce{Fe}$ to $\ce{Fe^2+}$; but non-oxidizing acids cannot further oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$:

$$\begin{alignat}{2}
\ce{2H+ + 2e- \;&<=> H2}\quad &&E^\circ = +0.000\ \mathrm{V}\\
\ce{[Fe(H2O)6]^2+ + 2e- \;&<=> Fe}\quad &&E^\circ = -0.440\ \mathrm{V}\\
\ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}\\
\end{alignat}$$

However, oxygen can oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$:

$$\begin{alignat}{2}
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad &&E^\circ = +1.229\ \mathrm{V}\\
\ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}
\end{alignat}$$

Therefore, iron is oxidized to $\ce{Fe^3+}$ in dilute hydrochloric acid when in contact with air.


Is a supersaturated solution still a solution? Given that a solution must be a homogenous mixture, and having crystallized solids at the bottom of a solution is most definitely not homogenous, can a supersaturated solution still be technically defined as a solution?

On that note, is it possible for a solution to have more solute than solvent and remain in solution form?

You seem to be getting it wrong. A solution with crystallized solids at the bottom is not the same as supersaturated solution, and has never been called such. It is called "solution with precipitate", "heterogenous mixture of solution and something", "water with mud", or anything in the same vein.

A supersaturated solution is a solution, not just technically, but in all sincerity. It is homogenous, clear, transparent, and you'll never tell it from ordinary solution - I mean, until you touch it, at which point a precipitate would start to form.

As for your second question: yes, there are quite a few compounds that would dissolve in water incredibly well. ZnCl2 comes to mind... Wait, it's simpler than that - common table sugar (sucrose) would do! It dissolves up to 2000 g/L, how'd you like that?


Difference between valency of nitrogen and phosphorus though both have same valence electrons Both nitrogen and phosphorus have 5 valence electrons, but nitrogen has valency of 3 while phosphorus has valency of 3 or 5. Why is this so?
Maybe phosphorus can gain 3 electrons or remove 5 electrons.But is it possible to remove 5 electrons? Or is it that phosphorus valence shell is farther away than nitrogen, so it is easier for phosphorus to also remove 5 electrons but nitrogen cannot.
I have searched but i need a concrete answer.

At your current stage, it's probably easiest to consider the octet rule which applies to nitrogen and the expanded octet rule which applies to phosphorus.

If nitrogen and phosphorus made five bonds, they would have 10 electrons total. This is inconsistent with the octet rule but valid under expanded octet.

In reality, the situation is governed by the fact that you need more orbitals to accommodate the extra electrons for the two extra bonds. Unfortunately, no orbitals of comparable energy to the $2s$ and $2p$ are available, so the stabilization of the electrons in the putative molecular orbitals would be poor in nitrogen. With phosphorus, you might more readily access the $4s$ and $3d$ orbitals which are closer in energy to the $3s$ and $3p$ orbitals.


Calculating charge on amino acid from pKa I have been given $\mathrm pK_\mathrm a$ values of an amino group, a carboxyl group and a side chain of cysteine. How can I find the ionic charge on it at different $\mathrm{pH}$ values?

For this, you can use the Henderson-Hasselbalch equation. Using the degree of dissociation, $\alpha$, this can be written as
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\alpha}{1-  \alpha}$$
Rewriting to solve for $\alpha$:
$$\alpha = \frac{1}{10^{\mathrm{p}K_\mathrm{a} - \mathrm{pH}} + 1}$$
As stated above, $\alpha$ is the degree of dissociation, meaning the degree at which $\ce{H+}$ is dissociated from the ionizable group. At $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$, $\alpha = 0.5$, meaning that $50\%$ of the ionizable groups in question are deprotonated. If $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + 1$, about $90\%$ of all groups are deprotonated and if $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + 2$, about $99\%$ of all groups are deprotonated. (If you want to, take a look at the shape of the $\alpha$ distribution for $\ce{-SH}$ in cysteine at Wolfram Alpha.)

From a textbook I found the following $\mathrm{p}K_\mathrm{a}$ values for cysteine:
\begin{align}
 \mathrm{p}K_\mathrm{a}(\ce{-COOH})   &= 1.9\\
 \mathrm{p}K_\mathrm{a}(\ce{-NH_3+}) &= 8.35\\
 \mathrm{p}K_\mathrm{a}(\ce{-SH})     &= 10.5
\end{align}

From these values, $\alpha$ can be calculated for each ionizable group at the desired pH and this will give you the net charge of the amino acid. Upon deprotonation, the following changes in charge occur for the ionizable groups:
\begin{array}{lclcr}
 \ce{-COOH} &:& 0 &\rightarrow &-\\
 \ce{-NH_3+}&:& + &\rightarrow &0\\
 \ce{-SH}   &:& 0 &\rightarrow &-\\
\end{array}

As an example, let's calculate the charge of cysteine at pH 10. Using the Henderson-Hasselbalch in a spreadsheet yields the following $\alpha$ values:
\begin{array}{lcr}
 \ce{-COOH}  &:& 0.9999\\
 \ce{-NH_3+} &:& 0.9781\\
 \ce{-SH}    &:& 0.2403\\
\end{array}

This gives the following total charge for cysteine at pH 10:
$$(-1)\cdot0.9999+(+1)\cdot(1-0.9781)+(-1)\cdot0.2403=-1.218$$


Enantiomers of 1,2-dichlorocyclohexane The other day in class we were discussing the enantiomers of 1,2-dichlorocyclohexane and my teacher told us that (R,R) and (S,S) are different isomers while (R,S) and (S,R) are the same.

I understand how (R,S) and (S,R) are the same (Because they can be rotated) but I don't understand why (S,S) and (R,R) are not the same, even though they are the same if you rotate them too.

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

The two pictures I posted are depicting two same molecules, except the one on the left was flipped around an axis in plane of the screen. With the second picture, it's easier to see that these two molecules are non-superimposable mirror images. Also notice how neither of these molecules have a plane (or point) of symmetry.

I recommend trying to build this with a model kit if you can't visualize it in your head easily.


Why are calorimeters often made of materials with low specific heats? When building calorimeters, a material with a low specific heat is often used (such as Styrofoam, aluminum, etc.). However, why would a material with a low specific heat be preferred? My thought process is below:

Scenario: An exothermic reaction occurs in a calorimeter.

Principle: In calorimetry, we are always trying to minimize the amount of heat transfer from the calorimeter to the surroundings. The amount of heat transferred from the reaction to the calorimeter doesn't really matter because we can calculate this quantity of heat using the calorimeter's calorimeter constant, which we can derive experimentally.

Case 1: Calorimeter made of material with low specific heat

If the calorimeter had a low specific heat, it would absorb less heat, but its temperature would increase more. Differences in temperature cause heat transfer, so a larger difference in temperature between the calorimeter and the surrounding air would mean more heat transfer from the calorimeter to the surroundings.

Case 2: Calorimeter made of material with high specific heat

If the calorimeter had a high specific heat, it would absorb more heat, but its temperature would not increase much. Differences in temperature cause heat transfer, so a smaller difference in temperature between the calorimeter and the surrounding air would mean less heat transfer from the calorimeter to the surroundings.

Therefore, a calorimeter with a high specific heat would be more effective because it would minimize heat transfer between the calorimeter and the surroundings.

Where is the flaw in my logic?

Additionally, according to http://www.greenspec.co.uk/building-design/insulation-materials-thermal-properties/,

  A good insulator has a higher specific heat capacity because it takes time to absorb more heat before it actually heats up (temperature rising) to transfer the heat.

Actually, a calorimeter with high specific heat would have less temperature change, which is harder to measure. Therefore, the error in the calculated enthalpy of the reaction would be higher.
Specific heat does not mean heat conductivity, as described by Fourier's Law.


Using Cv For Non Ideal Gasses So we are doing calculations with the first law of thermodynamics:
$$U = q + w$$

And we have been studying $\mathrm dU(T,V)$.
So it is true that for some equations of state other than the ideal gas equation, $\mathrm dU(T)$ only, so therefore it follows that $U= C_V(T_2-T_1)$. So I am wondering is it even possible that $C_V$ is used for non-ideal gasses? And would it also be equal to $\frac 32R$?
And if $C_V$ can be used, why?
(Also is this true for $C_p$?)

We usually don't use the real gas $C_V(T,p)$ directly in calculating the change in U in going from thermodynamics equilibrium state A to thermodynamic equilibrium state B, if both these states are beyond the ideal gas region.  Starting from state A, what we do first is use the real-gas p-v-T equation to determine the change in U in going from the initial pressure $p_A$ to a low pressure (or large specific volume) in the ideal gas region $p^*$ at constant temperature $T_A$.  Then, our second step is to hold the pressure constant at pressure $p^*$ and raise the temperature from $T_A$ to $T_B$, using the ideal gas heat capacity to determine the change in internal energy for this step (since we are now in the ideal gas region).  Finally, we hold the temperature constant at $T_B$ and raise the pressure from $p^*$ to $p_B$ using the real-gas p-v-T equation again to determine the change in U.  The sum of the changes in U for these three steps adds up to the change in U in going from $(T_A,p_A)$ to $(T_B,p_B)$.