Could sodium fluoride or aluminum accumulation in soils contribute to increased amounts of palmitic acid in food? I am interested in the effects of plant environments on their biochemistry in relation to their effects on the environment and animals(us).

Could sodium fluoride or aluminum  accumulation in soils contribute to increased amounts of palmitic acid in food?

Thus increasing the inflammatory effects of certain foods?

This article explains the effects of fluoride ion concentrations to Prosopis juliflora.

  Both catalase (3.2 folds) and peroxidase (2.7 folds) enzymes activity increased with increase in F concentration.

Checking out Wikipedia articles about these enzymes, neither catalase not peroxidase are involved in the fatty acid biosynthesis. So as far as I know, I don't think there's any effect of fluorides in the biosynthesis of palmitic acid.

Also, regarding aluminum ions, this article explains aluminum toxicity as a hindrance to cell division and elongation, but nothing is mentioned about fatty acid synthesis.

References:

Saini, P, et al. “Effects of Fluoride on Germination, Early Growth and Antioxidant Enzyme Activities of Legume Plant Species Prosopis Juliflora.” Journal of Environmental Biology., U.S. National Library of Medicine, 34 Mar. 2013, www.ncbi.nlm.nih.gov/pubmed/24620580.

Mossor-Pietraszewska, T. (2017). Effect of aluminium on plant growth and metabolism.. [online] Psjd.icm.edu.pl. Available at: http://psjd.icm.edu.pl/psjd/element/bwmeta1.element.bwnjournal-article-abpv48i3p673kz [Accessed 23 Aug. 2017].


What is the difference between graphite oxide and graphene oxide? I am considering making a sheet of graphene using the LightScribe technique and I have read that I need graphite oxide. However when I look up graphite oxide for sale, graphene oxide pops up.

Are these two oxides similar? What is the difference?

In short, graphite is several graphene sheets piled one above another.

Graphene is made up of one single sheet of carbon atoms arranged in hexagonal pattern (like a honeycomb), and graphite are several such sheets, each sheet linked to another by weak intermolecular forces, which gives the graphite its lubricative properties.

EDIT:
Similarly, graphene oxide is a layer of graphene with oxygen atoms, $OH$ groups and $COOH$ groups attached to it, while graphite oxide are layers of graphene oxide.

Because of the presence of $COOH$ groups, graphite oxide can disperse in basic solutions (or even water!) to give graphene oxide.

$CO_2$ is simply carbon dioxide, one carbon atom doubly bonded two two oxygen atoms.

By contrast, graphite oxide is like a huge field of carbon with the groups I mentioned earlier while carbon dioxide is like a small ball.


What is responsible for black tarnish on silver? Silver on being exposed to air, forms a black film. Now my question is whether the film is silver sulphide (due to reaction with hydrogen sulphide in air) or silver oxide.

Both compounds are black.

In my view it would be silver oxide because oxygen is more abundant and, intuitively, silver seems react to oxygen faster than hydrogen sulphide- oxygen is more reactive than hydrogen sulphide.

But a little google search loosely hints it is silver sulphide. Any suggestions on this too?

There is no such thing as "being reactive". Say, you have A and B; some compounds tend to react with A and not with B, and some would react with B and not with A. Which is more reactive? You can't tell; you can't even ask this question, for it's meaningless.

Black tarnish on silver is mostly silver sulphide, as confirmed experimentally. I'd even say that silver does not form a black film on being exposed to air, unless there is some source of hydrogen sulphide nearby (raw eggs, sewage pools, or something else).

It is just that $\ce{Ag}$ tends to prefer $\ce{S}$ over $\ce{O}$. Many other heavier metals are like that, and many lighter metals have different preferences. This is explained (kind of) by the hard/soft acids/bases concept.

So it goes.


Acidity of hydrochloric acid in acetic acid A question asked to find the $\mathrm pK_\mathrm a$ of $\ce{HCl}$ in acetic acid. The equation of $\ce{HCl}$ disassociating in water is: $$\ce{HCl + CH3COOH \leftrightharpoons CH3COOH2+ + Cl-}$$ I predicted that the $\mathrm pK_\mathrm a$ should be higher  as acetic acid is a weaker base than water so it would be less willing to accept a proton from a $\ce{HCl}$.

However when I finally calculated the $\mathrm pK_\mathrm a$, I got 8.56. This was much higher than I expected. Also doesn't this imply that $\ce{Cl-}$ is stronger as a base than $\ce{HCl}$ is as an acid in acetic acid, which doesn't sound right.

So I am asking if this value totally wrong or is it actually quite reasonable?

So basically I am asking, is this value totally wrong or is it actually quite reasonable?

It's quite reasonable. Compare to page 32 of this reference.

  This was much higher than I expected. Also doesn't this imply that
  Cl− is stronger as a base than HCl is as an acid in acetic acid, which doesn't sound right.

To answer whether Cl- is a stronger base than HCl is an acid, you need to know the self-dissociation constant of acetic acid, the equivalent of Kw for water. It is about $10^{-10}$. So whereas the middle of the pH scale is thought of as 7, the middle for acetic acid would be 5. So, yes, Cl- is a stronger base than HCl is an acid in acetic acid.


How do I find the oxidation number of a coordination complex I actually know how to calculate the oxidation number for coordination complexes but some complexes like the below are giving some reistance

$\ce{[Cu(NH3)_2][Fe(H2O)_3]}$ (diamminecopper(?) triaquoferrate(?))

I just wrote this example now and don't know wether it exists or not I am rather concerned about how do I find their oxidation numbers

Think of it like this. Ammonia and Water are both neutral molecules, therefore they won't contribute to the charge of the two coordination spheres. There are also no subscripts, which indicates that iron and copper both have the same magnitude of charge. Now copper here could be +1 and iron could be in -1, or copper in +2 and iron in -2. (Of course, this isn't possible since this molecule doesn't exist.) I cannot conclusively say which though, without any further data, such as maybe their spin only angular momenta.

For other complex compounds, i suggest you practice, and eventually you'll get the hang of what oxidation states are common for different metals. Also use the concepts you use for normal covalent compounds. Trust me, it's very easy!


How would I go about making an ethyl ester from vegetable oil? I'm teaching myself how to make lotion and a lot of lotions contain esters like isopropyl myristate or isopropyl palmitate. I'd like to make an ethyl ester from a popular skin-care oil (olive, apricot kernel, or castor).

Since there's so much controversy about using isopropyl alcohol in skin-care, I'd like to avoid that if possible. It's very difficult to find any info on the process, and the closest I could get was how to make biodiesel. All of those use methanol though.

How would I go about making an ethyl ester from these oils?

The biodiesel process should work well, you will need just slightly more aggressive conditions (i.e. slightly more catalyst, slightly higher EtOH excess, and/or slightly longer time) to make ethyl esters instead of methyl esters, as ethanol is slightly less reactive than methanol.

For skin care, you will need higher purity than biodiesel processes usually provides.  I would imagine that the cosmetics industry uses vacuum distillation to purify their esters.  This would separate the ethyl esters from unreacted triacylglycerols, any base (or acid) catalyst left over from the reaction and soluble in the organic ester phase, organic-soluble glycerol, and organic-soluble impurities.


Chilling water in copper vessel Does chilling drinking water in copper bottle result in any chemical reaction that turns it dangerous for consumption in any way?

Few days back, in summer, I carried kokum juice in copper bottle. I had a sip of it after some time when I reached office and I felt super dizzy. I felt that traveling might have an effect on me, so I stopped. I had another sip of juice after some hours I had a strong dizzy feeling. At first, the juice tasted ok but then it started tasting very very weird. When I reached home and poured the juice back out of the bottle, to my surprise, its green! The normal color of kokum juice is pinkish-reddish. So somehow copper reacted with the juice and possible resulted in some acid, even though it was at normal room temperature (or in fact cold one as the juice was prepared from cold water). I believe there shouldn't be an issue with cooling water in a copper bottle in the fridge, I just want to know if there is any slight drawback, maybe from Ayurveda point of view? Also, I hope you get the chemistry angle to this question.

Do not do it !! ( putting acidic, or rather any juice to copper bottles )

You are in danger of copper poisoning.

Generally, by food processing laws, copper is not allowed to be in direct contact with food, as there is danger of copper contamination.

Especially acidic liquids, like vinegar or citrus juices, directly slowly dissolve copper in presence of oxygen.

$$\ce{4 RCOOH + 2 Cu + O2 -> 2 (RCOO)2Cu + 2 H2O}$$

Note that cold liquids contain usually more oxygen than warm ones.

For similar reasons, even if there is not copper, it is very bad idea to keep the content of conserved fruit cans in the already  opened cans. You can otherwise see the juice corroding the can inner coating in air presence.

Soluble copper salts are usually blueish/greenish what with yellow/orange colour of juice gives green colour.

Long term usage of copper for keeping water is on the edge. It can cause chronical low dose sneaky poisoning, especially if water is treated by chlorine.

$$\ce{Cu + Cl2 -> Cu^2+ + 2 Cl-}$$

Oxidation can lead to partially soluble basic copper carbonates, soluble by traces of organic acids, related to minerals malachite and azurite.

$$\ce{2 Cu + CO2 + O2 + H2O -> CuCO3 \cdot Cu(OH)2}$$

$$\ce{3 Cu + 2 CO2 + 3/2 O2 + H2O -> 2 CuCO3 \cdot Cu(OH)2}$$

Plus, there is danger some people would want to store there other liquids than water as well, as the OP.


Why is the standard enthalpy of formation of diamond not zero? Why doesn’t diamond have $\Delta H_\mathrm{f}^\circ=0$, when graphite does? Is it something to do with the definition – diamonds can’t really form at STP, even though it is naturally occurring?

You are on the right track - diamond is not the thermodynamically stable carbon phase at STP. Taking two figures from A.T. Dinsdale, 'SGTE Data for Pure Elements', CALPHAD 15(4) 317-425 (1991) one sees:

and

Since graphite is the thermodynamically stable phase of carbon at STP, it is usually selected as the reference phase so it has $\Delta H^0_f = 0$. In the reference above, both absolute Gibbs free energies, as well as free energies with respect to the stable phase at STP, are given for most elements.


Given a theoretical combustion engine, how can I calculate the output of this engine depending on the fuel it uses? I'm writing a Minecraft mod (for those who don't know, it's sort of like a smaller "program" that loads into the Minecraft game and expands it, by adding other things to it).

My question is: consider a combustion engine that runs on anything that is flammable (even though Uranium has a lot of chemical energy, it can't be set on fire, so it's out).

The engine can burn, for instance, acetone, methane, ethanol, methanol, etc.

By burning these fuels the energy converts their chemical energy into mechanical or kinetic engine to move it's gears, pistons and stuff (whatever the engine uses to transmite the energy).

What is the simplest way of comming up with a "good enough approximation" function

energy = f(fuel)

What goes in this function? the volume of the engine doesn't change, so not that. pressure certainly is a factor (maybe the answer itself?) does it have anything to do with the Ideal Gas Law? What about the density of the fuel, molar mass, specific heat, etc?

The answer to this heavily depends on whether you have an internal combustion engine or an external combustion engine.

External

Because you're just heating the liquid from the outside, you can approximate the final output of the engine as the amount of energy gained from combusting the fuel multiplied by engine efficiency (as you like, but no greater than 100%). In other words:

$$E=kfV$$

Where E: Energy Output; k: Efficiency absorbing the energy; f: Thermal Energy obtained from burning a unit volume of the fuel; and V: volume of fuel combusted.

Internal

For an internal combustion engine, far more factors need to be taken into account. Wikipedia is always a good start to get a sense of the factors that play a role, and whether or not you want to consider them. De
pending on how dedicated you are, and how much the player can 'mess with' the settings of the engine, you could do something as simple as the External Engine (if all the user can do is select fuel and amount of fuel) or as complex as is required based on the variables the user can change.

Some Other Factors to Consider (Optional):

Compression Ratio (internal only)
Gear Ratio (for high torque applications)
Working Fluid (external only)
Ambient Temperature (mainly external only)
Heat Exchanger Material (external only)


Symmetry of an in-phase and an out-of-phase wavefunction "approaching" each other in an MO Hiho.

I have a question concerning this image (Chemistry, Catherine E. Housecroft et al., Pearson Education, 2010).

In the text, it states the following:

  "Each of the new molecular wavefunctions, $\Psi$(in-phase) and $\Psi$(out-of-phase), has an energy associated with it wich is dependent of the distance apart of the two hydrogen nuclei. As Figure 4.13 [provided image] shows, the energy of $\Psi$(in-phase) is always lower than $\Psi$(out-of-phase)."

The in-phase-function has a stabilized minimum, but the out-of-phase-function seems stable only at an infinite distance. For my intuition, the two graphs should be symmetric, since the two wavefunctions are alike and approach each other equally. Why do they behave like this?

Why are those functions not symmetric?

A following question would be, when I combine those two functions, I'll get a minimum at the x-value I marked with "minimum". Would this be the distance at which a nucleus in a stable diatomic molecule would be found like in this figure?

===

Thank you for your help and have a nice day :)

Considering this diatomic molecule, the total wavefunction $\Psi$ is formed via a linear combination of the atomic orbitals $\psi_1$ and $\psi_2$ (which is an approximation, but not a too bad one).
$$ \Psi^+ = c_1 \psi_1 + c_2 \psi_2 $$
$$ \Psi^- = c_1 \psi_1 - c_2 \psi_2 $$

The (electronic) energy of the system can be described using the following (stationary) Schrödinger equation:
$$ H_\text{el} \Psi_\text{el} = E_\text{el} \Psi_\text{el}$$

So somehow this all factors into each other and in the end gives you the curves (for the potential energy of the system vs. internuclear distance) that you posted. Where is the symmetry? Well, it is still there, with respect to the center of one atom. If you imagine rotating the functions around the $y$ axis, you will get a symmetric image. And you're allowed to do this because space expands in all three directions. So in theory (if you follow through and expand the function to everywhere) you get a 3D image of the energy at every point around one atom.

But: The wavefunction in itself is not all that interesting to us, because there is no "real" physical information that can be directly extracted from it. But fret not, we can at least get out the electron probability density by complex-squaring it:
$$ \rho = |\Psi|^2 = \Psi \Psi^* $$

And this is where those nice small bubbles come into play that we see in your second figure. And we have again found the symmetry that you seek, because both linear combinations have $D_{\infty h}$ symmetry (I think... I'm not all too good with point groups and the such).


Nomenclature of a cyclohexyl ether and locant position enumeration As far as I know, 'ethoxy' should be given the locant '1' as it is a functional group. The compound should thus be named 1-ethoxy-2,2-dimethylcyclohexane.

However, I was told that the correct name is 2-ethoxy-1,1-dimethylcyclohexane. Can anyone please explain why this is so? My teacher explained this to me in terms of the lowest locant sum rule, which I just discovered doesn't even exist.

The smallest sum of locants rule does not exist in the IUPAC recommendations. The application of this ‘rule’ can lead to wrong results in many instances.
The corresponding section in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) actually reads as follows:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
(…)

With regard to numbering of locants, simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl and ethoxy) are considered together with equal seniority:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Therefore, the example is named as 2-ethoxy-1,1-dimethylcyclohexane rather than 1-ethoxy-2,2-dimethylcyclohexane since the locant set ‘1,1,2’ is lower than ‘1,2,2’.


How can antibonding orbitals be more antibonding than bonding orbitals are bonding? In molecular orbital theory, the fact that a bonding and antibonding molecular orbital pair have different energies is accompanied by the fact that the energy by which the bonding is lowered is less than the energy by which antibonding is raised, i.e. the stabilizing energy of each bonding interaction is less than the destabilising energy of antibonding. How is that possible if their sum has to equal the energies of the combining atomic orbitals and conservation of energy has to hold true?

  "Antibonding is more antibonding than bonding is bonding."

For example, the fact that $\ce{He2}$ molecule is not formed can be explained from its MO diagram, which shows that the number of electrons in antibonding and bonding molecular orbitals is the same, and since the destabilizing energy of the antibonding MO is greater than the stabilising energy of bonding MO, the molecule is not formed. This is the common line of reasoning you find at most places.

Mathematical Explanation

When examining the linear combination of atomic orbitals (LCAO) for the $\ce{H2+}$ molecular ion, we get two different energy levels, $E_+$ and $E_-$ depending on the coefficients of the atomic orbitals. The energies of the two different MO's are:
$$\begin{align}
E_+ &= E_\text{1s} + \frac{j_0}{R} - \frac{j' + k'}{1+S} \\
E_- &= E_\text{1s} + \frac{j_0}{R} - \frac{j' - k'}{1-S}
\end{align} $$

Note that $j_0 = \frac{e^2}{4\pi\varepsilon_0}$, $R$ is the internuclear distance, $S=\int \chi_\text{A}^* \chi_\text{B}\,\text{d}V$ the overlap integral, $j'$ is a coulombic contribution to the energy and $k'$ is a contribution to the resonance integral, and it does not have a classical analogue. $j'$ and $k'$ are both positive and $j' > k'$. You'll note that $j'-k' > 0$.

This is why the energy levels of $E_+$ and $E_-$ are not symmetrical with respect to the energy level of $E_\text{1s}$.

Intuitive Explanation

The intuitive explanation goes along the following line: Imagine two hydrogen nuclei that slowly get closer to each other, and at some point start mixing their orbitals. Now, one very important interaction is the coulomb force between those two nuclei, which gets larger the closer the nuclei come together. As a consequence of this, the energies of the molecular orbitals get shifted upwards, which is what creates the asymmetric image that we have for these energy levels.

Basically, you have two positively charged nuclei getting closer to each other. Now you have two options:

Stick some electrons between them.
Don't stick some electrons between them.

If you follow through with option 1, you'll diminish the coulomb forces between the two nuclei somewhat in favor of electron-nucleus attraction. If you go with method 2 (remember that the $\sigma^*_\text{1s}$ MO has a node between the two nuclei), the nuclei feel each other's repulsive forces more strongly.

Further Information

I highly recommend the following book, from which most of the information above stems:

P. Atkins and R. Friedman: Molecular Quantum Mechanics, $5^\text{th}$ ed. Oxford University Press, 2011.


A Spectrophotometric technique to evaluate Vitamin C concentration in Citrus Fruit Is there a visible spectrophotometric technique that can gage the concentration of vitamin C in solution? I plan on boiling different citrus fruit in water, and taking aliquotes of the boiled solution to determine total leached vitamin C. I’ve found a couple of UV spectrophotometric techniques, but wasn’t certain if those techniques could carry forward using a visible spectrophotometer. Those sources also didn’t specify what the vitamin C concentration range was for a Beer Lambert’s Law assay.
Would spectrophotometry be possible? Or would it be more feasible to perform a redox titration of vitamin C (ascorbic acid) using iodine?

It is not quite correct of the statement that "ascorbic acid cannot be determined by Beer Lambert's law, because it does not absorb visible light." Although the second part of the statement is true, the highlighted part is incorrect since Beer Lambert's law applies for ultra-violet range of the spectrum. According to Ref.1, ascorbic acid (Vitamin C) absorbs light in UV range of the spectrum:

The result revealed that $\lambda_\mathrm{max}$ (peak wavelength of spectra) of L-ascorbic acid at $\mathrm{pH} \ 2.0$ was $\pu{289.0 nm}$ whereas at neutral $\mathrm{pH} \ 7.0$, $\lambda_\mathrm{max}$ was $\pu{295.4 nm}$. In alkaline $\mathrm{pH} \ 8.6$, $\lambda_\mathrm{max}$ was $\pu{295.4 nm}$ and at $\mathrm{pH} \ 7.4$ the $\lambda_\mathrm{max}$ of L-ascorbic acid remained the same as $\pu{295.4 nm}$.

Thus, if OP has access to a UV/vis spectrometer, assuming no interference from other components in the fruit extract in the UV range, OP can try to determine ascorbic acid percentage in fruits of interest using a direct method with calibration curve of known ascorbic acid concentrations.
In the event of having interferences, there is a better method of getting absorbance in visible region of the spectrum (Ref.2). This is a well-established method, which is called the 2,4-dinitrophenyl hydrazine method (Ref.3).
The method involves oxidation of ascorbic acid to dehydroascorbic acid by bromine water followed by preparation of the corresponding osazone using 2,4-dinitrophenylhydeazine (2,4-DNPH). The osazone upon treatment with 85% $\ce{H2SO4}$ forms red colored solution, which absorbs light at $\pu{521 nm}$. The procedure was successfully used to determine vitamin C content in various fruits and vegetables (Ref.3). The calibration curve of the red dye prepared from standard ascorbic acid samples is shown in the following diagram (LHS; the plots are from Ref.3):

This method need you to prepare a working solution (5% metaphosphoric acid-10% acetic acid): Fifteen grams of solid metaphosphoric acid were dissolved in mixture of $\pu{40 mL}$ of glacial acetic acid and $\pu{450 mL}$ of distilled water in a $\pu{500 mL}$ volumetric flask.
The standard vitamin C (ascorbic acid) solutions for calibration curve: First, prepare the $\pu{500 ppm}$ ($\pu{500 mg/L}$) standard stock solution by dissolving $\pu{0.05 g }$ of standard crystalline ascorbic acid in $\pu{100 mL}$ of working solution. From that stock solution, you can prepare $5, 10, 15, 20$, and $\pu{25 ppm}$ standard test samples by serial dilution (using working solution to dilute) to prepare the calibration curve.
In addition, you need 10% thiourea solution, 2,4-DNPH solution, and 85% $\ce{H2SO4}$ solution in hand.
Procedure: To the equal volumes of filtered fruit solution (prepared by using working solution to extract and dilute) or standard ascorbic acid solutions, add few drops of bromine water until the solution became colored (bromine water color; this is to confirm the completion of the oxidation of ascorbic acid to dehydroascorbic acid). Then few drops of thiourea was added to it to remove the excess bromine and the clear solution was obtained as result. Then 2,4-DNPH solution was added equal volume to all these oxidized ascorbic acid samples (both standards and fruit extract samples) and they are kept in $\pu{37 ^\circ C}$ water bath for $\pu{3 h}$ (Ref.4). When the coupling reaction of 2,4-DNPH dye with dehydroascorbic acid has been completed, add equal volumes of 85% $\ce{H2SO4}$ solution ($5\times$ volume of 2,4-DNPH) followed by spectrophotometric determination.
It should be noted that any other compound in the extract also reacts with 2,4-DNPH and 85% $\ce{H2SO4}$ solution, if they contain carbonyl groups. For example, common component in fruit is glucose (and other sugar), dye from which are giving absorbance at around $\pu{450 nm}$ (see the complete scan of dye prepared by D-glucose on the RHS of the above diagram). As indicated in the diagram, these interference is minimal at $\pu{521 nm}$.
Note: This method is good to determine total ascorbic acid content (ascorbic acid + dehydroascorbic acid). The full text of the relevant paper is attached for the convenience.
References:

Shaheen A. Maniyar, Jameel G. Jargar, Swastika N. Das, Salim A. Dhundasi, Kusal K. Das, "Alteration of chemical behavior of L-ascorbic acid in combination with nickel sulfate at different $\mathrm{pH}$ solutions in vitro," Asian Pacific Journal of Tropical Biomedicine 2012, 2(3), 220–222 (https://doi.org/10.1016/S2221-1691(12)60045-8).
R. Riemschneider, M. Z. Abedin, R. P. Mocellin, “Qualitäts und stabilisierung prüfung hitze konservierter Nahrungsmittel unter verwendung von Vit C als kriterium-Mitt 1 (Quality and stability evaluation of food preserved by heat utilizing vitamin C as a criterion),” Alimenta 1976, 15, 171-174.
M. Rahman Khan, M. M. Rahman, M. S. Islam, S. A. Begum, “A Simple UV-spectrophotometric Method for the Determination of Vitamin C Content in Various Fruits and Vegetables at Sylhet Area in Bangladesh,” Journal of Biological Sciences 2006, 6(2), 388-392 (DOI: 10.3923/jbs.2006.388.392)(Full Text).
Anal Parimal Desai, Shuchi Desai, "UV Spectroscopic Method for Determination of Vitamin C (Ascorbic Acid) Content in Different Fruits in South Gujarat Region," Int. J. Environ. Sci. Nat. Res. 2019, 21(2),  IJESNR.MS.ID.556056 (4 pages)(DOI: 10.19080/IJESNR.2019.21.556056)(PDF).


Acidity/basicity of oxides across the periodic table Recently while self-studying my chemistry book, it dawned on me that metals form basic oxides and nonmetals forma acidic oxides. Why is this?

It is one of the characteristics of the metals as for electrical conductivity, heat conductivity, mechanical properties. All these basically originate from the outer electrons to be "loosely" bound to the nucleus, when compared to non metals. Bonds in metal oxides are ionic, with positive charge on the metal. The same is for their hydroxides, i.e. their compounds containing an OH group. This leads to their basic dissociation in water  giving the stable hydroxide ion.

Contrary non metals have tightly bound electrons and are prone to fill their outer shell (Octet rule). On reaction with a non metal as oxygen the situation is therefore a bit more complicated depending on what specific element the non metal is. Non metals oxides are essentially molecular, although the covalent bond is polar to various extents. However, their compounds containing an OH group will retain electrons on the fragment consisting of the two non metals (of which, one is oxygen of course) leading to acid reaction, i.e. H+

(Note that HClO would be better wrote as ClOH).

It should be clear that things are going as ionisation potential - electron affinity - atom size (distance of the outer electrons from the nucleus). For a better explanation I would need more time, it is basically a big chapter of introductory general chemistry, which goes under the name of periodic law and properties of the elements, or something like that.


What library/program can draw molecules in 3D programmatically? Molecular graphics software is usually interactive.

But I am looking for a library or program that would be able to render a given molecule based on the commands from another program. I would like to give it a molecule description, the vantage point, options, etc, and it should produce the picture without opening any windows.

The input is the same data as contained in an XYZ file (or the file itself), the expected output is a 3D molecule drawing.

What is the easiest way to do this?

The "jmoldata" command (using JmolData.jar from Jmol) converts xyz/pdb and many other molecule formats to nicely rendered 3D images based on the user-provided Jmol scripts.

(answer based on the hint from the user @porphyrin)


Collision that governs kinetic model of gas From Atkins' Physical Chemistry 10th, it states that

The kinetic model is based on three assumptions:

The gas consists of molecules of mass $m$ in ceaseless random motion obeying the laws of classical mechanics.
The size of the molecules is negligible, in the sense that their diameters are much smaller than the average distance travelled between collisions.
The molecules interact only through brief elastic collisions.

I quite can't get the second assumption. The intermolecular distance between the molecules are greater than the diameter of the molecules, so we can neglect the volume. Given Boyle's law apply (as omitting size and volume of molecules is a property that ideal gas has), there will be many molecules collide with the walls and the molecules will hit another molecules after its collision with wall (right?), thus the pressure of the system should be low. How often the collision of what kind happens under what condition?
If the separation doesn't result any influence on each molecule (because third assumption said elastic collision, so any interaction (hydrogen bonding, van der Waals, so on) higher than the collision energy should be omitted), will the collision come purely from their collision with walls, not collision between each molecule?
Wikipedia states also the assumption that

The number of molecules is so large that statistical treatment can be applied.

There will be many molecules collide with the walls and the molecules will hit another molecules after its collision with wall (right?), thus the pressure of the system should be low. How often the collision of what kind happens under what condition?

The collision condition doesn't need to be strictly in Boyle's law condition, that the elastic collision is happened between anything and under any pressure condition. The requirement of large number of molecules but only collision interactions are applied implies that it is just not applicable to apply Boyle's law in kinetic model of gas.


In a sample of radioactive substance all the atoms don't disintegrate simultaneously. Why? Radioactivity though spontaneous all the atoms don't disintegrate at a time

The basic concept of radioactive decay kinetics is that every single nucleus has the same probability of decaying at any single given time, regardless of the physical-chemical surroundings (it does not know anything about the other atoms). This of course means that they do not disintegrate all at once, because the probability of all of them "deciding to decay" at the exact same time is infinitesimally small.


Estimate the concentration of HCl and volume of 6 M HCl required to complete precipitate each of the given salts In our chemistry lab we performed the following experiment below.

  Given that $K_\mathrm{sp}(\ce{AgCl}) = 8.2 \times 10^{-11}$ and $K_{\mathrm{sp}}(\ce{PbCl_{2}}) = 1.7 \times 10^{-5}$

  Procedure:
  $0.2\,\mathrm{mL}$ each of $\mathrm{0.05\: M \: \ce{Ag+}}$ ions and $\mathrm{ 0.05\: M \: \ce{Pb^{2+}}}$ ions were added into a centrifuge tube. The solution was diluted to $1\,\mathrm{mL}$ $6\,\mathrm{M}$ $\ce{HCl}$ was added to the solution till it completely precipitates. The quantity of $\ce{HCl}$ required was recorded. The precipitate and supernatant were separated and kept for subsequent experiments.

  Observations: Precipitate formation was observed almost instantly. It took 6 drops of $\ce{HCl}$ to completely precipitate the contents.

  Estimate the concentration of $\ce{HCl}$ and volume of $6\,\mathrm{M}$ $\ce{HCl}$ required to complete precipitate each of the given salts.

Here is my approach:

\begin{align} c_{1}V_{1} &= c_{2}V_{2} \\
\text{Let } c_{1}=0.05\:\mathrm{M}, V_{1}&=0.2\:\mathrm{mL}, \text{and} \ V_{2}=1\:\mathrm{mL}\\
[\ce{Pb^2+}] &= 0.2 \cdot 0.05\ \mathrm{M} \\
[\ce{Ag+}] &= 0.2 \cdot 0.05\ \mathrm{M}\\
K_{\mathrm{sp}}\:(\ce{AgCl}) &= [\ce{Ag+}] [\ce{Cl-}] \\
8.2 \times 10^{-11} &= 0.2 \cdot 0.05\ \mathrm{M} \cdot s_{1} \\
s_{1} &= 8.2 \times 10^{-9}\ \mathrm{M}\\
K_{\mathrm{sp}}\:(\ce{PbCl_{2}}) &= [\ce{Pb^{2+}}] [\ce{Cl-}]^{2}\\
1.7 \times 10^{-5} &= 0.2 \cdot 0.05\ \mathrm{M} \cdot (s_{2})^2\\
s_{2} &= 0.041\ \mathrm{M} \end{align}

Thus to precipitate $\ce{AgCl}$ and $\ce{PbCl_{2}}$ completely, the concentration of chloride ions in solution must exceed $8.2 \times 10^{-9}$ M and $0.041$ M respectively).

I am not sure of my computations up till this point, and I need some help on how to proceed further.

EDIT: My calculations  tell me that volume of $6\,\mathrm{M}\,\ce{HCl}$ necessary is $0.00683\,\mathrm{mL}$. In the lab we have estimated the volume of one drop of our pipette as $0.037\,\mathrm{mL}$, based on this I calculated that the number of drops of hydrochloric acid necessary to complete precipitation is $0.18$ drops. But in the actual experiment I performed 6 drops of $\ce{HCl}$ were dispensed and based on the computations we did, everything should have precipitated. But after centrifugation and separation the supernatant still had lead ions which were precipitated as lead chromate. Why is this so? (This is the main reason why I suspected my calculations and turned to this website for help).

EDIT2: I now see the flaw in my computations. what I have calculated is in fact that amount of HCl necessary to start precipitation. Thanks to everyone for their help.

Your calculations are looking good. I didn’t check the numerical values, I’ll just assume you’re able to do your algebra and use a calculator correctly.

What you got are two different concentrations required to start to precipitate all the cations in your solution. However, you are asked to completely precipitate the cations. Well — that is a nigh impossible task unless you define a concentration threshold which you want to reach. Your threshold could be $1\,\mathrm{mM}$, $1\,\mathrm{µM}$, $1\,\mathrm{nM}$ or whichever you see fit. Use this desired concentration to recalculate the concentration of chloride ions required for both cations. Then continue on.

I recalculated the value in the spoiler tag for $1\,\mathrm{µM}$ residual cation concentration.

   $4.12\,\mathrm{M}$ chloride, equivalent to $0.687\,\mathrm{mL}$ or 18 drops.
Backwards calculation tells me, that you had a residual lead concentration of ca. $9.58\,\mathrm{µM}$, explaining your second result.


Double bond crossed lines representation I have been studying organic chemistry for years, and I do not remember the spotting double bond represented with crossed lines, except maybe one ore two more times in addition to the following one

which is taken from the Aescin Wikipedia graphics here.

I guess it is a drawing or rendering mistake, or a representation of unspecific E/Z configuration. If the latter is true, where can I find more info on usage of this symbol?

UPDATE:

A comment by gilleain leads to GRAPHICAL REPRESENTATION OF
STEREOCHEMICAL CONFIGURATION (IUPAC Recommendations 2006). Relevant paragraph is:

  ...
  The most common convention for depicting double bonds with unspecified configuration in molecules of all sizes uses a wavy bond.
  ...
  A crossed double bond has sometimes been used to indicate unspecified double bond configuration or a mixture of configurations. This type of bond is not considered acceptable for general use, although it may still be required by some computer software.

There's still missing piece of information about which software and by what means it "requires" this representation.

It is, for example, used by Marvin JS -- one of the editors in the GUI of Reaxys by Elsevier to query the database for structures, where you do not impose either (E)-, nor (Z)-configuration of the double bond.  Below an example if you were looking for both isomers of stilbene:

(demo version here).

Beside ChemDraw, ChemDoodle's sketcher (here) and PubChem's sketcher are other examples of "some software" equally allowing such a representation, too.  I speculate it has something to do how databases (like ISISBase, Beilstein/Reaxys, Scifinder, for example) traditionally store(d) chemical structures and reaction schemes in a machine-readable format.

One may, for example, draw (cis)-stilbene with ChemDoodle "in the standard way" and save it as *.mol format and set this file in comparsion with the result of an undefined double bond with the using the wavy bond towards one of the phenyl rings.  In the chemical table, there is no significant difference spot:

If the "standard (cis) *.mol" file is compared with the one where the double bond is crossed -- as in the OP figure -- there is a difference on line #26 (fourth column now reads a three, instead of a zero), while maintaining the other entries:

This picture, naturally, differs from the *.mol files of classically specified (cis) and (trans) configuration:

used structures:


NMR Splitting for Adjacent, Equivalent Hydrogens? Why don't chemically equivalent hydrogens on adjacent carbons split the 1H NMR signal? There appears to be an answer here, but I'm having trouble understanding it because I cannot rationalize what happens with the magnetic field.

Let's say we're analyzing acetylene. When we subject the molecule to an external B-field, both (equivalent) protons have 2 possible spin states. For each proton, I would expect that the other proton's magnetic moment may either augment or oppose B0, causing a doublet. When we look at the signal for the other proton, it's the same story--I would expect 2 peaks identical to the 2 for the first proton (effectively doubling the integration).

Where have I gone wrong in extending spin-spin coupling to identical hydrogens? Why does the magnetic moment of adjacent, equivalent protons not affect the transition energy?

Edit: I asked my organic chemistry professor this question, and his response was that there does exist spin-spin coupling of identical protons, but the coupling constant is equal to 0. The reasoning involved quantum mechanics that went over my head.

There are two ways to look at this.

a) All transitions you can see are between a state where the spins are parallel and a state where they are antiparallel.

$aa <-->  ab$  or  $ab <--> bb$

Their distances are identical, as are the states $ab$ and $ba$, so, as porphyrin said in his comment, you only see one line.

b) If the two protons were chemically different, you would see two doublets.

If the difference in CS were small, you'd see the "roof effect": An AB spin system. The two inner lines come closer and grow at the cost of the outer lines.

But there is no difference in CS, and so the outer lines vanish, and the two inner lines coalesce into one singlet.


Should I write propanol or propan-1-ol? Should I write propanol or propan-1-ol?

Personally, I'd go for the former (it being the simpler of the two), but what's the answer according to the IUPAC system of organic nomenclature?

Propanol can refer to both propan-1-ol and propan-2-ol.

Earlier we used notations like 1-propanol or 2-propanol. But according to latest IUPAC recommendations, we prefer names like propan-1-ol and propan-2-ol.

For compounds like aldehydes, carboxylic acids and for other compounds involving carbon centered functional groups, we need not mention position because we always consider the carbon of these functional groups as the 1 carbon.

But, for alcohols, as in this case, position should be mentioned for being more specific.


Why is Cr²⁺ is a good reducing agent but Mn³⁺ is a good oxidising agent? The question is the same as the title states. In my textbook,[NCERT Chemistry I for Class 12 , pg no 217.] the following is written

$\ce{Cr^{2+}}$ gets converted to $\ce{Cr^{3+}}$ as the +3 oxidation state has half-filled $\mathrm{t_{2g}}$ orbitals; thus it is a good reducing agent.
On the other hand $\ce{Mn^{3+}}$ gets converted to $\ce{Mn^{2+}}$ as its +2 oxidation state has a half-filled d-subshell.

Now my questions are:

Why can't the $\mathrm{t_{2g}}$ argument be used for $\ce{Mn}$?

Why can't the half-filled $\mathrm{d}$-orbital argument be used for $\ce{Cr}$?

$\ce{Mn^3+}$ could be oxidized to $\ce{Mn^4+}$, which would have a half-filled $\mathrm{t_{2g}}$ set. As oxidation and reduction both can take place in aqueous medium, why wouldn't $\mathrm{t_{2g}^3}$ or $\mathrm{d^3}$ configuration be more stable than $\mathrm{d^5}$ configuration in aqueous medium?
A similar observation is made in case of $\ce{Fe^2+}$ and $\ce{Cr^2+}$, where it observed that $\ce{Cr^2+}$ is more powerful reducing agent than $\ce{Fe^2+}$ (due to reasons explained above).
If possible, I would like to see the crystal field splitting calculations that show this to be the case.

Why can't the $\mathrm{t_{2g}}$ argument be used for $\ce{Mn}$ ?

Because the number of $\mathrm{t_{2g}}$ electrons is not changing upon going from $\ce{Mn^3+}$ to $\ce{Mn^2+}$.  In both $\ce{Mn^2+}$ and $\ce{Mn^3+}$ there are 3 $\mathrm{t_{2g}}$ electrons.  I think the point they are trying to make with $\ce{Cr^3+}$ being half filled is that there are 3 $\mathrm{t_{2g}}$ electrons and no $\mathrm{e_g}$ electrons.  $\ce{Mn^3+}$ and $\ce{Mn^2+}$ both have $\mathrm{e_g}$ electrons.

You could make a similar argument for $\ce{Mn^3+}$ being oxidized to $\ce{Mn^4+}$, which would have half filled $\mathrm{t_{2g}}$ and empty $\mathrm{e_g}$.

  Why can't the half-filled d-orbital argument be used for Cr?

Because neither $\ce{Cr^3+}$ nor $\ce{Cr^2+}$ has a half filled d subshell.  $\ce{Cr^3+}$ has 3 electrons while $\ce{Cr^2+}$ has 4 electrons.


Why is fluorine the best oxidising agent, but also the most basic? (Please note that all comparisons I have made below are with respect to halide ions only.)

According to what I have learnt, the fluoride ion is the most basic ion, because it has the smallest size and thus the highest electron density. This implies that it has the highest tendency of sharing/giving away its electron density.

Why isn't the same logic applicable when reducing strength is compared? Fluorine should have the highest tendency to lose electrons, and hence should be the best reducing agent.

Fluorine is the most electronegative element.
But due to its high effective nuclear charge it holds up its electron density very tightly, moreover its a non-metal so preferably accepts electron to form stable (F–) ion which has configuration of Ne.
F2 ------> F- + e- is very favorable therefore it gets reduced very easily (reduced form F- is very stable than F2) and so it oxidises other elements and self gets reduced.


Why is the periodic table periodic? I am not a chemist, but I am interested in Science in a general sense. Can anybody explain why the periodic table is periodic in nature? I would appreciate links for further reading.

Elements interact with the rest of the world through their electrons. How those electrons interact with other atoms (or electromagnetic radiation) determines how that atom behaves.

You see this with isotopes. Different elements are distinguished from each other by the number of protons in their nucleus. Changing the number of neutrons has little effect on their properties. Why is this? It's not the protons themselves which change the properties, instead it's how the change in total positive charge of the nucleus changes how many electrons the atom can hold and how they behave that changes the properties. Different numbers of electrons - and specifically different numbers of electrons with respect to the net charge on the atom - change the properties of the atom. Changing the number of neutrons changes the nucleus, but has little effect on how the electrons behave, so has little effect on the chemistry of the atom.

So what does that have to do with "periodicity" of the periodic table? Well, there's a structure  to how the electrons are "stored" in an atom. These are orbitals. For a bunch of quantum mechanical reasons that are rather advanced for this question, there are repeating patterns in these orbitals. Most particularly, in the l quantum number which determines the shape of the orbital. Each period corresponds (roughly) to the n quantum number. For each n, there are a set of ls which go along. (Because of tradition, orbitals with different l quantum numbers get letter labels: s, p, d, f, ...)

This is why you see periodic trends. Having exactly filled s and p orbitals is the most stable, and each n has its own set of s and p. So each time you get a filled s and p, you get an inert noble gas. For halogens (like chlorine and bromine), you need just one extra electron to fill the s and p orbitals, so they really like taking a single electron from other elements. In contrast, the alkali metals (like sodium and potassium) can most easily get to a completely filled s and p orbital state by losing a single electron, resulting in their propensity for a positive charge. It's how close the atoms are to filled s and p orbitals - regardless of their n quantum number - and what it takes to get there which provides the bulk of the "periodicity" of the periodic table.

It turns out that the s and p orbitals are the most "exposed", so what's going on there contributes the majority of the behavioral change between elements. Elements that differ only in how the electrons in their d orbitals are situated have much less property differences (this is why most transition metals have similar properties). Elements which differ only in f orbitals (like the Lanthanides) are even more alike. There certainly are differences here, but they tend to rely on more subtle effects than those that cause the alkali/chalcogen/halogen/noble gas differences.

Note that this sort of orbital-based periodicity breaks down as you get lower in the periodic table. That's because you actually start to get relativistic effects, and how the orbitals behave starts to change. Periodic trends are just that - trends - and aren't absolute.


Formula for rate constant for the first order reaction My lecturer mentioned that the formula for the rate constant $k$ for the first order reaction is
$$k = \frac{2.0303}{t_{1/2}}\,\log\frac{[\ce{A}]_0}{[\ce{A}]_{t_{1/2}} - [\ce{A}]_0},\tag{1}$$
where $t_{1/2}$ is the half-life; $[\ce{A}]_0$ is the initial concentration; $[\ce{A}]_{t_{1/2}}$ is the concentration at half-life.
In which what I know recently is that the rate constant for the first order reaction is
$$
\begin{align}
&  & \ln\frac{[\ce{A}]_t}{[\ce{A}]_0} &= -kt \tag{2}\\
&\implies & [\ce{A}]_t &= [\ce{A}]_0\,\mathrm e^{-kt}\tag{3}
\end{align}
$$
I ask my teacher, but he said I should go and do more research, but I couldn't find it.

I am attaching a written answer as typing the log and other exponents is hard :(


Effect of side reactions on galvanic cell using iron rod as anode and sodium chloride as salt bridge We just made a galvanic cell on a chemistry lesson using iron, iron sulfate , copper and copper sulfate.
We used a sponge soaked in $\ce{NaCl}$ solution as the salt bridge. The issue is that we measured a voltage way below the expected value and it was dropping quite fast.
The iron sulfate solution was yellow which may have indicated that the iron ions were reacting with chloride ions.
I just can't see why this reaction would cause it to stop working from a charge perspective. What is happening here?

Since you state, "We measured a voltage way below the expected value and it was dropping quite fast," it indicates a few things that could lead to those observations.

How did you measure voltage? If you used a low impedance voltmeter, it would show much lower voltage because the impedance of the meter is low compared to that of the cell. Voltage is distributed in proportion to the impedance (here, the DC resistance) of the meter and the cell. For example a 1 volt cell with internal resistance of 20 kilohms, measured by a voltmeter with resistance of 20 kilohms, would show as only 0.5 VDC, because half the voltage is dropped inside the cell.
As a cell delivers current, the surface of the electrodes change. Corrosion, surface deposits and bubbles can cause current to drop. While the cell is being measured, observe the electrodes closely. Look for any change in appearance that progresses during measurement.
How well were the two half-cells isolated? Did the sponge effectively block liquid (as opposed to ion) flow between them? You can test that by putting some dye, such as food coloring or bromothymol blue, on one side and seeing if it migrates to the other. Flow might indicate the cell was "short circuited".


Nomenclature - Lanthanoids vs Lanthanides (and Actinides vs Actinoids) Apologies if this seems like a silly question.

Rare earth elements (the 4d series) are called by the names Lanthanides or Lanthanoids interchangeably. The same applies for the 5d series, which are called Actinides or Actinoids.

Which one of these two terms is more preferred? Is there an IUPAC directive preferring one over the other in scientific documents? If one is to write a paper or book regarding these elements, which term would one use? Or can both of them be used interchangeably?

According to p 52, Nomenclature of Inorganic Chemistry: IUPAC Recommendations 2005 (‘Red Book’):

  The ending ‘ide’
  normally indicates a negative ion, and therefore lanthanoid and actinoid are preferred to
  lanthanide and actinide.

In practice this is almost never followed, and consequently both terms are (for all intents and purposes) interchangeable, unless the context demands a differentiation.


Good choice for heat-conductive coil submerged in sea water I know only as much about chemistry as I can remember from a year in college.  So, please speak slowly and use short words.

I'm trying to decide on an appropriate material for a heat-transfer coil that is meant to remain submerged for extended periods in salt water, within a few feet of other metal parts.

I'm designing a solar still that will use a cooling/condensing coil to take advantage of the heat differential between the solar-heated evaporation chamber and nearby (cooler) water temperature.  It will be mounted on a sailboat.  The coiling coil will probably be about 12-18 inches long and 2-3 inches diameter.

Impulse was to use copper, but I'm not sure how this will react in a salt-water environment over time.  I know that sea life will stay off of it, since copper is the active element in many bottom paints designed to minimize subsurface growth.

Any help is appreciated.  Maybe the answer is to use a sacrificial anode on the coil?  Or, maybe some other material would be more suitable for long-term use?

My advice is to use copper and protect it with a sacrificial anode.  Copper would be the choice in a non corrosive environment and if you can protect it go for it.

However, if the coils will contact parts of the boat that are important than the coils should be aluminum.  Copper would cause the other metal parts to corrode faster.

Why not use rubber or plastic, something like the pipes used in drip irrigation?


Are electron shells probability distributions? I know that orbitals are probability distributions. Are electron shells probability distributions too?

Short Answer:

No.

Long Answer:

First, strictly speaking, the orbitals themselves in the quantum mechanical sense are not probability distributions.  They are eigenfunctions $\Psi_i$ of the Hamiltonian as defined by the time-independent Schroedinger equation $H\Psi_i=E_i\Psi_i$.

The probability distribution function $p(\vec r)$ for electrons is generated by summing the products of the eigenfunctions and their complex conjugates for all occupied eigenfunctions at a given 3-D point $\vec r$ and dividing by the total number $N$ of electrons in the system:

$$
p(\vec r) = \frac{1}{N}\sum_{occ}{\Psi_i^*(\vec r)\Psi_i(\vec r)}
$$

Note that $p(\vec r)$ is simply $\frac{1}{N}$ times the electron density function $\rho(\vec r) = \sum_{occ}{\Psi_i^*(\vec r)\Psi_i(\vec r)}$.

To be fair, prior to a detailed introduction to the quantum mechanical description of electronic structure, the distinction between the electronic wavefunction $\Psi(\vec r)$ and the electron density function $\rho(\vec r)$ is typically not clearly established in chemistry instruction, so your having conflated them is quite understandable.

Second, again strictly speaking, electron shells are not probability functions.  While Pushkar Soni is correct that the principal quantum number for a given electron approximately indicates the 'electron shell' to which it is assigned, the indistinguishability of electrons (among other factors) in general prohibits unambiguous translation of quantum number information into the real-space function $\rho(\vec r)$.  The best we can do is to interpret electron shells as corresponding to certain features of the structure of $\rho(\vec r)$. Offhand, I know of at least two ways of qualitatively identifying 'electron shells' from $\rho(\vec r)$.

First, as Geoff Hutchinson notes, the radial distribution function $r^2\rho(r)$ shows local maxima that can be interpreted as electron shells.  Note that  $\rho(r)$ in the r.d.f here is a function only of $r$, the distance from an isolated atom/nucleus; the angular dependencies (on $\theta$ and $\varphi$) have been integrated out.

Second, Bader's QTAIM describes how the Laplacian of the electron density, $\nabla^2\rho(\vec r)$, exhibits minima in 3-D space at positions corresponding to the notion of electron shells.


Standard Half-Cell Reduction Potential for Periodate to Iodine I'm looking for standard half-cell reduction potential $\ce{IO4-}$/$\ce{I2}$ in acid medium. Unfortunately I was unable to find the value in all comprehensible reference books. Hard to believe this values have not been measured or calculated so far.

Here is the standard half-cell reduction potential $\ce{IO4^{−}/I2}$ in acid medium depicted using Latimer diagram.

(source)

Note: Here $\ce{IO4-}$ (periodate ion) gets hydrated to form orthoperiodic acid which is $\ce{HIO4.2H2O}$  or $\ce{H5IO6}$.


What would Kp be if there is no gas reactant What would $K_\text{p}$ be if there is no gas reactant but the reaction produces gas. I found something confusing: $$K_\text{p}=\frac{P_\text{gas product}}{P_\text{gas reactant}}$$ If there is no gas in the reactant side, then the partial pressure of the reactant gas is 0 so the $K_\text{p}$ would be infinity.

Kp by definition is the product of the partial pressures of the gaseous products divided by the gaseous reactants raised to the power of their respective stoichiometric coefficient in the balanced chemical equation.

To answer your question, this definition by itself isn't enough. You need to go a little deeper. Since your reactants are not gaseous they got to be either liquid or solids. Compared to the gaseous products, the active mass of the liquid or solid does not change much. We assume the concentration of such substances to be constant value hence we shift them to the Kp to get a new Kp (actually we consider this Kp to be the correct constant). Or in other words, we can safely substitute 1 in place of all non-gaseous reactants while writing the Kp formula.

Therefore, Kp for your question is

= (P.Pressure of gaseous productA)$^a$ (P.Pressure of gaseous productB)$^b...$


What is the role of sulfuric acid and mercury(II) sulfate in the oxymercuration of alkynes? I have a homework problem where I'm given a terminal alkyne. The goal of the problem is to choose the reagent that will result in a particular ketone.

Here's a really general view of the reaction I drew out showing the reactants I chose (I don't want anyone to do my homework, just explain what's unclear to me).

$$\ce{some terminal alkyne ->[H2O][H2SO4, HgSO4] some ketone}$$

I've gone through the list of reagents given in the problem and all the others result in an aldehyde (which to me suggests anti-Markovnikov addition).

The book shows that $\ce{HgSO4}$ yields a ketone via the Markovnikov addition of water forming an enol (and undergoing tautomerization). My answer was correct, but I'm not satisfied that I really know this. The mechanism shown in the book involves only water (it doesn't show what the role of $\ce{H2SO4}$ or $\ce{HgSO4}$ is here, it only explains that it causes Markovnikov addition of water).

My only proposal is that the $\ce{HgSO4}$ adds first, in an anti-Markovnikov fashion, forcing an $\ce{H}$ from $\ce{H3O+}$ to add on the terminal carbon and thus water to add to the other side. I also am guessing that $\ce{H2SO4}$ makes the formation of $\ce{H3O+}$ more favorable.

Could someone show how this mechanism works or explain a bit about how the presence of $\ce{HgSO4}$ and $\ce{H2SO4}$ affect the reaction?

Let's first look at the acid-catalyzed hydration of an alkyne. Just like in the analogous reaction of alkenes, the new proton goes to the least substituted position, to give the most substituted carbocation. Water attacks the carbocation, and after deprotonation, an enol is produced. Under acidic conditions, the enol tautomerizes the the corresponding ketone.

The mechanism with mercury catalysis is very similar. The main difference is that the initial intermediate is a mercurinium ion, which is opened by the nucleophilic attack of water at the more substituted site. Beyond that, you can think of the mercury atom as a giant proton. The mechanism proceeds analogously, first through a mercury substituted enol, which is protonated and then demercurated.

So why is mercury typically used? We know from oxymercuration, that using mercury avoids carbocations that are prone to rearrangement. The sp2 (vinyl) carbocation shown in the first mechanism is less stable than sp3 carbocations, so we would expect rearrangement to be an issue. Avoiding the high energy carbocation also increases the rate of reaction.


Why does hydrogen phosphate act as a base? Let's look at question c:

  a) Write a balanced equation for the reaction.
  $$\ce{2 NaOH + H3PO4 -> Na2HPO4 + 2 H2O (l)}$$

  b) When some crystals of $\ce{Na2HPO4}$ were dissolved in water, the $\mathrm{pH}$ of the resulting solution was found to be $9.5$. Calculate the hydrogen ion concentration of this solution.
  $$[\ce{H+}] = 10^\mathrm{-pH}\, ;\qquad [\ce{H+}] = 10^{-9.5} = \pu{3.2e-10 mol L-1}$$

  c) Write an equation for the reaction of the $\ce{HPO4^2-}$ ion with water to account for the measured $\mathrm{pH}$.
  $$\ce{HPO4^2- + H2O (l) <=> H2PO4- + OH-}$$

So, I'm a bit confused with what's happening. It seems that water is acting as an acid in this reaction, and donating protons to the ion. Can someone just elaborate on this whole process please because I'm not really sure what is happening. Why is it $\ce{HPO}$ and not $\ce{H2PO}$ or $\ce{H3PO}$ (as is in the original equation?) Is it a step wise process and only one step included?

When hydrogen phosphate salts are dissolved in water there are two main equilibria formed. This is based on the fact, that hydrogen phosphate can act as a Brønsted–Lowry base, i.e. accept protons, or as an acid, i.e. donate protons. For water the same is true. In addition to this it can react with itself, which is known as the autoprotolysis of water:
$$\ce{H2O + H2O <=> H3+O + {}^{-}OH}\tag1$$

With this knowledge you can write
\begin{align}
\ce{HPO4^2- + H2O &~<=> PO4^3- + H3+O}\tag2\\
\ce{HPO4^2- + H2O &~<=> H2PO4- + {}^{-}OH}\tag3\\
\end{align}

To a lesser extent there is also the following equilibrium happening:
$$\ce{HPO4^2- + 2H2O <=> H3PO4 + 2 {}^{-}OH}\tag4$$

From the acidity constant you know that $\ce{HPO4^2-}$ will react mostly as a base, i.e. $\mathrm{p}K_\mathrm{a}(\ce{Na2HPO4})=12.35$, so the equilibrium $(3)$ will be predominant.


What would one have to do differently if they used NaMnO4 instead of KMnO4 for redox titrations? I am doing a chemistry report related to redox titrations (using $\ce{KMnO4}$ as the titrant), and one section of the report asks me to explain what would occur if I were to use sodium permanganate ($\ce{NaMnO4}$) instead of potassium permanganate ($\ce{KMnO4}$), and more importantly, what I would have to do differently in my experiment if I were to use this new chemical.

What I know as of now.

Sodium permanganate is not widely used because it is more expensive and is much more difficult to create than potassium permanganate.
Molar mass is different when performing stoichiometric equations, which leads to different results when calculating certain values.

I looked online and found that the chemical is "hygroscopic;" however, I don't know if this changes anything in the experiment.

Is there anything else I can "change" in my experiment if I were to use $\ce{NaMnO4}$ instead of $\ce{KMnO4}$?

You are right. Preparing sodium permanganate is more expensive. The second reason is, sodium permanganate is hygroscopic and occurs as a monohydrate. This means that sodium permanganate is more soluble than potassium permanganate. (Solubility of $\ce{KMnO_4}$ is $0.065$ g/ml and that of $\ce{NaMnO_4}$ is $0.9$ g/ml)

So if you require a larger concentration of $\ce{MnO_4^-}$ in the solution, you can go for $\ce{NaMnO_4}$.


Why is the probability of a didentate ligand to form a chelate with a central atom, high? When one ligating group of didentate ligands (like ethylenediamine) is coordinated with a central atom, then the probability of the other ligating group being coordinated, and hence forming a chelate, is high. This is seems to be because one of the ligating groups is bound to be closer to the central atom when the other is coordinated. Does this always hold true? If so, why?

Speaking in general, when a molecule is in a "energetic" environment (in this case, for instance, the energy comes from the temperature of the room, about 298 kelvins away from the "zero energy" point), the molecule moves A LOT. It bends, rotates, and changes its conformation.
When EDTA has no complexed metal, its "two arms" just keep moving randomly, until they catch a cation. When one of the arms "catches" a cation, the arms keep moving randomly. But this time, it's much more likely that the other arm will find the cation, being it so close (the cation won't move too far: the other "arm" keeps it close!).
When both arms find the cation, the rigid complex is formed.

To make a stupid example, think about a person with two magnets on his hands, moving randomly in a room in the hope of capturing with both his hands a metal ball: the "first arm" has to search around all the room, but the second one only has to find the first one


Is there relation between “Plane wave basis” “Correlation-Consistent” basis and “Fourier basis” This question is related to the following link I found while browsing the internet: http://www.gutenberg.us/articles/Basis_set_(chemistry) The question: Is there a relation between "Plane wave basis" "Correlation-Consistent" basis and "Fourier basis"?  Would anyone please add appropriate tags as well? Any comment would be appreciated.

One can think of Fourier basis as constructing the function with different frequencies. I was looking for constructing the function with vectors with same "correlation-consistent" . Is such concept possible? Is "correlation-consistent basis" is something similar?

This was supposed to be a comment, but there is too much text, so it will be an "answer".

  Is there a relation between "Plane wave basis"
  "Correlation-Consistent" basis and "Fourier basis"?

Well, mathematically, they are entities of the same type that serve for the exact same purpose:

they all are concrete examples of a well-known idea of linear algebra & functional analysis, the idea of a basis;
and they all are used to expand square-integrable functions.

Does this counts as a relation?

  I was looking for 'a kind' of basis (instead of Fourier) for some work
  and I personally gave the name for my basis as "correlation basis".
  When I typed my thoughts in google to my surprise I got the link
  mentioned in my question and I tried to understand at what context it
  is using the basis. From their I tried to relate the three basis to check whether I can use it for my work.

Are you just afraid of a terminological ambiguity out there? If so, don't worry too much. First, only the phrase "correlation-consistent basis", not just "correlation basis" has a particular well established meaning in quantum chemistry community. This is reflected even in the names of correlation-consistent bases that all starts with "cc" prefix. Secondly, "correlation-consistent basis" has a well established meaning only in quantum chemistry community concerning electron correlation and complete basis set extrapolation, so that unless you doing quantum chemistry, you shouldn't be interested in correlation-consistent bases.

Electron correlation is interdependency of the states of individual electrons in a many-electron system. Consider the simplest many-electron system, a two-electron one, and the following two events:

$1$ - the event of finding electron-one at point $\vec{r}_{1}$;
$2$ - the event of finding electron-two at point $\vec{r}_{2}$.

Then mathematically the idea of electron correlation can be simply expressed as the inequality of the joint probability of $1$ and $2$ on the one side and the product of unconditional probabilities of $1$ and $2$ on the other side
$$
    \Pr(1 \cap 2) \neq \Pr(1) \Pr(2) \, .
$$
Similarly, in terms of electron densities, we could indeed write
$$
    \rho_2(\vec{r}_{1}, \vec{r}_{2}) \neq \rho(\vec{r}_{1}) \rho(\vec{r}_{2}) \, ,
$$
where

$\rho(\vec{r}_{1})$ is the probability density of finding electron-one with arbitrary spin at $\vec{r}_{1}$ while having electron-two with arbitrary spin at arbitrary position.
Similarly, $\rho(\vec{r}_{2})$ is the probability density of finding electron-two with arbitrary spin at $\vec{r}_{2}$ while having electron-one with arbitrary spin at arbitrary position.
And $\rho_2$ is the joint probability density of finding a pair of electrons one and two with arbitrary spins simultaneously at $\vec{r}_{1}$ and $\vec{r}_{2}$ respectively.

I think $\vec{r}_{1}$ and $\vec{r}_{2}$ can indeed be thought of as random variables and that essentially electron correlation is about these variables being in a statistical relationship with each other (they are unlikely to be close to each other and likely to be as different as possible). But if you think that a special basis set is needed to account for electron correlation, that is wrong: taking electron correlation in account has very little to do with a basis set choice. There is no need to use a correlation-consistent basis set to treat electron correlation: Pople basis sets can be used to treat it, as well as correlation-consistent one.


Reaction of iron(III) oxide with nitric acid I'm currently working in a project in which I have to convert the iron oxides present in iron ore slime into ferric nitrate.

Iron ore slime is which has the particle size of below $\pu{0.15 mm}$ is being discarded as waste during the mining and processing stages iron ore and it will be stored at the tailing dam. It is estimated that $18~\%$–$25~\%$ of tailing will be generated during the processing of iron ore. The major compositions of iron ore slime are hematite, quartz, alumina, mica. So basically, it consists of ferric oxide and alumina, silica, magnesia, calcia, etc. impurities. I wanted to extract the iron part of the slime by converting it into water soluble ferric nitrate leaving behind the insoluble impurities. Even if some soluble impurities like clacium nitrate is formed then its not a problem.

I tried treating the slime with nitric acid both concentrated and dilute but no reaction was observed.
Please suggest me some useful ways to get ferric nitrate from iron ore slime — or any other steel industry waste product like mill scale, slag etc.

Yes, reacting iron(III) oxide with nitric acid would yield iron(III) nitrate.

$$\ce{Fe2O3 + 6HNO3 → 2Fe(NO3)3 + 3H2O}$$

  Iron(III) oxide react with nitric acid to produce iron(III) nitrate
  and water. Nitric acid - diluted solution. This reaction takes place
  slowly.(source)


Can a charcoal generating process be a low carbon footprint incinerator? Noting that cellulose, hydrocarbons and carbohydrates are mostly C, H and O, it occurred to me that one might take low grade recyclables no longer welcome, and trash in general, and convert them to energy and carbon using a sophisticated incinerator that uses advanced sensors and computer monitoring, air and heating controls to burn the waste with little carbon footprint.
Heat generated could be recycled to start future  batches, and excess heat could drive power generation.
An interesting side discussion would be how to use the carbon without releasing it to the atmosphere.  Carbon fiber, filtration?
Another aspect is what sorts of other pollution is a problem, like sulfur compounds, carbon monoxide?

If kept at 200°C with water under pressure, any organic molecules break down after a while, that's called hydrothermal carbonisation, basically how coal was formed geologically
With pure sugar, the gross reaction is
$\ce{4CHO -> 3C + CO2 + 2H2O }$.
Lignin is approximately $\ce{C31H34O11}$ (varies a lot of course), so
$\ce{C31H34O11 -> 31C + 6H2 + 11 H2O}$
or
$\ce{C31H34O11 -> 28C + 3CH4 + 11 H2O}$,
and other hydrocarbons, depending on the exact conditions.
You can also throw in finely shredded plastic, works the same.
The process obviously produces heat, and some $\ce{CO2}$, because once formed, it only very slowly reacts with hydrocarbon to carbon and water (you need higher temperatures and a potent catalyst for that).
Sulfur containing products can be handled by the same processes as in regular coal power plants, burn and wash out and turn into elementar sulfur.


pH value of HF, hydrogen fluoride or hydrofluoric acid? I've been looking for a source but I find different results when I try looking for an answer to this.
I'm looking for weight percentages of 1 to 10% HF (~0.5 to ~5.1M) and what their pH values are.
Now Honeywell says that around 5 to 10% it's pH is above 4.4
However if you calculate this using the $K_a$ of $6.8\times 10^{-4}$, you would get about 1.5 pH, which is quite the difference.
http://www3.imperial.ac.uk/pls/portallive/docs/1/7276108.PDF (Honeywell / imperial university)
This is just an example, because looking for these values online, for the same molar concentration, different answers are given of around pH 1 to 4.5.

So my question is can somebody explain how to properly get the correct pH values, or confirm what they are.

Also, secondary I would like to create a buffer solution with HF having a pH within the range of 4-7 (specific etching chemistry, $\ce{SiO2}$ vs $\ce{HfO2}$, not etching $\ce{HfO2}$). So other options to achieve this from stock $\ce{HF}$ (49%) with standard chemicals is also an option, however $\ce{NaF}$ is not readily available for me (mostly only access to solutions, such as $\ce{NH4OH}$).

I was able to get some rough pH indicators that are allowed in our lab (cleanroom environment). No clue why a supplier like Honeywell has completely wrong data.

Apparently some sources are completely off, doing the calculation based -log(1/H+) gives the correct values in the end, tested for 50, 10, 5, 1 and 0.5%.
for a quick calculation see:
http://www.endmemo.com/chem/phcal.php


What holds atoms together? I know atoms are held together by Ionic and Covalent bonds and i understand the ionically bonded atoms are held together by electrostatic forces. What about covalent bonds?
How are molecules of bonded atoms held together?Are ionic compounds just aggregates of oppositely charged ions and hence don't have any definitive "molecules". What about molecules of covalently bonded atoms?

In non-nuclear chemistry, everything is electrostatic interactions. This is why you can learn and predict so much just by "following the electrons"

Covalent bonds are also formed because of electrostatic interactions - they are just more complicated conceptually than ionic (actually, ionic bonds are more accurately described by wavefunctions, we just try to keep things simple in the beginning). Since electrons exist as waves, when you confine them they start to do weird stuff (or what seems weird to us). For example, the shapes of atomic orbitals are all the way they are because the electrons act like standing, 3-dimensional waves trapped between a nucleus with a positive charge and a "zero-point" at infinite distance. When you put two atoms close together, the electrons on different atoms interact with each other, and the wavefunction becomes much more complicated. The result is that in some pairs of atoms, the wavefunctions combine to form bonding orbitals.

So the short answer to your first question is: "Molecular orbitals hold atoms together in covalent bonds, and those are a result of electrostatic interactions and the quantum nature of electrons."

Yes, ionic compounds are large collections of ions, and you can't really define "molecules" for them - instead we talk about "formula units" which are the lowest possible whole-number ratio of elements that represent the compound. Groups of covalently bonded atoms are also held together by electrostatic interactions, but since the covalent bonds are so much stronger, a molecular compound can exist "on its own" as a single molecule. Collectively, the forces that hold collections of molecules together are called van der Waals forces if they don't involve ions. In any atom or molecule, there is never a completely uniform charge density on the surface. For some molecules, this is extreme (water is a good example) and we say it is very polar, or that it has a large dipole moment. This is just another way of saying that one part has a negative charge and the other has a positive charge. In water it looks like this (from wikipedia):

In this picture, red means "more electrons" and blue means "less electrons." Water can form hydrogen bonds, which are very strong electrostatic interactions. Some atoms and molecules have an almost uniform charge density on the surface. We call these "non-polar" molecules - noble gases are good examples. However, even noble gases have what is called an induced dipole due to statistically correlated fluctuations in electron density when the atoms are near each other. As a result, even noble gases can be cooled to the point where they become liquid - the very, very weak electrostatic interactions will hold them together at low temperature, when they are not moving very fast. These forces are called London Dispersion Forces - after the guy who first described them. London dispersion forces are important, because they are found in all molecules - polar or not. In fact, this is what makes most plastics solid. Polyethylene, for example, is made of very long chains of essentially non-polar molecules (from wikipedia):

Each chain is attracted to the others through weak London dispersion forces, however, since each chain has tens of thousands of atoms, those tiny forces quickly add up to make large forces holding the polymer together. This is why polyethylene is solid at room temperature, and can be used to make things like shopping bags!


Carcinogenicity of polynuclear hydrocarbons and benzene rings (1933) established that polynuclear aromatic hydrocarbons (PAHs) were
carcinogenic components of pitch. ... Although less is known about
their health effects than about those of PAHs, it is clear that some
nitro-PAHs are potent mutagens and carcinogens.(1)
Benzene and polynuclear hydrocarbons containing more than two benzene
rings fused together are toxic and said to possess cancer producing
(carcinogenic) property. Such polynuclear hydrocarbons are formed on
incomplete combustion of organic materials like tobacco, coal and
petroleum.(2)

From the two sources cited above, I keep seeing that polybenzenoid compounds like 1,2-benzanthracene, 3-methylcholanthrene and 1,2- benzpyrene are mentioned to be carcinogens.

In general, why are polynuclear compounds carcinogenic? I google searched on this topic and all I could find is their effects and concentrations in various substances.
My questions are:

Why are they carcinogenic

Is there a reason for a given organic compound to be carcinogenic?

References

Potential Carcinogenic Effects of Polynuclear Aromatic Hydrocarbons and Nitroaromatics in Mobile Source Emissions, STEPHEN S.HECHT(link)
pg403, NCERT Class XI part 2(pdf link)

1.Why are they carcinogenic:

From my point of view as a medicinal chemist, polycyclic aromatic hydrocarbons compounds tend to be metabolized by specific cytochrome P450 (CYP 450) isofroms to form highly electrophilic species, which are highly reactive and can react with the DNA of the cell and damage it. Specifically, they can react with the nucleophilic nitrogens in the DNA bases such as adenine.
To see more detailed mechanism of the formation of the electrophilic species from the compounds, check this article: https://socratic.org/questions/how-do-polycyclic-aromatic-hydrocarbons-cause-cancer

2.Is there a reason for a given organic compound to be carcinogenic?

Mechanism for carcinogenicity can vary, in general, any compound damaging or altering the DNA can be carcinogenic. For example, highly reactive electrophiles (such as formaldehyde).


AlCl3: ionic and covalent? My textbook states that AlCl3 is both an ionic and a covalent compound .How is it so?

The distinction between ionic and covalent compounds is arbitrary and many compounds fall somewhere in the middle. This is explained in a previous question.

In the case of $\ce{AlCl3}$ we have a metal 'cation' and a non-metal 'anion' so in a simplistic view we would expect this to be an ionic compound. However, it actually turns out that the $\ce{Al-Cl}$ bonds display a significant degree of covalent character. In the solid state it adopts an 'ionic lattice' structure with octahedral coordination for the $\ce{Al^{3+}}$ ions but in the liquid and gas phases it exists as a covalent compound, either as $\ce{AlCl3}$ or as a dimer $\ce{Al2Cl6}$.

We can rationalize this by considering the polarizing effect of the aluminium cation on the chloride anions. $\ce{Al^{3+}}$ is a small, highly charged cation and therefore has a high charge density. $\ce{Cl-}$ is a relatively large anion, with a low charge density and is easily polarized by the hard cation, giving the bond significant covalent character. The effect is even more pronounced if we move to a larger anion such as $\ce{I-}$ where the molecule, $\ce{AlI3}$ exists as a covalent dimer, $\ce{Al2I6}$, even in the solid phase.


How does DMSO act as a carrier for transdermal transport of large molecules? DMSO.org claims the following:

  "[DMSO] is more successful ferrying some drugs, such as morphine
  sulfate, penicillin, steroids, and cortisone, than others, such as
  insulin."

All of these drugs that it is said to carry are much larger than a single DMSO molecule.

So what is the general mechanism by which DMSO can bring these relatively high-molecular weight compounds into your body transdermally, where they would not do so otherwise?  Does it simply “wrap them up” in a large complex, or does it possibly just complex the portions of the other compound that prevent it from crossing transdermally?  What is it about insulin for example that prevents it from being carried through the skin?

A quick internet search will give pages of documents basically saying “this is what it does”  but I’ve never found a remotely technical explanation of how this works, particularly for large molecules.  That’s what I’m asking here.

NOTE 1: I have never used DMSO medicinally and currently have no intentions of doing so.
NOTE 2: I hope this is not considered a duplicate of this question as the answer is specific to the transport of small molecules.

Quote taken from the review (1) (and which is also duplicated in Jacob's 2015 "canonical DMSO" book (2, p. 16)):

  There is some evidence to suggest that DMSO can increase diffusion
  through the
  stratum corneum by disruption of the barrier function. This probably occurs through aprotic interactions with intercellular lipids and may
  also include reversible distortion of lipid head groups that produce a
  more permeable packing arrangement.  DMSO may also play a role in
  partitioning as well by forming solvent microenvironments within the
  tissue that can effectively extract solute from vehicle. Finally, DMSO
  can have a profound solubilizing effect on less soluble agents in a
  variety of vehicles, increasing penetration simply by delivering a
  higher concentration to the membrane barrier.

alongside with extensive molecular modelling by Gurtovenko in 2007 (3) are both suggesting quick pore formation and expansion alongside with the thinning of cellular membranes:

  The reported membrane thinning and, in particular, pore formation
  provide a credible molecular-level explanation of how DMSO promotes
  permeation of molecules, particularly those of hydrophilic nature,
  through lipid membranes as observed experimentally.

From what I understand, DMSO paves the way for bigger molecules, such as steroids, making inner skin layers and underlying tissue reversibly accessible, sometimes even on nanoscale level. For the precise mechanism I think it is worth reading through the entire paper (3).

(1) Capriotti, K.; Capriotti, J. A. The Journal of clinical and aesthetic dermatology 2012, 5 (9), 24.
(2) Jacob, S. W.; Jack, C. Dimethyl sulfoxide (DMSO) in trauma and disease; CRC Press, 2015.
(3) Gurtovenko, A. A.; Anwar, J. The Journal of Physical Chemistry B 2007, 111 (35), 10453–10460. DOI 10.1021/jp073113e


Why do we do electrolysis and electroplating using warm electrolyte? This is a pretty basic question, and I know it has something to do with conductivity, but I'm not quite sure how they are related.

ringo makes good points in his answer. Additionally, though, the increased temperature enhances mass transfer of ions to/from the electrode surfaces by at least two mechanisms:

Higher temperature results in lower electrolyte viscosity, leading to a thinner fluid dynamic boundary layer and concomitant greater mass transfer to/from the electrode surfaces.
Higher temperature also increases the diffusivities of the electrolyte solutes, which also contributes to increased mass transfer to/from the electrodes.


Molarity Numerical on Mixing Solutions If 3 L of 1 M NaOH solution is prepared by mixing properties of stock solutions of $x$ L of 2.5 M NaOH and $y$ L of 0.4 M NaOH and no water is to be used, find $x : y$.

My attempt:

$$3 = 2.5x + 0.4y$$

Now I am stuck how to calculate $x:y$.

In addition to the equation for the preservation of the amount of substance you've written, there is one more equation you can use for the system, namely the one reflecting preservation of volumes: $x + y = 3$:

$$
\left\{
\begin{align}
x + y &= 3 \\
2.5x + 0.4y &= 3
\end{align}
\right.
$$

$$x + y = 2.5x + 0.4y$$

$$1.5x = 0.6y$$

$$\frac{x}{y} = \frac{6\cdot 2}{10\cdot 3} = \frac{2}{5}$$

E.g. the answer is $x : y = 1 : 2.5$.


Sodium metal in acetone When I place a chunk of 99,99% pure sodium metal in a beaker of (100% chemically pure) acetone, the sodium metal starts reacting and dissolving into the acetone and it forms a yellow/brown liquid. When the acetone evaporates it leaves a dark-brown goo that is like a gel. And it smells like tar or something similar. What is this substance?

I can suggest only two possibilities:

Your acetone or sodium are not pure. Acetone contains water, or something.
Sodium might react like magnesium does, though it required different conditions. That's possible, but maybe you just discovered a new reaction or had some weird impurities that served as a catalyst.


Can you make deuterium-depleted water through a freezing method? Deuterium-depleted water (DDW) is sold commercially by some companies. It is apparently manufactured by fractional distillation, which leverages differences in boiling points of heavy and light water.

My question is this: can DDW be manufactured through a freezing method? Heavy water has a freezing point of 38.8 °F and normal water of 32 °F. If a jar of water were kept in a water bath of say a constant 35 °F, wouldn't the heavy water freeze and sink to the bottom of jar, allowing the DDW to be siphoned off of the top? I am surprised there appears to be no discussion of the merits or demerits of this idea anywhere on the internet.

Not easily. For all practical purposes, $\ce{D2O}$, $\ce{DHO}$ and $\ce{H2O}$ behave the same -- including complete miscibility. Though some small fraction of the liquid that freezes first would be enriched in $\ce{D2O}$, and to a smaller extent in $\ce{DHO}$, it would still be composed largely of $\ce{H2O}$.

According to Wikipedia, the Girdler process is the most energy efficient means of separation.


Reducing property of alkali metals In my book it is given

  Reducing property increases down the group in gaseous or molten state .
  The increasing order is  Li , Na , K , Rb ,Cs

In according to me reducing property is based on ionisation property . As Cs has highest ionisation property so it can easily loose elctron (oxidation potential)

  But in aqueous solution incresing order is given as
  Na , Cs ,Rb , k ,Li

For this what could be the reason .

Reducing property in aqueous phase depends upon (1)  enthalpy of sublimation (2)enthalpy of ionisation (3)enthalpy of hydration.
First two processes are endothermic while last one is exothermic.
For lithium enthalpy of hydration is very high because of its small size hence Li acts as stronge reducing agent in aqueous phase. For Cs enthalpy of hydration is very small hence overall process is less exothermic .


What does "a metallic oxide has 40 % oxygen" mean? Question
  If a metallic oxide has $40\,\%$ oxygen, find the equivalent weight of
  the metal.

This amounts to finding the atomic weight of the metal and the charge on the cation (in effect identifying the metal itself).

I didn't know how to do this, so I looked up the solution. It goes as follows.

  Solution
  Assume the oxide has the formula $\ce{MO}$. Let the atomic weight of $\ce{M}$ be $x$
  Then
  $$\frac{16}{x + 16} = \frac{40}{100}$$
  Simplifying and solving for $x$ yields
  $$x = 24$$
  Thus the metal is magnesium and the equivalent weight is $12$.

My problem is, how can we assume that the metallic oxide has the formula $\ce{MO}$? For all we know, the metallic oxide may be $\ce{M2O}$ or $\ce{M2O3}$. In fact, if we assume the latter formula we get

$$\frac{48}{2x + 48} = \frac{40}{100}$$

giving

$$x = 72$$

which I believe is Hafnium.

Your question is a good one, and no, you can't assume that the metal is $\ce{MO}$ necessarily.  However, you can find the equivalent weight of the metal without any assumptions about its formula.

The equivalent weight of a metal combining with oxygen is the mass that reacts with 8g of oxygen.  We know that the compound contains 40% oxygen, so it must contain 60% metal.  Suppose we had 100g of this compound; that would be 40g O and 60g M.

$40g\ \ce{O}:60g\ \ce{M}$ (Divide by 5)
$8 g\ \ce{O}:12g\ \ce{M}$

The equivalent weight of the metal must be 12g.

If you had been asked to identify the metal you would have needed the formula (or would have needed to test the likely formulas to see if they produced reasonable results.)

Given an equivalent weight for M of 12g:
$\ce{M2O-> 12g/mol, carbon}$ (not a metal)
$\ce{MO-> 24g/mol, magnesium}$
$\ce{M2O3-> 36g/mol, chlorine}$ (not a metal)
$\ce{MO2->48g/mol, titanium}$

Both magnesium and titanium fit the information that was given, $\ce{MgO}$ and $\ce{TiO2}$ are common oxides of those metals.


why do we obtain a sigmoid curve in vapour pressure versus temperature graph i have recently got a question in an assignment, which was somewhat like this

  what would be the shape of curve obtained in a graph between vapour pressure & temperature of a binary solution in a closed vessel

generally we do all the experiments regarding vapour pressure in a closed vessel, so the volume must remain constant and from the formula $PV=nRT$, the number of moles remains constant in a binary solution, as volume is remaining constant we get the equation as $Y=mX$, which is a straight line passing through the origin,
but the answer was a sigmoid curve, how did we get that, can i get an elaborate answer for this quiery.

the answer already present over here is absolutely excellent but, as an extention i would add some points to it

  the equation of $PV=nRT$ is made for gases and for a solution of gas and liquid in which the volatile solvent shows some vapour pressure on its solution

the curve sigmoid curve you get is through the Clausius Clapeyron equation which suits this condition [i.e. the situation of solutions]

after the whole solution is evaporated then you get pure solvent in the vapour phase after which your equation for an ideal gas $PV=nRT$ is applicable and you will get a striaght line there after


Tautomerism in alkenes Why doesn't but-2-ene exhibit tautomerism like alkenols?
By transfer of 1 hydrogen from terminal CH₃ to the middle carbon and forming a double bond between the two?

In order for tautomerism to occur, the enol must be deprotonated, so as to allow the negative charge to come down, kick the pi bond of the C=C bond, which will bond to a proton. Hydrogen atoms on oxygen are much more acidic than hydrogens on carbons because of the electronegativity differences between O and H. Thus, the deprotonation does not happen in alkenes.


What is the crystal structure of ammonium hydrogen sulfate? It's beautifully birefringent I make art using a microscope and camera. I have heated ammonium sulphate that decomposes to ammonium hydrogen sulfate, melts, and then recrystalises to produce exceptionally beautiful birefiringent colours.
I am interested in its typical crystaline structure as perhaps other chemicals with similar structure may also be interesting artistically.
I have tried looking it up in ChemSpider, but no luck. I took some university courses in chemistry, so I have some (out of date) chemistry knowledge.

Birefrigence is a result of structural anisotropy observed for crystals other than of the cubic Bravais class. Then, interaction of vectorial properties (e.g., light) with said material is anisotropic, too. It is a phenomenon seen both for inorganic, as well as organic materials (examples listed e.g., here). As seen e.g., by cooling films of molten polymers on a heat stage in polarization microscopy, local deviations from isotropy suffice to observe birefrigence, too (example). Thus, there is no «special functional group» which, if present, suddenly would switch on / switch off a sample to be birefrigent. It's observation and measurement by polarization microscopy allows to characterize the microstructure of e.g., minerals, and thus one of the  elder tools in geology (still used today, e.g., Stoiber's «Crystal Identification with the Polarizing Microscope» by 1994 lists some in its freely available back matter).  It may be used on organic samples, too (e.g., this tutorial).
The models to describe the crystal structure about ammonium sulfate, $\ce{(NH4)HSO4}$, depend on the temperature of the sample. This might increase or decrease birefrigence (I did not look out for a temperature-dependent recording of this about ammonium sulfate).  The explored variation of sample temperature in the work below (at ambient pressure) however do not change the unit cell symmetry this much that birefrigence suddenly would be cancelled out for symmetry alone.
At room temperature ($\pu{293 K}$), the best description of this material that there two hydrogensulfate groups per unit cell which are described independently from each other:

(credit 2012JChemPhysChemA223, Fig. 1a / $\pu{293 K}$)
In the illustration, these are marked by O (ordered) and D (disordered).  This is to say that the model describes the reality best assuming positions with $\ce{HSO^-4}$ with more than one orientation and local geometry.  (Similar to sitting either forward, or backward on a chair.)  This is, taking a different perspective on the sample (the representations omit hydrogen atoms):

(modified representation, based on 1972ActaCrystA445)
with ammonium ions, ordered and disordered $\ce{HSO^-4}$, the sites with the disordered would be either one of the following:

(credit 1972ActaCrystA445)
At lower temperatures, e.g. at $\pu{200 K}$ and at $\pu{125 K}$, it is energetically more favourable if both $\ce{NH^+4}$ and $\ce{HSO^-4}$ are in ordered position and orientation only, e.g.:

(credit 2012JChemPhysChemA223, Fig. 1b / $\pu{200 K}$)
This phase transitions and changes of unit cell symmetries (space group $P2_1/c$ (at $\pu{293 K}$) $\rightarrow$ $Pc$ (at $\pu{200 K}$) $\rightarrow$ $P1$ (at $\pu{125 K}$)) however will not cause a cancellation of  birefrigence because of symmetry reasons.

You do not state (this much) why knowing the the spatial arrangement of the atoms, and hence a model of the crystal structure is important for you.  However, because you mention to possess some command of chemistry and interest art, be aware among the crystallographers there are some equally with a sense for beauty.  One of the low-level entries is IUCr's Crystal growing competition for school children (note the their brochure with examples in the later section) considerably newer than their teaching pamphlet.
On the more artistically appealing side, though, selectively dyeing crystals may be of interest for you.   Work in this field was much popularized again by e.g. Bart Kahr:

(credit, includes protocol)

(credit 2001ChemRev893)
with protocols for chemicals easier to access (e.g., potassium dihydrogen phosphate [KDP] and amaranth dye, cf. here).  Beside beauty, the selective incorporation of dyes provides insight into the growth mechanism of crystals. You may find additional inspiration in other works by him, e.g., in 1996AdvMat941, 2001ChemRev893 including a 20 page long table of examples («reported
during the past 147 years»), or 2002CrystEngComm514 about «painting crystals» by an easy-to-follow protocol.


Raoult's Law confusion I have this confusion regarding the statement of Raoult's Law. It states that

The partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

Does mole fraction in the mixture mean the combined liquid-vapour mixture or only the liquid which is left after evaporation on attaining equilibrium?
I had this problem in my textbook, one mole of two volatile substances each is mixed. Pure pressures of both of them are given. Should I just use mole fraction as one-half? Aren't some of the moles gonna go into vapour phase?

It should be like:

The saturated partial pressure of each component of an ideal mixture of liquids is equal to the saturated vapour pressure of the pure component multiplied by its mole fraction in the liquid mixture.

These data are independent on the actual composition of the gaseous phase, that may, or may not be in the equilibrium with the liquid.
OTOH, composition of the liquid phase does depend on the amount and composition of the original liquid mixture put into a container, as evaporation changes liquid composition in favour of less volatile component.
If the pure liquid $\mathrm{A}$ has saturated vapour presure $p_\mathrm{0,A}$
and the pure liquid $\mathrm{B}$ has saturated vapour presure $p_\mathrm{0,B}$
and if the $\mathrm{A}$ in the liquid has molar fraction $x_\mathrm{A,l}$,
then the gaseous molar fraction of $\mathrm{A}$ ( in the saturated gaseous phase, with assumption of ideal behaviour given by the Raoult law ) is:
$$x_\mathrm{A,g} = \frac{x_\mathrm{A,l} \cdot p_\mathrm{0,A}}{ x_\mathrm{A,l} \cdot p_\mathrm{0,A} + (1-x_\mathrm{A,l}) \cdot p_\mathrm{0,B}}$$
For calculation for particular volumes of the container and the initial mixture of volumes, you would need more data, like these volumes and densities, and assumptions how to treat gases, like. e.g. as the ideal gas, but it goes behind the Raoult law. In all the complexity, it is solving a set of equations for set of variables, like molar amounts in both phases and phase volumes.
If the fraction of components in the gaseous volume is negligible, e.g. if we consider the total vapour pressure in a full or almost full container, we can consider total mixture composition is the same as liquid composition.


Is it incorrect to use Fisher Projections for carbons that don't have stereocenters? For example:

Is a cross specifically used for stereocenter representation?

According to IUPAC it is not incorrect for using it to depict carbohydrates being stereogenic or not. However here is an excerpt from the link listed below:

  The use of
  Fischer projections in non-carbohydrates is problematic. Any drawings
  of this type are necessarily ambiguous since it is unclear whether the
  author of the drawing intended to make a Fischer projection or simply
  happened to draw a diagram in that orientation. Molecules depicted in
  the style of Fischer projections are likely to be recognized as such
  when they contain at least two acyclic stereo- genic centers oriented
  vertically, although such interpretation is not guaranteed. Molecules
  containing a single stereogenic center pose even greater problems
  since it is often convenient to separate the four ligands by equal
  angles whether or not a single stereoisomer is intended. Accordingly,
  although Fischer projections of carbohydrates are acceptable, similar
  depictions of non-carbohydrates should be used only when the viewer of
  the diagram is certain to understand that the diagram was intended to
  represent a Fischer projection. Fischer projections should be avoided
  where ambiguity is possible, including when used in computer
  applications.

See more here (page 37-38):

http://www.iupac.org/publications/pac/2006/pdf/7810x1897.pdf