Magnitude of Lennard-Jones force evaluated with molecular dynamics simulation The intent is to evaluate the error source of a molecular dynamics (MD) simulation which is only based on the Lennard-Jones interaction between noble gases (helium in this case). What is the magnitude of the force?
For reference: $\varepsilon = \pu{1.49E-22 J}$ and $\sigma = \pu{2.967E-10 m}.$
The equation used to calculate the force is:
$$F = 4\varepsilon\left(\frac{6\sigma^6}{r_\mathrm{abs}^7} - \frac{12\sigma^{12}}{r_\mathrm{abs}^{13}}\right) \sim \pu{E-13 N}$$
Combined with the mass of helium this leads to an acceleration which seems to be too large. Would the above make sense?
Our step size is roughly $\pu{5E-3 s}$ and the velocity has a magnitude of $\pu{E13 m s^-1}.$ This is our problem at the moment because those values are obviously too large. We are using the Verlet algorithm.

The following plots the force for reasonable distances

Intuitively, if you are implementing the verlet algorithm then the step size will be $\delta x \propto a\delta t^2$ (ignoring velocities). If you want a step size of order ~$f \times \sigma$ (where f is a scaling factor) then you'd like to satisfy a relation such as $\delta t < \sqrt{\frac{f\sigma}{a}}$ (if ignoring velocities). The timestep cutoff $\delta t$ computed according to this equation with a force of $\pu{10^-12 N}$ and f=0.1 is 0.3 psec. But 0.3 psec is still a relatively long timestep. Clearly the problem is not with acceleration (provided the timestep is shorter than the cutoff just calculated), but maybe there is one with the velocities. You can proceed in the same way but solving the complete quadratic verlet equation for the timestep in terms of the step size, and inserting a value for the rms speed based on the desired T and step size, to arrive at a timestep cutoff. Using this reasoning I came up with a cutoff $\delta t < \pu{ 15 fsec}$
One problem with a large step size (and discretization of the trajectory in general) is that the energy is not conserved and can easily blow up (along with the temperature): if you take a step of order $\sigma$ from the potential well minimum toward the maximum in the potential then the atoms will suddenly find themselves in the strongly repulsive high energy region.
You may want to check Ref. 1 which discusses this to some extent, and provides this advice:

One wants to find the largest time step that will maintain the conservation of energy.

Reference

Sangrak Kim. Issues on the Choice of a Proper Time Step in Molecular Dynamics. Physics  Procedia 53 (2014) 60–62.
doi: 10.1016/j.phpro.2014.06.027


Why degree of dissociation/ionisation affected by dilution? What happens when we dilute, say, an acid which is weak, and then add water to it? What motivates or pushes it to dissociate more?

We are just adding some molecules of water, right? So, what exactly happens to a weak acid like $\ce{CH3COOH}$ so it starts to dissociate more?

Is it because of the fact that water being a polar solvent having high dielectric constant starts to split the $\ce{O-H}$ bond in $\ce{CH3COOH}$, or something else?

What motivates or pushes it to dissociate more?

Perhaps surprisingly, if you could look at an individual acid molecule, it does not (or hardly) changes its behavior upon dilution. It just dissociates with a certain rate, yielding the conjugate base and a hydrogen ion. There is no "extra motivation" to dissociate, and no "tapping on the shoulder" pushing it to react.

  Why is the degree of dissociation/ionisation affected by dilution?

The reverse reaction, where a dissociated acid (the conjugated base) reacts with a hydrogen ion to make the acid again, is affected by dilution. Here, two molecules have to come together, and this takes longer when the solution is more dilute.

This description is for the simplest case, where the net reaction not only describes the stoichiometry but also the mechanism of the reaction. However, even if there are intermediates, at equilibrium the forward and reverse rates are equal, and when you dilute solution, rates of the reverse steps would be significantly lowered, leading to a higher degree of dissociation.


Do soap molecules form liposomes? I know soap molecules form micelles. Do they also form liposomes in water? I need the reasons too.

Amphiphiles spontaneously form a variety of self-assemblies in water, such as
micelles, vesicles, and bilayers. The simplest form of self-assembly for single chain amphiphiles are micelles.

The concept of packing parameter proposed by J. Israelachvili allows a simple
and intuitive insight into the self-assembled aggregates of amphiphiles in water

The packing parameter $P$ is defined as:
$$P = \frac v{ l \times a_0}$$

where $v$ and $l$ are the volume and length of the surfactant hydrophobic tail, respectively, and $a_0$ is the optimal surface area per amphiphile.

Let me briefly explain the concept of optimal surface area per amphiphile ($a_0$). It is based on the principle of opposing forces; the hydrophobic interactions act to pack the soap molecules closer together to avoid exposing their hydrophobic tails to water, while the repulsive energy between the charged headgroups of the soap molecules increases as molecules are pushed closer together. The surface energy of the self-assembly is minimized when the amphiphilic molecules occupy the optimal area.

Single-tailed amphiphiles (like your soap molecules) have a large head-group area and have a packing parameter $P < \frac13$. Thus, they form spherical micelles. Ionic surfactants such as sodium dodecyl sulphate (SDS/NaDS) generally forms spherical micelles in solution at low salt concentrations. Single-tailed amphiphiles with a small head-group area ($\frac13 < P < \frac12$) will form elongated or cylindrical micelles. Non-ionic surfactants and ionic surfactants such as
SDS and CTAB in high salt are known to form elongated micelles. Double-tailed
amphiphiles with a large head-group area (phosphatidylcholine, phosphatidylserine, phosphatidylinositol etc) have a packing parameter approaching unity and tend to self-assemble into planar bilayers and vesicles/liposomes.

The figure here will explain this schematically:


How does adsorption work? I would like to know how adsorption works particularly between a gas and a solid. I know that there is an attraction of intermolecular forces between the gas molecules and the solid surface molecules. But how can this attraction suck all the gas molecules onto the solid surface? I thought the intermolecular forces cannot act at such long distances. Any idea?

Adsorption (with a d) is caused by intermolecular Van der Waals Forces allowing molecules to stick to a solid surface.  The solid surface can be described using a Langmiur model.  In this model a surface can bind with a number of gas molecules at any number of active sites. These active sites may be packed much closer together than molecules dispersed in a gaseous state. When dispersed gaseous molecules in a closed container interact with a surface packed with active sites, the surface may appear to "suck the gas molecules" as the gas molecules bind to the densely packed active sites.

This suction is particularly true of several activated carbons because the porous surface causes a large number of active sites to pack together.

The Langmuir model is only one model. In this model once all of the active sites are filled, the gas around the surface forms a monolayer. The monolayer is expanded upon in more sophisticated models such as the BET model.  The BET model is used to analytically determine BET surface area of a material like activated carbon.

Sometimes the behavior of the surface is described empirically such as in the Freundlich adsorption isotherm.  In this case, as with all empirical models, there is no explanation for the attraction sucking molecules to the surface.  It is simply accepted the model works.


What is the mathematical relation for the change in water density at different temperatures? I am curious to the relation of the density of water at different temperature and as to also how one would come to an equation that took the temperature of the water and would output it's density? I have graphed density vs temp from tabulated values and it seems to look logarithmic, but wanted to make sure. I am interested in this as I am currently taking analytic chemistry.

To do this one must measure density accurately at many temperatures and then fit a curve to the measurements. This has been done by several investigators in several industries. For example in the sugar industry we have

$$\rho =(((((-281.03006e-12*t +105.84601e-9)*t-46.241757e-6)*t-7.9905127e-3)*t+16.952577)*t +999.83952)/(1+16.887236e-3*t)$$

where $t$ is the temperature in °C and the density $g·cc^-1$. The coefficients were taken from the ICUMSA formula with the sugar concentrations set to 0.

And

$$\rho = 0.99984+t*(6.7715e-05-t*(+9.0735e-06-t*(1.015e-07-t*(+1.3356e-09-t*(1.4421e-11-t*(+1.0896e-13-t*(4.9038e-16-9.7531e-19*t)))))))$$

returns the density of water, also in $g·cc^-1$,  as a function of centigrade temperature (ITS 1990). It is based on a fit to data from
Bettin, H.; Spieweck,F.: "Die Dichte des Wassers als Funktion der Temperatur nach Einführung der Internationalen
Temperaturskala von 1990. PTB-Mitt. 100 (1990) pg 195-196
The referenced data set lists densities for each 0.1°C over the range (0,100). The data used in determining this
polynomial are the subset on integer degree values. Sporadic checking shows that the fit appears to be
accurate to 1 count in the 6th decimal place. The rms residual wrt the fit points is 3.1E-7 and the peak
residual 5.5E-7. The residuals appear noiselike.

You ought to be able to paste those formulas into a visualization program and plot values. I don't see any logarithmic behaviour. I've left the * in the formulas so you can do that. Note, when reading them, that 16.887236e−3∗t means (16.887236e−3)∗t.


How long did the most difficult compound to synthesize last? I heard that the most difficult compound to produce was a molecule that included Argon because it is a noble gas. I was wondering how long it was stable for.

There is no such thing as the most difficult compound to produce. Instead, there are some hundreds of thousands of "stable" compounds of which we know pretty much everything - melting point, boiling point, heat capacity over a wide range of temperatures, refractive index, speed of sound, magnetic susceptibility, electrical conductivity, crystal structure, and many more. Around that field is a much wider "grey zone" of the not-so-stable compounds (millions of them), which miss some of these properties because they can't be measured.

As we reach out to the far outskirts of that zone, we meet more and more compounds with less and less known properties in each one, because they are increasingly frail and short-lived, to the point that the very words "produce" or "synthesize" (or even "exist", for that matter) do not quite apply to them anymore. Some of them last but a fraction of second; some live longer, but require very peculiar conditions. There is absolutely no meaningful way to arrange them in order of increasing "difficulty".

Also, welcome to Chem.SE.


Clarification on the term: coordination number Based on what I've been reading the coordination number of a complex is defined as the number of atoms/molecules that the central atom is bonded to. However, it is simulaneously said that [FeEDTA] has a coordination number of 6. So which is it. Is the coordination number the number of atoms/molecules, or the number of dative covalent bonds?

Coordination number, also called ligancy is the number of atoms that a central atom or ion holds as its nearest neighbours in a complex or coordination compound or in a crystal.

In EDTA, there are two nitrogen atoms and four oxygen atoms, which act as neighbors and bind by sharing their electrons. So, even though EDTA is itself one single molecule, there are six atoms in EDTA which are sharing their electrons. So, the iron has the coordination number of 6.

The complex looks like this:

So, basically, coordination number depends on how many atoms are directly participating in the complexing.


Potassium permanganate oxidation of sugars I am told that all three sugars below can be oxidized to their corresponding carboxylic acids using potassium permanganate.

This strikes me as bloody wrong because only hemiacetal sugars interconvert in solution with their open chain forms, and we need that open chain form to access the aldehyde/ketone for oxidation. Potassium permanganate as far as we know creates C-O bonds while not breaking any C-C bonds ... hence ... the general rule in our class is that ketones cannot be further oxidized (further oxidation would rupture C-C bonds).

Each of your molecules also contains at least one primary alcohol appendage.  Primary alcohols are readily oxidized by potassium permanganate to the corresponding carboxylic acid (see this link for more information).


Can olivine be used to make glass? This concerns ongoing work to find a plausible approach to large-scale production of clear glass on the Moon. It's for an ultra-hard science fiction project and I really want to get it right.

Olivine of good purity can be gotten from deposits of dunite in the lunar highlands. You need to get to the bedrock, but it's there. If you melt the dunite you can use fractional crystallization to improve its purity - I have asked about that on Earth Sciences but it seems to be true. If that works it would also suggest you can do the same thing again to separate the olivine into forsterite ($\ce{Mg2SiO4}$) and fayalite ($\ce{Fe2SiO4}$).

If the forsterite is pure enough, could that be used to make decent clear glass?

Could the fayalite be processed into iron, water, and silicon dioxide with something like the reaction below?

$\ce{Fe2SiO4 + 2H2 -> 2H2O + 2Fe + SiO2}$

You could proceed from either end member of the olivine solid series and yield $\ce{SiO2}$ as you suggest.

However, I'd consider mechanisms that have been researched in the course of studying so-called mineral sequestration in addition to what you've written, especially considering the energy requirements you propose: the mineral sequestration reactions are (slightly) exothermic and would thus also yield heat as a by-product which I suspect would be desirable in a cold place like the Moon. That said, the reactions are slow at standard conditions and can be accelerated at the cost of supplying energy. How that balance works in your scenario is ultimately up to you, of course.

Additionally, carbon dioxide is a reactant (but needs to be in a supercritical state, which costs energy), which you'd get from human respiration and other organic sources "for free," which relieves you of having to have a lot of hydrogen gas on-hand.

The idea with mineral sequestration is to allow supercritical carbon dioxide to react with certain minerals to yield carbonates that are stable over some long time period:

  Mineral carbonation reactions are known to geologists and occur spontaneously on geological time scales. For example, the reaction of $\ce{CO2}$ with common mineral silicates to form carbonates like magnesite or calcite is exothermic and thermodynamically favored.

An example is:

$$\ce{Mg2SiO4 + 2CO2 -> 2MgCO3 + SiO2}$$

  (which) illustrates the transformation of forsterite, which is the end member of the common silicate mineral olivine. One ton of olivine can dispose of approximately two-thirds of a ton of $\ce{CO2}$. Again, the reaction is
  exothermic and releases 90 kJ/mole of $\ce{CO2}$.

In summary: There's nothing wrong with your chemistry or geology and there is more than one way to yield silica from mafic/ultramafic minerals.

The reference I have quoted above and cited below also details the process in general terms for minerals such as olivine and serpentine, and illustrates schematics for implementing such schemes on industrial scales, which might also be of interest to you.

Reference from the National Energy Technology Laboratory within the DOE:

  Goldberg, P. ,Chen, Z. Y. ,O’Connor, W. ,Walters, R., and Ziock, H. $\ce{CO2}$ Mineral Sequestration Studies in US. Technology. 1 (1): 1–10 (2000).


Why does iodine get oxidized to iodic acid and not periodic acid by nitric acid? Nitric acid oxidizes iodine to iodic acid, $\ce{HIO3}$ and not periodic acid, $\ce{HIO4}$, which is of a higher oxidation state. This happens although other nonmetals are oxidized to their highest oxidation state. How can this be explained?

Nitric acid is not strongly oxidizing enough to get iodine to +7.  Relativistic effects tend to make the $s$ subshell valence electrons more stable than the others in heavy atoms (http://www.annualreviews.org/doi/abs/10.1146/annurev-physchem-032511-143755), so to some extent iodine acts as if it has only five valence electrons.  To pull out those last two electrons and get periodic acid, electrochemical oxidation is commonly used with a very high oxidation potential (https://en.wikipedia.org/wiki/Periodic_acid).


Why boron trihydride is weaker lewis acid than boron tribromide and boron tri-iodide? There is no back-bonding in case of boron trihydride so it is a stronger lewis acid than boron trichloride ( even though chlorine atom is more electronegative than hydrogen atom).

Then  why boron trihydride is weaker lewis acid than boron tribromide and boron tri-iodide?

Predicting the acidic strengths of $\ce{BX_3}$ using back-bonding is one plausible theory. This article http://pubs.acs.org/doi/abs/10.1021/ic990713m hints us to use bond-strength data to predict the same:

  The weaker Lewis acid strength of $\ce{BF3}$ compared to $\ce{BCl3}$ with respect to bases such as $\ce{NMe3}$ is explained in terms of the ligand close-packing (LCP) model. The halogen ligands remain close-packed during the formation of the complex leading to an increase in the $\pu{B−Hal}$ bond length. Because a $\ce{B−F}$ bond is stronger than a $\ce{B−Cl}$ bond, more energy is required to stretch a $\ce{B−F}$ bond than a $\ce{B−Cl}$ bond. Hence $\ce{BF3}$ is a weaker acid than $\ce{BCl3}$.

Pulling up the bond strength data, I get:

Bond |  Strength (KJ/mole)
--------------------------
B-H  |  389
B-F  |  613
B-Cl |  456
B-Br |  377
B-I  |   ?

Building up on these two pieces of information, it can be said that since $\ce{B-H}$ bond is stronger than $\ce{B-Br}$ and $\ce{B-I}$ (see below), $\ce{BH3}$ is a weaker acid than $\ce{BBr3}$ and $\ce{BI3}$.

I'm afraid I couldn't find the data for $\ce{B-I}$ bond, but since we can compare the bond-strengths of boron trifluorides on the basis of back-bonding theory, we can be sure that the strength of $\ce{B-I}$ bond will be $\lt  \pu{377 kJ mol^-1}$.


What is the substance turning salty water green when electrolysed? I am a student and I take O.L. Chemistry, in my knowledge I know if I electrolysed concentrated aqueous NaCl solution I should get Hydrogen evolved at the cathode and chlorine at the anode.

I did an experiment, I add a lot of table slat to tap water to get the electrolyte ready (aq. NaCl) then I used graphite electrodes (just got it from a mechanical pencil) and passed electricity through the salty water. I used of course direct current using a 6 Volts battery I got.

Well my suggestions were half correct, Hydrogen really evolved and bubbled violently at the cathode, but at anode I didn't see any yellowish green gas indicating presence of chlorine, yet only colorless bubbles continue to evolve from the anode but was less violently than the cathode. I thought this is Oxygen evolving so I increased the concentration of solution by adding table salt but also only colorless bubbles evolves at the anode. Well I didn't really calculated how much precisely I should add salt but I added insanely a lot of table salt.

Only to discover after around 5 minutes of electrolyzing the salty water, it became darkish green in color, I didn't really knew what is this and this is my question, I used graphite electrodes so those doesn't react like if I used iron, the solution was just nice tap water and not really bad or muddy. My only suggestion is that this is actually chlorine dissolved in water ... So is this correct ?? I knew that chlorine is soulable in water but I didn't thought it will turn the whole water greenish in color....

Can anyone help me in that ? And please if you're going to answer me, answer me as if I am somebody who knows nothing in chimestry .....

Thanks in advance.

My speculation is based on the words specifying the use of 'tap water' during the electrolysis of aqueous sodium chloride.
In particular, the interaction of formed chlorine and existing ferrous bicarbonate (found in tap water) could form small amounts of aqueous FeCl2. [EDIT TO QUOTE SOURCE]

Bacterial Iron in Well Water – Iron bacteria is usually identified by slime in places such as toilet reservoirs or by the presence of a slimy mass fouling softeners or filters.
Ferric Iron in Well Water – Also known as red water iron, ferric iron in well water is essentially clear water iron that has been exposed to oxygen – usually from the air, thereby oxidizing.  Carbon dioxide leaves the water and the oxygen combines with the iron to form ferric ions (Fe+++).  This gives the water a red rust coloring.
Ferrous Iron in Well Water – Often called clear-water iron because it is clear when poured, this substance is found in water that contains no oxygen. Typically, it comes from deeper wells and groundwater sources. Carbon dioxide acts on iron in the ground to form soluble ferrous bicarbonate. In water this forms ferrous ions (Fe++)

Per Wikipedia on Ferrous chloride, to quote:

The compound is highly soluble in water, giving pale green solutions.

Repeat the electrolysis experiment with distilled water and see if you get the same coloration issue.
[EDIT][EDIT] Interestingly, low solubility Fe(OH)2 is clear and only acquires a greenish tint in the presence of oxygen (per Wikipedia, to quote: "Iron(II) hydroxide is a white solid, but even traces of oxygen impart a greenish tinge.). Further, well-water is associated with high levels of bicarbonate and low O2 levels (see this significant work). To claim that Fe(HCO3)2 is at all green is without basis, as I also could not find any source anywhere asserting a coloration, other than that it is, in effect, 'clear' as was claimed above per the comment: 'Often called clear-water iron because it is clear when poured'. Theoretical formation of the usual hydration sphere for Fe(OH)2 interacting with bicarbonate in low solubility conditions with high HCO3-, low O2, significant CO2 presence (and its associated partial pressure) suggest reasons why it may not be a valid argument. If I am wrong, please provide a single source citing color, else it should be likely assumed that my supposition is correct and Fe(HCO3)2 is clear.  By the way, Atomistry on Ferrous bicarbonate, to quote:

Ferrous hydrogen carbonate or ferrous bicarbonate, FeH2(CO3)2, exists in solutions obtained by dissolving ferrous oxide or carbonate in water charged with excess of carbon dioxide. On exposure to air carbon dioxide is evolved, and a precipitate is obtained consisting essentially of ferric hydroxide.

which one could interpret as that clear-water iron, upon exiting the pipe, liberates CO2 and forms Fe(OH)2, which, with further oxygen exposure, may display a green tint (or, per the introduction of electrolysis formed chlorine).


Using vapor mole fraction and pressure to determine liquid mole fraction I am studying Chemistry by Zumdahl and Zumdahl (9th edition) and I have come across a question in Chapter 11 that I can't figure out the explanation for:

Liquid A has vapor pressure x , and liquid B has vapor pressure y . What is the mole fraction of the liquid mixture if the vapor above the solution is 30% A by moles?

I know that Raoult's Law tells us that the total vapor pressure is χx + (1-χ)y in this case. I also assume that the total vapor must be 0.3x + 0.7y given the other constraint of the question. This would mean the answer should be:

0.3x + 0.7y = χx + (1-χ)y

However this leads to the answer χ = 0.3, whereas the text gives the answer as 0.3y/(0.7x+0.3y)

What have I got wrong? (Also sorry for not formatting it properly but I'm a noob here and I can't figure out the equation formatting).

Raoult's law tells us the relationship between the composition of the liquid and the vapor pressure above it:

$$P_A=\chi_A P^\circ_A$$

Where:

$P_A$ is the partial pressure of $A$ in the vapor above the solution
$P^\circ_A$ is the vapor pressure of pure $A$
$\chi_A$ is the mole fraction of $A$ in the solution

For a two component solution $A$ and $B$ (I don't want to use $x$ to avoid confusion with $\chi$) we can then use Dalton's Law to get:

$$P_T=P_A + P_B = \chi_A P^\circ_A +\chi_B P^\circ_B= \chi_A P^\circ_A +(1-\chi_A)P^\circ_B$$

However, from the definition of mole fraction we can derive a relation between the mole fraction in the vapor (from here on out $\phi$ to avoid confusion between vapor and solution, which is what your mistake is).

$$\phi_A=\frac{n_A}{n_T}$$

Where $n$ is the number of moles, and since at constant temperature and volume, $n\propto P$...
$$\phi_A = \frac{n_A}{n_T}=\frac{P_A}{P_T}$$

So, we can rewrite $P_A = \phi_A P_T$ and $P_B=(1-\phi_A)P_T$

We are given $\phi_A=0.3$. Thus $\phi_B=0.7$

Here we go. Substitute $P_A=\chi_A P^\circ_A$ and $P_T=\chi_A P^\circ_A +(1-\chi_A)P^\circ_B$ into $P_A=0.3 P_T$:

$$\chi_A P^\circ_A=0.3\left(\chi_A P^\circ_A +(1-\chi_A)P^\circ_B\right)$$
$$0.7\chi_A P^\circ_A=0.3(1-\chi_A)P^\circ_B$$
$$0.7\chi_A P^\circ_A + 0.3\chi_A P^\circ_B=0.3P^\circ_B$$

I think you can get it from here.


Melting point of aspirin, contradicting sources Querying wolfram alpha for the melting point of aspirin (link) returns $\pu{140 °C}$. The Wikipedia page for Aspirin lists $\pu{136 °C}$ instead, citing this book.

How am I supposed to know which figure to cite? And in general, what should I do when I encounter contradicting chemical data?

Wolfram company doesn't conduct any experimental determinations of physical constants for chemical compounds and uses literature data sources.
The webpage for ChemicalData Source Information lists numerous sources of chemical information used by the company's products, including Wolfram Alpha.

Wolfram Alpha Knowledge Database is linked with 87th ed. (2006) of CRC Handbook of Chemistry and Physics (m.p. $\pu{135 °C}$), whereas Wikipedia cites newer 92nd ed. (2011) of CRC Handbook of Chemistry and Physics (m.p. $\pu{136 °C}$).

Obviously, Wolfram's sorting algorithm preferred another data source despite it's quoting the CRC Handbook.
Why and how it did that cannot be answered as the server-side engine is proprietary.

Note that even newer 97th ed. (2017) of CRC Handbook of Chemistry and Physics adds measurement error of $\pu{4 °C}$ (m.p. $\pu{136(4) °C}$).
The difficulty of determining the exact value probably arises from the fact that at that temperature range thermal decomposition already begins, which is reflected in m.p. dependence on heating rate and atmosphere and melt solidifying at lower temperatures (down to $\pu{118 °C}$ [1, table 3.4]).

References

Aspirin and Related Drugs; Rainsford, K. D., Ed.; CRC Press; Taylor & Francis: London, 2004. ISBN 978-0-7484-0885-6.


Understanding the difference between physical and chemical changes Which of the following is a chemical change?
A. Hydrolysis of a dipeptide.
B. Stretching of a balloon.
C. Freezing of water.
D. Breaking a stick.

I'm pretty sure that the answer is (A), but I'm not 100% positive.
I know that, for (C), you can freeze water, and then you can melt it back to liquid water.
For (D), you can break a stick, and the stick is still now two sticks.
For (B), however, I think that it is a physical change, but I'm not entirely certain.
Can someone please explain this for me?

Yes, (B) is a physical change. Stretching the balloon simply rearranges the polymer strands in the balloon material, making them straighten out instead of being kinked up. When you let the balloon go, they just pull back to their former kinked shape.

Your reasoning on (C) and (D) is solid.

The answer is indeed (A). See Wikipedia's page on hydrolysis for more information.


Are the words "apparatus" and "equipment" synonyms? Are the terms "apparatus" and "equipment" synonyms, within a chemistry or physics laboratory? If not, what are distinguishing factors?

Within a lab they are synonymous. But if you consider in general, apparatus refer to non-machines. Equipment on the other hand can be divided into a couple of sections:

Non-mechanical which can be things like kitchen equipment, measuring cups, knives (all hand tools)
Mechanical machines which are things like sewing machines, mechanical diggers.
Powered machines that are powered by either gas, steam or electricity.


How to calculate the concentrations of the species in the carbonate equilibrium from a titration with hydrochloric acid? Lake water contains dissolved sodium carbonate and sodium hydrogen carbonate. The following equilibrium exists:
  \begin{align}
\ce{HCO3- &<=> H+ + CO3^{2-}}&
\frac{\ce{[CO3^{2-}]}}{\ce{[HCO3-]}}  &= 0.958
\end{align}
  When $\pu{10 cm^3}$ of lake water were titrated with  $\pu{0.2 mol/dm^3}$ $\ce{HCl}$, $\pu{22 cm^3}$ of acid were required to neutralise all the carbonate and hydrogen carbonate ions according to the following equations:
  \begin{align}
  \ce{H+ + HCO3- &-> H2O + CO2}\\
  \ce{2H+ + CO3^{2-} &-> H2O + CO2}\\
\end{align}
  Calculate the total amount of substance of acid used, and thus, by using the ratio quoted, calculate $\ce{[CO3^{2-}]}$ and $\ce{[HCO3-]}$ in the lake.

I was able to find the amount of substance of acid $\ce{HCl}$, which is $\pu{4.4E-3 mol}.$ How do I go on?

You have correctly identified the amount of substance $\ce{HCl}$ which were necessary to neutralize all $\ce{HCO3-}$ and $\ce{CO3^2-}$ in the sample. With the given sample volume of $\pu{10 cm^3}$, we can calculate the concentration of both ions in the sample:
$$
\frac{\pu{4.4E-3 mol}}{\pu{10 cm^3}}
= \pu{4.4E-4 mol//cm^3}
= [\ce{HCO3-}]+[\ce{CO3^{2-}}]
$$

Using the given ratio of $0.958$, we can express $[\ce{CO3^{2-}}]$ in terms of $[\ce{HCO3^{-}}]$:
\begin{gather}
[\ce{CO3^2-}]
  = 0.958\times[\ce{HCO3-}]\\
\pu{4.4E-4 mol//cm^3}
  = [\ce{HCO3-}] + [\ce{CO3^2-}]
  = [\ce{HCO3-}] + 0.958\times[\ce{HCO3-}]
  = 1.958[\ce{HCO3-}]\\
[\ce{HCO3-}]
  = \pu{2.25E-4 mol//cm3}
\end{gather}

We then use the calculated hydrogen carbonate concentration and the ratio to find $[\ce{CO3^2-}]$:
$$[\ce{CO3^2-}] = 0.958\times[\ce{HCO3-}] = \pu{2.15E-4 mol//cm3}$$


How does a mercury detector work? I am new to chemistry and saw a "portable mercury vapor monitor." How do these machines work?

Most commonly they work by pulling air through a UV absorption cell and simply measuring the absorption at 254 nm.  Often there is a pre-concentration step involved.  This usually involves a gold-plated region prior to the UV cell.  At room temperature mercury adsorbs to the gold, then after some period of time the gold-plated region is flash heated to release all of the mercury at once.


Are artificial ingredients better than natural ones because they are made in laboratory conditions? If we have something like an artificial vanilla flavouring, is it any worse than a natural one?

Will both the natural and artificial ones have the same chemical formula?

Won't the artificial one be purer since its made in lab conditions?

The question is by no means restricted to vanilla. It refers to any natural product that can be reproduced artificially. Ditto for the answer.

Sure, the artificial product might have the right chemical formula. Also, it can be made as pure as you want. The problem is that the natural product is not pure. It is a bunch of various compounds, and they all contribute to our perception of the flavour. You identify the compound that makes up 90%, investigate its structure, develop the synthetic route, and finally you hold it in your hands, and it feels just about right, but something is amiss. So you look for the next component, which constitutes 9% and hence is a lot harder to isolate, and after a long and tedious procedure you identify it, and repeat everything, and finally you end up with the product that feels totally right. Well, almost so. Is it the remaining 1% that is responsible for the difference? You never know until you try.

So it goes.


Why is oxygen paramagnetic? Paramagnetic molecules are molecules that have single electrons. When I draw the lewis structure of $\ce{O2}$, it appears to be a diamagnetic structure. What makes it paramagnetic?

To understand the paramagnetic nature of $\ce{O2}$, we must first understand how atomic orbitals mix together to form molecular orbitals. In the diatomic molecules of the elements in the second period, a phenomenon known as $\mathrm{s}$-$\mathrm{p}$ mixing  results in an increase in the energy of the $\sigma_\mathrm{2p_z}$ molecular orbital, and a decrease in the energy of the $\sigma_\mathrm{2s}$ orbital. This is also observed in the $\sigma^{*}_\mathrm{2p_z}$ and $\sigma^{*}_\mathrm{2s}$ orbitals.

The degree of $\mathrm{s}$-$\mathrm{p}$ mixing is determined by the energy gap between the $\mathrm{s}$ and $\mathrm{p}$ orbitals. The higher the energy gap, the less the orbitals mix. Because the $\mathrm{s}$-$\mathrm{p}$ gap increases across a period, $\mathrm{s}$-$\mathrm{p}$ mixing in $\ce{O2}$ and $\ce{F2}$ is actually so low that the $\sigma^{*}_\mathrm{2p_z}$ orbital is lower in energy than the $\pi_\mathrm{2p_{x,y}}$ orbital, unlike the preceding elements.

This can be seen here:

[source]

By constructing the molecular orbital diagram for $\ce{O2}$ and filling each  orbital according to Hund's rule, it becomes evident that $\ce{O2}$ is a diradical, with two unpaired electrons of the same spin. This is what gives oxygen its paramagnetism.


Chromium cyanide complexes I have rarely seen any chromium $(\ce{Cr})$ square planar complexes and I have been told that $\ce{[Cr(CN)4]-}$ is tetrahedral.

So, if the statement is actually correct, are there any other known examples of square planar complexes of chromium?

This doesn't appear to be a rare case and there are plenty $\ce{Cr^2+}$ square planar complexes.
I found 15 square planar complexes with well-determined crystal structures $(R_\mathrm{int} < 10\%)$ by applying the following geometrical restrains:

Chromium must have C.N. exactly 4 (T4).
Chromium is allowed to be coordinated by virtually any element of periodic table $\ce{El}$.
Dashed lines signify any bond type.
All four angles between neighbors should lead to nearly ideal square planar geometry: $∠(\ce{El-Cr-El}) = \pu{(90.0 ± 0.5)^\circ}$ (by allowing higher tolerance the number of structures grows in geometrical progression: for $∠(\ce{El-Cr-El}) = \pu{(90.0 ± 5.0)^\circ}$ there were nearly 100 structures).

List of compounds

CCDC ACACCS
[1]
bis(2,4-Pentanedionato)-chromium(II)
CCDC BHPZCR
[2]
bis(Dihydro-bis(1-pyrazolyl)borato) chromium(II)
CCDC CRPORT
[3]
α,β,γ,δ-Tetraphenylporphinato-chromium(II) toluene solvate
CCDC DAQDUV
[4]
hexakis(Tetrahydrofuran)-magnesium tetrakis(perfluorophenyl)-chromium(II) tetrahydrofuran solvate
CCDC DAYNEW
[5]
bis(Pyridinium) diaqua-dibromo-chromium(II) dibromide
CCDC EGINEN
[6]
bis((μ2-Chloro)-hexakis(tetrahydrofuran)-di-lithium) tetrakis(pentachlorophenyl)-chromium(II)
CCDC EQIVOP
[7]
trans-Dichloro-bis(2,6-dimethylpyridine)-chromium(II)
CCDC HARKIU
[8]
trans-bis(tris(t-Butoxy)siloxy)-bis(diethylammonio)-chromium(II)
CCDC KOMRAF
[9]
Tetrapyridine-chromium(II) bis(hexafluorophosphate) dipyridine clathrate
CCDC PITVAP
[10]
bis(Tetra-n-butylammonium) tetrakis(isothiocyanato)-chromium(II)
CCDC PUQPEY
[11]
bis(1-t-butyl-3-(di-t-butylphosphino)imidazol-2-ylidene)-dichloro-chromium
CCDC PUQPIC
[11]
dichloro-bis(1-(di-t-butylphosphino)-3-mesitylimidazol-2-ylidene)-chromium tetrahydrofuran solvate
CCDC PUQPIC
[12]
bis((Phenothiazinyl-N)-tetrahydrofuran)-chromium(II)
CCDC XEBZAF
[13]
trans-bis(4-t-Butylpyridine)-dichloro-chromium(II)
CCDC ZOYCUL
[14]
Dichloro-bis((trimethylphosphine)trimethylsilylamido)-chromium(II)

References

Cotton, F. A.; Rice, C. E.; Rice, G. W. The Crystal and Molecular Structures of Bis(2,4-Pentanedionato)Chromium. Inorganica Chimica Acta 1977, 24, 231–234. https://doi.org/10/bhjhh5.
Dapporto, P.; Mani, F.; Mealli, C. Vanadium(II) and chromium(II) Complexes with Poly(1-Pyrazolyl)Borate Ligands. Crystal and Molecular Structures of Potassium Tris[Dihydrobis(1-Pyrazolyl)Borato]Vanadium(II) Monoalcoholate and Bis[Dihydrobis(1-Pyrazolyl)Borato]chromium(II). Inorg. Chem. 1978, 17 (5), 1323–1329. https://doi.org/10/dzfdh8.
Scheidt, W. R.; Reed, C. A. Stereochemistry of the Toluene Solvate of α,β,γ,δ-Tetraphenylporphinatochromium(II). Inorg. Chem. 1978, 17 (3), 710–714. https://doi.org/10/cvqg73.
Forniés, J.; Martín, A.; Martín, L. F.; Menjón, B.; Zhen, H.; Bell, A.; Rhodes, L. F. The First Structurally Characterized Homoleptic Aryl-Manganese(III) Compound and the Corresponding Isoleptic and Isoelectronic chromium(II) Derivative. Organometallics 2005, 24 (13), 3266–3271. https://doi.org/10/bz3r8w.
Ladd, M. F. C.; Larkworthy, L. F.; Leonard, G. A.; Povey, D. C.; Tandon, S. S. Crystal and Molecular Structure of a Complex Bromide Containing Planar Diaquadibromochromium(II) Units. J. Chem. Soc., Dalton Trans. 1984, No. 10, 2351. https://doi.org/10/cj5z95.
Alonso, P. J.; Forniés, J.; García‐Monforte, M. A.; Martín, A.; Menjón, B.; Rillo, C. A New Series of Homoleptic, Paramagnetic Organochromium Derivatives: Synthesis, Characterization, and Study of Their Magnetic Properties. Chemistry – A European Journal 2002, 8 (17), 4056–4065. https://doi.org/10/cx9r27.
Gibson, V. C.; Newton, C.; Redshaw, C.; Solan, G. A.; White, A. J. P.; Williams, D. J. Low valent chromium complexes bearing N,O-chelating pyridyl-enolate ligands $\ce{[OC(Bu^t)(=2-CHN5H3Me-x)]^-}$ $(x = 3–6)$. Dalton Trans. 2003, No. 24, 4612–4617. https://doi.org/10/ffdkvr.
Terry, K. W.; Gantzel, P. K.; Tilley, T. D. chromium(II) and Chromium(III) Tri-Tert-Butoxysiloxy Complexes. Inorg. Chem. 1993, 32 (23), 5402–5404. https://doi.org/10/dfw68w.
Fochi, Giovanni.; Straehle, Joachim.; Gingl, Franz. Disproportionation of the Bis(Benzene)Chromium Cation Induced by Pyridine. Crystal and Molecular Structure of Two Tetrapyridine Complexes Containing chromium(II) and Chromium(III). Inorg. Chem. 1991, 30 (24), 4669–4671. https://doi.org/10/c3ffnj.
Larkworthy, L. F.; Leonard, G. A.; Povey, D. C.; Tandon, S. S.; Tucker, B. J.; Smith, G. W. Crystal Structures and Magnetic Behaviour of the Two Forms of Tetra-n-Butylammonium Tetraisothiocyanatochromate(II). J. Chem. Soc., Dalton Trans. 1994, No. 9, 1425. https://doi.org/10/cgtgbm.
Ai, P.; Danopoulos, A. A.; Braunstein, P. N-Phosphanyl- and N,N′-Diphosphanyl-Substituted N-Heterocyclic Carbene Chromium Complexes: Synthesis, Structures, and Catalytic Ethylene Oligomerization. Organometallics 2015, 34 (16), 4109–4116. https://doi.org/10/ggc5mx.
Edema, J. J. H.; Gambarotta, S.; Meetsma, A.; Spek, A. L.; Smeets, W. J. J.; Chiang, M. Y. chromium(II) Amides: Synthesis and Structures. J. Chem. Soc., Dalton Trans. 1993, No. 5, 789. https://doi.org/10/fcfcdg.
Cotton, F. A.; Dikarev, E. V.; Gu, J.; Herrero, S.; Modec, B. Alkylpyridine Complexes of Tungsten(II) and chromium(II). First Rotational Isomers of $\ce{W2X4L4}$ Molecules with $D_\mathrm{2h}$ and $D_2$ Symmetries. Inorg. Chem. 2000, 39 (23), 5407–5411. https://doi.org/10/c5cfr4.
Miekisch, T.; Mai, H. J.; zu Köcker, R. M.; Dehnicke, K.; Magull, J.; Goesmann, H. Silylierte Phosphanimin‐Komplexe von Chrom(II), Palladium(II) und Kupfer(II). Die Kristallstrukturen von $\ce{[CrCl2(Me3SiNPMe3)2]},$ $\ce{[PdCl2(Me3SiNPEt3)2]}$ und $\ce{[CuCl2(Me3SiNPMe3)]2}.$ Z. Anorg. Allg. Chem. 1996, 622 (3), 583–588. https://doi.org/10/bm3fq8.


Where did the rest of the elements come from? There were ninety original elements on the Periodic Table. Over time, new ones were founded and put on. However, they were all either the original elements experimented on. For example, Plutonium was discovered by deuteron bombardment of uranium. Therefore, I would like to know: How were the rest of the elements discovered? It would also be nice if an example could be given. Thanks!

By the "original periodic table" I assume you're talking about the "elements ocurring in measurable quantities in nature".--everything till Uranium sans Technetium and Promethium.

The method of preparation of these elements is pretty much the same -- bombardment. Just the projectile and substrate differ.For example, the (accidental) neutron bombardment of Molybdenum in old cyclotron parts led to the discovery of Technetium.

There are some other sources as well. Analyzing fission products can get you new elements--that's how Promethium was discovered. The process of fission creates smaller nuclei--but there is a shower of neutrons created as well (which can bombard everything in sight). Most of the discoveries have been due to the "shower of neutrinos", because generally all "smaller nuclei" are of already-known elements. Promethium was synthesized by the "formation of smaller nuclei", though.

Since fission occurs in nature (Usually it's just a little bit, though we sometimes have natural fission reactors), we actually have all the elements up till californium existing in minute quantities in nature (source).


Is it safe to look at a mercury gas discharge tube? As part of a chemistry course, I recently observed a demonstration in which the instructor electrified a mercury gas discharge tube (the tube looked something like this image). The glow from the tube appeared to be a light blue, but when I performed an internet search afterwards, I found that mercury gas discharge tubes apparently emit ultraviolet light, which can be harmful to the eyes and body. I was standing within one foot of the discharge tube (which looked something like this while lit) for several minutes. The tube was completely clear, which I am assuming means that it did not contain a phosphor to block the UV light (please correct me if I am wrong or if I am misunderstanding something).
The same demonstration was performed with a hydrogen gas discharge tube, which also was completely clear (see above).
Since I was looking at the tubes, could the higher frequencies of electromagnetic radiation (e.g. the ultraviolet light from the mercury) have damaged my eyes, skin, etc.?
After the demonstrations, my eyes seemed irritated, but I'm not sure if this was just a result of the placebo effect.

Could it have caused irritation or injury? Possibly, but not likely, depending on a few factors:

The intensity of the emitted ultraviolet.
Proximity to the light source.
Length of exposure.

You'd need to calculate the dosage integrated over time.
That said, it is very unlikely any harm was done, because of the following:

The power supplies are limited to to only a few mA current. The total power is the arc voltage drop, ~30 to 100 VAC, times the current. For the supply abvove, that would be 100 VAC * 0.01 A = 1 W. One watt is low power for a mercury lamp.
Demonstration spectral lamps, such as this capillary mercury vapor lamp, usually have thick borosilicate glass capillaries. Borosilicate glass does not transmit shorter wavelengths well. Most of the UV would have been absorbed by the thick capillary wall.
The length of time was given as "several minutes", so the integrated intensity over time (i.e., total energy absorbed) was low.
That you state you were, "within one foot of the discharge tube," ~30 cm, is of concern, because some older discharge tube power supplies had exposed high voltage terminals. Though limited to 10 mA, considered below a dangerous current for electric shock, in most cases, one could still get burnt by the ~10,000 VAC arc, should one come in contact with exposed terminals, or could be startled and hurt by a muscular reflex action. Likely, though, the ends of the tube and associated wiring were safely enclosed.

So, all things considered, there might have been fumes in the lab or other cause of irritation, but I believe the UV light was not likely an issue.


Standard list of odors? Is there such a thing as a list of "standard" odors?  In everyday context, there are various descriptions to odors such as "fruity" and "plastic-like".  But are there more accurate and normalized names of odors?

The problem with odors or taste is that it is a form of Qualia. Whether cyanide smells like bitter almonds or is odorless is subjective. Similarly, whether Phenylthiocarbamide taste extremely bitter or completely tasteless is also subjective (depending on genetics).

Even in everyday life, coffee/wine/spices taste different to different people. Even for fruits, the Durian fruit can taste and smell fragrant to certain people but utterly disgusting to others. This is probably why the perfume industry is making billions - if everyone agree that a certain smell is "nice", everyone would be using the same perfume. (Do Genes determine our perfume preferences?)

So I believe it is difficult to have a standard list of odors. Even in the event that the list is created, it will not be universally agreed upon. There are bound to be disputes. Colours on the other hand, (which is also a form of Qualia), are less subjective (everyone can agree that chlorine is green and bromine is brown).


ChemSpider versus SciFinder for finding chemicals without CAS number Just trying out a couple of difficult options for identifying chemicals without CAS number information, and wondering whether Chemspider or SciFinder is a better tool for doing these searches. Do they search the same databases/repositories? Also, which has better features and is easier to use?

SciFinder is a product of Chemical Abstracts Service (CAS), which is a division of the American Chemical Society. Its registry contains information on more than 111 million organic and inorganic chemical substances, and is updated with 15,000 new substances daily. Additionally, it has more than 66 million protein and nucleic acid sequences.[1] The chemical information is produced by CAS, and is prepared by the CAS Registry System, which identifies each compound with a specific CAS registry number, index name, and graphic representation of its chemical structure.[2]

ChemSpider is a free chemical database from the Royal Society of Chemistry. Its registry contains information on 49 million chemical structures, coming from 513 sources. You can search the database by systematic names, synonyms, trade names, database identifiers, and also using structure-based queries, with the option to draw structures in the web page or use structure files from your computer.[3]

Both registries allow you to find literature references, experimental and predicted property data (boiling and melting points, etc.), commercial availability, and spectra,[3] but the CAS registry also offers preparative methods and regulatory information from international sources.[1]

SciFinder may offer more, but it's extremely expensive. In fact, I'm not even sure if CAS offers SciFinder to individuals. You can see their pricing options here. Taking into consideration everything ChemSpider can do and the fact that it's free, it's an excellent resource, and I definitely recommend it.

References

CAS, Chemical Substances
Wikipedia, CAS Registry
ChemSpider


is the adsorption of resveratrol within wine or grapes linked with the time it was kept in your mouth? This question is about resveratrol in wine or grapes and not in some special delivery system.

If one is drinking some wine or eating grapes, does it matter if they keep it in their mouth a bit so the resveratrol within the wine or grapes get absorbed better?

I know now all nutrients can be gotten into onces body via buccal method so I am asking the experts what is the case with resveratrol from wine or grapes.

This article on the bioavailability or resveratrol seems to partly support your assumption: Keeping it in the mouth for a bit longer rather than swallowing directly might help to facilitate the uptake.

However, the solubility of resveratrol in water is apparently as low as 30 mg/L.


What happens with the boiling point of a liquid in a vacuum This may be a silly question, but when I was reading my chemistry textbook, there was this following statement:

  "A liquid boils when its vapor pressure equals the external pressure acting on the liquid
  surface. At this point, bubbles of vapor form within the liquid. The temperature at
  which a given liquid boils increases with increasing external pressure."

But what would happen if that liquid was sealed in a vacuum, or at least submitted to very low pressure? Would the boiling point be a lot lower?

"Sealed in a vacuum" is an oxymoron, a contradiction in terms, much like "frozen with fire". If you seal a liquid in a flask containing nothing else but vacuum, then a part of the liquid will quickly evaporate and fill the flask with vapor, so it would no longer be a vacuum.

The said vapor will exert some pressure, depending on the temperature and the nature of the liquid. For example, with water at room temperature, it will be a lot lower than the normal atmospheric pressure. Indeed, the boiling point of water at that pressure will be a lot lower than the usual $100\;^\circ\rm C$. Guess what? It will be the same as ambient temperature. That's how equilibrium works.

Now what if we just keep the liquid at vacuum by pumping out the vapor as soon as it forms? Well, the simplified† answer is: the liquid will evaporate gradually, bit by bit, until nothing is left. There will be no more liquid, and consequently no more boiling point. That's what happens with the boiling point in a vacuum.

† In fact, it is more complicated. As the liquid evaporates, the rest of it gets progressively colder. At some point, it will get cold enough to freeze, and then you'll be dealing with ice. The molecules will still be leaving its surface, but slower and slower, and the complete sublimation might well take forever, or even longer. That's why the space rocks are still there after all these years.


Can we separate positive ions from negative ions in a solution? Can we make a solution containing only positive ions by using electrodialysis but instead of using 2 kinds of membrane (one blocks the positive ions and one blocks the negative ions) we can use only one kind (the one which blocks the positive ions) so the positive ions wont reach the electrode and wont be electrolysed while the negative ions will be electrolysed leaving the positive ions in the solution

While it is true that solutions are not always locally electroneutral, that non-neutrality is small, and does not extend to larger scale. Thus, it does not impact chemical concentrations: you cannot separate positive and negative charges in a solution at a macroscopic level. See for example this excerpt from Ionic Transport Processes:in Electrochemistry and Membrane Science:


How to calculate valence factor (n-factor) for a element? Suppose I am given some reaction in which
$$\ce{C6H12O6 -> CO2}$$
and I want to calculate n-factor for this reaction to ultimately calculate equivalent weight of carbon for this reaction. Since $$E=\frac{M}{n}$$
I know that n-factor is no of electron gained/lost by one atom of a compound.

So Initial Oxidation state of carbon is $$6x + 12-12=0$$ $$x=0$$ meaning
and final oxidation state of carbon is $$-4$$ Hence, n-factor of this reaction should be $4$ but it's given to be $28$. How is this possible? Can someone explain how to correctly calculate n-factor with few more examples?

Hint :

n-factor of a molecule/compound is defined as the change in oxidation state per molecule.
You have correctly calculated the change of one carbon atom as 4.
But how many carbon atoms are there in the glucose molecule?

Note:

The average oxidation state of carbon in glucose is zero while in reality the different carbons have different OS. (Reference)
n-factor of a reaction is not defined but it is defined for a single species participating in that reaction.
n-factor of glucose is not equal to the n-factor of the product formed in that reaction, that is, $\ce{CO_2}$.


Which chiral carbon do L and D in L- or D-glucose refer to? I see and D- and L- glucose are mirror images of each other. and I understand D and L are for chirality. But it is not clear which carbon do D and L refer to. Could anybody let me know? Thanks.
What are the differences between D and L glucopyranose?

I think your confusion arises from trying to locate a particular carbon atom, which is D or L. This is not the case.
Points to remember:
The reference for D and L notation is glyceraldehye molecule not a particular carbon atom.
D and L are not equivalent to R and S which are individually applicable to each chiral center in a molecule.
Similarly, D and L cannot tell you the direction of rotation of light before hand.
Capital “D” and “L” notations have no connection with small d and small l, which tell you the direction of rotation of light.


What could be causing my bath salts containing baking soda to explode out of their containers? I run a small herbal wellness business and one of the products i sell are bath salts containing baking soda. The base ingredients are Epsom salts, Dead Sea salts, magnesium chloride, baking soda, coconut milk powder, and fractionated coconut oil. I’ve made and sold this recipe for over a year and because i buy my ingredients in bulk i haven’t bought new ingredients recently. All of a sudden, some kind of chemical reaction is happening (which i assume is coming from the baking soda), and after i seal the bags they blow up like a balloon and then eventually explode. I cannot for the life of me figure out what is causing the baking soda to react. I’ve noticed when I’m mixing up batches that there is a fizzing sound that has never been there before, and I’ve wondered if the salts have somehow absorbed water. But water doesn’t really cause baking soda to react like this. Thanks for the help!

I suspect the seasonal introduction of humidity is the root of the problem, which is readily taken up by the magnesium chloride. Then, the following possible reaction:
$\ce{ MgCl2 (aq)  + 2 NaHCO3 (s) -> 2 NaCl (aq)  + Mg(HCO3)2 (aq)}$
Now, aqueous Magnesium bicarbonate exists only in solution (so decomposes on drying), is not stable on standing (per my experience), or warming (whereupon it also decomposes):
$\ce{ Mg(HCO3)2 (aq) ->  MgO (s) + H2O (l) + 2 CO2 (g)}$
where the pressurized buildup of carbon dioxide gas is likely causing a problem.
Solution: Try adding a drying agent or removing the very hygroscopic Magnesium chloride from your formulation.


what is a pseudohalogen/pseudohalide? What do you mean by a pseudohalogen/pseudohalide and how do you tell whether a molecule/ion is a pseudohalogen/pseudohalide ?

Pseudohalides are "fake halides."   Their chemistry bears some resemblance to true halides ($\ce{F-}$, $\ce{Cl-}$, $\ce{Br-}$, $\ce{I-}$).

Examples of pseudohalides include $\ce{CN-}$ and $\ce{N3-}$. These are generally weak Lewis bases that bear a formal negative 1 charge.

In my opinion, it is rather subjective as to what can be labeled as a pseudohalide. In general, it is in the eye of the beholder, and for a certain set of chemical circumstances.

In short, there is no "dividing line" for being a pseudohalide. If we assume that halides undergo certain (characteristic) types of chemistry, and the species X in question also generally does those same types of chemistry with a similar mechanism, then you may be able to label X a "pseudohalide."

(Note that within the true halides themselves, they can have very different characteristics, and so there is a bit of a moving target for what constitutes a pseudohalide.)


Homemade "milk of magnesia" (suspension of magnesium hydroxide in water) I want to make homemade "milk of magnesia" as a suspension containing only magnesium hydroxide and water.
I see that I can purchase a bag of magnesium hydroxide powder. Can I use this to make a suspension or is there a multi-step process?

Milk of magnesia is typically just a suspension of about $\pu{8\%}$ wt/v (i.e. $\pu{8g / 100 mL}$) magnesium hydroxide in water. To get a good suspension you will want to have very fine magnesium hydroxide. Ideally, it should feel like flour when rubbed between you fingers. Regardless, there will be some settling and you will need to shake it thoroughly prior to each usage. No other preparation should be necessary.


Have we found stable, super-heavy elements? Have we found a stable super heavy element or created one in a lab, or are they still hypothetical?

There already is an element, heavier  (higher atomic and mass numbers) than all naturally occurring ones, that is stable enough for everyday use.  You are probably using it, well, everyday, in your smoke detector.

It's americium, element 95, with three isotopes having 100+ year half-lives.  Americium-241 has a half-life of over 400 years, for americium-243 it's over 7000 years and there is even an odd-odd isomer, with mass number 242, having a half-life over 100 years.  (https://en.wikipedia.org/wiki/Americium).

This is not perfect (and whether it's called "superheavy" depends on definition), but the americium is out-lasting other components of smoke detectors, or industrial gauges where americium is also used.  In that sense we can indeed call the most long-lasting americium isotopes stable.


Is there a formula to tell how many conformers of a molecule to generate? Let's say we want to generate conformers of molecules (e.g., ligands for docking in a protein pocket).

Is there a formula, like a function of the number of rotatable bonds, to decide how many conformers (maximum) to generate for a given molecule ?

I haven't seen any rigorous benchmarking or guidelines.

Consider that the total number of conformers goes up like $3^n$, where $n$ is the number of rotatable bonds, and ~3 is the approximate number of symmetrically unique conformers for each rotatable bond.

For small molecules with <3-4 rotatable bonds, using 100 conformers seems to be enough (i.e, only a small deviation from the crystal structure geometry). There's a more complete discussion in Ebejer's paper "Freely available conformer generation methods: how good are they?": J Chem Inf Model. 2012 52(5) pp. 1146.

(Slide from Jean-Paul Ebejer)

Beyond 3-4 bonds, it's less clear if sampling more conformers is a clear win. I would guess the answer is "yes," but I have not seen any papers (yet) dealing with such flexible molecules. One may need better tools, and also better force fields or computational methods.


Are all alkene geometrical isomers achiral? I was reading a book and it tells that alkene geometrical isomers are achiral.
I wanted to know that is it applicable for alkene geometrical isomers?
Thanks in advance.

While in general, you should consider chirality and alkene geometry as orthogonal concepts, the statement in your question isn't even true.

Consider the case of 2,3-pentadiene, aka, 1,3-dimethylallene. This compound is chiral. Any change in the geometry of either of the double bonds provides the other enantiomer.


Is the following paragraph about steam distillation accurate? I've been struggling with the various methods of purification of organic compounds. If a question comes up in my examination, asking me to explain steam distillation in my own words, I'd like to know if this applies and if this is accurate.

i) Steam distillation is utilized to separate volatile constituents which are immiscible with water but steam volatile.

ii) Steam from a steam generator is passed through a heated flask which contains the liquid to be distilled.

iii) The volatile component separates and thusly, one obtains a mixture of steam and the volatile organic compound.

iv) This is then condensed and separated by means of a separatory funnel.

In steam distillation, the liquid boils when the sum of vapour pressured due to organic liquid and that due to water become equal to the atmospheric pressure.
i.e Patm=Pw+Pc

Since $Pc<P$ ; liquid vaporises at temperature lower than its boiling point. That is, if water insoluble substance and water are the components, the mixture boils at a temperature close to, but below 373K.
Example: Aniline-Water

Please be highly specific. There should be no error, however minute.
Many thanks.

You have it right on all counts. But, to clarify...

i) Steam distillation is used to separate steam-volatile organic components of a mixture from non-volatile ones (not from each other, however). It is most useful for higher boiling, relatively non-polar organic compounds that are immiscible with water.


Why do NMR samples require a solvent? I am running a proton NMR on a liquid organic product. I understand that we wouldn't want to use anything that contains hydrogen atoms as a solvent, but I'm not sure why we need a solvent at all.

There is a minimal volume you need to fill into the NMR tube to be able to get a good shim, this is around 500 microliters with regular 5mm tubes. Not using a solvent would require a lot more of your actual sample just to fill the tube. In many cases you might not even have enough of your product to fill an NMR tube.

The solvent does have some effect on the spectra, it could be harder to compare spectra of pure product to the more typical spectra measured in a common solvent. You're measuring in a different environment, product in product is different than product in solvent.

Modern NMR spectrometers use the deuterium signal from the solvent for the lock, which corrects a drifting magnetic field. You could measure without the lock if your experiment is short or your signals are very broad anyway. But this is not something that every NMR facility has just set up for all users, you might have to manually set this up at the spectrometer.

I suspect that the automatic shimming methods typically used now would not be able to deal with a sample without solvent. But I never tried that.


Is there an easy way to find number of valence electrons" I want an high-school level answer. What I mean with "valence electrons" is the outermost electrons in that atoms' electronic arrangement? For example, 3 in an atom with an electronic arrangement of 2, 8, 3.)

Without actually writing the electron configuration, or orbitals, and knowing electron number, is there any way to find this number easily?

Also, is there an easy way to find if certain atom fits "atomic/orbital symmetry" given the numbers of electrons? (I don't know the english words for this and couldn't find them. So I translated it literally, but what I mean is, if orbital is half-full or full i.e  $\mathrm{s^1}$ or $\mathrm{2p^3}$ or $\mathrm{2p^6}$ and so on... hope you understand what I am trying to say)

I am asking this because sometimes questions give me electron numbers in a range of 40+ and just writing it out doesn't seem like a good way.

You can get the valence electrons in an atom's electronic arrangement by consulting the periodic table:

The Group 1 atoms have 1 valence electron.
The Group 2 atoms have 2 valence electrons.
The Group 3 atoms have 3 valence electrons.
The Group 4 atoms have 4 valence electrons.
The Group 5 atoms have 5 valence electrons.
The Group 6 atoms have 6 valence electrons.
The Group 7 atoms have 7 valence electrons.

The atom with electric configuration 2, 8, 3 is Aluminium and you can find it in Group 3 of the periodic table.

The periodic table is usually given in exams so...

The first two groups of the periodic table are the groups concerned with the filling of the $s$ orbital. So, all Group 1 atoms will have an outermost $s$ orbital filled with a single electron.

Similarly, Group 3 - Group 8 (Group 8 is also referred to as Group 0) are where the $p$ orbital is being filled up. All the Group 5 atoms thus all have an outermost $p$ orbital filled with 3 electrons.

The Periods in the periodic table can be used as an indication of the number of shells the atom has. Sulfur for instance is in Period 3 and thus has 3 shells.


clarification on background limits in electrochemistry I’m teaching myself electrochemistry, and this question is from the general introduction of the book I’m using. According to Bard’s Electrochemical Methods book, in a given cell, the background limits ‘are the potentials where the cathodic and anodic currents start to flow at a working electrode when it is immersed in a solution containing only an electrolyte added to decrease the solution resistance (a supporting electrolyte).’ I don’t fully understand the significance of the italicized, but my main question comes from the following figure, which shows that the onset of H+ reduction and Br- oxidation (Where the background limits are) occur at different potentials. This is counter-intuitive for me, since, according to introductory electrochemistry from gen chem courses, one half of a reaction cannot occur without the other for a given applied potential.

How can this occur?

the background limits ‘are the potentials where the cathodic and anodic currents start to flow at a working electrode when it is immersed in a solution containing only an electrolyte added to decrease the solution resistance (a supporting electrolyte).

Bard's book is notoriously complex. Some bright scientists are not good at communication. Richard Feynman was an exception.

All Bard is saying that the background limits define the potential window [see flat line in your figure] where electrolysis does not occur on the working electrode. The background limits are dependent on the electrolyte used, it is also dependent on the solvent and the electrode material as well.

Now keep in mind that modern potential sweep experiments there are three electrodes in a cell. There is a reference electrode, in your case, an Ag/AgBr electrode. Then you have a working electrode, Pt/H+ and then the writer is quiet about the third one, which is called the counter electrode. Most of the time it is understood that it is their. It could be another Pt electrode but not necessarily a hydrogen electrode i.e., nobody is bubbling hydrogen gas over the counter electrode.

Now the reference electrode has no role in the cell. There is no redox reaction taking place at the reference electrode because the circuitry is such that no or very little current passes through the reference electrode. Electrochemists don't care what is happening at the counter electrode. The potential difference is measured between the reference electrode and the working electrode. This is the only purpose of the reference electrode.

When you are sweeping (=continuously changing) the potential of the working electrode with respect to the reference electrode, it can become a cathode or an anode during scanning of the potential. You can scan in any direction. You are right, isolated reduction or oxidation cannot occur. If the working electrode was a cathode at a particular moment, the counter electrode is the anode and vice versa.


Safe way to extract hydrogen from water through electrolysis I've done some research on google but most guides I've seen recommend adding salt to the mixture for better conductance which results in the release of chlorine.

Is there a way to perform electrolysis on water that releases hydrogen and at the same time does not release any gases that are harmful to humans?
By that I mean can I use another electrolyte that does not produce harmful side effects (so sulfuric acid is also out, and I do realize it's next to impossible to do electrolysis on distilled water.)

I'm looking for a way to show this experiment to kids in enclosed space without good ventilation.

Some sort of advice, since you are concerned with safety and demonstrating the experiment to children. You should generally not perform an experiment in a badly ventilated space. Please be aware, that the generated hydrogen gas is highly flammable and with the addition of the right amount of oxygen, it can make very loud booms: How explosive is hydrogen gas?

Here are two questions, that are very closely related and already contain the answer to your question:

Can I perform electrolysis without electrolyte? (with distilled water)
How can I produce hydrogen gas efficiently and cheaply?

In general any kind of salt as an electrolyte will do.

In low concentrations about 1-2 mol/L they are usually quite harmless. I personally would prefer diluted sulfuric acid. Any addition of group 2 salts may cause passivation of the electrodes by trapping carbon dioxide from the air.

Diluted sodium hydroxide or potassium hydroxide are also safe to handle.

Probably the safest is sodium sulfate, but it might not be the best performer.

Please wear safety goggles at all times during the experiment, best case: The children wear them, too.


Is Van der Waal's force the real source of friction? I'm currently taking a physics class, and my teacher explained that friction is the result of intermolecular interactions, and that energy must be applied in order to break these bonds.

This got be thinking about what kinds of things have a high coefficient of friction, and immediately my mind went to sandpaper. The atomic surface of these mineral crystals would be really smooth and not easily polarizable, as they are held tightly in place by the other ions. So does surface area have more to do with friction than actual Van der Waals forces? This would also seem to explain how geckos are so adept and clinging to things like glass, as the tips of their fingers are so highly ridged.

While van-der Wall forces contribute to the frictional force, they are not the only reasons. Other secondary inter-molecular forces like electrostatic attractions, polar interactions, hydrogen bonding and more also play a significant part. Apart from this features on the surface of materials, present as abrasions are also responsible for locking together and resisting change.
The case of a gecko sticking to wall is not just about these forces and interlocking, but more about how the wrinkles on feet of a gecko act as several small vaccum cups which provide it the stability of sticking upside down as well(think how friction plays almost no role here)


Retro Diels–Alder reaction in synthesis of corallopyronin A I was looking at the synthesis of corallopyronin A in the following paper by Rentsch and Kalesse: Angew. Chem. Int. Ed. 2012, 51 (45), 11381–11384.
There is one reaction which is described as

oxidation and a retro Diels–Alder reaction generated pyrone

I understand that the DMP oxidises the secondary alcohol to a ketone, and at the end there is an intramolecular esterification to produce the lactone, but I don't quite understand how the retro Diels–Alder reaction works. Could someone explain the reaction steps?

Here is my assessment of the mechanism. You are correct about the Dess-Martin periodinane (DMP) oxidation of 1 to 2. The resulting ketone 2 can also exist as the enols (E)-3 and (Z)-4. Any one of the three can undergo a retro-Diels-Alder-like thermolysis, but I have chosen (Z)-4 for the convenience of cyclizing ketene 5, which has lost acetone in the thermolysis. Finally, the ketene 5 affords pyranone 6.
ADDENDUM: Citation for the "retro-Diels-Alder". A. R. Katritzky, Z. Wang, M. Wang, C. D. Hall and K. Suzuki, J. Org. Chem. 2005, 70, 4854. https://doi.org/10.1021/jo050307m


Effect of an inert gas on steam vapor pressure at equilibrium Suppose we have a closed container with mobile walls, so it will change its volume in order to equal the external pressure(constant) to the internal pressure.
This container has inside it a liquid phase only of component 1 and a vapor phase of 1 and the inert gas(2) that is insoluble in the liquid phase.
If the external pressure is constant how will change the vapor pressure of the component 1 changing the temperature?  (suppose we know the molar partial enthalpy of the the components in gas phase and of 1 in liquid phase at the required temperature)
$$\left( \frac{\partial P_{vap}}{\partial T} \right)$$

I have tried to find this relation by equaling the chemical potentials, and proceeding in the same manner of the demonstration of the Clausius-Clapeyron equation. However I am not sure of the result. Can anybody help me in the search of this partial derivative?

If the gas phase can be treated as an ideal gas mixture and the Poynting correction for the liquid can be neglected, then the partial pressure of component 1 in the gas phase will be the equilibrium vapor pressure of component 1 at the temperature of the system.  The partial pressure of the non-condensible component 2 will be determined by the ideal gas law, based on the number of moles of this component, the temperature of the system, and the volume of head space.

If the gas phase cannot be treated as an ideal gas mixture but can still be regarded as an ideal solution, then the Lewis Randall law can be used to determine the fugacity of component 1 in the vapor, and this can be matched to that of component 1 for the liquid, with the Poynting correction included.

Does this come anything close to answering what you are asking?


Is sign convention for delta H and enthapy change of a solution different? In the Ncert Chemistry Tb, it is given that

  $\Delta H$ is negative for exothermic reactions which evolve heat during the reaction and $\Delta H$ is positive for endothermic reactions which absorb heat from the surroundings.

In enthalpy change video lesson which I got access to online, it is given that
for $\ce{NH4NO3(s) -> NH4NO3(aq)}$ since the temperature decreases during the reaction, it is positive enthalpy of solution.

How is it positive? I guess both $\Delta H$ and enthalpy change of a solution is almost change. Since the temp decreases here, it is endothermic, so it should be positive right?

Are both of the above processes different from each other and are the sign conventions vice versa for each other?

The reaction is endothermic, meaning that it takes energy from the surrounding. The "surrounding" in your example is the solvent, so in the dissolution process, energy is taken from the solvent,  thereby decreasing its temperature.


Free radical bromonation of 1-ethyl-4-nitrobenzene In the above reaction , I am confused between . Among them which is major product and why .

This is would be a free radical bromination reaction. Bromine is very selective. It is going to replace the hydrogen from the highest substituted carbon. So according to me second one should be the answer i.e. Br attached to secondary carbon.

Hope it helps


How long does it take for nitric oxide to turn into dinitrogen and dioxygen? I'm new to chemistry and am studying nitric oxide as a combustion accelerant. Nitric oxide is said to be a free radical.

People say that nitric oxide, by itself, is pretty stable. But I saw that $$\ce{2NO -> N2 + O2}$$ so in a room of nitric oxide, will it decompose to nitrogen and oxygen?

Theoretically, it will decompose, but the reaction is rather slow, on a human timescale. One experiment ran for almost forty years (J. Phys. Chem. 1958, 62 (3), 360–361), from 1918 to 1957, without noticeable decomposition.

That said, in aqueous solutions it breaks down more rapidly. In vivo, the half-life is a few seconds.

It is amazing to me that a free radical is so stable!


Finding the major contributing tautomer of 1-(propan-2-yl)-1,2,3,5-tetrahydro-6H-pyrazolo[3,4-b]pyridin-6-one I am having a problem generating the correct tautomer for 1-(propan-2-yl)-1,2,3,5-tetrahydro-6H-pyrazolo[3,4-b]pyridin-6-one. I am told by my collaborator that the one I am generating is wrong.
I am not able to figure out why the structure below is the wrong tautomer.

In keto-enol tautomerism, the equilibrium is mostly towards the keto form, as it should be, since $\ce{C=O}$ is a very strong and stable bond. In some cases however we find that the equilibrium sometimes shifts to enol form. This happens in cases where some additional benefit is done by the enol structure. here are two examples:
A) Removing anti-aromaticity or adding aromaticity to a ring.
B) Extended conjugation formation, providing additional stability.
In the structure you have shown, the enol form is more stable because it adds aromatic character to the ring. This additional stability drives the equilibrium
towards the enol form.


What is difference between H⁺ and proton? What is difference between $\ce{H+}$ and a proton?

There is no chemical difference, only a psychological one: how do you think about it. They are both the same thing, but many people associate $\ce{H+}$ ions with chemical reactions and protons with particle physics.
A hydrogen atom has one electron and a proton, no neutron. Therefore $\ce{H+}$ is just a proton.

That is why acids are sometimes referred as proton donors as they donate $\ce{H+}$ to a base (also known as proton acceptor).


Why are chlorides, sulfates, and bicarbonates of Ca and Mg solely responsible for the hardness of water? When soap is dissolved in water - irrespective of the anion - $\ce{Ca^2+}$ and $\ce{Mg^2+}$ form precipitates. Why are just chlorides, sulfates, and bicarbonates the only specie considered as "hardness" in water?

If I remember correctly, hardness is defined by the Mg and Ca content of water. Temporary hardness is the mineral content which can be removed by boiling and permanent is the total Mg and Ca content - this is what counts when the precipitate with the fatty acid is being formed.

Now typical potable water will only contain sulphide, bicarbonate, halide and perhaps nitrate anions. Other anions either don't occur naturally or they are harmful so they are not considered in the context of drinking water.


Stability of ferrous ions and iron(II) hydroxide in acidic conditions I was reading a chemisty textbook when I came across the following statement.

  $\ce{Fe(OH)2}$ is not stable, even more so in basic conditions. [---] $\ce{Fe^2+}$ ions are also less likely to oxidise in the low pH range.

First attempts

No explanation was given. At first, I was hoping for a simple case of Le Châtelier – Braun principle. However, the equilibrium

$$\ce{4[Fe(OH2)6]^2+ (aq) + 4H3O^+ (aq) + O2 (g) <=> 4[Fe(OH2)6]^3+ (aq) + 6H2O (l)}$$

would suggest the opposite. Next, I considered a simple approach to complex stability. As a general guideline, metals in lower oxidation states favour cationic complexes. The opposite is also true. Therefore, one might conclude then that if anionic complexes are not preferred, $\ce{Fe^2+}$ ions ought to be stable.

One example would be that the hydrolysis is more hampered in acidic media.
$$\ce{[Fe(OH2)6]^2+ (aq) + H2O (l) <=> [Fe(OH)(OH2)6]^+ (aq) + H3O^+ (aq)}$$ Furthermore, in very basic conditions the hexahydroxidoferrate(II) would form. $$\ce{[Fe(OH)2(OH2)4] (s) + 4OH^{-} (aq) <=> [Fe(OH)6]^4- (aq) + 4H2O (l)}$$

However, this rule breaks down for $\ce{[Fe(CN)6]^4-}$ which is rather stable with $K_\mathrm{instability} \approx 10^{-37}$.

Half-reactions

As you might imagine, I was trying to avoid using half-reactions and the Nerst equation. Iron chemistry is rich and there are a myriad of possibilities. Before we move forward, let us note that the corresponding ion electronic structures are
$$\ce{Fe^2+:[Ar]{3d^6}};\\
  \ce{Fe^3+:[Ar]{3d^5}}.$$
So it seems that ferric ions are indeed more stable. Yet the reduction half-reaction
$$\begin{align}
  \ce{Fe^3+ + e- &-> Fe^2+} &
  E^\circ &= +0.77~\mathrm{V}
\end{align}$$
 implies the opposite. (Is this due to solvation?)

I did find a useful Pourbaix diagram (Andel Früh: Pourbaix diagram of Iron; $c(\ce{Fe}) = 10^{-6}~\mathrm{mol/l}$, $T = 25~^\circ\mathrm{C}$; wikimedia.org):

What we can tell is that ferric ions themselves are not favoured at a higher pH. In fact, the opposite is true. It just requires a strong oxidizer. This does not mean ferrous ions are not oxidized in basic conditions, it is just that the stable species are iron(III) oxide hydrates (probably iron(III)oxidehydroxide hydrates as well).

Problem

I have not really thought about this particular issue before, so I would definitely love to hear your ideas.

Why are ferrous ions and iron(II) hydroxide more stable at lower pH?
Why is the ferrous ion generally more stable, even though the electronic structure suggests otherwise?
This Pourbaix diagram does not take into account possible complex formation. If someone could explain the issue via complexes as well, I would greatly appreciate it.

The stability of iron(II) hydroxide J. Chem. Educ., 1957, vol. 34 pages 178-179 shows that the book isn't entirely correct.

  ...Iron(II)
  hydroxide would thus appear to be unstable in neutral
  solutions and to possess increasing stability as the
  basicity of the medium increases.

  The foregoing reasoning led to the attempt to prepare
  iron(II) hydroxide from the chloride under such
  conditions that the hydroxyl ion concentration was not
  allowed to fall below approximately one molal until the chloride ion had been removed. The hydroxide was precipitated from strongly basic solution by the addition of dilute iron(II) chloride to a saturated solution of
  sodium hydroxide in an air-free system with vigorous
  stirring. The precipitate was repeatedly washed with
  molal sodium hydroxide (oxygen and carbonate-free)
  until a test for chloride ion with silver nitrate could
  no longer he obtained. Five additional one-liter washings with normal base completed the preparation of the
  iron(II) hydroxide used for the solubility studies.
  This technique led to the formation of a coarse-grained
  precipitate, white when viewed by reflected light and
  slightly greenish tinged by transmitted light. Dilutions
  of the material were made to one-tenth molal base
  without evidence of darkening when maintained under
  an oxygen-free atmosphere.


Materials used in cooktop heating element, and reason for black appearance I have been interested in the structure of heating elements/coils used in cooktops. I am referring to something such as in the following picture:

I have a cooktop at home with such elements, and at parts where there was more scraping, I can see a silver-ish colour rather than the black, suggesting the black material is a thin layer around some metal resembling stainless steel, in appearance at least.

I have read on Wikipedia that often such heating elements would have an outer layer made "of stainless steel alloys, such as Incoloy, or copper" (and similar statements elsewhere). This seems to suggest that the silver-ish layer I have seen at the scraped parts in my cooktop is indeed stainless steel.
What I am interested in is what the black layer is.
My question is why these heating elements appear black in appearance; what the black layer is.

I have a possible guess, which I would also be interested to affirm or negate with you, which is that this is due to oxidation of the metals, which perhaps is done by the manufacturer(s) in advance because they expect that the high temperature that the metal would get to during cooking would itself speed up  oxidation and the resulting discoloration, and doing the oxidation in advance (and in a controlled manner which does that uniformly, so the colour seems uniform) prevents that from happening by forming a protective layer of oxidised metal. One reason I am guessing that, is that I looked on Wikipedia at possible oxides of iron, chromium, and nickel, and all of these seem to have at least one oxide with a black appearance. Iron for example has $\ce{FeO}$ and $\ce{Fe3O4}$.

The heating element ( on the inside where you cannot see it ) is Nichrome or similar alloy ( 10 to 20 % Cr, bal. Ni ) . The surface sheath is typically Inconel 601 ( 15 Cr, 10 Fe, 0.2 Al, bal. Ni). They were Inconel 600 ( and maybe some still are 600 ) until the addition of a small amount of Al was found to increase oxidation resistance. There are several similar alloys and there are different names for the same alloy. So the black surface is oxides of Ni. Cr with some Fe. Some compositions and temperature exposure will produce a green tint in the black ( in high temperature furnace tubes of the same family of alloys). Diffraction analyses of the oxides ( from furnace tubes) often do not give specific compositions as Cr, Ni and Fe can substitute for each other.


Why isn't Mg deposited on the cathode after electrolysis? Well here's a question from my book:

A solution containing one mole per liter of each $\ce{Cu(NO3)2}$, $\ce{AgNO3}$, $\ce{Hg2(NO3)2}$ and $\ce{Mg(NO3)2}$ is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are:

$\ce{Ag+/Ag} = +0.80$,
$\ce{Hg2^2+/Hg} = +0.79$,
$\ce{Cu^2+/Cu} = +0.34$,
$\ce{Mg^2+/Mg} = -2.37$

The sequence of deposition of the metals on the cathode will be?

The answer to the above question is: $\ce{Ag}$, $\ce{Hg}$, $\ce{Cu}$.
Why isn't Mg deposited on the cathode? Is it because the reduction potential is negative?
If yes, what exactly is the reason?

You are in aqueous solution. Everywhere where there is an aqueous solution, you have a certain concentration of a further cation: $\ce{H+}$. It has a potential of $\pu{\pm0 V}$ by definition, thereby being nobler than $\ce{Mg^2+}$.

Thus, hydrogen gas is ‘deposited’ on the cathode first and bubbles away while the solution gets removed.

It is not possible to deposit considerably less noble metals on the cathode in normal, aqueous solutions.


What is the reason behind neglecting & putting the overlap integrals $0$ in Simple Hückel Theory? This is an excerpt about Simple Hückel Theory from Elements of Physical Chemistry by Peter Atkins:

  The first step that Hückel took was to ignore the $\sigma$-bonding framework and focus solely on the $\pi$ electrons. That is, he assumed that the atoms had taken up the positions they have in the actual molecule, then calculated the properties of the $\pi$ orbitals that matched that framework.

  [...] we show in Derivation 14.3 that the Schrödinger equation for the orbitals $\hat H \psi = E\psi\;,$ then becomes the following pair of simultaneous equations for the coefficients: $$(H_{AA}- E)c_A- (H_{AB}- ES)c_B= 0\\ (H_{BA}- ES)c_A+ (H_{BB}- E)c_B= 0$$ ... . These equations are called the secular equations.

  Hückel then made further approximations. First he neglected all the overlap integrals and set $S= 0$ wherever it appear. .....

I didn't quite get why he made this approximation. There must be some reason for neglecting the overlap integral, I suppose. Also, Atkins didn't mention why Hückel only made calculations for $\pi$ orbitals only; does his equations not work for $\sigma$-bonding?

So my questions are:

$\bullet$ What is(are) the reason(s) behind making the approximation of putting all the overlap integrals $0\;?$ After all, making $S= 0$ would mean there is no overlap.

$\bullet$ Why did Hückel work only on $\pi$ orbital? Doesn't his calculation work for $\sigma$-bonding?

What is(are) the reason(s) behind making the approximation of putting
  all the overlap integrals 0?

I am not 100% sure here, but using such a simplification, the mathematics becomes significantly easier. In principle, Hückel theory would still work if there would be overlap. In fact, it is possible to create a completely new basis by applying a transformation to the basis set by which a new set of orthogonal basis functions is created. The advantage of setting $\hat{S}$ (the overlap matrix) to $\hat{I}$ (the identity matrix) is that the eigenvectors of the matrix equation can be directly interpreted in terms of the $\pi$ orbitals in the system. You do not have to back-transform your solution as is the case in the Hatree-Fock method.

  Why did Hückel work only on π orbital? Doesn't his calculation work
  for σ-bonding?

Only $\pi$ electron molecular orbitals were included because these determine the properties of the molecules that Hückel was studying (conjugated molecules). The $\sigma$ electrons were ignored. This is known as $\sigma$-$\pi$ separability and is justified by the orthogonality of $\sigma$ and $\pi$ orbitals in planar molecules. As a consequence though, the Hückel method is in principle limited to planar systems.


Effecting Chemical Change Apologies in advance as I am not a science person by any definition. By most, I'm probably not a person but we'll let that be.

I am studying the notion of "change" in things and focusing on chemical change. Could someone please let me know if:

the change in the color of a photogray (photosensitive, photochromic) lens is a "chemical" change?
the change in a baby's diaper when it gets wet is a chemical change (and if it is similar in any nature to the change in #1) and if that is the same as the change in a pregnancy test or a urinalysis stick
if the change in a thermochromic/temperature sensitive item (a beer can that changes color in the cold, a color changing thermometer or spoon or mood ring ) is the same type of chemical change

Sorry if this is too much for one question, or if it can't be explained to a lay person, but I appreciate any responses. I'm too newbie on this board to post all the links as links so the questioned items are connectable through the list of urls.

TIA.

Lenses

Diapers

Pregnancy test

Blood detection

Coors Light Cold-Activated Bottles

Color changing thermometer

Frozensolutions

Mood ring -wiki

-----edit---------------
The question of "chemical" change would be, in my mind different from "physical" but I don't know if that distinction comes into play here, or is an issue when the change is permanent or temporary.

For starters, I think most chemists would agree that a chemical change is different from a physical change by involving the formation or breaking of chemical bonds, or changes to the charge, coordination, or bonding state of some of the atoms involved. A chemical change can be reversible, however, so reversibility is not a determining factor. Either a physical change or a chemical change could result in a change of color, or not--as the case may be.

In your examples, the absorption of water by a disposable diaper, as well as the color changing of liquid crystal-based displays (like the beer temperature indicator, thermometer, and mood ring) are clearly physical changes. The liquid crystals change their optical properties due to the physical alignment of the molecules, and not because of making or breaking of bonds. On the other hand, the pregnancy test and blood detection test probably involve chemical changes to produce a color readout (sometimes initiated by a relatively loose binding between an antigen and an antibody, or formation of a metal complex). The photochromic lenses turn dark as a result of a chemical change induced by light (according to Wikipedia, anyway--this is not my area of expertise).


What are the concentration changes when diluting an equimolar acetic acid: acetate buffer? When a buffer (for example, a mixture of 100 mM acetic acid and 100 mM acetate), is diluted 10-fold, all the concentrations change drastically. Surprisingly, after re-establishment of equilibrium, the pH is not much different from the starting pH.

How can you visualize how diluting a buffer disturbs the equilibrium of a buffer  and how the system returns to equilibrium? To make things simple in a thought experiment, we will posit that dilution is faster than acid/base chemistry, i.e. the first step is dilution without any reactions, and the second step is acid base chemistry without further dilution.

The explanation/visualization could be at the molecular or bulk level, and could use kinetic or K vs. Q or Gibbs free energy arguments.

Consider an acid buffer with $pK_a=5$. Ignore the hydroxyl concentration, assuming it is negligible.

Dilution leads to an identical (proportional) reduction in the concentration of all solutes: undissociated acid, complementary base (anion), and protium. Because the concentration of anion and protium is reduced, encounters between the two will occur less frequently. On the other hand, dissociation of the acid into component ions continues unabated, since this process is (we are ignoring activities) a first order reaction. The result is an increase in the concentrations of anion and protium, thus reestablishing a value of the protium concentration close to that prior dilution.

Other  important factors should be remembered in the context of ideal buffers and the Henderson-Hasselbach equation: (1) the concentrations of $\ce{A^-}$ and $\ce{HA}$ are set equivalent in an ideal buffer (2) the $\ce{H^+}$  concentration is set equal to the $K_a$ and (3) the concentrations of $\ce{A^-}$ and $\ce{HA}$ are much higher than those of $\ce{H^+}$ . That means that in both absolute and relative amounts, the concentrations of $\ce{A^-}$ and $\ce{HA}$ need to change by only a small amount in order to bring the pH back near the desired set point. This is very clear from the following plots (concentrations are molar and $pK_a = 5$). Every 0.1 time units the concentration is diluted by 17%. The pH spikes up but then settles back again as the acid dissociates. You practically don't see the changes in the undissociated acid and anion on the plot during the recovery because they are tiny.

You can compare this to the case when your buffer is not as good (same $pK_a$, but lower capacity, 100 $\mu M$ concentration). I altered the kinetics too so the pH would stabilize after each dilution:

I should add: ignoring $[OH^-]$ is admittedly bad near neutral pH/low buffer capacity, but then I was attempting to provide a visualization. Adding terms to cope with the hydroxyl equilibrium etc would complicate things without adding much value or accuracy (since the kinetics are not meant to be accurate anyway, just illustrative).

Addendum: Here's the code:

K = 1e-5;  % M
kd=10;     % arbitrary, set to alter convergence speed, numerical stability
dt = 1e-7; % arbitrary, alter to change convergence speed, numerical stability
v0 = [0.2; 0.2 ; K]; % M [HA], [A], [H],

%% for dilute conditions
%kd=80;
%dt = 1e-5;
%v0 = [0.0001; 0.0001 ; K]; % [HA], [A], [H]

Nstep = 10000;
t = [0:dt:(Nstep-1)*dt];
tdilute = t(round(Nstep*[0.1:0.1:0.9]));

dv_dt = @(kd,K,v) [ -kd*v(1)+kd/K*v(2)*v(3); kd*v(1)-kd/K*v(2)*v(3)] ;
dv_ = @(v,dt) dv_dt(kd,K,v)*dt;
dv = @(v) dv_(v,dt);
v = v0;
conc = v0;
for ii=2:Nstep
    vstep = dv(v);
    v = v + [vstep(1); vstep(2); vstep(2)];
    conc(:,ii) = v;
    if any(t(ii) == tdilute)
        v = v/1.2;  % <-- change factor to dilute by here
    end
end

figure
subplot(2,1,1)
plot(t,conc(1,:),'r')
hold on
plot(t,conc(2,:),'g--')
plot(t,conc(3,:),'b--')
ylabel('[c]')
legend('[HA]','[A^-]','[H^+]')
subplot(2,1,2)
plot(t,-log10(conc(3,:)),'b--')
ylabel('pH')
xlabel('time (au)')


Definition of ionisation energy My lecturer has defined ionization energy or ionization potential of a species A as the molar internal energy change, ΔU, for the reaction: $\ce{A (g) -> A+ (g) + e-}$ at $\pu{0 K}$, with reactants and products being in their standard states.

How can reactants and products be in their standard states when they must be gaseous?!

He then goes on to say that we often assume  $\Delta U (\pu{0 K})\approx \Delta H (\pu{298 K})$. I presume he means that the molar internal energy change of the above process at $\pu{0 K}$ is approximately equal to the enthalpy change of the above process at $\pu{298 K}$? Why is this approximation valid?

It is common practice to define the standard state of a gas when describing its thermodynamic properties. For instance, the free energy of an ideal gas can be defined relative to the free energy in its standard state (the gas at a partial pressure of 1 bar and at the same temperature), as

$$G = G^\circ + RT\log\left( \frac{p}{p^\circ} \right)$$

where $p^\circ= \pu{1 bar}$.

In practice it may be impossible to achieve the gaseous state described in the OP, not least because absolute zero is not experimentally accessible. However, there are workarounds that might allow one to extrapolate to $\pu{0 K}$ and to standard conditions (a pressure of 1 bar). First the easy part, the pressure: if you have a very dilute gas, then intermolecular interactions become unimportant in describing its behavior, it becomes practically ideal. As such, its behavior at any other pressure can be described by using standard equations applicable to ideal gases. It does not matter that the state described is not real, it can be used as a hypothetical reference point. For instance, if we know the free energy under ideal gas conditions at a particular pressure, we can compute the free energy in the standard state as follows:

$$G^\circ = G - RT\log\left( \frac{p}{p^\circ} \right)$$

For an ideal gas, the internal energy $U$ is independent of pressure, so in the given example there is no need to perform any computation at all, provided we are at the right T. The problem is of course that we are not at the right T. In practice it may be possible to either assume that the experiment was conducted sufficiently close to $T=\pu{0 K}$, so that we can assume the equality holds, or to extrapolate to that temperature by performing experiments over a range of T.

As for the approximation $\Delta U (\pu{0 K})\approx \Delta H (\pu{298 K})$, I'll give a roundabout justification, since I am not an expert on the subject. There are two points that can be made. First, it is probably ok to assume $\Delta U (\pu{0 K})\approx \Delta U (\pu{298 K})$ if it can be assumed that the electronic transition is not sensitive to differences in the distribution of (mainly vibrational) thermal motions at $\pu{0 K}$ and $\pu{298 K}$.

Second, $$\Delta H = \Delta U + \Delta (pV) = \Delta U + \Delta n RT $$ However here we have a conundrum. The reported IE ignores the kinetic energy of the electron (it assumes that the emitted electron has no kinetic energy). But the thermodynamic expression assumes thermal equilibrium, so which is it? Well, at zero Kelvin there is no translational energy, so when considering $\Delta U(\pu{0 K})$ we can safely ignore any thermal contribution from the electron. So we are ignoring the thermal properties of the released electron, and then since $\Delta n =0$ (ignoring the electron) we have that $$\Delta H = \Delta U$$