Ramachandran Plot for Phi and Psi angles of peptides

A remarkable aspect of the Ramachandran plots is that the implementation of a hard-sphere potential with flexibility limited to the two backbone dihedrals proved sufficient to largely explain the conformational range sampled by real polypeptides. Take for instance the following map from a seminal article by Ramachandran (from Ref. 1; dashed line encloses the partially allowed region):

The map was generated with the following set of restraints:

In the article Ramachandran et al. explain how they selected the range of vdW distances used to generate their maps:

The next step towards working out the allowed conformations is to choose a set of permitted minimum contact distances between the different types of atoms. Only those conformations (4,4') which do not have any of the contact distances less than these minimum values now become allowed. Two such sets are given in Table II, termed "normally allowed" and "outer limit" contact distances. These are essentially the same as those given by Ramachandran, Ramakrishnan, and Sasisekha-ran (1963) arrived at from a study of the contact distances observed in structures of various organic compounds. Contact distances inbetween the two limits have also been observed in actual crystals, but not as frequently as the normally allowed values.

van der Waals radii are meant to define distances of closest approach. They are useful to identify conformational regions forbidden due to steric clashes. However, tabulated vdW radii are representative values. In real molecules the internuclear potential can vary somewhat depending on atomic types and the local electronic environment.
In addition, molecules are fairly flexible and can to a degree circumvent potential clashes by changing internuclear distances and angles, but a rigid model does not capture that. That a simple rigid model suggests a steric clash, because one or more vdW distance restraints are violated, does not guarantee that a conformation (phi/psi pair) is forbidden. Therefore, if you use a rigid model to generate a Ramachandran plot, that is, if only phi and psi are allowed to change, then you should allow "wiggle room", for instance by setting a lower threshold for the allowed values of the vdW distances. The use of "tolerances" in vdW radii should therefore not be surprising.
Close contacts can be energetically costly even with compensating rearrangements, so the associated configurational space - marked as "partially allowed" in a traditional diagram - is more sparse. In a modern approach a more descriptive potential for the protein backbone is employed, and ideally degrees of freedom other than the dihedral angles are relaxed, and energy isolines are displayed. The following, from Ref. 2, shows roughly how such a diagram might look:

References

G.N.Ramachandran, C.M. Venkatachalam, and S.Krimm. Stereochemical Criteria for Polypeptide and Protein Conformations II. Helical and Hydrogen-Bonded Polypeptide Chains. Biophysical Journal  6 (1966) 849-872.
ibid, 909-933.


Why Cu looses 2 electrons to become stable instead of gaining one electron to fill its 4s2 3d10 configuration?

Copper can gain 1 electron to become Cu-.

Copper has an electron affinity of 1.236 eV, so Cu- is stable to electron loss by 1.236 eV.

See also "The Dissocationi Energy and Heat of Formation
of the Molecule NaCu" Z. Naturforsch. 28a, 316-317 which says:

  But unlike as for NaAg a noticable ionic contribution
  to the bonding is observed for NaCu

Also there is Resonant two-photon ionization spectroscopy of LiCu :

  The unusual shape of this curve is thought to derive from avoided
  crossings between the $\ce{Li^+Cu^-}$ ion pair state and covalent states, with the $\ce{Li^+Cu^-}$ ion pair state
  ultimately correlating to the ground electronic state of the molecule.

So if Cu is to make a compound with an electronegative element, Cu should lose electrons, but with an electropositive element it can gain an electron.


(How) Does ICP-OES determine the quantity of detected elements, or just their occurance in the sample?

Although fp's answer is absolutely correct and I see that you have something more than high school chemistry, I'd like to give a more elaborated, yet simplistic answer, if you or anyone else reading this has any interest on the basic principles of an ICP-OES - you only need high school chemistry to understand it.

I'm not certain if it really does give quantitative information

It does, and very "accurately". Detection limits can be in ppb, that is, it can detect 0.0000001g (0.1 µg) of something in a 1L sample. That's really, really impressive if you stop to think about it.

how it does so

Wizardry! Well, almost. Read on and you'll see.

First of all the sample must be prepared, you can't simply toss something in it. Depending on what your sample is, it must be mixed with a known amount of this or that chemical (you'll see later why). This answers your next question:

and whether it can do so on its own.

Although the actual analysis and preparation can indeed be done purely by the machine in some cases (here's a video), and there are methods such as laser ablation which make it possible to almost just toss something in the ICP, the know-how of a trained analyst is still require to tell the machine what to do, to interpret the results it outputs, and to tell what chemicals to mix in the preparation process, if there is one at all. Equipments such as gas chromatographs can (and might need to) be coupled to assist in the sample preparation.

Back to how it does so:

When you introduce the prepared sample into the machine (it basically sucks it) some processing is done to it, and at some point it reaches the atomizer - the plasma torch. The atomizer does what it's name suggests: It turns stuff into separate atoms (by heating them to absurd temperatures). Here's a picture I took from an ICP atomizer:

Note: Not 100% of the sample is atomized. In fact, only a small percentage is.

When it's turned on it looks like the picture you saw at Wikipedia. You can see by the size of the parts that only a few milliliters of sample - if not less - is atomized at a time. Now that it got our sample atomized, it heats the sample up just a bit more, so the electrons get exited and go from their current level of energy to a higher one. While the sample passes through the equipment it gets cooler, so the electrons get back to the lower energy level, and by doing so, the atom emits light, hence the name optical EMISSION spectroscopy.

(source: wikimedia.org)

But what elements are emitting light, and at what wavelength? The principle of every spectroscopy analysis is that each element emits and absorbs specific, known, predictable wavelengths ranges, and only when heated to specific temperatures (in the case of emission). Because the temperature was set to a specific one for the analyte and a specific set of chemicals were mixed with the sample, only the atoms of the element we want to quantify will emit light at a certain element-specific wavelength range. We set the equipment to detect only that wavelength range and ignore others and that's how we know how much of that element there is in the sample.

But if you kept track of the proportions you might be thinking: "if it detects ppb amounts in only a few ml of sample where only a few % of it is atomized and only certain elements will emit only a certain wavelength, how the hell can it detect that little light?". That's the photomultiplier's job. It's a fantastic piece of equipment and you should definitively read about it.

After it detects the light emitted by the analyte in the sample, it compares it to a blank, does some calculations and displays the results in the form of a graph. Here's a screenshot of an ICP-OES software:

As I mentioned, this is a simplistic explanation, most of the process isn't that simple in reality. So as you can see, it's basically witchcraft.


To make it clear, the question is: do both molecules have to contain hydrogen in order for hydrogen bonding to occur?

No, as noted by Prakhar in his answer, tertiary amines are one class of compounds that can participate in hydrogen bonding (as an acceptor) when there is no hydrogen atom directly bound to the participating atom.

As an example, trimethylamine can hydrogen bond just fine with water:

More information can be found, e.g., here at ChemGuide (scroll down to "Solubility in water").


How to calculate the number of protons in a sphere given the volume and the pH?

As pointed out in the comments, the concept of pH as a thermodynamic property of nanodroplets does not quite make sense unless the nanodroplets are in equilibrium (stable). If they are in equilibrium you can still apply the equation
$$\textrm{pH}=-\log_{10}(a_{\ce{H+}})$$ but now you have to be aware that the activity will be influenced by surface (and other) effects and the approximation $$\textrm{pH}=-\log_{10}(c_{\ce{H+}})$$ may not hold. But assume it does. Then as you point out, the average number of hydronium ions per nanodroplet is

$$\begin{align} n_{\ce{H+}} &=\pu{10^{-6} M}\cdot(\frac{4}{3}\pi\times\pu{20^3 nm^3}\times\pu{10^{-27} {(m/nm)^3}}\times\pu{10^3 L/m^3})\times \pu{6.022\times10^{23} mol^{-1}}\\&=0.020 \end{align}$$

Which is ok. That just means that on average only 1 out of 50 nanodroplets contains a solvated proton. Note by the way that one nanodroplet also contains about a million water molecules.

This is a real "rabbit hole" of a question, given all that is not specified, which is why I used sleight-of-hand and wrote that I am ignoring surface effects and assuming the particles are at equilibrium. It doesn't address how to prepare nanodroplets with the desired average proton concentration or what equilibrium means (are nanodroplets exchanging matter or are they isolated?). As noted in comments, composition (buffering molecules if any, salts and other ions including hydroxyl), droplet charge etc, are being ignored and you can expect statistical fluctuations in concentrations between different droplets. However if I stick to the assumption of equilibrium then I can say (for what it's worth) that ~2% of droplets have ~pH 4.3, and most of the rest are at neutral pH.


Differences between miscible and soluble

Your definition is basically correct, except using two liquids might provide for a clearer example than a solid and a liquid.  For example, diethyl ether is soluble in water to about 7% at $\mathrm{20^oC}$. This does not mean that diethyl ether is partially miscible in water, but that it it partially soluble. This Wikipedia article provides a good distinction for miscibility:

  Miscibility is the property of substances to mix in all
  proportions (that is, to fully dissolve in each other at any
  concentration), forming a homogeneous solution.


If ammonia gas is bubbled into pure water, the following reaction takes place:

$$\ce{NH3(aq)  +   H2O(l)   <=>     NH4+(aq)  +    OH−(aq)}\tag{1}$$

How to decrease the percent of ammonia that is converted to the ammonium ion in water?

Le Chatelier's Principle indicates that adding $\ce{OH-}$ would push the equilibrium to the left.

I lifted the following figure from:

  HYDROPHOBIC MEMBRANE TECHNOLOGY FOR AMMONIA EXTRACTION FROM WASTEWATERS by Airton Kunz and Saqib Mukhtar Eng. Agríc. vol.36 no.2 Jaboticabal Mar./Apr. 2016
  https://doi.org/10.1590/1809-4430-Eng.Agric.v36n2p377-386/2016

As can be seen in the figure, an increase in pH favors formation of $\ce{NH3}$ and a decrease in the pH favors $\ce{NH4+}$.

  Since the OP asked only for the percent of $\ce{NH4+}$ to be reduced, there is another way. Dilute the solution.

Will $\ce{Fe(NO3)3}$ decrease the percent of ammonia that is converted to the ammonium ion in water?

Iron (III) forms an relatively insoluble hydroxide as shown by the following equation:

$$\ce{Fe^{3+} + 3OH- <=>[K_{\mathrm{sp}} = 2.79\times10^{−39}] Fe(OH)3(s)}\tag{2}$$

So again according to Le Chatelier's Principle removing $\ce{OH-}$ from reaction (1) favors the reaction moving to the right. Thus the precipitation of $\ce{Fe(OH)3}$ will increase the formation of $\ce{NH4+}$.

  In a comment on the question user Poutnik indicated the reaction $\ce{Fe^{3+} + 3NH3 + 2H2O -> FeO(OH) + 3NH4+}$.  I'm sure there is a study somewhere on $\ce{Fe(OH)3 \pu{ vs } FeO(OH)}$ formation, but I didn't find it in 15 minutes of fooling around. I'd expect that strong $\ce{OH-}$ concentrations and higher temperatures (boiling solution) would favor $\ce{FeO(OH)}$.


Does syn-dihydroxylation form racemic mixture?

Oxidation  of alkene to 1,2-syn-diol by cold basic $\ce{KMnO4}$ is a common classification test for alkene in undergraduate organic chemistry laboratory courses. OP had correctly recognized that this reaction is a case of syn-dihydroxylation, and OP's conclusion of the product to be meso is also correct since the alkene in hand is a cis-but-2-ene. The schematic representation of the proposed mechanism is given below:

Source of the scheme
Since but-2-ene structure is planer and a prochiral molecule, it has two faces (Si- or Re-face; above or below the plane) to get attacked by a reactant, here it is $\ce{KMnO4}$. The above mechanism displays the attack by only one face. The oxidation gives butan-2,3-diol by syn-attack on that face. The product has two stereocenters and the stereochemistry of that product is $(2R,3S)$. As indicated in the diagram, the product also has a plane of symmetry, and hence, the $\ce{C}$2 stereocenter is the mirror image of the $\ce{C}$3 stereocenter. Therefore, the product is a meso-compound.
If the $\ce{KMnO4}$ attack is done on the opposite plane, the product would be $(2S,3R)$-butan-2,3-diol, if numbering has maintained the same. This is essentially the same meso-product, as we discussed above.
If the starting substrate is trans-but-2-ene, the $\ce{KMnO4}$ attack from one face gives you $(2S,3S)$-butan-2,3-diol, while the $\ce{KMnO4}$ attack from opposite face gives you $(2R,3R)$-butan-2,3-diol. Since possibility of attacking from the Si-fase and/or Re-face is $0.5$ (or $50\%$), you get essentially a racemic mixture.
Based on above facts, OP's textbook description of the reaction is erroneous, not only on stereochemistry, but also on the notation of the reaction (it is syn- not sym-addition).


How do water models fit into force field methods?

I guess for starters, try finding a torsion angle on a 3 point molecule.

Is it fast enough, accurate enough?

The real issue is why do some force-fields have intra-molecular bonds and others do not.

Including bond and angle forces in water models make their computation more expensive. There are classes of flexible forcefields for water. There are even more indepth forcefields that are polarizable.

Why don't we use the most sophisticated forcefield models for water?

The more sophisticated the model, the longer our simulations take. Often, rigid water models are good enough for certain properties. So, we use the simplest model that gives us the right answer for the property we are after. This is just a general good rule.

Parameters

Force-fields use parameters, and proteins use ALOT of parameters. These parameters are often fit for a specific water model. Hence, you are kind of stuck using that water model if you are using that Force-Field for the protein/solute molecule.

Complexity

As our interests move towards more involved calculations that involve water it does become important to include more effects such as flexibility and polarizability. This is increasingly the case in calculating Free Energies, whether it is the Free Energy of hydration or the Free Energy of binding a ligand to a protein. Another example is modelling supercritical water - only polarizable models have been successful. Rigid models such as SPC/E and TIP3P simply are not good enough for many properties under many conditions. You can of course fudge parameters to make anything fit a particular problem, but we want to use generalizable force-fields as much as possible. In this case, models with more parameters (flexible, polarisable) give us additional accuracy for the difficult problems.

Bandwagon

Another reason many people use one model more than another is because their colleagues use it, so they use it because other people use it and it likely won't be questioned since other people use it. This is not a good reason for using a certain model! usually the first people to use a certain model had a reason, it solved their problem. However often the following bandwagon users don't read the strength/weaknesses of the model but instead just use it because they see it used lots for seemingly similar areas of research. Welcome to molecular modelling.

Mixing Rules

Generally atoms have dispersion and electrostatic terms. In the case of dispersion, when the force or energy between two atoms is being calculated a single parameter for each term is required for the calculation. Take the well known Lennard-Jones for instance

$$U(r_{ij}) = 4 \epsilon_{ij} \left[ \left( \frac{\sigma_{ij}}{r_{ij}}\right)^{12} - \left( \frac{\sigma_{ij}}{r_{ij}}\right)^6 \right]$$

different forcefields may have different numeric values for $\epsilon_i$, $\epsilon_j$, $\sigma_i$,$\sigma_j$ but they will all have a value. You need to use mixing rules to take the respective $i$ and $j$ parameters and create an $\epsilon_{ij}$ and $\sigma_{ij}$. We call these mixing rules a.k.a. combining rules. Wikipedia lists several of the common ones. https://en.wikipedia.org/wiki/Combining_rules

For instance for the Lorentz-Berthelot
$$\epsilon_{ij} = \sqrt{\epsilon_{i} \epsilon_{j} }$$
and
$$\sigma_{ij} = \frac{\sigma_i + \sigma_j}{2}$$

Finally, in SPC and SPC/E the hydrogens do not have Lennard-Jones parameters. It is not uncommon for hydrogens to have no dispersion parameters. This can lead to problems though, since dispersion parameters keep other atoms/particles from getting too close.

So a specific example of an SPC water molecule interacting with a Sodium ion, the electrostatics just needs to know that the Oxygen has a negative charge of $q_o =−0.82$ and the sodium has a charge of $q_{N_a} =+1$. A naive approach would be to treat this as a bare coulomb interaction which is just

$$ U(r_{ij})^{\rm electrostatic} = k \frac{q_o q_{N_a}} {r_{ij}}$$

where

$$k = \frac{e^2}{4 \pi \beta}$$

$e$ is the charge of an electron and $\beta$ is the permittivity and a global and constant value for the simulation.

A better approach is to use EWALD's summation for electrostatics. It is more complicated, but the only charge specific parameters it needs to know is still just $q_o$ and $q_{N_a}$. The other parameters it uses for damping can be considered global parameters. Note: also do this for each of the hydrogen charges interacting with the Sodium ion.

For Lennard-Jones you would need to use mixing rules for the oxygen (there is no LJ parameters for the hydrogens)

$$\sigma_{(o-N_a)} = \frac{\sigma_o + \sigma_{N_a}}{2}$$

and

$$\epsilon_{(o-N_a)} = \sqrt{\epsilon_o \epsilon_{Na}}$$

then do the standard LJ calculation for force and or potential energy.


First Order Degradation Constant (λ)(d^-1)

Based on a bit more than a cursory search, there doesn't seem to be a first order degradation constant readily available for 1,3,5-trimethylbenzene (mesitylene) in the open literature. This leaves us with your second option, the Title 35 Illinois Administrative Code 740$^{[1]}$, Table A (herein referred to as Table A).

By far the most similar compound to 1,3,5-trimethylbenzene that can be found in Table A, based on both chemical structure and physical properties, is the total xylenes (CAS # 1330-20-7).  This is actually a mixture of the three possible isomers of xylene.

Both the images below and all physical and chemical properties given for the total xylenes were taken from this Wikipedia page unless otherwise stated. The 1,3,5-trimethylbenzene image and all of its chemical and physical properties were taken from here.

1,3,5-trimethylbenzene (mesitylene)

In order: o-xylene, m-xylene, p-xylene

Selected properties:
Mesitylene: density = 0.86 ($\mathrm{20^oC}$), B.P. = $\mathrm{165^oC}$, M.P. = $\mathrm{-45^oC}$, and water solbility = 50 mg/L at $\mathrm{25^oC}^{[2]}$
Xylenes : density = 0.86 ($\mathrm{20^oC}$), B.P. = $\mathrm{138^oC}$, M.P. = $\mathrm{-47^oC}$, and water solbility = 110 mg/L at $\mathrm{25^oC}^{[2]}$

1) TITLE 35: ENVIRONMENTAL PROTECTION, SUBTITLE G: WASTE DISPOSAL, CHAPTER I: POLLUTION CONTROL BOARD PART 740, SITE REMEDIATION PROGRAM

2) Yalkowsky, S.H., He, Yan., Handbook of Aqueous Solubility Data: An Extensive Compilation of Aqueous Solubility Data for Organic Compounds Extracted from the AQUASOL dATAbASE. CRC Press LLC, Boca Raton, FL. 2003


What does "lit." mean in boiling point or melting point specification?

Yes, you are correct: “lit.” refers to the literature value.
Usually one would also expect the corresponding reference, but it looks like Merck doesn't attach any to the provided data directly in place.
From The ACS Style Guide, section Reporting Analytical Data [1, p. 274] (emphasis mine):

Melting and Boiling Points
mp 175.5 °C (lit.25 mp 175–176 °C)
mp 225 °C dec
bp 127 °C
Abbreviations: mp, melting point; bp, boiling point; lit., literature value; and dec, decomposition. A full space is used between the number and the unit °C; the
degree symbol is closed up to the C. A superscript number after “lit.” denotes the
number of the reference.

References

The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. ISBN 978-0-8412-3999-9. https://doi.org/10.1021/jm068034q


I can't imagine anyone arguing that ionic resonance structures are invalid.  The often contribute significantly to the structure of a molecule and help explain the various physical aspects of the molecule.  The following ionic resonance structure contributes significantly to the description of the carbonyl compound.  $\ce{NaCl}$ is not totally ionic and is described by a blend of both the ionic (major) and covalent (minor) resonance structures shown below.

Resonance structures with a positive hydrogen (proton) are important in explaining hyperconjugative effects.


OK, here is the difference: formation enthalpy refers to the formation of a compound from elementary compounds, while bond energy is applicable when considering the formation of a molecule from isolated atoms. The relation between the two is pretty much explained by the Hess's law: the enthalpy of a reaction does not depend on its pathway, but only on the initial and final states.

Now consider the following two pathways for your example.

Pathway 1: we burn graphite in oxygen, just like $\ce{C + O2 -> CO2}$. There is some change of enthalpy associated with this reaction.

Pathway 2: we vaporize graphite to get atomic carbon (see how many C-C bonds are broken, take that into account), atomize $\ce{O2}$ to get elementary oxygen (again, some bonds are broken), and finally produce $\ce{CO2}$ (this time some bonds are formed). Now if you'll look up the energies of all bonds involved and carefully add/subtract them all together, you'll have a decent approximation of the reaction enthalpy obtained in the pathway 1.

(Normally it works the other way around, that is, we use reaction enthalpies to calculate bond energies, but... whatever.)


Calculating the percentage of (+)-2- butanol in the sample

A different, more logical solution gives the correct answer as well.
Since the optical rotation of (+)-2-butanol is 12.5 degrees, the optical rotation of (-)-2-butanol is -12.5 degrees. If the mole fraction of the (+) isomer is $x$ and the mole fraction of the (-) isomer is $1-x$, then
$$-2.8 = 12.5x -12.5(1-x)$$
Solving this simple linear equation gives $x = 0.388$, which when converted to mole percentage, is $38.8\%$.


Why is the melting point of BaCl2 > BeCl2 while CsCl<NaCl

Sodium chloride and caesium chloride are both primarily ionic compounds. Within ionic compounds, a greater melting point can be caused by a higher ionic charge or a higher charge density/better interaction. Since both have the same charge while sodium has a smaller ionic radius, the sodium-based crystal is held together better due to more interaction than the caesium based one.

When comparing barium chloride and beryllium chloride, you are essentially comparing apples and oranges since beryllium chloride is a primarily molecular compound. In its gas phase, it forms $\ce{BeCl2}$ monomers and $\ce{Be2Cl4}$ dimers. The intramolecular attractions in molecular compounds are typically rather strong with the intermolecular ones are weaker. For ionic compounds such as barium chloride, we cannot distinguish between the two since we cannot define molecular entities.

When comparing ionic compounds to molecules, the ionic compounds typically have a much higher melting point (with exceptions; both low-melting ionic compounds and high-melting molecular compounds).


During electrolysis, why are the products attracted to the cathode?

You have to think about what exactly a cathode and anode are. You are falling victim to this:

A widespread misconception is that anode polarity is always positive (+). This is often incorrectly inferred from the correct fact that in all electrochemical devices, negatively charged anions move towards the anode and positively charged cations move away from it. In fact anode polarity depends on the device type, and sometimes even in which mode it operates, as per the above electric current direction-based universal definition. Consequently, as can be seen from the following examples, the anode is positive in a device that consumes power, and the anode is negative in a device that provides power. - Wikipedia

It's your conditioning from other chemical nomenclature--"cation" and "anion"--which are absolute about their charges.

So beyond misleading nomenclature, what does cause this to happen?

Take this picture for example: http://img.sparknotes.com/figures/0/02480ae8fc1a41b131a3fdb5a698e9a3/compare.gif

If E°cell > 0, then the process is spontaneous (galvanic cell)
If E°cell < 0, then the process is nonspontaneous (electrolytic cell)
where E°cell = E°(V) reduction + E°(V) oxidation
In your case,

Pb → Pb2+ + 2e- = 0.13

2Br- → Br2 + 2e- = -1.06

E°cell = .13 - 1.06

E°cell = -.93

In your case, the process is nonspontaneous (makes sense, think about trying to reduce ionized bromine), so you're looking at an electrolytic cell, which is battery driven (it needs work put in to operate.)

So, again, to directly answer your question:

  In the electrolysis reaction, lead is formed at the cathode and bromine is liberated at the anode. But why is it so? I mean, before gaining or losing electrons to become neutral, lead is positive and bromide is negative. So, naturally they should be attracted to the cathode and the anode respectively.

Lead is formed at the anode because work from a battery was put in that pushed the reaction in that direction. Before work was put in, the reaction would have proceeded in the opposite direction (think exothermic vs endothermic, but in this case spontaneous cell potential vs nonspontaneous cell potential.)

  But why do they become neutral after they reach the electrodes.

They become neutral at the electrodes because that is where the reaction happens, and that is where the electrons are transferred.

  Why do they gain electrons at only the electrodes to become neutral?

Again, where else could they gain electrons? They are sitting in a solution contain themselves and their respective anions, which they have already reacted with, and will no longer be transferring electrons with.


What am I assuming when I make this approximation?

The approximations you can do are the following :

1 - Practically all silver atoms are included in the complex ion. So that the concentration of that ion is equal to [Br-]

2 - The amount of AgBr dissolved is negligible with respect to the amount of Na2S2O3 available. So that, at the end, $\ce{[S2O3^{2-}]}$ is equal to the concentration M of the thiosulfate.

With these approximations, you may rewrite $\beta{_2}$, so that [Ag+] is :
$$\ce{[Ag+]= {[Ag{_2}(S_2O_3^{3-}]}/{(M·{\beta_2})}} = [Br^-]/(M·{\beta_2})$$
With the solubility product of AgBr , you have a system of two equations with two unknowns. The first solution [Ag+] will be very small. But the second solution, namely [Br-] gives you the solubility of AgBr you are looking for.


Why is the melting point of magnesium oxide higher than aluminium oxide?

Two ideas to consider:

Alumina is not fully ionic.  Neither is magnesia, but magnesia has more ionic character and we may see the greatest amount of ionic attraction in magnesia versus alumina which is less ionic, or soda which has only singly charged cations.

When metal oxides are melted they do not necessarily produce free metal and oxide ions.  As described by Shi et al. [1], molten alumina retains mostly four-and five-coordination of oxygen to aluminum, rather than forming "free" aluminum and oxide ions and also contrasting with the six-coordination of aluminum in the solid conundrum phase.  Presumably magnesia would behave similarly to alumina when melted, but the greater ionic bonding character in magnesia makes reduced coordination less favorable and thus enhances retaining the fully octahedrally-coordinated solid phase.

Reference
1.
Caijuan Shi, Oliver L. G. Alderman, Diana Berman, Jincheng Du, Joerg Neuefeind, Anthony Tamalonis, J. K. Richard Weber, Jinglin You and Chris J. Benmore,  "The Structure of Amorphous and Deeply Supercooled Liquid Alumina",
Front. Mater. 6:38 (19 March 2019), https://doi.org/10.3389/fmats.2019.00038.


Why does potassium react the most vigorously with water?

I interpret "reacts most vigorously" to mean which metal reacts the fastest.  This is a question of kinetics, not thermodynamics (e.g. the answer is not based on the enthalpy of the reaction).  Kinetic rates are determined by the height of the energy barrier that needs to be surmounted, we need to determine and compare the activation energies in this process for the various metals.  The activation energy in this process is determined by the following two steps: $$\ce{M(s) -> M(g)}$$ $$\ce{M(g) -> M^+(g) + e^-}$$ $$\ce{or}$$ $$\ce{M(g) -> M^{+2}(g) + 2 e^-}$$ depending on whether we are analyzing a group I or group II metal.  The following link provides a nice analysis and tabulation of the energies required for these two steps with the group I metals

Group I link

and this link does the same for the Group II metals

Group II link

I'll summarize their findings

Metal ...   Activation Energy (kJ/mol)

K   ...         +508

Na  ...        +603

Mg  ...       +2200

Ca  ...       +1950

CONCLUSION: Potassium has the lowest activation energy and should react the fastest.


Why iron reacts differently with concentrated and dilute sulfuric acid?

In the first equation, $\ce{Fe}$ is oxidised to $\ce{Fe^2+}$ by $\ce{H+}$; the $\ce{SO4^2-}$ ions play no role in the reaction. You could replace $\ce{H2SO4}$ with $\ce{HCl}$ and get the same reaction.

In the second equation, the same oxidation of $\ce{Fe}$ to $\ce{Fe^2+}$ by $\ce{H+}$ also occurs. However, on top of that, some of the $\ce{SO4^2-}$ ions are reduced to $\ce{SO2}$, and some of the $\ce{Fe^2+}$ is further oxidised to $\ce{Fe^3+}$. Clearly, in the dilute acid, $\ce{SO4^2-}$ was not strong enough to act as an oxidising agent - but in the concentrated acid it is.

The reason for this can be somewhat explained by Le Chatelier's principle. Consider the redox reaction:

$$\ce{2Fe^2+ + SO4^2- + 4H+ -> 2Fe^3+ + SO2 + 2H2O}$$

In the dilute acid, this equilibrium lies far to the left because the concentrations of $\ce{SO4^2-}$ and $\ce{H+}$ are rather low. However, making the acid more concentrated shifts the equilibrium to the right so that you get some of the products.

As to why you get a mixture of iron(II) and iron(III), the simple answer is that the equilibrium above doesn't lie entirely to the right. $\ce{H2SO4}$ is not that strong an oxidising agent to begin with.

More quantitatively, the redox reaction above can be investigated using standard reduction potentials and the Nernst equation. The following data is taken from Wikipedia (you will get pretty similar values from more "reliable" sources):

$E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) = +0.77 \mathrm{~V}$
$E^\circ(\ce{SO4^2-}/\ce{SO2}) = +0.17 \mathrm{~V}$

Under standard conditions, $E^\circ$ for the full reaction is therefore $0.17-0.77 = -0.60 \mathrm{~V}$; the negative value indicates that the forward reaction is non-spontaneous.

The Nernst equation for this reaction can be written:

$$E = E^\circ - \frac{RT}{nF}\ln Q = E^\circ - \frac{RT}{nF}\ln\left[\frac{(a_{\ce{Fe^3+}})^2(a_{\ce{SO2}})(a_{\ce{H2O}})^2}{(a_{\ce{Fe^2+}})^2(a_{\ce{SO4^2-}})(a_{\ce{H+}})^4}\right]$$

I will make several assumptions here:

Activity coefficients are taken to be unity, meaning that the activities can be replaced by concentrations.
The activities of all the products, as well as $\ce{Fe^2+}$, are unity.
The only source of $\ce{H+}$ is from the dissociation of the $\ce{H2SO4}$, and the dissociation is complete. That means that if $c$ is the formal concentration of $\ce{H2SO4}$, we have $[\ce{H+}] = 2c$ and $[\ce{SO4^2-}] = c$.

Substituting in the values of $E^\circ = -0.60 \mathrm{~V}$, $R = 8.314 \mathrm{~J~K^{-1}~mol^{-1}}$, $T = 298 \mathrm{~K}$, $n = 2$, and $F = 96485 \mathrm{~C~mol^{-1}}$, this allows us to simplify the Nernst equation:

$$E / \mathrm{V} = -0.60 + 0.0128\ln (16c^5)$$

Strictly speaking, I should use $c/c^\circ$ instead of $c$ in order to make the logarithm dimensionless, but I'm fed up of technicalities. Now for the actual calculation. Let's assume that you're using $10\mathrm{~M}$ sulfuric acid, such that $c = 10$ (or $c/c^\circ = 10$, if you want to insist). This gives a value of $E = -0.42 \mathrm{~V}$ - as expected, it is less negative than before, which corresponds to a larger value of $K = \exp(nFE/RT)$. The fact that it is still rather negative just means that there is a difference between the real world and my theoretical treatment. Presumably, those assumptions above are not entirely valid.


If a chiral molecule reacts with an achiral molecule will the product be chiral or achiral?

The key here is to look at the transition state through which the reaction proceeds.

The short answer is that the transition states are enantiomeric, i.e., also mirror images of each other, and in an achiral environment, enantiomers have the same energy. If the transition states have the same energy, then the reactions proceed at the same rate.

The long answer is, unfortunately, convoluted by the fact that a new bond may be formed in the transition state, creating a stereogenic center. In other words, you might have something like:

$$\ce{($S$)-A + B -> [($S$,$R$)-AB]^{\ddagger} + [($S$,$S$)-AB}]^{\ddagger}$$

These are diastereomeric and the rates through these two transition states is different.

But $\ce{($R$)-A}$ does the exact same thing. So the rate of reactivity is controlled via 4 transition states:

$\ce{[($S$,$R$)-AB]^{\ddagger}}$
$\ce{[($S$,$S$)-AB]^{\ddagger}}$
$\ce{[($R$,$S$)-AB]^{\ddagger}}$
$\ce{[($R$,$R$)-AB]^{\ddagger}}$

Here, 1 and 3, and 2 and 4, are enantiomeric, respectively, and the reaction will proceed through pairs of enantiomeric transition states, but since they are enantiomeric, the two sets transition states provide matched rates from the enantiomeric reactant pair.

So, we can safely conclude that indeed enantiomeric reactants react at the same rate with achiral reagents.


I imagine it's called something like $\ce{Na2SiO4Mg}$.  Tomorrow I will try some combinations of sodium silicate with Aluminum Oxide, and magnesium oxide. What would these be called?

Much simplier than that. Sodium hydroxide dissolves incredibly exothermally:

$$\ce{NaOH (s) -> NaOH (aq)}\qquad{\mathrm{\Delta} H = \pu{-44.5 kJ mol-1}}$$

This is what heated your mixture.


It depends on what kind of reactivity you mean. Nitromethane is indeed flammable and is used as a racing fuel. It can also detonate, especially when mixed with an oxidizer. But it is generally quite stable to atmospheric oxygen at room temperature, perhaps because the nitro group is electron-withdrawing and reduces the likelihood that the hydrogens on the methyl group will react with oxygen or other oxidizers.

In other ways nitromethane is indeed reactive: for example, the nitro group makes the hydrogens on the methyl group acidic, so it reacts with an aqueous base like sodium hydroxide to give the nitromethide anion, which is useful in organic synthesis.

Consider the splitting of nitromethane into methyl radical and nitrogen dioxide. It has a normal sigma bond between carbon and nitrogen, and to break that bond you have to put a lot of energy in. You are right that NO2 is indeed stable on its own, but it is still a reactive, high energy species. Methyl radical is even more reactive, and the energetics of this bond-breaking reaction are overall unfavorable, at least at lower temperatures. Higher temperatures, such as might occur when sparking a detonation, may give the kind of splitting you are talking about, plus a lot of other reactions as well.


I think it would be more correct to say that pi bonds tend to exist in conjunction with sigma bonds, rather than that they need sigma bonds to exist - as pointed out in the comments.

This can be explained rather easily by basic molecular orbital (MO) theory, which seems like an appropriate framework for the question. Let's consider the simple case of a homoatomic bond such as C=C. When you have two C atoms coming into contact so as to form MOs, you form sigma MOs (bonding and antibonding) from the set of atomic S orbitals, and pi MOs (again, bonding and antibonding) from the Px and Py atomic orbitals.

Among the (molecular) orbitals, the sigma bonding MO generally has the lowest energy, followed by the pi bonding MOs. This means that the sigma bonding orbitals will 'fill' before the pi bonding orbitals. Because electrons preferentially occupy MOs from lower to higher energy, as long as the sigma bonding MOs are lower energy than the pi bonding MOs you will see sigma bonding orbitals populated before pi bonding MOs. Thus, to have a pi bond (occupied pi bonding MOs), you generally also have a sigma bond (occupied sigma bonding MO).

As pointed out in the comments to your question, this is somewhat of a generalization - but I think it's an accurate description of the phenomenon your instructor described.


The most general way of naming a graph is by writing,

Y vs X graph

An example is, Distance vs Time graph. In which the Y axis represents Distance and X represents time.

In order to know which comes in which axis you have to have a clear understanding about independent and dependant. Thus is easy to understand. Once you have clearly understood the axes you have to make the Y axis the dependant variable and X the independent one.

Then the name of the graph becomes ,

Dependant variable vs Independent variable.

For instance consider the Distance vs Time graph.
We see that Distance depends on how long has the time passed. Thus D becomes the dependant variable


According to the Cahn-Ingold-Prelog system assign a priority to each of the following groups. (Highest priority is 1)

The keys are:

going outwards, only look one bond at a time
look for the first point of difference
to break ties follow the chains of highest priority first

So, going out from the point of attachment, the first atom in each chain is a carbon, so we expand from there to the next three atoms in each case (in red). Draw these out in priority order (ie. highest to lowest atomic number), top to bottom. Now compare the highest priority atom on each chain, ie. C vs Cl vs Br vs C. Bromine has the highest atomic number, so this makes that whole substituent number 1 priority, chlorine is next, so that whole group is number 2 priority, the remaining two groups are a tie, so we go down their list of attached atoms, looking for the first point of difference. The second atom on each is a carbon - ie. a tie. The final atom is a hydrogen on the first group, but a carbon on the final group, so the latter has the higher overall priority. In this particular example, there is no need to go further out along the chains to determine priority order for the substituents.

So, the final priority order is: (1) -CH2Br, (2) -CHCl2, (3) -C(CH3)3, (4) -CH(CH3)CH2F


skeletal structure of dimethylamine? HN(CH3)2

Please look at the skeletal structure of different types of amines:

I hope it's clear now.


Generally, nitric acid is a strong oxidizing agent which get reduce to nitrogen oxides on reacting with metals which act as reducing agent thus forming a redox reaction. But in this case, only nitric acid is involved which get reduced to nitrogen oxides on reaction with air/light. So, this may be called oxidation reaction where nitric acid gets oxidized. From wikipedia article of nitric acid:

  Nitric acid is subject to thermal or light decomposition and for this reason it was often stored in brown glass bottles:

  $$\ce{ 4 HN^{V}O3 → 2 H2O + 4 N^{IV}O2 + O2}$$

  This reaction may give rise to some non-negligible variations in the
  vapor pressure above the liquid because the nitrogen oxides produced
  dissolve partly or completely in the acid.

  The nitrogen dioxide ($\ce{NO2}$) remains dissolved in the nitric acid
  coloring it yellow or even red at higher temperatures. While the pure
  acid tends to give off white fumes when exposed to air, acid with
  dissolved nitrogen dioxide gives off reddish-brown vapors, leading to
  the common name "red fuming acid" or "fuming nitric acid" – the most
  concentrated form of nitric acid at Standard Temperature and
  Pressure (STP). Nitrogen oxides ($\ce{NO_x}$) are soluble in nitric
  acid.

There are some old papers(here and here) which discuss the mechanism of nitric acid decomposition. Give a quick read on the papers and you will understand the mechanism of the reaction.


Reaction of secondary alcohol with ammoniacal silver nitrate (Tollens' reagent)

This question refers to the use of Tollens' Reagent (ammoniacal silver nitrate). The oxidising species here is the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$.

Tollens' reagent is usually used as a test for aldehydes, with which it gives a characteristic silver mirror:

It is often said that Tollens' reagent does not react with alcohols. However, from personal experience, gentle heating of a primary alcohol with Tollens' will cause a small amount of oxidation and result in a fine black precipitate. Again, in my experience, secondary alcohols (such as in the question) do not react with Tollens' reagent and so I would predict that no reaction will take place here.

EDIT: Your proposed mechanism certainly looks plausible but whether it will take place in reality I can't say for certain (my experience of Tollens' reagent says no).


Why does ethylamine have two infra-red spectra? Are both correct?

When looking at IR spectra of hydrogen-bonding groups, always check how IR spectra was recorded. It appears that NIST spectra is recorded using gas phase, while the first one used liquid film. In liquid film $\ce{NH2}$ groups form a network of hydrogen bonds, resulting in shape distortion and widening of $\ce{NH2}$-related bands. Same is true for carboxylic acids, alcohols, phenols, and many other group.  Comparison of water spectra in ice, liquid and gas is particularly enlightening.
If the spectrum is recorded in a solution, you also have overlapping spectrum of the solvent in your recording. If you look at water spectrum in liquid phase (nist webbook), you can see some wide peaks suspiciously similar to wide bands in the first spectrum you posted, so it might be that 'liquid film' is actually a water solution. In this case the situation is even worse, because in addition to ethyl amine spectrum you also have overlapping ethyl ammonium spectra.


How to find chemical similarity between two compounds using Tanimoto method

Apparently, it is more generally known as the "Jaccard Index":

  In "A Computer Program for Classifying Plants", published in October 1960,[5] a method of classification based on a similarity ratio, and a derived distance function, is given. It seems that this is the most authoritative source for the meaning of the terms "Tanimoto similarity" and "Tanimoto Distance". The similarity ratio is equivalent to Jaccard similarity, but the distance function is not the same as Jaccard distance.

In terms of chemistry, this calculation is carried out on a fingerprint of the molecule. That wikipedia page has a nice reference to this paper, although it is not open access. There are no doubt many resources out there about fingerprints/Tanimoto, even open source implementations...

This also looks interesting: "Why is Tanimoto index an appropriate choice for fingerprint-based similarity calculations?".


Amount of Cl- needed to precipitate AgCl given Ksp and [Ag+]

First, solve for [$\ce{Cl-}$] using the expression for K$_{\mathrm sp}$ and the given molarity for $\ce{Ag+}$:

$$K_{\mathrm sp} = 1.8\times 10^{-10} = [\ce{Ag+}][\ce{Cl-}]$$

which rearranges to

$$[\ce{Cl-}] = {1.8\times 10^{-10}\over 7.2\times 10^{-5}} = 2.5\times 10^{-6}\, \mathrm{M}$$

Note that the units for K$_{\mathrm sp}$ are actually $\mathrm{M}^{2}$: I have omitted that from the math above but dimensional analysis tells us that the answer for $[\ce{Cl-}]$ is in units of molarity ($\mathrm{M}$).

Because we have 100 $\mathrm{mL}$ of $\ce{Ag+}$ solution, divide the $[\ce{Cl-}]$ molarity by 10 to get the actual number of moles of $\ce{Cl-}$ needed (here, I multiply by 0.1 to achieve the same result):

$$\mathrm{mol}\,\ce{Cl-} = (2.5\times 10^{-6}\,\mathrm{mol}\cdot\mathrm{L}^{-1}) \times 0.1\,\mathrm{L} = 2.5\times 10^{-7}$$

We are given a solution that is 0.001 $\mathrm{M}$ $\ce{Cl-}$, so we can get the volume needed in liters:

$${2.5\times 10^{-7}\,\mathrm{mol}\,\ce{Cl-}\over 10^{-3}\,\mathrm{M}\,\ce{Cl-}} = 2.5\times 10^{-4}\,\mathrm{L}$$

Finally, convert the above quantity to $\mathrm{mL}$:
$$\left({1000\,\mathrm{mL}\over\mathrm{L}}\right)\cdot\,2.5\times 10^{-4}\,\mathrm{L} = 0.25\,\mathrm{mL}$$


Disregarding negative externalities and other barriers, what are some fuels that have a higher power density than gasoline?

I would suggest you take a look at what they use as rocket fuel.  There are a lot of fuels with higher power densities but they wouldn't work well on the highway.

The Navy has had great success with powering vessels on nuclear energy.  That is the ultimate top of the viable fuel density list.  Maybe one day fusion will become viable, then it will be at the top of the list.

As for the second part of your question, the characteristics of chemical fuels have to with free radicals.

Any viable fuel needs to be readily split into component parts (triggered by a spark or sudden compression or the like) that then combine in other configurations in an exothermic reaction so you get a net energy output.


How to estimate HCN content in a volatile mixture?

From a quick search of the literature, it seems that as you correctly thought, gas chromatography mass spectrography is a method that can be used to quantitate small amounts of cyanide in vapour.

While I am certainly no expert on this field or the precise method that you would use, this paper, which describes quantitation of cyanide and other chemicals in tobacco smoke may be of interest: A simple GC–MS technique for the analysis of vapor phase mainstream cigarette smoke


Each liquid produces some vapours. When you leave wet clothes, they eventually dry. Same here. It is true, that vapour pressure for mercury at room temperature is extremely low, but since mercury accumulates and has quite high toxicity, it is still a trouble. Same for other liquids.

Mercury must be kept in sealed container and/or under layer of water. Any works with it must occur in large pan, so no tiniest drop of mercury escaped. Escaped drops must be collected using thin copper plate, washed in concentrated nitric acid. All solutions of mercury compounds and remaining mercury should be utilized according to local procedures (there is an agency for it at my city). Metallic mercury may be transformed into salts using $\ce{FeCl3}$ concentrated solution. While not removing the mercury, this transforms it into compounds that does not produce vapours. However, in case of mercury leak a special agency must be informed to perform all necessary procedures. Consult local regulations on the cases.

Dichloromethane has regrettably low boiling point. It must be kept away from heat sources and in case of high temperatures it must be cooled or kept in sealed container. It evaporates swiftly on open air, carrying away much heat.

All the liquids should be used under fume hood. In case one is unavailable, an acceptable alternative is open air as long as there is a gentle wind. The chlorinated hidrocarbons mentioned has strong odor, strong enough to make handling them without good ventilation impossible. None of the liquids should contact skin. It is not immediately fatal, but still. In case the contact occurred, the chloroalkanes will evaporate quite quickly, though washing skin with soap and water is highly encouraged. Again, one-time breathing-in of the vapours is not fatal with this compounds, but repeated breathing-in guarantees health problems.

Good thing: none of the liquids is flammable, so flammability is not a trouble here. Unfortunately, heating them at presence of air is still highly unrecommended as well as continuous exposure to sunlight: it may produce phosgene.


Palin, D. E.; Powell, H. M. "Hydrogen Bond Linking of Quinol Molecules," Nature  156, 334 (1946), and DOI: 10.1039/DF9694800181

White aqueous starch/iod$ide$ tests for oxidizing agents by turning brown, liberated iodine and ionic oligomers intercalated within amylose starch (linear).  Amylopetcin (branched) gives red.  The mix is purple-brown.  Purple-brown aqueous starch/iod$ine$ tests for reducing agents by the reverse.


Understanding how electrons are counted for lone pairs vs. bonds especially when understanding resonance?

Count 2 electrons per bond and you'll be fine. The partners of a bond usually share two electrons in their mutual orbitals. While you are correct in thinking 2 divided by 2 is 1, thats not how the octet rule works - count them all.

Left w/ dbl bond: 4 lone pair electrons + 4 bond electrons = 8.

Left w/o dbl bond: 6 lone pairs electrons + 2 bond electrons = 8.

Octet rule satisfied.


Which chemical is responsible for making marker ink permanent?

The "permanence" is just a reference to it being waterproof.

  In general, the ink comprises a main carrier solvent, a glyceride, a pyrrolidone, a resin and a colorant.

This is mostly the result of solubility. There are other markers sold on the market with labels such as "buff proof" (slang for removal proof), that are really just pens that use a different carrier solvent. With those, typically organic solvents would be the correct solvent for removal.


Is the equation dG = Vdp − SdT valid only for a reversible process? Can it be applied for an irreversible one too?

The easiest way to avoid confusion is to start from the expression for the $pV$ work associated with a reversible process:

$$\mathrm dw_\mathrm{rev} = -p_\mathrm{ext}\,\mathrm dV = -p\,\mathrm dV$$

where $p$ is the internal pressure of the system, and here the system is subjected to $pV$ work only. Yes, it is absolutely true that for reversible $pV$ work the condition $p_\mathrm{ext} = p$ holds. While you came to that conclusion, you did so in a flawed way because your expressions for $\mathrm dH$ are not correct.

The above definition of reversible $pV$ work leads to the expression for $U$ in terms of $pV$ work and entropy $(\mathrm dU = -p\,\mathrm dV + T\,\mathrm dS),$ and ultimately, through the definition $H = U + pV,$ to the expression $\mathrm dH = V\,\mathrm dp + T\,\mathrm dS.$

  But my teacher argued that because this is a state function, it should be valid for all types of processes.

The laws of thermodynamics hold irrespective of path. Provided the end points are the same, two different paths will result in the same changes in state variables.

In practice, when attempting to determine the change in a state variable by evaluating a path integral, you want to choose a convenient path. This is usually a reversible one because then the work can be related to equation of state variables, and the entropy change can be related to the heat via the second law.


Why must basis sets used in computations be non-orthogonal?

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


Pressure dependency in Haber Bosch ammonia synthesis

According to Le Chatelier's principle, a pressure increase should result in the equilibrium shifting to the right, but it should not - at least to first approximation - affect the equilibrium constant.

$$K = \frac{\ce{[NH3]}^2}{\ce{[H2]}^3 \ce{[N2]}}$$

Using partial pressures, this can be rewritten as

$$K = \left(\frac{RT}{P}\right)^2 \frac{y_{\ce{NH3}}^2}{y_{\ce{H2}}^3 y_{\ce{N2}}}$$

Then for $K$ to be constant, an increase in the denominator, given by the pressure for example, has to be followed by an increase in the numerator, which means the reaction working in the direction of formation of ammonia.


Definitely you need to remember it. Potassium Chromate  is IUPAC name of $\ce{K2CrO4}$. There are other names such as Chromic acid and dipotassium salts. But mostly IUPAC names are accepted.

I don't think you should try to mug up all 24 names instead you should keep getting familiar with them when they come in reactions or somewhere else. After seeing them at few different places you will be remember them all. They will come very often while you are learning chemistry.


In Rutherford's experiment to show the existence of nucleus in an atom, the alpha-particles were exposed on the surface of certain metal i.e. gold. He observed that more than 99% of these particles were able to go straight and pass through the gold atoms. Thus he was able to conclude that atoms are mainly empty space.
But why did these particles not collide with the electrons present in the gold atoms? I think that the reason behind it is that the speed of the electrons travelling in orbits did not cause any obstacle in the travelling path of the alpha particles.


Mixing of powders to make stock solutions

Once the powders are mixed, it is unlikely that they will separate unless the container is subjected to vibrations.

As for mixing, companies that need to mix batches of material use a mixer to make their batches. The second photo in the article is similar to what I have seen.


What physical properties are manifested when an electron leaves Na and moves to Cl to form an ionic bond?

The light and heat are created by the exothermic nature of the reaction.  If you look up "Born Haber cycle sodium chloride" you will see information on why the reaction is exothermic.

The reaction "needs help" because of the fact that the sodium will likely be coated with a layer of sodium oxide or hydroxide that will not allow chlorine to collide with the sodium metal.  Even then, with pure sodium metal the reaction will be faster if heated as any reaction needs activation energy to proceed.  Adding water causes an exothermic reaction that can provide activation energy for the reaction with chlorine.


How can I know when I have added an excess of magnesium oxide?

$\ce{MgO}$ is insoluble in water (solubility: 0.0086 g/100 mL). On the other hand, $\ce{MgSO4}$ is soluble in water (solubility: 25.5 g/100 mL).

The reaction that is taking place is an acid-base reaction:

$$\ce{MgO + H2SO4 -> MgSO4 + H2O}$$

Now for your question, how would you know when excess $\ce{MgO}$ is present? Here you are trying to make $\ce{H2SO4}$ the limiting reagent. When all the $\ce{H2SO4}$ has reacted, no more $\ce{MgO}$ can further react. Therefore, you will see solid in your reaction mixture that cannot be further dissolved.

At the end of your reaction,

$$\ce{MgO_{excess} + H2SO4 -> MgSO4 + H2O + MgO_{remaining}}$$


Your definition of enantiomer is correct.

This simple further classification might be helpful.  Let's take 2 molecules with the same molecular formula.  These molecules are either constitutional (or structural) isomers or stereoisomers.  If they have different structures (or connectivity patterns)  and are separated by an energy barrier, then they are constitutional isomers.  If they have the same structure or connectivity pattern, then they are stereoisomers.

Among stereoisomers there are 3 classes that we can place a pair of molecules into.  Take any two stereoisomers and compare them, they are either

the same (2 identical molecules)
mirror images of one another (enantiomers)

If they are not the same or enantiomers, then they are

different, and we call them diastereomers.


it would actually produce a faster reaction to have the $\ce{LiOH}$ in a aqueous solution rather than having it as a solid. all you would do is bubble the $\ce{CO2}$ through the solution and it will go cloudy. you've probably done a similar thing but with $\ce{Ca(OH)2}$ to form calcium carbonate as a test for $\ce{ CO2}$.


According to the IUPAC, the is not such thing as groups IV A and B anymore.
All the elements you are analyzing belong to the 14th group of the periodic table, each one in one period, from the 2nd to the 6th.

In regards to your other question, you need to look for a trend on the elements of the group.
What happens as you go down the group?
I think you've already figured out that the number of electrons around the nucleus significantly increases, and this has an important effect on the radius of the atom.


Chemistry of solids: allowed wavevectors

In a crystal, electron waves are reflected off the periodic lattice and this (Bragg) reflection is the cause of the energy gaps also called a band structure. If there is a weak periodic potential in a 1D crystal then the energy gaps occur at wavevector $k=\pm \pi/d$ where d is the lattice constant.

If $\bar G$ represents the reciprocal lattice vector then the general Bragg condition for diffraction of a wave with wavevector $\vec K$ can be written as
$(\vec k + \vec G)^2=k^2 $

In one dimension this condition becomes

$$k=\pm \frac{1}{2}G = \pm n\pi/d$$

and $G=2\pi n/d$ is a reciprocal lattice vector and $n=1,2,3..$  an integer. This is where your factor of 2 appears.

The reflection at $k=\pm \pi/d$ occurs because one wave is reflected of an atom and interferes constructively with phase difference $2\pi$ with one reflected from the nearest neighbour atom. The Brillouin zone is the $\bar k$ space between $-\pi/d$ and $+\pi/d$.

(The reciprocal lattice vectors have the form $\vec G =h\vec A+k\vec B + l\vec C$ for integers h, k, l. The vectors are defined as

$$\vec A= 2\pi \frac{b~ \times~ c}{a\cdot b~ \times~ c}~~
\vec B= 2\pi \frac{c~ \times~ a}{a\cdot b~ \times~ c}~~
\vec C= 2\pi \frac{a~ \times~ b}{a\cdot b~ \times~ c}$$
for primitive lattice vectors a, b, c)


Theory behind Experiment: Extraction and Identification of DNA

1) Sodium dodecyl sulfate (SDS) is a solid anionic detergent. It can solubilize proteins and lipids that frame the cell membranes by degrading the cell (the proteins from the cell membrane get damaged and cell gets broken) and nuclear membranes that protect the DNA. This will help the cell membranes to separate and expose the chromosomes that contain the DNA. SDS can also release the DNA from histones (a group of basic proteins found in chromatin) and other DNA binding proteins by denaturing them. Breaking the nuclear membrane exposes the DNA to reactive and possibly dangerous chemicals in the cell. To slow down these chemical reactions, which might damage or break up the DNA, you must use cold ingredients and add cold alcohol. Also, it is also known fact that the colder the alcohol is, the more DNA it will extract.

2) Once the DNA has been released, adding papain extract helps untangle and unfold the DNA from the other parts of the cell. The enzymes in papain break down protein attached to (or tangled with) DNA molecules, making them free from proteins.

3) DNA is not soluble in absolute alcohol but soluble in water. If you pour alcohol rapidly it will mix with water and you may not have DNA precipitation. Therefore, you should slowly pour ethanol (or isopropanol) down the inside of the test tube with a Pasteur pipette, so alcohol layer build up on the extract.

4) DNA is soluble in water, but not in salty water. All the other cell contents are also soluble in water and salt water. This difference in solubility allows you to separate DNA from the rest of the cell material. When the DNA leaves the salty water mixture, it comes into contact with the alcohol but cannot dissolve (DNA is not soluble in alcohol either). When DNA comes in contact with alcohol, it uncoils and precipitates. All the other cell materials stay in the salty water solution. Note that an uncoiled piece of DNA can be $\pu{2.8 inch}$ ($\pu{7.2 cm}$) long, and visible when few of them tangled each other.

5) Diphenylamine will react with deoxyribose sugar in an appropreate condition to give blue-colored complex. This is te principle underlying estimation of DNA using diphenylamine. This is called Dische Diphenylamine Test For DNA:

One of the acid degraded products, $\omega$-hydroxylevulinylaldehyde, under acidic conditions, reacts with diphenylamine to produce a blue-colored complex that absorbs at $\pu{595 nm}$. The intensity of the blue color of this complex is proportional to the concentration of DNA.

About using UV-vis-Spectroscopy to determine the purity: The fact is that nucleic acids have an absorption maximum at $\pu{260 nm}$ UV wavelength. However, this absorbance suffers from interference from othe molecules such as single nucleotides, proteins, and single-stranded nucleic acids (as well as RNAs). Thus:
DNA concentration (in $\pu{μg}$ DNA/$\pu{mL}$ Solution) $= (A_{260} - A_{320}) \times 50\times$ dilution factor where $50$ is DNA extinction coefficient, $A_{260}$ and $A_{320}$ are absorbance of DNA at $\pu{260 nm}$ and $\pu{320 nm}$, respectively. In general, since the DNA has limited absorbance at $\pu{320 nm}$, this is simplified as:
DNA concentration (in $\pu{μg}$ DNA/$\pu{mL}$ Solution) $= A_{260}  \times 50\times$ dilution factor

Yet, the DNA purity is determined with the help of software. This is done by measuring the absorbance of the DNA against protein:

$$\text{DNA Purity} = \frac{(A_{260} - A_{320})}{(A_{280} - A_{320})}\approx \frac{A_{260}}{A_{280}}$$

If $\frac{A_{260}}{A_{280}}$ is in the range of $1.7-2.0$, the content of the solution is considered to be pure DNA. In other words,, constructing an absorbance ratio between these two absorbance wavelengths can provide an estimate of sample purity. As a general rule any preparations with a $\frac{A_{260}}{A_{280}}$ ratio greater than approximately $1.7$ is called pure.

Sources:

http://www.biotech.iastate.edu/publications/lab_protocols/DNA_Extraction_Onion.html

https://openlab.citytech.cuny.edu/bio-oer/chemistry/biologically-important-macromolecules/nucleic-acids/nucleic-acids-dna-extraction-and-disches-diphenylamine-test-activity/

https://www.news-medical.net/whitepaper/20140203/DNA-Analysis-with-UV-Visible-Microspectroscopy.aspx

https://www.bmglabtech.com/fileadmin/06_Support/Download_Documents/Application_Notes/AN263.pdf


Evaluate indicator choice (phenolphthalein)

Easy marks for phenolphthalein usability are:

$99.9\%$ of acetic acid is titrated at about
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{1000} = 7.75$$

$100\%$ of acetic acid is titrated at
$$\mathrm{pH} = 7 + \frac12 \cdot (\mathrm{p}K_\mathrm{a} + \log c) = 9.385 + \frac12 \log c$$

Phenolphthalein color transition is $8.2-10.0.$


Hydrogen and Helium ideal gas behavior in room temperature

For inert gas molecules, intermolecular interactions would be by Van der Waals forces, which are based on the total volume and polarizability of the gas molecule.

The diatomic hydrogen is larger and more polarizable of the two.


The activity series summarizes the experimental behavior of metals and their cations. A more active metal will displace the cation of a less reactive metal from solution.

Let's consider the case of zinc and copper. Zinc is higher on the activity series than copper. Zinc will displace copper ions from solution:

$$\ce{Zn(s) + Cu^2+ (aq) -> Zn^2+ (aq) + Cu(s)}$$

If we place a strip of zinc metal into a solution containing $\ce{Cu^2+}$ ions, the some of the zinc will dissolve into the solution, while some of the copper will plate onto the zinc.

What happens if we put a copper strip into a solution of $\ce{Zn^2+}$? Very little. Copper is less active than zinc. Copper metal will not displace zinc ions from solution.

$$\ce{Cu^2+ (aq) + Zn(s) -> No\ reaction}$$

If you did not know which was more active, copper or zinc, these two experiments would give you the information you need. Zinc displaces copper from solution. Copper does not displace zinc. Zinc is more active than copper $(\ce{Zn>Cu})$.

How can you apply this logic to your question?


Is there any difference between isotonic and isosmotic solution?

Yes, there is a difference. Isotonicity implies a biological compatibility, whereas isoosmoticity implies similarity of chemical and/or physical composition. Solutions that are isoosmotic to biological fluids/blood are not necessarily isotonic as tonicity refers to a given cell membrane [1, p. 230]:

  The 0.9% sodium chloride solution is said to be iso-osmotic with physiological fluids. In medicine, the term isotonic, meaning equal tone, is commonly used interchangeably with iso-osmotic. However, terms such as isotonic and tonicity should be used only with reference to a physiological fluid. Iso-osmotic actually is a physical term that compares the osmotic pressure (or another colligative property, such as freezing-point depression) of two liquids, neither of which may be a physiological fluid, or which may be a physiological fluid only under certain circumstances. For example, a solution of boric acid that is iso-osmotic with both blood and lacrimal fluid is isotonic only with the lacrimal fluid. This solution causes hemolysis of red blood cells because molecules of boric acid pass freely through the erythrocyte membrane regardless of concentration. Thus, isotonicity infers a sense of physiological compatibility where iso-osmoticity need not. As another example, a chemically defined elemental diet or enteral nutritional fluid can be iso-osmotic with the contents of the GI tract, but would not be considered a physiological fluid, or suitable for parenteral use.

References

Gennaro, A. R. Remington: The Science And Practice Of Pharmacy, 21st ed.; Lippincott Williams & Wilkins: Philadelphia, 2005. ISBN 978-0-7817-6378-3.


Like Dr J said there is no single polycarbonate polymer, the most common one is formed from phosgene (or another carbon dioxide reagent) and bisphenol A. The repeating unit is $\ce{C6H4C(CH3)2C6H4-O-C(=O)-O -}$

In Nagahata, R.; Sugiyama, J.; Goyal, M.; Goto, M.; Asai, M.; Ueda, M.; Takeuchi, K. X-ray analysis of the bisphenol-A-type macrocyclic carbonate trimer. Polym. Adv. Technol. 2000, 11 (6), 294–300. DOI: 10.1002/1099-1581(200006)11:6<294::AID-PAT979>3.0.CO;2-C. (OCAMUA) a tetramer of the repeat unit of this polymer has been reported. This is a small cyclic polymer (oligomer).


is there any hydrophobic material to be used on paper cups?

The paper cups I have seen already have wax on the inside to prevent water from soaking the paper.


Acid-Base Reaction in MacGyver Episode "Countdown"

The answer to your first question is yes, to the second maybe.

You are right that milk can act as a buffer, as explained for instance by this page from the Food Science Department of the University of Guelph:

  This leads to the concept of buffer capacity, which is an important principle in cheese making. The effect of protein removal on the TA of whey, is related to the ability of protein to 'buffer' the milk against changes in pH. That same buffer property is the reason it helps to take acidic medication, like aspirin, with milk.
  [...]
  The two most important buffer components of milk are caseins (buffer maximum near pH 4.6) and phosphate (buffer maxima near pH 7.0). The buffer maximum near pH 5.0 is extremely important to cheese manufacture because the optimum pH for most cheese is in the range of 5.0 - 5.2. As the pH of cheese is reduced towards pH 5.0 by lactic acid fermentation, the buffer capacity is increasing (i.e., each incremental decrease in pH requires more lactic acid). The effect is to give the cheese maker considerable room for variation in the rate and amount of acid production. Without milk's built in buffers it would be impossible to produce cheese in the optimum pH range.

On the other hand denaturation of the milk proteins might act to form a gel that could perhaps help stabilize the trigger fluid, but milk gelation appears to display a complex pH dependence and a mechanism of stabilization is unclear.

The electrolyte concentration might be diluted by milk, since its ionic strength is surprisingly moderate, see Ref (1) for instance:

  Taking into account these different associations, the diffusible fraction of milk at pH 6.6–6.7 appears to be supersaturated in calcium phosphate (Tab. V), and to have an ionic strength of about 80 mM.

Much of the salt is bound to protein which forms part of the buffering system. So much depends on the composition of the sinister trigger fluid.

The substantial concentration of divalent calcium (~1 g/L or ~50 mM) might be useful, but how will presumably also remain a well-kept secret of the writers privy to the composition of the very unstable trigger electrolyte fluid.

Reference

Frédéric Gaucheron. The minerals of milk. Reproduction Nutrition Development, EDP Sciences,2005, 45 (4), pp.473-483.


C14 Nuclear Diamond to AA Battery Apples-to-Apples Engery output

Let me address this part of your question first:

  "[...] how many Joules per gram would a C14 nuclear diamond produce if it weighed 20 grams"

I think it's important to understand that the bolded/italicized phrase is redundant, because your are discussing (as did the article) the energy output per gram.  So the per-gram output of a 20 gram battery is the same as that of a 1 gram battery.  The 20 gram battery would have a total output of 14000 Joules in it's lifetime, where the one gram battery would have a total output of 700 Joules.  The fact that the article, and your question, normalize to a per-gram basis is how they've maintained an apples-to-apples comparison.

The semantics get tricky when they talk about the mass of the diamond battery, and by my thinking this is where they may fail to give an apples-to-apples comparison.  The diamond would not be pure $\ce{C^{14}}$, and at a minimum would contain a thin protective coating of non-radioactive $\ce{C^{12}}$ (protecting those on the outside from the radiation on the inside).  So when they say "A diamond beta-battery containing 1 gram of $\ce{C^{14}}$", it is not made clear what the total mass of diamond would be (e.g. we aren't given the ratio of $\ce{C^{14}}$ to the $\ce{C^{12}}$ that would also be present).

That said, I think that the apples-to apples answer to your question, based on the information given in the article, is as follows:

Conventional 20 gram battery: 14000 Joules/day, for 1 day.
Nuclear 20 gram $\ce{C^{14}}$ battery: 300 Joules/day, for 5,730 years.

or, equivalently:

Conventional 1 gram battery: 700 Joules/day, for 1 day.
Nuclear 1 gram $\ce{C^{14}}$ battery: 15 Joules/day, for 5,730 years.


There are 1,000,000 mg in a kg (this is, mg/kg = ppm), so all you need to do is divide the mass of $\ce{Hg}$ by the mass of the person as-is (the units work out to ppm):

$$\left({30\,{\rm mg}\over 63.5\,{\rm kg}}\right)\cdot\left({{\rm ppm}\over {\rm mg\cdot kg^{-1}}}\right) = 0.47\,{\rm ppm}\,\ce{Hg}$$


An equivalent is the factor of the number of molecules you need, at minimum, to carry out a reaction. The number of equivalent depends on your choice of what you consider is your main reactant.

Simply put, if you take a chemical equation where A + B → C, you need 1 eq. of B compared to A, i.e. you need one molecule of B for each molecule of A.

But sometimes, you can get something like 2A + B → C, and you want to know the "number of equivalent of B you should add to A". The quantity of B is one half that of A. If you consider your main reactant to be A, then you need to add only 1/2 (aka 0.5) equivalent of B.


I just thought I'd add a visualisation of this concept. Electrostatic maps show the electrostatic potential at a given point of the molecule. We know water is very polar, and this is evident from its electrostatic map:

If we look at octanol though, we see that the vast majority of the molecule is non-polar (evenly distributed electrostatic charge):

The hydroxy-group of octanol is indeed hydrophilic (polar), but the majority of the molecule wants to stay in the non-polar phase.

Molecules with strongly hydrophilic "heads" and non-polar tails have unique properties in aqueous solutions, including some very important for life. Take phospholipids. In water, they self assemble into the basic structure of the cell membrane, the so-called phospholipid bilayer:

You can see how the phospholipids arrange themselves to have their polar portion in contact with water (hence "hydrophilic" - "loving water"), while their non-polar portion is in contact only with the non-polar portions of the surrounding phospholipids (hence "hydrophobic" - "fearing water").


What is the mechanism for the oxidation of benzene to maleic anhydride using vanadium pentoxide?

A talk given by Ken Waugh named The Detailed Kinetics of the Partial Oxidation Reactions: Benzene/Butane to Maleic Anhydride, Ethylene to Ethylene Oxide,[1] proposes the following mechanism on page 4:

[1]: Acessible as pdf via rsc.org


How can I- be both a nucleophile and a weak base?

As Jori pointed out, nucleophilicity is a kinetic property while basicity is a thermodynamic property. Nucleophilicity in part is about the rates of chemical reactions; thermodynamics is in part about the favorability of chemical reactions.

A reaction may be very favorable, i.e. it has a very negative free energy change, but it might take a long time to actually complete. For example, the rusting of iron is favorable from a thermodynamic standpoint. The products, iron oxides, are very stable. However, the process is relatively slow from a kinetic standpoint; putting water on iron isn't going to instantly result in a familiar reddish-brown hue. It might take a day or so.

On the other hand, if you squirt something such as tert-butyllithium into the air, you'll probably catch yourself or something on fire. This is a carboanion salt. Carboanions are strong (Bronsted, and therefore, Lewis) bases. Acid-base reactions also tend to be relatively fast (kinetics) since they involve the transfer of something so small - a hydrogen proton. A graduate student unfortunately exposed some of this carbocation salt to the air and the resulting fireball instantly caught her sweater on fire. The carboanion salt likely grabbed protons from atmospheric moisture - a very favorable process from a thermodynamic standpoint - and kinetically, a very fast process. And so the resulting heat likely ignited the resulting protonated carboanion (alkane in this case) on fire.

In your case, $\ce{I-}$ is a good nucleophile in both protic and non-protic solvents. It's a rather big ion - the biggest of the halide anions. That means its negative charge is relatively dilute. This in turn means it's not as strongly solvated by the solvent molecules; it doesn't have as many tenacious solvent molecules hindering its trek toward a carbocation. So $\ce{I-}$ can reach carbocations relatively quickly. Note that terms dealing with time usually indicate that the writer is speaking about a chemical's kinetic properties.

For the same reason that $\ce{I-}$ is a good nucleophile it's a poor base. It's dilute negative charge (due to its size) means that $\ce{H+}$ has better things to do than be attracted to $\ce{I-}$. $\ce{I-}$ has a dilute amount of negative charge so it cannot effectively stabilize the hydrogen proton as well as many solvents might - especially water, which extensively associates with $\ce{H+}$; $\ce{H3O+}$ only begins to describe how well the proton is associated with water molecules; the proton has been found to be associated with 2, 3, 4, or more water molecules!


What's a good measure of the Eigen-ness of a wavefunction?

$\hat{F}\psi_i=\varepsilon_i \psi_i \implies \langle\phi_k|\hat{F}|\psi_i \rangle = \varepsilon_i\langle\phi_k|\psi_i \rangle$.  So the Fock matrix $F_{ki}$ can be approximated by the product of orbital energy $\varepsilon_i$ and the overlap matrix $S_{ki}$. The closer these two values are, the higher the "Eigen-ness".

Of course $\phi_k$ and $\psi_i$ have to be non-orthogonal, so $\phi_k$ could be, for example, a basis function or an MO for another molecule.  I've used this to approximate intermolecular Fock matrix elements here.


Experimental melting-point of Arm-and-Hammer sodium bicarbonate is remarkably high(?)

See the MSDS for Arm and Hammer Baking Soda here, from the manufacturer (Church & Dwight Co.) (it will link to a PDF) in which it is stated:

CONDITIONS TO AVOID: Temperatures above 65°C (150°F).
INCOMPATIBILITY WITH OTHER MATERIALS: Reacts with acids to yield carbon dioxide. May also yield free caustic in presence of lime dust (CaO) and moisture (i.e., water, perspiration). Dangerous reaction with monoammonium phosphate or a sodium-potassium alloy may occur.
HAZARDOUS DECOMPOSITION PRODUCTS: Heating above 100°C may cause dangerous levels of carbon dioxide gas to be present in confined spaces. Yields sodium oxide if exposed to temperatures above 850°C. Avoid inhalation, eye and skin contact with sodium oxide.

The ingredients are listed as 100% sodium bicarbonate with CAS number 144-55-8.
Your choice of thermometer is questionable in the present case. See this Wikipedia entry which reliably reports that the upper limit on temperature measurement from an alcohol-based thermometer is 78 $^{\circ}$C. A mercury thermometer has a higher upper temperature range (see this Wikipedia article) and is preferable to the alcohol thermometer. Alternatively, use a gallium thermometer or a digital model with an operating temperature range that suits your needs.
You ask:

I've heard rumors that some baking-soda products contain aluminum and this has raised some alarm about ingestion hazards. Is this valid?

Probably not in the case of the specific brand in question here, based on the MSDS.


Why is -OCH3 more strongly activating than -CH3 in electrophilic aromatic substitution?

Since we are talking about the rates of these reactions, we really want to examine the structures of the transition states in the rate determining step in these reactions, rather than the starting materials. For electrophilic aromatic substitution, there is a high energy carbocation, the Wheland intermediate, formed when the benzene derivative attacks the electrophile. Formation of this intermediate is normally the rate determining step. Since this carbocation is high in energy, the transition state leading to it is going to be similar in structure, according to the Hammond Postulate. So, looking at the influence of varying substituents on the stability of these Wheland intermediates is a good approach to justifying differences in reaction rates (not to mention selectivities).

If we take the resonance theory approach, we can draw many resonance structures for eg. the para-substitution intermediate, putting the positive charge ortho and para to the "E" substituent we just added. Looking at the methyl-substituted benzene substrate (toluene), resonance theory would allow us to draw an additional "no bond" resonance structure, where we break the C-H bond and have a proton associated with a neutral methylidene cyclohexadiene. This is the resonance theory approach to hyperconjugation. This resonance contributor can justify why toluene is more reactive than benzene. However, how important is the contribution of a resonance structure where we have one less σ-bond and a positive charge on a tiny hydrogen atom?

Looking at anisole (methoxybenzene) the same way, resonance theory would lead us to draw a resonance structure where the non-bonding (lone) pair of electrons on oxygen is used to make an additional π-bond and the oxygen ends up with the formal positive charge. It may not feel nice to put a formal positive charge on the oxygen, but this resonance contributor has a lot going for it - we have gained an additional π-bond relative to the other resonance structures and all non-hydrogen atoms satisfy the octet rule. The "loss" of the lone pair is no problem - as a non-bonding pair of electrons they were already relatively high in energy. For these reasons, this is an excellent contributor to the overall structure of this intermediate - probably the most important.

Taken together, and using resonance theory, the above can help justify the differences in reactivity. Another approach would be to use molecular orbital theory. Hyperconjugation would be viewed as overlap of the C-H σ-orbital with the adjacent carbocation π-system. The efficiency of this overlap is poor because there is a big energy difference between a low energy C-H σ-bond and π-system of the carbocation. In contrast the energy difference between the oxygen non-bonding (lone) pair and the π-system is relatively small and overlap is therefore efficient.


Does relative amount of lithium and cobalt oxide change in recharged battery?

About maximum $20~\%$ (as I remember, perhaps more now) of the $\ce{Li}$ goes (as $\ce{Li+}$) out of the structure of $\ce{LiCoO2}$, through the elctrolyte, and into the electrode on the other side of the battery. That electrode consists mainly of graphite, which intercalates the $\ce{Li}$ in it's layers in a quasi-metallic form.

So it is $\ce{Li_{1-x}CoO2 + x~Li}$ after charging. $\ce{LiCoO2}$ has a special structure that does not break down when some $\ce{Li}$ is missing, except if you keep a charged battery at over $\mathrm{50~^\circ C}$ or so.

By the way, the related wikipedia article tells you all this. Two days of researching? ;-)


Mechanism of initiation in contact explosives?

My guess would be cavitation, presumably 'non-inertial cavitation':

Non-inertial cavitation is the process in which a bubble in a fluid is forced to oscillate in size or shape due to some form of energy input, such as an acoustic field. Such cavitation is often employed in ultrasonic cleaning baths and can also be observed in pumps, propellers, etc.

The energetics of cavitation are astonishingly out-sized with respect to the tiny size of bubbles, probably leading to more than enough energy to cross the initial activation barrier of the combustion reaction.  As one example, the abstract here summarizes other reports' results indicating that the temperature inside collapsing cavitation bubbles may reach between 10,000 K and 100,000,000 K.
I don't have access to the full text to determine whether or not they met with success in attempting to ascertain cavitation's role in nitroglycerin initiation, but these authors apparently investigated the question almost fifty years ago.  The abstract (references excised):

It is known that in some circumstances liquid explosives can be initiated with unexpected ease, and in other circumstances only with great diffculty.
Thus, Winning has shown that nitroglycerine (NG) free of gaseous inclusions and poured into a vessel so as to leave no wall surfaces free of liquid is not exploded even by the action of a fairly strong shock wave from a detonator immersed in the NG.
On the other hand, in handling liquid explosives there have been quite a few cases in which relatively weak vibrations or impacts have led to unexpected explosions, which have sometimes had serious consequences. For example, a British report describes an unfortunate accident that resulted from dropping a polyethylene bottle containing NG. Upon hitting the ground the NG exploded.
The initiation of explosion by “hot spots” resulting from the adiabatic compression of gaseous inclusions even before the arrival of the shock wave has been reliably demonstrated in numerous experiments. However, some cases of initiation of liquid explosives simply cannot be attributed to the heating of such gaseous inclusions, since in these cases the adiabatic compression temperatures of the gas are so small that it is not possible to talk of a “hot spot. “Such puzzling cases include, for example, the above-mentioned explosion of NG in a polyethylene bottle. In other experiments the role of gaseous inclusions has been completely eliminated by first subjecting the liquid explosive to a constant high pressure. This so reduced the degree of compression of the gaseous inclusions by a weak shock that strong heating of the gas in the bubbles, if any were present in the liquid explosives, was completely excluded. Nonetheless, there was no reduction in the sensitivity of the explosive to weak shocks.
In attempting to explain such puzzling cases it is usually pointed out that explosion can be initiated by cavitation, which may develop in a liquid even as a result of a weak impact or vibration. So far, however, no one has offered any direct experimental evidence of the possibility of cavitational initiation of explosion in liquid explosives. The object of our research was to fill that gap.

Clearly they published their results, so I figure the odds are good that they reached some sort of affirmative conclusion.


Elution Rate - Alkanes VS Alkenes in Gas Chromatography

Chromatography in general is a separation method in which the partition equilibrium of a compound between two phases is permanently disturbed since one of the phases is changing/exchanged (mobile phase).

Apparently, the mobile phase in gas chromatography is the carrier gas.
The stationary phase isn't just a hollow tube but a capillary covered with various differently functionalized materials, that may be rather unpolar (squalane), more polar (polyethylene glycol) or polar and shape-selective (cyclodextrines). Other very common materials are alkyl-substituted polysiloxanes, such as polydimethylsiloxane.

Using the latter stationary phase, Alan Katritzky and coworkers published QSPR: the correlation and quantitative prediction of chemical and physical properties from structure, Chem. Soc. Rev., 1995, 24, 279-287 (DOI, PDF). Here, the retention times for dodecane and 1-dodecene were given as 4.59 and 4.45 min, respectively.

Ladislav Sojak published a freely available 35-page review Separation and identification of isomeric hydrocarbons by capillary gas chromatography and hyphenated spectrometric techniques in Petroleum & Coal, 2004, 4(3), 1-35 (PDF). You might want to have a look at this article on the effect of having non-terminal $\ce{C=C}$ double bonds and cis-trans isomers of longer alkenes, which therefore have significantly different shapes.


Should I include solvation in quantum chemical calculations of a simplified protein-ligand interactions?

This is difficult to answer without full knowledge of the system, but some tips:

1) The overall energy of the system is arguably irrelevant.  Rather, the relative difference between different conformations, and perhaps, different ligands is of utility.

2) The level of theory and basis set your using is quite low.  If you have the time and the resources, scale it up.  For biomolecules I've found the minnesota functionals to be most accurate (if insisting on DFT) but I still believe MP2 is king.

3) Referring back to #1, if you solvate, then your comparison should too be solvated.  Perhaps your approach would be to run a solvated and unsolvated model for both your ligand, and perhaps, another known ligand. This could too help convince you of the utility of the solvation model.


Half-Life of Radioactive Isotopes: Why? How?

As commenters have pointed out, there is nothing about half-life that makes it special; we could as easily measure and tabulate one-tenth lives or one-fifth-lives.

Half-life has two marginal advantages: the half that has decomposed and the half that hasn't are equal, so whether we are referring to remaining or already decomposed nuclei is unambiguous, while for a third-life we could be confused as to whether we are talking about the time it takes for a third of the nuclei to decompose or for all but a third of the nuclei to decompose; and experimentally speaking, half is convenient because it's distinctly smaller from the initial amount of nuclei but large enough to be easily measured compared to the initial amount. However, these two advantages are fairly unimportant, and if we had decided to measure and tabulate one-tenth lives for all radioactive isotopes, we would be fine.

What's conceptually more important is that we can use half-lives as a proxy for isotopic disintegration rates, because radioactive decay follows first-order kinetics. Nothing prevents us from using half-life to describe any kinetic process, such as a chemical reaction rate. However, most of the time that's of limited value, because in general, half-lives are not constant, but change depending on the amount of reagents. It is only for first-order kinetics (i.e. when the amount of nuclei decays exponentially with time) that half-lives (or any fractional life) are constant.

For a general kinetic equation with only one reagent $\ce{A}$,

$$ -\frac{\mathrm{d} [\ce{A}]}{\mathrm{d} t} = k [\ce{A}]^n$$

where $n$ is the kinetic order of the process, we can always define the half-life of the process by integrating from an initial amount of $\ce{A}$, $N_\ce{A}$, to half that amount:

$$ \int\limits_{N_\ce{A}}^{N_\ce{A}/2} \frac{\mathrm{d} [\ce{A}]}{[\ce{A}]^n} = \int\limits_{0}^{t_{1/2}} - k\,\mathrm{d} t $$

which means that the half-life can be expressed in terms of the integral on the left side:

$$ t_{1/2} = - \frac{1}{k} \int\limits_{N_\ce{A}}^{N_\ce{A}/2} \frac{\mathrm{d} [\ce{A}]}{[\ce{A}]^n}  $$

This integral will in turn depend on the kinetic order $n$. While in general the half-life will take the form

$$ t_{1/2} = \frac{1}{k} \frac{1}{n-1} \left( (N_\ce{A}/2)^{1-n} - N_\ce{A}^{1-n} \right) $$

and therefore will be dependent on the starting conditions $N_\ce{A}$, and will change as $\ce{A}$ is consumed, for first-order kintetics ($n=1$)

$$ t_{1/2} = - \frac{1}{k} \left( \ln{|N_\ce{A}/2|} - \ln{|N_\ce{A}|} \right) = \frac{\ln{2}}{k} $$

so half-life does not change with the amount of $\ce{A}$.

As long as the decay is first-order (or exponential, which is the same), half-life is such a constant that we can measure and tabulate, as it will be the same under all possible circumstances; no matter if we are talking about tonnes of fissile material or a handful of nuclei, the value for half-life will be the same.

If we start with $N_\ce{A}$ radioactive nuclei, after a time $t_{1/2}$ you'll have half that amount, $N_\ce{A}/2$; since now you have $N_\ce{A}/2$, after another $t_{1/2}$ you'll have $N_\ce{A}/4$; after a third interval equal to $t_{1/2}$ you'll have $N_\ce{A}/8$, and so on.

This makes tabulating half-lives equivalent to tabulating kinetic coefficients $k$, as these can be transformed into each other quickly and unambiguously, while for other kintetic orders we'd need additional informations to perform that transformation.

Note as well that the last step also shows why we could arbitrarily decide to use any other fractional lives in the same way: if, for instance, we used one-tenth lives, we'd get

$$ t_{1/10} = \frac{\ln{10}}{k} $$


Yttrium -- rare earth or transition metal?

Comparing "transition elements" to "rare earth elements" (REE) is a bit like comparing apples and oranges. It all boils down to the different ways in which the elements can be classified.

Let's take a look at the periodic table:

(modified from this)

The rare earth elements are marked with a red outline.

Why are all rare earth elements are technically transition elements?

Strictly speaking, transition elements are the elements in groups 3 to 12 of the periodic table. The rare earth elements, in their widest sense, are Sc, Y and the lanthanides, La to Lu. You can see that if you group the lanthanides as one element, they fit nicely below Y in group 3 of the periodic table. The lanthanides are occasionally called the "inner transition elements".

Why is yttrium a rare earth element?

There are chemical and historical to the inclusion of yttrium in the rare earth elements.

Historically, yttrium was among the first elements to be isolated from the REE-bearing minerals of the Ytterby mine in Sweden. What was though as first to be "yttria" was later shown to contain also the rest of the heavy REE. It just took a longer time to purify them.

Chemically, yttrium commonly occurs with the other REE in the same mineral deposits around the world. You probably know the the chemical properties of the lanthanide series vary continuously from La to Lu. Yttrium also follows that pattern, and in terms of chemical properties it sits very nicely somewhere between Dy and Ho (that's why it's associated in mineral deposits with the heavy REE).

Yttrium is commonly used in the same high-tech applications as the other REE: it is commonly alloyed with other REE for speciality materials. It is also commonly used in combination with the other REE in lasers and various detectors in electron microscopy and analytical chemistry.

A different look

If you take a look at this figure, taken from here: https://earthscience.stackexchange.com/a/2951/725

You can see that most transition elements, as they appear in nature, usually have a cationic charge between 60 and 80 pm. The REE (including Y) are somewhat larger than that, resulting in slightly different properties.

You may also be interested in reading the answers to the following questions, also here on Stack Exchange:

What are rare earths and why do they cluster near alkaline magmatism?

Why are the rare earths erbium (Er), terbium (Tb), yttrium (Y) and ytterbium (Yb) named like that?


Reduction of N,N-dimethylbenzamide with LiAlH4

Here's the mechanism. The key is the second step, it's the lone pair of nitrogen that attacks the carbon, and not the lone pair of oxygen.

The reasoning is that the lone pair of nitrogen is more reactive because of the two electron-donating methyl groups attached to the nitrogen atom. The electron density is higher on the nitrogen atom than the oxygen atom.

The second reason is that the oxygen anion as a leaving group is much more stable than the nitrogen anion R2N- you would have gotten if the reaction led to your product.


Does single-layered Aluminium Phosphide exist?

I suspect that it does not.  As far as I know, all trivalent phosphorous is highly pyramidal--or at least not planar. This would rule out any graphene-like structure.

The reasoning that I have seen is based on hybridization. In the second row (third period) the 3s and 3p atomic orbitals are well separated, implying that s and p can not mix and that the outer valence is unhybridized. If bonding is with unhybridized p's, then bonding would occur at right angles.

(Kutzelnigg, W. Angew. Chem., Int. Ed. Engl. 1984, 23, 272.)

(Brown, E. Organometallics 2000, 19, 2208-2214.)


It's a bit of a blunt instrument, but I'd tend to just go straight for hydrogen peroxide if it's available where you are. It breaks up all sorts of organic dyes via oxidation, and it slowly decomposes by itself, especially in sunlight, so you can just rinse (thoroughly) and leave it to dry in the sun.

(NB: I have never owned a wine decanter, and take no responsibility if your decanter dissolves, explodes, or spontaneously gains the ability to sing operatically.)


Why are there distortions in electron cloud in molecules? How does increase surface area amplify these distortions?

The distortions are a quantum effect

Yes, branched chain hydrocarbons tend to have lower boiling points and this depends on their surface area. But the "distortions" in the electron clouds don't depend on the surface area.

The ultimate reason why there are "distortions" (more accurately described as temporary dipoles caused because electrons are not evenly distributed) is because of quantum effects. The uncertainly principle says we can't know the combined location and momentum of quantum things beyond a certain limit. When you work through the mathematics of this, this means that electron clouds always contain random fluctuations that appear as short lived temporary dipoles. The fluctuations in one molecule's cloud can be felt by the electron clouds in other molecules (and can cause second order effects that alter their electron clouds as well as interacting with their random dipoles). When you work out the (fairly complicated) mathematics of this you get a weak, short range, attraction between even non-polar molecules. We call those London Dispersion forces (or Van Der Waals forces).

But the underlying electron fluctuations don't change because of the shape of a molecule. However, the total force of all the fluctuations added up depends on the surface area of the molecule (how can they interact with other molecules except at the surface?) More surface, more opportunity for the fluctuations to interact with other molecules.

So it isn't that the distortions themselves depend on the surface area, it is that in a larger surface there is more opportunity to have them interact and hence the total force caused by the focus is larger.