How was the diatomic nature of many common gaseous elements originally determined?

Here I'll only provide the reasoning for $\ce{O2}$ and $\ce{H2}$ diatomic structure, I think similar reasoning was to deduce that of $\ce{Cl2}$.
The divalency of oxygen, shown by Avogadro's hypothesis.
It is known that two volumes of hydrogen gas and one volume of oxygen gas combine to form water. From Avogadro's hypothesis (equal volumes of gas contain equal numbers of molecules) it can be concluded that water must have two atoms of hydrogen and one atom of oxygen (or conceivably some multiple). Hence, oxygen is divalent (valence of 2) while hydrogen has a valence of 1:
$$\ce{2H + O \to H2O}$$
This would indicate that $\frac12$ gram of hydrogen atoms reacted with 8 grams of oxygen atoms. Since the
atomic weight of hydrogen was defined as $H = 1$, then the respective atomic weights of hydrogen and oxygen would be $H = 1$ and $O = 16$ (the values accepted today).
The diatomic nature of hydrogen and oxygen, as shown by Avogadro's hypothesis.
It is observed that when two volumes of hydrogen gas and one volume of oxygen gas combine, two volumes of water gas are formed. If hydrogen and oxygen were monatomic (as shown in the equation above), then only one volume of water would be formed. The only explanation is that hydrogen and oxygen are diatomic:
$$\ce{2H2 + O2 \to 2H2O}.$$
(Copied almost verbatim from this page)


There is no such thing as unified atomic radius. An atomic radius is a class consisting of van der Waals radii $R_\mathrm{vdW}$ (steric interactions), covalent radii $R_\mathrm{cov}$, and ionic radii $R_\mathrm{i}$ (and some other as well).

From the recent edition of CRC Handbook [1, p. 9-57]:

  $$
\begin{array}{llrr}
\hline
\text{Element} & \text{Symbol} & R_\mathrm{vdW}~(Å) & R_\mathrm{cov}~(Å)\\
\hline
\text{Chlorine} & \ce{Cl} & 1.75 & 1.00\\
\hline
\end{array}
$$

Ionic radii are given for the crystallographic data separately [1, p. 12-12]:

  $$
\begin{array}{llr}
\hline
\text{Ion} & \text{C.N.} & R_\mathrm{i}~(Å)\\
\hline
\ce{Cl^{+5}} & 3~\text{(pyramidal)} & 0.12\\
\ce{Cl^{+7}} & 4 & 0.08\\
\hline
\end{array}
$$

Reference

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC handbook of chemistry and physics: a ready-reference book of chemical and physical data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.


Would ionic compounds dissolve in methanol?

You don't need to do compare the said enthalpies. Maybe the process of dissolution of $\ce{KI}$ in $\ce{CH3OH}$ is energetically favorable, maybe it isn't (the latter seems more likely to me); that's irrelevant. Thanks to entropy, everything dissolves in everything, if only in minuscule amounts.

You just need to compare solubilities of $\ce{KI}$ in methanol and hexane, which basically boils down to the comparison of solvation enthalpy in methanol vs. that in hexane, and here the difference in their polarity provides a clear and obvious answer.


Software for determining symmetry elements and point group of drawn molecules

Coordination compounds / complexes may be a bit tricky in terms of «what do you define as a bond» (covalent / dative bond) and the convention what a stroke between atoms expresses (e.g. in the representation in compounds like ferrocene).  But if you already have a .mol or .pdb file in available, you could try Jmol:

(credit, includes some details how to use the program this way)
While the example is about methane as a simple organic molecule, Jmol is well capable to interact with crystal structure models, and displaying symmetries of the unit cell, too:

(credit, Hanson, J. Appl. Cryst. 43, 2010, 1250-1260 [open access])
including interactive galleries like Symmetry@Otterbein both about molecular and crystal symmetries.


Yes, you will definitely need to account for common ion effect.

However, use Kw to compute the pH. Assume that there is a complete dissociation by $\ce{HCl}$. However, you would need to solve the according quadratic for Kw.

as such, $\ce{H+= concentration~~ H+ of~~ HCl + x ~~and ~~OH- = x}$.

You should get a result around 6.79=pH.


How to condense a melted wax in to small aggregate forms like stones of different sizes?

Part of your problem is that wax is an amorphous solid and a mixture of various compounds.  This causes it to have no specific temperature that can be defined as a melting point, but rather a range of temperatures where it transforms from a hard brittle solid to a soft plyable solid to a liquid.
The best I can recomend is that you first pour very hot wax into a much larger amount of ice cold water to form wax shot.  Then take the shot and a propane torch, torch the wax beads as they fall to partially melt the surface of the shot and that may get you your desired effect.


Yes, the two structures you've drawn are the same; both substituents are in axial positions.  The two structures shown in the link you provided are not the same; they are conformational isomers.  The one on the left is the same as the one you've drawn with both substituents in the axial position.  The one on the right has both substituents in the equatorial position.  These two conformational isomers exist in equilibrium with the (e,e) isomer favored.  The two conformers interconvert by a cyclohexane ring flip.


How to tell the difference between octane-1-ol and oct-1-ene?

Alkene is best characterized, then, by its property
of decolorizing both a solution of bromine in carbon tetrachloride  and
a cold, dilute, neutral permanganate solution (the Baeyer test). Both
tests are easily carried out; in one, a red color disappears, and in the other, a purple color disappears and is replaced by brown manganese dioxide.


Yes, valency can change, just like that. Some elements (in fact, most of them) can manifest different valencies; that's their nature. If you look again at this very reaction, you might notice that Fe is not the only element to change its valency here; S does the same. Reactions like this one are called redox reactions. There is a huge body of knowledge that explains what valencies (or rather, what oxidation states) an element can have, and which of them are more stable; the answer by @Raghav Gupta touches on that in more detail.

As to the number of protons (if you mean the protons in the element's nucleus), it has nothing to do with valency. Really, Fe contains 26 protons; is its valency 26? Absolutely not, it is 2 or 3. Also, the number of protons in an element can never be changed at all. If you change it, you'll get another element; also, that would be a nuclear reaction, a thing which is out of scope of chemistry textbooks.


How to derive this temperature-pressure-specific volume relationship?

For an adiabatic system like a piston where $\delta Q = 0$, using the first law of thermodynamics gives you the following expression:

$$\mathrm dU = \delta Q + \delta W$$

$$\mathrm dU = - p\,\mathrm dV$$

This expression is pretty much useless however, in that you can't integrate it, since $T$, $V$, and $p$ are all constantly changing interdependently in the system. However if you assume the gas is ideal and making a few substitutions...

$$C_V\,\mathrm dT = \frac{-nRT}{V}\,\mathrm dV$$

$$C_V \frac{1}{T}\,\mathrm dT = -nR\frac{1}{V}\,\mathrm dV$$

$$C_V \int_{T_1}^{T_2} \frac{1}{T}\,\mathrm dT = -nR\int_{V_1}^{V_2} \frac{1}{V}\,\mathrm dV$$

$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -nR \cdot \ln\left(\frac{V_2}{V_1}\right)$$

$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -(C_p - C_V) \cdot \ln\left(\frac{V_2}{V_1}\right) = (C_p - C_V) \cdot \ln\left(\frac{V_1}{V_2}\right)$$

$$\ln\left(\frac{T_2}{T_1}\right) = \left(\frac{C_p}{C_V}-1\right) \cdot \ln\left(\frac{V_1}{V_2}\right)$$

$$\ln\left(\frac{T_2}{T_1}\right) = \ln\left(\frac{V_1}{V_2}\right)^{\gamma-1} ; \gamma = \frac{C_p}{C_V}$$

$$\frac{T_2}{T_1} = \frac{V_1}{V_2}^{\gamma-1}$$

From here it's just more cycling through variables; see if you can work the mathematics a bit so to get your second relationship.


What is the difference between molar conductance and molar conductivity?

I can understand your frustration. The use of terminology is often inconsistent and confused (much to my chagrin). I think you've got the general idea, the conductance ($G$) can be defined as follows:

$$G = \frac{1}{R}$$

i.e. the ease with which a current can flow. As you said,  $$R = \rho \frac{l}{A}$$

one can now identify, $$G = \kappa\frac{A}{l}$$ where the conductivity $$\kappa = \frac{1}{\rho}$$

Molar "any quantity" always has the dimensions (it is helpful to think in terms of dimensions) $\text{"quantity" } \mathrm{mol^{-1}}$

so, it follows molar conductivity $$ \Lambda_m = \frac{\kappa}{c}$$ where $c$ is the molar concentration. It is useful to define molar conductivity because, as you already know, conductivity changes with concentration.

Now, to address the effect of change of concentration on molar conductivity, we need to consider the case of weak and strong electrolytes separately.

For a strong electrolyte, we can assume ~ 100% disassociation into constituent ions. A typical example is an $\ce{MX}$ salt like $\ce{KCl}$

$$\ce{MX} \rightleftharpoons \ce{M^+} + \ce{X^-}$$ The equilibrium constant for this reaction is
$$ K = \frac{[\ce{M^+}][\ce{X^-}]}{[\ce{MX}]}$$ and thus, with decreasing molar concentration of the electrolyte, equilibrium shifts towards the disassociated ions.

At sufficiently low concentrations, the following relations are obeyed:

$$\Lambda _{m}=\Lambda _{m}^{0}-K{\sqrt  {c}}$$

where $ \Lambda _{m}^{0}$  is known as the limiting molar conductivity, $K$ is an empirical constant and $c$ is the electrolyte concentration (Limiting here means "at the limit of the infinite dilution").

In effect, the observed conductivity of a strong electrolyte becomes directly proportional to concentration, at sufficiently low concentrations

As the concentration is increased however, the conductivity no longer rises in proportion.

Moreover, the conductivity of a solution of a salt is equal to the sum of conductivity contributions from the cation and anion.

$$ \Lambda_{m}^{0}= \nu_{+}\lambda_{+}^{0}+\nu _{-}\lambda _{-}^{0}$$
where:
$ \nu _{+}$ and  $\nu _{-}$ are the number of moles of cations and anions, respectively, which are created from the dissociation of $\pu{1 mol}$ of the dissolved electrolyte, and $\lambda _{+}^{0}$, $\lambda _{-}^{0}$ are the limiting molar conductivity of each individual ion.

The situation becomes slightly more complex for weak electrolytes, which never fully disassociate into their constituent ions. We no longer have a  limit of dilution below which the relationship between conductivity and concentration becomes linear.
We always have a mixture of ions and complete molecules in equilibrium. Hence, the solution becomes ever more fully dissociated at weaker concentrations.

For low concentrations of "well behaved" weak electrolytes, the degree of dissociation of the weak electrolyte becomes proportional to the inverse square root of the concentration.

A typical example would be a monoprotic weak acid like acetic acid (again, from your graph):

$$\ce{AB} \rightleftharpoons \ce{A^+} + \ce{B^-} $$

Let $\alpha$ be the fraction of dissociated electrolyte, then $ \alpha c_0$ is the concentration of each ionic species. And $(1 - \alpha)$, and $(1 - \alpha)c_0 $ gives the fraction, and concentration of undissociated electrolyte. The dissociation constant is:

$$K = \frac{\alpha^2 c_0}{1-\alpha}$$

for weak electrolytes, $\alpha$ is tiny, so the denominator is nearly equal to one so, $$ K \approxeq \alpha^2 c_0$$ and $$ \alpha = \sqrt{\frac{K}{c_0}}$$ (like I said earlier)

for conductivities, one can now write the following relation
$$\frac{1}{\Lambda_m} = \frac{1}{\Lambda_m^0} + \frac{\Lambda_mc}{K (\Lambda_m^0)^2}$$ This fits the curve seen in your graph.

Caveat, all these arguments hold for dilute solutions. Things get out of hand at high concentrations, and one has to account for some additional phenomenon (for example, acetic acid will form hydrogen bonded dimers).

Anyway, long story short, conductivity increases with increasing dilution (though differently for strong and weak electrolytes). From the definitions I outlined at the very start, guessing how conductance changes is trivial.


How to find out oxidation state when compound has multiple electronegativities

You can assign oxidation states with a method similar to assigning formal charges.

Draw the Lewis structure.
Assign each atom all of its lone pair electrons.
For bonding electrons assign all of the electrons to the more electronegative element.
For each atom perform the following calculation:
number of electrons in the atom in its elemental state - assigned electrons = oxidation state.

In your example, the oxidation number of each $\ce{H}$ is +1 and of $\ce{F}$ is -1.  The oxidation number of $\ce{C}$ is -2 and of $\ce{O}$ is 0.


Let’s start off with the things you thought you got right: Oxygen will not be $\mathrm{sp^3}$ hybridised but somewhat closer to $\mathrm{sp^2}$. This is because that is how the oxygen lone pair in a p-type orbital can take part in π conjugation. I’m not sure what $\angle(\ce{C-O-H})$ I would expect but it’ll be somewhere between $109^\circ$ and $120^\circ$, that’s for sure.

To the entire molecule’s structure: Yes, at first glance it seems like a potato. However, there is still something that you can say: for proper conjugation of the double bonds (usually favourable), you need a planar structure. Thus, the entire molecule will be planar with all atoms in the same plane. That is about all that you can safely derive, since there can still be rotation around the $\ce{C-C}$ single bond.

Also, as Ivan pointed out, the molecule is likely to prefer the keto-tautomer rather than the enol one. The conjugation argument may apply, but it is rather weak.


Why NaHCO3 added to the water used (Ion Chromatography Suppression)

Ion chromatography usually means chromatography where aqueous (and usually inorganic) ions are the analytes.  They are separated from each other by the use of ion exchange resins.

If your analytes are anions, then they will be separated on an anion exchange resin.  These resins have a crosslinked organic polymer that contains permanent positive charges.  The negative charges that neutralize these resin-bound positive charges come from aqueous solution.  Different aqueous anions will have slightly different propensities to bind to and temporarily neutralize the solid-phase cations.

If you don't use an anion-containing eluent such as bicarbonate or hydroxide, the analytes that you would like to eventually detect through conductometric detection will permanently stick there!  So adding e.g. $\ce{NaHCO3}$ as an eluent will force them off of the column and ensure that your analytes eventually flow to the detector.

A decent diagram of the process is found in this PDF from Thermo (formerly Dionex). $%edit$


How to plate zinc evenly on a threaded bar

I think the easiest way is to sputter coat the bars, if you can access one. One that can rotate your bars such that the rotation is vertical to the direction of sputtering would be the best. If not, switch sides manually.

Depends on the composition of your bars, you might want to coat something else before coating zinc. A wild guess: nickel.  I suggest so because I don't recall zinc being sticky. So the coating might come off easily.

And I'd try sputter slow but for a long time. I think low sputtering rate tends to produce more densely packed and uniform coating. I am not sure. But dense and uniform coating provides good foundation of the zinc coat.

For electric coating, I think low voltage and uniform coating is still nice. I don't think you want formation of gas at the bar electrode, as that disturbs the structure of the coating. Not sure what's influencing the nucleation sites on the bars. I would guess that the current is not uniform on the bars, say there more on the hill tops. Still consider coating something else first. I would try whatever is available. I would try to get the solutions as pure as possible though. Organic compounds can act as surfactants which tend to aggregate the particles.

Whichever your method is, search for some journal articles characterizing similar processes. That might help.


According to “Aqueous Solubility of Inorganic Compounds at Various Temperatures”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL., the solubility (expressed as mass fraction) of potassium bromide $(\ce{KBr})$ in water is $39.4\ \%$ (at $T = 20\ \mathrm{^\circ C}$) and $40.4\ \%$ (at $T = 25\ \mathrm{^\circ C}$).

According to “Concentrative Properties of Aqueous Solutions: Density, Refractive Index, Freezing Point Depression, and Viscosity”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL., the properties of a potassium bromide solution of with a mass fraction of $w_{\ce{KBr}} = 40.0\ \%$ at $T = 20\ \mathrm{^\circ C}$ are as follows.

molality: $b_{\ce{KBr}} = 5.602\ \mathrm{mol\ kg^{-1}}$
concentration: $c_{\ce{KBr}} = 4.620\ \mathrm{mol\ l^{-1}}$
density: $\rho = 1.3746\ \mathrm{g\ cm^{-3}}$


How does a simple alkaline battery work?

An alkaline battery (or more precise alkaline manganese cell) consists of a zinc anode, a $\gamma\!-\!\ce{MnO2}$ cathode and $\ce{KOH}$ (in paste form) electrolyte. Zinc is the negative electrode and $\ce{MnO2}$ the positive electrode.

During battery discharge, zinc at the anode is oxidized and dissolves in the electrolyte as hydroxozincate. When the electrolyte is saturated with zincate, zinc oxide starts to precipitate. With continued discharge and decreasing $\ce{OH-}$ concentration, zinc hydroxide is formed, which can slowly dehydrate to zinc oxide.

\begin{aligned}\ce{
Zn + 4OH- &->~ [Zn(OH)4]^{2-} + 2e- \\
[Zn(OH)4]^{2-} &->~ ZnO + 2OH- + H2O\\
Zn + 2OH- &->~ Zn(OH)2 + 2e- \\
Zn(OH)2 &->~ ZnO + H2O\\
}\end{aligned}

At the cathode, $\ce{MnO2}$ is in turn reduced, first to $\ce{MnO(OH)}$. When the battery is discharged slowly, this can be further reduced to manganese(II) hydroxide.

\begin{aligned}\ce{
MnO2 + H2O + e- &->~ MnO(OH) + OH- \\
MnO(OH) + H2O + e- &->~ Mn(OH)2 + OH- \\
}\end{aligned}

The overall reaction, when only considering the first reduction step at the cathode, is:

$$\ce{Zn + 2MnO2 + 2H2O + 2OH- ->~ [Zn(OH)4]^{2-} + 2MnO(OH)}$$

During discharge, zinc is irreversibly oxidized and water is consumed. As a consequence, the battery is "dead" when the electrolyte is dried out and/or all zinc has dissolved.

Sources: 1, 2


The main ring of a polycyclic hydrocarbon ring system is selected so as to include as many skeletal atoms of the structure as possible. The main bridge is the bridge that includes as many of the atoms as possible that are not included in the main ring. Bridges other than the main bridge are called secondary bridges. Thus, tricyclic hydrocarbons consist of the main ring, the main bridge, and one secondary bridge.
The nomenclature of tricyclic hydrocarbons is based on the nomenclature of bicyclic systems. The additional steps for the naming of tricyclic hydrocarbons are described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as follows.

P-23.2.5 Naming and numbering tricyclic alicyclic hydrocarbons
P-23.2.5.1 Tricyclic hydrocarbons having an independent secondary bridge are named on the basis of a bicyclic system (…). Rings not described by the bicyclic system are defined by citing the number of atoms in the independent secondary bridge as an arabic number. The locants of the two attachment points of the independent secondary bridge to the main ring are cited as a pair of superscript arabic numbers (lower number is cited first) separated by a comma.
The name of the tricyclic system is then constructed by citing:
(a) the prefix ‘tricyclo’, in place of ‘bicyclo’, indicating the presence of three rings in the polyalicyclic system;
(b) numbers indicating the bridge lengths, starting with the two branches of the main ring (…), followed by the main bridge (…), and the secondary bridge (with superscript locants separated by commas indicating its points of attachment to the main ring), all separated by full stops and placed in brackets,
(c) the name of the acyclic hydrocarbon having the same total number of skeletal atoms.

For example tricyclo[2.2.1.02,6]heptane

(The main ring is shown in red, the main bridge is shown in black, and the secondary bridge is shown in blue.)


Mixing hydrochloric acid and sulphuric acid, both concentrated

Three things occurred. First, sulfuric acid as an high affinity for water (it is indeed used as a drying agent in some cases. For instance, if you wanted pure and anhydrous bromine, you could bubble wet bromine gas in concentrated sulfuric acid). So, one thing is the "stealing" of water by sulfuric acid, which took the solution one step forward the saturation point (with the release of gas).

The release of gas was aided by the exothermic reaction of sulfuric acid with water: (self) heating of the solution helped releasing the gas.

Edit: the following is an hypothesis, made by assuming that the "extreme" reaction conditions could have aided an unlikely redox reaction, but I am not sure about the oxidation of HCl:

And finally, sulfuric acid is a oxidizing acid: it oxidized part of the hydrochloric acid to chlorine, which made the solution green (which probably was released as a gas, together with HCl).


Electrolytes are ionic compounds that conduct electricity

Sucrose is an covalent compound and it does not ionize.
$\ce{KNO3}$, $\ce{NaOH}$ are good ionic compounds hence the can form electolytic solutions.
Ammonium acetate also can ionize.
Hence the answer is 2


How do you find the bond enthalpy from an equation?

Option D is the only one corresponding to the definition "energy required to break one mole of a specific bond" and nothing more.


Is carbonyl-enol tautomerization only intermolecular or can it be intramolecular?

tl;dr: The intramolecular variant is an impossible rearrangement. Only intermolecular proton transfer can be observed.

This is actually a very nice example to discuss sigmatropic rearrangements, because the desired [1,3]-rearrangement would be one. In general, these can proceed in four possible ways, depending on the configuration of the migrating group and whether it remains on the same side of the π system or not. We would have to draw orbital schemes to determine whether or not the orbital of the migrating group is able to interact both with the point of origin and the point of arrival favourably.

With regards to the configuration, the two possibilities are:

migration under retention
migration under inversion

A migration under retention is always possible because the corresponding orbital’s lobe is already pointing towards the π system it is migrating along. A migration under inversion is only possible if the (atomic) orbital whose bond is broken and later reformed has lobes on two different sides; typically for migrating atoms, this means a $\mathrm p$ orbital.

In the case of a proton shift, we must consider a hydrogen atom. Its only occupied orbital through which it bonds to the carbon chain is $\mathrm{1s}$. This orbital does not have two lobes, therefore a rearrangement mechanism under inversion is impossible; we need only consider the retention mechanisms.

With respect to the π system, the two possibilities are:

suprafacial (remaining on the same side)
antarafacial (switching sides)

Again, a suprafacial migration is practically always possible. For an antarafacial migration to be considered, the chain must be flexible enough so that both sides can interact with the migrating group; ideally at the same time.

Since the thought experiments are rather hard to perform, here are some depictions to make it easier.

Figure 1: Suprafacial [1,5] (left) and antarafacial [1,7] (right) hydrogen migrations. The depicted π orbital corresponds to the HOMO of a full π system including the $\ce{C-H}$ bond.

Generally, the full π system and the neighbouring $\ce{C-H}$ bond that is to be broken must be considered as I have done in figure 1. This π system can be drawn up with the standard rules for linear π systems. In the [1,5]-rearrangement, the system is penta-2,4-dien-1-yl while in the [1,7] case it is hepta-2,4,6-trien-1-yl. The $\ce{C-H}$ bond is considered part of the π system with the proton being the migrating group. Therefore, the HOMO (the orbital of importance) will be the one in question. (If the $\ce{C-H}$ bond is considered part of hydrogen, we must consider the π system’s LUMO, meaning that the orbital will remain the same. We must always have an occupied and an unoccupied orbital for consideration.)

The image shows us that the [1,5]-migration — over two double bonds — can only occur in a suprafacial manner with hydrogen. If we had a group capable of inversion, an antarafacial manner would be possible given the correct geometry. In the [1,7] case — over three double bonds —, hydrogen can only migrate in an antarafacial manner under retention; the only possible suprafacial manner would be under inversion which hydrogen cannot do.

This has been turned into a general rule of thumb: If your migration is [1,$4n+1$], it may proceed suprafacial under retention but if it is [1,$4n+3$] it must proceed in an antarafacial manner. In the [1,3]-case, only antarafacial migration would be possible, but there is no way that the π system can twist so it would allow for antarafacial migration. The π orbital in question is shown in figure 2 below.

Figure 2: The π orbital relevant for a [1,3]-migration.

Therefore, an intramolecular proton transfer is impossible and an external source would have to protonate.

One thing to note: the molecule itself is able to deprotonate/reprotonate another molecule of its type intermolecularly. So the compound would not be stuck in the enol form but a second molecule would provide the much desired proton to allow the keto form to be regenerated, even in pure hexane (if it is even soluble in that).


What is the balanced chemical equation for a redox reaction between a lithium-cobalt battery and a graphite electrode?

Below we have a simpler way of the $\ce{Li-Co}$ cell reactions.

Cathode half-reaction:

$$\ce{CoO2 + Li+ + e- <=> LiCoO2}$$

Anode half-reaction:

$$\ce{LiC6 <=> C6 + Li+ + e-}$$

Net-cell reaction:

$$\ce{LiC6 + CoO2 <=> C6 + LiCoO2}$$

Over-discharging supersaturates lithium cobalt oxide, leading to the production of lithium oxide.

$$\ce{Li+ + e- + LiCoO2 -> Li2O + CoO}$$

Similarly overcharging (up to $\pu{5.2 V}$) leads to following reaction:

$$\ce{LiCoO2 -> Li+ + CoO2 + e-}$$


Do polymerized cooking oils release free radicals

If you use fresh cooking oil containing unsaturated fats (like olive or canola oil) to season your pan, it will thermally polymerize to form a stable layer that gives the cooking surface its nonstick properties. Note that a saturated fat like coconut oil, lard, or ghee will not work very well or at all for this, since it does not have any double bonds to undergo polymerization. The radical formation you are concerned about results over time from air oxidation of unsaturated fats to form peroxides--these in turn break down to give aldehydes and acids that cause the bad flavor of rancid fat, and have adverse health effects. (Not a problem for coconut oil.) Heating fresh oil in a pan to create a nonstick coating may result in some air oxidation in addition to polymer formation, but I think it is not a big cause for concern. High polymers are generally nontoxic because the body does not absorb them.


Is atom-clean aluminum active enough to react with alkali salts?

Even normal metallic aluminum (with its spontaneously-formed corrosion-resistant oxide coating) reacts with concentrated sodium or potassium hydroxide to give the aluminate (e.g. NaAl(OH)4), plus hydrogen gas. Lithium hydroxide should also react (lithium is likely functioning as only a spectator ion) whereas alkali metal carbonates may not be sufficiently reactive.

This reaction is not what you refer to in your question as "substitution" reactions, where the alkali metal ion would be reduced to metallic form. Note that aluminum chloride is reduced by sodium or potassium metal to give aluminum metal, so the reverse reaction is unlikely to occur, but the energetics may be different for the hydroxides or carbonates in a non-aqueous environment.


What is Molar Conductivity of an electrolytic solution

I will answer your question about the definition of molar conductivity and why it increases with decreasing concentration.

As Jerry pointed out, you can probably learn what you need about equivalent conductivity using a Google search.

The molar conductivity of an electrolyte solution is the conductivity of the solution divided by the molar concentration of the dissolved electrolyte.

Conductivity ($\kappa$, in $\text{S/cm}$) is proportional to the square root of the concentration of ions ($\sqrt{[\text{ions}]}$, withy $\ce{[ions]}$ in $\text{mol/L}$). In the following equation, A is an empirical proportionality constant with units needed to convert from $\text{M}^{1/2}$ to $\text{S/cm}$, namely $\frac{\text{S/cm}}{\text{M}^{1/2}}$

$$\kappa = A\sqrt{\ce{[ions]}}$$

The ion concentration is related to the electrolyte concentration ($C$) by the van't Hoff factor ($i$).

$$[\ce{ions}]=iC$$

$$\therefore \ \kappa = A\sqrt{iC}$$

The molar conductivity $\kappa_M$ is the conductivity divided by the molar concentration of electrolyte ($C$).

$$\kappa_M=\frac{\kappa}{C}=\frac{A\sqrt{iC}}{C}=A\frac{\sqrt{i}}{\sqrt{C}}$$

As $C$ decreases, so does $\sqrt{C}$, and thus $\frac{1}{\sqrt{C}}$ increases, while $A$ remains constant. The van't Hoff factor changes a little with concentration, but not dramatically. ($A$ probably also varies a little with concentration. It certainly varies with temperature.)

For example, if you have a series of solutions of sodium chloride, concentration decreases faster than conductivity, so conductivity divided by concentration increases. Note that the conductivity data are fabricated with an arbitrary value of $A$. I could not find the specific value of $A$ for sodium chloride. The van't Hoff Factors come from the "Colligative Properties" Chapter of Chemistry: Atoms First by Burdge and Overby, McGraw Hill, 2012.


Balancing reaction between copper and nitric acid given the ratio of formed nitrogen oxides

The final arbiter of formal correctness of chemical reaction enumeration  are laws of mass and charge conservation. If total counts of charge and atoms of every element are not the same on each side, the equation is wrong. Reaction enumeration by following these laws may be troublesome, as general solution leads to resolving a set of linear equations. That can be useful for educational purposes, or rather to implant terror and despair in eyes of young chemistry students.
Chemists do not go that way. They usually use various heuristics to speed up the process, enumerating equations in several steps. For redox reactions, the heuristic is inventory of exchanged electrons, which is tracked by formal oxidation numbers (ON). These numbers are rather just arbitrary convention, but extremely useful.
For enumeration of your reaction ( with expressed ONs of relevant atoms):
$$\ce{Cu^{0} + HN^{+V}O3 -> Cu^{+II}(N^{+V}O3)2 + N^{+II}O + N^{+IV}O2 + H2O}$$
the first step is matching components undergoing redox change.

Copper changes ON from 0 to +II.i.e. by 2.
Nitrogen changes ON from +V to +IV ( $\ce{NO2}$), or from +V to +II($\ce{NO}$), i.e. by 1 resp. 3.
There is given $\ce{NO/NO2}$ ratio 2:3, so there would be as products $\ce{2 NO + 3 NO2}$, or their multiples.
$\ce{2 NO}$ exchange $2 \cdot 3 = 6$ electrons.
$\ce{3 NO2}$ exchange $3 \cdot 1 = 3$ electrons.

For simplicity, we can take formally $\ce{2 NO + 3 NO2}$ as a single entity, created from $\ce{HNO3}$ by exchanging totally 9 electrons.
Now comes the "magical" formula $9 \cdot 2 = 2 \cdot 9$.
9 Cu exchanging 2 electrons each gives 18 electrons totally.
2 ( $\ce{2 NO + 3 NO2}$  ) exchanging 9 electrons each has accepted 18 electrons.
So the intermediate result is:
$$\ce{9 Cu + HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + H2O}$$
This is enough for the question answer, but we have to finish the equation. It may sometimes happen some coefficients would get fractional, so all would get multiplied to have all integer.
We need supply enough nitrogen for the salt and oxides: $2\cdot 9 + 4 + 6 = 28$
$$\ce{9 Cu + 28 HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + H2O}$$
We need to create water from hydrogens:
$$\ce{9 Cu + 28 HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + 14 H2O}$$
And we are done. We can formally check the item conservation:
$\ce{Cu}$: $9=9$
$\ce{H}$: $28=28$
$\ce{N}$: $28=28$
$\ce{O}$: $3\cdot 28=84=54+4+12+14=84$
charge: $0=0$
For a quick check,we may just check items we were not enumerating before, like just oxygen and charge.


What is left over when water and sodium hypochorate decompose, i.e. household bleach?

It depends on temperature. At higher temperatures, hypochlorite has tendency to disproportionate: $$\ce{3 ClO- -> 2 Cl- + ClO3-}$$ Otherwise it mostly releases atomic oxygen : $$\ce{ClO- -> Cl- + O}$$
In reality, it would be combination of both in various ratio, depending on relative reaction rates,  all ending as mixture of sodium hydroxide, carbonate, chloride and chlorate.


Does spectrochemical series trend depend on the metal center?

The spectrochemical series of ligands, our instructor clarified, gives only a rough idea of the relative strengths of ligands, and may vary for different Central Metal Atoms (CMAs).
An important example that I have encountered is of $\ce{[Co(H2O)6]^+^3}$, wherein $\ce{H2O}$ acts as a Strong Field Ligand (SFL), despite its lower position in the series.
Also, JD Lee mentions in his Concise Inorganic Chemistry (5th ed., Wiley publications) that there is an increase in CFSE as we go down the group in d-block, and several instances may be found where the ligands designated as WFL begin to act as SFL for 4-d or 5-d members.
Overall, I believe there are certain deviations from the series, but it generally works well.


How to calculate molar concentration from equilibrium equations with no initial concentration given?

I will continue with the data of $[\ce{H2CO3}] = \pu{10^{-4.97} M}.$
Now, as the $K_2$ of $\ce{H2CO3}$ is very small as compared to its $K_1,$ we can assume that all the $\ce{H+}$ will come from the first dissociation of $\ce{H2CO3}.$
$$
\begin{array}{lccc}
& \ce{&H2CO3 &<=> &H+(aq) &+ &HCO3-(aq)} \\
&\text{Initial} & 10^{-4.97} && 0 && 0 \\
&\text{Final} & 10^{-4.97}(1-\alpha) && 10^{-4.97}\alpha && 10^{-4.97}\alpha
\end{array}
$$
$$K_1 = 10^{-6.35}$$
$$10^{-6.35} = \frac{(10^{-4.97}\alpha)^2}{10^{-4.97}(1-\alpha)}$$
As $K_1 \ll 1,$ $(1 - \alpha)\approx 1:$
$$\alpha = \sqrt{\frac{10^{-6.35}}{10^{-4.97}}}$$
$$
\begin{align}
[\ce{HCO3-}] &= \ce{[H+}] = 10^{-4.97}\alpha \\
  &= \sqrt{10^{-6.35}\times 10^{-4.97}} \\
  &= 10^{-5.66} \approx \pu{2.19E-6 M}
\end{align}
$$
Note, that I have not considered self ionization of water because the $\mathrm{pH}$ of this solution will be slightly less than $6$ $(\mathrm{pH} = 5.66).$ Which means even after accounting for self ionization of water, the results won't change much. This is a good approximate of what  we needed. And if you want to calculate $[\ce{CO3^2-}]$ than just calculate it in a similar way to which you calculated $[\ce{H2CO3}],$ because now you have $K_2,$ $[\ce{HCO3-}]$ and $[\ce{H+}].$


The equilibrium constant should be dimensionless, it has to be to be used in equations such as the free energy for reaction at equilibrium
$\Delta G^0_R = -RT\ln(K_p)$.

If the reaction is
$\ce{\alpha A + \beta B \<=>\ \gamma C + \delta D} $
 if we write

$$K_p= \frac{P_A^{\alpha} P_B^{\beta}}{P_C^{\gamma}P_D^{\delta}} $$

where $P_a$ etc are partial pressures at equilibrium then $K_p$ is going to  have dimensions in general depending ion the values of $\alpha , \beta $ etc. as you state in your question.

In the past it was sort of implicitly assumed that each pressure was divided by 1 atm so then $K_p$ becomes dimensionless (or the units were simply ignored and numerical values used) but nowadays this is explicitly added so we obtain

$$K_p= \frac{\left(\frac{P_A}{P^0}\right)^{\alpha}  \left(\frac{P_B}{P^0}\right)^{\beta}}{\left(\frac{P_C}{P^0}\right)^{\gamma}\left(\frac{P_D}{P^0}\right)^{\delta}} $$

where $P^0 = 1 \pu{ atm}$  and so makes things dimensionless no matter what the values of $\alpha, \beta $ etc are but does not change any numerical value.

Although I have used partial pressures the same applies to concentrations $\ce{[A], [B]}$ etc where now each concentration is divided by  $\ce{[C^0]}$ where this is  $ \pu{1 mol/dm^3}$.

I'm not quite sure what your notation $K^0P$ etc. means but I suspect it refers to the two cases I have described. Hope this helps.


Relaxed scan of Cp-M torsion angles of piano stool compounds

I would use the dummy, define the 3 X 10 11 (16 and 18 too) dihedrals in the connectivity matrix, run the optimization and later write a shell script for running subsequent calculations in separate files replacing the value of the previously defined dihedrals. So I guess my solution is only slightly above doing it by hand but so far I haven't found any software that does this and G09 has no workaround.

I hope this helps


Is there an example of a closed shell molecule with a permanent ground state magnetic dipole?

Updated 4/6/2017: It turns out that singlet ($^1\Delta_g$) molecular oxygen is paramagnetic due to orbital angular momentum only, as all electrons are paired. It has a magnetic dipole moment of $1\hbar$ and can be determined experimentally. (This makes sense from a computational perspective, given that the lowest restricted Hartree Fock solution of singlet oxygen is always complex, and only complex wave functions yield non-zero expectation values of orbital angular momentum.)

For closed shell (even number of electrons) molecules the answer is no.

After a little more reading, I found the answer. The reason is that the magnetic dipole is odd under time reversal (Table 4.1, p. 218 in Barron, 2004).

Apparently, (Barron, p. 221) "the expectation value of a time-odd operator [e.g. magnetic dipole] vanishes for states invariant under time reversal, which can always be constructed for an even electron system and hence any nondegenerate state".

I do not believe this applies to transition properties, however, as we can see magnetic dipole allowed transitions.

Citation: Barron, Laurence D. Molecular light scattering and optical activity. Cambridge University Press, 2004.


How can a protein folding transition state have zero lifetime?

I don’t know where the person from the Google Groups you quote got the information that a transition state is "presumed to have zero lifetime", but a far as I can tell it is wrong. On the contrary, the existence of transition states is postulated in transition state theory (TST), which evidently implies that within the TST framework transition states have non-zero lifetime. Existence clearly contradicts zero lifetime: if some object has zero lifetime, what do you mean then by its existence?

Next, while the lifetime of a transition state in TST is postulated to be not zero, it is also proposed to be quite small, on the order of the time of a single molecular vibration ($10^{-13}$ seconds). Why do we have to make such a proposition? Well, transition state can be defined as the state corresponding to the highest potential energy between the reactants and the products (i.e. along a reaction coordinate) such that the probabilities of forming the reactants or products are equal. On potential energy surface (PES) then transition states correspond to the so-called saddle points (first-order saddle points, to be more precise), which, loosely speaking, can be thought of as a maximum in one direction (along the reaction coordinate) and minimum in all other directions. And structures that correspond to maxima (even in one direction) on PES) are unstable: once a system is at a maximum on PES it will almost instantly fall from it to one of the two neighbouring minima (either to reactants or to the products side).

Due to small assumed lifetime of a transition state, it is often said that "no physical or spectroscopic method is available to directly observe the structure of the transition state" (quote from Wikipedia article on TST). Although, with femtosecond IR spectroscopy one can actually observe molecular structures extremely close to the transition state structures. The practical problem is that quite often along the reaction coordinate in addition to a maximum corresponding to transition state between the reactants and the products there exist a minimum (or few) corresponding to the so-called reactive intermediates, and it is quite difficult to distinguish between transition states and reactive intermediates.

Note here that reactive intermediates correspond to minimum in all directions on PES, and are thus stable, or, to be more precise, metastable, since they are high-energy molecules which are quickly converted into more stable molecules (either another reactive intermediate or eventually the products). Reactive intermediates are also short-lived but with lifetimes appreciably longer than a molecular vibration, so they cannot usually be isolated, but they can be observed with the already mentioned femtosecond IR spectroscopy.

To summarise:

One should not in the first place think about transition states being real (and consequently detectable). A transition state is a part of a model description of a chemical reaction introduced by TST, where we just postulate their existence, and say no word on are they really exist or not. Irregardles of their real existence, transition states exist in our model description, and their existence implies non-zero lifetime.
Besides, with femtosecond IR spectroscopy it is possible to directly observe a transition states for some chemical reactions and it was actually done.


In a reaction A(s) giving B(g) + C(g) Increasing the surface area of A increases the rate of the forward reaction, but the backward reaction is between two gases and is unaffected.

This is an assumption that you're making and it is thermodynamically and geometrically questionable.  If increasing the surface area of $\ce{A}$ increases the rate at which it sublimes/gasifies to form $\ce{B + C}$, it means that the reactive molecules of $\ce{A}$ are only those at the surface, in which case it is misleading to denote the reactant as $\ce{A(s)}$.  It is really $\ce{A(surface)}$ which is reacting.  Similarly, if the rate of the reverse reaction is not really dependent on the surface area and surface geometry of $\ce{A(s)}$, then it implies that very quickly molecules of $\ce{B}$ and $\ce{C}$ would condense to form a nanocrystallite of $\ce{A}$ that is one molecule big.  These nanocrystallites of $\ce{A}$ would have, collectively, nearly infinite surface area.  So the backward reaction creates surface area under the assumptions you have made.  You have to account for that created area.

Solid-phase reactions are complicated by phenomena such as crystal nucleation and other mechanisms of crystal growth.  Treating solids as uniformly reactive materials with unit activity is inappropriate for understanding such phenomena.  Another way to say it is that the assumption that solids have unit activity and don't appear in expressions for $K$ or $\Delta G$ is an assumption, one that is inapplicable when dealing with reactions whose rates depend on surface area.


Part of your confusion lies in the all-too common belief that it's the ratio of ligand to binding partner (e.g. substrate and enzyme) which determines the amount of binding. That's not the case. Instead, it's the absolute concentration of free ligand which determines how much is bound.

Let's take a microscopic view, and look at a single enzyme molecule. The enzyme doesn't know or care about what the other enzymes are doing, it only knows about its own local environment. To bind a substrate molecule, it needs to encounter (collide with) a substrate molecule in the correct orientation. How frequently it does that is directly proportional to the amount of free, bind-able substrate molecules there are in solution. For bound enzymes, (ignoring catalytic turn over for the moment), transitioning to the unbound state is a first-order process, independent of the concentration of anything else in the solution. (The total macroscopic rate of unbinding events depends on the number of bound complexes in solution, but for an individual bound enzyme it's concentration independent.) So for a single enzyme molecule, the fraction of the time it spends in the bound state is only dependent on the concentration of free ligand, so the time average of a large number of molecules is going to reflect that same distribution, by the ergodic hypothesis.

If binding is difficult but release is easy, then you're going to need a large amount of substrate in order to push the rate of binding to be comparable to the rate of release. If binding is easy but release takes a long time, you won't need as much substrate. That's what the KD of a binder (and roughly KM of an enzyme) measures. It's the concentration of the free ligand which makes the effective on-rate an the off-rate identical, and thus results in enzymes binding about half the time (or equivalently, at any one time half the enzymes are in the bound state).

So it doesn't matter if you have 100 molecules of substrate for every molecule of enzyme - if you're dilute enough, the concentration of substrate will be too low, the rate of enzyme/substrate encounters will fall off, and any enzyme-substrate complexes which form will fall apart long before some other one can form.

Enzyme/substrate (and ligand/binder) ratios are important, but mainly because it's the concentration of free substrate, rather than the concentration of total substrate, that's involved in the calculation. As your enzyme binds substrate, it depletes the pool of free substrate. If your enzyme concentration is large in comparison to your concentration of substrate, substrate going to the enzyme-bound state could cause a significant drop in free substrate concentrations, leading the nominal substrate concentration to be different from the actual free substrate concentration. (This is also part of why people primarily work with initial rates - the amount of free substrate drops over time due to catalysis.)

Caveat: KM is not quite the binding affinity of an enzyme for the substrate. In addition to the standard off rate of dissociation, there's also the "off rate" due to catalysis. There's also the lifetime of catalytic intermediates and the product-bound state which comes into play. With typical Michaelis-Menten assumptions - single step catalysis which is very slow compared to binding, with rapid product dissociation - the difference is minimal, but for many enzyme systems those assumptions don't hold.


Reaction of Benzene with 2-methylpropene?

Sulfuric acid protonates the alkene to give the highest substituted carbocation, tert-butyl.  That electrophilically adds to a benzene carbon atom, to give the obvious next cation.  A proton is eliminated to obtain tert-butylbenzene.  Akyl groups activate benzene rings toward ortho, para attack (mild electron donation via  hyperconjugation), thus 1,6-di-tert-butylbenzene, certainly if the addend is in excess.  Especially stable carbocations can dissociate and wander around the benzene ring under Friedel-Crafts conditions.  If you push it, you get 1,3,5-tri-tert-butylbenzene.

Steric hindrance strongly discourages further reaction.  You can go to the literature to find how many adjacent tert-butyls can be jammed onto a benzene ring - and how it is done.

With due respect, you are in deep trouble for passing the course if the Friedel-Crafts reaction throws you.   Either look at things with the LCAO model, look in a few more textbooks, or talk with folks who have grasped it.  You must understand it.  You cannot survive substituent effects if you cannot flip bonds.  This is easy stuff compared to what will follow.  Chess can be well-played by computer.  Organic synthesis is done in wetware, though computers help help.

http://www.chemtube3d.com/model/electrophilic_aromatic_substitution/friedel-crafts/alkylation/benzene%2Btbu/mechanism.png
Close enough


There are 3 phases at the critical point. Sub-critical vapour, Sub-critical liquid and critical fluid. 3 phases -> DoF = 0

That is one way to see it - and for convenience that is how I see it. It might be a bit wobbly to regard that as rigorous analyse though.

You can also say, more rigorously, that there is 1 phase and 2 constraints:
$$\left(\frac{\partial\rho}{\partial V}\right)_T = 0 $$
$$\left(\frac{\partial^2\rho}{\partial V^2}\right)_T = 0 $$
Subtract the number of constraints from the DoF and you get 0 again.

This defines a single point on the P/T curve.


How do you change the cis-trans isomerism of compounds?

In general, any setup that gives enough energy to the compound can isomerize it. This can be done by heating it, irradiating it with visible or UV light, or by using a catalyst (reducing the energy required for transition rather than supplying more energy). Excitation of electrons from the $\pi$ to the $\pi^*$ (bonding orbital to anti-bonding orbital) allow free rotation about the $\sigma$ bond. Relaxation back to the $\pi$ orbital locks the configuration. The specific procedures depend on the compound of interest.

FYI, photoisomerization (light-induced isomerization) is the principle on which sight is based.


What exactly is hydrogen bonding (and does it really need fluorine, oxygen, or nitrogen)?

There are two main ways to look at hydrogen bonding.  The first is electrostatic, where the electronegativity of the atoms is used to describe the interaction.  Your argument about chlorine being more electronegative than nitrogen is a good one suggesting that the electrostatic argument is only part of the story, and there is at least one study that suggests that hydrogen bonding does occur with chlorine in polar molecules such as chloroform.

We can increase dramatically the number of atoms considered to undergo hydrogen bonding if we take a molecular orbital approach.  My reasoning here is a summary of what can be found in Inorganic Chemistry by Miessler, Fischer and Tarr.  They, in turn, rely heavily on a new definition of hydrogen bonding recommended by the IUPAC Physical and Biophysical Chemistry Division.

A hydrogen bond is formed when an $\ce{X-H}$ (where X is more electronegative than H) interacts with a donor atom, $\ce{B}$.  The attraction $\ce{X-H...B}$ can be described as consisting of the following components:

An electrostatic contribution based on the polarity of $\ce{X-H}$
A partial covalent character which arises from the donor-acceptor nature of the interaction
Dispersion forces

The first bullet is typically the only phenomenon discussed in General Chemistry classes, and this is not unreasonable since the second bullet requires the introduction of Lewis Acid/Base concepts which may not have been covered.

It is important to note that the new definition includes what I like to call "the proof is in the pudding" where the existence of hydrogen bonding requires experimental evidence, which can be found using a number of methods:

The $\ce{X-H...B}$ bond angle: a bond angle of 180 degrees indicates strong hydrogen bonding and would be accompanied by shord $\ce{H...B}$ bond distances.
A red shift in the IR frequency upon formation of $\ce{X-H...B}$
Hydrogen bonding results in deshielding of the H atom, which can be observed in NMR spectroscopy.
Thermodynamic studies should indicate that $\Delta G$ for the formation of $\ce{X-H...B}$ should be larger than the thermal energy of the system.

While electrostatics is the dominant contributor to hydrogen bonding, an analysis of frontier orbitals can also be insightful.  A thorough answer requires a fair amount of MO theory background, but for those interested, looking at the MO diagram of $\ce{FHF-}$ will be helpful.  In brief, the base ($\ce{F-}$ in this case) has a filled $p_z$ orbital that gains access to a delocalized orbital with lower energy, thus validating the formation of a strong hydrogen bond.


Gel filtration separation of cellulose, starch, oxytocin and palmitic acid

I think by using a gradient mobile phase that changes by time. At the beginning, the mobile phase would have higher affinity to cellulose and lower to starch and then the gradient changes where it would have higher affinity to starch than cellulose.

Specifically, which mobile phase you would use and what gradient would take some time and research to find the answer but it think you get the idea


Does anyone know what program was used to make this molecular structure picture?

The first type of image is very easy to make. I would suggest Avogadro with POV-ray. Avogadro can be used to draw the structures (or open a file containing the information, like .mol2, .sdf etc.) and then high-quality images can be generated using POV-ray.
To do this, install Avogadro, then draw a structure, and then File> Export > POV-ray. A new subwindow will open, where you can specify the location of the POV-ray executable.

In my case, the executable is pvengine64.exe, it might be different in different platforms, or in different versions.
You can also change the resolution here. Finally, click "Render", and a new windows with POV-ray should open, and it should display the image, after a few seconds.

This is the image of the molecule that you showed in your question. The background looks like that because the "Alpha transparency" option is turned on in the Avogadro subwindow, which makes POV-ray render only the molecule, and there is no background.
As for the second picture, it looks like something that would be used in crystallography to represent probability ellipsoids of atoms. I don't know much about this, so I can't help you there.


This is a nice question. Although I do not speak with authority on diamonds, most crystal surfaces have several imperfections which mitigate the hanging valencies that would otherwise appear for the atoms (constituents in general) at the surface.

Impurities. This is the primary method which helps satisfy the unfulfilled valencies at the surface. Many metallic, non-metallic or compound impurities are usually incorporated throughout the crystal, but are usually high at the surface. By bonding with these excess impurities, the surface carbon atoms can fulfill their valencies. An extremely pure crystal surfaces (lots of "hanging" valencies) will quickly adsorb most materials that it comes in contact with. Chemisorption or physisorption depends on the nature of the crystal, metals prefer physisorption while non-metals have at least unimolecular chemisorption usually.
Another way to mitigate surface valencies is distortion. The surface layer might not follow the exact space-filling Bravais lattice structure and may form short range distortions in structure to try to make the unfulfilled valencies satisfied. There might be kinks, depressions, steps etc which at least help in completing the valencies of a few atoms (at the boundary of distortion, say, a "step" structure).
If none of these operate, there is an inherent surface energy or surface tension between any two phases which quantifies the desire of the surface atoms to fulfill their hanging valencies to resemble the atoms in bulk which are stably attracted. In case of metallic and ionic crystals, this is comparatively low, but may be high in a covalent crystal like diamond.

In all cases, the surface layer of constituents (actually first few layers) are different from the atoms in bulk in energy, stability, forces acting (unbalanced on the surface and balanced in the bulk) and their structure/orientation.


Ionization isomerism in coordination compounds

I think it should be -

tetraamminechloridoiodoplatinum(IV) chloride iodide.

The rules of the naming of counter ions are pretty much the same as in general Inorganic Chemistry. I do know of a compound AgClI - named as Silver chloride iodide. So, I think the counter ions here - chloride and iodide should be named as by the nomenclature rules of anions i.e with an -ide suffix and alphabetically if there is the presence of more than one anion

Proof of Reference - Silver chloride iodide (NCBI PubChem)

EDIT: My teacher did give an example of it being a valid complex. I think it is not generally stated in the preliminary books of this topic to avoid confusion among students.I imagine it would be present in some more in-depth books of this topic.


Chemical Equilibrium - Why do changes in pressure cause a shift in the ratio of products and reactants?

With gasses, what you're doing by changing the pressure is you change the partial pressures or the reactants. As long as there's the same moles of gas on either side, the equilibrium is unaffected, but if there's an un unequal number, the reaction quotient is changed. The same would happen if you added water to an aqueous reaction. You can play with the numbers yourself, I'll give you an example to use:

$$\ce{N2(g) + 3H2(g) <->2NH3(g)}$$

We can use the reaction quotient with partial pressures, but it's more clear if we use the one with concentrations:

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

Using $c=\frac{n}{V}$:

$$Q_c\ce{ = \frac{\frac{n(NH3)^2}{V^2}}{\frac{n(N2)}{V}\times \frac{n(H2)^3}{V^3}}}$$

Take notice of how this fraction depends on volume! So it's really just the system reacting to attempt to reach equilibrium again (making it so that K = Q).

As for temperature. My understanding is that it's not to do with activation energy. It IS related to the enthalpy of the reaction though, and your understanding of what a temperature change means for a particular reaction is correct.

Concentration is the same deal as I've talked about in the beginning of this post, and incidentally it matches your own understanding of it.

You've got catalysts right too. Have a look at the following diagram:


Where do things connect in EAS reactions?

The position of substitution in EAS is determined by the electron-donating/-withdrawing effects of existing substituents. The effects can be inductive (the electronegativity of heteroatoms polarize the bonds), or they can be resonating (free electron pairs are donated to a double bond in the transition state).

1. rule; resonance effects overrule inductive effects.

If a ring has both a resonating and an inductive substituent, or one substituent with both effects, the resonating effect will control the reaction.

2. rule; donating effects activate the ring (more electrons for the electrophile), withdrawing effects deactivate the ring (less electrons for the electrophile).

Strong donating groups: -OH, -OR, -NH2, -NHR, -NR2, -SH, -SR, -F, -Cl, -Br, -I, -CH=CH-X (X is any donating group,

Neutral/weak donating groups: -CH3, -(CH2)R, -CHR2, -CR3 (R is any hydrocarbon), Ph.

Strong withdrawing groups: -(C=O)R, -NO2, -SO3,

Weak withdrawing groups:  -(CH2)X (X is any atom more electronegative than carbon),

3. rule; activated rings favor para and ortho positions, deactivated rings favor meta position (para and ortho are deactivated).

Deactivated rings can still react, but much higher temperatures are required, and the reaction is much slower.

4. rule; para is favored over ortho.

Because the existing substituent is "in the way" of the attacking electrophile in ortho position.

An Example

In electrophilic aromatic substitution the aromatic ring has a positive charge in the transition state (because an electrophile (E+) has "borrowed" to two electrons). The electrophile can, and will, attack all three possible positions, but this first step is reversible, and any charge on a carbon is highly unfavorable. Therefore, if the charge cannot be placed on in the bottom path in the example below, the electrophile will be ejected quickly, while the other two paths are stabilized long enough for a base (B-) to extract the proton, leaving the electrons used to bond with the proton to the ring.

rule; Oxygen is inductivly withdrawing, but at the same time donating by resonanse. => donating.
rule; Donating => activated ring.
rule; Activated => ortho/para.
rule; both ortho and para are available. => para favored

Your example

Question (i), the -(C=O)R is a withdrawing group on the left ring, while the RO- group has two electron pairs that it can share with the right ring. ==> para position on the right ring.

Question (ii), the -NO2 on the left ring is deactivating for that ring, while -CH2R is weakly activating for both ring. ==> para position on the right ring.

Question (iii), the ring has a strong and a weak withdrawing group, and is thus highly deactivated. Also, as ssavec pointed out, NaI is no electrophile.

Still let us assume that you had some strong electrophile; Both substituents deactivate their ortho positions, while having little effect on their meta positions. Since -NO2 is a stronger withdrawing group, it rules, and the reaction would happen meta to the -NO2 group (ortho to the -CH2Br).


Role of zinc chloride in reaction of ether

I think that it reminds Fischer esterification at the beginning. Zinc dichloride acts as a Lewis acid by coordination at the carbonyl oxygen in AcCl. It activates C=O bond because of larger contribution of carbocation form. And this cation undergoes coordination of Et2O with formation of zwitterionic tetrahedral intermediate. This particle eliminates our catalyst and cloride-anion. The product of this step is acetyl diethyl oxonium. Cloride-anion is quite nucleophilic, but we have an excellent leaving group without charge (AcOEt), so Sn2-reaction occurs and lead to the final products


Which factors best account for the increase in boiling point?

You're definitely right on London. Between the three molecules, the only thing that's changing is the carbon chain on the right hand side of the Oxygen. That plus the corresponding increase in boiling temperatures together form a good indicator that London forces are at play here.

Dipole-dipole (or Hydrogen bonding for that matter) are non-significant in this case, because you're not changing the electronegativity of anything here, only adding more of the same repeating atoms.

Source: This was on the ACS 2011 local exam, question 55.

Further information here


A mixture of 4 compounds is extracted with HCl. Which one would go to the aqueous layer?

This is a classic example of an acid-base extraction. If a mixture of organic compounds in some organic solvent is treated with an aqueous acid, any bases in the original mixture react to form their conjugate acids. These conjugate acids tend to have ionic functional groups that increase their solubility in water.
The key skills required to solve this problem are 1) identifying the principal functional group in each structure and 2) recognizing the characteristic acid-base properties of each function group.
These compounds are (in order), a sulfide, an amine, an ether, and a ketone. While you could envision perhaps three of them being polar and suitable hydrogen bond acceptors (sulfuer is approximately as electronegative as carbon, and sulfur does not participate in hydrogen bonds), only one is a base - the amine. The characteristic acid-base behavior of functional groups could be constructed each time you need it, but may best be memorized. There are only a few common functional groups that typically behave as acids or bases in aqueous solutions.

Acids: carboxylic acids, phenols, thiols, sulfonic acids, 1,3-dicarbonyl compounds, imides
Bases: amines, imines, amidines, guanidines


Octahedral Crystal Field Splitting Orbital Energy Levels

The total energy of the d orbitals before and after the application of the octahedral (cubic) field has to be the same. As three levels fall and two rise, the ratio is as shown. The $\Delta_0$ arises from the solution of the equations. (In older literature the total gap is $10Dq$ with the levels $6Dq$ and $-4Dq$.)

If $V_0$ is the interaction energy $3\Delta_0/5=\int\psi(e_g)^*V_0\psi(e_g)d\tau$ and $-2\Delta_0/5=\int\psi(t_{2g})^*V_0\psi(t_{2g})d\tau$. In practice $\Delta_0$ is found empirically.

The order of levels is reversed in a tetrahedral field and the size of $\Delta_0$ is also changed to $-9/4\Delta_0$.


If all you want to know is the relative amounts of calcium and magnesium carnbonates disolve it in vinegar (it will fizz). Now add lye until the pH is about neutral (use pH test strips) in order to neutralize any excess acid. Obtain a test kit for determining water hardness that will give both the magnesium and calcium hardness values for a sample. Dilute with DI (or nearly DI i.e. RO) water to be within range of the kit Be sure you understand what units the kit is reporting in. Usually it is units related to the equivalence of the calcium and magnesium so that if calcium harness turns out to be 1/5 magnesium hardness that tells you the original mix has 1/5 of a mole of CaCO3 for each mole of MgCO3.


How do I get MarvinSketch to use the ACS style of drawing?

You can find some information on JACS style here.

If you are using Marvin Sketch 5.1, first clean your structure, then got to edit then format. Click the Apply changes for all the atoms / bonds and then Load Style. Pick J. Am. Chem. Soc and OK.

In Marvin Sketch 6 open File then Document Style and continue as above

For more information see the documentation here.

That should work


How to compare SN2 rate of cyclopropane with bridged cyclic compound?

Concerning the rates of SN2 reaction, the rate of reaction for cyclopropyl iodide will be relatively higher than 1-iodonorbornane (name of given bridged compound).

The cyclopropyl iodide (same as 1-iodocyclopropane) has one hydrogen atom substituted by the iodine atom. If you look into the proper stereochemistry of the compound, this $\ce{I}$ atom can be above the plane of the cyclopropane ring or below the plane. Angle between the above-planar atom and below-planar atom is approximately 115 degrees. So, the attacking nucleophile can attack from above the plane if $\ce{I}$ is below the plane, or vice-versa. This attack would be best if leaving group $\ce{I^-}$ and the nucleophile are in almost 180 degrees position. Here the repulsion due to lesser bond angle and with the hydrogen will decrease the rate, but SN2 is still possible to a low extent in the cyclopropane ring.

Now to the bridged compound. Firstly, there is very very little chance for backside attack. Secondly, for occurrence of SN2, there should be a Walden inversion of configuration. But in the briged bicyclo compound, the ring is in a very compact structure, and if it tries to undergo any inversion of configuration, a huge strain will be created in the bicyclo-compound, which is thermodynamically, very unstable. So, the rate of SN2 is almost nil for 1-iodo norborane.


How does NaCl maintain its crystalline structure?

Crystals have inspired a great many chemists because they are fascinating for a good reason. Not only are they aesthetically pleasing, but they serve as an excellent subject to tour a variety of theoretical subjects important for understanding high-level chemistry.

Crystalline materials are made up of periodic structures. We’re only going to primarily focus on binary compounds where there is not a high degree of covalency. There are several ways to think about this problem, but let’s start with the melting of a crystal.

We say that at some definite temperature a highly ordered crystal will melt into a liquid. Those of us familiar with the language of equilibrium thermodynamics might recognize that the change in free energy for this phase change can be written, at constant temperature, as,

$$ G_\text{liquid} - G_\text{crystal} = H_\text{liquid} - H_\text{crystal} - T ( S_\text{liquid} - S_\text{crystal} ) $$

$$ \Delta G = \Delta H - T \Delta S $$

If we suppose that this process is spontaneous then we would say that the change in Gibbs’ free energy is negative, i.e. $\Delta G < 0$. This is true if and only if,

$$\Delta H < T \Delta S$$

Traditionally we interpret this as saying that there is a thermally-driven increase in entropy when we melt a highly ordered crystal into a liquid which more than offsets the energy cost associated with the enthalpies of the interactions holding that crystal together. A chemist tends to learn early on that the reverse is not necessarily true: at some definite temperature a perfect crystal rarely forms from the liquid. This inability to just heat up any substance and always produce a perfect crystal by cooling illustrates how crystal formation is a case of kinetic- rather than thermodynamic- control. So the process by which you form your crystal could possibly result in a different crystal structure. Sometimes crystal structures change just by altering the temperature of the chamber you’re measuring the crystal structure in! Now neither of these cases apply to sodium chloride to the best of my knowledge.

The formation of an ionic crystal such as sodium chloride is a delicate balance between electrostatic attraction and Pauli repulsion. Electrostatic attraction says that between two different charges, $q_+$ and $q_-$, there is a Coulomb force given by, $$F= \frac{k q_+ q_-}{r^2}$$

where $r$ is the distance between the two charges. If one plays with the numbers then it’s easy to see that at short distances the force is strongest, but there is a limit to how close they may come together. Eventually a repulsive force due to a quantum mechanical principle called the Pauli Exclusion Principle overpowers the attraction. An equilibrium results in which the atoms sit a certain distance from one another so that, if you will humor me, the “forces” between them balance out. This is why we traditionally represent crystal packing using marbles with a unique radii. The radii of the hard marble represents where the Pauli repulsion overpowers the attraction.

You might say, “Sure, we have these kinetic, electrostatic and quantum mechanical factors to consider, but how do these help with the final crystal structure?” Hold your horses, we’re getting there.

A famous mathematician and scientist thought about the most efficient ways to pack spheres of the same size together. By most efficient I mean this in terms of what FedEx considers efficient, fitting things together into the smallest possible volume. This is also what electrostatics want. Kepler suggested that the best way to pack spheres with this in mind can be maximized in two ways: fcc and hcp.

This conjecture, however, rests on all the spheres being uniform. We can’t assume this is the same for the atoms in table salt because the cation, sodium, is said to be smaller than the chloride anion. A tool that is useful as a guide for helping predict the structure is based off of the relative size of the cation, $r_\text C$, to the anion, $r_\text A$. This radius ratio, $r_\text C / r_\text A$, is mostly only useful for simple binary species. These $r_\text C$ and $r_\text A$ are tabulated as ionic radii in many chemistry books; they are contrived by various rules proposed by researchers, such as Linus Pauling, applied to the experimentally determined interionic distances. In this system we find it useful to discuss how many neighbors an anion or a cation has of the opposite charge.

The coordination number ($\mathrm{CN}$) tells us how many neighbors a type of atom has around it. A certain $r_\text C/r_\text A$ gives us a feel for the probable CN. From one source I was able to calculate the ratio for $\ce{NaCl}$ as $r_{\ce{Na+}}/r_{\ce{Cl-}} = 0.564$. Most textbooks will say that if your ratio is $0.414{–}0.732$, then you have $\mathrm{CN}=6$. I’ll show you how to calculate the minimum value for $\mathrm{CN}=6$. The easiest way to obtain the maximum is to obtain the minimum for $\mathrm{CN}=8$.

Briefly, if we put a small cation on a $XY$ plane and surround it in the manner shown below then we could also place two circles in the $Z$-axis above and below the central cation for $\mathrm{CN}=6$. (I won’t show those two in the $Z$-axis for clarity.) We would say that a given $\mathrm{CN}$ is stable only if the spheres are all touching each other. We would say that for the ratios drawn that a $\mathrm{CN}=8$ is not stable because our cation’s sphere would be too small and wouldn’t touch all eight of its neighbors. A higher density is always preferred, so $\mathrm{CN}=4$ is not preferred when $\mathrm{CN}=6$ is stable.

Clearly, $\angle \mathrm{DAC} = 45^{\circ}$. Moreover, we can say that $\overline{\mathrm{DC}} = r_\text A$ and $\overline{\mathrm{AC}} = r_\text A + r_\text C$. Trigonometry will tells us that $$\sin 45^{\circ} = \frac{\overline{\mathrm{DC}}}{\overline{\mathrm{AC}}}$$ Substituting in those values just determined,
$$\frac{\sqrt{2}}{2} = \frac{r_\text A}{\overline{r_\text A + r_\text C}}$$

Solving for $r_\text C/r_\text A$ gives,

$$r_\text C/r_\text A=\frac{2}{\sqrt{2}} -1 = 0.414$$

which is the lower bound desired.

So we can say that the $\mathrm{CN}=6$ for the cation. Similarly you can figure out that $\mathrm{CN}=6$ for the anion. This is why we say it has a 6:6 coordination pattern, which is indicative of a sodium chloride type structure. Cesium chloride, on the other hand, is 8:8 and indicative of its structure type. We tend to say that these patterns are consistent with a certain crystal structure because all the pieces fit together consistently. This geometric concept is based purely on sphere packing and is not without its limitations.

A more quantitative way to figure out which geometry is preferred is to calculate the electrostatic energies for the different geometries. A Russian by the name of Kapustinskii derived a formula to do this,

$$E_\text{lattice} (\mathrm{kJ/mol}) = \frac{\alpha}{r_\text C + r_\text A} $$

From here you need to put it in terms of the radius ratio with a constant $r_\text A$,

(I should note that this is evil algebra: $ \frac{a/b}{d/b + 1}= \frac{a}{b+d} $ where $a=\alpha$, $b=r_\text A$, $d=r_\text C$.)

$$E_\text{lattice} = \frac{\alpha/r_\text A}{\frac{r_\text C}{r_\text A} + 1} $$

Basically $\alpha$ is different for a change in the crystal structure type and from there you compare which has the preferred energy. You can reproduce the geometric trends using this more quantitative model. There are problems with this approach too, such as the neglected dipole and quadrupole and covalency. Yes, even in $\ce{NaCl}$ there is some covalent character. I think we say that $\ce{NaCl}$ $67\ \%$ ionic character. These become important when two different crystal structures are close in energy or when more exact calculations are required.


What makes a compound a dehydrating agent?

To answer your question, what makes a compound a dehydrating agent is, in my opinion, stability, a good affinity for water and possibly inertness (or in special cases, selective activity but in reforming a designated salt, see example below).

Now, $\ce{P2O5}$ is a white powder that loves water, in fact, it reacts violently and exothermically with water forming $\ce{H3PO4}$.  Phosphorus pentoxide is the anhydride of orthophosphoric acid, $\ce{H3PO4}$ , as dehydration of the latter yields $\ce{P2O5}$.

A safe, effective and inexpensive dehydrating agent is dry Calcium chloride. Again a strong affinity for water and a stable salt.

Per one preparation, one can dehydrate aluminum chloride hydrate by gentle heating in a stream of DRY hydrogen chloride gas (which is the gaseous anhydride of Hydrochloric acid), see here. Just heating the hydrate in air can result in basic aluminum chloride formation, but, I suspect, the dry hydrogen chloride gas can convert any created basic salt back to just $\ce{AlCl3}$ and water (removed by heating and contact with the dry gas). Note, gaseous $\ce{HCl}$ is inert with respect to chloride and stable.


You are correct. Another way to calculate it is: $$X_{\mathrm{L}} = \frac{s-b}{e-b} = \frac{0.18-0}{0.31-0}=0.58$$
You can also inspect the diagram visually and see that the $bs$ line is longer than the $se$ line. The lever rule dictates that the line length is proportional to the percentage of the phase opposite from the line. Thus, it is not possible that there is more solid than liquid at this point.

There is definitely a mistake in your textbook's solutions.


Can dimethylmercury originate from cleaning up a broken thermometer with disposable wet wipes?

According to the US EPA, disposable wet wipes are a recommended tool for cleaning up a broken CFL. Although having a number of adverse health hazards at high concentration BZK is a common ingredient in many sanitizers, wet wipes and disinfecting wipes, where it occurs at very low concentrations; typically much less than 1%. These ingredients are not likely to react adversely with the released mercury.

For future reference to cleaning up broken CFLs, be sure to read this guide.


It is possible that carbonic acid will form; however, only in small amounts because the equilibrium of the hydration of $\ce{CO2}$ to $\ce{H2CO3}$ lies mainly on the educt side. As more $\ce{CO2}$ is soluble in cold water, more carbonic acid will be present in the solution at lower temperatures.

$$\ce{CO2 + H2O \rightleftharpoons H2CO3}$$

At $25~\mathrm{^\circ C}$ and in pure water, the quotient of the concentrations is $[\ce{H2CO3}]/[\ce{CO2}] \approx 1.7 \times 10^{-3}$, so most carbon dioxide remains physically dissolved in water. This makes carbonic acid - despite a $\mathrm{p}K_\mathrm{a}$ of $3.6$ for the first dissociation step - a fairly weak acid. It is not a dangerous compound, otherwise it would not be used in carbonated drinks. It is also a natural constituent of blood, where it acts as a pH buffer. In red blood cells, it is produced as an intermediate by the enzyme carbonic anhydrase, which catalyzes the hydration of carbon dioxide.


Regular decrease in the atomic radius of 3d series

Periodic Table Website confirms these atomic radii.
So why does the atomic radius decrease? The nucleus of the atom gains protons moving from left to right, which increases the positive charge of the nucleus which attracts electrons. Although electrons are also added as the elements move from left to right across a period, these electrons reside in the same energy shell and therefore do not offer increased shielding.

Source: Periodic Trends


What do typical results from inductively coupled plasma mass spectrometry (ICP-MS) look like, and how do I interpret them?

I am not an expert in ICP-MS, but an ICP-MS should be considered an ICP ion source coupled with a mass spectrometer. There are some key issues raised by the coupling between both instruments due to the high temperature of the ICP source as well as due to the fact that one wants to analyse atomic ions and remove any interfering molecular ions (mostly oxides, but other molecules and clusters are possible).

Therefore, to attempt an answer to your question, the standard output of an ICP-MS will be presented as a mass spectrum (intensity versus m/z or intensity versus amu). There are a couple of sample spectra in the brochures published by instrument makers, see for instance the Bruker brochure or a Thermo application note. How this mass spectrum is obtained from the detector will lead to the type of data that is acquired:

For a quadrupole mass filter, as well as for magnetic sector instruments, two modes of operation are possible. Either a single mass is selected, and therefore ont will only get a trace of the abundance of a given mass as a function of time (chromatography or on-line sample analysis). Otherwise the mass filter can be scanned, leading to a mass spectrum.
For time of flight detectors, the mass spectrum is built by the measurement of a flight time for an ion to reach the detector. Therefore the mass spectrum is converted from a time dependent signal to a spectrum.
Finally there have been some attempts to couple an ICP ion source with a Fourier transform mass spectrometer. As far as I know, these are not commercial instruments. In such instruments, the ions oscillate at m/z dependent frequencies, and the recorded image current is converted to a spectrum through the use of Fourier transform.


The pH of acidic solutions does not change much with increasing temperature. The pH of basic solutions does change more significantly with increasing temperature.
Ex. : The pH of $0.05$ M Potassium tetroxalate is $1.67$ at $0$°C to $1.81$ at $95$°C
The pH of $0.05$ M potassium hydrogenophtalate is $4.012$ at $0$°C and $4.23$ at $95$°C.
The pH of pure water is $7.47$ at $0$°C and $6.13$ at $100$°C
The pH of $0.1$ M borax is $9.46$ at $0$°C to $8,83$ at $95$°C.
The pH of $0.02$ M calcium hydroxyde is $13.42$ at $0$°C, and $11.45$ at $75$°C
Ref. : D. R. Lide, Handbook of Chemistry and Physics, CRC Press, 43rd Ed. 1961, and 83rd Ed., 2002.


How to name the compound with the formula Br3C-CHCl2?

The first name is correct (1,1,1-tribromo...).  Counting the substituents from either end, you would get 1 for the first substituent going from either direction.  As a tie-breaker, look at the next substituent, which would give you 1,1 going from the bromine end, and 1,1 going from the chlorine end.  Since the tie is still not broken, you go to the third substituent.  Going from the bromine end you would have 1,1,1 but 1,1,2 going from the chlorine side.  So, you number the carbon with the bromines number 1, giving you 1,1,1-tribromo-2,2-dichloroethane.

As a side note, if the carbon on the right had three chlorines instead of two, this method wouldn't work (you'd get 1,1,1,2,2,2 numbering from either direction).  In this circumstance you would use alphabetical order to break the tie, with bromo- having higher priority than chloro.  That would give you 1,1,1-tribromo-2,2,2-trichloroethane.  I know you didn't ask this, but it could prove useful to know!


How is Carboxymethyllysine formed from the heating of olive oil?

Given that the levels are expressed in arbitrary units (based on the results of an immunoassay), it's difficult to say how much CML there actually is and accordingly, how much lysine is needed to produce it. It may be that only a small quantity of lysine is required) In fact, a different study found no CML in raw olive oil.

CML is produced during cooking from glycation of the side chain amine, starting with the Maillard reaction between the amine and a reducing sugar's carbonyl at high temperature. However, in the paper you cited, there was already significant CML in raw olive oil that was only somewhat increased by cooking. This paper suggests that CML formation is related to the oxidation of fatty acids and found that oils with increased concentrations of unsaturated fatty acids (olive oil is in the middle of the 5 oils they tested) were more prone to forming CML, but they deliberately added lysine to test it. It could be that this oxidation is something that happens in storage and the study you linked was using old oil or something. Also, just as with any other plant derivative, there could be significant variation in lysine content from different olives. This site indicates a lysine content of 3 mg/8.3 g whole olive, so there is some there. How much actually makes it into oil, I don't know.


Why do seemingly reliable sources give NFPA 704 Health ratings ranging from 1-4 for elemental mercury?

I believe that the radical rage of possible hazard ratings for elemental mercury arise from different assumptions about how the mercury is handled. I can't prove this, but it makes sense when you consider the possible hazards arising from mercury use.

Properly handled liquid mercury is not particularly harmful. There are no acute risks and a competent person could drink it and handle it with little risk. In a chemical laboratory, for example, it is common to use mercury manometers to provide limited gas overpressure for supplies of nitrogen or argon. This involves bubbling gas through mercury. As long as the vapour is safely vented to a scrubber, this is not a hazard to people in the lab.

Equally, chemists can use mercury safely with few precautions as long as they clean up any spills thoroughly. The hazard primarily comes from vapour buildup which, if it is allowed to happen, will create serious long term chronic problems.

One example of potential danger is if non-experts come into contact with mercury. Some old gas-flow meters for domestic gas supplies used mercury which could be discharged from the meter in certain rare situations, leaving droplets of mercury around the floor near the meter. This can (and has) created a great deal of hazard in a domestic environment. This hazard is magnified enormously, for example, if the householder tries to clean the mercury spill by vacuuming it up. The result is a fine spray of mercury droplets inside a warm vacuum cleaner, which vastly increases the vapour emission and may spread fine droplets widely through the house. If badly ventilated, this can be a serious chronic issue.

The two situations span a wide range of hazard. But the difference is the context: in experienced hands mercury is not very hazardous; in other hands it is potentially very dangerous.

In short, the different rating may reflect different assumptions about who is using the mercury: someone who knows what they re doing or someone who does not.


Why am I getting higher conversion values of isopropanol dehydration over a zeolite catalyst than n-propanol dehydration?

In a typical acid-catalyzed dehydration, the reaction goes by an SN1 mechanism, and the rate-limiting step is
$$\ce{R-OH2+ -> R+ + H2O}$$
The rate of this step is typically a function of the stability of the cation product. In your case, the secondary cation formed from isopropanol is more stable than the primary cation formed from n-propanol.
I'm using "typical" and "typically" here, because these are results from homogeneous reactions, and it sounds like you are doing heterogeneous catalysis with a solid zeolite. That can certainly change the reaction behavior, but in your case it appears that the zeolite is not selective for one molecule over the other, and the trend is the same as what is observed for homogeneous systems.


Origin of Virtual Molecular Orbitals in Hartree-Fock equations

Yes, they result from the HF equations just like the occupied orbitals. The diagram should also be show them as present in the initial set of equations, unless the basis  (used for the LCAO) was changed between the two states shown. The number of occupied orbitals plus the number of virtual orbitals is equal to the basis set size because they are the eigensolutions of
$$
 \mathbf{F}\mathbf{C} = \mathbf{\epsilon}\mathbf{S}\mathbf{C},
$$
where $\mathbf{F}$ is the Fock matrix, $\mathbf{C}$ is the matrix of orbital coefficients, $\mathbf{\epsilon}$ are the eigenvalues which are the orbital energies, and $\mathbf{S}$ is the atomic orbital overlap matrix (because the basis set used is typically not orthogonal). The dimension of all matrices involved is determined by the size of the basis set, which will be finite because otherwise the computational effort becomes infinite.


What is the chemical basis for superglues?

This is a very general answer to your question.

According to the Chemistry World podcast transcript, Cyanoacrylate and the article The Chemistry of Superglue (Brubaker, 2000), the main ingredient cyanoacrylate, which is

  a type of plastic known as an acrylic resin. It exists in the tube as single molecules in liquid form. But when they come into contact with water, the molecules react with hydroxide ions to form long polymer chains that set into a hard solid. As there is a thin layer of moisture covering most surfaces, cyanoacrylate can bond them together very effectively.

This also explains why superglue tends to harden very quickly when in contact with skin.

A diagram of cyanoacrylate polymerisation in the presence of water is shown below:

from the University of Rochester, who also state that asides from water, that the presence of

  Weak bases (such as those on the proteins in skin) work even better

for polymerising the cyanoacrylate.


How do I make a dysprosium chloride solution from dysprosium oxide?

DyCl3 is often prepared by the "ammonium chloride route," starting from either Dy2O3 or hydrated chloride or oxychloride.[2][3] or DyCl3·6H2O.[4] These methods produce (NH4)2[DyCl5]:

    10 NH4Cl + Dy2O3 → 2 (NH4)2[DyCl5] + 6 NH3 + 3 H2O

    DyCl3·6H2O + 2 NH4Cl → (NH4)2[DyCl5] + 6 H2O

The pentachloride decomposes thermally according to the following equation:

    (NH4)2[DyCl5] → 2 NH4Cl + DyCl3

The thermolysis reaction proceeds via the intermediacy of (NH4)[Dy2Cl7].

Treating Dy2O3 with aqueous HCl produces hydrated chloride (DyCl3·6H2O). This salt cannot be rendered anhydrous by heating. Instead one obtains an oxychloride.

source: wikipedia


How to find the number of stereoisomers of a tetra-substituted allenes, exhibiting both axial and central chirality?

Axial and planar are just adjectives we can use to describe what type of chirality a molecule possesses.  Usually, just saying the molecule is chiral is good enough.  Being planar or axially chiral does not create any new enantiomers or diastereomers compared to if we just said the molecule is chiral.

Let's analyze your problem using standard methods - no special techniques just because the molecule happens to be axially chiral.

I get the same 4 diastereomers that you've drawn above (good job!).  When analyzing allene stereochemistry I've always found it convenient to analyze them using Newman projections.  I've redrawn the 4 diastereomers below using Newman projections.

In the top diastereomer, all 4 chiral substituents are the same; the molecule has a non-superimposable mirror image (enantiomer).  Therefore this diastereomer is chiral.

In the second diastereomer, 3 chiral substituents are the same, one is different; again, the molecule has a non-superimposable mirror image (enantiomer).  This diastereomer is also chiral.

In the third diastereomer, two chiral substituents with different chirality are attached to the same carbon; again, the molecule has a non-superimposable mirror image (enantiomer).  This diastereomer is also chiral.

In the fourth diastereomer, two chiral substituents with the same chirality are attached to the same carbon.  This diastereomer belongs to the interesting point group $\ce{S_4}$.  I say interesting because this point group does not contain a $\ce{\sigma}$ plane, but is still nondissymmetric (achiral).  Look at my Newman projection for this diastereomer and picture it lying behind the plane of the screen.  Reflect it in the plane of the screen and then rotate it 90° (n=4, 360/4 = 90°) and you have recreated the original molecule. This molecule is achiral (all molecules belong to point group $\mathrm{S}_n$ [n is an even integer] are achiral).

So what do we wind up with?  4 diastereomers, 3 of which are chiral (enantiomeric pairs) and one is achiral.


Order of Lewis acid character among sulfur hexafluoride, sulfur tetrafluoride and tetrachloromethane

S is in its highest oxidation state (+6) it has no electrophilicity hence not a Lewis acid.
Also, CCl4 can't have more than 8 electrons in its outermost orbit and it can not also increase it's coordination number also so it can't behave as a Lewis acid.
SF4 can accept electrons as S isn't in its highest oxidation state.
Hence order of Lewis acid character is : SF4 > CCl4 = SF6


What happens when pyridine N-oxide reacts with dichlorocarbene?

Dichlorocarbene has been shown to deoxygenate pyridine N-oxide producing phosgene as an initial byproduct. In this case, since methoxide is present, dimethyl carbonate is ultimately formed.[1]

The proposed mechanism involves nucleophilic attack on the carbene by the oxygen, followed by an elimination:

(1) Schweizer, E. E.; O'Neill, G. J. A Novel Deoxygenation Method for Pyridine N-Oxide. J. Org. Chem. 1963, 28 (9), 2460–2461. DOI: 10.1021/jo01044a516.


The mole is used extensively in chemistry, why not elsewhere?

Because the mole is not a convenient basis for people and TV sets (for instance, one mole of people would be approximately 86 trillion (short-scale) times the population of earth, or put another way the earth contains a human population concentration of approximately 12 femtomoles/planet), and because there's no connection back to atomic mass, which is what makes the mole useful in the first place.

This was not part of your original question, but addresses Manishearth's modifications:

Because atoms and molecules react in discrete units (cf. the law of definite proportions), the number of particles is more fundamental than the masses of those particles. The mole is useful in chemistry because it is defined such that the mass of Avogadro's number of particles (atoms or molecules) of a substance (the molar mass) is equal to the sum of the numerical values of the atomic masses of its constituent elements. The value of the mole is calibrated such that the molar mass of carbon-12 is exactly 12 grams per mole. On the periodic table you will note that the atomic masses given are non-integral (often markedly so, e.g. chlorine at 35.45 g/mol). This is because the atomic masses represent a weighted average of isotopic masses, and because the proton and neutron masses aren't quite equal.

The mole makes stoichiometric calculations particularly convenient as we can measure out, for instance, 16 grams (one mole) of $\ce{^{16}O}$ and 2 grams (two moles) of $\ce{^1H}$ and obtain one mole of $\ce{H2O}$. In the case of water, oxygen makes up 89% of the molecule by mass, but is only one of the three atoms present in the molecule. The mole is necessary to transpose stoichiometries involving single atoms and molecules (with their tiny masses) into the everyday world of grams and litres.