Find the displacement of mercury in a vertical tube

I assume the Hg is a competent barrier between the parcels of air and the parcels both have an initial pressure $P_0$ of 1 bar (760mm of Hg). Each air parcel has an identical initial energy content of $$ P_0V_0 = P_0A(0.45\mathrm{m}) $$

With hydrostatic equilibrium, $\Delta P = P_{\mathrm{bot}} - P_{\mathrm{top}}$ would all be due to the weight of the Hg, IOW, in mm of Hg terms, would be 100mm (0.1m) of Hg. -- This is your second equation. $$\Delta P = P_{\mathrm{bot}} - P_{\mathrm{top}} = 100\, \mathrm{\scriptstyle mmHg} $$From Boyle's Law we know that $$ P_{\mathrm{top}} = P_0 \frac{V_0}{V_{\mathrm{top}}} \quad\quad\text{and}\quad\quad P_{\mathrm{bot}} = P_0 \frac{V_0}{V_{\mathrm{bot}}} $$ Substituting, $$ \Delta P = P_0(\frac{V_0}{V_{\mathrm{bot}}} - \frac{V_0}{V_{\mathrm{top}}}) $$ or $$ (\Delta P) V_{\mathrm{top}} V_{\mathrm{bot}} = P_0V_0(V_{\mathrm{top}} - V_{\mathrm{bot}}) $$ Substituting again, $$ (\Delta P) A^2 ( 0.45\mathrm{m} - x )( 0.45\mathrm{m} + x ) = P_0A^2(0.45\mathrm{m}) ( (0.45\mathrm{m} - x) - (0.45\mathrm{m} + x) ) $$ eliminating $A^2$ and simplifying $$ \Delta P ( (0.45\mathrm{m})^2 - x^2) = -2P_0 x(0.45\mathrm{m}) $$ or in quadratic form $$ x^2 - 2\frac{P_0}{\Delta P}(0.45\mathrm{m})x - (0.45\mathrm{m})^2 = 0 $$ which would be $$ x^2 - (6.84\mathrm{m})x - 0.2025\mathrm{m}^2 = 0 $$ You can take it from there.


What is the pOH of a 0.0100 F solution of dipotassium phtalate?

So, we know that dipotassium phtalate is a conjugate base of a diprotic acid, so the addition of it into water will remove some $\ce{[H+]}$s from $\ce{[H2O]}$ and increase the level of $\ce{[OH^{-}]}$ with means a lower pOH.

We calculate the pOH by first converting to $\mathrm{K_a}$
$$\mathrm{pK_{a2}=10^{-5.408}=3.908*10^{-06}}$$
$$\mathrm{K_{b1}=\frac{K_w}{K_{a2}}= \frac{1*10^{-7}}{3.908*10^{-06}}=2.559*10^{-2}}$$
$$\ce{\frac{[HA][OH^-]}{[A^-]}=\mathrm{K_{b1}}}$$
$$\mathrm{\frac{x^2}{F-x}=2.559*10^{-2}\rightarrow 1.60*10^{-2}}$$
We did that last one assuming that x was negligibly small.
We now know
$$\ce{[HA]}=\mathrm{x=1.60*10^{-2}}$$
$$\ce{[H+]}=\mathrm{\frac{K_w}{\ce{[OH^-]}}=\frac{K_w}{x}=6.25*10^{-6} \rightarrow pH=5.20_4}$$
$$\mathrm{pOH=14-pH \rightarrow 14-5.20_4 = 8.79_6}$$


Best place to take off vacuum for distillation

Assuming that you are using a typical distillation set up (boiling container, fractionating column, distillation head, condensation tube, collecting adapter, and collecting vessel), and the the "three-way" you are talking about is the distillation head. It is much better to put the vacuum take off adapter right before your collecting vessel. The reason is this set up creates a vapor flow that goes through your condensation tube, and most of your fractions will be condensed and collected.

If you attach the vacuum take off adapter on top of your distillation head, most of your vapor will take shortcut to your pump through your adapter instead of going to the condensation tube and the collecting vessel (this direction is a dead end).

Actually the top of the distillation head is the place to put your thermometer to measure the temperature of vapor, not for anything else.


Why do we have glycogen in the muscles and not glucose?

I think you already answered your own question.  If the glycogen, either in liver or muscle, were all converted to glucose at once it would have a disruptive osmotic effect. But muscles also readily break down glycogen, when they need glucose to contract.  They just don't ship the glucose to other organs the way liver does.


Why are higher oxides of non-metals more acidic?

It is expected oxides with more polar $\ce{M-O}$ bonds are stronger Lewis bases than those with less polar bonds.

Furthermore, for the same element, the polarization gets weaker/Lewis alkality weaker when more $\ce{O}$ atoms are attached.

When hydrated to acids, another phenomena plays a role in acidity: a charge delocalization.

It is well observable that acids like $\ce{HNO3}$, $\ce{HClO3}$, $\ce{HClO4}$, $\ce{H2SO4}$, $\ce{HBF4}$ are strong, as the delocalization  has a good sandbox to play in.

It also plays role in solubility of salts in organic polar solvents, where there is a large  organic cation with one of the above small anions.


Producing soap from sodium carbonate and oil

In nature fat is stored as triglycerides (TGs): Three fatty acid (FA) chains bound together by a glycerol backbone. Olive oil is mainly composed of TGs, though it also contains small amounts of free fatty acids (FFAs).

An example of Triglyceride (from Wikipedia)

Soap is commonly made via the reaction of triglycerides  with a strong base: usually Sodium Hydroxide ($\ce{NaOH}$) or Potassium Hydroxide ($\ce{KOH}$). Though $\ce{Na2CO3}$ is a base, it is too weak of a base to be efficiently used in the traditional production of soap: It may saponify triglycerides but only at a very slow rate, if at all. So you are unlikely to produce much soap by mixing $\ce{Na2CO3}$ (of any source) with oil.

Soap can however also be made by reacting a weaker base with free fatty acid (FFA) chains; that is, FAs that have already been broken away from the TGs. $\ce{Na2CO3}$will form salts with FFAs, and it will do so at a much faster rate than it would saponify TGs. It can thus be used to make soap that way instead (and will, for instance, react with the small amount of FFAs in your oil).

You could try to find a way to separate the FFAs from the rest of the oil (though it would be a highly inefficient process as you'd be left with the dominant TG portion), you could in theory also separately break down TGs into FFAs prior to saponifying them (using enzymes for instance), or (highly recommended) you could go the traditional route of using a strong base which will efficiently and (relatively) rapidly saponify both TGs and FFAs in your oil.

Splitting triglycerides is the "hard work" from a chemical point of view.

Sources

https://en.wikipedia.org/wiki/Saponification
https://en.wikipedia.org/wiki/Triglyceride
https://en.wikipedia.org/wiki/Fatty_acid
https://en.wikipedia.org/wiki/Sodium_carbonate
https://www.soapmakingforum.com/threads/how-much-sodium-carbonate-to-use-for-soapmaking.60798/


I believe that you are mixing up the concepts of electronegativity and of electron affinity.
The first statement you present is related to Electronegativity.

  If two similar sized atoms have different electronegativity values, the atom with higher electronegativity holds the electrons more tightly, and thus is more stable.

Yes. This is generally true.
Electronegativity establishes how well an element can pull electron density towards itself.

The second statement is about electron affinity.

  If two atoms have different sizes (down the periodic table), the larger atom has electrons more spread out and thus is more stable. Example: fluoride vs iodide anion

Generally true as well. Flouride anion is less stable than iodide anion due to the repulsion of electrons in the outermost octet of fluoride. But this rule is not absolute as chloride has the highest stability (in the gaseous state) though it is above bromine and iodine.

Edit:
Considering some confusion in the comments I shall put this here.

The electronegativity is generally not compared based on inter-electron repulsion because the shared pair doesn't completely enter the orbits of the electronegative atom but is shared and so electron density is between the two atoms and not concentrated on one atom for repulsion to take much effect.


Effect of +M and +R effect in basicity of 4-methoxy aniline

Since you mentioned the $\mathrm +M$ effect, I'm assuming you're familiar with resonance and aromaticity.
This is a structure of cyclohexane amine:

Here, the lone pair of electrons present on the Nitrogen atom make it a Lewis Base (an electron donor).
This is a structure of 4-methoxy aniline:

Here the lone pair of Nitrogen is delocalized due to resonance. Hence it's not readily available for donation. That's why 4-methoxy aniline is a weaker base.


Use of aqueous KOH and alcoholic KOH in dehydrohalogenation reactions

Aqueous $\ce{KOH}$ is alkaline in nature i.e. it dissociates to produce a hydroxide ion. These hydroxide ions act as a strong nucleophile and replace the halogen atom in an alkyl halide.

$$\ce{RCl + KOH (aq) -> ROH + KCl}$$

This results in the formation of alcohol molecules and the reaction is known as nucleophilic substitution reaction.

Alcoholic, $\ce{KOH}$, specially in ethanol, produces $\ce{C2H5O-}$ ions. The $\ce{C2H5O-}$ ion is a stronger base than the $\ce{OH-}$ ion. Thus,the former abstracts the ß-hydrogen of an alkyl halide to produce alkenes. This reaction is known as elimination reaction.

$$\ce{CH3CH2Br + KOH (alc) -> H2C=CH2 + KBr + H2O}$$


Reaction in Zinc-Carbon-Ammonium Chloride battery

Surprisingly this battery uses Zn without any oxidant like $\ce{MnO_2}$ and it  produces a voltage as high as $1.1$ V. In order to establish its characteristics, the two compartments should be studied independently.
First, the anode, with zinc. $\ce{Zn}$ should not produce simple $\ce{Zn^{2+}}$ ions, as there are ammonia and ammonium chloride in the solution. So the anode reaction is probably :  $$\ce{Zn  + 4 NH_4^+ ->  [Zn(NH_3)_4]^{2+}  +  4 H+  +  2 e-}\label{rxn:zn4}\tag{1}$$ The reduction potential for this anodic equation is not given in the tables. But it may be obtained from the next equation, whose standard reduction potential is known to be $\ce{E_2° = - 1.03 V}$.  $$\ce{Zn  + 4 NH3 ->  [Zn(NH_3)_4]^{2+}  +  2 e-}\label{rxn:zn}\tag{2}$$  In a $1$ M solution of $\ce{NH4Cl}$, $\ce{[NH_3]}$ = $2.4 ·10^{-5}$ M. So the corresponding reduction potential of ($2$) is : $$\ce{E_2 = - 1.03 \mathrm{V} + 0.0296\mathrm{V}· log (2.4 ·10^{-5})^4} = - 1.03~ \mathrm{V} - 0.54~ \mathrm{V}  = -1.57 ~\mathrm{V}\label{rxn:nh}\tag{3}$$
Now the cathodic reaction should be studied. Suppose the cathode reaction is  : $$\ce{2 H2O  + 2 e- -> 2 OH- + H2}\label{rxn:z}\tag{4}$$ with a redox potential equal to  $- 0.83$ V. Couping these two expressions ($2$) and ($3$) are never sufficient to get $-1.1$ V as the overall voltage. So, as the ammonium chloride solution is acidic, the cathode reaction must rather be : $$\ce{2 H+ + 2 e- ->  H2}\label{rxn:znn}\tag{5}$$ In a $1$ M  $\ce{NH4Cl}$ solution, the pH is $4.62$, so the corresponding potential of the cathode is E$\ce{_5}$ = - $0.059$ pH = - $0.059·4.62$ V = - $0.27$ V.
To summarize, the anode reaction is ($1$) or ($2$), at will, and the cathode rection is ($5$). As a consequence, the theoretical voltage of the cell is : $\ce{E_5}$ - $\ce{E_2}$ = - $0.27$ V + $1.57$ V = $1.3$ V, if the concentration of ammonium chloride is $1$ M.
Ref.: Standard electrode potentials have been taken from "Chemistry Data Book" from J. G. Stark and H. G. Wallace, John Murray, London, $1994$, Table $43$, p. $70$.


The NEW Periodic Table Song (Updated)
AsapSCIENCE
https://www.youtube.com/watch?v=VgVQKCcfwnU


Why is electron gain generally exothermic?

Energy of electron in vacuum is zero. If it can attach (however weakly), the energy is gained and process is therefore exotermic. It is not an atom which releases energy, it is the whole system.
Exactly opposite of previous. The atoms you mentioned gladly release electrons to other atoms, but not to vacuum (which is how ionization energy is defined).
In some cases, adding of extra electron to neutral atom is not an exothermic process. See Periodic table of electron affinities. For example, a helium atom has filled shell with $n=1$. The next electron would be accepted to the shell $n=2$ which has much higher energy, and therefore is not stable. For the half-filled orbitals, similar reasoning applies, but spicing it with Hund's rule.


Synthesis of ethyl methyl ketone from 2-(4-hydroxyphenyl)ethyl methyl ketone

There are a number of ways to accomplish your goal, but they all involve several synthetic steps. As mentioned in the comments, it's always easier to buy a compound then run a reaction, but if you had a small amount of some exotic derivative, you could convert it to methyl ethyl ketone.

One possible route would be to convert the raspberry ketone to an ester using the Baeyer-Villager reaction.  In this reaction the secondary (ethyl) branch migrates in preference to the primary (methyl) side.  The resulting ester can then be converted to the desired ketone by any number of methods such as the Weinreb route.


Polarity of alcohols and their miscibility in water

If your book says exactly that, then it says nonsense. Sure enough, the addition of CH2's makes a molecule less polar. As for the solubility in water, methanol is miscible - i.e., infinitely well soluble in it. So is ethanol. How would you compare the two infinities?

That being said, the bigger molecule of ethanol is indeed more polarizable (not more polar, mind you!), and has more intermolecular interactions per molecule than methanol, which accounts for the higher boiling point.


Can protons be swapped in a chemical reaction?

Roughly speaking, no.   This would amount to nuclear fission/fusion, and is generally not considered "chemistry" in any colloquial sense of the term, but rather physics.

In general, chemical reactions consider nuclei to be tightly bound by the strong force, and immutable over the course of the reaction.

Disclaimer: there is an ill-defined border between physics and chemistry, and so one could argue that I am wrong about this.  Indeed, alchemists had/have been trying to e.g. convert lead to gold for centuries, and this is related to the reaction that you propose.


Would nodal planes of participating Atomic orbitals be nodal planes of molecular orbital?

AFAIK: The nodal planes are where the wavefunction sign inverses. If you drew the colors according to the sign value then you would see that for antibonding sigma MO there is a virtual nodal plane, since the sign changes between + and - between the two centre orbital lodes. Even if they by some magic grew larger, they can never touch, they interfere destructively. There will always be a plane between them where the wavefunction returns 0. (when the two spheres meet they will form a planar interface. Imagine soap bubbles.)

I found this, it is more descriptive; https://socratic.org/questions/how-can-i-draw-antibonding-orbitals


Change in pH when boiling off acidic water

$\rm H^+$ ions (as well as any other ions, for that matter) do not boil off. Acid does. Depending on the nature of the acid, it might boil off faster or slower than water, and hence your pH might shift up or down.

$\ce{HCl}$, in particular, forms a high-boiling azeotrope with water at 22%, so the solution of pH 2 (which is pretty dilute) would form a vapor enriched with water, and the liquid phase would get enriched with acid. So despite the seemingly volatile nature of $\ce{HCl}$, it will mostly stay in the solution and make it progressively more acidic as it boils.


How to convert micromolar concentrations into molar quantities, when the volume is also known

Molar concentration, or molarity, is moles per unit volume. If you know concentration and sample volume, you can calculate the moles of a particular component in the sample. Since your concentrations are already in micro-molar, when you multiply by the sample volume (in liters), you will have the micro-moles of each component.

For the Mn result, you will get the mass of Mn (in micrograms) when you multiply the test result by the sample volume, so you can convert that to moles using the molar mass of Mn.


Meaning of the symbol “$\Delta_\mathrm{sub}H^\circ[\ce{I2,s}]$”

$\ ΔH_{sub}(I_2,s)$ can be interpreted as the change in enthalpy when one mole of solid $\ I_2$  converts to gaseous $\ I_2$  at a constant temperature. So the answer should be reaction (b).


What comes after f-orbitals, and in what order are they filled?

After s, p, d, and f, the next letters follow in alphabetical sequence: g, h, i, k, l, and so on. Note that j is omitted from the sequence: see Why is the letter J omitted in the spdf... sequence? for more information. Since g-subshells would have $l = 4$, the lowest g-subshell would be the $\mathrm{5g}$ subshell, bearing in mind $l$ must be strictly less than $n$.

According to the aufbau principle, orbitals are filled in increasing order of $(n+l)$. Amongst orbitals with similar $(n+l)$, those with smaller $n$ are filled first:

$$\begin{array}{ccc}
\text{Orbital} & \hline (n+l) & n \\ \hline
\mathrm{1s} & 1 & 1 \\ \hline
\mathrm{2s} & 2 & 2 \\ \hline
\mathrm{2p} & 3 & 2 \\
\mathrm{3s} & 3 & 3 \\ \hline
\mathrm{3p} & 4 & 3 \\
\mathrm{4s} & 4 & 4 \\ \hline
\mathrm{3d} & 5 & 3 \\ \hline
\end{array}$$

Following this principle you would theoretically have the order

$$\mathrm{\underbrace{1s}_{\text{Period 1}} \to \underbrace{2s \to 2p}_{\text{Period 2}} \to \underbrace{3s \to 3p}_{\text{Period 3}} \to \underbrace{4s \to 3d \to 4p}_{\text{Period 4}} \to \underbrace{5s \to 4d \to 5p}_{\text{Period 5}} \to \underbrace{6s \to 4f \to 5d \to 6p}_{\text{Period 6}}} \\
\mathrm{\to \underbrace{7s \to 5f \to 6d \to 7p}_{\text{Period 7}} \to \underbrace{8s \to \mathbf{5g} \to 6f \to 7d \to 8p}_{\text{Period 8}} \to 9s \to \cdots}
$$

However, at such large atomic numbers, the aufbau principle starts to fail. The subshells are much more closely spaced in energy and the exact electronic configurations are not so clear-cut: this is already evident within the d- and f-blocks where some anomalies are present.

On top of this, relativistic effects (which become significant at large atomic numbers) cause subshells to split into two distinct energy levels (except for the s-subshell). See also: Shape of the P1/2 Orbital.


Maybe you would understand better with an image, or if you admit that all metals "are wanting" to loose electrons. And in order to do that, they must find another atom which is weak enough or "humble" enough to accept to work "against its will". In other words, some atoms are more "willing" than others. It is like a fight, with a "strong" and a "weak" atom. In your copper-silver cell, copper is more assertive : it imposes its electrons to the "weak" silver atom. Silver cannot emit its own electrons. It is even obliged to accept electrons from copper and have these electrons react with its proper ions. producing more silver atoms. Some atoms are stronger than copper. Zinc for example, opposed to copper, imposed its electrons to copper. This order of "strength" is experimental, and cannot be derived by calculations.
Now to speak more scientifically, you may replace the present notion of strength by the reduction potential. The more the reduction potential is negative, the more the metal has a tendency to loose its electrons.
Anyway, a salt bridge is absolutely necessary for the cell to work. If there is no salt bridge, it is not a cell, and it produces no chemical reaction, and no electric currant.

Complement, required by Manit.
If you want a local phenomena to explain the cell, you may consider the following development. You may admit that copper "wants" to create more $\ce{Cu^{2+}}$ ions in solution. These ions are positively charged. So the copper compartment must attract negative ions to compensate the new positive charges appearing around the copper plate. As a consequence, the new $\ce{Cu^{2+}}$ ions attract negative ions from where they are available, namely in the other compartment. And as $\ce{Ag}$ is not able to create $\ce{Ag+}$  ions with the same force, Silver plate must admit that it looses its negative nitrate ions to send them to the copper compartment. And as a consequence, it must admit that the positive ions $\ce{Ag+}$ should also "disappear" from the compartment. And the only way to make this charge disappear from the compartment is to accept electrons, so that the $\ce{Ag+}$ ion is transformed into metallic silver.
OK ?


Do I use the Nernst equation when the concentrations of electrolyte in both half cells are equal?

Yes, you still use the Nernst equation, but one of your implicit assumptions is not correct.

A point about the Nernst equation that often confuses people is that, at first glance, it doesn't predict any change in $E$ as temperature changes, as long as $Q$ remains equal to $1$, and $\ln Q = 0$.

$$E = E^\circ - \frac{RT}{\nu F}\ln Q \tag{1}$$

However, a slightly more careful analysis reveals that if you set $\ln Q = 0$, it only implies that $E = E^\circ$. The conclusion that $E$ doesn't change is based on an assumption that $E^\circ$ doesn't change, which is not true.

"Standard conditions" do not stipulate any temperature. So $E^\circ$ itself is a function of temperature. The standard cell potential $E^\circ$ at $\pu{298 K}$ is going to be different from the standard cell potential $E^\circ$ at $\pu{250 K}$, and hence $E$ will vary with temperature, even if $\ln Q = 0$.

So, how does one determine the temperature dependence of $E^\circ$? The best way is to measure it, but if that is not possible, then you have to somehow find an expression for it. The Nernst equation alone has no answer for this. Instead you would have to turn to thermodynamics. We know that

$$E^\circ = -\frac{\Delta_\mathrm rG^\circ}{\nu F} \tag{2}$$

Note that $\Delta_\mathrm rG^\circ$ is a function of temperature, and hence so is $E^\circ$. So, the problem becomes one of determining the variation of $\Delta_\mathrm rG^\circ$ with temperature. The most primitive way is probably to look up the data for $\Delta H_\mathrm{f}$ and $S_\mathrm{mol}$ of each of the compounds, calculate $\Delta_\mathrm{r} H^\circ$ and $\Delta_\mathrm{r} S^\circ$ for the reaction (using Hess's law), and then find

$$\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T\Delta_\mathrm{r} S^\circ \tag{3}$$

Of course, when you look up data, the chances are that the data you find are specified for $T = \pu{298 K}$. So, when you use these data to calculate $\Delta_\mathrm{r} H^\circ$ and $\Delta_\mathrm{r} S^\circ$, you are finding the values of these quantities at $T = \pu{298 K}$. If you plug these values into equation $(3)$, then you also assume that the values of $\Delta_\mathrm{r} H^\circ$ and $\Delta_\mathrm{r} S^\circ$ at your desired temperature are equal to their values at $\pu{298 K}$.

Depending on how much accuracy you need, this may or may not be tolerable, and there are certainly more thorough ways of calculating it. However, that discussion is best left for another question.


Well, @NilayGhosh already commented on stability of $\ce{KI3}$ in aqueous solution, also $\ce{K+}$ is not bulky enough. But as long as OP didn't mention explicitly that it must be water we isolate potassium triiodide from, there is a workaround, if one desperately want to get its crystals (in an altered form though).

From isopropanol one relatively easy obtains red diamond-shaped crystals of (18-Crown-6)-potassium tri-iodide (1) as the result of reaction between potassium isopropylxanthate and iodine:

There are also many other examples in the literature showing how one can man make potassium "bulkier", but I found this particular example the most straightforward.

Of course, there is always a fallback option to crystallize ammonium triiodide $\ce{NH4I3}$ (2) as ammonium is probably the closest cation resembling potassium for this case.

(1) Sievert, M.; Krenzel, V.; Bock, H. Zeitschrift für Kristallographie - Crystalline Materials 2010, 211 (11), 794–797. DOI: 10.1524/zkri.1996.211.11.794
(2) Cheesman, G. H.; Finney, A. J. T. Acta Cryst B, 1970, 26 (7), 904–906. DOI: 10.1107/S0567740870003357


Isopropyl lithium and ethyl bromide would be my go-to if I were to try to prepare this stuff (which I will call TAA). This requires a solid fume hood, and the resulting mixture of salt, TAA, and isopropyl lithium solvent (usually n-hexane) should be easy to separate. The hexane can be stripped from the TAA via fractional distillation, with more distillation passes resulting in a more pure product. TAA boils at $\pu{102 ^\circ C}$ (Sigma Aldrich), and n-hexane boils at around $\pu{70 ^\circ C}$ - this $\pu{30 ^\circ C}$ gap in BP passes the rule-of-thumb "can I separate with distillation?" question.

All that said, even pricey Sigma Aldrich can get you 99% for $115 a liter.

I am aware that TAA is an experimental recreational drug, and a possible replacement for ethanol. I advise caution in sourcing this material for that purpose, as the impurities present are likely to be more toxic than TAA itself. I believe the industrial synth is from 2-methyl-2-butene (2m2b) with the addition of water (this would be cheapest); TAA prepared this way can be purified further by refluxing with an open condenser, enabling any unreacted alkene to evaporate (Boiling point = $\pu{39 ^\circ C}$). Acetone and ethyl Grignard would also form TAA, but with much more hassle and hazard, so the remnant ethyl chloride and acetone from such a synthesis are not likely impurities in any cheap sample.


Which one is the most stable canonical structure?And why?

You're supposed to recognize that oxygen which has more valence electrons than carbon can take on the positive charge without giving up its octet, thus favoring Structure III.

Now, is it really true that the oxygen atom has more positive charge than the formally positive carbons in the other two structures, when oxygen is also more electronegative versus the octet rule?  I suspect that is beyond the scope of your course material.  You have to accept a higher priority for the octet rule in this situation whether that is real or not.

I hate oversimplified textbook problems :-( .


Is there an experimental way of measuring the critical concentration of pigment in paint?

An image of the CPVC effect is shown in the diagram below. The red color is the binder, the black is the pigment and the white is air (because there is not enough binder to coat all the pigment). The lines depict the change in properties you would expect, and you already have the formula to correlate the paint properties with the raw material properties.

BTW, economics generally suggests maximizing the inexpensive pigment/filler concentration, but adding too much increases permeability to water, which is bad in the case of metals which might corrode. On the other hand, blistering can be caused by corrosion of the substrate which then disadheres some of the film if it is too rich in binder; blistering is minimized by allowing the film to break a little. Of course, then corrosion continues - you just don't see it as much until it is too far gone. The best paint will protect from corrosion with low permeability and without disadhering.
So, how to measure a property related to CPVC if you can't pick one of these? First, you need some of the paint film; I assume you can start with the plain binder and add different amounts of the pigment/filler. Mix well, do not incorporate unnecessary air bubbles when mixing. Cast a film of constant thickness on aluminum foil by using a threaded rod; let the paint dry.

Then do your experiments!

Flexibility. If you bend the painted aluminum foil over a rod of defined radius, the rich binder film will not crack, but as more filler is added more and more cracking tendency will be noticed.

Permeability. If you put a drop of water (or perhaps alcohol) on the dried film, it will wet and spread. You may be able to differentiate between different levels of added pigment. If the aluminum foil substrate does not give good results, you can apply the paint to a decent grade of paper. Then a drop of water or other solvent might wick all the way thru and appear on the other side of the paper.

Glossiness. Ideally, you should have some kind of electronic measurement, but ranking the specimens subjectively (with your own eyes) is better than nothing. In practice, flat paints have a high pigment concentration and are not durable when washed; semi-gloss paints have less pigment, more binder, and can be washed more often. (I always wondered why they didn't go all the way to "full gloss", but I guess the binder was never so good as to give much better performance.)

Washability. Apply a water stream to the painted aluminum foil (like a dripping faucet: constant height, constant volume per minute), watch the paint film disappear.

And try coming up with some more tests!


Why is a alkyne anion more stable than a benzene anion?

Aromaticity enters the reckoning only if the excess electron associated with the negative charge enters the conjugated pi electron system.

In your case that does not happen.  The negative charge is on a formally nonbinding pair on the carbon like other carbanions, but in a conjugated ring when you do not have a ligand on that carbon the electron pair takes the place of the ligand and isn't in the pi system.  So not stabilized by aromaticity.  Here you don't have the ligand, the top carbon in A where they put the negative charge has no hydrogen atom.  If a hydrogen atom were there the molecule would be neutral benzene.

You can have a negative charge stabilized by aromaticity in two ways represented by the benzyl anion and the cyclopentadienyl anion.

Benzyl anion:  This has the formula $\ce{C6H5CH2^-}$, a phenyl ring attached to a methylene group.  The methylene group acts as the necessary ligand that's missing in your $\ce{C6H5^-}$, so the excess electron pair enters the pi system and conjugates with the ring.  Note that the $4n+2$ rule for aromaticity does not work because with that extra pi-bonded carbon you don't have all the conjugated electrons in a ring anymore.  But you do have a highly de-localized structure that inherits the stability of the ring.

Cyclopentadienyl anion:  this ion, $\ce{C5H5^-}$, has the negative charge and all the ligands on the ring atoms, so the extra electron pair and negative charge are already in the pi system without having to conjugate with anything else like the case above.  You have a pure aromatic ring with the negative charge built in.  Pyrrole, furan, and thiophene are a variation on this theme where the ring is negative but neutralized by a positive formal charge on the nitrogen, oxygen or sulfur atom.


What chemicals are used in so-called urinal "odor eaters"?

There are 1,4-dichlorobenzene-based and non-PDCB products. In the latter, the chemical mixture from the linked MSDS is:
SODIUM SULPHATE, SODIUM DODECYL SULPHATE, COCAMIDE MEA, DIPENTENE, CITRAL, LINALOOL,BENZYL ALCOHOL

  Do they actually destroy odors, or do they just mask them?

They mask odors.

  Also, are these chemicals subject to any environmental laws or regulations?

Yes, as detailed in the MSDS sheets linked above. Please note that my knowledge of urinal cakes and related products is not exhaustive.


Strength of substituted Carbon Double Bonds and C-halo bonds

Out of the 2 bromo compounds, the first one has has stronger $\ce{C-Br}$ bond, because the lone pair of $\ce{Br}$ is in conjugation with $\ce{C=C}$.
In first case I think Compound 1 is more stable for the reason that it has more substituted $\ce{C=C}$. In reasoning 2 I think when you think about decrease in bond order, the tendency of pulling electrons of $sp^2$ carbon has not decreased. Me- group itself donates electrons because Carbon is $sp^2$ hybridised. The tendency will not increase or decrease. I think it will remain in the form of slight polarisation.


Metal-Ligand-Complex and bond directions: What defines coordinate system of the complex ion?

The answer lies in symmetry.  When placed in an octahedral field, the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals are directed at the surrounding ligands because they have $e_g$ symmetry.  The remaining three orbitals have $t_{2g}$ symmetry and are directed between the ligands.  This explanation can be found in an undergraduate inorganic chemistry textbook such as that by Miessler, Fischer and Tarr.  (In the 5th edition, that's section 10.2.1 - Crystal Field Theory).


What can cause artesian well water to smell?

As mentioned in the comments, water is often treated by ozonation in order to sterilize it and ensure it can be drank with no ill effects. Ozone is a very reactive gaseous allotrope of oxygen, produced in situ by exposure of atmospheric oxygen to UV-B light (in fact this happens naturally in the stratosphere, generating a thin ozone blanket covering the world). Ozone can be produced artificially by using a mercury lamp, which emits large amounts of UV light. The water is then brought into contact with the ozone in order to be purified.

When water is collected after being ozonated, typically a small amount of ozone is still dissolved in the water. If the water is allowed to rest for a few minutes after being collected, the residual dissolved ozone diffuses out. Ozone has a very strong, characteristically sharp odour, which can be detected even at very low concentrations. This is likely why the asker's water had an odd smell. Ozone is toxic, though, so exposure should be avoided.


$99.9\%$ of the acid $\ce{HA}$ is titrated at  $\mathrm{pH}=7.9$ resp.$ 7.5$.

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+\log \frac{[\ce{A-}]}{[\ce{HA}]}=\mathrm{p}K_\mathrm{a}+3$$

Considering the final concentration 0.05M,
titrating 0.1M acid by 0.1M hydroxide,
100% of the acid is titrated at:

$$\begin{align}
\mathrm{pH}&=7+\frac12 \cdot ( \mathrm{p}K_\mathrm{a} + \log  c_\mathrm{acid, total}) \\
&=9.45(9.25) + \frac12 \cdot (-1.3) \\
&=8.8(8.6)\\
\end{align}$$

Titration with $100.1\%$ of used hydroxide we can consider
as $0.05 \cdot 0.001=5\cdot 10^{-5}\ \mathrm{M} \ \ce{NaOH}$
with $\mathrm{pH}=-\log ( 5\cdot 10^{-5})=9.7$

The indicator  phenolphthalein with range $8.2-10.0$ works well in both cases.


What reaction occurs when etching copper to create circuit boards?

Well your equation balances but that isn't the right reaction.

First lets break this down into half cell reactions.

Oxidation reaction

  $\ce{Cu_{(s)} <-> Cu^{2+}_{(aq)} + 2e^{-}}$

Reduction reaction

  $\ce{H2O2 + 2 H^+ + 2e^{-} ->  2 H2O_{(l)}}$

Decomposition reaction

Hydrogen peroxide spontaneously decomposes in acid solution.

  $\ce{2H2O2 ->  2H2O_{(l)} + O2_{(g)}}$

Stoichiometric reaction between copper and hydrogen peroxide

So the stoichiometric reaction between copper and hydrogen peroxide is:

  $\ce{H2O2 + 2 H^+ + Cu_{(s)} -> Cu^{2+}_{(aq)} + 2 H2O_{(l)}}$

the $\ce{2 H^+}$ ions come from the vinegar. (Let's use $\ce{HOAc}$ for vinegar).

  $\ce{H2O2 + 2 HOAc + Cu_{(s)} -> Cu^{2+}_{(aq)} + 2OAc^{-} + 2H2O_{(l)}}$

Sodium chloride's Role

Finally the role of the $\ce{NaCl}$ plays in the etch. The $\ce{NaCl}$ must be fairly high to drive the $\ce{Cu^{2+}}$ to a copper chloride complex. This thus reduces the "free" $\ce{Cu^{2+}}$ in solution which keeps the electrochemical potential of copper high so that the solution keeps dissolving copper. (Think of the half-cell copper reaction as if it is in a battery. The voltage of the half-cell would drop as the copper builds up.)


Is the thermodynamic stability against reduction/oxidation of an electrolyte dependent on the electrode material?

Technically you are correct. The thermodynamic potential of electrolyte decomposition should not be affected by the electrode composition.

  which just influences kinetics, not thermodynamics.

The word "just" here is doing way more work than you're giving it credit for. There are situation where the kinetics of electrolyte decomposition are so slow that there is effectively no reaction. The stability window of electrolytes commonly used in Lithium-ion batteries, for example, is smaller than the potentials needed to charge the materials. However, once the reaction begins, initial decomposition forms an inter-phasic layer on the anode known as the solide-electrolyte-interphase (SEI). The kinetics of decomposition on this SEI are slow, which allows the battery to be operated at high potentials[1]. In fact, damage to the SEI has been linked to capacity loss in Li-Ion batteries over extended cycling[2].


It's true that all odors come from compounds with at least some volatility.

However, some of the smells we associate with certain materials (like some metals) aren't the metal themselves but products of reactions caused by those materials, so you have to be careful about that.


Correlation between resonance and coordinate covalent bond

Based on the examples you have given, that does appear to be the case but if you look for more examples you'll find that this is not particularly true. Two well known compounds can be used to demonstrate:

(1) Carbon Monoxide: Coordinate Covalent Bond from O -> C, but no resonance structure

(2) Benzene: Resonance Structure but no Coordinate Covalent Bonds


Yes, lithium ion batteries works with the principle of intercalation and deintercalation of ions from cathode to anode and anode to cathode respectively. To be specific for intercalation, lithium (in cathode) having high oxidation potential oxidize, loosing an electron which goes to the current collector to anode. So lithium ion get positive charge. Since the battery source you use also has positive terminal hence repulsion between ion and positive terminal occurs. Thus positive ion of lithium goes towards anode, which has layered structure thus result in intercalation. Another way of seeing it is because of difference in the concentration the diffusion of ions occur or you can also consider the fact charge neutrality.


Can the rate of reaction of a Voltaic/Galvanic cell be determined theoretically?

Since battery scientists got Nobel this week, this question is worth pondering. It is not a trivial question. However consider this question:

A car battery has 12 V, whereas 9 V cells are also common. If we connect two 1.5 V cells and one 9 V cell on series, one can generate 1.5+1.5+9 =12 V, yet this arrangement cannot start a car despite producing the same voltage.

What is so magical about the lead-acid car battery and this series arrangement? All this shows is that voltage and current produced by battery (which is the rate, hence kinetics and hence the rate of reaction) are two separate things. The main problem in batteries, the rate of electron transfer, is always a limiting factor as to how much current can be drawn from them. So far the lead acid battery is still one of the best because it has a very fast electron transfer rate unparalleled by other batteries. I do remember there was another Nobel Prize winner (Marcus?) who did a lot of work on electron transfer kinetics on the electrodes.

The Nernst potential you see for batteries is the potential when a very very small amount of current is drawn (ideally none) from the battery. This way of measurement of the potential is called null point potentiometry. Under those conditions, the rate of reaction is nearly zero because no current is being drawn from the battery.

In short one cannot measure the rate of the electrochemical reaction just from the Nernst potential because it heavily depends on the amount of current being drawn from the battery. Indeed a non-trivial problem worth someone's PhD.


Oxides are acidic or basic based on their reaction with a base or an acid.
Here, in group 14 elements, $\ce{CO2}$ is acidic ie; it reacts with a base. When the oxides are amphoteric, they react with both acids and bases.
Oxides of higher oxidation states are more acidic because of higher positive charge. On going down the group the ability to attract electron decreases and electropositive character increases. Therefore an element becomes less acidic and more basic down the group.
I did not understand your last question, but each period has a trend for their acidic and basic nature of oxides.


An example of an endothermic dissolution is dissolving potassium iodide in water.

Added: Here is a table of Enthalpy of Solution for various compounds. Since this is the enthalpy of solution, some compounds will release energy when dissolve when the table indicates it would require energy to dissolve, and vice versa. If the number is negative, it is likely to be exothermic, and positive number are likely to be endothermic.


Why did life end up with the big 4 elements, and specifically nitrogen?

Because chemistry built on carbon, hydrogen and oxygen would be very dull and building complex structures from modular components would be impossible

The key point to remember is that life depends on complex chemistry. Living things are, by definition, things that can replicate and things than can build structures that can do complex chemistry. Metabolism is the process of taking simple things from the environment and making them into complex things. This requires complex machinery, built from simpler chemicals available in the environment.

While it might, in principle, be possible to build very complex carbon-oxygen compounds that could metabillise and reproduce themselves, there is no easy way to do so from simple components. But throw nitrogen into the mix and simpler routes to complexity are available.

Take the molecules that make up most of the complex machinery of all living things: proteins. Proteins make up most of the complex machines that do the job of metabolism and they also constitute many of the structural components that make up living things. But they are built from a small number (about 20) simpler chemicals: amino acids. Amino acids are fairly simple chemicals that might be thought of as the equivalent of Lego bricks for building an incredible variety of living structures. Most consist of only carbon, hydrogen, oxygen and nitrogen. All have the core structure $\ce{NH2CH(R)CO2H}$ (where R can be a variety of organic chunks) which constitutes the equivalent of the standardised bricks in Lego allowing all pieces to fit together so more complex structures can be built. In the case of proteins, that core functionality allows a variety of different amino acids to be "strung" together end-to-end (the amino end reacts with the carboxylic acid end on each acid).

The variety of different structures than can be created by changing the sequence of amino acids is staggeringly, unimaginably large. Many proteins contain hundreds of amino acids and you only need about 60-70 for the possible number of different sequences to exceed the number of atoms in the universe. This means that, by exploiting combinations of simple chemicals (which are not that hard to create with simple chemistry), life can suddenly explore an unimaginably wide range of possible complex structures. Perhaps such structures could be built from fewer elements, but not in a modular way that uses a small range of available components and a single way of joining them together.

The availability of this level of complexity from things that (mostly) consist of C,H, N and O is why the use of nitrogen is key to life. The constraint isn't the availability of the elements (though simple nitrogen-containing compounds are fairly available), it is the accessibility of complex structures from simple components.


Distillation of methanol and water - Temperature Change

The thermometer that records the temperature is in the air high above the solution (usually above a Vigreux column). Your distillation was under ambient conditions, so there was air in contact with everything.

The thermometer can only records a rising temperature if the temperature around it rises. During the distillation, this is due to methanol vapour (around $\pu{70 °C}$) which can also condense on the thermometer to transfer heat.

Once all methanol has been distilled away, there is nothing coming from below that could transfer any additional heat: water, sulphuric acid and benzoic acid all have far too high boiling points and low vapour pressures. Thus, the thermometer is again in contact with the surrounding air and cools down.

While I have described this with air, basically the same thing will happen in a vacuum distillation, too.


Why don't free radicals undergo rearrangement like carbocation/carbonium ions do?

First off, your observation is correct, rearrangements involving free radicals and carbanions are noticeably less common than rearrangements involving carbocations.

Many common carbocation rearrangements involve 1,2 shifts.  If we look at the MO diagram for a 1,2 shift and compare shifting a substituent to an empty p orbital (carbocation, case A in diagram) or to a filled p orbital (carbanion, case B in diagram), we see that the carbocation process involves a 2-electron system whereas the carbanion process is a 4-electron system.  The 2-electron process is allowed to proceed themally according to orbital symmetry, but the 4-electron process would need to proceed with inversion of configuration (bonding to the other lobe of the p orbital as we transform from reactant to product) in order to be thermally allowed - and this would be geometrically difficult.

(Source - V8rik at the English Wikipedia via Wikimedia Commons)

If we consider hydrogen shifts and replace the p orbital in our digram with an s orbital, we see that a 1,2 hydrogen shift in a carbocation is thermally allowed, but the corresponding shift in a carbanion is thermally forbidden (with an s orbital, there is no "other" lobe and proceeding with "inversion" is not possible)

Since thermal 1,2 shifts with retention at the migrating center are not allowed in carbanions by orbital symmetry, and a 1,2 shift with inversion at the migrating center is geometrically difficult, such processes would involve higher energy transition states.  As a consequence of the higher energy, 1,2 shifts in carbanion systems are less common than orbital symmetry allowed, thermal 1,2  shifts involving carbocations. One might reasonably expect that a radical process would be somewhere in between these two cases; and again the transition state would be of higher energy (since it is not orbital symmetry allowed) making such rearrangements less common.

While radical and carbanion shifts are not as common as carbocation shifts, they do occur.  In the case of radicals it requires high temperatures (hydrocarbon cracking) or weak bonds (peroxides for example). Carbanionic rearangements are more common when the carbanion is stabilized.  This can be achieved by having an electronegative heteroatom attached to the carbanionic center (as in the Wittig rearrangement) or by placing the anion on an $\ce{sp^2}$ or $\ce{sp}$ hybridized carbon, rather than an $\ce{sp^3}$ hybidized carbon (the Fritsch rearrangement provides an example here).


What's the difference between graphite, nanotubes, and fullerenes?

All three carbon modifications indeed contain covalently bond carbon atoms. The $sp^2$-hybridized C atoms are connected by $\sigma$ and delocalized $\pi$ bonds in all cases. It is the geometric arrangement of the atoms that differentiates the modifications: Stacked planes in graphite, fullerenes: ball-shaped molecules (hence the name buckyball), and for carbon nanotubes, the name is quite self-explanatory.

A single layer of the graphite structure:

A $\ce{C60}$ fullerene molecule:

And finally some carbon nanotubes:


Strange batch of potassium nitrate with lower than expected melting point

It might contain $\ce{NaNO3}$ as well as $\ce{KNO3}$. This paper by Foong et. al. cites some thermal measurements, such as melting point and heat capacity, that you could make with a thermometer and heater. Note that there are inflections in the heat capacity curve as the different components ($\ce{NaNO3}$, $\ce{KNO3}$ and the eutectic of the two) melt.

If you have access to a spectroscope, or even a hand-held diffraction grating, you could compare the spectrum with that from samples of known substances, such as $\ce{NaNO3}$, $\ce{KNO3}$, $\ce{NH4NO3}$ and $\ce{Ca(NO3)2}$.


How does the treatment of resin R-SO3H with sodium chloride lead to formation of RNa?

It is a typing mistake. Cation exchange process can be written as
$$\ce{R-SO3^-H^+ (s) + NaCl (aq) -> R-SO3^-Na+(s) + H^+(aq) +Cl^-(aq)}$$
You can immediately see why the resin must be in sodium form for water softening. If the acid form of the resin is used for softening water, what would you get? Extremely strong acids!!
For a divalent ion, the sulfonic acid groups should be in close proximity to "catch" ions such as $\ce{Ca^{2+}}$. Each calcium ion would quantitatively release two hydrogen ions from the resin.


why maximum (w/w) percent concentration of commercial acids is same to their molar mass?

It's a coincidence as HBr is 48% (MW ~81), phosphoric acid is 85% (~98), hydrofluoric acid 97% (MW ~20).Look here for more http://www.csudh.edu/oliver/chemdata/acid-str.htm


How to measure and transfer suspensions?

Stir the dilute suspension rapidly, e.g., using a magnetic stirrer. Let a  vortex be generated whose size you can use as an estimate for vigor of stirring. With the same mixing vigor, and using the same size vessel and same original volume for the suspension, and withdrawing only a small portion (say, ~10%) for the other container(s) should enable fairly repeatable measurements.

For obsessive accuracy, you could do three withdrawals: #1 for reference, #2 for experiment, and #3 for reference. If #1 and #3 show the same concentration of solid, #2 will certainly be the same, and the concentration will be verified while sedimentation will be proven to be absent.


How to calculate Gibbs free energy from pKa?

To find the change in Gibbs free energy $\Delta G$ for the reaction of an acid, you can simply use $K_\mathrm{a}$ in the formula you gave in place of $K_\mathrm{eq}$ ($K_\mathrm{a}$ is an equilibrium constant).


It gets complicated. A drop of water (30 mg) will evaporate completely in several hours under standard conditions. You need to consider these factors:

(1) Temperature
(2) Relative humidity
(3) Wind
(4) Surface area
(5) Thermal flux to water (container walls -> water heat flux)

By all means you can do it, but it is a question for physics forums. The main idea would be to write to equations:

(a) Heat flux +walls_to_water + air_to_water -heat_of_evaporation = 0
(b) Diffusion balance. Water pressure at its surface, vs water pressure at the top of the flask is governed by diffusion laws.
(c) Material balance: amount_of_water(t)=amount_of_water(t$_o$)-evaporated_water.

Those are differential equations, so it would be hard to work with them without a solid understanding of calculus.


Why does silver sulfate not precipitate in this double displacement reaction?

Since the lab is past due now, I'll give what I think is the answer.

$\ce{Ag2SO4}$ is somewhat soluble in water. It is most likely that you simply didn't get enough to cause precipitation.

The other possibility, which I don't think applies here, is that you have a supersaturated solution. There are some precipitates for which crystals are just stubborn to nucleate.


You should probably be cautious in general applying the concept of bond dipoles to predict molecular dipole moments, and it does seem ad hoc if not downright flimflam to invoke a dipole to the nitrogen lone pair to predict dipole moments in the examples you present (particularly in the case of nitrogen trifluoride).

Now if you read the wikipedia page on nitrogen trifluoride, it explains that

  $\ce{NF3}$ is slightly soluble in water without undergoing chemical reaction. It is nonbasic with a low dipole moment of $\pu{0.2340 D}$. By contrast, ammonia is basic and highly polar ($\pu{1.47 D}$).[9] This difference arises from the fluorine atoms acting as electron withdrawing groups, attracting essentially all of the lone pair electrons on the nitrogen atom. $\ce{NF3}$ is a potent yet sluggish oxidizer.

The low basicity of $\ce{NF3}$ supports the argument that electron density associated with unpaired electrons on N is withdrawn by the fluorine atoms. So it would appear that in the case of $\ce{NF3}$ there might not be a significant dipole moment between the $\ce{N}$ nucleus and an electron pair (or the charge opposite the three fluorines). Instead the electron density (relative to ammonia as a reference) is polarized toward the fluorines. This makes the application of a method of linear summation of bond dipoles somewhat sketchy.

Continuing along similar sketchy lines, the website to which you provided a link argues that the difference in molecular dipoles cannot be explained simply in terms of the order of electronegativity ($\ce{F > N > H}$). But note that electronegativities are computed from dissociation energies, electron affinities, etc, and not from partial charges or dipole moments, and the relation between charges or dipole moments and energies is bound to be highly non-linear, making quantitative prediction of dipole moments based on electronegativities difficult.

What seems necessary is to perform some ab initio electronic structure calculations.


Simply put, the energy costs for a sextet carbocation are too high to allow chlorine's strong inductive effects to dominate the situation.

The driving for for a carbon octet is very high.


How to find a symmetry group of a system if all the symmetry transformations do not obey closure and don't form a group?

Hmm, so I'm not entirely sure what notation you're using ($D_4$ sounded like Schönflies to me), but I'm sure you can figure out the equivalent.
In general I agree with Andrew's point that molecular symmetry is determined by atomic positions and not orbitals. But at least to me as of now, this seems to be a thought exercise more than anything (after all, no real atom has only two p-orbitals), so I'll play along. In Schönflies notation your system would be labelled $D_\mathrm{2h}$. The three $C_2$ axes and three mirror planes are illustrated below. I left out the identity operation $E$ and the centre of inversion $i$.


What you are looking for does not — to the best of my knowledge — exist and likely won’t ever exist.

Methylating agents are usually classified as toxic as they are usually able to penetrate cells and nuclei and methylate DNA bases — although they will also methylate most nucleophiles on the way there. This can lead to all sorts of things will the shadow of cancer overarching it all. Off the top of my head there are a few methylating agents that might not penetrate the skin such as methyl Meerwein salt. These, however, are so reactive that they just react before they reach the skin.

Secondly, methylating agents will likely never work in water. Water itself is a nucleophile and a (Brønsted) acid. If you use an electrophilic methylating agent such as methyl iodide or methyl Meerwein salt, water is able to nucleophilically attack these to generate methanol and then dimethyl ether, thereby deactivating your methylating agent. If you choose to use a nucleophilic methylating agent such as methyllithium instead, water will supply a proton to liberate methane (and lithium hydroxide in the case of methyllithium) instead. So practically no methylating agent will survive in aquaeous solution.

Finally, your proposed reaction pathway is skewed from the beginning: you suggest using a deprotonated amine in water as a reactant. However, the $\mathrm{p}K_\mathrm{a}$ of amines is typically around $30$, depending on the substituents. ($\mathrm{p}K_\mathrm{a}(\ce{LDA}) \approx 35; \mathrm{p}K_\mathrm{a}(\ce{LiHMDS}) \approx 25$; source: Evans $\mathrm{p}K_\mathrm{a}$ table.) On the other hand, the $\mathrm{p}K_\mathrm{a}$ of water is $14$ or $15.7$, depending on the source you consult. This is a difference of $10$ to $20$ logarithmic units depending on the type of amine — meaning that amines are practically never deprotonated in aquaeous solution.[1]

If you mistyped and are using a deprotonated amide, things won’t get that much better ($\mathrm{p}K_\mathrm{a}(\ce{RCONH2}) \approx 25$).

Therefore, you will likely have to give up your plan and go for a more traditional methylation method.

[1]: Note you might argue that the free base, $\ce{R3N}$ is enough, so all you want to do is lower the $\mathrm{pH}$ of water to below $\approx 10$ so that the protonated form $\ce{R3NH+}$ is neglegible. However, at these $\mathrm{pH}$ values, water is also deprotonated to a non-neglegible amount and a hydroxide ion is a better nucleophile than a free amine base.


tl;dr: two different definitions. Astronomy: multiple carbon atoms in molecule. Chemistry: polymer

Interestingly enough, after reading about COMs here, as well as reading the Wikipedia page and the corresponding arXiv paper, it seems like chemists and astronomers have different definitions of what a complex organic molecule should be!

As far as I knew, in chemistry complex organic molecules were long polymers, such as proteins, which were composed of thousands upon thousands of amino acid units. In the astronomy paper, however, they cite other types of molecules.

$\ce{CH3OH,
CH3CHO, HCOOCH3 and CH3OCH3}$, all cited as "complex" (haha) organic molecules in the paper, would appear to chemists as relatively simple molecules. (I read the paper, because it piqued my interest that something like a protein could be found in space). I then read the Springer article.

  The term “complex organic molecules” is used differently in astronomy and chemistry. In astronomy, complex organic molecules are molecules with multiple carbon atoms such as benzene and acetic acid. These molecules have been detected in interstellar space with radio telescopes. In chemistry, “complex organic molecules” refer to polymer-like molecules such as  proteins.


Both of those explanations include part of the puzzle, but are not really complete on their own. To get a better understanding, you need to combine them, as well as understand something called Coulomb's law.

The basics of Coulomb's law are this:

The closer two oppositely charged things are, the stronger the attraction
The stronger the charge on either of the things, the stronger the attraction

With those points in mind, let's combine it all. I'll keep my explanation similar to the way you phrased yours:

Each step across a row on the periodic table adds both one nuclear proton and one valence electron. So, that is increasing both the total positive charge of the nucleus and the total negative charge of the valence shell, but not the distance between them (because electrons are added to the same shell).
The force of attraction between two charged objects depends on how strong each of them are (Coulomb's law). Since both the nucleus and the valence shell are stronger than in the previous step, and the distance between them has not increased (the electrons are added to the same shell) there is a stronger force of attraction between them, and the two are pulled closer together.
So, the atomic size of elements decreases across a period.


How to set up an equation to solve a rate law computationally?

It should be set up as a series of first order ordinary differential equations like so: $$\frac{d[A]}{dt}=-k_f[A][B]^2+k_r[C]$$$$\frac{d[B]}{dt}=-2k_f[A][B]^2+2k_r[C]$$
$$\frac{d[C]}{dt}=k_f[A][B]^2-k_r[C]$$
If we substitute $[C]=[C]_0+x$, $[A]=[A]_0-x$, and $[B]=[B]_0-2x$ into these equations, they all reduce to the single equation:$$\frac{dx}{dt}=k_f([A]_0-x)([B]_0-2x)^2-k_r([C]_0+x)$$


What is the origin of the names P-1 and P-2 of nickel boride catalysts?

The actual differences between the two forms are well-documented on Wikipedia. As for the names: as far as I can tell, the choice of "1" and "2" seems to be arbitrary. The choice of the letter "P", though, is described by C. A. Brown (who carried out investigations into the synthesis and reactivity of these catalysts).1,2 In footnote 13 of reference 1, Brown writes:

  The designations are derived from Purdue University, where this work was initiated.

So, it is within reason to assume that the P stands for Purdue.

References

Brown, C. A. Catalytic hydrogenation. V. Reaction of sodium borohydride with aqueous nickel salts. P-1 nickel boride, a convenient, highly active nickel hydrogenation catalyst. J. Org. Chem. 1970, 35 (6), 1900–1904. DOI: 10.1021/jo00831a039.
Brown, C. A.; Ahuja, V. K. Catalytic hydrogenation. VI. Reaction of sodium borohydride with nickel salts in ethanol solution. P-2 Nickel, a highly convenient, new, selective hydrogenation catalyst with great sensitivity to substrate structure. J. Org. Chem. 1973, 38 (12), 2226–2230. DOI: 10.1021/jo00952a024.


Can non-deuterated solvents be used for 13C-NMR (and in fact other nuclei)?

If using a deuterium lock, then normally you would require, as minimum, about 10% deuterated solvent. Without any deuterated solvent, it is not possible to obtain a spectrometer lock (unless using a 19F lock channel, but that is unlikely, and is a hardware requirement).
It is, however, perfectly feasible to acquire spectral data for all nuclei without any deuterated solvent, and hence, without a spectrometer lock. When running without a lock, you need to turn the lock sweep off, to hold the spectrometer field constant. This does not provide the same level of field stability as a lock, but is more than adequate for short-medium 1D experiments that do not require very high resolution. Using gradient shimming, it is also possible to shim on non-deuterated solvents and achieve the same lineshape specifications.
Running long experiments (such as an overnight 13C{1H} experiment), without a lock will give you linewidths that are slightly broader than usual; the consequences of this are that (i) your resolution is reduced (rarely a problem with 13C), and (ii) your signal-to-noise will be reduced (and therefore longer experiment times may be required...leading to further broadening and longer experiment times...)
Not claiming to be the first to report about this technique, the Hoye group published two papers in Organic Letters (2004OL953 and 2004OL2567) with examples of application of no-D $\ce{^1H}$-NMR spectroscopy, covering sample preparation and data acquisition, too.  Especially the second one, outlining how to determine the concentration of LDA or Grignard reagents by addition of 1,5-cyclooctadiene as internal standard may be a reliable and faster alternative to the titrations once you are used shim manually for samples of this type.  But speak and obtain clearance in advance with your NMR manager before running your no-D sample; some interfaces to the spectrometer tend to be less puzzled to record NMR with a lock after your sample (e.g., elder Varian), than the more modern (e.g., Topspin Bruker) to prevent unnecessary delays in the queue of samples following yours.


What is meant by 'electrons of like/unlike rotation'?

It simply refers to electrons in orbitals with the same sign of $m_l$.

The quick and easy way to determine the ground-state term symbol of a lanthanide is to fill the 4f orbitals from most positive $m_l$ to most negative. This is in order to get the largest possible value of $M_L$:

$$M_L = \sum m_l$$

This is not valid by quantum mechanical arguments due to the indistinguishability of the electrons. However, it always leads to the correct ground-state term symbol because of Hund's second rule, which essentially states that the term with the largest value of $L$ is the most stable (after the first rule has been taken into account). Since the value of the projection quantum number $M_L$ ranges from $-L, -L+1, \ldots, +L$, a ${}^{2S+1}L$ term necessarily contains a term with $M_L = +L$. Therefore, by maximising the value of $M_L$, you will automatically generate the term with the largest possible $L$.

For example, consider the $\mathrm{f^3}$ ion $\ce{Nd^3+}$. You would want to "fill the orbitals" in this way:

where you have first and foremost put the electrons all with the same spin in order to satisfy Hund's first rule, and then filled them in decreasing order of $m_l$ in order to satisfy Hund's second rule.

The classical argument behind Hund's second rule is that electrons that have the same sign of angular momentum "collide" into each other less and as a consequence repel each other less. It is like two people running around a track; if they ran in the same direction, they would bump into each other less than if they ran in opposite directions.

Therefore, removal of an electron with "like rotation" (as you would if you were to ionise $\ce{Nd^3+}$ to $\ce{Nd^4+}$) is marginally more difficult than expected. Do not imagine it as a huge effect. It is very small and only leads to minor deviations from the ionisation energy trends (quarter-shell and three-quarter-shell effects).


Well, think of a gas that is expanding against a piston shown below:

We want to find the total work which is given by: $$w = F\Delta d$$ So we need to know what is the force that is applied by the gas onto the piston. To do this, we must realise that, the force that is stopping the gas from expanding freely into the atmosphere is the external atmospheric pressure. It is this pressure which is pushing downwards onto the gas, preventing it from expanding. Therefore, the force that is required for the gas to expand must be equal to the total pressure pushing down it and has nothing to do with its internal pressure.

Consider this example. I want to push a box across my room which is covered in carpet. To move the box, obviously I need to apply a force to it. The force that I need to apply to must be equal or greater than the resisting force which is the friction between the box and carpet. This force that is required has nothing to do with the speed of the box.

So now that we know the force, we get that $$w = -p_{ex}A\Delta d$$ but the area times the displacement is actually equal to the volume, therefore we get: $$w =-p_{ex}\Delta V$$

So, as you can see, we have derived the equation for work done by a gas. The main thing that you need to recognise is that the internal pressure is irrelevant when considering the force required for the gas to expand. All you need to consider is the force that is preventing it from expanding which is the external pressure.


I like the suggested treatment suggested here which is to basically clean/scrub the pan, dry and coat with vegetable oil, invert the pan, place over aluminum foil on an oven rack and bake. To quote:

Now that you’ve returned your skillet to its base layer, you’ll need to season it for use. Drop a tablespoon of vegetable oil on the pan and spread it around with your fingers or a paper towel. (Take care not to burn yourself on a warm skillet!). Oil the outside and the handle, too. Next, take a folded napkin and wipe off the excess oil. You’ll want the skillet to be lightly greased. Set your oven to 350 degrees and put some aluminum foil on the bottom rack. Then pop the skillet in, upside down, on the upper rack for about 1 hour. Turn the oven off and leave the pan inside until it’s cool. The oil will bake into the pores of the pan, providing a non-stick finish.
Repeat this step a second time and you’ll have an almost glassy surface of seasoning.
Ta-da! Your rusted cast-iron skillet is good as new.

The created carbonized layer is safe, chemically inert and durable.


Will this series of reactions work to produce potassium phosphate?

Getting phosphoric acid by heating ammonium phosphate probably won't work since it requires temperatures above 200°C where phosphoric acid will start to vaporize and decompose.

A better approach might be the reaction of ammonium phosphate with potassium carbonate. The ammonium carbonate formed by this reaction decomposes at much lower temperatures (> 60°C). However, heating will produce corrosive, irritant and toxic products (ammonia, phosphoric acid, phosphorus oxide), so I do not recommend doing such experiments.


Energy of hydrogen bonds vs. kinetic energy of water molecules

It looks like they calculated the average thermal energy of water molecules, which by classical equipartition theorem is $3RT/2$ kJ mol$^{-1}$. This works out at approx 4 kJ mol$^{-1}$ at 40 C. This amount of energy includes translational, vibrational and rotational energy. The vibrational quanta are large in water and almost all molecules will have only zero point vibrational energy. Rotational quanta are smaller and many will be excited at these temperatures. The smallest rotational quantum value for water is approx 0.2 kJ mol$^{-1}$ the next 6 times larger and the next 12 times (increasing as $J(J+1)$ for quantum number J) so a few rotational levels will be populated. Energy will flow between translation and rotations as random collisions occur between molecules of water. A full calculation involves using the partition functions. Calculating these and their contribution to heat capacity is treated in most undergraduate phys. chem. textbooks.


Using a calcium metal anode in aqueous solution?

Basically, the reaction of calcium with water short-circuits your battery; the electrons you wanted to go through the external circuit end up going directly into the electrolyte instead.

Magnesium will work better for your anode; it reacts little or none with water when cold and it's anodic potential still packs a pretty good punch.


The term “negative pressure” is often used in engineering to refer to a situation in which an enclosed volume has lower pressure than its surroundings. If the a region is surrounded by more pressurized area around it would cause substances to flow inwards and thus the term negative pressure.


1: sp3 with 1 electron withdrawing group
2: sp3 with 2 electron withdrawing groups
3: sp3 with 0 electron withdrawing groups
4: sp3 with 1 electron withdrawing group

I would say your earlier analysis was correct.  If you look at the resonance structures that can be drawn for an amide linkage, you find there is considerable positive charge both on carbon and nitrogen.  This should serve to make them both electron withdrawing, relative to an attached carbon, in your structure.

EDIT: To clarify this answer with regards to some of the comments posted above, I'd like to add that typically when an amino nitrogen is situated on a double bond or aromatic ring it can donate electrons through resonance.  In fact, that is what it is doing in the resonance structure on the right - it is donating its lone pair of electrons via resonance.  However, the amino group is also electron withdrawing in an inductive sense relative to carbon.  Further the $\ce{-NH3^+}$ group can no longer interact via resonance and is simply a strong electron withdrawing group.  The resonance structure on the right is analogous to the $\ce{-NH3^+}$ situation and since it is a significant contributor to the overall description of the amide (cf. there is significant double bond character in the amide $\ce{C-N}$ bond) it acts inductively as a strong electron withdrawing group for the attached $\ce{sp^3}$ carbon.


Bonds actually have negative potential energy. Think about it this way: in order to break a bond you have to expend energy so when a bond is created, energy must therefore naturally be released. Hence internal energy is lower.


How does the diffusion coefficient of a species affect the limiting current for a RDE voltammagram?

I would recommend you take a look at the book Electrochemical Methods fundamentals and applications.

For the Koutecý-Levich equation (totally irreversible one-step, one-electron reaction):

$\Large \frac{1}{i}=\frac{1}{i_K}+\frac{1}{i_{l,c}}=\frac{1}{i_K}+\frac{1}{0.62nFAD^{2/3}\omega^{1/2}\nu^{-1/6}C^*_O} $

The higher diffusion coefficient is, the higher is the current.

The diffusion coefficient can be computed from the Koutecý-Levich equation if you know all other coefficients.


If an $\alpha$-$\beta$ unsaturated heteroatom isn't already participating in a $\pi$-bond and it has a lone pair of electrons, these electrons will be delocalized and in the $\mathrm{p}$-orbital.

Let's compare pyridine and pyrrole:

$\hspace{5cm}$

The nitrogen in pyridine is already participating in a double bond, so it is clearly $\mathrm{sp^2}$ hybridized. The nitrogen in pyrrole is $\alpha$-$\beta$ unsaturated, so its lone pair is delocalized. Both of these molecules are heterocyclic aromatic rings, but these are not the only cases in which a nitrogen atom's lone pair are delocalized. They are also delocalized in amides (which resemble the $\alpha$-$\beta$ unsaturation of pyrrole):

$\hspace{4.6cm}$

If you want to know more about criterion for aromaticity, read about Hückel's rule.


How to calculate what mass of halothane is inspired in one minute given the partial pressure and temperature?

Volumetric Flow Rate (VFR) is given to you.
By applying ideal gas equation to the gas, in terms of volumetric flow rate,
$$\frac{PV}{t} = \frac{nRT}{t}$$

And in an ideal gas mixture (the assumption),
the partial pressure of a gas is what is exerts alone in same volume in same temperature as of the mixture.
$$\frac{P_\mathrm{gas}\cdot V_\mathrm{tot}}{t} =
\frac{n_\mathrm{gas}\cdot R\cdot T_\mathrm{total}}{t}$$

amount of substance = mass of the compound/molar mass (or molecular weight of the compound)
$$n = \frac{m}{M}$$

$$\frac{P_\mathrm{gas}\cdot V_\mathrm{total}}{t} =
 \frac{(\frac{m_\mathrm{gas}}{M})\cdot R\cdot T_\mathrm{total}}{t}$$

Apply the values, mind the units:
\begin{align}
V &=\pu{0.1 L}\\
T &= (273+21)\pu{ K}\\
P_\mathrm{gas} &= \pu{0.01 atm}\\
R &=\pu{0.082057 L atm mol^-1K^-1}\\
M_\mathrm{gas} &=\pu{197.4 g mol^-1}\\
t &= \pu{1 min}\\
\end{align}

$$\frac{\pu{0.01 atm}\cdot\pu{0.1 L}}{\pu{1 min}} =
\frac{
  \frac{m_\mathrm{gas}}{\pu{197.4 g mol^-1}}
  \cdot\pu{0.082057 L atm mol^-1 K^-1}\cdot\pu{294 K}}{\pu{1 min}}$$

For $m_\mathrm{gas}$ I can see the value in your multiple choice.


According to MO theory, do 1s orbitals in dioxygen interfere?

It is always possible to define a bonding and antibonding orbital regardless of how small the overlap is. Because we solve the homonuclear diatomics in the symmetry of the molecule ($D_\mathrm{\infty h}$), the MOs we choose must respect that symmetry. It wouldn't matter if I put the oxygen atoms 10 light years apart, if I am treating them as part of the same quantum system, there are bonding and antibonding molecular orbitals. (Of course, at large distances bonding and antibonding orbital pairs becomes very near degenerate.)

That said, because oxygen has a doubly occupied 1s orbital, the corresponding sigma 1s bonding and antibonding orbitals will both be filled. Within MO theory we are always free to 'rotate' any two occupied MOs and the observable properties will all be the same.
So, there's no difference between saying that each oxygen has a non-bonding 1s orbital, and saying that there is a filled bonding and antibonding orbital. When both orbitals are filled you can't even ask the question of what the electrons are 'doing' down there, because the concept of a distinct orbital has no meaning.


Finding new pH after NaOH added to buffer solution

Start by calculating $\mathrm{p}K_\mathrm{a}$ using the Henderson–Hasselbalch equation and the initial data given. $$\mathrm{p}K_\mathrm{a} = 3.35 - \log\frac{0.15}{0.2} = 3.47$$

Calculate the amount of substance of $\ce{HA}$ and $\ce{A- }$ after addition of $\ce{NaOH}$. $$n_{\ce{HA}} = 0.15\times0.5 - 0.0015 = 0.0765$$ $$n_{\ce{A- }} = 0.2\times0.5 + 0.0015 = 0.0985$$

Calculate the final $\mathrm{pH}$. The final volume is not needed because it is the same for both concentrations and so it cancels in the equation. $$\mathrm{pH} = 3.47 + \log\frac{0.0985}{0.0765} = 3.58$$


How does molecular structure affect physical properties?

Predicting (approximate) physical properties starting with molecules is complicated, below.  Nobody has ever produced a way to accurately predict how molecules will pack in a crystal structure.  Big Pharma is desperate to predict crystal polymorphs and their properties. Slightly vast amounts of grant funding pursued predicted super-explosive octanitrocubane.  When it was finally synthesized it crystallized in a less dense crystal structure than expected, deeply disappointing the DoD.

http://www.chemspider.com/ACDLabs.aspx
http://pubs.acs.org/doi/abs/10.1021/ed064p575
http://molsoft.com/mprop/

If you have a way to predict crystal structures given complex formula units, you're a big winner.


Rust consists of hydrated iron(III) oxides Fe2O3·nH2O, iron(III) oxide-hydroxide (FeO(OH), Fe(OH)3.

The rusting of iron is an electrochemical process that begins with the transfer of electrons from iron to oxygen. The rate of corrosion is affected by water and accelerated by electrolytes. The key reaction is the reduction of oxygen:

$$\ce{O2 + 4 e- + 2H2O -> 4OH-}$$

The following redox reaction also occurs in the  presence of water and is crucial to the formation of rust:

$$\ce{2Fe2+ + 0.5 O2 <=> 2 Fe3+ + O2−}$$

Additionally, the following multistep acid-base reactions affect the course of rust formation:

$$\ce{Fe2+ + 2 H2O <=> Fe(OH)2 + 2H+}$$

$$\ce{Fe3+ + 3 H2O <=> Fe(OH)3 + 3 H+}$$

as do the following dehydration equilibria:

$$\ce{Fe(OH)2 <=> FeO + H2O}$$

$$\ce{Fe(OH)3 <=> FeO(OH) + H2O}$$
$$\ce{2 FeO(OH)  <=> Fe2O3 + H2O}$$

From the above equations, it is also seen that the corrosion products are dictated by the availability of water and oxygen. With limited dissolved oxygen, iron(II)-containing materials are favoured, including FeO and black lodestone (Fe3O4). High oxygen concentrations favour ferric materials with the nominal formulae Fe(OH)3-xOx/2.

Thus you can actually calculate the value of x (water of hydration) in the compound. Hope this works

Credits: Chemical formula of rust


Please explain the mystery of the equilibrium constant expression for chemical reactions

I think the most intuitive way to think about this is to view it through a kinetic lens. That is, recall that at equilibrium, the rate of the forward reaction and the rate of the reverse reaction must be equal. When we write expressions for the rates of reactions, we multiply the reactants because the probability of collision is a function of the product of the two concentrations, not the sum. This only really makes sense for an elementary step, but the result applies to the whole reaction.

Let's consider the simpler reaction $\ce{A + B <=> C}$.

If there's is only one step in the reaction, the rate of the forward reaction will be given by $k_\mathrm{fwd}[A][B]$ and the rate of the reverse reaction by $k_\mathrm{rev}[C]$.

At equilibrium, these are equal, so

$$k_\mathrm{fwd}[\ce{A}][\ce{B}] = k_\mathrm{rev}[\ce{C}].$$

Rearranging,

$$\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}=\frac{[\ce{C}]}{[\ce{A}][\ce{B}]},$$

and we see the product of the concentrations in the equilibrium constant.

You can also go the route of deriving the expression from the free energies and Boltzmann distribution, but I don't think that's nearly as intuitive as thinking in kinetic terms.


Does sanding the oxide layer of a magnesium sample break bonds?

I upvoted the question because it is actually very insightful. We may think of simply removing the MgO layer, and voila, there is bare magnesium. I think not.

If you took bare Mg (no oxide at all) in a vacuum, and dropped powdered MgO onto the Mg, then exposed the metal to an argon atmosphere, I think you could blow off the oxide powder because there would be no significant adhesion.

In other words, the unreactivity of the oxide-coated magnesium arises because the oxide coat is strongly bonded to the metal. This transition from metal to metal oxide occurs over a few atoms and is probably very irregular. After all, even the surface of the metal won't even be flat to within 1 or 2 atoms.

Visible scratches on the metal after sandpapering prove that some mechanical effect has removed metal. If you remove metal, you must have broken some metal-metal bonds; and in the turmoil, at least some metal - oxygen bonds must also have been broken - even tho they might soon reform with O2 in air.

A similar, but chemical, film removal occurs when stainless steel, which is likewise protected by a very thin oxide coat, is treated with hydrochloric acid. The film is removed, and the steel corrodes. Because the reaction is so thorough, removing the whole oxide coat and a few atomic layers of metal as well, we don't stop to think about the transition layer. Different thicknesses can produce different colors as well as different degrees of protection.


Why is the probability for the signal at 160 in the bromine mass spectrum twice as high as for the signals at 158 and 162?

All possible arrangements of $\ce{Br2}$ molecule:

$\displaystyle 79 + 79 = 158$
$\displaystyle \color{red}{79 + 81} = 160$
$\displaystyle \color{red}{81 + 79} = 160$
$\displaystyle 81 + 81 = 162$

The amount of $\ce{^{79}Br}$ and $\ce{^{81}Br}$ in nature is roughly the same, thus each permutation is equally probable. There are two arrangements that lead to $160$. While $158$ and $162$ each have only one arrangement. Therefore $160$ is twice as likely to be found compared to other masses.


It simply spreads the temperature change over a longer distance, enabling separation of distillates with boiling points close to each other. "Generally the component parts boil at less than 25 °C from each other under a pressure of one atmosphere."

Imaging reading a mercury thermometer one cm tall as compared with one 100 cm tall: it would be difficult to distinguish temperatures on the smaller one. In the fractionation column below, it helps keep the outputs separate.


How can -G/T and -A/T be thought of as 'disguised' entropies?

The fundamental equation of (chemical) thermodynamics $\ce{\Delta G} = \Delta H – {T\Delta S}$ may be divided by  (-T):

$$\ce{-\frac{\Delta G}{T}} = -\frac{\Delta H_{sys}}{T} + {\Delta S}_{sys}$$

In this equation $\frac{\Delta H_{sys}}{T}$ is the entropy change of the system by “heat” exchange with the surroundings, hence $\frac{\Delta H_{sys}}{T} = -\Delta S_{surroundings}$. On this basis, the above equation may be identified as:

$$\Delta S_{total} ={\Delta S_{surroundings} + \Delta S}_{sys}$$

$\Delta S_{sys}$ and $\Delta S_{surroundings}$ strive mutually to a maximum of $\Delta S_{total}$.

$\ce{\Delta G}$ is a disguised entropy change, because $\ce{\Delta H}$ is intrinsically an entropy change too, as explained above.


Does diamond have a pungent smell like graphite?

None of the carbon allotropes have a smell. If you smell something from a pencil lead (graphite), then it will be from the impurities in it.

Graphite is made from impure organic compounds that frequently contain sulphur. It is likely that you are smelling this. Mercaptans and other sulphur compounds can be very smelly. on its own


USNCO Question - Determining solute using spectrophotometer

An aqueous solution of $\ce{Mn^2+}$ is only faintly pink, whereas that of $\ce{Co^2+}$ is much more colored:


To reflect on the comments below:
In literature this reaction seems to actually produce product A and synthesis of product B seems to require much more exotic conditions than described in the question. As such I have to conclude that the correct answer would have be A.

Old answer:

Due to the polar aprotic solvent (DMF) one would think Sn2 is favoured as the reaction mechanism, however if the product is indeed B it would suggest an Sn1 mechanism where the aromatic ring stabilises the cation. One of the resulting resonance structures would indeed react with fluoride. Due to the stability of aromatics the product is formed by re-aromatisation.


Why do electronegative atoms depress coupling constants?

Let's set some starting parameters so we are hopefully working from the same page. Interestingly, the summary that an increase in electronegative substitution leads to a decrease in coupling constant is not strictly correct per se.

For geminal coupling, electronegative substitution at the alpha position leads to a more positive coupling constant. H-H Geminal couplings are negative, and hence we observe a coupling that becomes smaller in magnitude. (There are a few examples where they actually go the other way, but still become more positive).
 On the other hand, H-F geminal couplings are positive, and so an increase in 2JHF coupling is observed with an increase in electronegative substitution.
Electronegative substitution at the beta position has the opposite effect, leading to a more negative coupling; this actually gives a coupling that is larger in magnitude. Effects from substituents at the beta position are usually smaller than for the alpha position, but the effect is nevertheless opposite.

For vicinal couplings, electronegative substitution leads to a reduction in magnitude of the coupling constant.

The magnitude of scalar spin coupling depends on a few parameters:

the angles between overlapping bonds between coupled nuclei
electronegativity of substituents
bond length

The traditional accounts for the mechanism of spin-spin coupling might go something like...

In the simplest terms, spin coupling comes about because the magnetic moment of nucleus A interacts with the electrons surrounding it, and these electrons produce a local magnetic field at the neighbouring nucleus B. Spin coupling is dominated by the Fermi contact over a short number of bonds. The Fermi contact describes the magnetic interaction of an electron within the nucleus. Only s-orbitals can contribute electron density at the nucleus (all other orbitals have nodes at the nucleus), and so bonding to entities that will promote s-character of molecular will strongly influence the coupling constants.

The more stable orientation of for the nucleus and electron are antiparallel (because they have opposite sign magnetogyric ratios), and for electrons in a bond also to be antiparallel. For the most stable arrangement we get something like (lower case = electron spin, upper case = nuclear spin, u=up, d=down, - are bonds):

Ud-uDu-dUd-uDu-.....

Essentially, the observable nuclear transitions depend on the molecular orbitals, and how well the molecular orbitals stabilise the antiparallel, or spin-polarised configuration.  There are some comprehensive explanations of the molecular orbital model given in Gunther, as you have already encountered, and first published by Pople and Bothner-by. You will also find an excellent account, as always, on Reich's NMR page.

A coupling constant is positive, by definition, if it stabilises the anti-parallel electron spins, as shown above. So, a 1J will be positive, 2J negative (usually) and 3J positive. If the probability of finding an electron in the up state is the same as for the down state, the coupling constant will be zero. For a simple 2 nucleus, 2 electron bond system, if there is an up-electron at one nucleus, then there is a very high chance that there will be a down electron at the other nucleus, and so the coupling will be large. For 2 nuclei across multiple bonds, the spin state of an electron at nucleus A will have a negligible effect on the spin state of electron at nucleus B, and so the coupling will be very small.

...Pause for a breath....

So, to the crux of your question...
If a substitution element can help stabilise the anti-parallel arrangement of the electron spins, by reducing the energy gap for the HOMO-LUMO of the molecular orbitals then the coupling constant will become more positive. Electron-withdrawing groups serve to reduce the energy gap between these orbitals (arguments based on symmetry-allowed transitions - read the above references). σ-acceptors will reduce the LUMO, while π-donors raise the HOMO; both have the same outcome of making the coupling more positive.

This explains the geminal trend, but what about vicinal? There are some theories about alternating polarisation along the bond pathway, so that stabilising the anti-parallel alignment in one molecular orbital will destabilise the adjoining orbitals, hence the trend towards reducing J. It all gets a bit vague about here, and not very convincing. There was some good work by Altona et al in the 80s/90s (see for example) extending the Karplus equation to account for electronegativities, and improved Karplus equations were published, but nothing universal. At this point, we begin to accept that the ability to describe the physical nature of what is going on here is not really going to work using an emperical model.

More recent work by Cremera and Gräfenstein, for example, pretty well put these old 'textbook' explanations back in their boxes, and describe the mechanisms for spin coupling as composites of 4 contributing factors: Fermi contact, spin dipole, diamagnetic spin orbit and paramagnetic spin orbit. It all gets very Hamiltonian early on, but essentially there are co-dependent interactions of all these terms in the electronics. Fermi contact coupling is propagated over multiple bonds and occurs through space via a molecular orbital vector involving the tails of the orbitals, and these tail interactions are strongly influenced by the substituents. An electronegative substituent on a carbon will increase the local Fermi contact, but lead to a decrease in the tail orbital overlap, leading to a reduction of vicinal coupling, for instance.

Some very useful references for you might include:

Phys. Chem. Chem. Phys., 2007, 9, 2791–2816. Calculation and analysis of NMR spin–spin coupling constants
Annu. Rep. NMR Spectrosc. 2000, 41, 55–184. Advances in theoretical and physical aspects of spin-spin coupling constants


Is the rate of desorption dependent on pressure?

Without knowing exactly what you were told its hard to be exact, but here is a description of the situation.

The Langmuir model makes assumptions (a) adsorption is complete when the surface is filled with one gas molecule  per site, (b) all adsorption sites are equivalent (i.e. the same) and (c) adsorption and desorption are separate processes, i.e. one does not depend on the other, i.e if one site is occupied it does not affect adjacent sites an any way. Note that this is a model and this determines how we look at the problem.
The model is then $$M(g) + S(surface) = GM(surface) $$ where G is gas and S surface and rate constant for adsorption is $k_a$ and for desorption $k_d$.

We let $\theta$ be the extent of adsorption which varies from 0 to 1, i.e. how full the surface is, then the rate of change of $\theta$ depends on the rate constant for adsorption , pressure and fraction of vacant sites or $N(1-\theta)$ for N vacant sites. Thus for adsorption $$ \frac{d\theta}{dt} = k_aPN(1-\theta) $$

and for desorption $$ \frac{d\theta}{dt} = -k_dN\theta $$

At equilibrium we make the rate of change equal to zero, i.e. rate of adsorption equals rate of desorption and with a bit of algebra find that $$\theta=\frac{KP}{KP+1}$$  where K is the equilibrium constant or $K=k_a/k_d$ This equation has the fancy name as the Langmuir Isotherm.

You can see that the rate going onto the surface must be equal to the pressure (no gas, no adsorption) and the area free to adsorb onto $(1-\theta)$. (We assume in this model than no molecule double adsorbs; this is dealt with in more advanced treatments). The rate leaving has to be proportional to the amount already there , i.e. proportional to N the total number of surface sites multiplied by the fraction covered.

Ask more questions in the comments if this is not clear.


Number of pi-electrons in an energy state

Your calculation is indeed correct. However, because each electron in the $\pi$ system can have two spin states, there are two electrons per energy level of the “particle in a box” Hamiltonian. Hence, $N=2k$.


The clearest answer is that an analyte is retained by both the mobile and the stationary phase and the retention factor is a measure of the ratio between retention in the mobile phase and in the stationary phase.
So you could think of it as a higher Rf value indicates a higher retention in the mobile phase.

(Also, in answer to some of the comments, the retardation factor is the inverse of the retention factor)


That explains a eutectic but does not answer the question, which is a good one for solid-state chemists.
A cocrystal is a crystalline, multiple component solid with a distinct stoichiometry (molar ratio) between components where a single component is neutral overall(can exist alone without a counter ion, typically non-ionic but a zwitterion counts as a single neutral component) and the components interact by non-covalent interactions that can be partially ionic and has distinctive crystalline characterization data, i.e. to be crystalline a solid the mixed solid phase needs to have X-Ray or electron diffraction data that distinguishes it from its starting components.
So, the eutectic phase of 2 solids, or the solid mixture with the lowest melting pt, can be a cocrystal, but a cocrystal is not necessarily a eutectic. A solid mixture with no regular repeating crystal lattice and no consistent molar ratio (stoichiometry) is called a solid solution. So a eutectic phase could be a cocrystal, or a salt, or a mix of those, or physical mix (solid solution).


How to index molecules in computer programs?

A molecule is a graph, where nodes have labels (chemical element type), and edges are chemical bonds.

while fairly common and acceptable for an initial approach, this models fails to catch many types of isomers and tautomers. There is no well-established general approach because the basic definitions are kinda wonky and the subjects itself is extremely complex. For example, monomethylbulvalene would for all intents and purposes work as a single compound yet a set of different molecules would be present in all reasonably analyzed samples.

  By extension, how to efficiently search a set of molecules and, given a particular molecule, find a set of the closest structural matches?

That's a part of QSAR big problem: to identify "similar" molecules. It doesn't work reliably.

  What characteristics of the molecules should be memorized that would allow such search to be efficient?

There are specialized graph databased that aim to solve the problem of efficient search on graph-like structures, but see above.


Does someone know the name of this equipment?

There are different equipment used to measure force like pivot balance, strain gauge, spring balance, etc.

By seeing the image, we can easily figure out it is spring balance.


Technical name for the holes (missing electrons) in a p-type semi-conductor?

Yes, the technical name is, in fact, an electron hole (typically referred to as simply a hole). It is considered suitable to describe these holes as electric charge carriers with a positive charge that is equal in magnitude to the negative charge of an electron. So, you have actually answered each one of your questions!


Effect of periodic acid on cyclohexane derivatives

I think your problem with this reaction lies in the easily confused term syn and cis. Both appear to mean that two functional groups are oriented in the same direction. On a cyclic molecule, syn and cis are the same. However, cis is a stereochemical descriptor and syn is a conformational descriptor. Simplifying this reaction to an acyclic case may help. Consider the meso isomer of  2,3-butanediol shown.

This isomer is drawn in the anti conformation, which is probably the lowest energy for steric reasons. This isomer has a syn conformation, which looks like the following. Note that the two -OH groups are not both oriented forward with solid wedges. That drawing implies a high energy eclipsed conformation.

(source: nist.gov)
Now back to the reaction with HIO4. In order to form the cyclic intermediate, the -OH groups need to be close in space. On an acyclic molecule, this requires a syn conformation, which most acyclic molecules can achieve even if it is not the lowest energy conformation. In cyclic molecules, particularly cyclohexanes, the syn conformation might not have the -OH groups close together. For cis-1,2-cyclohexanediol, which is also syn respective to the "plane" of the ring, to move the two -OH groups into the correct arrangement, it would need to wiggle into the higher energy boat or twist-boat conformations. However, trans-1,2-cyclohexanediol, which is anti has the two -OH groups oriented close together. In a sense, the trans isomer has the two -OH groups oriented syn with respect to the C-C bond that they share. Thus, trans-1,2-cyclohexanediol should react more readily with HIO4 that cis-1,2-cyclohexanediol, because the lowest energy conformer of trans-1,2-cyclohexanediol has the close arrangement of two -OH groups. The cis isomer would have to change conformation, which increases the energy input required, increasing the activation energy and slowing the reaction rate.
The diaxial case (whether boat or chair), would definitely be unreactive.


A commercial solution would be a dye like IRF940A (from QCR Solutions) with an excitation maximum at $\lambda_{\mathrm{exc}} = 405\,\mathrm{nm}$ and an  emission maximum at $\lambda_{\mathrm{em}} = 940\,\mathrm{nm}$. The supplier does not provide any information on the structure of the dye.
What compounds are promising candidates?
Ytterbium(III) complexes!
In a recent article,published in J. Photochem. Photobiol. A, 2015, *309, 65-71 (DOI), J. Legendziewicz and coworkers prepared a $\ce{Yb(III)}$ phthalocyanine with acetate as a chelating axial ligand and examined the optical properties in solution and in PMMA polymer. In PMMA, the compound shows a Soret band with a maximum at $\lambda_{\mathrm{abs}} = 350\,\mathrm{nm}$. Excitation in the Soret band result in emission with a maximum around around $\lambda_{\mathrm{em}} = 700\,\mathrm{nm}$.
Other examples for even larger wavelength differences, resulting from ligand-to-metal energy transfer are known  for $\ce{Yb(III)}$ complexes bound to eosines or fluoresceines as the primary chromophor.


$\ce{C=C}$ average bond energy is somewhere around $\pu{614 kJ/mol}$ so "k" probably indicates a double bond


Why is the equilibrium constant unaffected by a change of the initial concentration?

Clearly the math tells you that the equilibrium constant is independent of the initial compositions:

For $a\mathrm{A}+b\mathrm{B}\rightarrow c\mathrm{C}+d\mathrm{D}$,
$$K_c=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$$
where $[\mathrm{C}]$, $[\mathrm{D}]$, $[\mathrm{A}]$, and $[\mathrm{B}]$ are concentrations at equilibrium.

This makes some sense intuitively: the rate of the reaction is dependent only on the present concentrations of compounds (and some other stuff like pressure and temperature), not previous concentrations. And equilibrium is defined as a position where the rate of forward and backward reactions are the same.