Difference required for compound to be a phenol vs benzene compound

In organic nomenclature, when you have a compound with multiple functional groups, you follow the priority rules that have been nicely summed up on Wikipedia:

  When compounds contain more than one functional group, the order of precedence determines which groups are named with prefix or suffix forms. The highest-precedence group takes the suffix, with all others taking the prefix form. However, double and triple bonds only take suffix form (-en and -yn) and are used with other suffixes.

These rules apply to benzene derivatives as well as aliphatic and other cyclic compounds.

Continuing,

  Prefixed substituents are ordered alphabetically (excluding any modifiers such as di-, tri-, etc.), e.g. chlorofluoromethane, not fluorochloromethane. If there are multiple functional groups of the same type, either prefixed or suffixed, the position numbers are ordered numerically (thus ethane-1,2-diol, not ethane-2,1-diol.) The N position indicator for amines and amides comes before "1", e.g. $\ce{CH3CH(CH3)CH2NH(CH3)}$ is N,2-dimethylpropanamine

Here's a shorter (and quite handy) version of the Wikipedia table (shamelessly copied from masterorganicchemistry.com, I've taken a liking to their color-scheme :-))

So in the example you've posted:

You give the carbon atom that has $\ce{-COOH}$ attached to it the first locant (because $\ce{-COOH}$ takes precedence over $\ce{-OH}$, according to the priority rules), and you continue numbering clockwise so the the two $\ce{-OH}$ groups take locants 3 and 4 respectively (Lowest set of locants rule).

That's how you end up with 3,4-dihydroxybenzoic acid :-)


Determining the reduction potential of a half cell reaction at a given pH

The K you refer to relates to the overall cell reaction and should be Q, the reaction quotient. This is only equal to K when the potential difference between the two 1/2 cells has fallen to zero.

This question does not refer to a full cell, but to a 1/2 cell. The form of The Nernst Equation you must use, assuming we are at 298K, is:

$\rm E=E^\circ+\dfrac{0.0591}{z}log\dfrac{[oxidised\,\, form]}{[reduced\,\, form]}$

Where z is the number of moles of electrons transferred which, in this case = 2.

This becomes:

$\rm E=E^\circ+\dfrac{0.0591}{2}log[Cu^{2+}]$

The 1/2 cell in question is:

$\rm Cu_{(aq)}^{2+}+2e\rightleftharpoons Cu_{(s)}\,\,\,\,\,\,\,\,\,\,E^\circ=+0.34\,V$

At such a high pH the copper(II) ions will precipitate.

$\rm pH=14\,\,\,so\,\,\,pOH=0$

So $\rm c[OH^-]=1\,\,mol/l$

This will precipitate out the copper(II) ions as the hydroxide.

So we have:

$\rm Cu(OH)_{2(s)}\rightleftharpoons Cu_{(aq)}^{2+}+2OH_{(aq)}^-$

$\rm K_{sp}=[Cu^{2+}][OH^-]^2=10^{-19}$

$[\ce{OH-}]$ is 1 (pH is 14), so $[\ce{Cu^2+}]$ is equal to $\pu{e-19}$.

As you can see this is a very low concentration. Applying Le Chatelier's Principle to the 1/2 cell couple we would predict that reducing the concentration of Cu(II) will drive the position of equilibrium to the left. This would push out electrons and make the electrode potential of the 1/2 cell more negative.

Putting in the numbers into the Nernst Equation:

$\rm E = 0.34+\dfrac{0.0591}{2}\log \pu{e-19}$

$\rm E = 0.34- 1.12\,\,\,\,\,V$

$\rm E = -0.78\,\,\,\,V$

As predicted, it has been reduced considerably.


In answer to your last statement about the formation of Hydrogen. The actual amount of hydronium ions doesn't matter because it is at equilibrium ($\ce{H2O <=> OH- + H+}$) and so once you use an $\ce{H+}$ you'll simply have more $\ce{H2O}$ dissociate into $\ce{OH-}$ and $\ce{H+}$ (Le Chatlier states) so if you have battery current running through the system you should experience a healthy, visible stream of hydrogen bubbles. I built a similar thing myself back in high school.

The hydroxide ions will come from the water also. As mentioned above, the $\ce{H2O}$ will dissociate into $\ce{H+}$ and $\ce{OH-}$ and you are using both. The $\ce{H2O}$ will keep dissociating and try to keep the $\ce{H+}$ and $\ce{OH-}$ concentrations at $10^{-7} \, \small\text{M}$  as you keep using them.

Feel free to question my knowledge.


Why do some solid chemicals dissolve better in colder water?

As with every process, dissolving can be described by the thermodynamical equation:

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

Which the process being spontaneous if $\Delta G$ is negative.

We have two components here, enthalpic and entropic. The enthalpic component $\Delta H$ is a constant of the salt in question. The entropic component $\Delta S$ can usually be considered positive because a highly ordered crystal lattice is broken to give dissolved, much less ordered fragments — hence why for most salts they dissolve easier in hot water.

If the solubility decreases with higher temperatures, then $\Delta S$ must be negative. The salt will then only dissolve if $\Delta H$ is negative enough to give an overall negative result. This seems to be the case for $\ce{Na2SO4}$

It is impossible to predict $\Delta S$ and $\Delta H$ a priori without any computer quantum calculations. One has to resort to experiments to determine whether a salt will dissolve better or worse in the heat.

While I have used the term ‘salt’ most of the time, the same is true for molecules. Take sucrose: I was asked in a physical chemistry lab to guess with my colleague whether sugar would dissolve endothermicly or exothermicly. We decided, that the degree of order be a lot higher in dissolved sugar due to a high amount of water surrounding it rather tightly, that the entropy must decrease, so that the enthalpy has to be negative and sugar has to dissolve exothermically. Hence sugar should dissolve more easily in cold water, too.

According to a comment by Nicolau Saker Neto, our reasoning was false and sucrose displays increasing solubility at higher temperatures. Apparantly the testing faculty member approved of it, though.


Why is the composition of a liquified vapor the same as that of the vapor?

There is a misconception in using this diagram. You cannot go from Point a to Point b This diagram should not be read as you do. It should be done this way.

Suppose you start from a liquid made of 25% A + 75% B. It is a mixture. Its representative point on the diagram is on the edge of the rectangle between "Composition" and "100% B". When heating this mixture, the corresponding point moves upwards, but stays liquid. When reaching the curve "Liquid", the first bubble of gas is formed. We are at the point a (red) But this bubble has another composition. The composition of the gas corresponds to b (in red). So it is practically pure A. This is the beginning of the vaporization phenomena.

If you continue heating the liquid, more gas will be produced. But, as A is more eliminated than B by vaporization, the remaining liquid looses some A and practically no B. So in percentage, the liquid will contain more B. Its composition will soon be 76% B and 24% A, then 77% B and 23 % A, etc. But the temperature has to be elevated to allow this richer mixture to vaporize. The representative point of the liquid will move but stay on the sloping curve. After some time, the point of the liquid will be just above the capital L from the word "Liquid" At this moment the gas is mainly A but it starts containing 1 or 2% B. The composition of the gas could be determined by drawing a horizontal line from the point above the letter "L". This horizontal line will cut the curve "Gas" at an abscissa equal to about 1% or 2%. This defines the composition of the gas.

If you continue heating, the liquid will contain more and more B. Its point will stay on the curve "Liquid". The temperature must be higher and higher. The amount of liquid decreases. Simultaneously the composition of the gas which is eliminated contains more and more B. Its composition is always determined by drawing a horizontal line from the liquid to the curve on the left-side of the diagram.

The last drop of liquid disappears when the liquid reaches the Point "bp B".

Hopefully you have understood that passing from point a (in red) to point b (in red) has no meaning. a and b are always present : one describes the composition of the liquid, and one the composition of the gas.


How does the atomic structure of crystals determines its shapes?

It’s all about minimizing the energy of the system. Imagine a 2D example:

You can imagine the green circles as the cations ($\ce{Na+}$) and the blue ones as anions ($\ce{Cl-}$). Where is the next chloride you add going to go?

Not a great choice – it doesn’t really balance. The nearest blue ion is already balancing its charge with two other green ions and so can’t help the new ion balance its charge as well.
This is on the grid but still has all the same problems where the new ion doesn’t get its charge nicely balanced because the nearest blue ion is already donating part of its charge to two other ions.
This is the best option because it can share partial charges with two other blue ions. It gets more shared charge and the whole thing ends up being lower energy.

This pattern will keep happening with every addition until you get a macro-scale cube. It is always lower energy to fit new ions into gaps then to branch out and make new structures.

Ions that aren’t part of a complete unit cell will develop surface charges. These surface charges are better at attracting ions that are precipitating onto the surface from solution.


How do the atomic orbitals look like in reality?

Atomic orbitals are just the solution of Schrodinger's wave equation. They help us to understand better the behaviour of electrons. Electrons behave more like waves and atomic orbitals can be thought of as places where the probability of finding those electrons is maximum. In reality, there is no such place as an atomic orbital.


Direct connection in a zinc copper battery

A close circuit is needed for electrons to flow. The electrolyte and salt bridge serves to complete the circuit. Additionally, the salt bridge also serves to function of maintain electrical neutrality in each half-cell, preventing the accumulation of charges, aiding the flow of electrons.

If charges were to accumulate in each half-cell, the potential difference between the two half-cells, which is the key driver of electrons, would decrease over time to 0 V and no electrons would flow.

There have been many discussions on this site. To cite a few:
What is the real purpose of the salt bridge in an electrochemical cell? (look at the comments)

Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

electrochemical (Galvanic) cell solution purpose


In an SN1 reaction, 1-bromohexane reacts with NaOH, but would the Na or H be the nucleophile that substitutes the Bromine?

SN1 is very less likely to be followed in 1-bromohexane, rather a SN2 pathway will be followed. OH- replaces Br as Br- as Br is a good leaving group. This has nothing to do with product stability. The haloalkane substrate is reactive so it reacts, as simple as that.


Argon is a noble gas. Meaning, it does not simply "bond" with anything. These special gasses have a VERY strong resistance to joining with other atoms. So, to answer your question, no, you can not bond Argon and Lithium under any circumstance. Anyone who says they have done so is a LiAr ;)

Also, since Lithium is highly reactive and flammable and Argon is used in neon, I would not suggest trying to combine these two without a proper location and gear.


How right is defining elements with the number of protons?

Most of chemistry doesn't care a whit about all of the properties you named: they're in the realm of nuclear physics, as noted by user badjohn.  Even radioactive decay is something of a black sheep, included in physical chemistry as a convenient example of a first-order kinetic process. To my knowledge, it's not substantially used in conjunction with any other sub-discipline of chemistry, as it'd be more of an annoyance than anything ("Gah, the cobalt center of my complex keeps changing into nickel!").  It does of course have plenty of applications (medical imaging, nuclear power, etc.), but they're arguably in the realm of physics, with any attendant chemistry being a secondary consideration.

Magnetic moment, neutron count (isotopes), and nuclear quantum states do have some slight influence over certain spectroscopic properties (primarily NMR, as far as I know), but other than the somewhat special cases of the spin isomers of hydrogen and the kinetic isotope effect of deuterium vs. protium I'm not aware of any appreciable effect of these on chemical reactivity.

For the chemist, the identity of a nucleus is sufficiently defined (within an experimentally measurable perturbation, at least) in virtually all cases by its atomic number.


If capsaicin is insoluble in water, what makes chilli water taste spicy?

You got me curious, so I poked around a bit on this.

First there is insoluble and then there is insoluble. The Wikipedia article on capsaicin lists its solubility as $\pu{0.0013 g}/\pu{100 mL}$ which is $13$ parts per million. So capsaicin is "relatively insoluble", but not wholly so. Second the Wikipedia article also points out that capsaicin itself is just one of the various capsaicinoid chemicals which give peppers their "hot" taste.

At the end of the (old) Wikipedia article on the Scoville scale, a taste test for hotness, it states

  [...] A measurement of one part capsaicin per million corresponds to about 16 Scoville units. [...]

So, 13 pmm multiplied by 16 Scoville units per ppm is about 200 Scoville units.

A table on hot sauces and peppers lists the Original TABASCO® brand Pepper Sauce from McIlhenny Company as being 2,500 to 5,000 Scoville units. So 200 Scoville units would be about one tenth strength sauce. That should still be enough to tingle your tongue.


This is one of those questions where I think you'll need to know more chemistry than you probably have been taught.

The freezing point depression formula is only usable (i.e. linear) for "dilute" solutions. As the concentrations of solutes become larger, the relationship becomes non-linear.

So when $\ce{NaCl}$ is dissolved in water all the sodium and chloride ions are truly dissociated. There are no "molecules" of $\ce{NaCl}$ floating around in the aqueous solution.

Imagine the solution this way. Instead of the water molecules bouncing around in the solution more or less randomly, the positive charged $\ce{Na^+}$ ion attracts the more negative oxygen atom of the water molecule leaving the positive hydrogen atoms on the outside of this first "shell" of ligands. Now a second "shell" of water molecules will be attracted to the first (Oxygen to hydrogens) and so on. Now each shell has more and more water molecules but the binding energy decreases as the shell number increases. So the gist is that there is an exponential decrease in the ordering of the water molecules. Ultimately the ordering becomes random again as in the bulk of the water.

Now in dilute solutions these spheres of water around each ion are separated by so much water that they never interact. But as the solution becomes more concentrated then these spheres of water do start to interact. The interaction is measured by what is known as an activity coefficient.


Desorption instead of adsorption from solution using activated charcoal

In such experiments you typically run a blank. So add the homemade activated charcoal to just some water, filter, then titrate that filtrate as a blank.

There are two possibilities for the increase.

    (1) Your batch of homemade activated charcoal was contaminated with who knows what soluble compound. (This seems the most likely...)

    (2) Quite possible that some of the charcoal particulates are small enough to not be trapped by the filter.

Another way to run the experiment would be to add the homemade activated charcoal to just some water, filter, then wash the wet charcoal into the oxalic acid solutions (this would remove soluble compounds from the charcoal). Then dilute all the solutions to some co0nstant volume, filter, and titrate the filtrates. (You should still run a blank.)


Why doesn't the Bessemer process oxidize the iron?

I thought you asked a great question, one that I have wondered about myself.

The answer, as best I could determine, is that the iron does oxidize.  A good description of the Basic Oxygen Steelmaking process, today's descendent of the Bessamer process, says so anyway.  The BOS process uses pure oxygen instead of the original Bessamer processes's air.  It reports that some iron does oxidize.

The oxidation of carbon to form carbon monoxide is more thermodynamically favorable than the oxidation of iron, so it occurs preferentially.  The page I found (once again the link is here) mentions using on the order of 1800 scf of oxygen per ton of metal.  A ton of steel is about 18 kmol of iron.  1800 scf of oxygen is about 2 kmol of oxygen.  So there is not enough oxygen to oxidize even close to all the iron in the average BOS run.  That fact, combined with the fact that carbon, silicon, and other impurities in the pig iron oxidize preferentially, mean that not much iron is oxidized during BOS or Bessamer.

But "not much" does not mean zero.  My guess is that iron oxides are more soluble in the slag than they are in molten iron and so are also removed with the slag after BOS.


Why does increasing number of salt bridges increase voltage of electrochemical battery?

The voltage you measure between the terminals of a voltaic cell will depend on two factors:

The intrinsic maximum voltage $(V_\mathrm{max} = E_\mathrm{cell})$ that the cell could produce, depending on the $E^o_{red}$ of each half cell, the ion concentrations and the temperature. This is calculated from the Nernst equation:
$$E_\mathrm{cell} = E^⦵_\mathrm{cell} - \frac{RT}{zF}\ln{Q_r}$$

Resistive terms that come from within the cell, including overpotentials at the electrodes and other factors that influence the internal resistance of the cell.

Overall, the observed voltage $(V_L)$ measured at the terminals of the cell will be lower than the theoretical $V_\mathrm{max}$ because the internal resistance of the cell involves a voltage drop, in the same way that you would see an external voltage drop across a resistor in series in the circuit.
$V_L$ can be calculated in the following way:
$$V_L = V_\mathrm{max} - V_I$$
where $V_I$ is the internal voltage drop from the internal resistance of the cell. This drop in voltage depends on the internal resistance of the cell ($R_I$) and the total current $(I)$ running through the cell (and the circuit, see farside.ph.utexas.edu):
$$V_I = I \cdot R_I$$
Your question seems to revolve around how the salt bridge affects the internal resistance of the cell. It would make sense that the more salt bridges you add, the less internal resistance would result, so the smaller the voltage drop and the larger the measured voltage from the cell.
It is possible to think of the salt bridge, which involves the flow of ions, as being analogous to a wire, which involves the flow of electrons. The resistance of a wire $(R)$ is dependent on it cross-sectional area (see http://hyperphysics.phy-astr.gsu.edu):
$$R = \frac{ρL}{A}$$
where $ρ$ is resistivity; $L$ is length; $A$ is cross sectional area.
So it follows that the same may be true for a salt bridge. By adding more salt bridges, you are increasing the cross-sectional area, decreasing the internal resistance and increasing the observed voltage, which would approach the theoretical maximum voltage as the internal resistance approaches zero.


How to determine force constant of C-C bond in ethanol?

You can estimate the force constant (as I have done here) and calculate the frequency of the stretch or, as you request, go the other way. Regardless, you have 2 unknowns and thus the problem you present isn't solvable. Below, I work it from the assumed value of the force constant: to go the other way, you'll need to locate the absorption spectral line for the C-C stretch in ethanol and go from there. Either way, you'll need to make some assumptions regarding reduced mass.

From page 496 in Organic Chemistry by Brown, Iverson, Anslyn, and Foote, the formula

$$\tilde{\nu} = 4.12\;\sqrt{{k\over\mu}}$$

is given, where $\tilde{\nu}$ is the frequency in wavenumbers, $k$ is the force constant in units of dyne/cm, and $\mu$ is the reduced mass (in amu, or atomic mass units). They state that for a single bond, the force constant can be taken as $5\cdot 10^{5}$ dynes/cm (an approximation, and we do have to start somewhere). The 4.12 differs from the constant in your given equation due to unit conversion between dynes/cm and N/m.

Now, on to your problem. The reduced mass term is (generally)

$$\mu = {m_{1}m_{2}\over m_{1} + m_{2}}$$

which is for two masses. I posit that in ethanol, one carbon has a "mass" of 15 amu (that's the carbon with 3 hydrogens attached to it) and the other has a "mass" of 31 amu (the carbon plus 3 hydrogens and the oxygen). To go beyond this involves a significant amount of advanced work that is beyond the scope of your problem (but if you're interested, Molecular Vibrations by Wilson, Decius, and Cross is the place to start).

We calculate

$$\mu = {15\times 31\over 15+31} = 10.1$$

and substitute into the first equation

$$\tilde{\nu} = 4.12\;\sqrt{{5\cdot 10^{5}\over 10.1}} = 916\;\mathrm{cm}^{-1}$$

From some notes for a class at Caltech, it appears we can expect a carbon-carbon single bond to appear anywhere in the range of 800 to 1200 cm$^{-1}$.

The calculated number for the C-C stretch here (towards the lower range we'd expect) is due to our "bloated" reduced masses, and that makes intuitive sense (if we do the same calculation for ethane, with "lighter" reduced masses for each carbon, we get $\tilde{\nu} = 1063$ cm$^{-1}$).


Why is free radical chlorination exothermic whereas bromination is endothermic?

First off, we can see that both reactions are exothermic If we do the thermodynamic analysis.

Given that your starting materials have the same heat of formation:

$$\Delta H^\circ_f(Products) = \Delta H^\circ_f (\ce{R-H}) + 0, \quad{\Delta H^\circ_f(\ce{Cl}) = \Delta H^\circ_f(\ce{Br})= 0}$$

$$\ce{R = Et}\\ \Delta H^\circ_f (\ce{EtH}) = \pu{-84 kJ mol^{-1}}$$
$$
\begin{array}{cccccc}
\hline
\text{Product 1} & \Delta \text{H}^\circ_f [\pu{kJ mol^{-1}}] & \text{Product 2} & \Delta \text{H}^\circ_f [\pu{kJ mol^{-1}}] & \Sigma\ \Delta \text{H}^\circ_f\  [\pu{kj mol^{-1}}] &\text{Result}\\
\hline
\ce{HBr} & -\text{36.45} & \ce{Et-Br} & -\text{97.6} &-\text{134.05}<-\text{84} &\text{Exothermic} \\
\ce{HCl} & -\text{92.31} & \ce{Et-Cl} & -\text{137} & -\text{229.31}<-\text{84}  &\text{Exothermic}\\
\hline
\end{array}
$$

Back to your question:

  Why is there a difference?

The graphs you are looking at only consider the reaction:

$$\ce{R-H + X^. ->  R^. + HX}$$

For which we get:
$$
\begin{array}{cccccc}
\hline
\text{X}
& \Delta \text{H}^\circ_f(\ce{X^.})
& \Delta \text{H}^\circ_f(\ce{RH})
& \Delta \text{H}^\circ_f(\ce{R^.})
& \Delta \text{H}^\circ_f(\ce{HX})
& \Delta \text{H}^\circ_{rxn}
& \text{Result}\\
\hline
\ce{Br} & -\text{96.94}^{**} & -\text{84} & -\text{119}^\dagger &-\text{36.45} &\text{25.49}&\text{Endothermic} \\
\ce{Cl} & -\text{121.29}^{**} & -\text{84} & -\text{119}^\dagger & -\text{92.31} &-\text{6.02}&\text{Exothermic}\\
\hline
\end{array}
$$
*all units in $[\pu{kj mol^{-1}}]$ all values from Wikipedia unless otherwise stated
** Table of Bond Dissociation Energies
$^\dagger$ Values from NIST WebBook

Here the enthalpy of radicalization of the alkane is different for chlorine or bromine, but the halogenation in total is exothermic


Why is the buckminsterfullerene the purest form of carbon?

Diamond has dangling bonds on the outer surface of the crystal for pretty much the same reason as graphite. If you understood graphite differently, then you understood it wrong.

See, a molecule of oxygen contains 2 atoms, a molecule of sulfur has 8; but how many atoms are there in a "molecule" of diamond or graphite? Try drawing one to the end, so as to count them. You won't be able to do that. There is no end. The thing is infinite. But the real-world objects are finite, which means that at some point you have to say "Enough" and crop your ideal structure, and in doing so, you leave dangling bonds which attract other atoms. Fullerene lacks those, and hence is "more pure".

There is an altogether different dimension to the problem. Our thought experiment implied that we are able to produce a huge crystal without defects except maybe some on the surface. This is not true. Real-world compounds always contain impurities, and once you have a wrong atom built into the crystal lattice of graphite or diamond, it is stuck there forever. You'll never remove it, short of destroying the entire crystal. Fullerenes, on the other hand, are molecular compounds. They can be dissolved. They can be put through chromatography, sublimation, and other purification techniques. We can always remove any impurity (not that we can remove all of them, because nothing is ideal).

Either way, fullerenes win.


Reaction of phosphorous acid (H3PO3) and potassium hydroxide (KOH)

You have made the assumption that all the hydrogens are attached to oxygens. This is not a good assumption; one of the hydrogens is bonded directly to phosphorus and is not removed by OH-.
A better description of phosphorous acid is HPO(OH)2 with one oxygen doubly bonded to phosphorus  wikipedia


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

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

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


Gibbs free energy where T, V, and P change simultaneously

Your original equation is incorrect.  It should read $\Delta G=\Delta H -\Delta (TS)$, and $\Delta (TS)\neq T\Delta S+S\Delta T$ unless the changes are infinitesimal.

That said, for an ideal gas, irrespective of path, $$\Delta S=S_f-S_i=C_p\ln{(T_f/T_i)}-R\ln{(P_f/P_i)}=C_v\ln{(T_f/T_i)}+R\ln{(V_f/V_i)}$$and$$\Delta H=C_p(T_f-T_i)$$So, $$\Delta G=G_f-G_i=C_p\Delta T-\frac{(T_f+T_i)}{2}\Delta S-\frac{(S_f+S_i)}{2}\Delta T$$Clearly the 3rd term on the rhs depends on either $S_i$, $S_f$, or both, so $\Delta G$ depends on these.  It then all comes down to how you assign your reference state values for G, H, and S, such that $G_{ref}=H_{ref}-T_{ref}S_{ref}$.


Why is hydrogen sulfate put together as it is?

Hypervalency is an obsolete concept - it was used to explain bonding situations of molecules, that seem to exceed the octet rule. It was accomplished by the help of $\ce{spd}$ hybrid orbitals. It is now known, that for almost all molecules that were described this way the contribution of $\ce{d}$ orbitals is negligible (<1%).

Focussing on the question at hand, the bonding situation of the core $\ce{SO4^{2-}}$ anion is isoelectronic to the phosphate anion $\ce{PO4^{3-}}$.

For the methylsulfate this situation may be extended. A qualitative analysis of the natural charges (BP86/cc-pVDZ) reveals no difference. All of the oxygen atoms carry a negative charge, while sulfur is highly positive charged. Further more, NBO analysis confirms that the terminal oxygens have three lone pairs, while the other oxygens have two lone pairs. A more accurate Lewis structure would represent that fact and only use single bonds.


Why is the thione tautomeric form of 2-Mercaptopyridine favoured?

Assumption (1) doesn't hold too much precedence in this case, just like in the case of 2-pyridone, so it is a minor thing to consider. In the case of the assumption (2), electronegativity is only a part of it. Sulfur atom can stabilize the negative charge better than oxygen due to it's size. For the same reason, $\ce{CH3SH}$ is a stronger acid than $\ce{CH3OH}$: $\ce{CH3S-}$ is a more stable conjugate base than $\ce{CH3O-}$. So, although sulfur has a lower electronegativity than oxygen, it is still preferred to have a negative charge on it. Thus, altogether, the thione form is more stable than the thiol.

BTW, thank you for a very detailed question and thorough explanation of your reasoning!


Why do Cu⁺ ions spontaneously form copper metal and Cu²⁺ ions in solution?

Answer:
Both reactions involving Cu+ have the lowest and highest Standard Electrode Potentials respectively. Thus, Cu+ has to be the strongest oxidizing/reducing agents, and therefore it undergoes disproportionation.

Because the formation of Cu has a higher SEP than Cu+, it is more likely to get reduced and therefore is a better oxidizing agent - NOT a better reducing agent - than Cu+ (which is more likely to get oxidized as it has a higher SEP).


What is the purpose of trimming the bottom corners of TLC plates?

Yes, your question contains the answer. Think about capillarity.

Without the little cuts the eluant would be "sucked" and flow from both the bottom and the left ( and right) edges of the thin plate, resulting in a tilted flow that

tends to move the elute faster along the vertical edges while clumping the spots to the center;
decrease the resolution at the given conditions by deforming and enlarging the spots.

Think of it as a weird bad executed two dimensionally-developed plate, if it helps.

Note that beside resolution, the cuts should be there to guarantee a levelled horizontal and straight flow front, as requisite to obtain reliable and useful ratio front values.

(Good practice: draw the real appearance of the developed plate or even post it to the lab journal, so that a future user get a feeling of the separation whatever it happened. The latter not recommended if a product spots out to be irritant/allergenic).

Left: correct - Right: not horizontal flow frontline due to capillarity at vertical edges.

As OP mentions an example with cutted corners but not levelled flow, obviously cutting the corners but plunging the plate in too much eluant is useless. The idea is to have eluant sucked up just by the bottom edge.


Can an ant etch a hole in a tent with formic acid?

I would be VERY surprised if it did. The answer depends on what material the tent is made of. Most tents now days are made of polyester and according to the table below (taken from: http://rfelektronik.se/manuals/Datasheets/solventguide.pdf) polyester has good resistance in formic acid (Note, S stands for satisfactory in table).

However, even if it was nylon, unsatisfactory resistance means that the material will deteriorate in hours to weeks when used with the chemical. I dont think an ant or some ants would be able to produce enough acid to seriously damage the thick fibers of the tent. Maybe if a lot of ants were attacking the same spot one after the other you would eventually dissolve some of the nylon to make a small hole, you never know.

Also, an ant attacking a tent is a very unlikely scenario (ants use formic acid to defend themselves).


How semi-permanent hair dye stays in place?

According to this: http://questions.sci-toys.com/node/72, hair dyes work by creating pockets within the hair. The colored dye molecules then get stuck in these pockets - requiring multiple washes to escape. Short term hair dyes use larger dye molecules so that they cannot get stuck in the hair. On the other hand, longer term dyes actually react with you hair to stay in there for an extended duration of time. For the most part, all of these hair-dye interactions are the result of intermolecular forces.

Semi permanent hair dyes tend to be small compounds that get stuck in your hair. This means that they are not permanently bound to your hair, but they will require a lot of washes to escape. I hope this helps!


Palladium hydride forming water and heat at room temperature

I'm not an expert, but typically there are a series of surface reactions where the reactive species are all adsorbed on the surface. So atomic hydrogen wouldn't "leave" the palladium before it reacts, it just diffuses from an internal crystal site to a surface site.

Likewise, the oxygen would adsorb onto the Pd surface, I'll assume dissociatively. So now there is atomic O and H diffusing around the surface sites of the Pd. These can react and form adsorbed OH, which can further react and form adsorbed H2O, which can then desorb and give you your water.

Your enthalpy calculation is off, the equation for water already includes the fact that an oxygen molecule must break apart to form it.

If you include the hydrogen gas reaction with Pd in your picture, all you're doing is turning hydrogen gas and oxygen gas into water, using Pd as a catalyst. So the total enthalpy change is just -285.83kJ/mol. You don't have enough information to determine how much heat is released (or absorbed possibly) in the hydrogen absorption reaction and how much is released in the water production reaction, but the total of the two will be -285.83kJ/mol.


Why does diamond have no molecular formula?

To quote my answer to Is a diamond a single molecule?:

Diamond is a covalent network solid, like a number of other common materials (quartz, graphite, glass, and a whole bunch of stuff).
Because they are not discrete molecules - there is no 'diamond' molecule the same way there are molecules of caffeine, benzoic acid, citric acid, N,N-dimethylaminopyridine, etc.

Another way to think about it, since diamond is a covalent network solid, each diamond is a single giant molecule. Since each diamond is one molecule, there is no universal diamond molecule with the exact same number of carbon atoms every time it occurs.
Fullerenes form distinct molecules with the same number of carbon atoms in each species every time. Thus buckminsterfullerene is always comprised of molecules of $\ce{C_{60}}$.


Size consistency in computational chemisty methods

Size consistency is essentially the idea that if you were to determine the energy of two noninteracting systems A and B, you should obtain the same result if you sum the energy of A and B separately and if you calculate the energy of A and B as one supersystem (usually with A and B separated by a large enough distance where they don't interact). Expressed mathematically, we might write: $$E(A+B)=E(A)+E(B)$$ provided that A and B don't interact.
As to the second part of your question, I want to address a finer point. Full CI is size consistent; the issue comes when you have to truncate CI to make it computationally feasible. Truncated CI is only exactly size consistent up to the point of truncation. For example, CISD (CI with single and double excitations) would only be exactly size consistent for systems with $N=2$ electrons.
The issue becomes a problem for large systems because this lack of size consistency leads to the wrong limiting behavior of the correlation energy. Specifically, truncated CI leads to the false conclusion that $\lim_{N\rightarrow \infty}\frac{E_{corr}}{N}=0$, while the correlation energy should vary linearly with $N$.
For more information on this, I would read chapter 4, section 6 of Szabo and Ostlunds Modern Quantum Chemistry. The chapter as a whole gives a thorough primer on CI, with this particular section discussing the size inconsistency problem.


White hot hydrogen gas, like that during a big bang

(This might get more/better answers on Physics Stack Exchange)

At the high temperatures of the early Big Bang, you have individual protons and electrons, not hydrogen atoms and molecules.  This is because the $kT$ thermal energy is much, much more than the binding energy of the hydrogen atom.

In that case, the moving individual charges create a continuous black body spectrum:

(from Sun.org)

The light will look like a 10,000K source: a bit on the blue side.  But most of the energy will be at wavelengths much shorter than visible light.  The Wien displacement law shows that:

$$\lambda_{\rm{peak}} = {{b}\over{T}} = {{3\times 10^{-3}m \cdot K}\over{1\times 10^6 K}} = 3\times 10^{-9}m$$

A 3nm wavelength corresponds to soft X-rays (or perhaps the very hardest ultraviolet light).

You'll note that the vertical axis below is logarithmic.  At 100X temperature to the 10000K like, the peak will be at the shorter (left) wavelengths of X rays, but also much higher.  The energy flux goes as $T^4$, so a 100X temperature is a factor of $10^8$ in intensity:  There will be a lot of X rays at these early temperatures.


Below is a picture of the 4 blocks in the Periodic Table.  The elements in a block all use the same orbital to hold their valence or outer shell electrons.  For example, all of the elements in block 1 have their valence electrons in an $\ce{s}$ orbital, those in block 2 have their valence electrons in a $\ce{p}$ orbital, block 3 the $\ce{d}$ orbitals and block 4 the $\ce{f}$ orbitals.


Why do heavier transition metals show higher oxidation states?

The inert pair effect is based on the fact that main group elements’ oxidation states depend on s and p orbitals (and only them). When going down the periodic table, the energy difference between s and p orbitals changes leading to some elements losing their valence s electrons more easily than others.

Transition metals’ chemistry happens in the d orbitals primarily — unless you count the copper and zinc groups wherein a significant part of the chemistry is in fact only s orbital chemistry. The d orbitals — at first approximation of the free ion — are degenerate, i.e. they all have the same energy. Thus, these electrons are typically much more accessable. Furthermore, going down the periodic table increases the number of electrons counted as core electrons meaning that the outermost valene electrons experience a weaker effective nuclear attraction. It is therefore easier (i.e. requires less energy) to remove valence electrons and higher oxidation states are much more accessible.

If it weren’t for the inert pair effect, this would also be visible in main group chemistry — except it is, if you compare the relative stabilities of iodate and chlorate.


Why can wood burn all the way to charcoal instead of just creating a passivating charcoal layer?

Charcoal doesn't passivate and the results you see depend on the physical structure of the fire

The difference between a match and a larger wood fire has little to do with the formation of charcoal and more to do with the circumstances of the reaction. And the charcoal isn't really passivating in the reaction on a match.

Wood consists of a complex mix of chemicals as well as the core cellulose, which is a carbohydrate. Some of the ingredients, like the sap and other volatile components, burn easily. Cellulose burns in multiple steps one of which involves pyrolysis where some components are driven off quickly leaving mostly charcoal. Charcoal itself will burn but it takes more energy to ignite and therefore more time to get going. It isn't so much that charcoal passivates, it is just that it is the last component to burn.

The difference between a match and a larger wood fire isn't about passivation but about the scale of the fire. There isn't enough energy in a match head to drive combustion all the way so only the easy-to-burn components ignite (unless you hold the match upside down so the flame spreads up the matchstick when the heat from the tip is enough to cause much more extensive combustion). In a larger wood fire there is enough energy in the flame to cause more complete combustion. So the volatile components burn quickly but leave enough heat for the charcoal to burn as well. Since heat penetrates charcoal very well, the volatile components under the charcoal will tend to ignite before the charcoal itself (it isn't doing much to passivate the wood underneath). In a badly organised fire the overall heat may never be enough to set off the charcoal.

So the issue isn't passivation just a reflection of the order the components of wood burn and the physical structure of the fire. Charcoal isn't passivating, just the last component to burn and, if the fire isn't well constructed, there may never be enough heat to burn it. Much more of a match will burn if you drive the flame along the stalk by holding it upside down and a well organised fire will burn the whole way to ash.

Also controlled pyrolysis (which is used to create charcoal deliberately) relies on controlling the amount of air available to the work being pyrolysed. If there were enough air, the charcoal would burn but the conditions for pyrolysis involve restricting the air so the end-product cannot burn. Charcoal doesn't passivate anything when there is enough air (which is why you can cook on a barbecue).


Which substance has the higher oxidation strength?

Look right here ... .

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/electpot.html

Hypochlorite ion (+0.90 V) has a lower reduction potential in basic solution than elemental chlorine (+1.36 V), so the hypochlorite ion is less strongly oxidizing and more stable in basic solution (which is what you write the reaction for).

Household "chlorine" bleach, which really uses hypochlorite, is made alkaline as one way to stabilize the hypochlorite ion and prevent evolution of chlorine gas.


Why are glycosidic bonds always 1->4 in polysaccharides?

The premise of the question is false.

  Why not 1->3 or 1->2 ?

1,3-linked glycosidic bonds certainly exist.  In fact there's a whole class of molecules called beta glucans of which the 1,3-linked variety forms a major part.

  Could it be that, the formation of glucose chains in plants and animals always make use of enzymes and that it is these enzymes which, by the shape of their active site, dictates what the bonds will be?

I think what you say here is true, but you should keep in mind that there are many different enzymes, and some will specifically form one linkage, and others will form alternate linkages.


Does the hydrogen taken in an E1 reaction have to be antiperiplanar?

In an E2 reaction, there is a strong preference for the hydrogen to be coplanar with the leaving group and pointed in an anti-direction to the leaving group. In an E1 reaction, the hydrogen does not have to be anti-periplanar to the leaving group because in the E1 reaction the deprotonation event happens in a separate mechanistic step, after the leaving group has left. So there's no need to define the relative orientation of the hydrogen and leaving group in the E1 reaction.

However, for the E2 reaction, remember that the anti-periplanar arrangement is preferred because it places the C-H bonding orbital anti to the C-LG antibonding orbital. In the transition state, those are the orbitals that reorganize to create the C-C pi bond.

With respect to the E1 reaction, there is no C-LG antibonding orbital, but there is an empty p-orbital on the carbocation. The C-H bond should be parallel to the p-orbital to create a similar (but strictly speaking, not anti-periplanar) alignment of orbitals.


If we take an atom of any element and then supply heat energy to it
  then what will happen?

A single atom can move in three dimensions (x, y, z) through space, it cannot rotate about an external axis like a diatomic molecule can, nor can it vibrate.  As you supply heat (or any form of energy) to the system containing the atom, it's kinetic energy will increase and it will move faster through space.  If you supply enough energy, then the electrons surrounding the atom will absorb that energy and undergo electronic transitions from the ground state to various excited states.  This process can be diagrammed using a Morse potential diagram , I've drawn a simple one below.  The diagram is meant to show promotion of an electron from the ground electronic state to higher and higher electronic levels as more energy is applied to the system.  Note that as the electron is promoted to higher energy levels, the average separation distance between it and the nucleus increases.  If enough energy is supplied, an electron will dissociate from the atom and we will be left with an ion.  Continue to supply even more energy and more electrons will be removed from the atom. Pumping even more energy into the system will ultimately lead to transformations in the nucleus of the atom.


What does the subscript of atomic orbital mean?

For each azimutal quantum number $l$, the magnetic quantum number $m$ ranges from $-l$ to $+l$. Those give you the number of atomic orbitals "subsripts" you should obtain.
Those are then expressed using spherical harmonics. If you look at the table of spherical harmonics at the $l=1$, you will find that they contain the cartesian axes as $x-y$, $z$ and $x+y$. Therefore it is advisable to form the linear combination out of them, so you get $x$, $y$, $z$. If you go to higher angular momentum, things get more complicated, but there is some tradition what linear combinations to use.

To answer your second question, all this formulas are not some inevitable result you obtain by juggling the numbers. Contrary, it is carefully crafted in a way that the formulas are so simple.


Acid-Base determination of organic compounds

There is a slight error in your sentence. The nitrogen does not have more electrons, but it has a higher electron density because of it's neighbours. But I think this is how you meant it. Often it is described in terms of donating electrons, whilst these electrons are not donated in a cation/anion sense, so I think electron density is a better to grasp description.

Anyway, with being "better electron donors", it is meant that they are more readily able to donate part of their electron density, just like you describe with your description of polarization.

The inductive effect is the ability of an atom or functional group to donate part of it's electron density through $\sigma$-bonds. So in this we are talking about the electronegativity of a group. It is something you have already encountered in chemistry, but this time it has received the proper name 'inductive effect'.

What you would want to be reading, is about Hammett plots. Hammett plots are a tool to picture the ability of substituents on benzoic acid to influence the acidity. Originally Hammett used hydrogen as the reference substituent. The cool thing is that you can compare the effect of substituents on a different molecule, with the effect they have on benzoic acid. I can advise you to get to the library and find 'Modern physical organic chemistry' by Anslyn and Dougherty. Especially chapter 8.2 about Substituent Effects and chapter 8.3 about Hammett Plot's as a tool will be useful.


How do stoichiometric coefficients effect the half-life of a reaction?

Consider what happens to your argument if I rewrite the equation in the following equivalent form:
$$\ce{A -> \frac{3}{2}B +\frac{1}{2}C +D}$$
That is, you write

  ... since rate of consumption of A is double the rate of reaction as can be seen from the stoichiometric coefficients

In contrast, the rate of the reaction, in this case, is defined to be the rate of decay of $\ce{A}$,
$$-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k[\ce{A}]$$
When you look at it like this, it is more clear that the stoichiometry should not have an effect on the decay process as you claimed.

Furthermore, if we integrate the equation above after separating by variables,
$$\int_{[\ce{A}_0]}^{[\ce{A_t}]}\frac{1}{[\ce{A}]}\mathrm{d}[\ce{A}] = \int_0^t-k\mathrm{d}t$$
we get,
$$\ln\left(\frac{[\ce{A_t}]}{[\ce{A}_0]}\right)=-kt.$$

If we try to find the time at which the original concentration has halved, we get the formula you have written above which is,
$$t_{\frac12}=\frac{\ln(2)}{k}.$$

The important point here, and this isn't true for other reaction-orders, is that the half-life is independent of initial concentration. This is not what one would intuitively expect, and indeed it's not true in general, but it is characteristic of all exponential-decay processes.

Indeed, this is a very deep result which doesn't tend to get emphasized enough in my opinion.

It is true that sometimes you can say that a reaction will be second order based on stoichiometric coefficients, but you are told this is first order, so that is probably the source of your confusion.

So, your textbook is right.


How were the absolute masses of atoms calculated before the mass spectrometers?

Absolute mass can be determined electrochemically, once the coulomb is defined in terms of numbers of electrons, as through the Millikan oil-drop experiment, to be  $6.24×10^{18}$ electrons per coulomb.

See Determination of Avogadro's Number and Absolute  Determination  of  the  Electrochemical  Equivalent and  the  Atomic  Weight  of Zinc.

One thing that caused confusion in the determination of atomic mass is natural isotopic separation, e.g. sulfur derived from condensation from vulcanism vs. from mineral compounds, so that there were slightly different experimental results depending on source.


Selectivity in acetonide formation from a polyol

Acetonide formation (isoproylidine ketal formation, to be more correct) is indeed a thermodynamic process, and under acidic conditions acetonides will tend to migrate to form the lowest energy product.

If you look at the penta-ol you've been given, there are three possible permutations containing two acetonides (theres also the possibility of just forming a single acetonide, but these kinds of reactions are usually ran with an excess of the acetone/dimethoxypropane, so its likely to go on as many times as possible):

The green one (which is the one labelled A in your image) is the thermodynamic product, which becomes obvious if you consider the three-dimensional conformation of 1,3-syn and 1,3-anti acetonides (note that this analysis only holds up then there is no substituent at the 2-position. If there is a methyl group there, as is common in polyketide type motifs, then the acetonides don't necessarily adopt chair conformations):

In the syn acetonides, there are three equatorial and one axial groups (minimal strain), whilst in the anti acetonides (which both of the un-observed products contain one of), there are two equatorial and two axial groups, giving a 1,3-diaxial clash across the ring (hence raising the energy of these products).


Why are the calculated NMR values on the DFT & GIAO level of cyclopentane-1,3-dione so far off?

Brief Summary

The computational results given below indicate that there is something inherently problematic with this seemingly simple system.

Computational Methodology

Full optimizations and corresponding harmonic vibrational frequency computations of cyclopentane-1,3-dione was were at two levels of theory:

B3LYP/6-31+G(d,p)
MP2/aug-cc-pVTZ

Each optimized geometry was subjected to GIAO-NMR computations at the following levels of theory:

mPW1PW91/6-311+G(2d,p)
mPW1PW91/IGLO-II
PBE0/IGLO-II
mPW1PW91-SMD/6-311+G(2d,p)
mPW1PW91-SMD/IGLO-II
PBE0-SMD/IGLO-II

The NMR levels of theory were chosen specifically to look at basis set effects as well as implicit solvation (SMD) effects.  The solvent chosen was chloroform.

All computations were carried out using the Gaussian 09 software package.  All DFT computations employed an ultrafine integration grid.  The frozen-core approximation was invoked for the MP2 computations.

Results

Table 1 presents the absolute isotropic shielding constants ($\sigma\ce{C}$) as well as the computed chemical shifts ($\delta\ce{C}$; determined using Equation 1 given by Martin in the comments) with all values given in ppm.  The carbons are labeled in figure 1 for cyclopentane-1,3-dione.  Despite the level of theory used for the optimization  or NMR computation, the $\delta\mathrm{C}^2$ values are consistently off by ≈50 ppm with respect to the 'experimental' values obtained from equation 1.

Figure 1: Cyclopentane-1,3-dione with labeled carbons

Equation 1:
$$\delta=\frac{\text{intersection}-\sigma}{-\text{slope}}= \frac{186.5242-\sigma}{1.0533}$$

I am actually quite surprised by these results as I did not expect this system to be difficult to characterize with straightforward NMR computations.


Coupling constant (Jab) - Why more unpaired electrons means a smaller coupling?

I hope I have understood your question correctly.

For illustrative purposes, I will restrict myself to the Hartree-Fock picture. In a broken spin system, e.g., UHF, the spin state is not a good quantum number. That is to say that the UHF wavefunction need not be an eigenfunction of $S^2$ or $S_z$. For example two complexes with a single unpaired electron each will combine three singlet states (two ionic states and one open-shell) and three triplet states ($m_s\in \{-1,0,1\}$).

The open-shell singlet ($|\alpha\phi_1 \beta\phi_2|-|\alpha\phi_2 \beta\phi_1|$) and the $m_s=0$ triplet ($|\alpha \phi_1 \beta \phi_2|+|\alpha\phi_2 \beta\phi_1|$) states are not single Slater determinants. The singlet states need not be of the same energy, whereas the triplet states have to be degenerate for a Hamiltonian without explicitly spin-dependent components such as spin-orbit coupling. Hence, you have to accommodate for the multiple counting.

For higher spins, the degeneracy grows with $S_{max}$ and must be corrected accordingly. This does not imply that coupling necessarily decreases with higher spins.


Why does a liquids height in a Barometer depend linearly on pressure?

In a barometer at equilibrium, the pressure of the mercury equals the pressure of the air (at the air/mercury interface).

The compressibility of mercury in principle does make it non-linear.  The density would be 3.8 parts per million higher at the air interface than at the vacuum interface.

Certainly there are other greater sources of error: the vacuum isn't really a vacuum but contains mercury vapor, density varies with temperature, vapor pressure varies with temperature, the size of the tube varies with temperature, there are surface tension effects.

Just like any experimental technique, you need to evaluate sources of error and report a measurement with appropriate uncertainty.


Meaning of "ad" in a pharmaceutical patent: "Acetic acid ad pH 5.0"

Most likely yes. In latin,(1) ad means "to, towards, until", like in "ad libitum" (aka ad lib) which means "keep going until I am satisfied", or "ad nauseam".

(1) Pharmaceuticals very often use latin.


In my field, protein crystallography, we often use seeding to improve our crystallization experiments. Typically seeding speeds up crystallization by introducing well defined nucleation cores for crystal formation.

It can often be observed that seeding influences the crystal formation, as crystals grown from seeded crystallization conditions often have better diffraction. This is very important as all our crystals must be formed in their native solvent enviroment (water, lipids/detergents, ions etc).

Sometimes the act of seeding can influence the crystallographic symmetry, it is not uncommon for seeding to cause crystallographic symmetry to turn into non crystallographic symmetry. Personally I have had P12(1)1 crystals generate P1 crystals due to small changes in the asymmetric unit. However my case were also rather extreme as seeding cut my crystallization time from ~3-6 months to 3-6 days.

Also note that crystal seeds, does not necessarily have to be crystalline in nature. Small glass fragments, dust particles and hairs, can also induce crystal growth (I have personally seen examples of all three on a project I have worked on). This probably occurs when the protein being crystallized is highly stable, but cannot form strong inter-molecular interactions in the crystal packing. Then having a foreign surface to interact with may help keep the protein molecules close together and allow weaker indirect interactions (such as hydrogen bonds via water molecules) to form and stabilize the crystal structure.

So in my field, seeding can have an enormous impact on crystallization experiments.


What is the difference between the mean radial distance and the most probable radial distance of an orbital?

In multi-electron atoms the energy of the s, p and d orbitals for a given $n$ is determined primarily by shielding by inner electrons and penetration of the orbital towards the nucleus. As s orbitals have some probability of being at the nucleus their energy is lowest, next is p then d orbitals in increasing energy. Also there is spin-orbit coupling which splits sub-levels of p and d orbitals or any orbital which has angular momentum so this does not effect s orbitals. This effect is much smaller in d than p orbitals and generally smaller than the separation of s to p to d.

The statements in the book extract you give seem contradictory. The mean radial value of an H type orbital is calculated using $\displaystyle \int_0^\infty r^3R_{n,l}^*R_{n,l} dr $ where $R_{n,l}$ is the (normalised) radial part of the wavefunction for an atom with quantum numbers $n,\;l$. The $r^3$ arises because the volume element in the integration is $r^2\sin(\theta)d\phi d\theta$ in spherical coordinates. The other $r$ is how the average is defined, i.e. $\int r\psi^*\psi d\tau$ for some normalised wavefunction $\psi$.

The radial parts of the wavefunction for the 3s,3p and 3d are define with $q=Zr/a_0$ ($a_0$ is the Bohr radius) as
$$\begin{align}
R_{3,0}&=N(27-18q+2q^2)exp^{-q/3}\\
R_{3,1}&=N(6q-q^2)e^{-q/3}\\
R_{3,2}&=Nq^2e^{-q/3}
\end{align}$$

and $N$ is the normalisation which is different for each wavefunction. The integrals are not hard to evaluate but very tedious as there are many terms. However, there is a general formula for the average of H atom type orbitals and is

$$\langle r\rangle =\frac{n^2a_0}{Z}\left( 1+\frac{1}{2} \left( 1-\frac{l(l+1)}{n^2}  \right) \right ) $$

The average values are

$$3s \;\langle r \rangle = 0.476 \text{ nm,}\quad 3p \;\langle r \rangle = 0.441\text{ nm,}\quad 3d\; \langle r \rangle = 0.370\text{ nm}$$

The s orbital value is the same as in your question but the others differ both in value and in order of size. Perhaps there are typos in the text you give or some other unstated factors producing these values.

The maximum values have to be calculated by making the derivative (of $q^2R_{n,l}^2$) zero and solving the equation. Alternatively the value can be taken with sufficient accuracy  from a plot of the function which is far easier. The values are

$$3s \;r_{max} = 0.46 \text{ nm,}\quad 3p \;r_{max} = 0.43\text{ nm,}\quad 3d\; r_{max} = 0.32\text{ nm}$$

the plot below shows the radial functions $q^2R_{n,l}^2$  each normalised to the same maximum height. The mean is greater than the maximum in each case as expected as the main part of the radial probability is 'lopsided' towards larger distances.

(In evaluating the integrals $\displaystyle \int_0^\infty r^ne^{-\beta r}= \frac{n!}{\beta^{n+1}}$ is useful.)


Orbital mixing in square planar D4h complexes

All orbitals of the same symmetry will interact (mix) to some degree.

So the ligand SALC will interact with both $\mathrm{a_{1g}}$ orbitals on the metal. The degree of interaction will depend on how close in energy the ligand $\mathrm{a_{1g}}$ is to the $\mathrm{4s}$ or $\mathrm{3d}_{z^2}$ orbitals on the metal.

Generally, the ligand orbitals are closer in energy to the $\mathrm{3d}_{z^2}$ orbital than to the $\mathrm{4s}$.


Why is the voltage produced by a voltaic cell affected by temperature as described by the Nernst Equation?

Consider that even when exist a "driving force" to move ions to electrodes, it is necesary that particles stay not static, they need some capacity to move to change direction and take way to electrodes, and what gives this capacity to move, Energy, and in any particle system this energy distribution follows Maxwell-Bolzmann and this means that temperature counts.. e^-(1/kT)


How does the human body process Bromelain in pineapple? (A meat tenderising/dissolving enzyme)

The enzyme is slow acting, so it does not dissolve your mouth, just slightly tenderises it, which is the sandpaper feel that you get on your tongue after eating it. When the pineapple reaches your stomach the enzyme gets dissolved by your stomach acids, so cannot harm you any longer.


@TAR86 was correct, I mistook the volume integral as simple 1d integral. Also integrating $1/|r_2 - r_1|$ would result in a singularity at $r_1 == r_2$ which would yield really wrong results.
As per @user1271772 suggestions, materials modelling stack exchange was quite useful (see question here). There as per Susi Lehota's suggestion I was able to get correct result by integrating Legendre Expansion.
My octave code for two electron integral (11|11) follows:
zeta1=1.45;
chi1111 = @(r1,r2) 16*zeta1^3*zeta1^3*exp(-2*zeta1.*(r1)).*exp(-2*zeta1.*(r2)).*r1.*r1.*r2.*r2./max(r1,r2);
dblquad(chi1111,0.000001,10,0.0000001,10,1E-3)
ans =  0.90623

Similarly other integrals can be encoded, and I get results exactly as Levine example.
Thank you everyone


From the abstract to ref 1:

  We investigate both experimentally and theoretically the structure and photodynamics of nitrophenol molecules and clusters, addressing the question how the molecular photodynamics can be controlled by specific inter- and intramolecular interactions. Using quantum chemical calculations, we demonstrate the structural and energetic differences between clusters of 2-nitrophenol and 4-nitrophenol, using phenol as a reference system. The calculated structures are supported by mass spectrometry. The mass spectra of 2-nitrophenol clusters provide an evidence for a stacked structure compared to a strong O–H···O hydrogen bonding for 4-nitrophenol aggregates.

The key words in the context of 2-nitrophenol are "stacked structure". This suggests that Mithoron's comment might be correct. However "stacking" is not evidence of a particular attractive interaction mechanism, it doesn't tell us of the relative importance of classical pi-stacking versus dispersion interactions between 2-nitrophenol molecules.

I'll throw up my hands and claim ignorance, deferring explanation to the Wikipedia:

  In chemistry, pi stacking (also called π–π stacking) refers to attractive, noncovalent interactions between aromatic rings, since they contain pi bonds. These interactions are important in nucleobase stacking within DNA and RNA molecules, protein folding, template-directed synthesis, materials science, and molecular recognition, although new research suggests that pi stacking may not be operative in some of these applications. Despite intense experimental and theoretical interest, there is no unified description of the factors that contribute to pi stacking interactions.

References

Photochemistry of Nitrophenol Molecules and Clusters: Intra- vs Intermolecular Hydrogen Bond Dynamics. Kateryna Grygoryeva, Jakub Kubečka, Andriy Pysanenko, Jozef Lengyel, Petr Slavíček, and Michal Fárník. The Journal of Physical Chemistry A 2016 120 (24), 4139-4146. DOI: 10.1021/acs.jpca.6b04459


Yes, you can form a nitroamine (nitramine) from a secondary amine and nitric/sulfuric acids. Some explosives have a nitramine structure (e.g. tetryl, RDX, HMX, HNIW).


What Causes Atomic Orbitals to Hybridise?

Hybridization is a mathematical concept developed to cope up the flaws of Valence Bond theory. It hinges solely on the linear combination of wavefunctions of atomic orbitals having more or less same energy. Hybridized state is a purely theoretical concept & cannot be detected even spectroscopically.

There is nothing like "collapse." The wavefunction of a hybridized orbital is "mixing" of the atomic orbitals only. The number of hybrid orbitals are same as the number of atomic orbitals participating in the "mixing up". The wavefunctions interfere constructively or destructively to give rise to equivalent orbitals known as hybrid orbitals.

The Methane Case:

Carbon has valence configuration: $2s^2 2p_x ^1  2p_y^1$. So, VB theory predicts that it can form two bonds. This is one of the main defect of VB theory. Then promotion comes to the rescue! One electron from $2s$ gets promoted to unoccupied $2p_z$ orbital, so as to make the configuration: $2s^12p_x^12p_y^12p_z^1$ . So, $\ce{C}$ can form four bonds. But still there is a problem: experiments show that the four bonds in methane are exactly equivalent which is not possible by this configuration since $s$ & $p$ orbitals are not  same. To deal with the problem, concept of hybridization was developed. Hybridization gives rise to four equivalent orbitals but having different orientations (four corners of the tetrahedron.) The non-normalized wavefunctions of four orbitals are as follows: $$\psi_1 = \psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_2 =\psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\\\\\\ \psi_3=\psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_4 = \psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}$$


Carbocation formation in electrophilic addition of hydrogen halides to alkenes

The rate determining step is carbocation formation. After which, depending on conditions, the trapping of the cation may or may not be slow enough for [1,2]-shifts.

The [1,2]-shifts are probably under thermodynamic control if there is time for them to take place. This means that if there's time for the shifts, we're hunting for the most stable cation. Otherwise, you trap the most readily formed cation in a kinetic sense (which is not necessarily the most stable).

Assuming shifts are fast, I think your solution is better because the tertiary cation is more stable. Dichloromethane doesn't undergo substitution as easily as t-butyl chloride, so I would have to argue that the tertiary carbocation is more stable.


What compounds exist in the bitter peel of specific fruits?

The reason behind the bitterness of most fruits is due to presence of Tannin in them.

Tannin is a naturally occurring polyphenol found in plants, seeds, bark, wood, leaves and fruit skins.

Tannin tastes bitter,dry and astringent and you can feel it specifically on the middle of your tongue and the front part of your mouth. Besides the peels of fruit, we can feel bitterness in other food products as well like,

High-Tannin Foods =
 Tea Leaves
Walnuts, Almonds and Nuts with Skins
Dark Chocolate
Cinnamon, Clove and other spices
Pomegranates, Grapes
Quince
Red Beans

The  bitterness and astringency are caused by tannins binding with proteins in our saliva. The chemical constituent of tannins are: gallic acid , ellagic acid and their derivatives(present in hydrolyzable tannins) , catechin and its derivative (present in condensed tannins)

As for your specific question about the the pomegranate fruit pith, it contains ellagitannins, pelargonidin, punicalin, punicalagin, anthocyanins, cyanidin, ellagic acid which makes it bitter.


Would it be possible to create a (non-toxic) alcoholic drink that, when zapped with laser, changes color?

I see a couple of problems here:

Beam diameter vs volume of the beverage

The volume of the sample that is actually hit by a narrow laser beam is very small, as compared to the total volume of the sample (= the drink). As a consequence, only a very small amount of the dye in drink will undergo a transformation. A wide beam of a switchable light source definitely is the better option.

Is is possible in general?

Yes, a number of compound classes undergo transformations upon iradiation and change their colour (spiropyranes, fulgides, etc.). Often, the reverse reaction can be induce thermally.

So what?

Food dyes are strictly regulated, you can't add a photochromic dye to a drink just because the colour change looks nice in the lab. If the dye isn't approved as a food dye, there's no legal way! To my knowledge, food dyes typically do not show photochromism but are photostable. Consumers usually don't like that their food changes the colour in the light.

In summary, the idea sounds funny but I rather doubt that it can be realized.

The only option to add a dramatic effect would be the use of fluorescent and food-compatible ingedients. Think in quinine (tonic water).


No.  Most molecules of air have two atoms.

So if you want to be quantitatively accurate, if you have 100 molecules in a graphic, it would show

78 dinitrogen (N-N) molecules

21 dioxygen (O-O) molecules

1 argon atom (Ar)

http://eesc.columbia.edu/courses/ees/slides/climate/table_1.html

And you could add some water (H-O-H) molecules if you want to depict humidity.


What is the pKa Range for weak acids and bases?

$\mathrm pK_\mathrm a$ is negative log of the acid dissociation constant ($K_\mathrm a$).

$$\mathrm pK_\mathrm a = -\log K_\mathrm a$$

Acid dissociation constant is the equilibrium constant of the dissociation of ions of an acid in an aqueous solution. Consider a weak acid $\ce{HA}$. Weak acids do not dissociate completely in aqueous solution. The equilibrium for the dissociation of such acids can be expressed as

$$ \ce{HA + H2O <=> H3O+ + A-}$$

The equilibrium constant for this reaction(dissociation) will be

$$K_\mathrm a = \frac{\ce{[H3O+][A^{-}]}}{\ce{[HA]}}$$

It can be seen that the greater is the extent of dissociation, the greater will be the value of $K_\mathrm a$. Or, the stronger is the acid, the greater will be its $K_\mathrm a$.

Since, $\mathrm pK_\mathrm a$ is negative log of $K_\mathrm a$, it's values will be greater for weaker acid.

There is no sharp boundary between weak and strong acid. Wikipedia defines strong acids as acids which ionize completely in aqueous solution. So, an acid, say $\ce{HA}$ , is said to be strong if one mole of this acid dissociates in aqueous solution to give one mole of $\ce{H+}$ and one mole of $\ce{A-}$. Now $K_\mathrm a$ of such an acid will be 1.

Usually, $\mathrm{pH}$ is used to measure the acid strength, which is negative log of $\ce{H+}$ ion concentration. $$\mathrm{pH} = -\log\ce{[H+]}$$

The concentration of $\ce{H+}$ ions in water is $10^{-7}$ (It has been found out experimentally). Hence, its $\mathrm{pH}$ is $-\log(10^{-7}) = 7$.
For acids stronger than $\ce{H2O}$, the concentration of $\ce{H+}$ ions, $\ce{[H+]} > 10^{-7}$. So, $\mathrm{pH} < 7$ for acids and $\mathrm{pH} > 7$ for bases.

What you are referring to may actually be $\mathrm{pH}$ and in that case the ranges you've given are may be correct. But once again there is no clear difference between weak and strong.

There is a simple relation between $\mathrm{pH}$ and $\mathrm pK_\mathrm a$,
$$\mathrm{pH} = \frac{1}{2}[\mathrm pK_\mathrm a - \log c]$$
where $c$ is concentration of the acid. You can derive this relation using Ostwald's dilution law.

Similarly, we have $K_\mathrm b$, $\mathrm pK_\mathrm b$ and $\mathrm{pOH}$ for bases which are basic analogues of $K_\mathrm a$, $\mathrm pK_\mathrm a$ and $\mathrm{pH}$. These variables show the strength of a base.

Conclusion

The greater is the value of $K_\mathrm a$, the stronger will be the acid and the weaker will be the base.
The greater is the value of $\mathrm pK_\mathrm a$, the weaker will be the acid and the stronger will be the base.
The greater is the value of $\mathrm{pH}$, the weaker will be the acid and the stronger will be the base. For acids, $\mathrm{pH} < 7$ and for bases, $\mathrm{pH} > 7$.
The greater is the value of $K_\mathrm b$, the stronger will be the base and the weaker will be the acid.
The greater is the value of $\mathrm pK_\mathrm b$, the weaker will be the base and the stronger will be the acid.
The greater is the value of $\mathrm{pOH}$, the weaker will be the base and the stronger will be the acid. For bases, $\mathrm{pOH} < 7$ and for acids, $\mathrm{pOH} > 7$.


From what I know, you can add the solution of unknown to the solution of known concentration. The only thing that matters is you have a known volume of the solution to which you're adding the other solution.

For example, suppose we're titrating $\pu{0.15 M}$ $\ce{Na2CO3}$ with an unknown concentration of $\ce{HCl}$, if we need to add $\pu{say 15 mL}$ $\ce{Na2CO3}$ to a known volume of say $\pu{30 mL}$ of $\ce{HCl}$ to reach the endpoint, we can calculate the concentration of the HCl solution.

$$\ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2}$$

$n(\ce{Na2CO3}) = \pu{0.15 mol/L} \cdot \pu{0.015 L}$

$n(\ce{Na2CO3}) = \pu{0.00225 mol}$ (i.e. the known concentration times 15 mL change to L)

We know that 2 moles of $\ce{HCl}$ will react with 1 mole of $\ce{Na2CO3}$ to form the products on the right hand side of the equation. So,

$n(\ce{HCl}) = 2 \cdot n(\ce{Na2CO3})$

which we calculated above using the relation of moles as product of concentration times volume. So

$n(\ce{HCl}) = \pu{0.0045 mol}$

Since we know that we had $\pu{30 mL}$ of a $\ce{HCl}$ solution with $\pu{0.0045 mol}$ of $\ce{HCl}$ in it,

$[\ce{HCl}] = \frac{\pu{0.0045 mol}}{\pu{0.03 L}}$

$[\ce{HCl}] = \pu{0.15 mol/L}$

So, we had a $\pu{0.15 M}$ solution of $\ce{HCl}$.

If we're now adding $\ce{HCl}$ to $\pu{15 mL}$ $\ce{Na2CO3}$ and it takes $\pu{30 mL}$ of $\ce{HCl}$ to reach the endpoint, we can use similar calculations to once again find the concentration of $\ce{HCl}$ (and get the same answer).

It doesn't matter whether you add the known solution to the unknown one, the only thing you need to know is the volume of the thing you're adding to and how much of the other one you need to add to reach the endpoint.
Think of it like this,
We have 4 variables so to speak;
- concentration of the known solution (we know)

concentration of the unknown (what we want to work out)
volume of the known solution (either adding or added to) which we measure
volume of the unknown solution (either adding or added to)which we also measure

So we know 3 things no matter what we add what to and these 3 values are enough to work out the concentration of the unknown solution. You should clarify with your teacher what she meant. She may have said always add the acid to the base because if we place the base in the burette it may react with the glass, $\ce{SiO2}$ which acts as an acid and that might degrade up the precision for future use.


How can I generally determine which bonds are longest/shortest in a molecule?

Because atomic radii of elements tend to decrease across a period (source), and bond length increases with atomic radius, the bond lengths are expected to decrease in the following order: C (C-C bond) > B, D (C-N) > A (C-F).

The difference between B and D is that in D, the nitrogen is $sp^3$-hybridized, while in B, it is $sp^2$-hybridized due to conjugation of its lone pair with the aromatic ring 2. This results in partial double bond character of the C(ring 2)-N bond, similar to aniline:

More $s$ character leads to shorter bonds, and therefore the C-N bond B is shorter than D.

In summary, the bond lengths decrease in the following order: C > D > B > A.


Let us suppose that your assumption is correct and nitrogen has already $\ce{2p^3}$ configuration. Let's further assume that it doesn't undergo hybridisation. Instead, the pure atomic orbitals undergo overlap with 3 $\ce{Cl}$ atoms and the filled $\ce{2s}$ shell electron pair acts as a lone pair and thus $\ce{NCl3}$ is formed. This consideration creates problems.

If pure atomic orbitals of nitrogen, i.e. $\ce{2p_x}$, $\ce{2p_y}$ and $\ce{2p_z}$ undergo σ-bond formation with the chlorine atoms, then the bonds should be $90^\circ$ to each other, as $\ce{p_x}$, $\ce{p_y}$ and $\ce{p_z}$ orbitals are mutually perpendicular. But $\ce{NCl3}$ has 3 σ-bonds with bond angles between $\ce{Cl-N-Cl}$ close to $107.1^\circ$, which is close to the tetrahedral bond angle of $109.5^\circ$. Therefore the pure orbital undergoing bond formation doesn't support this observation.

If we consider $\ce{sp^3}$ hybrid orbitals of $\ce{NCl3}$, this tetrahedral arrangement of three σ-bonds and a lone pair (disregarding  the lone pair, it's pyramidal) is nicely explained. That's why, the hybridisation is $\ce{sp^3}$ which supports every observed chemical property of $\ce{NCl3}$.


Is there a method for finding the net-ionic equation?

But apparently the correct molecular equation is $\ce{6KCN\+AlBr3 \to 3KBr\+Al(CN)6}$

That's not correct, what happenend to the other 3 potassium atoms?

It could form $\ce{Al2(CN)6}$ but you can't exclude $\ce{Al(CN)3}$ without further information beyond balancing of equations.

There are two main steps to balancing equation:

Balancing of each type of atom
Balancing number of electrons

If you've done both of those correctly, you've done all you can with balancing alone.

  $\ce{Cl2\+2NaOH \to 2NaCl\+Cl2O\+H2}$

Above is clearly wrong because number of oxygen atoms doesn't balance

  $\ce{Cl2\+2NaOH \to 2NaOCl\+NaCl\+H2O}$

Above is clearly wrong because number of oxygen atoms doesn't balance


There are 2 methyl groups on the second carbon of the main chain, and 1 methyl group on the fourth carbon of the main chain, so 2,2,4-trimethylpentane is correct! If there was a methyl group on the 5th carbon of a pentane, it would be a hexane, so it would not make sense to name it the way your book did.


The picture shows parts of an arsine generator.

The apparatus can be used to determine trace amounts of arsenic by the silver diethyldithiocarbamate photometric method.

A typical method description can be found here (this is also the source of the drawing):

https://hmc.usp.org/sites/default/files/documents/HMC/GCs-Pdfs/c211.pdf

or

http://hmc.usp.org/sites/default/files/documents/HMC/GCs-Pdfs/GC_pdf_USP38/c211.pdf


Even though your question get downvoted, I think it is an legitimate question for people to understand the trend. I believe I-VIII Roman numeral nomenclature belong to CAS system while 1-18 Arabic numeral groups are recommended by IUPAC (Ref.1). The IUPAC recommended Periodic Table is given below (ignore the Roman numerals, which I put in there to explain):

According to Ref.1, the confusion of Periodic Table of Elements has been addressed as follows:

H.G. Deming used the long periodic table in his textbook General Chemistry (See following diagram with Roman numerals only; Ref.2), which appeared in the USA for the first time in 1923, and designated the first two and the last five Main Groups with the notation "A", and the intervening Transition Groups with the notation "B". The numeration was chosen so that the characteristic oxides of the B groups would correspond to those of the A groups. The iron, cobalt, and nickel groups were designated neither A nor B. The Noble Gas Group was originally attached by Ueming to the left side of the periodic table. The group was later switched to the right side and usually labeled as Group VlllA. This version of the periodic table was distributed for many years by the Sargent-Welch Scientific Company, Skokie, Illinois, USA. $[\cdot\cdot\cdot]$
The designations A and B have been extensively and rather arbitrarily used in the meantime in textbooks and in publications. An investigation of the application of the subgroup designations A and B in all articles, which appeared between 1972 and 1981 and covered by Chemical Abstracts, revealed a completely arbitrary use of the designation. Moreover, more than 10% of the articles it was nearly impossible, from the wording of the text, to recognize which elements were being discussed without further information (Ref.3).

The IUPAC Commission on the Nomenclature of Inorganic Chemistry (CNIC) proposed the designations of groups by Arabic numerals in 1984 (Ref.4), which was approved by the Nomenclature Commission of the American Chemical Society (ACS) and rest is history.
Thus, it is safe to say that this nomenclature is now outdated and not worth discussing.

References:

E. Fluck, “New notations in the periodic table,” Pure and Applied Chemistry 1988, 60(3), 431-436 (https://dx.doi.org/10.1351/pac198860030431).
Horace Grove Deming, General Chemistry; J. Wiley & Sons, Inc.: New York, NY, 1923.
W. C. Fernelius, W. H. Powell, “Confusion in the periodic table of the elements,” J. Chem. Educ. 1982, 59(6), 504-504 (https://doi.org/10.1021/ed059p504).
K. L. Loening, “Recommended Format for the Periodic Table of the Elements,” J. Chem. Educ. 1984, 61(2), 136-136 (https://doi.org/10.1021/ed061p136).


What is the meaning of the '2' on some Compressed Gas Hazard signs in the lab?

The 2 refers to DOT (Department of Transportation) hazard class 2. This reference describes the different DOT hazard classes.

As summarized in the comments:

  Class 2 (compressed gas) is divided into three subclasses (2.1 - flammable, 2.2 - non-flammable, non-toxic and 2.3 - toxic). I imagine the diamonds for Class 2 don't contain the subclass number because there is a different diamond colour/symbology for each subclass (as opposed to for instance 5.1 (oxidising agent) and 5.2 (organic peroxide), which have the same symbol and colours).


Can formaldehyde be stored at a higher temperature than room temperature?

Trimerization of formaldehyde can happen at high temperatures, but appear to require catalysts. There is a patent for synthesizing 1,3,5-trioxide at high temperature. This patent uses phosphoric acid catalysts.
According to the patent:

  The process of the invention is advantageously carried out at reaction temperatures of from 80 to 160° C., preferably from 90 to 145° C., at an inlet partial pressure of the formaldehyde of from 0.5 to 5 bar absolute, preferably from 0.5 to 2 bar, and at an inlet concentration of formaldehyde of from 1 to 100% by volume, preferably from 20 to 100% by volume

There is also an early study which explored the kinetics of the reaction but for a different catalyst. In "Vapour-phase trimerization of formaldehyde to trioxane catalysed by 1-vanado-11-molybdophosphoric acid" by Kern and Emig, they use a molybdophosphoric acid catalyst. They used a tube reactor and varied the temperature from
102.5 to 110° C. According to their study:

  The
  conversion
  observed
  increases
  extremely
  with
  decreasing
  temperature.

However, they comment that

  the
  temperature
  dependence
  cannot
  be
  attributed
  to
  thermodynamic
  effects
  only.
  It
  has
  to
  be
  related
  to
  the
  properties
  of
  the
  heteropoly
  compound.

Another study is "Synthesis of Trioxane Using Heteropolyacids as
Catalysts" by Mazamoto et al. Their process uses heteropolyacids as catalysts and they were able to obtain a conversion % of 22.9 at 110° C. Where
$$\text{Conversion %}=\frac{3[\text{trioxane}]+2[\text{methyl formate}]+[\text{methylal}]}{[\text{feed formaldehyde}]}\cdot 100$$


Synthetic route for the conversion of ethanol to but-1-yne

The first reaction is the conversion of the alcohol to an alkyl chloride using thionyl chloride. The mechanism of the reaction starts with nucleophilic attack of the alcohol oxygen on sulphur, followed by displacement of $\ce{Cl-}$. Chloride then substitutes the alkyl chlorosulfite group in an $\ce{S_{N}2}$ reaction, and the latter decomposes into $\ce{SO2}$ and $\ce{HCl}$.

(Image source)

In the second reaction, chloroethane is alkylated with sodium hydrogen acetylide to yield the desired product.

$$\ce{C2H5-Cl + HC#C^{-} ->~ C2H5-C#CH + Cl-}$$

This is a simple $\ce{S_{N}2}$ reaction, in which $\ce{Cl-}$ is substituted by $\ce{HC#C^{-}}$. Acetylide salts of alkali metals can be prepared by deprotonation of ethyne with a strong base, for example, with butyllithium in THF or sodium amide, which can be generated in situ by dissolving sodium metal in liquid ammonia.


Why can't a galvanic cell be a single cell?

Perhaps this isn't the full answer, but it probably plays a role. It is critical for galvanic cells that there not be any electronic conductivity through the cell between the two electrodes. If there is a path for electrons, they will take it which means the cell will spontaneously discharge and not do any useful work through the external circuit. Pure water is a very poor conductor but becomes conductive once ions are dissolved.

By comparison, the construction of a lithium ion cell is close to what you describe for a single chamber galvanic cell. There is a separator involved but mostly to keep the two electrodes from coming into direct contact. One reason this is possible is that the electrolyte is an organic solvent (usually ethylene carbonate, dimethyl carbonate, etc) and doesn't conduct electrons, even when the appropriate salt (eg. $\ce{LiPF6}$) is dissolved in it. There is no salt bridge in this cell, though it is clearly galvanic.

Regarding the electrolytic cell, I don't see any particular reason that you couldn't build one with a salt bridge, provided the salt-bridge let you balance the charges generated on each side. Maybe they're all presented as one beaker just because it's easier to make and there's not really any advantage to a salt bridge in this case.


If I want to sample rate constants of chemical reactions, what the distribution will be appropriate?

Since you have accepted my proposal of using a free-energy calculation, you can use that in a combination with classical Transition State Theory (think of it as an extension to the Arrhenius equation).

More practically, you can read-up, for example on

GROAMCS
NAMD

For more options, see this overview presentation.


Which functional group is more polar in a drug: primary amide or a carboxylic acid?

It depends on your exact definition of polarity
The measured dipole moments of formamide (3.7 D)1 and formic acid (1.4 D)2 in the gas phase indicate that the amide is intrinsically more polar.

However, this neglects additional factors relevant to the discussion: ionization and solvation. Under physiologic conditions, carboxylic acids are generally deprotonated resulting in a formal charge separation with a -1 charge delocalized across two oxygens. Depending on what parameters you are concerned with, this may be a more profound effect than the difference in gas phase polarity.


Let's look at following table from a Yale website:
$$
\begin{array}{c|ccc}
 & \ce{(CH2)3} & \ce{(CH2)4} & \ce{(CH2)5} & \ce{(CH2)6} & \ce{(CH2)7} \\
\hline
\Delta H_\mathrm{Combusion} \ (\pu{kcal/mol}) & -499.8 & -656 & -793.5 & -944.6 & -1108.3  \\
\Delta H_\mathrm{Combusion} \text{ per } \ce{-CH2}-  \ (\pu{kcal/mol}) & -166.6 & -164 & -158.7 & -157.4 & -158.3 \\
\text{Ring strain per } \ce{-CH2}-  \ (\pu{kcal/mol}) & 9.2 & 6.6 & 1.3 & 0 & 0.9 \\
\text{Total ring strain}  \ (\pu{kcal/mol}) & 27.6 & 26.3 & 6.5 & 0 & 6.3 \\
\hline
\end{array}
$$
Now, just concentrate on ring strain values for cyclopropane and cyclobutane only:

Ring strain per $\ce{-CH2}-$ (X-values) are $9.2$ and $\pu{6.6 kcal/mol}$, respectively.
Total ring strains (Y-values) are $27.6$ and $\pu{26.3 kcal/mol}$, respectively.

According to Y-values, cyclopropane and cyclobutane are equally unstable (relatively speaking), while according to Y-values, cyclopropane is significantly less stable than cyclobutane. Which valuse should be more reliable? Let's look at experimental findings:

Although cyclopropanes are far less reactive than alkenes, they react with chlorine and bromine under polar conditions to form corresponding addition products, 1,3-dihalopropanes, as the predominant product (Ref.1). To my understanding, cyclobutane and higher cycloalkanes do not give this ring opening reaction.
Cyclopropane also reacts with HCl and HBr to form corresponding addition products, 1-halopropanes (Ref.2; Reactions of bromocyclopropane with hydrobromic acid has given mixture of 1,1-, 1,2- and 1,3-dibromopropanes). Again, to my understanding, cyclobutane and higher cycloalkanes do not give this ring opening reaction.

So which one you think should be more reliable and why?
References:

Joseph B. Lambert, Erik C. Chelius, William J. Schulz, Jr., Nancy E. Carpenter, “Polar bromination and chlorination of cyclopropane,”  J. Am. Chem. Soc. 1990, 112(8), 3156–3162. (https://doi.org/10.1021/ja00164a043).
Choi Chuck Lee, Bo-Sup Hahn, Kwok-Ming Wan, D. J. Woodcock, “Reactions of cyclopropane with hydrochloric acid and of bromocyclopropane with hydrobromic acid,” J. Org. Chem. 1969, 34(10), 3210–3211. (https://doi.org/10.1021/jo01262a097).


A glass electrode is usually stored in a 3 or 4M KCl solution, depending on whatever the manufacturer advices you to do.

But the pH that you measure from this solution is absolutely not the pH. It's the measured potential of this concentrated KCl solution that is (in this case wrongly) converted into the pH scale.

The normal pH measurements with glass electrodes cannot be done without the calibration of your device through references solutions, which are common buffer solutions with acidic, neutral and basic pH values like, e.g., pH 4, 7, 9.

In order to calibrate the pH electrode, you or actually your device saves the measured potential for each of these known buffer solutions and calculates a regression function that should give you something near: $$E = 414.12 - 59.16~pH~~~~~(\text{at}~25~^\circ C)$$

(The time when I had to do this in practical courses gave us $E = 401.93 - 56.711~pH$ ... I have no idea what the temperature was :D )

For further informations on all this you can read Mettler-Toledo's pH guide.

Now, as your KCl solution has a certain potential the device converts it through this regression equation into a pH value. But as I said, that is wrong - I guess that pH indicators like the universal indicator would show no big difference from pH 7 (I've never done this actually, so this is seriously a guess).

Your actual question should rather be about what the potential of a 3M or 4M KCl solution should be. And that, as can be found here, is:


Can a reaction have carbocation resonance followed by a hydride shift?

Substituted cyclopentadienes are an unique molecules, which are not stable at room temperature (recall dimerization). For example, 1-methylcyclopent-1,3-diene (1) or 2-methylcyclopent-1,3-diene (2) or 5-methylcyclopent-1,3-diene (3; diene in the question) would rearranges to a mixture of all 3 positional isomers if leave at room temperature for 1-2 days by hydrogen migration (Ref.1 & 2). The abstract of Ref.1 states that:

  A number of substituted cyclopentadienes display characteristic properties which are representate of all members in the series: (a) the endocyclic double bond isomers are in the thermodynamic equilibrium; the process of establishing the equilibrium being retarded; (b) the transformation of isomeric cyclopentadienes proceeds by steps, i.e. a proton from the methylene group (5th position) passes mainly to the adjacent C-atom (1st position). The proton-cyclopentadienyl anions are probably transitional states in the transformation of isomers. It is assumed that this transformation is a first order reaction. As a result of this investigation it has been shown that the structure of the substituted cyclopentadiene (or the composition of the isomer mixtures) depends mainly on the character of the substituent and not on the method of synthesis.

The ratio of 1 : 2 : 3 would be 45:54:1. Kinetically, the process is first order and activation energy is $\pu{20.4 kcal/mol}$ over temperature range of $\pu{298-313 K}$ (Ref.3). The kinetic studies by NMR has been suggested that rearrangement is due to 1,2-hydrogen shift (Ref.4).

Yet, these cyclopentadienes are also suseptible to protonation by strong acids (Ref.5). Thus, it is very possible that protonation at $\ce{C1}$ 1,5-addition after 1,2-hydride migration. However, as explained above, 5-methylcyclopent-1,3-diene (3) is relatively the least stable positional isomer. At room tempeature, it would stay mostly as 1-methylcyclopent-1,3-diene (1) and 2-methylcyclopent-1,3-diene (2) with $\approx 1:1$ ratio. Therefore, addition would be either 1,4- and 1,2-.

References:

V. A. Mironov, E. V. Sobolev, A. N. Elizarova, "Some general characteristic properties of substituted cyclopentadienes," Tetrahedron 1963, 19(12), 1939-1958 (https://doi.org/10.1016/0040-4020(63)85008-5).
Sigmund M. Csicsery, "Methylcyclopentadiene Isomers," J. Org. Chem. 1960, 25(4), 518-521 (https://doi.org/10.1021/jo01074a009).
Raymond K. Crandall, “Investigation of the electronic and thermal rearrangement properties of spiro[2,4]hepta-4,6-dienes,” M.S. Thesis, Rochester Institute of Technology, Rochester, NY, 1982.
S. McLean, P.Haynes, "Hydrogen migration in cyclopentadienes," Tetrahedron 1965, 21(9), 2329-2342 (https://doi.org/10.1016/S0040-4020(01)93887-6).
Ronald F. Childs, Maung Zeya, “Diprotonation of 5-Acyl-1,2,3,4,5-pentamethylcyclopentadienes,” Can. J. Chem. 1975, 53(22), 3425-3430 (https://doi.org/10.1139/v75-490)


Why do lithium and sodium corrode so easily?

$$\ce{2Li + 2H2O -> 2LiOH + H2}$$
$$\ce{2LiOH + CO2 -> Li2CO3 + H2O}$$
$$\ce{4Li + O2 -> 2Li2O}$$
$$\ce{Li2O + CO2 -> Li2CO3}$$
$$\ce{6Li + N2 -> 2Li3N}$$
Lithium and sodium are reacting with the gases in the air; a destructive physical process.


acid alpha glucosidase vs glycogen phosphorylase

Acid alpha glucosidase  (sometimes called α-1,4-glucosidase or acid maltase) is not among those enzymes involved the pathways of glycogen synthesis and breakdown. It however occurs in lysosomes,where it functions to hydrolyze the disaccharide maltose  and linear oligosaccharides, as well as the outer branches of glycogen,thereby yielding free glucose.

This "second pathway" of glycogen metabolism is not quantitatively important as it concerns a minor fraction of glycogen. The reason that lysosomes normally take up and degrade glycogen granules is unknown perhaps for the disposal of structurally aberrant glycogen molecules.

Glycogen phosphorylase (a dimeric enzyme)
catalyzes glycogen phosphorolysis (bond cleavage by the
substitution of a phosphate group) to yield glucose-1-
phosphate (G1P).

$\ce{Glycogen + Pi <=>  glycogen + G1P}$

This enzyme will only release a glucose unit that is at least
five units from a branch point. (Note: I am not going to describe the reaction mechanism of glycogen phosphorylase here is much more complex than it sounds)

Having briefly described the enzymes, the most important thing to note is that:

structually these two enzymes are different,
active in the different locations
the mechanism of glycogen breakdown are different (i.e phosphorylation, hydrolysis)
different regulation mechanisms


after revisiting the problem I realised where I went wrong. The main mistake with not being careful with units and not reading the qurstion and having incorrect values for hole mobility.
$$\text{mass}_{\ce{Fe(III)}}=((0.24)(55.85))/((N_\mathrm{a})(10^3))$$
$$=2.225838592\times10^{-26}\: \mathrm{kg}$$
$$\rho_{\ce{Fe(III)}}=2.225838592\times10^{-26}/(4.37\times10^{-10})^3$$
$$=\mathrm{266.7\: kg\: m^{-3}}$$
$$\text{No. of holes}=\rho_{\ce{Fe(III)}}/1\:\text{AMU}_\ce{{Fe(III)}}$$
$$\Rightarrow \text{No. of holes}=266.7/(9.296\times10^{-26})$$
$$=2.868975904\times\pu{10^{27}\:holes/m^3}$$
Conductance for a p-type semiconductor is given as: $$\sigma=p(q)(\mu_{h})$$
$$\implies\sigma=(2.868975904\times10^{27})(1.6\times10^{-19})(3.1\times10^{-5})$$
$$\approx1.4\times10^{4}\:(\Omega.m)^{-1}$$


How many coordinates should be used for the potential energy surface of a non-rotating linear molecule? (3N-6) or (3N-5)?

The Cartesian Hessian ($\mathbf{H}_C$) can be obtained by from the Hessian computed in internal coordinates ($\mathbf{H}_i$) through a non-linear transformation
$$\mathbf{H}_C = \mathbf{B}^T\mathbf{H}_i\mathbf{B}$$
where $B_{ij}=\frac{\partial R_i}{x_j}$ and $R_i$ and $x_i$ is an internal and Cartesian coordinate, respectively.

In the case of the degenerate bending mode you can obtain the components of $\mathbf{H}_C$ that you need from $\frac{\partial^2 E}{\partial \theta^2}$, where $\theta$ is the bond angle, through suitable choices of $B_{ij}$'s.

So the Hessian generated in the three internal coordinates has all the information you need.


Which two substances react spontaneously?

From the faint outline of electrochemical series I remember $Cu$ is above $Ag$. The substance which is above is able to displace the substance below from its aqueous solution(obviously lower substance is in ionic form).

Hence $(B)$ is true.

We were made to learn this outline in lower class :

K Na Ca Mg Al Zn Fe Pb H Cu Hg Ag Au Pt