Simple answer is consider the stability of carbocations. You already got the tertiary carbocation in hand that is much stable than secondary carbocation, which would be resulted by ring expantion. Sure, a methide migration would give you another tertiary carbocation, but before you get that, the reaction progress has to climb another actvation barrier, which would be steep ($3^\circ \rightarrow 2^\circ$). Bottom line is rate of clorination of carbocation is faster than rate of this rearrangement followed by methide migration ($E_\mathrm{a2} \lt E_\mathrm{a3}$):


Confusion on enantiomers and meso compounds

You are right, the 2 structures are equivalent and the substance has a plane of symmetry and is a meso compound. When in doubt, make a 3D model if possible.


Does the water in hydrates affect final concentration?

Firstly, copper would have to be in the 4+ oxidation state for there to be $\ce{Cu(SO4)2.5H2O}$, so this isn't a realistic question.

But, no it won't be 5M, because you need to know the final volume of the solution to calculate molarity.  If you add 345.55 grams of just about anything I can think of to 200 mL of water, the volume will increase substantially.  So not only will the 90 grams of hydrate water add to the volume of the final solution, generally you need to consider how all added substances contribute the final volume.

In some cases, such as magnisium sulfate, a small amount added to water can actually decrease the volume.


So long as we assume that this is an ideal situation in which there is really insignificant solubility of the gas in ethanol but significant solubility in water, then yes, you can assume that the concentration in the mixture is linear with the water fraction of the liquid. In other words:

$${H_m = H_w * x_{w}}$$

Where:
${H_m}$ = Henry's law solubility of the gas in the water-ethanol mixture
${H_w}$ = Henry's law solubility of the gas in water
${x_{w}}$ = mol fraction of water in the liquid phase


Why does symmetry have to be maintained in molecular orbitals?

The orbital and geometrical symmetry are closely related. You know that $\ce{XeF_4}$ is square planar, therefore $\ce{D_{4h}}$ symmetric. That also means that the four fluorine atoms are indistinguishable.

So if you perform some manipulation of the molecule, e.g. rotation by $\ce{90^{\circ}}$, you must end up with the same picture (or just with opposite sign), therefore ruling out our suggestion of 3:1 different orbitals. This holds for all orbitals, not only bonding.

One could possibly argue, what is the physical meaning of the unoccupied orbitals, but this is out of scope of the question.


Why does iron have an abnormally high ionization energy?

"I mean "abnormal" in that it defines a general trend" makes no sense to me at all. Did you mean "defies"?  Take a look at the chart here: https://en.wikipedia.org/wiki/Ionization_energy.
By my count, for the s-block elements the "trend" always holds.  For the p-block elements, it holds in 5/6ths of the cases. And for the d-block, I can't tell from the chart, but claiming it is increasing for them seems agross exaggeration to me; I'd say they might have a very slight tendency to increase, but they're so close to being flat-lined, that it seems to be not a useful claim. If you count the exceptions to the rule, and you expect your "rules" to hold at least 90% to 95% of the time (this isn't the Social Sciences, after all), then it is a just plain lousy rule. Nothing unusual about that: I think the Periodic Table is best when used to "predict" Atomic Number (they increase moving right and down, without exception), and so untrustworthy with the other periodic trends as to not be something in my "mental toolbox". Use it at your own peril. (The at. no. "trend" was a joke, of course)


Why does the Gibbs free energy only correspond to non-expansion work?

The association of the Gibbs free energy with “additional”, or “non-expansion” work, is simply a mathematical result of its definition: $G = H - TS = U + pV - TS$.

From the First Law, for a closed system, we have $\mathrm{d}U = \delta q + \delta w$.

For a reversible change, the Second Law tells us that $\mathrm{d}S = \delta q_\text{rev}/T$; so $\delta q_\text{rev} = T\,\mathrm{d}S$.

We can split $\mathrm{d}w$ into two parts, expansion work and additional work: expansion work is given by the formula $\mathrm{d}w_{\text{exp}} = -p\,\mathrm{d}V$, and for now we can just label additional work as $\mathrm{d}w_{\text{add}}$. In this context, additional simply means anything that isn’t expansion work. Putting all this together, we have

$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}}$$

and from the definition of the Gibbs free energy in the first paragraph,

$$\begin{align}
\mathrm{d}G &= \mathrm{d}(U + pV - TS)\\
&= \mathrm{d}U + (p\,\mathrm{d}V + V\,\mathrm{d}p) - (T\,\mathrm{d}S + S\,\mathrm{d}T)\\
&= T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}} + p\,\mathrm{d}V + V\,\mathrm{d}p - T\,\mathrm{d}S - S\,\mathrm{d}T\\
&= V\,\mathrm{d}p - S\,\mathrm{d}T + \delta w_{\text{add}}
\end{align}$$

Now, if a closed system is kept at constant pressure and temperature, $\mathrm{d}p = 0$ and $\mathrm{d}T = 0$. Therefore $\mathrm{d}G = \delta w_{\text{add}}$.

The simplest application of this interpretation of $G$ is in electrochemistry, where one could move $|\mathrm{d}n|$ moles of electrons through a certain potential difference $E$; the “additional work” done in this case is given by $-FE\,|\mathrm{d}n|$ (you could consult a physics text for details). This gives us $\mathrm{d}G = -FE\,|\mathrm{d}n|$, and (after several more steps) the familiar equation $\Delta_\mathrm{r} G = -zFE$.


When we talk about a molecule being linear or planar, we only consider the positions of the nuclei and ignore the electron orbitals. Sure, π-bonds might be out of plane, but that does not count. If it would be otherwise, we would have to consider that any atom except hydrogen has p-orbitals, and some of those are out of plane, and so no molecule ever would have the right to be called planar.

So yes, $\ce{C3O2}$ is kinda like an allene, with all its carbons having sp-hybrid orbitals, which inevitably makes the molecule linear and hence planar.

One might point out that the molecule is relatively easy to bend, but that's another story.


Mono-addition of HCl to unsymmetrical dienes?

Do we agree that the first step ist the addition of a proton to the diene? This is bit different from the addition of HCl to an alkene, where you would rely on the Markovnikov rule.

Here, stabilization of the cation through resonance comes into play.

  I don't know how to draw it [...]

Draw stepwise!

In principle, a proton could add to one of the positions a to d.

How do the resulting cations look like and which of them are stabilized by resonance?

Is it now a bit more clear?


Priority order of attack of grignard reagents on functional groups

For the compound given in the question, the priority order of Grignard reagent attack follows,
$\ce{R-OH > R-CO-R' > R-COOR'}$
Reason:

Acid-base neutralisation are very fast.

Carbonyl carbon is more nucleophilic than carboxylic carbon, as the latter one is stabilized by resonance.

Therefore, first alcohol is attacked and hence P is (a), then the carbonyl part is attacked and hence Q is (b).
Finally, one more mole of RMgX gets used up on the carboxyl part, hence R is not (d).


Why are solid phase organics often conjugated to HCl?

Does it aid crystallization?

Yes.

All of the drugs that you see as "hydrochlorides" are alkaloids - a class of basic natural products containing one or more amine functional groups. The dimethylamino group in the upper right of the structure of diphenhydramine (below) is a prime example.

Many of these compounds are not crystalline (at best amphorous, at worst waxy or liquid). These compounds are expensive to purify, since the inexpensive method of recrystallization is not an option. Chromatography is an expensive and wasteful process on the industrial scale.

Reacting alkaloids with HCl (or other acids - for example dextromethorphan HBr) usually produces a crystalline compound. These crystalline compounds are easier to purify.

There is a second more important benefit associated with the hydrochloride salts - they have higher bioavailability than the freebases. The freebase is nonpolar and poorly soluble in water. The hydrochloride salt has an ionic group, which increases water solubility.


According to collision theory, for a chemical reaction to occur, the reacting molecules must collide with each other.

For a bimolecular reaction (as given by you), collision occurs in the single step and thus becomes the rate determining step, as there are no other steps involved.

If you consider reactions like bromide ions reacting with bromate ions in the presence of acid, it involves 12 reacting particles. If all the 12 reacting particles collide simultaneously you can write rate law expression considering it to be elementary, but those events are extremely rare. In reality such reactions are found to be complex i.e reactions which take place through a sequence of two or more consecutive steps. Here, there are many steps involved, and one should consider slowest step to write rate law expression. But, why? That will become another SE question. In case of reactions taking place in single step, as there are no other steps left to compare with, whether it is slower or not, so you need to consider  the theoritical equation.

One more thing to be noted is that elementary reactions are associated with single transition state and not single activated complex. For more info read this page: Activated complex


Why trans-1-chloro-4-methylcyclohexane have a higher relative energy than trans-1-chloro-2-methylcyclohexane?

trans-1-Chloro-4-methylcyclohexane (1) has two chair conformations, 1-ee and 1-aa. Conformation 1-ee has two non-interacting functional groups, chlorine and methyl, both of which have anti-relationships with the $\ce{C2-C3 and C5-C6}$ bonds of the ring. Accordingly, the energy of this conformation is set at 0 kJ/mol. There is no need to try to set some absolute value to the energy because the only issue is how 1-ee compares with conformation 1-aa. Conformation 1-aa has both substituents axial in gauche-relationships with the aforementioned C-C bonds. This conformation behaves as a composite of axial methylcyclohexane and chlorocyclohexane. Its energy is merely the sum of their respective A-values, 9.7 kJ/mol.
Diaxial conformation 3-aa of trans-1-chloro-2-methylcyclohexane has the same energy value of 9.7 kJ/mol because both groups are axial and non-interacting with each other. However, the diequatorial conformation 3-ee has a gauche interaction between the chlorine and the methyl group. The magnitude of this interaction may be estimated.
The value of the $\ce{CH3/CH3}$ interaction in conformation 4-ee of trans-1,2-dimethylcyclohexane may be estimated by dividing the energy of 4-aa by 4 or by halving the A-value for a methyl group. The value of 3.8 kJ/mol is in accord with the difference in the heat of combustion of dimethylcyclohexanes 2 and 4. Indeed, the heat of combustion of 4 is greater than 2 by ~5 kJ/mol.  [https://nvlpubs.nist.gov/nistpubs/jres/39/jresv39n1p49_A1b.pdf
]. Similarly, division of the energy of 3-aa by 4 gives an estimated value for the $\ce{CH3/Cl}$ gauche interaction of 2.4 kJ/mol. The heat of combustion of 1,2-isomer 3 should be slightly greater than that of 1,4-isomer 1 by ~2.4 kJ/mol. In the event, trans-1-chloro-4-methylcyclohexane is more stable than trans-1-chloro-2-methylcyclohexane by ~2.4 kJ/mol. By more stable is meant, more negative heat of formation and smaller heat of combustion.


Is the ionic solution of a half-cell part of the electrode in a electrochemical cell?

Your question is about definitions.

I haven't read the Collins book, but, in general, the electrode is the metallic conductor that connects a circuit to the "non-metallic" part of the circuit. The electrolyte is this non-metallic part - dissociated ions that conduct electricity. Note electrolyte has broader uses than electrochemistry.

The term electrode can be used in reference to the whole half-cell since it's the site at which the reaction is taking place. At other parts, all we have is storage of reactant and product for the reversible reaction.
Consider the Standard Hydrogen Electrode(SHE), the whole half cell is referred to as an "electrode", though this is more convention than technical.

-

I suspect the Collins book is for UK A-levels; they should use the clear general definitions;
the electrode is the solid metal connecting circuit to non-metallic part, & the electrolyte is the solution of ions:

Here are some additional brief notes on contextual half cell setups:

SolidMetal/MetalSolution half-cell: Metal Electrode in metal ion solution.
All AqueousIons half-cell: Platinum electrode in solution.
Gaseous/AqueousIons half-cell: Glass/Platinum Deliver tube (for gas) bubble into solution.

The salt bridge is an electrolyte; A-level questions on electrode, electrolyte and salt bridge definitions aren't asked too often, consult your spec & exam papers. Importantly, solutions need to be $1dm^3$,  at $298K$ and $100kPa$, & salt bridge is typically $KNO_3$


How to rationalise that when we dilute a weak acid, the pH increases but also the dissociation increases?

It's simply due to dilution.

If it's true that adding water increases dissociation (Gerard has done a nice effort in Does the number of H+ ions in solution go up on dilution of a weak acid?), concentration of $\ce{H3O+}$ inevitably decreases. As a consequence, $\mathrm{pH}$, defined as $-\log[\ce{H3O+}]$, increases.

Start with $\pu{1 L}$ acetic acid concentration $\pu{1 M}$,
$$\ce{[H3O+]} = \sqrt{K_c} = \pu{4.22E-3M},$$
that corresponds to $\pu{4.22E-3mol}$ of $\ce{H3O+}$.
If you double the volume ($\pu{2 L}$),
$$\ce{[H3O+]}=\sqrt{K_c/2}=\pu{2.98E-3M},$$
that corresponds to $\pu{5.96E-3mol}$ of $\ce{H3O+}$.

The amount of substance is higher, but concentration is less: dilution does play a role here.


Note that -Cl and -CH(CH3)2 are on equatorial positions(The slant lines suggest that)

And the -H is on axial position(The perfectly vertical line suggest that)

Note that it is not just a coincidence, rather it is intentional. I think it would suffice further.


Finding a substance which has a continuous gradient in refractive index

It's not entirely true that no such substance exists, though you won't find this property in a homogeneous substance.

It's well known that changing the concentration of solute in a solution can change the index of refraction; a concentration gradient would then provide a changing refractive index, though having this mimic a square root function would be difficult. Typically, the refractive index changes linearly with the concentration of solute, so a nonlinear function may be difficult to achieve.
A similar approach could be applied to solids, forming a solid solution, though again there would be some difficulty in finding two compounds with similar lattice parameters that would allow to a continuous gradient between compositions to product the refractive index gradient.

Your best bet may be with polymers, where changing proportions of the monomers used could change the refractive index; organic dopants can also be used to effect* this gradient in polymers (see, e.g., Polym. Adv. Technol. 2014, 25, 1099). Again, the difficult part would be matching a square-root function, but you could at least have a varying gradient.

*https://xkcd.com/326/


Why are electrons shared equally when calculating formal charge, but unequally when calculating oxidation state?

The oxidation state is a form to distribute the charge of atoms in molecules considering to have only ionic bonds. In contrast, the formal charge is the distribution of the electrons considering to have only 100% electron-pair covalent bonds.

Actually, the formal charge is not very realistic and is more or less a bookkeeping tool. Now there exists theoretical and computational methods to calculate more reasonably charges on atoms from electronic structure calculations like Hartree-Fock, post-HF, semi-empirical and density-functional methods.

One example, taken from Anslyn's Modern Physical Organic Chemistry, is the case of the tetramethylammonium. One would place a positive formal charge on the nitrogen atom, but this is more electronegative than carbon, so there is a contradiction. The calculations mentioned before, indicate that the nitrogen is essentially neutral and the positive charge resides on the methyl groups. So the formal charge is more adequate for indicate the charge on the molecule and not on the atoms.


According to the principle of Microscopic Reversibility all elementary reactions, i.e. those that proceed in a single step, are reversible.
And you are right: For a reversible reaction $\Delta G$ is very small while for an irreversible reaction $\Delta G$ is large. This can be most easily described using a simple schematic energy curve diagram for your situation.

Each reaction is microscopically reversible. So either your reactants $\ce{A}$ and $\ce{B}$ or your products $\ce{C}$ and $\ce{D}$ can combine forming the transition state ($\ce{TS}$). As your text book says, the $\ce{TS}$, being an energy maximum, is very unstable and collapses very fast. Since the transition state represents a maximum of the energy curve the collapse may go in either the direction of the reactants or the direction of the products with equal probability, pretty much in the same way a ball placed at the top of hill may roll down on either side of it. So, once you got to the transition state you have a 50:50 situation.
The thing that determines which way your reaction goes is how easy it is to make the transition state or to "get the ball to the top". In order reach the transition state you have to overcome a thermodynamic barrier, which is the difference of the free energies of products and reactants $\Delta G$, and a kinetic barrier, which is the activation energy $E_{\mathrm{a}}$. If $\Delta G$ is a very large and negative then the reaction will proceed in the direction from $\ce{A}+\ce{B}$ to $\ce{C}+\ce{D}$ and will be irreversible because then it will take considerably more effort to "climb" the energy hill coming from the $\ce{C}+\ce{D}$ side than coming from the $\ce{A}+\ce{B}$ side. This means that in a given time a lot more $\ce{A}$ and $\ce{B}$ molecules are "elevated" to the $\ce{TS}$ than $\ce{C}$ and $\ce{D}$ molecules. So, even if $\ce{TS}$ will revert to $\ce{A} + \ce{B}$ 50 % of the time the reverted molecules will quickly be back "on top of the hill" in order to have another try while for $\ce{C}$ and $\ce{D}$ this takes much longer - the reaction rate for the forward reaction is much higher than the reaction rate for the backward reaction. So over time more and more $\ce{A} + \ce{B}$ is converted to $\ce{C}+\ce{D}$ until equilibrium is estabilished and the ratio of reactants/products is equal to the ratio of the reaction rates.
When you have a reversible reaction and $\Delta G$ is very small the "hill to climb to the transition state" is nearly the same height irrespective of whether you come from the product or from the reactant side. So, forward and backward reaction proceed nearly at the same rate.

Concerning your second question: I'm sorry, but I couldn't find a non-mathematical explanation as to why $G$ depends on the concentration. The Gibbs free energy is the thermodynamic potential of an ensemble at constant pressure and temperature. For variable particle numbers $N_{i}$ its fundamental relation is:

\begin{equation}
  dG = -S dT + pdV + \sum_{i} \mu_{i} dN_{i} \ .
\end{equation}

So, it is connected to the particle numbers (and thus to the concentrations) of the species in the system.


Find the stoichiometric coefficient of a reactant given Kp, Kc, and the stoichiometric coefficient of the product

Using the relation of $K_\mathrm{p}$ to $K_\mathrm{c}$:

$$\frac{K_\mathrm{p}}{K_\mathrm{c}} = (RT)^{\Delta n}$$

where $\Delta n$ is the change in the number of moles of gas in the reaction (here, $\Delta n = 3 - x$). Plugging in (using $R = \pu{0.0821 L atm mol-1 K-1}$ and $T = \pu{523.15 K}$) we get:

$$\frac{0.0105}{0.45} = 0.0233 = 42.950615^{(3-x)}$$

To solve we take the natural logarithm of both sides:

$$\ln{(0.0233)} = (3-x) \ln{(42.950615)}$$
$$\frac{\ln{(0.0233)}}{\ln{(42.950615)}} = -1 = (3 - x) \implies x = 4$$

And so $x = 4$. We can check to see if our answer is correct by plugging the calculated value for $x$ into the second expression from the top of this post:

$$0.0233 \buildrel{?}\over{=} 42.950615^{-1}$$

which does turn out to be correct.


Why is the tropylium anion non-aromatic?

The rules in determining the aromaticity of a compound are as follows:

The system must have 4n+2 π electrons
The system must be planar
The system must be completely conjugated
The molecule must be cyclic.

The cycloheptatriene anion does not follow two out of these four rules. Namely,

The anion system has 4n electrons (whereas it should have had 4n+2 electrons in it π system)
The anion system has an sp3 carbon and is hence non-planar (where it should have been planar)

Due to these two issues (second issue mainly), there is no ring current in the complete ring and so no conditions need to be considered whether the compound is antiaromatic or aromatic in nature.
Cyclopentadienyl anion also has an sp3 carbon as an anion. However the anion formed is aromatic if planar and so this overcomes the ring strain and so retains aromaticity.


Should the 'a' in the acid dissociation constant (Ka or pKa) be capitalised?

TL;DR: for the documents written in English use $\mathrm{p}K_\mathrm{a}$ (upright lowercase "a").

Current (2017Q3) chemmacros' documentation (p. 9) includes a reference for how acid-base module handles \pKa[...]:

  \pKa: $\mathrm{p}K_\mathrm{a}$, \pKa[1]: $\mathrm{p}K_\mathrm{a1}$, depends on language settings, see section 6.5 starting on page 30. The translations can be adapted. [...]

  As you can see the default subscripts of \Kw, \Ka and \Kb are lowercase letters. The literature is inconclusive about if this is the right way or if uppercase letters should be preferred. In textbooks the uppercase variant usually seems to be used while journals seem to prefer the lowercase variant. chemmacros’ default follows the usage in The IUPAC Green Book [Coh+08].

From IUPAC Green Book [1, p. 59]:

  The equilibrium constant of dissolution of an electrolyte (describing the equilibrium between excess solid phase and solvated ions) is often called a solubility product, denoted $K_\mathrm{sol}$ or $K_\mathrm{s}$ (or $K_\mathrm{sol}^⦵$ or $K_\mathrm{s}^⦵$ as appropriate). In a similar way the equilibrium constant for an acid dissociation is often written $K_\mathrm{a}$, for base hydrolysis $K_\mathrm{b}$, and for water dissociation $K_\mathrm{w}$.

Additionally, "a" should be typed as an upright letter [1, p. 5]:

  Subscripts and superscripts that are themselves symbols for physical quantities or for numbers should be printed in italic type; other subscripts and
  superscripts should be printed in Roman (upright) type.

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen R. E., Ed.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.


What is the criteria for determining structure of a molecule?

No, not symmetry per se but generally the lowest energy arrangement of  the atoms.  What structure this is is not always obvious by just combining atoms according to the valency rules; if it were, experimental (analytical) methods would not be needed. This is particularly true of 'organic' molecules where one formula can give many similar structures, e.g. isomers, but also of the small molecules you list.

The rules we use are the result of numerous experiments over the last 150 years or so reacting one species with another and trying to make a consistent set of rules to explain things. But ultimately structural studies are needed, the definitive method is x-ray diffraction from a crystal of the material but of course this is a not always possible. In this case infra-red, Raman and NMR spectroscopies have been used as well as mass spectrometry.  These methods do not give complete structures directly (bond lengths and angles as does x-ray diffraction) but allow confirmation of one structure (as topology) over another, for example $\ce{N2O}$ is  NNO rather than  NON. More latterly computational methods are being used to predict structure but ultimately these will have to be backed up by experiments.


Why is Finkelstein reaction not conducted using NaF?

Further to my comment - the solubility of Sodium Fluoride in Acetone is 2.4 microgrammes per 100g of solvent i.e. essentially insoluble source. There will be no F- in solution so the reaction will not occur.


Activating effects of amino and hydroxyl groups in different pH

Both the amino and hydroxyl groups activate benzene rings, as the lone pairs on N or O are in conjugation with the benzene π system.

However, in acidic media, the amino group is protonated to form an ammonium cation. Now, the lone pair on nitrogen is no longer available: it has been used to make a new N–H σ bond.

Consequently, the $\ce{-NH3+}$ group no longer acts as an electron donor via resonance. Instead, it is an electron-withdrawing group via the inductive effect (neutral amines are, too, but the resonance donation far outweighs the inductive withdrawal). For further details, you may wish to read this question: Is the ammonium substituent (-NH3+) really meta-directing in electrophilic substitution?

Phenols, on the other hand, can be deprotonated in basic media to form phenolates. Perhaps unsurprisingly, this has the opposite effect from before: the negatively charged substituent $\ce{-O^-}$ is even more electron-donating than the neutral hydroxyl group.

It's tempting to think that this is because "there are more lone pairs on oxygen", but that's not the truth. In order to have maximal overlap between a lone pair and a π system, they must be parallel to each other: on this oxygen, there can only be one lone pair which is parallel to the benzene π system. So, whether a substituent has two or three lone pairs doesn't make all that much of a difference.

The explanation is, instead, to do with the availability of the lone pair. We could think of the neutral phenol as being the same as the phenolate, but with an extra proton which is pulling one lone pair on oxygen towards it. This stabilises the lone pairs, and therefore makes them "less willing" - in a sense - to be delocalised into the benzene ring. (Technically speaking, the lower energy of the lone pair makes for a poorer energy match with the benzene ring LUMO.) In the deprotonated phenolate, though, there are three lone pairs around oxygen which repel each other. This means that they are higher in energy, and are more delocalised into the benzene ring.

As a result, phenolate reacts with very poor electrophiles such as $\ce{CO2}$ (Kolbe–Schmitt reaction) quite well.


The value you have reported is the standard Gibbs free energy change of formation, which has a specialized definition. A standard free energy change of formation is the free energy change that occurs when 1 mole of substance is formed by reaction of its constituent elements in their standard states at 1 atmosphere of pressure and 25 degrees Celsius.

Thus, for sodium chloride:

$$\ce{Na(s) +\frac{1}{2} Cl2(g) -> NaCl(s)}\ \ \ \Delta_f G^\circ =-384.1 \text{ kJ/mol}$$

Now, all we need to be able to do is measure $\Delta G$ for this reaction at standard temperature and pressure.

In practice, we will separately detemrine $\Delta_f H^\circ$ and $\Delta_f S^\circ$ and plug them into $\Delta_f G^\circ = \Delta_f H^\circ -T\Delta_f S^\circ$.

At constant pressure (which we are by definition operating at),  $\Delta H$ is equal to the heat transfer  $\Delta H = q = -C_p\Delta T$, which is equal to the heat capacity of the surrounding (in practice the calorimeter, which is reasonably isolated from the universe) multiplied by the change in temperature.

$\Delta S$ is still tricky to measure. At equilibrium, for a reversible process at constant temperature $\Delta S = \frac{q}{T} = \frac{\Delta H}{T}$. If the process is not reversible or you are not at equilibrium, then... then it is challenging.


Why ammonium hydroxide precipitates metal salts?

It's simply because of the concentration. In aqueous solution, the formation of a precipitate is favored over the formation of a complex ion, and it can be shown using solubility equilibria and $K_\mathrm{f}$ (but it won't be pretty).
This is just like the QA test for copper(II). After a small amount of aqueous ammonia is added, $\ce{Cu(OH)2}$ is observed. Add more, and you get the deep blue $\ce{[Cu(NH3)4]^2+}$ complex.


Qualitative Comparisons of Dielectric Constants of Organic Solvents

Yes, but it is much more complicated. The molecule does not have to have a dipole, in which case the (static) dielectric constant (relative permittivity) depends on the molecule's polarisability and if it does have a dipole it depends on both factors. The equation is $\displaystyle \epsilon_r=\frac{1+2b}{1-b}$ where $b=\left(\alpha+\mu^2/(k_BT)\right)N/(3\epsilon_0)$ and $N=N_A\rho/M$ where $\alpha$ is polarisability, $\mu$ dipole, $\rho$ density,$M$ molar mass, $\epsilon_0$ permittivity of free space.  See Atkins, & Friedman, 'Molecular Quantum Mechanics'. The dielectric constant also depends on frequency of the applied electric field, but the value quoted is normally the static value measured by comparing the capacity of a capacitor with and without the solvent. The name relative permittivity arises from the potential a charge produces. In a vacuum at a distance $r$ this is $\displaystyle \phi=\frac{q}{4\pi\epsilon_0 r}$, in a dielectric medium of permittivity $\epsilon$ this becomes $\displaystyle \phi'=\frac{q}{4\pi\epsilon r}$ and $\epsilon_r=\epsilon/\epsilon_0$.

As an aside you can see why a polar molecule is generally insoluble in a non-polar solvent. If the permittivity is small (say hexane) the electric potential of a charge is higher at any given $r$ than for a solvent with a large $\epsilon$, such as acetonitrile. This means that polar molecules are less soluble in low dielectric solvents than they are in high dielectric ones.


Why would this reaction occur? Won't it be unfavourable?

You're right: $\ce{NO}$ is a free radical and the reaction between $\ce{N2}$ and $\ce{O2}$ can be achieved - supposed that enough energy is provided.

This may happen through an electrical arc discharge in the lab or, in the mesosphere (50-85km above earth) through ionizing cosmic rays. Note however that nitric oxide react with oxygen to eventually furnish nitrogen dioxide $\ce{2 NO + O2 <=> 2 NO2}$.


Quoting from Hollas, Modern Spectroscopy, 4th ed. (p 236):

  There is a convention, which is commonly but not always used, for labelling electronic states. The ground state is labelled $X$ and higher states of the same multiplicity are labelled $A, B, C\ldots$ in order of increasing energy. States with multiplicity different from that of the ground state are labelled $a, b, c \ldots$ in order of increasing energy.

As such, for the oxygen molecule, where the lowest three states are ${}^3{\Sigma}_g^-$, ${}^1{\Delta}$ and ${}^1{\Sigma}^+_g$ (in increasing order of energy), the triplet ground state would be labelled $X$ and the next two states $a$ and $b$ respectively.

A word of caution: it seems that this convention is not always obeyed. On the next page, Hollas writes (for $\ce{I2}$, where the ground state is a singlet)

  The transition $B\,{}^3{\Sigma}_{0_u^+} - X\,{}^1{\Sigma}_g^+$ involves the $\Omega = 0$ component of the $B$ state [...] The labelling of the $B\,{}^3{\Sigma}_{0_u^+}$ state follows general usage rather than convention, which would label it $b\,{}^3{\Sigma}_{0_u^+}$.

Wikipedia also adds, although without a citation:

  In polyatomic molecules (but not in diatomics) it is customary to add a tilde (e.g. $\tilde{X}$, $\tilde{a}$) to these empirical labels to prevent possible confusion with symmetry labels based on group representations.


I have a solution of citric acid, tartaric acid, water and glucose. How do I find the concentration of the citric acid present in the solution?

Citric acid has $\mathrm{p}K_\mathrm{a}$-values of 3.1, 4.7, and 6.4, while those of tartaric acid are 3.0 and 4.3. If you adjust the $\mathrm{pH}$ to 6.4, tartaric acid would be roughly 100% deprotonated, while 50% of citric acid still has one proton to give off.

If you titrate this solution with $\ce{NaOH},$ you could estimate the buffer capacity, giving you an estimate of the concentration. The biggest systematic error is that the apparent $\mathrm{p}K_\mathrm{a}$ is dependent on ionic strength, which you don't know unless you know the concentrations in the first place.


Winkler's method in the absence of iodide

Bad news is that Winkler's method will fail in the presence of bleach or any other oxidizing agent besides oxygen. Neither sodium thiosulfate nor potassium iodide are compatible with bleach, whether it is a oxygen bleach or a chlorine bleach.

  I used $\ce{NaOH},$ $\ce{MnSO4},$ $\ce{H2SO4}$ in this experiment (and $\ce{Na2S2O3}$ as the titrant) and I'm not too sure whether I used potassium iodide. However, I've obtained a result and found out a relationship where the dissolved oxygen content increases as I add more bleach into the water.

As a good scientist of future, you always write whatever you are doing at the time of experiment in your notes.

Your results are skewed towards higher numbers because sodium thiosulfate is being consumed by released iodine not only from the oxygen reaction but also from the bleach.


Inductive effect and hyperconjugation - one elephant, different parts?

Are the inductive effect and hyperconjugation both different ways of
  looking at the same phenomenon

I think so, both inductive and resonance (that's what hyperconjugation is after all) effects move electrons around.  I tend to separate the two effects as follows:

inductive effects are associated with electron movement through sigma bonds due to electronegativity differences
resonance effects are associated with electron movement through p orbitals due to electronegativity differences

For example, we often say that a methyl group stabilizes a double bond (carbon-carbon or carbonyl) or a carbocation.  The methyl group is roughly $\ce{sp^3}$ hybridized while the double bond carbon or carbocation carbon is roughly $\ce{sp^2}$ hybridized.  An $\ce{sp^2}$ orbital is generally lower in energy than an $\ce{sp^3}$ orbital because it contains more s-character.  Therefore electrons will generally prefer to flow from $\ce{sp^3}$ to $\ce{sp^2}$ orbitals.  The direction of this electron flow also indicates  that $\ce{sp^2}$ orbitals are more electronegative than $\ce{sp^3}$ orbitals.  How can the $\ce{sp^3}$ methyl group shift electrons to the lower energy, more electronegative $\ce{sp^2}$ carbon?  It can do it inductively (through sigma bonds) or through resonance (through p orbitals).

I've added a picture showing resonance structures involving hyperconjugation, just to be clear as to what hyperconjugation involves.


No, 2-methylpropan-2-ol cannot give iodoform test, as it cannot be converted into a methyl ketone.

Generally, alpha hydroxy methyl type (this is not the actual nomenclature) groups can give iodoform on oxidation.


What is the reason for more stable keto form to tautomerise to a less stable enol form?

We can think of it an an equilibrium that is disturbed by another, irreversible reaction.

Think of a more familiar reaction.  When you try to dissolve magnesium hydroxide in water you get only a little solubility; the dissolved magnesium and hydroxide ions are only a minor species at equilibrium.  But when hydrochloric acid is added the acid reacts selectively with the hydroxide ions, neutralizing them; in response more magnesium hydroxide keeps dissolving into the water (to form magnesium and hydroxide ions, the latter then reacting with more acid) until you have a salt solution containing the accumulated magnesium ions and the chloride from the acid.  Reaction continues until you've dissolved all of the magnesium hydroxide or neutralized all of the acid.

Similarly, the enol form is only a minor species in the equilibrium, but when you add the selenium dioxide it reacts selectively with the enol form.  This draws more of the carbonyl compound into the enol form in the same way that the acid in the example above draws more magnesium hydroxide into its dissolved form.  Thereby the reaction keeps going until you run out of one reactant or another.


I think you are most of the way to the answer, but I will start the process from scratch for the sake of a full explanation for future readers.
First, we determine the symmetry group of $\ce{BF3}$, which by quick inspection we can determine to be $D_{3h}$ (since it has a principle $C_3$ axis, $3\perp C_{2}$ axes, and a horizontal mirror plane of symmetry).

Knowing this, we can now make use of the $D_{3h}$ character table (thank you @orthocresol for compiling these in mathjax format) to determine the irreducible representation of the atomic orbitals (or their symmetry adapted linear combinations SALCS) within the group.
$$\begin{array}{|c|cccccc|c|c|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & \text{linear/rotations} & \text{quadratic}\\ \hline
\mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_2'} & 1 & 1 & -1 & 1 & 1 & -1 & R_z & \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0 & (x,y) & (x^2-y^2,xy) \\
\mathrm{A_1''} & 1 & 1 & 1 & -1 & -1 & -1 & & \\
\mathrm{A_2''} & 1 & 1 & -1 & -1 & -1 & 1 & z & \\
\mathrm{E''} & 2 & -1 & 0 & -2 & 1 & 0 & (R_x,R_y) & (xz,yz) \\ \hline
\end{array}$$
Using this, we can quickly determine the irreps of boron's orbitals using the 2nd column from the right.

$\ce{B}$: $2s=a_1'$, $2p_x,2p_y=e'$, and $2p_z=a_2'$.

Next, we will determine the irreps for the $3\ce{F}$ atoms, which will require that I consider combinations of $s,p_x,p_y, \text{and } p_z$ orbitals.
Below, I show how the symmetry operations of the group affect each set of orbitals and show the orbitals that result (image source).
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_s & 3 & 0 & 1 & 3 & 0 & 1 \\ \hline
\mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0  \\ \hline
\end{array}$$
$\hspace{20ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_{p_x} & 3 & 0 & 1 & 3 & 0 & 1 \\ \hline
\mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0  \\ \hline
\end{array}$$
$\hspace{12ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_{p_y} & 3 & 0 & -1 & 3 & 0 & -1 \\ \hline
\mathrm{A_2'} & 1 & 1 & -1 & 1 & 1 & -1 \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0  \\ \hline
\end{array}$$
$\hspace{20ex}$
$$\begin{array}{|c|cccccc|} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\Gamma_{p_z} & 3 & 0 & -1 & -3 & 0 & 1 \\ \hline
\mathrm{A_2''} & 1 & 1 & -1 & -1 & -1 & 1 \\
\mathrm{E''} & 2 & -1 & 0 & -2 & 1 & 0  \\ \hline
\end{array}$$
$\hspace{21ex}$
These reductions to irreducible representations can be done via the reduction formula $$n_i=\frac{1}{h}\sum_c g_c \cdot \chi_{i_c} \cdot \chi_{r_c}$$ where $n_i$ is the number of times a particular irrep $i$ occurs in the reducible representation, $g_c$ is the number of symmetry operations $c$, $\chi_{i_c}$ is the character of the irrep $i$ for the symmetry operation $c$, and $\chi_{r_c}$ is the character of the reducible representation $r$ for the symmetry operation $c$.
This allows us to write the irreps of the three $\ce{F}$'s orbitals.

$3\ce{F}$: $a_1'+a_2'+a_2''+2e'+e''$

With all the formalities out of the way, we can construct the qualitative MO diagram for $\ce{BF3}$. Like much of this post, my source for the MO diagram is the pdf of lecture notes for a Dartmouth chemistry course located at http://www.dartmouth.edu/~chem64/64%20pdf%20files/PS3A.pdf

Working our way up through this diagram, we see that we start by assuming that the SALCs of the $s$ orbitals of the three $\ce{F}$'s are assumed to be low enough in energy where they do not interact with the orbitals of $\ce{B}$.
The next point of interest is the $e' (\sigma)$ MO, which we form from the $p_x$ SALC of the $\ce{F}$ atoms and the $p_x$ and $p_y$ of the $\ce{B}$. We could additionally consider the interaction of the $e'$ $p_y$ SALC to form a $\pi$ bonding interaction, but we will assume that, with the $\sigma$ interaction already present, this $\pi$ bonding interaction will be weak and thus can be neglected.
This previous point leads us to have a set of nonbonding orbitals $a_2'+e'(y)+e''$. We then also have antibonding orbitals for each of the bonding orbitals below.
As a check on our work, we can compare this to the somewhat more quantitative diagram given in Principles of Inorganic Chemistry by Brian PFennig using the Student Editon of Spartan:

The only noticeable difference here is that the nonbonding MOs are now placed with respect to their computed energy rather than being lumped together. The classification of orbitals as bonding, nonbonding, and antibonding seems to fit with the diagram we have above, with the caveat that this more quantitative diagram does suggest some $e'(y)$ bonding contribution, as its energy is somewhat lower than the p orbitals it came from.
Now that we have the correct diagram, we can then compute the bond order for each $\ce{B-F}$ using: $$\frac{\text{Bonding electrons } -\text{Antibonding electrons}}{2\cdot\text{number of bonds}}$$ which for this diagram gives a value of $\frac{4}{3}$ per bond. This gives a sense of why $\ce{BF_3}$ is a strong Lewis acid, since it can accept a pair of electrons into its antibonding LUMO and still leave a compound with a BO of $1$ per bond.
As a final point, it is worth noting that this method of classifying orbitals is far from perfect and in general we should really look at individual bonds to classify if the orbital is anti/bonding with respect to that bond. However this energy based approach is what is commonly taught in school and happens to work well in this simple case; just be aware that it can fail for even seemingly simpler molecules like $\ce{CO}$.


Trouble with Converting from Structural Formulas into Newman Projections

this is how you can convert the str.s into each other. It should have had some wedge/dash or sawhorse information... for the conversion.


In polar solvents such as water, the water molecules will solvate the carboxylic acid. Thus, hydrogen bonds will form between the 2 molecules and the carboxylic acid will dissolve.

The water is in large excess as it is the solvent so it will solvate most of the carboxylic acid molecules.

In a non-polar solvent such as benzene, no such hydrogen bonds can take place with the carboxylic acid and benzene molecules. As a result, the carboxylic acids simply interact with themselves and form dimers.


Wattage to produce 6L of air from Water in less than a second?

I guess it would have to be the amount energy released by the reaction $$\ce{2H2 + O2 -> 2H2O}$$ where you have four liters of hydrogen gas and two liters of oxygen gas, all divided by one second (or less, increasing the wattage as you decrease the time).

You asked about "air" — but you can't get normal air from electrolysis of water.

So, the heat of combustion for hydrogen is $286\ \mathrm{kJ/mol}$. $$286\ \frac{\mathrm{kJ}}{\mathrm{mol}} \times \frac{4\ \mathrm L}{22.4\ \frac{\mathrm L}{\mathrm{mol}}} = 51.1\ \mathrm{kJ}$$ and that is assuming atmospheric pressure which is certainly low, but how low depends on your depth below the surface of the water. So with the "barely under the surface" energy of $51.1\ \mathrm{kJ}$, we can then divide that by seconds: $$\frac{51.1\ \mathrm{kJ}}{1\ \mathrm s} = 51.1\ \mathrm{kW}$$ so you would need a bare minimum of $51.1\ \mathrm{kW}$ of power to convert water into a 67:33 mixture of $\ce{H2:O2}$. That seems like a lot of energy in a small amount of time to me: think about a six-liter balloon with the perfect combustion ratio of hydrogen and oxygen. That reaction is a lot faster than one second, but it's still a lot of energy. Tough to get into a small, portable battery and have it fill a capacitor with that amount of e energy every few seconds.

I'll be honest, I'm just doing these calculations on-the-fly, and I may have made a mistake or five, but it seems reasonable to me. If anyone takes any issue with my answer, please, feel free to edit/give suggestions.

Edit: This Wikipedia page has some information on hydrogen-oxygen diving air. They use a 96:4 ratio. Not sure why, but if that's the best ratio, this would take significantly more energy per breath, since a lot of oxygen would just get dumped.


How many grams of sodium phosphate must be added to precipitate as much of one ion as possible?

In an answer that is now hidden user Maurice pointed out the complexities of the chemistry. So let's assume:

All Al present in solution as $\ce{Al^{3+}}$ and any hydoxy complexes or precipitates will be ignored.
All Ca present in solution as $\ce{Ca^{2+}}$ and any hydoxy complexes will be ignored.
All $\ce{PO4^{3-}}$ added to solution stays as that ion and doesn't hydrolyze to $\ce{HPO4^{2-}}$
Two salts that will be produced are $\ce{AlPO4}$ and $\ce{Ca3(PO4)2}$
$K_\mathrm{sp} (\ce{AlPO4})=9.83\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\times 10^{-33}$

Concentration of $[\ce{PO4^3-}]$ that will cause $\ce{Al^{3+}}$ to precipitate:
$$[\ce{Al^3+}]=\dfrac{\pu{0.002 mol}}{\pu{200 mL}}=0.01M$$
$$[\ce{PO4^3-}]_\ce{Al}=\dfrac{K_\mathrm{sp}}{\ce{[Al^{3+}]}} = \dfrac{9.83\times 10^{-21}}{\pu{0.01M}}=9.83\times 10^{-19}$$

Concentration of $[\ce{PO4^3-}]$ that will cause $\ce{Ca^{2+}}$ to precipitate:
$$[\ce{Ca^2+}]=\dfrac{\pu{0.001 mol}}{\pu{200 mL}}=0.005M$$
$$[\ce{PO4^3-}]_\ce{Ca}=\sqrt{\dfrac{K_\mathrm{sp}}{\ce{[Ca^{2+}]^3}}} = \sqrt{\dfrac{2.07\times 10^{-33}}{(\pu{0.005 M})^3}}= 1.2890\times 10^{-13}$$

$\therefore \ce{AlPO4}$ will precipitate first.

The molecular mass of $\ce{Na3PO4}$ is 163.941 g/mol.

Note here that you can't just assume a stoichiometric amount of $\ce{Na3PO4}$ to say three significant figures. The solubility of the $\ce{Ca3(PO4)2}$ is so low that the $\ce{Na3PO4}$ would have to be weighed to an absolutely unrealistic precision.

So we'll go father down fantasy lane and suppose that:

There are $0.02\bar0$ moles of  $\ce{Al^3+}$
That we are using God's balance which can weigh an infinitely precise amount of $\ce{Na3PO4}$.
That the molecular mass of $\ce{Na3PO4}$ is exactly 163.941 g/mol (this is so precise that isotopic ratios limit the value...)
$K_\mathrm{sp} (\ce{AlPO4})=9.83\bar0\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\bar0\times 10^{-33}$
That the volume changes in the solution when adding $\ce{Na3PO4}$ or precipitating $\ce{AlPO4}$ are insignificant.

So if the $[\ce{PO4^3-}] = 1.\bar0\times 10^{-13}$ then no $\ce{Ca3(PO4)2}$ will precipitate. At that concentration the amount of $\ce{Al^3+}$ left in solution will be:

$$\ce{[Al^{3+}]}=\dfrac{K_\mathrm{sp}}{\ce{[PO4^{3-}]}} = \dfrac{9.83\times 10^{-21}}{1.\bar0 \times \pu{10^{-13} M}} =9.83\times 10^{-8}$$

So $\pu{0.001999998034000 mol}$ of $\ce{AlPO4}$ will precipitate.

For 200 ml of a $1.\bar0 \times \pu{10^{-13} M}$ solution there are $2.\bar0 \times \pu{10^{-14} moles}$

So the total moles of $\ce{Na3PO4}$ needed is $\pu{0.001999998034020  mol}$ or $\pu{0.327881677695273 g}$.

% Al left in solution $= \dfrac{0.002\bar0 - \pu{0.001999998034000 mol}}{0.002\bar0} =0.0000983\% $

Again the problem requires such a precise amount of $\ce{PO4^{3-}}$ to precipitate "all" the $\ce{Al^{3+}}$ yet leave "all" the $\ce{Ca^{2+}}$ in solution that it is totally unrealistic.


Why is the boiling point of water and ammonia so different?

If I interpret your edited question correctly, it now boils down to why is the difference between the boiling points of $\ce{HF}$ and $\ce{NH3}$ so large, even though they have almost the same molecular masses ($M_\mathrm{r}(\ce{NH3}) = 17, M_\mathrm{r}(\ce{HF}) = 20$).

Ammonia can self-ionise much like water can according to the following equation $(1)$. And much like water, the process is important to determine a $\mathrm{pH}$ value, but rather neglegible when it comes to determining attractive forces between ammonia molecules where ‘traditional’ hydrogen bonding predominates.

$$\ce{2 NH3 <<=> NH4+ + NH2-}\tag{1}$$

On the other hand, self-ionisation in hydrogen fluoride, especially in the liquid and solid states, is much more important and leads to the two fragments shown in equation $(2)$.

$$\ce{3 HF <=> H-F-H+ + F-H-F-}\tag{2}$$

These two linear fragments both feature de facto four-electron-three-centre bonds and are thus much more prevalent than the corresponding ammonia-derived ones. (Indeed, something analogous to $\ce{FHF-}$ doesn’t exist for ammonia.) Since these also introduce ionic interactions into the equation, the intermolecular forces between $\ce{HF}$ molecules are much stronger than those between ammonia molecules contributing to $\ce{HF}$’s higher boiling point.


Yes, you are right. The reaction depends on with which ion it is accompanied with. Note that these species are in equilibrium and present in the solution  and that particular reaction proceeds according to which ion is present in the solution in concentrated form.

Aluminum hydroxide is formed when soluble salt of $\ce{Al^3+}$ i.e
$\ce{[Al(H2O)6]^3+}$ when it is reacted with base.

$$\ce{[Al(H2O)6]^3+(aq) + 3 OH-(aq) → Al(H2O)3(OH)3(s) + 3 H2O(l)}$$

This insoluble metal hydroxide can act as a base, since it can be
redissolved by reacting it with an acid.
As Brønsted base

$$\ce{Al(H2O)3(OH)3(s) + 3 H3O+(aq) → [Al(H2O)6]^3+(aq) + 3 H2O(l)}$$

Alternatively, the metal hydroxide can act as an acid, since it can
react with a base.
Lewis acid

$$\ce{Al(H2O)3(OH)3(s) + OH-(aq) → [Al(H2O)2(OH)4]-(aq) + H2O(l)}$$

Thus, all aluminium compounds are said to be amphoteric.


Why does period 3 of the periodic table contain 8 elements instead of 18?

You could say that the period number tells you about the largest value of principal quantum number in which a electron is present.

Electrons are filled according to $n+l$ rule. It states as $n+l$ increases, the energy of the orbital increases. If $n+l$ is the same, then the bigger $n$ has larger energy.

So $\mathrm{4s}$ would be filled before $\mathrm{3d}$. Therefore elements containing $\mathrm{3d}$ electrons would have their $\mathrm{4s}$ orbital filled, and therefore, come in 4th period. There 3rd period contains elements having $\mathrm{3s}$ and $\mathrm{3p}$ electron as their last electron which are equal to 8.


Why aren't solid pure transition metals colourful?

Colour is a property of electronic transitions: the transitions in metals are different to those in their compounds
In many discrete compounds of transition metals the colour arises because there are accessible electronic transitions between molecular orbitals in the molecules (usually involving metal d-orbitals) matching the energy of wavelengths of visible light. These transitions are often fairly narrow so giving a wide range of possible colours.
Pure metals are not like that. To simplify a lot, the definition of a metal is having a continuous band of electronic states (not a simple molecular orbital) for the electrons involved in conduction of electricity (the conduction band). This band involves the whole bulk substance, not just isolated atoms of the metal. So the transitions that might involve colour (d-d transitions in a molecule) are not present in the same way: instead transitions occur inside the conduction band or between orbitals and the conduction band. Either way, the narrow transitions giving colour in compounds are now much broader because  a wide range of continuous electron energies exist in the conduction band.
This is a gross simplification, but it is good enough to explain the key observation if you don't want to get into the theory behind the electronic structure of metals.


An ionic compound dissociates into 3 ions of 3 different elements?

Yup, they are definitely possible. Such salts are called as mixed salts. They furnish more than two ions when dissolved in water. A popular example is Mohr's salt, popularly used as a standard titrant to measure concentrations of oxidizing agents. It's (anhydrous) formula is:

$$\ce{Fe^{II}(NH4)2(SO4)2}$$

This salt on dissolving in water would split as:

$$\ce{Fe^{II}(NH4)2(SO4)2 <=>> Fe^2+ + 2NH4+ + 2SO4^2-}$$

This is one such example. There are many others, notably alums. Alums are a general type of mixed salts, of the form:

$$\ce{A^{I}M^{III}(SO4)2}$$

where $\ce{A^{I}}$ is a monovalent metal ion, and $\ce{M^{III}}$ is a trivalent metal ion. Clearly, these are also mixed salts.

These are just a few practical examples. There's literally quite nothing that can stop imagination of course.


Arrange these compounds: CO2, CH3OH, RbF, CH3Br in order of increasing boiling points

You know $\ce{CO_2}$ is gaseous at room temperature, so let's put that at the bottom. Methanol  forms hydrogen bonds, so that will be above bromomethane which does not. At last we have rubidium fluoride which is a salt. Salts generally have a very high boiling point (> 1000 °C, much higher than molecular structures) because of the ionic (electrostatic) interaction between the ions, so that one will be at the top. Ionic forces can be seen as extreme dipoles in a certain way, there is a grey area when electronegativity becomes large enough, that it can be seen either as a molecular structure or ionic structure. Consulting online information about the boiling points of these compounds (i.e. just check Wikipedia or some MSDS site) confirms the theory.


Why will internal energy decrease in this reaction?

If the reaction is exothermic, and the system you choose includes both the insulated chamber and its mixture of reactants and products, then the change in internal energy of this system will be zero.  However, if the system you choose includes only the mixture of reactants and products, then heat will be transferred from the mixture to the chamber, and the internal energy of your mixture will decrease.


Reagents/solutions that are acidic enough will protonate the urea group, just not where you think. The most basic site is the oxygen. Perhaps this is a surprise.

The lone pairs on the nitrogen atoms are delocalized (see resonance depiction below).

The oxygen atom has the most electron density. Below is an electron density map, where ${\mathrm{\color{red}{red}}}$ represents areas of high electron density and $\mathrm{\color{blue}{blue}}$ represents areas of low electron density. You can see the area around the oxygen atom is red and yellow, while the areas around the nitrogen atoms are green.

This image is from "The effect of intermolecular hydrogen bonding on the planarity of amides" by James A. Platts, et al. published in Phys. Chem. Chem. Phys., volume 2012, issue 14, page number 11944-11952, DOI:10.1039/C2CP41716B. The image is copyright the Royal Society of Chemistry and released under a Creative Commons Attribution 3.0 Unported license.

Protonating the oxygen atom leads to a cation where the positive charge is delocalized (better, more stable).

Protonating one of the nitrogen atoms leads to a cation where the positive charge is localized (worse, less stable).


Order of rate of acid catalysed dehydration of alcohols

Your basic assumptions are correct.
It can be observed as such:

In Q the carbocation is stablised by resonance, inductive and hyperconjugative effects so it is quite stable and will be formed fastest.
Comparing P and R: In P, a secondary carbocation is formed and there are 4+1 or 5 hyperconjugative structures possible. Moreover, the inductive effect is also greater in P than in R. Hence, it forms carbocation faster.
Both R and S form primary carbocations. Inductive effect is more prevelent in R than in S, while S has 3 hyperconjugating structures.
But R offers capacity to rearrange and stablise the carbocation, by methyl migration to produce highly stable tertiary carbocation, also stablised by resonance over the adjoining phenyl ring.
But, S also forms a carbocation which is primary, but offers no stablising effect other than 2 +1 or 3 hyperconjugating structures.

Similarly, this rearrangement may also be attributed to the comparison of P and R, where hydride shift and methyl shift occur. The migration amplitude of hydride, being lower, leads to rearrangement of positive charge quickly and hence is kinetically also favoured.
Thus finally the order may be:
Q>P>R>S


Sodium-Fusion test or Lassaigne test was originally introduced by Lassaigne in 1843 (Ref.1), and was subsequent modified by Jacobsen in 1879 (Ref.2). The major modification of Jacobson was the replacement of potassium by sodium metal as the active reagent. Since then, the test has altered very little, and basically is still in use today. The difficulties associated with the test have been reviewed in 1969 (Ref.3).
The sodium-Fusion test or Lassaigne test is a general test for the detection of halogens, nitrogen, and sulfur in an organic compound. These elements are usually covalently bonded to the organic compounds such as halo, nitro, amino, and thio compounds. In order to identify their presence, each covalently bonded element has to be converted into their ionic forms. This task is achieved by fusing the organic compound with melting sodium metal. The ionic compounds formed during the fusion are then extracted into aqueous solution and subsequently identified by simple chemical tests. The aqueous extract is called sodium fusion extract or Lassaigne's extract.
The reference 3 gives good procedural practice for preparation of Lassaigne's extract, thus I'm not going to elaborate too much unless it is necessary and going to address the question directly: Test for nitrogen usually involves boiling an alkaline sample of
the water-extract from the sodium fusion with ferrous sulfate and then cooling and acidifying with dilute sulfuric acid. The development of a blue coloration or a dark blue precipitate is confirmatory for nitrogen.
The theory: The carbon and nitrogen present in the organic compound on fusion with sodium metal gives sodium cyanide ($\ce{NaCN}$), which is soluble in water. This is converted into sodium ferrocyanide ($\ce{Na4Fe(CN)6}$) by the addition of sufficient quantity of ferrous sulphate ($\ce{FeSO4}$). Ferric ions ($\ce{Fe^3+}$) generated during the process react with ferrocyanide to form $\color{blue}{\text{prussian blue}}$ precipitate of ferric ferrocyanide ($\ce{Fe4[Fe(CN)6]3}$):
$$\ce{Na + C + N -> NaCN }$$
$$\ce{6NaCN + FeSO4 -> Na4Fe(CN)6 + Na2SO4}$$
$$\ce{3Na4Fe(CN)6 + 4Fe^3+ -> Fe4[Fe(CN)6]3 + 12Na+}$$
There are two issues with this test, according to Ref.3:

It seems that the development of a blue coloration and/or precipitate in a positive test depends on pH and on temperature. With $\mathrm{pH}$ only just below 7, the development of coloration is inhibited. It is, thus, necessary to ensure the presence of an excess of acid. Similarly, cooling of the solution is advisable because development of the blue coloration appears to he inhibited in hot solutions.

During the Lassaigne's extract, if the compound has too much nitro groups and too little carbons, it is advisable that to add a little paraffin or a hydrocarbon before heating with sodium. Carbon in them provide enough carbon to make cyanide with nitrogen. In this praise, there is a answer for your question: Nitro group contains oxygen but did not interfere with the test.

References:

J. L. Lassaigne, "Mémoire sur un procédé simple pour constater la présence de l'azote dans des quantités minimes de matière organique (Memoir on a simple procedure for confirming the presence of nitrogen in minimal quantities of organic matter)," Comptes rendus 1843, 16, 387-391.
Oscar Jacobsen, "Ueber die Oxydation der Parasulfamintoluylsäure (On the oxidation of 3-methyl-4-sulfamoyl-benzoic acid)," Berichte der Deutschen Chemischen Gesellschaft 1879, 12, 2316-2320.
R. P. Gower, I. P. Rhodes, “A review of techniques in the Lassaigne sodium-fusion,” J. Chem. Educ. 1969, 46(9), 606-607 (https://doi.org/10.1021/ed046p606).


The conversion of energy during electroplating

That's a good start, but there is also ohmic loss, as much of the energy is also converted to heat in the electrolyte, electrodes or even wiring.

BTW, during WWII, silver was used to replace copper bus-bars in stationary machinery such as electrolysis apparatus, freeing the copper for use in war materiel and slightly improving the efficiency of the apparatus due to its better conductivity then commercial copper.


Is oxalic acid dihydrate considered a primary acid standard in analytical chemistry?

Jander-Blasius (14. Ed., 1995) uses nonhygroscopic sodium oxalate, dried at 230-250°C (it decomposes above 250 according to wikipedia), to standardise permanganate titer solution against. They give no other recommended standard for manganometry, so I assume this is it.

I have no idea why anybody would want to use (or recommend using) the free acid instead, except perhaps to insult his first-year students' intelligence. Can't imagine it's much cheaper, in analytical grade.

For NaOH titer solution, Jander recommends using a secondary standard, e.g. HCl solution. HCl itself is standardised against freshly precipitated and dried sodium carbonate, or HgO ($\ce{+ 4 KI  + H2O -> K2[HgI4] + 2 KOH}$)  dried over conc. sulfuric acid.


What is the voltage of an electrochemical cell at equilibrium?

There is a cell equilibrium and 2 independent half-cell equilibrii.
There are 2 half-cell equilibrium potentials and a cell open voltage as EMF as the difference of 2 half-cell equilibrium potentials, because the cell in the whole is not in equlibrium.
When a cell gets into equlibrium as the whole system, the cell voltage is zero.


which starting material on heating with concentrated sulfuric acid generate cyclic ester

The correct answer is glycolic acid, B.

It forms the cyclic di-ester glycolide (1,4-dioxane-2,5-dione).


Acid promoted Enol ring formation Mechanism

This reaction proceeds in a few steps. In acidic media, the upper carbonyl oxygen is proton. This results, in greater electron deficiency of the upper carbonyl carbon, which strongly attracts nucleophilic attack. Due to strong activation by the methoxy substituent, the arene's $\pi$ electrons can attack the electron-deficient carbon in the carbonyl group. This nucleophilic addition to the carbonyl group gives a tetrahedral intermediate with a positive charge delocalised in the neighbouring ring.

Note that two types of reactions are happening here: Electrophilic aromatic substitution at the position para to the methoxy substituent and also, a nucleophilic addition to the upper carbonyl group.

A chloride ion now comes in to remove the proton from the para position of the arene. We then get back the aromatic ring with the methoxy group. In the next phase of the reaction, we now want to remove the $\ce {-OH}$ that we have created on the previous carbonyl carbon. This would entail us first protonating it, to make it a good leaving group. How do we create the double bond then?

Well... It is now important to note that $\alpha$ hydrogens in carbonyl compounds are relatively easily lost due to the electron-withdrawing effect of the carbonyl oxygen. Furthermore, the negative charge generated from this loss of the proton can also be delocalised into the the neighbouring aromatic ring on the right of the figure, enhancing the stability of this conjugate base, increasing the acidity of this hydrogen. What happens now is simply elimination to form the double bond. The protonated $\ce {-OH}$ first leaves as water to generate a tertiary carbocation. Then, a chloride ion acts as a base to remove the acidic $\alpha$ hydrogen, with the electrons going into forming a $\pi$ bond between the two carbons. This is essentially an $\ce {E 1}$ process. The desired product is now obtained.


The only thing that's different about ${Fe^2}^+$ and ${Fe^3}^+$ is the number of electrons, which in turn results in different properties of these species.

${Fe^2}^+$, aka ferrous, is pale green and turns violet when added to water. ${Fe^3}^+$, aka ferric, is yellow-brown in solution. You can use several methods to distinguish these two ions. I will mention some of them.

${Fe^2}^+$ can oxidised to ${Fe^3}^+$ with the help of oxidizing agents like $KMnO_4,\space HNO_3$, etc.. $KMnO_4$ is better because its pink color decolorizes on addition of ferrous ions.

Another method to distinguish them is by using colored complexes. For eg., ${Fe^2}^+$ forms a red-orange compound with amine lignads. ${Fe^3}^+$ forms blood red solution with thiocyanate ions, which is used as fake blood in movies, dramas and so on.

Ferrous ions normally have paramagnetic properties, but can become diamagnetic due to the formation of low-spin complexes. A ferric ion is only paramagnetic due to the presence of only one lone electron.


Meso and achiral does both term means the same?

From the Gold book:

meso-compound:
A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members

Achiral compound:
A compound that can be superimposed on its mirror image.
The difference lies in the presence of the chiral centre. All meso compounds are achiral (due to internal compensation as you have mentioned) but all achiral compunds aren't meso.

Look carefully and you'll be able to spot chiral centres in all the above compounds but you'll also notice that they have a plane of symmetry (internal compensation) which makes them achiral.

Bromocylohexane has a  plane of symmetry which makes it achiral. But it has no chiral centre, that's why it doesn't qualify to be a meso compound.


What is the process or material that can enhance the conductivity of the graphene oxide?

Addressing the problem of high resistance, we can identify that the resistivity derives from the oxidation of graphene, changing $sp^2$ carbons to $sp^3$ in the lattice, which increases the band gap, and creates an insulator. There are a number of answers to this question that could be correct, including simply mixing another, more conductive material in the synthesis, but since the question seems to imply that the material has already been synthesized and simply needs to be reduced, removing the oxygen, creating reduced graphene oxide (rGO). One simple approach is through thermal annealing, where GO is reduced by a heat treatment.

You can read more about the reduction of graphene oxide here.


I will point out another equation:

$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S$$

For adsorption $\mathrm{d}S<0$, as molecules go more organized.

Therefore also $\mathrm{d}H < 0$ must be true, otherwise $\mathrm{d}G < 0$, needed for a spontaneous process, would not be achieved.


I don't know how you call it in English but you can apply this approximation $\frac{d[D]}{dt} = 0$ because (3) is fast and $D$ have a really short lifetime, so you can say that

$$k_2[B]^2[C]=k_3[D][E]$$

This approximation is why the teacher is always saying that only the slow reaction is important

Fast equilibrium means that the equilibrium is quickly governed by thermodynamics laws and so the constant

$$K = \frac{[B]}{[A]}$$

is not variating, so you can express $B$ in function of $A$

$$\frac{d[F]}{dt} = k_2K^2[A]^2[C]$$

I don't know how far you need to go but if you really want to solve for $[F]$ you need to write an equation of conservation

$$[N] = [A] + [B] + [C] + [D] + [E] + [F] = cst$$
$$d[A] + d[B] + d[C] + d[D] + d[E] + d[F] = 0$$

$$d[D] = 0$$
$$d[E] = 0$$

You can also say that $d[C] \approx 0$ because it looks like it's just a partner of collision, (it's the solvent) because the solvent is in much larger quantity you can say the concentration stay pretty the same

$$d[A] + d[B] = - d[F]$$

$$[A] - [A]_0 + [B] = -[F]$$

$$[A](1+K) = [A]_0 - [F]$$

$$[A]^2= \frac{([A]_0 - [F])^2}{(1+K)^2}$$

$$\frac{d[F]}{dt} = K_0\frac{([A]_0-[F])^2}{(1+K)^2}$$

with $K_0 = k_2K^2[C]_0$


Boundaries are defined with respect to the system in observation.

Imagine you're in a room with the door closed. Say, you are measuring the temperature of the room, i.e. the room is your system. In that case, it's easy to understand that the boundaries are real. In this case, the walls, the ceiling, the closed door comprises the boundary of the system.

Now, say somebody opens the door. Understand that this does not change the system; you are still interested in measuring the temperature of the room only. So, we can say that at the place of the closed door, there is an imaginary boundary. By specifying the boundary, you are now able to define your system more clearly.

This concept of imaginary boundary helps us to clearly define our system.


According to the IUPAC gold book, oxidation state is:

  A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules: (l) the oxidation state of a free element (uncombined element) is zero; (2) for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;...

Let's consider your reaction $Cu + 2Ag^+ \longrightarrow Cu^{2+} + 2Ag$ The superscripts represent charges. Reaction equations typically don't include oxidation states. In the reactants, $Cu$ has no charge and thus an oxidation state of 0 (rule 1). $Ag^+$ has a charge of +1 and thus an oxidation state of +1 (rule 2). In the products, $Cu^{2+}$ has a charge of +2 and thus an oxidation state of +2 (rule 2). $Ag$ has no charge and thus an oxidation state of 0 (rule 1).

In molecules, oxidation state can be thought of the charge of an atom when all bonds are treated as ionic. For example, in $H_2O$, Both $O-H$ bonds are polar and, since $O$ is more electronegative, all bonding electrons can be considered to belong to $O$. Thus, $O$ has 2 bonding pairs and 2 lone pairs for a total of 8. $O$ has 6 protons so it's oxidation state here is $6-8=-2$. Each $H$ has no electrons and 1 proton for an oxidation state of +1. The total oxidation state of the molecule is $2(+1)+-2=0$, which matches its charge.


Max's comment sums it up. Beware, I'm gonna have to use many simplifications in order to explain this.
A bit of introduction
The species you mentioned are ionic compounds. For your level it's useful to take ionic bonds as bonds between atoms that have such different electronegativities (ChemGuide typically provides useful introductions, I recommend clicking on that link and reading it carefully) that we can imagine them as different ions. Hence, this bond is the attraction between charged particles. What does that mean? It means that one of the species will have a positive charge and the other a negative charge.
Electronegativity increases in diagonally, as we move from bottom left to top right.
$\hspace{5ex}$
$\hspace{30ex}$ Electronegativity (Periodic Trend) – Source
For now just don't apply the electronegativity concept to noble gases. If you really want to know how their electronegativities are determined, take a look at interesting posts like this.
The more electronegative the atom, the less the chance that it'll bond by losing its electrons. Thus, we can conclude two important rules that occasionally become useful, while they're not highly accurate at higher levels:

Metals in general, and especially alkali and alkali earth metals, will bond ionically by losing their valence electrons.
Non-metals in general, and especially halogens, oxygen and nitrogen, will bond ionically by gaining electrons.

I repeat: The two bullet points above are just rules of thumb. Don't expect them to work all the time. But they're good starting points, so I included them in the answer. Don't hesitate to resort back to the first sentence whenever you ran into cases where you don't get a definitive answer by them.

Lithium oxide's case
So let's apply what we learnt to the $\ce{Li2O}$'s case. Lithium will lose its electrons in order to gain the configuration of a noble gases, in which case it would be helium.
$$\ce{Li(s)->Li(g)}^{~1}$$
$$\ce{Li(g)->Li+(g)}$$
Oxygen, however, doesn't favor losing its electrons at all. It will, instead, gain electrons and its electronic configuration would arbitrarily look like that of neon, a noble gas.
$$\ce{O2(g)->2O^2- (g)}$$
Finally, the two ions will form a lattice. Billions of billions of these ions form $\ce{Li2O}$.
$$\ce{Li+ (g) + O^2- (g) -> Li2O(s)}$$
Hence the overall reaction is
$$\ce{2Li (s) + O2 (g) -> 2Li2O(s)}$$
So let's review what we understood: In lithium oxide, oxygen will gain electrons while lithium will lose electrons, since their electronegativities differ greatly. So we should see how many electrons it will lose in order to get to the configuration of the previous noble gas in the periodic table, while oxygen will gain electrons to reach the configuration of the next noble gas in the periodic table. See this Q/A to gain more insight on how to figure this out using their electronic configuration or group number if you don't already know how.
Note that the formula for ionic compounds is the smallest relation between its ions. By $\ce{Li2O}$ we indicate that for every two lithium ions, one oxygen ion is present in the lattice, crudely, and assuming the crystal is without defect. Think of it like this: There was a need for two lithium ions to provide a single oxygen ion with the electrons it needed to fulfill its stable electronic configuration.
Silver hydroxide's case
Let's solve the case of silver and hydroxide.
The hydroxide ion has a charge of $-1$, hence the trouble of determining the charge it will have while bonded ionicallly with silver is gone! Unfortunately, for now, the best way to find the charge out is memorizing it. You may get some fuzzy speculations by drawing Lewis structures, but poly-atomic ions are sometimes exotic chemical species and unpredictable, since many effects are at play to determine the charge of the ion. You can find a list of them here, or more just with a simple Google search.
You have no choice but to memorize the charge of transition metals though, since they tend to assume abnormal electronic configurations. Over time, you'll face it so much you'll learn by heart that $\ce{Ag+}$ is silver's most stable ion.
Now try to apply the logic and reasoning we used back in $\ce{Li2O}$'s case. What would the formula be?

 $\ce{AgOH}$

Useful tips:

Don't forget, that if the poly-atomic ions are more than one in the ionic formula, you should indicate their number with them in parenthesis.
$\Longrightarrow \rm\color{red}{Wrong:} \quad \ce{(NH4)F,~Ca3PO42}$
$\Longrightarrow \rm\color{green}{Correct:} \quad \ce{NH4F,~Ca3(PO4)2}$
Beware of poly-atomic ions that only consist of one atom. These make for the biggest traps and confusions of the students. Good examples are peroxide, superoxide, and azide. Hence $\ce{Li2O2}$ and $\ce{LiO2}$ are correct! Simply because, the ions in them are not the oxide ion, $\ce{O^2-}$, but $\ce{O2^2-}$ and $\ce{O2^-}$, respectively. 2
The oxidation states of the elements, if mentioned, give clue as to what charge their corresponding ion will have. $\ce{Pb(II) \Longrightarrow Pb^2+ ,~Fe(II) \Longrightarrow Fe^2+ ,~Fe(III) \Longrightarrow Fe^3+}$ and so on.

1: It's simplified if I recall correctly, but we take the lattice as formed by ions in the gas state. It helps in defining lattice enthalpy, and I haven't read anywhere that it's not what's really happening.
2: Actually, you can't necessarily make any ionic compounds with any ions you want. Some of them are too unstable to reasonably exist, like lithium superoxide. But it seems the examiners want you to be a magician anyway.


What makes charcoal a good adsorbent to organic and non polar substances while alumina and silica are best for polar substances?

Charcoal (although activated carbon, a specially processed microporous charcoal product is better) is a material that is mostly carbon. The black allotrope of carbon is graphite, which is made of overlapping planar sheets of graphene. Graphite is very nonpolar (since it has only carbon atoms). Thus, nonpolar substances will adsorb. Most organic substances impurities are nonpolar substances (though not all, for example, sugars). Additionally, the graphene sheets are aromatic. The ring current generates a dipolar magnetic field and a quadrupolar electric field that attract other aromatic substances through $\pi$-stacking. Thus graphite is best at removing aromatic substances, which include all organic dyes. Carbon filtering can also remove contaminants like chlorine, although this is likely a reaction leading to chlorination of the surface.

Silica $(\ce{SiO2})$ and alumina $(\ce{Al2O3})$ are polar materials. They are polar for three reasons. 1) They are composed of polar $\ce{Si-O}$ and $\ce{Al-O}$ covalent bonds. 2) The geometries of these minerals especially around the oxygen atoms are not perfectly in-line with VSEPR predictions, so polarity is not canceled by symmetry (see the silica unit image below).

3) Both materials are hygroscopic and absorb/adsorb water from the atmosphere. The $\ce{Si-O-Si}$ and $\ce{Al-O-Al}$ covalent network on the surface is partially hydrolyzed so that the surfaces of these materials actually present $(\ce{-OH})$ groups, increasing the polarity of the surface. Polar substances are attracted to silica and alumina by dipole-dipole interactions and hydrogen bonding.

$$\ce{-Si-O-Si - + H2O -> -Si-OH + HO-Si -}$$


Differentiating between Titration, Buffer, diluted buffer, Deionized H₂O

What is the expected pH of deionized water and what would the pH be after adding a few drops of acid? For both buffer solutions, there is an equivalence point, how would you use that to tell them apart? Usually deionized water is slightly acidic. I kind of mentioned that water doesn't have an equivalence point. Is there a graph without one?

What is the purpose of a buffer? A diluted buffer would have the same characteristic, but in a smaller amount. Water would be missing this characteristic. You should be able to figure out which graph is which now.


According to Huckel's rule Four Criteria for Aromaticity
The molecule needs to be (1) planar, (2) cyclic, (3) fully conjugated, and has (4) 4n+2 electrons.
Your molecule does not seem to be planar (b/c of N and O atoms) but assuming that it has a nearly planar conformation it satisfies all other criteria.
You only need to count one of Oxygen's lone pair and the only Nitrogen lone pair:
10=4*2+2
So it could be aromatic (if it is nearly planar)


The name ‘5-methylbenzyl-3-carbaldehyde-1-oic acid’ is not correct for several reasons. In the following, I focus on the question in the title concerning the choice of ‘carbaldehyde’ or ‘formyl’.

Both, ‘carbaldehyde’ and ‘formyl’, denote the substituent group $\ce{-CHO}$ and can be used, among other ways, to systematically name some aldehydes.

The part ‘carbaldehyde’ is a suffix. It can be used when the $\ce{-CHO}$ group is attached to a ring, for example cyclohexanecarbaldehyde. However, the preferred IUPAC name (PIN) for the analogous benzene derivative is ‘benzaldehyde’, which is derived from the retained name of the corresponding carboxylic acid.

The part ‘formyl’ is a prefix. It can be used in the presence of a characteristic group having priority to be cited as a suffix or when the $\ce{-CHO}$ group is attached to a side chain.

The compound given in the question contains two characteristic groups, $\ce{-CHO}$ and $\ce{-COOH}$. The principal characteristic group must be expressed at the end of a name by means of a suffix or a class name, or implied by a trivial name. Since $\ce{-COOH}$ is the senior characteristic group in this case, the compound is named ‘3-formyl-5-methylbenzoic acid’ (not ‘3-carboxy-5-methylbenzaldehyde’). Note that the $\ce{-CHO}$ group cannot be expressed at the end of a name; it has to be expressed as a suffix (i.e. ‘formyl’).


Prediction of surface atoms in molecule from its graph

The Quantum Theory of Atoms in Molecules (QTAIM) provides a fairly unambiguous answer. First, I would like demonstrate the concept of an Atomic Basin. A region of space belongs to a given atom if the steepest ascent path through the electron density terminates at that atom.  Consider a plane that contains the nuclei of the water molecule:

and the atomic basins could also be represented in 3D:

These atomic basins are the QTAIM definitions of ATOMS in MOLECULES. Note that there are gradient paths for each atom that extend to infinity, hence each atom could be considered an "external atom."  (This is a quite trivial case, of course.)

Now consider formamide in this configuration:

The atomic basins of form amide look like:

You can see that the carbon atom is fairly "inaccessible" in the plane of the molecule, but it does have some extent "peeking" out of the plane of the molecule (here, the maroon-colored basin, sorry for change-up in color):

This is a completely general concept, and so can be extended to any scenario. Here is a molecular graph of thiol on a silver cluster (meant to mimic a surface) (note the bond paths unambiguously show which atoms are bonded to what, even for C-Ag, S-Ag, H-Ag interactions where these "cooked-up" rules of hybridized bonding and "valency" are totally out of their league):

Plotting the atomic basins is just an exercise:

The final answer: you can define "interior atoms" as QTAIM atomic basins that do not have steepest ascent trajectories starting at infinity that terminate at that atom, such as an interior silver atom in the above example. If you are conceptually separating a system into a substrate and an adsorbate, then you could say "exterior" atoms of the substrate include the ones that have bond paths connecting the two.

I must caution that as two or more atoms approach each other, their atomic basins change their shape.  (Of course they do!)  And so one must consider the possibility that atomic basins of "exterior" atoms move out of the way and lead to exposure of what was, in isolation, an interior atom.

(PS:  the atomic basins of peripheral atoms actually extend to infinity.  Here, I have applied cutoffs so that one can actually plot the molecule in a finite region of space.]


When is the Assumption that Enthalpy is Constant over Temperature Change Invalid?

The $G{-}H$ equation describes the effect of $T$ on $\Delta G$ at constant pressure. But, along the fusion contour of $T$ vs $p$, $T$ is a unique function of $p$. So, along this contour, $p$ is changing. So the $G{-}H$ equation is not valid to use for a change of phase such as fusion. The $G{-}H$ equation was developed to describe the effect of $T$ on $\Delta G$ for chemical reactions at constant pressure, usually for the change from separate pure reactants to separate pure products at the standard pressure of $1\ \mathrm{bar}$.


How do we represent a molecule in text form?

The representation of a molecule in text form actually has a name: the chemical formula (pl. formulae).

The formula we use for everyday purposes is an empirical formula. That article should give you more information, but in a nutshell, it is a simple way of identifying a compound by the ratio of its constituent elements. Since ionic compounds and other compounds such as macromolecules can have huge absolute numbers of atoms which are technically bonded but in practice are subdivided into their simplest components, empirical formulae are useful for unambiguously identifying most compounds. Note, however, that we run into trouble when talking about isomers or other compounds with the same elemental ratio but different structures.

For these compounds, we must disambiguate by using a structural formula, which is basically a graph of the molecule. Translating this into text can be done many different ways, using a formal system for specifying branches and stereochemistry. The three most widely used formats for chemical database applications are SMILES, PDB, and CML. You can get more information on all these from Open Babel (alternately OpenBabel).

There is also a middle ground, using a formula similar to SMILES, variously called a molecular formula or condensed formula. It implies structure by counting every atom and splitting individual elements apart to show functional groups like hydroxyl ($\ce{-OH}$) or amino ($\ce{-NH2}$) bonded to the main structure.

I think this is what your question is really about. Yes, in a condensed formula, the order of elements and functional groups matters, because you are trying to communicate information about a compound's structure with a minimum of overhead. It wouldn't make sense to say that butyraldehyde is $\ce{C2H4CHOCH3}$ because the formyl group is at the end. Therefore we say butyraldehyde is $\ce{CH3(CH2)2CHO}$ to better represent the compound's structure. This helps distinguish isomers (but not stereoisomers). Even using the Hill system, one must still be careful.

I hope this clears things up for you on how chemists use different formulae in different situations, and how a condensed formula works.


ChemDraw temporary label for chemnum LaTeX

The ChemNum package documentation that can be found on CTAN specifies:

  Create the scheme and place temporary tags like TMP1, TMP2 and so on where you want the compound labels to be.

A question on TeX.SX specifies:

  When you create a text box and type into it, it will store it as a string correctly in the EPS. However, it was (sometimes) automatically putting it into chemical formula mode - so that in TMP1, the 1 was subscript (TMP1). Even if you fix this manually, it now saves the TMP and 1 separately in the EPS, meaning CHemNum fails to find TMP1.

Thus, it seems all you need to do is create textboxes with the appropriate labels.


Possible erratum in solutions to Orbital Interactions in Chemistry 2ed

I ran the calculations through the MATLAB command window and obtained the same answers as you:

>> H = [-13.60 -14.18 0; -14.18 -13.60 -14.18; 0 -14.18 -13.60]
>> S = [1 0.596 0; 0.596 1 0.596; 0 0.596 1]
>> energies = eig(H,S)

energies =

  -18.2615
  -13.6000
   41.0717

On the other hand, if you assume that $S_{12} = 0$ i.e. $S = I_3$ (3x3 identity matrix), then you get the same result as the authors:

>> energies = eig(H)       % or eig(H, eye(3))

energies =

  -33.6535
  -13.6000
    6.4535

I find it slightly telling that the authors explicitly comment that the bonding orbital is stabilised by exactly as much as the antibonding orbital is destabilised. If you solve the secular equations without explicitly substituting in numbers, you should find that this can only be true if the overlap matrix is set to the identity matrix.

The same is true of part (e); you only get the authors' answer by setting $S = I_3$.

I would consider this an erratum in the provided answers.

To find the eigenvalues, use instead [energies, MOs] = eig(H,S). You can try this block of code online. Note that this website has Octave instead of MATLAB, so energies and MOs need to be reversed:

H = [-13.60 -14.18 0; -14.18 -13.60 -14.18; 0 -14.18 -13.60];
S = [1 0.596 0; 0.596 1 0.596; 0 0.596 1];
[MOs, energies] = eig(H,S);
disp(energies);
disp(MOs);

energies is a 3x3 matrix with the eigenvalues as the diagonal elements. MOs is a 3x3 matrix, where each column vector is the MO corresponding to the relevant eigenvalue (i.e. the left-most column is the MO corresponding to the first eigenvalue in energies).


Absolute limits on solubility of immiscible solutes?

If you speak about solubility, you are out of domain of simple approximations and the only correct answer is the experiment. For the example given, see http://dx.doi.org/10.1111/j.1365-2621.1992.tb01192.x

The problem is the difficulty to quantify the effect of temperature, which is very important here.


Purify gas produced after reactions with sulfuric acid

Pass the gas mixture through Anhydrous Calcium Chloride or Silica Gel. Calcium Chloride is quite hygroscopic and absorbs water vapors from the gas mixture. Alternatively you can use Silica gel, which adsorbs water due to its nature as a good adsorbent.

Edit: To get rid of the acids present in the vapor, bubble the mixture through a concentrated base solution. Acids would react with the base, while $\ce{CO}$ being a neutral oxide, simply doesn't react and can be collected.


Why are amidines more basic than amines?

Note that when you are comparing amidines with amines, you are comparing 2 nitrogen atoms with 1 unless you are comparing it with diamines.  Amidines which have 2 ionizable nitrogen atoms are stabilized  by 2 resonant structures, therefore making them more likely to accept protons (that is more basic).  Similarly, guanidine (not to be confused with guanine which is a nucleic base) which has 3 ionizable nitrogen atoms are stabilized by 3 resonant structures thereby making it even more basic.
For additional reading, check out the basicity of the following drugs.
Betrixaban (which has 1 amidine group)
Pentamidine (which has 2 amidine groups)
Guanethidine (which has 1 guanidine group)
Metformin (a biguanide, that is, having 2 guanidine groups fused together)
Altretamine (which has 3 guanidine groups fused together into a cyclic structure)
Famotidine (which has 1 amidine group and 1 guanidine group).


How to recover C18 column after prolonged exposure to phosphate buffer?

The bad news is that HPLC columns often die because this ill-treatment. We have had the same issues and eventually several expensive columns was discarded due to the negligence of a student. Acetonitrile, when used above 90% with a phosphate buffer, leads to precipitation inside the column. I have great deal of experience with column manufacturing technology. You never add antibiotics to mobile phases. So don't worry about that. The solvents themselves, like ACN, and MeOH ensure that bacteria don't survive inside the column. This is the typical storage solvent after washing the column and ensuring all the buffer is out.

The salvaging procedure can be the following:

Run the column at really low flow rates say, 0.1 mL/min with a high aqueous content (90/10 H2O/ACN). If the pressure is extensively high change the flow direction. Columns have a flow direction marked on them. Try the direction which shows a lower pressure.
Record the pressure, if the pressure slowly goes down, this is a very good news.
If the step 2 is going well, wash overnight at 0.1 mL/min. If all went well, pressure is back to normal in the morning, you are good to go.
Never sonicate the column. It will damage the bed permanently.

P.S. Phase collapse has nothing to do with problem of precipitation inside the column. This is a completely different story. Please don't use that suggestion from LC-GC magazine (link in the comment above). Ron Majors is very experienced but here he talks about another issue.

Phase collapse is not a well understood phenomenon, a better word is de-wetting. Since the C18 phase is very hydrophobic, with highly aqueous mobile phases, the pores (where all the retention occurs) exclude water. Chromatographers, especially those who write in commercial magazines for public readership, tend to do a lot of imaginations.

If the column pressure becomes normal, repeat and old separation. If the retention time is similar, no need to worry. If it is reduced, then try the LCGC suggestion. Pressure is the main test to confirm if all ppt is gone or not.


Is there a difference between equilibrium and steady state?

Yes, equilibrium and steady-state are distinct concepts.

A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant.

Steady-state implies a system that is not at equilibrium (entropy increases). A species is said to be at steady state when the rate of reactions (or more general, processes) that form the species is equal to the rate of reactions (or processes) that remove the species.

In both cases, there are rates ($\mathrm{rate}_1$ and $\mathrm{rate}_2$) that are equal. For an equilibrium, the forward and reverse rate of the same reaction are equal to each other. For a steady state, the rates of processes leading to increase of the concentration of a species are equal to the rates of processes leading to decrease of the concentration of the same species.

$$\ce{A <=>[rate_1][rate_2] B}\ \ \ \ \    vs \ \ \ \ \  \ce{source->[rate_1]C->[rate_2]sink}   $$

For an equilibrium, all concentrations are constant over time. For a steady-state, there is a net reaction, so some amounts change (the amount of source and sink), while at least one species - the one at steady state - has a constant concentration as long as the conditions of steady state prevail.


These values are consistent with the slight solubility of water in octanol and octanol in water.

For example, the solubility of 1-octanol in water is 0.46 g/L. Assuming the density of a saturated solution of octanol in water is approximately that of water, then the concentration in grams per liter can be converted to moles per liter:

$$\dfrac{0.46\ \mathrm{g\ octanol}}{\mathrm{L\ water}}\times \dfrac{\mathrm{mol\ octanol}}{130.23\ \mathrm{g\ octanol}}=3.53\times 10^{-3}\ \mathrm{M}$$

The molar concentration of pure 1-octanol can be calculated from its density:

$$\dfrac{824 \ \mathrm{g\ octanol}}{\mathrm{L\ octanol}}\times \dfrac{\mathrm{mol\ octanol}}{130.23\ \mathrm{g\ octanol}}=6.33\ \mathrm{M}$$

And the $\log P$ is:

$$\log P = \log \left( \dfrac{\mathrm{[octanol]_{octanol}}}{\mathrm{[octanol]_{water}}} \right) \log \left( \dfrac{6.33\ \mathrm{M}}{3.53\times 10^{-3}\ \mathrm{M}} \right)=3.25 $$

Likewise, if we knew the solubility of water in 1-octanol, we could do the same calculation.


Are homemade batteries a practical alternative during a blackout?

... are homemade batteries ... a practical alternative for powering your appliances ...?

No, those type of batteries don't have enough energy to power more than a small light bulb or a transistor radio. If fact the light bulb probably won't stay lit for more than 10 minutes. An LED would be much better.

You need at least 30 car batteries to power your home for a significant amount of time; this is why most people going off the grid have a significant infrastructure cost.

  Are there other, better, non-produce-dependent methods for producing use-able electricity from household materials and minimal expertise?

Other than the alkaline batteries you buy in stores, not that I can think of.


How to get 3D molecular structures for 270 compounds?

There are a few resources that can get you what you want, but you'll probably want to consider some level of scripting automation. A search for 270 molecules isn't enormous, but doing it by hand would still be tedious.

PubChem: Run by the NIH, it offers many types of searching, and 3D coordinates for most compounds.
ChemSpider: Run by the RSC, it allows searching by name and has 3D structures for many (if not most) compounds. I'm less sure on quality, since they don't clarify how they generate their 3D coordinates.
NIH Chemical Resolver: Run by the NIH as well, it converts between different representations, including name to 3D structure.

Each offers automation tools, including Python interfaces called PubChemPy, ChemSpiPy, and CIRpy.

My choice would probably be the chemical resolver, partly because the access is very simple to automate:

https://cactus.nci.nih.gov/chemical/structure/"structure identifier"/"representation"

So you'd set up a URL like: https://cactus.nci.nih.gov/chemical/structure/aspirin/sdf?get3d=true

The result will be in SD format, but it's easy to convert that to XYZ or Gaussian files with Open Babel.


In a half cell reaction, Is the potential halfed?

Those $\pu{-0.74V}$ are measured against a standard hydrogen electrode. That's the universally accepted but totally arbitrary standard.

You don't have a second electrode, hydrogen or other, so the absolute number is totally meaningless. If someone had decided to use the zinc electrode instead, as zero point for the electrochemical series, then the voltage you mean would be zero. Makes no sense.

There is an absolute cell potential (absolute electrochemical series), with a different zero point, but it's totally uncommon, because it's very tricky to measure absolute electrostatic potentials of only a few hundred millivolts. I believe the problem is the calibration, while a platinised platin electrode bubbled with reasonably pure hydrogen is probably the simplest reference standard to reproduce at all.


How does changing volume affect concentration of solutions?

The concentration of $\ce{H2O2}$ doesn't change but when you have more of it in the reaction (50 ml vs. 12.5 ml) there are more moles of $\ce{H2O2}$ meaning more collisions speeding up the reaction.


How are intermolecular forces between non-metallic substances affected when the substance is heated to its melting point?

First of all, multiple intermolecular forces could be present between molecules, and, yes, all forces influence attraction, binding and etc. However, if you study any chemical system, you will see that is usually dynamic.
It all depends on conditions like temperature, pH and especially compound properties, etc. For example, the drug captopril (an antihypertensive drug) could form multiple bonds with the protein angiotensin-converting enzyme (ACE).:

How about one specific atom of a molecule? In theory, one atom could form multiple intermolecular forces at a specific time (let's say picosecond), however, it is not practical - energetically unfavorable. Imagine that atom should be in such a position, an electronic configuration that reduces its potential energy and also form multiple intermolecular bonds. It is possible that such systems exist, but they are not common.
Second answer:
Yes, intermolecular forces have specific values of bonding energy:

For example, it takes less energy to break London dispersion forces than to break hydrogen bonds.
Third answer:
It all depends on the level you want to know... Basically, when you heat molecules (give them more energy) they absorb it and move faster.  Vigorous movements make bonds break. For example, water molecules at 100 °C start to move so fast that almost all its intermolecular forces are broken. Fun fact: you can smell lemons at room temperature because it has volatile molecules (molecules that form very few intermolecular bonds) called Limonene.