Properly speaking, a "metal" is any substance with certain specific properties; chiefly, that it presents the charge transport behaviour of an electrical conductor. Since these properties are phase-dependent, it'd be more accurate to talk about "metallic phases" rather than "metallic substances" - as a given substance can present both metallic and non-metallic phases (to cite a famous, even if still not completely settled, example, hydrogen is thought to present a metallic phases, but not all hydrogen phases are metallic).

There are other ways to conceptualise metallic properties (such as a focus on band structure, or in terms of bonding) but these are pretty much equivalent to the definition that focuses on electrical conductor behaviour, as these are the underlying mechanisms that explain electrical conductors.

In any case, is it correct to identify certain elements as "metals", beyond talking about a specific metallic phase?

The answer is that no, it's not formally correct (IUPAC Gold Book prefers the term "transition elements"), but it is a well-established term, so when we refer to "metals" there is usually little ambiguity. Note that the term metal is also used in accepted definitions that focus on the properties and not on the elements (such as "molecular metal").


Is organic carbon the same thing as organic compounds?

Organic compounds typically refers to particular, identified compounds and/or compound classes.

Organic carbon, ususally expressed as total organic carbon (TOC), is a relevant term in waste (water) management.

In this context the total carbon (TC) is expressed as the sum of total inorganic carbon (TIC) and total organic carbon (TOC).

TIC and TOC are often measured as the amount of $\ce{CO2}$ liberated from a sample.

UPDATE

While Ivan Netetin's remark on the broader context is certainly true, terms like TOC and COD (in German CSB), the chemical oxygen demand, play an important legal role in the question whether and under which conditions waste water may be submitted to the public sewer systems. This involves water directly from procdution processes, as as the elution of material from from landfills.

In Germany, this is specified in the Abwasserverordnung - AbwV. Here, §6(3) states

  Ein nach dieser Verordnung einzuhaltender oder in der wasserrechtlichen Zulassung festgesetzter Wert für den Chemischen Sauerstoffbedarf (CSB) gilt unter Beachtung von Absatz 1 auch als eingehalten, wenn der vierfache Wert des gesamten organisch gebundenen Kohlenstoffs (TOC), bestimmt in Milligramm je Liter, diesen Wert nicht überschreitet.

Translated to non-legalese English, this means: Business-specific limits are considered to be met (= it is legal to submit the water to the sewer) if the fourfold TOC does not exceed the COD given by law.


Why do gases coexist as liquid and vapour?

Vapor-liquid coexistence.

This is possible because changing the system's molar volume $V_\mathrm m$ does not affect the Gibbs free energy of either phase, so equality of the Gibbs free energy (that is, phase coexistence,) can be maintained for a range of $V_\mathrm m$; this is the coexistence region $\mathrm{XCEBY}$.

On the other hand, changing $p$ and $T$ will affect the Gibbs free energy, and if instead we plot this phase diagram on the $p,T$-plane, the entire coexistence region $\mathrm{XCEBY}$ will reduce to a single curve, the liquid-vapor coexistence curve.

In other words: Fix $T$. Only one value of $p$ results in phase coexistence; phase coexistence is a curve on the $p,T$-plane. A range of $V_\mathrm m$ results in phase coexistence; phase coexistence is a region on the $p,V_\mathrm m$-plane.

Phase transitions without coexistence.

This is made possible because the liquid and vapor phases share continuous translational symmetry. In contrast, a phase transition from the liquid phase to a solid, crystalline phase requires breaking this continuous symmetry, resulting instead in the discrete translational symmetry possessed by a crystalline lattice. This shared symmetry permits the possibility of continuously going from one phase to the other without an explicit phase transition.

This possibility is realized past the critical temperature, the temperature represented by the critical point $\mathrm E$. Past the critical isotherm identified by line segment $\mathrm{HE}$ is the supercritical fluid "phase", which shares properties of both liquid and vapor. Isotherms like $\mathrm{GF}$ do not represent a liquid-vapor transition, because the species remains as a supercritical fluid throughout. A liquid-vapor transition that bypasses the coexistence region, like $\mathrm{AFGHD}$, is a continuous transformation from liquid to supercritical fluid to vapor, rather than a discrete phase transition.


Tetragonal (z-in/z-out) in tetrahedral complexes?

Since $D_{h4}$ is not a subgroup of $T_d$, the z-in/z-out distortion you talk about does not give rise to a tetragonal distortion for a tetrahedral structure. It is possible to have tetragonal distortions in tetrahedral complexes, and has been observed in some cases (Bates and Chandler 1975) and (Sharnoff 1965). There is an illustration in Figure 1 of (Virot, Hayn, and Boukortt) which shows what turns out to be the easiest and most common tetragonal distortion for a tetrahedron, the one to the point group $D_{2d}$. So while z-in/z-out is associated with octohedral groups (and is usually the easiest way for them to have a tetragonal distortion), there are other distortions which occur in tetrahedrons.


Acid catalysed dehydration of unsymmetric alcohols

Zaitsev's rule clearly addresses your query. Or you could look at it this way. The product with more hyper-conjugation is more stable. Hence but-2-ene is more stable than but-1-ene. Always see hyper-conjugation for more stable, major product.


Calculating phosphorescence quantum yield using the given datas'

The triplet can only be formed by intersystem crossing from the singlet. The fluorescence yield is $\phi_f=k_f^0/(k_f^0+k_i)$ where $k_i$ is the intersystem crossing rate constant to form the triplet and $k_f^0$ the radiative rate constant for fluorescence. The fluorescence lifetime is $\tau=1/(k_f^0+k_i)=10\cdot 10^{-9}$. The triplet yield is $1-\phi_f$. The triplet lifetime is not important and as the triplet only phosphoresces (you mention no other processes, so no, non-radiative, pathway from triplet) all the  molecules formed must phosphoresce (and it does not matter how long this takes as its the only pathway out of the triplet.)  However, you still have one piece of evidence missing, namely $k_f^0$.


Can I equalise osmotic pressure using two different solutes?

Yes, I think so. Based on our comments above, I believe you want to prepare two solutions that are isotonic with each other (have the same concentrations of different solutes) and allow them to communicate across a membrane that is selectively-permeable to water only.

Tonicity is described here as:

  Tonicity is a measure of the effective osmotic pressure gradient (as defined by the water potential of the two solutions) of two solutions separated by a semipermeable membrane. In other words, tonicity is the relative concentration of solutions that determine the direction and extent of diffusion. It is commonly used when describing the response of cells immersed in an external solution. Unlike osmotic pressure, tonicity is influenced only by solutes that cannot cross the membrane, as only these exert an effective osmotic pressure. Solutes able to freely cross the membrane do not affect tonicity because they will always be in equal concentrations on both sides of the membrane.

I found this nice explanation with illustrations and one example where it shows that you could prepare a 10 mM solution of albumin and a 10 mM solution of glycine, separated by a water-permeable membrane, and no net flow of water occurs in either direction.


Second ionization potential of Gadolinium

The transition corresponds to $[Xe] 4f^{7} 5d^{1} 6s^{2} \rightarrow [Xe] 4f^{7} 5d^{1}$.

Well actually, the second ionization energy being plotted in the graph is only the energy to lose the second electron so the transition is

$[Xe] 4f^{7} 5d^{1} 6s^1 \rightarrow [Xe] 4f^{7} 5d^{1}$.

For all the other elements Pr through Yb, there is no 5d electron, and instead an additional 4f electron. In all cases, it is the 6s electron that is being lost in the second ionization.

The more the nuclear charge is screened, the easier it is for the 6s electron to be lost.

The 5d electron is less effective than a 4f electron at screening the 6s electron from the nuclear charge, because 4f electron density is concentrated much closer to the nucleus (see Fig. 2 of below reference).  Therefore it takes more energy to loss the 6s electron for gadolinium.

For more information on nuclear charge screening see Atomic Screening Constants from SCF Functions. II. Atoms with 37 to 86 Electrons J. Chem. Phys. 47, 1300 (1967).


The functional group in question is a vinyl alcohol, and it is a mistake to consider these groups as isolated. The oxygen is in conjugation with the vinyl group, meaning that resonance favors the entire system attacking. There is no way to leverage conjugation as an energetic driving force if the oxygen attacks by itself.

This is almost the same as your other question:

Hydrolysis of 3,4-dihydropyran


Why is EDTA most soluble at a pH 8? (addition of NaOH)

A disodium salt of EDTA also has two remaining carboxylic-acid groups, which can be deprotonated by addition of base, like $\ce{NaOH}$. By the way, higher pH is generally better especially if you plan to use it for chelation, i.e. cation-capture. The labs I've taught in the past use buffers of pH ~10 with EDTA for complexometric titration experiments.

Deprotonating an organic acid in general improves solubility in water because the electrostatic interactions of $\ce{-O^-}$ and the solvent (the hydrogens on water) are stronger than interactions of $\ce{-OH^0}$ and water.

A similar example would be 4-t-butylphenol, which is nearly impossible to dissolve in water without addition of a base, e.g. $\ce{NaOH}$. In experiments I had to reach pH of about 12 to strip the compound of the alcohol-proton, yielding a negative charge which interacts more strongly with water.

Nonetheless you can still make a decently concentrated (~0.1M) solution of EDTA in water by stirring alone, if you allow a stir-plate to do the work for ~30 minutes.

In short, it's not the addition of $\ce{Na+}$ that matters, but rather the addition of $\ce{OH-}$, to increase solubility.


How can we define a non-spontaneous reaction?

It sounds like your confusion arises from not making a distinction between $\Delta G$ and $\Delta G^\circ$ when describing a reaction as spontaneous or not.

The $\Delta G^\circ$ is the free energy change for the reaction at the defined "standard" conditions of 1 M solute concentrations and/or 1 bar gas partial pressures of both the reactants and products. When we make the general statement that a reaction is spontaneous or not, we are usually referring to whether this $\Delta G^\circ$ is greater or less than 0.

If $\Delta G^\circ < 0$, then the reaction is described as spontaneous, and it will proceed in the forward direction when starting from a mix of 1 M reactants and 1 M products until equilibrium is reached.

In contrast, $\Delta G$ is the free energy change for the reaction in whatever state is of interest. In your case, you have chosen a state with no product present at all. When we say that a reaction is spontaneous based on $\Delta G$, we mean whether it will go in the forward or reverse direction from this specific state of interest, not the standard state.

Obviously, if there is no product present, the reaction cannot go in the reverse direction (forming reactant from product), so all reactions will go in the forward direction if started in a state of $0$ product. That is what you have shown quantitatively using $Q=0$ so that $\Delta G <<0$.

So you are absolutely correct that every reaction in a state in which no product is present is spontaneous. That does not mean, however, that the reaction will necessarily happen anytime soon. Very often, the activation energy barrier is high enough that the reaction rate is extremely slow. Or it is possible that the equilibrium constant may favor reactants so much that even one molecule of product is sufficient to reach equilibrium and the reaction stops very quickly.


Mechanism of the oxidation of alcohols with KMnO4

$\ce{KMnO4/OH-}$ Oxidation

$\ce{H2CrO4}$ Oxidation

The two main differences are

$\ce{H2CrO4}$ Oxidation occurs in acid, $\ce{KMnO4}$ oxidation occurs in base
The source of the nucleophile in the removal of the metal ester. In $\ce{KMnO4}$ oxidation, the $\ce{Mn}$ ester itself extracts the $\ce{H+}$ from the alcohol carbon, while in $\ce{H2CrO4}$ oxidation the nucleophile is the solvent.


Why rotate orbitals in a CASSCF calculation?

The reason for rotation of orbitals (really the reordering of orbitals) is that the starting orbitals are usually not in the correct order for any sort of MCSCF calculation (CASSCF, RASSCF, GASSCF, adding correlation, ...). Doing a single CASSCF calculation with no "preparation" beforehand should force most programs to start with Hartree-Fock-like orbitals; this is the reference wavefunction $\Psi_0$. Hartree-Fock includes almost no static correlation to speak of, which can manifest in an incorrect ordering of the orbitals when compared to MOs that do have static correlation accounted for.

From here:

  Reorder your orbitals if necessary before running the MCSCF. Except for the most trivial cases, the orbitals that belong in the active space are not the orbitals that come straight out of your single-reference calculation. If they were, then you might not be running MCSCF to begin with.

Poor convergence of an MCSCF procedure is a good sign that the active space orbitals are not the correct ones. It is not a matter of the ordering within a space, since the energy is invariant to rotation/swapping within a space (inactive/frozen, active, or secondary/unoccupied). It is that an orbital Hartree-Fock predicts as a frozen core orbital should be a valence-type orbital in the active space, etc.

One of the reasons that MCSCF calculations are multi-step calculations and shouldn't be done in one shot is the generation of (hopefully) better orbitals than HF ones, inspection and any necessary rotation, then the MCSCF calculation. The best starting orbitals for MCSCF are other MCSCF orbitals, but you probably don't have those. Usual choices are MP2 or CI natural orbitals, which gives an indication of the total size of the active space and adds some dynamic correlation to "patch up" the lack of any real correlation in HF orbitals.

In my limited experience, HF orbitals are a fine starting point as long as the order is correct.


How can I synthesize an (E)-configured enone using the Wittig reaction?

The Wittig reaction allows for the construction of alkenes from a phosphorus ylide and an aldehyde (or ketone). Shown in the retrosynthetic sense, keeping in mind that there are two possible Wittig disconnections:

Applying this idea to (E)-5-methylhex-3-en-2-one gives the following possibilities:

Which option is chosen often comes down to which gives the most readily accessible starting materials. However, in this case, the stereochemical outcome is critical, and that is controlled by the nature of the ylide used. Looking now in the forward direction, the top option uses an unstabilized ylide, which tend to give the cis-alkene as the major product. The bottom option uses a stabilized ylide, which changes the stereochemical outcome to favor the trans-alkene.

In practice, many chemists would opt for the Horner–Wadsworth–Emmons variant, which gives the trans-alkene as the major product and avoids the production of triphenylphosphine oxide, a byproduct that can be difficult to remove.


Reviving a 6 Volt lead-sulfuric acid battery

Depending on how long the battery has been sitting empty even adding electrolyte might not help.

Your best bet is to fill the battery with distilled water and put it on a charge, charge it for a couple of days if possible.

Get yourself a hydrometer (used for checking the specific gravity of your electrolyte) and check the water's gravities. After you charge the battery for awhile check them again. If the gravities start to rise, and the liquid is reasonable clear stop charging the battery and let it sit 12-24 hours. (If however the water turns cloudy/grey the battery is scrap get a new one).

After the battery has sat and cooled down, empty the water into a container, and neutralize with soda ash, or baking soda. BE CAREFUL even though what you put it was water, what will come out might/will be corrosive. Dispose of properly.

Now go to your local auto parts store and buy a box of battery electrolyte its usually sold as 1.28 spg, fill the battery with electrolyte re-charge until your gravities come up, and voila, use just re-conditioned a battery.

Be sure to wear rubber gloves / goggles or face shield and any other appropriate safety gear. Remember that when a battery charges it releases hydrogen sulfide gas which is explosive so no open flames sparks etc....

Of course the battery may only last your a couple of months or another year or so, and it may not even come back at all. But I have used this procedure with great success to revive old batteries....and squeeze a bit more life out of them.

Jeff
Industrial battery specialist for 30 years.


The reaction is demethylation of methoxybenzene. You can do it by using an acid which has a nucleophilic counter ion.

It is not actually obvious why the iodine nucleophile attacks the methyl group and not inside benzene ring. The other option would be:

This reaction doesn't happen because otherwise the iodine ion would appear inside the benzene ring and resulting product would be in a very strange conformation (high energy).
More information:
 Clayden, Jonathan; Greeves, Nick; Warren, Stuart; Wothers, Peter (2001). Organic Chemistry (1st ed.). Oxford University Press. p. 434, 589.

As it was correctly stated in the other answer there are also other reagents which can be used to cleave the methyl ether. $\ce{BBr3}$ is a Lewis acid which also can be used in this reaction. A relatively new paper describes a computationally calculated bimolecular mechanism for this reaction.


What's an "amagat"? How is it different from the Loschmidt Number?

I haven't seen any of these particular units before so I could be mistaken, but I imagine it is similar to how we don't say "I have 3 Avogadro's Numbers of water" we would say "I have 3 moles of water". We very well could describe amounts by explicitly naming multiples of the constant, but we instead defined a new unit so that we could describe things more easily.


Finding the frequency factor when raising temperature

This question has no answer.

The Arrhenius equation is:

$$ k = A e^{-\frac{E_a}{RT}}$$

A linearized form of the Arrhenius equation is

$$ \ln{k} = \ln{A} - \frac{E_a}{R}T^{-1}$$

This equation linearly relates $\ln{k}$ to $T^{-1}$: the intercept is $\ln{A}$ and the slope is $-\frac{E_a}{R}$.

To fully define a line, we need two parameters.  This can be two completely specified points that lie on the line, or any single point on the line plus a slope for the line.  For this problem that would mean either (a) two temperatures and two rates, or (b) one temperature, one rate, and one slope.

Using the information we are given:

$$ \ln{k} = \ln{A} - \frac{E_a}{R}T_1^{-1}$$
$$ \ln{2k} = \ln{2} + \ln{k} = \ln{A} - \frac{E_a}{R}T_2^{-1}$$

Any way we combine those two equations will only yield an equation equivalent to

$$\ln{2} = -\frac{E_a}{R}\left(T_2^{-1} - T_1^{-1} \right)$$

in which $\ln{k}$ and $\ln{A}$ have both cancelled out.  That is because the starting two linear equations have the same coefficients for $\ln{k}$ and $\ln{A}$ in each equation.  Similarly, the two equations $2x=y$ and $2x+2=y+2$  can't be solved for $x$ and $y$.

The problem as stated gives us only a slope, but not even a single point that lies on the line.  The rate could double by going from 1,000,000 $\text{s}^{-1}$ to 2,000,000 $\text{s}^{-1}$ (a very fast reaction!) or by going from 0.1 $\text{yr}^{-1}$ to 0.2 $\text{yr}^{-1}$ (pretty slow).  There is no way to find the intercept of a line when we are given only the slope.  Thus, there is no way to solve for $A$ using the information given.


Making elemental silicon is easy, but not for amateurs. Since it is cheap (~$2/kg) there is little reason to do this yourself.

Industrial processes do indeed mix silica with a source of carbon like coal. These react when very hot to give silicon. But the process usually uses an electric arc furnace and controlled conditions with carefully controlled use of gasses (including extra oxygen late in the process to enhance the amount of impurities that gather in the surface slag). It is unlikely that an amateur can easily reproduce this control.

You might get a result at lower temperatures using magnesium. But you will struggle to do this safely without some control of the atmosphere above the mix as magnesium will burn violently in air (or even in carbon dioxide). You might even need to use argon to blanket the mix as even nitrogen will react with magnesium.

So, basically, this is not something to contemplate as an amateur outside a well-equipped lab.


Is the Self Consistent Field an actual field?

The notion of a self-consistent field comes from the orbital approximation - that a multi-electron wavefunction can be approximated as multiple one-electron wavefunctions.

At that point, the question remains - how does each of those electrons interact with the remaining $N-1$ electrons?

Well, there's an electric field from the electron density, right? Maybe you want to call this an electrostatic potential -- but Hartree considered it a spherically-symmetric electric field.

The problem, of course, is that since you don't know the final charge density, you have to make a guess and iterate until it's "self consistent."

Thus, SCF has been associated first with HF theory, and other related methods, including Kohn-Sham DFT.


Will this SN1 reaction result in an equilibrium situation?

Thinking about the SN1 reaction generally, we know that the rate is first order, being proportional to the rate of decomposition of the C-LG bond of the reagent, with the identity/concentration of the nucleophile being completely irrelevant. Your issue, therefore, becomes one simply of the relative leaving group ability of water vs chloride.

Quantitatively, you could think about this on an energy diagram (below). For both the forward and reverse reactions, there is an intermediate carbocation (formation of this is rate limiting), once this carbocation forms there is then a fast recombination of cation with a nucleophile to form a 'stable' species. So, in essence, absolutely the reaction can go in both directions, but in practice often goes in one direction (actually microscopic reversibility dictates that this is almost always the case, but the energetics of one direction are such that reactions aren't realistically reversible).

With respect to your actual example, you might like to take a look in Clayden/Warren, Organic Chemistry. The actual example you're interested in is discussed, along with a discussion of relative leaving group/nucleophile ability.

  Ref: Clayden/Warren, Organic Chemistry.

As a general side point, the website you cited isn't great. There are many errors and general mistakes. You'd be far better looking in a textbook (theres a meta post with good suggestions if you're interested).


Why doesnt the overall rate of a reaction increase if you increase the concentration of a reactant in a non rate determining step?

Let's think about this reaction:

$\ce{ A + B->[k_1][] C + D ->[k_2][] E }$

let's say $k_1 << k_2$ so the first reaction is much, much slower than the second step (at equal concentrations)

Now it get's a little bit tricky since the reaction rates depend on the concentration. Let's assume for now that the concentrations of $A$, $B$ and $D$ are in a way that the reaction rate of the first step is really much faster than the second step.

What happens is that $A$ and $B$ react to $C$ quite slowly, but once you got any $C$ it reacts very fast with $D$ to $E$. This means your concentration of $C$ will always be quite low, since everything that is produced will react to $E$. But this also means that the concentration of $E$ is basically increasing with the rate of the first step. The slow reaction part is limiting the overall rate.

What happens if we increase the concentration of $D$? Well, the second step will be even faster, so all the $C$ that comes out of the first step will be converted to $E$ even faster. But the first step is still slow, so your $E$ concentration will still increse approximately with the rate of the first step.

Only if we would reduce the concentration of $D$ to very low levels (much lower than $A$ and $B$) we would see a difference, in that case the second reaction will get slower than the first one (reaction rate depends on concentration) and will be the limiting step.


How do we count charges in this example of a Redox reaction equation?

If you look at the charges in step 4, the left side adds up to 7 and the right side adds up 2. The MnO4 on the left has a -1 charge, and the 8 hydrogens add a +8 charge, adding up to a +7 charge on the left overall. The Mn on the right has a +2 charge so that is +2.

To balance out this charge, they must be equal. How can +7 be made equal to +2? By subtracting 5. In this case, adding 5 electrons creates a -5 charge that means both sides of the equation have a +2 charge.
You just need to add up the charges like this for a given equation to balance it.


When aluminium is ignited in atmosphere, will it form only aluminium oxide or also aluminium nitride?

According to the paper by Gromov and Vereshchagin [1], it is indeed formed when burning aluminium, but you could have predicted this simply by considering the standard enthalpy of formation of aluminium oxide $\Delta_\mathrm{f}H^\circ (\ce{Al2O3}) = \pu{−1675.7 kJ/mol}$ which will be liberated upon ignition and this energy helps in breaking the strong nitrogen-nitrogen triple bond (bond energy is $\pu{−945 kJ/mol})$ and finally this would form aluminium nitride, which also produces some energy (standard enthalpy of formation is $\pu{−318 kJ/mol}),$ but the combustion here is not a simple one it requires high temperature as mentioned in the paper, I think this is to sublimate $\ce{Al2O}$ which will react with nitrogen and give rise to aluminium nitride.

As HYDR0GEN stated, this combustion isn't simple like throwing a match on aluminium sheet, it needs high temperature. According to the work by Trunov et al. [2, p. 38], as the size of aluminium particles decrease the needed temperature for combustion decreases and that's why researchers who wrote the paper used superfine aluminium powder:

  Figure 5. Experimentally determined temperatures of aluminum
  ignition as a function of the used sample size.

References

Gromov, A.; Vereshchagin, V. Study of Aluminum Nitride Formation by Superfine Aluminum Powder Combustion in Air. Journal of the European Ceramic Society 2004, 24 (9), 2879–2884. DOI: 10.1016/j.jeurceramsoc.2003.09.020. (PDF)
Trunov, M. A.; Schoenitz, M.; Dreizin, E. L. Ignition of Aluminum Powders Under Different Experimental Conditions. Propellants, Explosives, Pyrotechnics 2005, 30 (1), 36–43. DOI: 10.1002/prep.200400083.


your question is good yet lacks the specifics.
there are different so called "energy fields" in which changes occur (entropy, potential ,gibes and so forth).
the thermodynamic field is endless and can be spanned with any variables (using maxwells equations and Legendre transformation)
there is
a sort of an axiom that states that all matter will try to find a way to a lower state of energy.
but that said its not to scientific.

now for the question:
every reaction creates some amounts of electromagnetic waves (as a price for electron "movement") some can be seen with the naked eye as they emit waves in the Uv-vis spectrum some are just infrared radiations of inner states of electron decay to a lower energetic state via vibrational relaxation.

ideally no reaction will occur in the ideal zero as there is no energy to "pay" for any kind of a reaction.

for further read:
quantitive chemical analysis by harris
chapter 17 fundamentals of spectrophotometry.


A thermos bottle containing milk is shaken vigorously

A conceptual approach might work better, since the shaking system is not going to convert mechanical energy into thermal energy as neatly as something like Joule's Paddle Wheel.  As the bottle is shaken, the liquid in it is swirled around into more or less randomly oriented vortices.  These cancel one another out and slow to rest via friction with their container, and the energy they contained has to go somewhere.  As you wrote, it winds up increasing the average velocity, and thereby temperature, of the particles in the milk.

This basic mechanical to thermal energy conversion is the source of the conversion factor from joules to calories.  Joules are used to measure mechanical energy and calories were used to measure thermal energy.  It was the paddle wheel experiment I linked above that first related the two to one another.

The challenge presented by trying to relate the work done on the bottle of milk to the thermal energy that winds up going into the milk is less mathematical than practical.  On any given shake we would first put energy into the bottle (move it upward, let's say) then take energy back out (slow it to a stop) before putting energy back into the bottle (moving back downward) and taking energy out again (slowing to a stop a second time.)  This repeats until we get tired of shaking, and if we measured the difference in all of the energy inputs and outputs that ought to equal the thermal energy added to the milk, but actually doing that for real would be a pain.


Permeability of containers to poop smells

The problem likely is not (so much) the plastic material of your container, either solid, or bag.*)  Assuming there is no contamination on the outer side of the container, most likely the smells are noticed because of closing the container by either knot or lid is not tight enough.
After some search on sibling sites LifehacksSE and OutdoorsSE I learned hikers found multiple solutions to this problem (example).  Assuming you stay on ground and don't climb, a) a poop tube may be suitable for you. In this tube, you carry the waste with you until you find back to a place connected to a waste management / waste water facility.

(credit, source describes both construction and use)
Step #3 of this guide includes an important clue how to reduce the smell.  To quote:

«Before you go out camping, hiking, or climbing, preload the airtight bags with 2-3 small scoops of kitty litter or powder detergent to remove moisture. This will also keep obnoxious odors from leaking out of your poop tube.»

b) The idea to catch liquids and thus smell in a bag filled partially with some of (advanced?) cat litter is commercialized elsewhere, too (example).
Maybe this avenue is working well enough that additional sealing of the lid by an O-rings of suitable diameter and material (viton is a material resistant to many chemicals, but I don't dismantle a vacuum deposition to check this field of application) is not necessary.
*) Although permeabilities of liquids and gases across plastics may be determined experimentally.


Why is water evaporation not more commonly used to cool buildings

"This question seems rhetorically posed: how can you justify calling your solution better and cheaper without having run the numbers? – Aesin 1 hour ago"

A central air conditioning system for a house takes about 3500 watts.

http://michaelbluejay.com/electricity/howmuch.html

1 watt = 1 J/sec so that's about 1,260,0000 J/hour

A gallon of water weighs about 3.7854118 Kg

so at 2260 kJ per kg you can extract 6,780,000 J of energy per gallon of water which is more energy than the air conditioning system even uses and the air conditioning system doesn't extract the amount of heat energy from a building than the energy it uses.

If you run the numbers even with a gallon of water the energy efficiency of water usage is much superior than that of an air conditioning system.

The high heat of vaporization of water is how sweat keeps you cool. Ancient civilizations used to pour water on their tents or tepees or whatever to keep themselves cool. It's cheap and efficient. The superiority of this method when opposed to air conditioning, in dry areas at least, is very obvious. I have yet to see one good reason why it's not used more often.


Material that doesn't melt at 800 deg F, but can be dissolved?

Aluminium and Magnesium
react vigorously with mineral acids.
A lot of heat is evolved.
Try iron/steel wire, which is cheaper and
available in any diameter.
Georg


What is the substance in the new water cooler filter that give bubbles and taste a bit like soap

The most likely cause of a "soapy" taste and sudsing from filtered water is a reverse osmosis filter, for which the usual membrane preservative used to keep the filter from degrading before its first use is plain old glycerin, a common ingredient in and byproduct of traditional soap (though most cleansing products nowadays, even those marketed as soaps, are technically anionic surfactant "detergents", based on alkyl-sulfates). The glycerin is hygroscopic, meaning it will attract and hold water on the membrane surface to keep it from drying out.

An activated carbon filter sometimes has a similar preservative to improve the material's shelf life, or else a binder that allows the manufacturer to mold a block of the carbon rather than loading loose powder into filter housings, which reduces carbon dust in the factory (a major fire and health hazard of these types of plants).

In all cases, the manufacturer usually recommends flushing any new filter with between one and five gallons of source water to wash out these substances, along with any loose filter media or other byproducts of manufacturing. This obviously didn't happen in your case. The glycerin in the water won't kill you; it's an FDA-approved food additive used as a sweetener and "plasticizer" (it's used in cake frosting for instance, to avoid it hardening after it sets). In sufficient doses it's an irritant to the mucous membranes and the GI tract, which can cause nausea and diarrhea, but you'd have to eat almost a kilo of the pure stuff for it to become dangerously toxic.


How many ether metamers are possible for C4H10O?

"Metamerism" is an extremely old, obsolete, and ill-defined property. It is also not educationally or synthetically useful. An additional, and pointless, way to categorise chemical compounds.

See this answer to the question "What is metamerism": https://chemistry.stackexchange.com/a/104576/17952

If you have to answer this in a test, do your best with whatever definition your teacher has provided. And tell him that he should stop teaching such nonsense.


The video shows the "purification" of zinc metal (probably covered with an oxide layer) by heating/reacting it with powdered carbon.

This is a very common process to reduce metal oxides to the corresponding metals. In turn, carbon is oxidized to carbon monoxide and carbon dioxide. The most prominent example is the production of iron from iron oxide in a blast furnace.

However for some metals, this fails and metal carbides are formed instead.

For tin oxide, the reduction with carbon (or coke) is an established industrial process.

Alternatively, the reduction of tin oxide can can be achieved using hydrogen, see Materials Transactions, 2011, 52, 1814-1817 (PDF).


Do chaotropic and kosmotropic affects occur in polar non-aqueous media?

Chaotropic and kosmotropic effects of solutes can occur in any solvent in which interactions between solvent molecules lead to a more ordered structure, like a hydrogen bonding network in water. Such a network will also be present in alcohols like ethanol, although to a lesser degree, because the alcohol molecules have only one proton which can act as a hydrogen bond donor. The Hofmeister (or lyotropic) series for alcohols might still be similar to that for water (not identical), but it will be more different with increasing structural difference between the solvent molecule and water. In fact, the Hofmeister series for water is still not fully understood on a molecular basis, as it is described in this post, and this makes it difficult to predict Hofmeister series for other polar, non-aqueous solvents.


Tip of my tongue: what is the name of this process?

The general term for the process would be "encapsulation".

The Wikipedia article on micro-encapsulation lists different processes by which such encapsulation is achieved. Two which sound similar to your described process are interfacial polycondensation and interfacial cross-linking:

  In interfacial polycondensation, the two reactants in a polycondensation meet at an interface and react rapidly. ... Condensed polymer walls form instantaneously at the interface of the emulsion droplets.

Alternatively, the process your class did might have been ionotropic gelation:

  Ionotropic gelation occurs when units of uric acid in the chains of the polymer alginate crosslink with multivalent cations.


Pyrite does not turn into sulfuric acid. You can burn it with oxygen to give iron oxide and $\ce{SO2}$.

$\ce{4 FeS2 + 11 O2 -> 2Fe2O3 + 8 SO2}$

$\ce{SO2}$ with water gives sulfurous acid. To really get sulfuric acid, $\ce{SO2}$ needs to be oxidised to $\ce{SO3}$ (Contact process), and then reacted with water. Actually with more sulfuric acid to give oleum, which decomposes into two equivalents of sulfuric acid upon dilution with water.

The surface of the pyrite will react very slowly in the acidic conditions in your hand. Very, very slowly. There might be a faint smell of $\ce{H2S}$ recognisable if you handle the pyrite for a while in your sweaty hands.

(Also with water and oxygen from the air, pyrite will slowly react, to finally give sulfuric acid, which will speed up further reaction/degradation. But this typically needs decades of very bad ventilation.)


What are the half reactions for the redox reaction between SO2 and iodine?

Your initial comment exchange made me realize that you have some difficulties in balancing redox equations. Thus, I decided to include some clues for your benefit.

An important part of writing half-reactions is making sure they're balanced by mass and charge. Since most redox reactions are done in aqueous medium, you can always balance $\ce{O}$ by $\ce{H2O}$. By doing so, you contribute extra $\ce{H}$ to the equation so you can balance them by $\ce{H+}$ since it is mostly an acid medium reactions (if they are in base or neutral conditions, you may cancel $\ce{H+}$ by adding the same amount of $\ce{OH-}$ to both sides of the reaction). Finally, cancel the net plus charge by adding $\ce{e-}$s to the appropriate side.

Let's see the easy one first, in your case reduction half reaction where $\ce{I2}$ reduce to $\ce{I-}$: $\ce{I2 -> I-}$. Balance its mass, which gives you: $\ce{I2 -> 2I-}$. Now, balance the negative charges by $\ce{e-}$s. So, you got balanced reduction half-reaction ($\ce{e-}$s are in LHS):
$$\ce{I2 + 2e- -> 2I-} \qquad \mathrm{E^\circ = \pu{0.536 V}} \qquad \text{(1)}$$

Now, see the more difficult second equation, the oxidation half reaction where $\ce{SO2}$ oxidizes to $\ce{SO4^2-}$: $\ce{SO2 -> SO4^2-}$. Its $\ce{S}$ is already balanced, but $\ce{O}$ is not. So, balance it with $\ce{H2O}$, which gives you: $\ce{SO2 + 2H2O -> SO4^2-}$. Now, balance additional $\ce{H}$ by $\ce{H+}$, which gives you a mass-balanced equation: $\ce{SO2 + 2H2O -> SO4^2- + 4H+}$. Now, balance the negative charges by $\ce{e-}$s. So, you got balanced oxidation half-reaction ($\ce{e-}$s are in RHS):
$$\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-} \qquad \mathrm{E^\circ = \pu{0.157 V}} \qquad \text{(2)}$$
If you add (1) and (2) together in order to cancel $\ce{e-}$s, you get the redox reaction you are looking for:
$$\ce{SO2 + 2H2O + I2 -> SO4^2- + 4H+ + 2I-} \qquad \mathrm{E^\circ_{cell} = \pu{0.693 V}} \qquad \text{(3)}$$

The value of $\mathrm{E^\circ_{cell}}$ is positive means the reaction is spontaneous.

(Note: The value of $\mathrm{E^\circ_{\ce{SO2/SO4^2-}}}$ is from Ref.1)

Reference:

J. A. O’Brien, J. T. Hinkley, S. W. Donne, “Electrochemical Oxidation of Aqueous Sulfur Dioxide II. Comparative Studies on Platinum and Gold Electrodes,” J. Electrochem. Soc. 2012, 159(9), F585–F593 (DOI: 10.1149/2.060209jes).


How is the standard electrode potential measured without allowing the electrolyte concentrations to change?

A voltimeter has very high resistance and consumes a very small amount of current, so if you make the measurement quickly, the concentrations are essentially constant for the duration of the measurement.


Frost diagram and acidic/basic conditions.

Write out the half-reactions and you will see that oxidation of S almost always creates lots of protons.  So you can think of Le Chatelier's principle: if there are already lots of protons around, it will be harder to create more of them.  Here's the example for complete oxidation of sulfide to sulfate at pH ~7:

$$\ce{HS- + 4 H2O -> 8e- + 9H+ + SO4^{-2} }$$

Conversely, under alkaline conditions, the equilibrium favors the products even more, because of the "removal" of the protons by the alkaline conditions.


Does BET theory apply to type H3 and H4 isotherm for calculating surface area?

The sample that shows loop of H3 type have (very likely) no micropores (small uptake in low p/p0 range). In this case you can apply BET theory in common p/p0 range (0.05-0.3). The sample that shows H4 hysteresis have micropores (large uptake in low p/p0 range). The BET theory can be applied here, but in a much lower p/p0 range. For activated carbon, for example, one would look for a linear(!) range between p/p0 = 0.02-0.1. Generally, the BET theory is not suitable for microporous materials.


Distillation of diethyl ether and tetrahydrofuran

Sodium and potassium are both drying agents. Being alkali metals they react (often vigorously) with water.

Sodium wire is often used in diethyl ether stills, whilst potassium is used in tetrahydrofuran stills. In general, potassium is more reactive, and hence a better drying agent.

The reason different desiccants are used is due to temperature. THF boils at a higher temperature, meaning a lump of potassium metal can be used in the still ( the lump of K melts once the THF reaches reflux ). Diethyl ether boils much lower, meaning potassium would not melt, for this reason sodium is often used in the form of a wire (giving a much larger surface area).

In respect to the use of benzophenone (or other indicators such), it generally indicates the presence (or absence) of oxygen in the solution. In the past, it was thought that the purple (blue is actually not great) colour was related to the moisture content, however we now know that the colour is related to the oxygen content. For this reason, using the indicator allows us to determine how oxygen-free the solvent is. A purple THF/ether still in which metal is still present indicates that the solvent is both dry and oxygen free,


Why distinguish between polarity and hydrophobicity?

Hydrophilicity/hydrophobicity/lipophilicity/lipophobicity and polarity/non-polarity are terms that overlap to a great extent in simple substances, but which are not strictly equal. The former terms can have a much more involved explanation depending on several factors, even if polarity is usually the largest by far.

Philicity and phobicity are determined by contact angles between substances. It is possible to vary how a same substance gets wetted by exploiting surface patterning, which fiddles with the surface tension between the substances in contact.

Another interesting example I can think of is that some large perfluorocarbons are non-polar, but they are also lipophobic. It is possible to mix water, oil and a perfluorinated solvent, shake, and then watch as three clearly distinct phases separate.


Free energy and kinetic energy for reacting gases

My initial thoughts on this are that you are viewing the Maxwell-boltzmann $velocity$ distribution the wrong way. Let me explain,

In nature there are three main distributions that "things" follow:

1) Boltzmann

2) Bose-Einstein

3) Fermi-Dirac

The first one is a classical result while the latter two are the result of quantum mechanical effects (symmetry of the wavefunction with respect to permutation). These three distribution functions describe the "way" energy is distributed among the energy levels of a molecule/atom/generic particle.

Perhaps a better way to phrase it would be that the macrosystem can be separated into lots of tiny microsystems. Each of these tiny microsystems can have their own particular "state". The individual microstates will combine in a collection or $ensemble$ to give the overall macrostate observed on our scale.

At this point we can vary our definition of how to define the ensemble and indeed there are quite a few buzz terms to learn. A $canonical$ ensemble is one in which the number of particles in each microstate ($n_i$), the volumes of the microstates, and their temperatures are all the same. If you choose the $micro-canonical$ ensemble then you would have $n_i$, $V_i$ and the energy of the $i$th state as being kept constant but not the temperature. A $grand-canonical$ ensemble has constant $n_i$, $V_i$ and chemical potential $\mu _i$.

Now depending on what you choose you will get slightly different equations, but they are all variations on a general unifying theme:

"The frequency distribution of a particular state or outcome out of all those states or outcomes possible, is proportional to
\begin{equation}
e^{-\frac{E}{KT}}
\end{equation}
Which is know as the Boltzmann factor. $E$ is the energy of the state, which varies from state to state."

You have already come across this factor in kinetic theory. !!AS A PARTICULAR EXAMPLE OF A NATURAL PHENOMENA THAT OBEYS THE BOLTZMANN DISTRIBUTION!!.

You can clearly see that the velocity distribution function, the frequency of having a particular velocity, is proportional to the Boltzmann factor, in this case the particular energy we are considering is the kinetic energy, so let $E=\frac{1}{2}mv^2$.
\begin{equation}
f(v)=\sqrt{\frac{m}{2\pi KT}}e^{-\frac{mv^2}{2KT}}
\end{equation}

The Maxwell velocity distribution is just a special case of a Boltzmann distribution.

With this in mind I'm sure you can see that many different phenomena will be described by this distribution. If you wanted to put the Gibbs energy in there (indeed as we do in transition state theory (a personal favourite of mine)) then we can.

This brings me to your next question which is a little easier to answer. The Activation energy can readily be described by a Gibbs potential as indeed it usually is. I think this inherently answers your third question.

If you have any questions I will try to answer them when I can! :)


The comments on the question discuss this may not be about chemistry.
One could argue that colors are chemistry related because
they are associated with chemical substances: like minerals, chlorophyl, rust, dyes etc.
So colors seem to be properties of substances.
And that's fundamentally wrong.
Color is a function of the visual system of the viewer!
If that sounds surprising, think of color blind people, which see different colors.
Also, there are not just us - there are other animals, with different visual spectrum, partly including what we call UV.
Colors are not well defined
Actually, there are more than enough ways, even standards, to define which part of the rainbow is green, exactly - people working with visual media like print design can tell details.
And there are way more details than anyone could expect.
So, without a common definition of the colors, it's not the wavelengths that vary in different sources - it's the colors themselves.

To illustrate that color is more complex when looking closely, below you see a diagram defining color names. It is part of a formal standard which is in heavy use:

From https://en.wikipedia.org/wiki/CIE_1931_color_space


Why are 'sec-' and 'tert-' prefixes printed in italics?

From the ACD Labs page about IUPAC italicization,it is stated that

  Italicizing mainly serves to mark letters which are not involved in the primary stage of alphabetical ordering.

According to the American Chemical Society Style Guide (p. 235), the use of italics for

  positional, stereochemical, configurational, and descriptive
  structural prefixes when they appear with the chemical name or formula.

(There are several prefixes in the list).

These are separated from the chemical name with a hyphen and are not capitalised


Elimination of HCl from isomeric chlorocyclohexadienes

That's a fair question,

Assuming that you use a strong base for dehydrohalogenation you will be deprotonating the most acidic $\ce{H}$ resulting in formation of a carbanion which has the highest stability in compound number 2

You can see that the carbanion formed in compound numer 1 will have resonance with only one of the double bond and hence will be less stable then the carbanion in the compound 2 which has resonance from both the double bond.

Mechanism can be as follow :-

Also;


Thermal decomposition of magnesium bicarbonate

I can quite easily remember from my class XI studies that carbonates of alkali metals decompose on heating to give carbon dioxide and the corresponding metal oxide.

Moreover the thermal stability of the alkali metal carbonates increases with increasing cationic size because carbonate ion is big in size and increased cationic size leads to better bonding and hence greater stability.

In this regard the beryllium carbonate is the least stable alkali carbonate or in other words, an unstable alkali carbonate and readily decomposes into $\ce{BeO}$ and $\ce{CO2}$. $\ce{BeCO3}$ is so unstable that it can only be kept in an atmosphere of $\ce{CO2}$. $\ce{MgCO3}$ is also a bit unstable taking into consideration the above fact.

So the reaction you mentioned basically gives $\ce{MgCO3}$ as the product but $\ce{MgCO3}$ decomposes readily to give $\ce{MgO}$ and $\ce{CO2}$.
$$\ce{Mg(HCO3)2 -> MgCO3 + H2O + CO2}$$
and
$$\ce{MgCO3 -> MgO + CO2}$$
giving
$$\ce{Mg(HCO3)2 -> MgO + H2O + 2CO2}$$


Do formic acid and phosphorus pentoxide react at room temperature?

As noted, the dehydrating action of H2SO4 on HCOOH liberates CO gas. To quote a source:

The sulfuric acid removes the elements of water (i.e., H2O) from the formic or oxalic acid and absorbs the water produced.

As P2O5 is at least as dehydrating as Sulfuric acid, expect the same per the reaction:
$$\ce{P2O5 + 3 HCOOH -> 3 CO (g) + 2 H3PO4}$$


Can triethylamine/pyridine be used in the synthesis of Labetalol?

I disagree with IT Tsoi's answer.  Triethylamine absolutely will compete, and form ammonium salts.  Those salts may react with 4-phenylbutan-2-amine to lead to product, but more likely they will precipitate.  And be general gunk in the reaction flask.  Pyridine will probably be less of an issue.

In general direct alkylation of amines is a pretty ugly reaction with overalkylation products likely.  A better reaction in this vein would be to make the acetamide and alkylate that with the chloroacetophenone, and then deacetylate, or similar strategy.

If you must alyklate the amine, use Hunig's base (N,N-diisopropylethylamine) which is too sterically hindered to alyklate.  Also, add the electrophile to excess nucleophile, if the reaction is faster this will decrease polylaklyation probabilistically, and the carbonyl alpha to the amine may have enough electron withdrawing character to slow down the increase in nucleophiliciity that you get from a more substituted amine.

Further mucking up this route is the phenol with two carbonyls on the ring, that is probably quite acidic as well, and may be deprotonated with the base scavengers.  Use a phosgene protecting group to cover up both the phenol and the aramide, and this problem will be side-stepped.


How does the pH of solutions of salts in which the cation is acidic and anion basic, depend only on the relation of their constants?

Disclaimer: as we will see later, the formula

$$[\ce{H+}] = \sqrt{\frac{K_\mathrm{w}K_\mathrm{a}}{K_\mathrm{b}}} \label{1}\tag{1}$$

is only applicable to highly soluble salts with equimolar rests of monovalent weak acid and weak base such as $\ce{NH4OAc}$, $\ce{NH4HCO3}$, $\ce{NH4HSO3}$, $\ce{NH4NO2}$ and cannot be used for polyvalent electrolytes, poorly soluble salts and when hydrolysis is irreversible (e.g. $\ce{Al2S3}$).

Let's derive formula \eqref{1} in for the salt of the general $\ce{BHA}$ formula. As we agreed, the salt is highly soluble (also true for your textbook example, ammonium acetate), so the main approximation here is that we assume this salt dissociates completely in solution:

$$\ce{BHA → BH+ + A-} \label{R1}\tag{R1}$$

Each ion participates in hydrolytic equilibrium (\eqref{R2}, \eqref{R3}), plus there is autoprotolysis of water, \eqref{R4}. Subtracting \eqref{R4} from \eqref{R2} and \eqref{R3} we get a brutto equation for hydrolysis \eqref{R5}:

$$
\begin{align}
&+\quad&\ce{BH+ + H2O &<=> B + H3O+} \label{R2}\tag{R2}\\
&+\quad&\ce{A- + H2O &<=> HA + HO-} \label{R3}\tag{R3}\\
&-\quad&\ce{2 H2O &<=>[$K_\mathrm{w}$] H3O+ + HO-} \label{R4}\tag{R4}\\
\hline
&&\ce{BH+ + A- &<=>[$K_\mathrm{h}$] B + HA}, \label{R5}\tag{R5}\\
\end{align}
$$

where $K_\mathrm{h}$ is hydrolysis constant. According to the law of mass action, it can be written for \eqref{R5} as such:

$$K_\mathrm{h} = \frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]} \label{2}\tag{2}$$

Now let's multiply both divider and denominator in \eqref{2} by $[\ce{H3O+}][\ce{HO-}]$ so that

$$
\begin{align}
K_\mathrm{h} &= \left.\frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]}\qquad \right|\cdot \frac{[\ce{H3O+}][\ce{HO-}]}{[\ce{H3O+}][\ce{HO-}]}\\
 &= \frac{\color{blue}{[\ce{B}]}\color{red}{[\ce{HA}]}[\ce{H3O+}][\ce{HO-}]}{\color{blue}{[\ce{BH+}]}\color{red}{[\ce{A-}][\ce{H3O+}]}\color{blue}{[\ce{HO-}]}}\\
 &= \frac{[\ce{H3O+}][\ce{HO-}]}{\color{red}{K_\mathrm{a}}\color{blue}{K_\mathrm{b}}}\\
 &= \frac{K_\mathrm{w}}{K_\mathrm{a}K_\mathrm{b}}\label{3}\tag{3}
\end{align}
$$

where $K_\mathrm{a}$ and $K_\mathrm{b}$ are dissociation constants of the conjugated acid and conjugated base, respectively:

$$
\begin{align}
\ce{HA + H2O &<=>[$K_\mathrm{a}$] A- + H3O+} \label{R6}\tag{R6}\\
\ce{B + H2O &<=>[$K_\mathrm{b}$] BH+ + HO-} \label{R7}\tag{R7}\\
\end{align}
$$

Since the dissociation of the salt is complete (see \eqref{R1}), its initial concentration $c_0$ equals to the concentrations of $\ce{BH+}$ and $\ce{A-}$ in solution:

$$c(\ce{BH+}) = c(\ce{A-}) = c_0 \label{4}\tag{4}$$

and the corresponding equilibrium concentrations of the ions participating in hydrolysis are

$$[\ce{BH+}] = [\ce{A-}] = (1-α)c_0, \label{5}\tag{5}$$

where $α$ is the degree of hydrolysis showing the ratio between the amount of hydrolyzed salt $n_\mathrm{h}$ to the total amount of salt $n_0$ ($α = n_\mathrm{h}/n_0$; $α\in [0;1]$). The equilibrium concentrations of the weak acid and base formed during hydrolysis are

$$[\ce{B}] = [\ce{HA}] = αc_0 \label{6}\tag{6}$$

and eq. \eqref{2} for $K_\mathrm{h}$ can be rewritten using degree of hydrolysis:

$$K_\mathrm{h} = \frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]} = \frac{α^2c_0^2}{(1-α)^2c_0^2} = \frac{α^2}{(1-α)^2} \label{7}\tag{7}$$

Note that hydrolysis constant doesn't depend on salt's concentration. Since we are interested in $\ce{pH}$, it is convenient to express $[\ce{H3O+}]$ from $K_\mathrm{a}$:

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

$$[\ce{H3O+}] = K_\mathrm{a}\frac{[\ce{HA}]}{[\ce{A-}]} = K_\mathrm{a}\frac{αc_0}{(1-α)c_0} = K_\mathrm{a}\frac{α}{1-α} \label{8}\tag{8}$$

From \eqref{7} $α/(1-α) = \sqrt{K_\mathrm{h}}$ so that \eqref{8} finally becomes the first equation you asked about:

$$[\ce{H3O+}] = K_\mathrm{a}\sqrt{K_\mathrm{h}} = K_\mathrm{a}\sqrt{\frac{K_\mathrm{w}}{K_\mathrm{a}K_\mathrm{b}}} = \sqrt{\frac{K_\mathrm{w}K_\mathrm{a}}{K_\mathrm{b}}} \label{9}\tag{9}$$

An equation for $\mathrm{pH}$ is obtained trivially by taking negative decimal logarithm on both parts of \eqref{9}.


Free rotation of sigma bonds; diastereomer formation; thermodynamic and kinetic products

You have only considered reaction from the s-trans conformation of 1,3-butadiene, where the 2 double bonds are in a cis relationship about the central single bond.  The s-trans conformation is in equilibrium with the s-cis conformation.  The s-cis conformation can react also react to the terminal olefin, but also  (by the mechanism you described) it can react to produce the z 2-butene as one of its kinetic products.  Hence, all 3 butenes (1-butene and e and x 2-butene) can be formed as the kinetic products.


Is it customary to express heat loss as a negative number?

Your assumptions are correct, and from the wording of the question, the "correct" answer doesn't make sense, unless like you said, the object gained heat, which is possible; people have negative profits all the time. Sign conventions are not flawed in chemistry, but you just have to be careful when wording a question. To answer the title of your question, yes it is not only customary, it is necessary to express heat loss as a negative number, but the question still has to be phrased correctly--using the term "lost" serves as a negation in this use of language, the same way the unary operator in -1 serves as negation for a number. If heat loss were not expressed negatively, it wouldn't make sense because the system of an exothermic reaction must lose energy for its surroundings to gain it.


Choosing Parent Chain, Complexity of Side Chains

You have already reduced the problem to two possible parent chains:
A: 5-(2-methylpropyl)-6-(3-methyloctan-4-yl)dodecane
or
B: 5-(butan-2-yl)-6-(2-methyloctan-4-yl)dodecane

In order to identify the preferred name, we have to read further in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book). In particular, the following potentially relevant rules are actually not helpful in this case:

P-45.2.1 maximum number of substituents cited as prefixes
P-45.2.2 the lower locant set for substituent groups cited as prefixes
P-45.2.3 the lower locant set for substituent groups in order of citation

However, we finally find:

  P-45.5.1 The preferred IUPAC name is the name that is earlier in alphanumerical order (see P-14.5). Alphabetic letters are considered first in the order that they appear in the name; all Roman letters are considered before any italic letters, unless the latter are used as locants or are a part of a compound or composite locant, for example, ‘N’ and ‘4a’. Then, if still there is a choice, numerical locants are considered in the order of their appearance in the name.

Therefore, the preferred name is 5-(butan-2-yl)-6-(2-methyloctan-4-yl)dodecane since ‘butan-yl’ is earlier alphabetically than ‘methylpropyl’.


There are plenty that are smelly and colourless, or are odourless and colourless but I would agree that there are none that are coloured and odourless. Surely it is just coincidental if any of these pairs occur because the chemistry involved in vision and smell are totally different. (Whatever the chemistry of smell is it does not involve photons!) If you were less restrictive in your conditions then most gases will be 'coloured' in the sense that they have electronic absorption in the near ultra-violet and also infra-red at longer wavelengths. Perhaps some insects would see gases such as N$_2$O or O$_3$ as coloured. Also if you were to allow vapours then many compounds can be considered if heated, for example azulene (an isomer of naphthalene) is blue and has a small but measurable vapour pressure if warmed slightly. I don't think I want to smell any, however. The ideal of filling ones nose and lungs with aromatic or dye vapours that then condense is not an attractive one!


Is there a way to efficiently estimate the enthalpy of formation for an organic molecule in computational chemistry?

I would say yes, you could estimate standard enthalpies of formation for molecules more or less accurately by application of quantum and statistical mechanics, although there is a 'but': the usual procedure works for gas-phase heats of formation only.

And thus, if the standard state of the material of interest corresponds to the condensed phase, you have to resort to some procedure which allows you to calculate a condensed phase heat of formation given the corresponding gas-phase one. Such procedures are based on a trivial application of the infamous Hess's law, which for the case of solids can be written as follows
$$
   \Delta H_{\mathrm{solid}} = \Delta H_{\mathrm{gas}} - \Delta H_{\mathrm{subl}} \, ,
$$
where $\Delta H_{\mathrm{subl}}$, the heat of sublimation, is estimated usually from experiementally available data on simillar compounds by a correlation scheme. However, as with any empirically fitted scheme the reliability and the accuracy of any such approach are doubtful.

Alternatively, in many cases, it is enough to compare the gas-phase heat of formations, even if the standard state of material is solid or liquid. To calculate the gas-phase heat of formation for a substance in Gaussian all you need is to run frequencies calculations on the previously optimised geometry of a single molecule.

If you would like to give the gas-phase heats of formations a try, I suggest running G4 calculations first. In the route section just write

# G4

and that would do it. Alternatively, in GaussView's Gaussian Calculation Setup window first choose "Energy" at the "Job Type" tab and then at the "Method" tab choose "Compound..." in the second combobox and "G4" in the third one as shown below.


How to get Jmol to automatically determine and display bond order?

The current definition of the *.cif file does not care about bond order, rather than indicating connectivity of atoms within a molecule.  This is why representing organometallic complexes (like ferrocenes) drawn in sight of what is important for a crystallographer may differ from the one drawn by an organo(metallic) chemist. Among other data, *.cif include the position of the (hopefully correctly assigned) atoms, but nothing about hybridization ($sp^3$, $sp^2$, ...) -- which is one reason why you offer a drawing of the molecular structure of your product to your crystallographer when it comes to characterization by single crystal diffraction or powder diffraction analysis.  In the style of a the ball-and-stick representation, the above drawing is correct.

It is however possible to request Jmol to discriminate between single, double, triple bonds.  One approach to solve such a question is based on tables which relate typical bond lengths to single, double, triple bonds.  To document this, I used the data set indicated by you, as well as Jmol (version 14.13.1, 2017-04-09) under Xubuntu 16.04 LTS with openjdk version "1.8.0_131".  As far as I know, one has to proceed manually, individually for each bond in question.

As a preparation, be sure that Jmol may calculate the bonds, the corresponding menu is accessible via edit -> preferences:

If the option previously was disabled, restart Jmol now.  Subsequently, open your *.cif file and activate the model kit module

which will enable a little drop-down menu on the left hand side, marked by the ribbon in magenta. For now you do not need to access the options there, but hover the mouse over the first, subsequently over the second atom along the bond you are interested.  You will see that first the first atom, then the second atom is encircled.  If you now return hovering from the second atom in question to a position between the two atoms in question (and still on top of the bond in question), and left-click, the bond order may be recalculated, and its display will change if deemed suitable -- as shown in the example of the oxygen atoms of the sulfonamide.

This takes some practice, and frankly, the operator needs to stay vigilant.  In the case of aromatic cycles, the model kit module may correctly recognize the $sp^2$ character of the carbon atoms, but on the expense of hydrogen atoms attached to it.  Hence a result like the following may be expected, where I intentionally halted to "re-assign" the bond order on the last two carbon atoms of this cycle:

Be aware that if you left-click on one of the atoms in question, you may alter its designated element.  In the following, I did so while tapping on the carbonyl oxygen of the acid residue:

In such an instance, you should access the pull down menu of the model kit, and re-assign the atom type. Aberrant hydrogen atoms will be removed by this automatically, too:


Calculating bond energy given the enthalpy value of burning reaction

How can I calculate the enthalpy for those balanced reaction given the data of $\ce{NO +O2 -> NO2}$ enthalpy?

The way the enthalpy data is presented is sloppy (is this the original text of the exercise?). For some types of enthalpy, it is unnecessary to specify the chemical reaction equation because there is a convention. For example, for the enthalpy of formation, it is understood that the substance in question is the only product, the stoichiometry coefficient is one, and that the reactants are elements with coefficients so that the equation is balanced. I am not aware of any such convention for the enthalpy of combustion.

What is the enthalpy of $\ce{2NO(g) + O2(g) -> 2NO2(g)}$?

We can look up the enthalpies of formation for the nitrogen-containining compounds:

$$\Delta H_f(\ce{NO}) = \pu {90.25 kJ/mol}$$
$$\Delta H_f(\ce{NO2}) = \pu {31.18 kJ/mol}$$

From this, we can calculate the reaction enthalpy:

$$\Delta H_r = 2 \cdot \pu {31.18 kJ/mol} - 2 \cdot \pu {90.25 kJ/mol} =  \pu {-118.14 kJ/mol}$$

Which combustion reaction has an enthalpy of -57 kJ/mol?

If we divide the balanced chemical equation by 2, we get:

$$\ce{NO + 1/2 O2 -> NO2}$$

The reaction enthalpy is one half of that of the original reaction, -59.07 kJ/mol. So for the reaction describing the combustion of one mole of NO, the reaction enthalpy is about -57 kJ/mol.

What about the bond dissociation energy?

Everything else in the OP's calculation is correct. If you plug in the correct combustion enthalpy, the calculated bond energy will be correct.


You did the right thing. However the method you used sounds confusing sometimes, it gave me a hard time too. Hess's law says that the resultant enthalpy change in a reaction is same whether it occurs in one or several steps .
We can use that to our advantage by assuming a hypothetical situation where the given reaction proceed in a number of steps for our convinience.
In your example, we disassociated $\ce{SO_2}$ molecule to corresponding standard state constituents and the enthalpy change associated with it will be $-\Delta H_f(\ce{SO_2})$.
$$\ce{2SO_2(g) + O_2(g) ->[-2\Delta H_f(SO_2)] 2S(s) + 2O_2(g) + O_2(g)->[2\Delta H_f(SO_3)] 2SO_3(g)}$$
I took the following reaction through above mentioned 2 steps $$\ce{2SO_2(g) + O_2(g)->[\Delta_r H] 2SO_3(g)}$$
According to Hess law, the Enthalpy of reaction $\Delta_r H$ will be the summation of the all enthalpy changes occuring in the hypothetical intermediate steps.
$$\Delta_r H = -2\Delta H_f(\ce{SO_2}) + 2\Delta H_f(\ce{SO_3})$$


What are radicals and how do they act in radical chain reactions?

free radical is a molecule with one or more unpaired electrons. Despite this definition, d-element compounds with unpaired electrons on d-element, like $\ce{CrCl3.6H2O}$ are not usually considered as free radicals because their unpaired electrons are relatively unreactive. Free radicals are especially reactive, because their reactions usually go through activated complex with relatively small energy. The reasons behind it are not that easy do describe.

In general most reaction with free radicals in organic chemistry falls into two categories: one, where free radical reacts with exposed hydrogen/heteroatom and one where it reacts with exposed multiple bond. In both cases, orbitals of interests are three orbitals of the radical in consideration and two atoms of the organic molecules. MO theory says, that such interaction results in three MOs: bonding, nonbonding and antibonding. With three electrons occupying this orbitals, the energy of the system is mostly same as for the radical and the molecule separated, so the activation energy of the reaction is near zero.

In case of nucleophilic/electrophilic reactions in , the interaction in activated complex are similar, with two electrons on bonding and zero or two electrons on nonbonding orbitals. However, in case of aliphatic derivatives, the reaction usually goes throw sterically hindered state with 5-coordinated carbon  and/or requires generation of hi-energy particles that are not preserved after the molecule reacts. In cases when hi-energy intermediate may be preserved (for example, ionic polymerization), and/or reaction goes through sterically unhindered intermediates (for example, nucleophilic substitution in acyl halides), and/or one of the reagents easily produces active intermediates (for example, nitrating mixture in aromatic substitution) reactions without free radicals as intermediates may proceed swiftly.

Example of chain reaction for aliphatic radical substitution:

$\ce{Cl2 + CH4 -> CH3Cl + HCl}$

in case if $\ce{CH4}$ is presented in huge excess, resulting in reduced rate of multichlorination, the reaction proceeds as following:

Chain initiations: $\ce{Cl2 -> 2.Cl}$
Chain growth 1: $\ce{.Cl + CH4 -> HCl + .CH3}$
Chain growth g 2: $\ce{.CH3 + Cl2 -> CH3Cl + .Cl}$

In case of ionic polymerization

Chain initiation: $\ce{H2SO4 + CH2=CH-Ph -> HSO4- + H-CH_2-\stackrel{+}{C}H-Ph}$
Chain growth: $\ce{H-(CH2-CHPh-)_{n}-CH2-\stackrel{+}{C}H-Ph + CH2=CH-Ph ->} $  $\ce{H-(CH2-CHPh-)_{n + 1}-CH2-\overset{+}{C}H-Ph}$

Both reactions proceed smoothly at room temperature if source of active particles is available (UV-light for breaking $\ce{Cl-Cl}$ bond for the former and concentrated sulfuric acid for the latter)

Compare nucleophilic substitution $\ce{KOH + CH3I -> CH3OH + KI}$

$\ce{KOH -> K+ + OH-}$
$\ce{OH- + CH3-I -> (HO-CH3-I)- -> HO-CH3 + I-}$
$\ce{I- + K+ -> KI}$

As you see, the reaction proceeds through sterically hindered state with 5-coordinated carbon. Compare to acyl halide case $\ce{KOH + CH3COCl -> CH3COOH + KCl}$

$\ce{KOH -> K+ + OH-}$
$\ce{OH- + CH3-(C=O)-Cl -> CH3-C(O^{-})(OH)Cl -> CH3-COOH + Cl-}$
$\ce{Cl- + K+ -> KCl}$

in this case the reaction proceeds through sterically unhindered state with 4-coordinated carbon, so acetylchloride reacts with water at room temperature.

TL;DR : free radicals are particles with unpaired electron, readily available for covalent interactions, they reacts with molecules with easy because they are ones of big family of 'active particles' with low energy of activation of many reactions, and, unlike many other active species, they usually regenerate themselves in reaction, allowing chain reactions to proceed. Other active particles or plainly active substrates also exists, so free radicals are not special. The fact of presence of unpaired electron does not necessary make the particle an active free radical, many relatively inactive free radicals exists, like $\ce{NO}$.

Further reading: MO theory (case of 3-atomic interactions) and kinetics (case of chain reactions) parts of your favorite advance chemistry textbook, parts on reaction mechanisms and active particles of your favorite advanced organic chemistry textbook.

Despite above answer based exclusively on organic chemistry examples, everything stated above is true for inorganic chemistry. However the set of possible kinds of elementary acts, active particles and interactions is much wider than for basic organic chemistry, leading to many special and unique cases and consequently to much less predictable reactions in complicated cases.


Formamide and Hoffmann Bromamide Reaction

I have spent a lot of hours to find the evidence to support OP's idea of formamide may undergoes Hoffmann degradation reaction, but I can't find any. However, I find at least one publication, which may suggest formamide may decompose to formic acid and ammonia (ironically) in hot alkaline medium, before exchanging amide $\ce{N-H}$ to $\ce{N-Br}$ (please keep in mind that $\mathrm{p}K_\mathrm{a}$ of $\ce{HC(O)NH2}$ is about 24 at $\pu{25 ^{\circ}C}$ in DMSO, Evan's pKa Table):

  Ref.1 has described the rates of decomposition rates of acetamide and formamide, which have been studied by individually decomposing them in pressurized hot water. According to the studies, while apparent decomposition rates for both amides were more represented by the second order rather than the first order reaction kinetics, the rate constant for formamide was more than 10 times higher than that for acetamide (Ref.1).

Even though, the work has been done in high temperature and high pressure, the authors observed the decomposition without using acids or bases, and throughout wide range of concentrations. They also found that rate increases with increasing concentrations. Thus, one can expect that in highly alkaline media used in Hoffmann degradation, some amide may hydrolyzed before rearrangement. According to a mechanistic study of the rearrangement, hydrolysis of even aromatic amides under alkaline conditions has been observed:

  It has been found that in the presence of excess alkali at $\pu{30 ^{\circ}C}$, p-nitrobromobenzamide gives approximately as much p-nitrobenzoic acid by hydrolysis as p-nitroaniline by rearrangement; at $\pu{90\!-\! 100 ^{\circ}C}$, however, a $90\%$ yield of p-nitroaniline is obtained.

All of these evidence directed to the theory of hydrolysis of formamide (which is aliphatic and most susceptible to hydrolyze under the conditions due to its size) before it converted to N-bromformamide. Further more, the Ref.3 described a theoretical study of the Curtius rearrangement, which also believed to be going through same mechanism as Hoffmann degradation. Although authors have discussed various pathways of rearrangement including going via oxazirene ion, and identified the most stable intermediate is isocyanic acid during the rearrangement, their conclusion wasn't clear on the hydride ion shift.

References:

M. Okazaki, T. Funazukuri, “Decomposition of acetamide and formamide in pressurized hot water,” Journal of Materials Science 2006, 41(5), 1517–1521 (DOI: 10.1007/s10853-006-4616-1).
C. R. Hauser, W. B. Renfrow, Jr., “The Removal of $\ce{HX}$ from Organic Compounds by Means of Bases. III. The Rates of Removal of Hydrogen Bromide from Substituted N-Bromobenzamides and their Relative Ease of Rearrangement in the Presence of Alkali. The Hofmann Rearrangement,” J. Am. Chem. Soc. 1937, 59(1), 121–125 (DOI: 10.1021/ja01280a029).
A. Rauk, P. F. Alewood, “A theoretical study of the Curtius rearrangement. The electronic structures and interconversions of the $\ce{CHNO}$ species,” Canadian Journal of Chemistry 1977, 55(9), 1498–1510 (https://doi.org/10.1139/v77-209).


Find number grams of the solute in a dissolution given its density and Molarity

Using 1L of solution, you have 16M of HNO3, each mole of which weighs 63g. Therefore, in 1L you'd have 63 X 16 = 1008g. Mass of 1L is given as 1420g from the density.

%mass/mass = 1008/1420 = 70% HNO3


What can possibly turn litmus paper white if there is even no chlorine present?

Chlorine turns litmus paper white by oxidizing it.  There are, of course, other oxidizing agents (like bromine) that can do the same thing; chlorine is just the best known because we associate it with commercial bleaches.


Generally a blank means no sample and just solution. So it would not contain the 10g of soil in this case. I think for soil analysis it's used as a control so you can have something to compare to.

Here's some links I found by doing an internet search.
http://asl.irri.org/lims/index.jsp?page=about&feature=soilanalysis

http://www.soiltest.vt.edu/PDF/lab-procedures.pdf


Why does nucleophilic aromatic substitution occur at C-2 and C-4 of pyridine?

Nucleophilic aromatic substitution on pyridines (or related heterocycles such as quinolines or pyrimidines) regioselectively occurs at the 2- and 4- positions. These can be thought of as being "ortho" or "para" to the nitrogen.

In the mechanism for nucleophilic aromatic substitution, the initial step involves attack of the nucleophile (in this case, methoxide anion) on the π-system of the aromatic ring. Because this step generates a high-energy anionic intermediate which has its aromaticity broken, it is also typically the rate-determining step. In general, the stability of this intermediate is what dictates whether a nucleophilic aromatic substitution is possible or not.

If we consider the resonance forms for this intermediate resulting from attack at each of the three positions, we can understand why attack at C-2 and C-4 is favoured.

When the nucleophile attacks at C-2 or C-4, the resultant intermediate possesses one resonance form where the negative formal charge is located on nitrogen. On the other hand, this is not possible when the nucleophile attacks at C-3.

In general, resonance forms involving negative charges on electronegative atoms such as N and O are much more important contributors than resonance forms with negative charges on carbon. Thus, the anionic intermediate resulting from attack at C-2 or C-4 is better stabilised by resonance, allowing the nucleophilic substitution to take place.


Why is this molecule named 1,1-chloroiodoethane, rather than 1-chloro-1-iodoethane?

The correct name is 1-chloro-1-iodoethane since the locants are placed immediately in front of the prefixes that represent the substituent groups.

Furthermore, since the release of the IUPAC recommendations of 1993, generally all locants are placed immediately in front of the part of the name to which they refer (e.g. cyclohex-2-en-1-one). The corresponding rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

  P-14.3.2 Position of locants

  Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are placed at the front of their names.

  P-15.1.7.1.4 The longest chain is numbered from one end to the other by arabic numbers, the direction being chosen so as to give the lower locants to the substituent groups (side chains) [see P-14.4(f)]. The lower set of locants is defined as the set that, when compared term by term with other locant sets cited in order of increasing magnitude, has the lower term at the first point of difference (see P-14.3.5). The locants are placed immediately in front of the part of the name to which they refer. Identical simple substituent groups are indicated by multiplicative prefixes, such as ‘di’, ‘tri’, etc. [P-16.3.3(b)]. For compound or complex substituent groups (see P-29.4 and P-29.5), the multiplicative prefixes ‘his’, ‘tris’, ‘tetrakis-‘, etc. (P-14.2.2) are used as described in P-16.3.5(a).


Why is the equilibrium constant of water with its vapour equivalent to its vapour pressure?

Lets start by writing the definition of the equilibrium constant, for some general reaction:

$$\sum_i^{n_\text{reac}} a_iA_i = \sum_j^{n_\text{prod}}b_jB_j$$

The above is a general reaction, where $a_i$ and $b_j$ is the stoichiometry and $A_i$ and $B_i$ is the molecules. For this reaction we can write the equilibrium constant:

$$K=\frac{\prod_j^{n_\text{prod}}\frac{B_j^{b_j}}{U_j^{\circ}}}{\prod_i^{n_\text{reac}}\frac{A_i^{a_i}}{U_i^{\circ}}}$$

Here $\prod$ is the symbol that denotes we take the product over a range. The $U_i^\circ$ are a standard unit, to ensure that our equilibrium constant is well defined.

Now if we look at your reaction:

$$\mathrm{H_2O\left( l \right)} \rightleftharpoons \mathrm{H_2O\left( g \right)}$$

We can from the above equations identify that; $a_1 = 1$, $b_1=1$, $A_1=\left[ \mathrm{H_2O} \right]$ and $B_1=p_\mathrm{H_2O}$.

We can thus write our equilibrium constant as:

$$K=\frac{\frac{p_\mathrm{H_2O}}{1\text{ bar}}}{\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}}$$

As you might note, I have also inserted the $U$s. For pressure $U=1\text{ bar}$ and for concentrations $U=1\text{ M}$. Now we need one last ingredient. You might have learned that the activity of liquids is $1$, i.e. the means that we have to set $\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}=1$, this is due to the definition of activity. If interested I would suggest to take a look at this. As our final result we can see that:

$$K=\frac{p_\mathrm{H_2O}}{1\text{ bar}}$$

I.e. we have that the equilibrium constant equals the partial pressure of water in standard units.


Do I have to systematically show the protons in my reaction equation to find the Pourbaix diagram's line equations?

the standard potentials are given in the literature generally in an acidic medium at pH = 0. In this case, we must therefore write the equations in an acidic medium with $ H ^ + $ and not with $ HO ^ - $


How can bromine water be used to distinguish between benzoic acid and salicylic acid?

Salicylic acid is partially activated towards electrophilic aromatic substitution by the hydroxyl group, and slightly deactivated by the carboxylic acid. Overall, it is activated towards the substitution. Benzoic acid is deactivated by the carboxylic acid, so overall, it is deactivated towards substitution. Hence, salicylic acid will react without the need of a catalyst (and upon decarboxylation, the reaction will actually become faster), whereas benzoic acid needs a catalyst, such as $\ce{FeBr3}$, to react.

Hence, salicylic acid will react with bromine water, but benzoic acid won't. Physically, bromine water is orange due to $\ce{Br2}$, but if the bromine reacts, then the orange color will no longer be present. Thus, the solution will turn colorless upon contact with salicylic acid, and will stay orange in contact with benzoic acid. Furthermore, depending on the scale and concentrations, you may also observe the evolution of $\ce{CO2}$.


Biphenyl symmetry group dependence on torsional angle

The $20$ degree structure exhibits the $D_2$ point group and the $90$ degree $D_{2d}$.

The figure below shows an $D_{2d}$ structure (a spiropentadiene) but the similarity in symmetry should be clear (I hope ). (From molecule-viewer.com)

The biphenyl at an angle not $90$ degrees is shown below


Why does activation energy depend on temperature?

Although the question has been already answered, I will like to add a bit more of detail.
The expression you are quoting come from the thermodynamic formulation of the TST that expresses the dependence of the rate constant of temperature as
$$k(T)=\frac{k_\mathrm bT}h\exp\left(-\frac{-\Delta G^\ddagger_0}{RT}\right)$$
and can be partitioned in an entropic and enthalpic contributions
$$k(T)=\frac{k_\mathrm bT}h\exp\left(-\frac{-\Delta S^\ddagger_0}R\right)\exp \left(-\frac{-\Delta H^\ddagger_0}{RT}\right)$$
From this equation you can obtain the activation energy $E_\mathrm a$ using the relation
$$\frac{\mathrm d\ln k(t)}{\mathrm dT}=-\frac{E_\mathrm a}{RT^2}$$
Using this equation you will obtain
$$E_\mathrm a=\Delta U^\ddagger_0+RT$$
Since $H=U+pV$
$$\Delta H^\ddagger_0=\Delta U^\ddagger_0+p\,\Delta V^\ddagger_0+RT$$
For a condensed phase reaction $\Delta V^\ddagger_0\approx0$ and
$$E_\mathrm a=\Delta H^\ddagger_0+RT$$
For a gas phase reaction, $p\Delta V^\ddagger_0=\Delta n^\ddagger RT$.
For a gas phase unimolecular reaction $\Delta n^\ddagger=0$, the result is equal to that the obtained by a condensed phase reaction, while for a second-order reaction $\Delta n^\ddagger=-1$, and
$$E_\mathrm a=\Delta H^\ddagger_0+2RT$$
You can find a more detailed discussion in Keith Laidler, Chemical kinetics 3rd. ed., Harper Collins Publishers, (1987); Jeffrey I. Steinfeld, Joseph S. Francisco, William L. Hase, Chemical Kinetics and Dynamics, Prentice Hall (1989); or Paul L. Houston, Chemical Kinetics and Reaction Dynamics, McGraw-Hill (2001).


Will a copper anode and hydride ion cell work?

TL;DR - What you know from the activity series of metals is still true. You made a mistake in your interpretation of standard electrode potential. The rationale follows:

Standard electrode potentials are written in the reduction direction:

$$\ce{M}^{x+} + x\ce{e- -> M(s)}$$

Standard electrode potentials are measured against the standard hydrogen electrode:

$$\ce{2H+ + 2e- -> H2(g)}$$

Thus for the $\ce{Cu(s)|Cu^{2+}}$ electrode, $E^\circ$ is defined for the followed redox reaction (everything at 1 molar, 1 bar, and in standard states):

$$\ce{Cu^2+ + H2 -> Cu(s) + 2H+}\ \ \ E^\circ = +0.34$$

For the oxidation of copper with acid, the electrode potential would be negative in the direction you wrote it, which translates to a positive $\Delta G^\circ$, since:
$$\Delta G^\circ = - nFE^\circ$$


Is sulfuric acid or hydrochloric acid stronger?

When using $\mathrm{p}K_\mathrm{a}$, one typically does not consider multiple dissociations for polyprotic acids, as the acidity of the conjugate base ($\ce{HSO4-}$ in the case of sulfuric acid) can and should be measured (or calculated) separately. Using that metric, $\ce{HCl}$ is stronger (at least per Wikipedia's acid strength page). One needs to keep in mind that $\mathrm{p}K_\mathrm{a}$ is solvent-dependent, and that the values typically given are for relatively dilute solutions. Acidity can sometimes vary widely with concentration ($\ce{HF}$ being a notorious example). It should also be noted that experimentally measuring acidity for very strong acids is actually quite difficult for numerous reasons.

On the other hand, if you evaluate acid strength by, say, the $\mathrm{pH}$ of the resulting solution, then you'd need to take subsequent dissociations into account. Using the crude approximation that an aqueous "strong acid" dissociates completely, then any diprotic "strong acid" is going to be stronger than a monoprotic one for equal concentrations (assuming a non-zero $K_\mathrm{a}$ for the second dissociation). Of course, this is not accurate, and a proper calculation would take into account not only $K_\mathrm{a}$ values for all dissociations, but also water's auto-ionization. I suspect that $\ce{HCl}$ would still be stronger, given that the $K_\mathrm{a}$ for the dissociation of $\ce{HSO4-}$ is fairly small, but one would need to actually crunch the numbers to be sure.


Is the inductive effect always measured relative to hydrogen?

Your definition is falling apart because you are over-applying it. In the definition, you can see that it says:

  ...transmission of charge through a chain of atoms in a molecule...

It doesn't make sense to apply this in the case of a two-atom molecule, because the inductive effect is used to measure how one bond effects the electron density of other bonds in the molecule. You can assign any bond a dipole moment, but inductive effects are used specifically to explain phenomena observed in organic chemistry, where chains of molecules are (exceedingly) common.

To answer your other question: yes, inductive effects are always measured relative to hydrogen. Linear free-energy relationships are a quantifiable way of measuring a reaction's sensitivity with respect to a certain parameter. The Hammett equation is used to study how sensitive a molecule is in a given reaction to a change in its subsituent's inductive and resonance effects, for which the reference subsituent in all cases is a hydrogen atom. This is perhaps because the $\ce{C-H}$ is the most common bond in all of organic chemistry.

Linear free-energy relationships are particularly useful because similar as to what you said, substituent groups will always have the same relative inductive effects no matter the molecule. A quaternary ammonium group will always have a larger $-I$ effect (and larger $\sigma$ value) than a nitro group than a trifluoromethyl group. A molecule's response to these changes can give great mechanistic insight into organic reactions and the pathways through which they proceed.

A table of the different types of Hammett relations$^{[1]}$:

Their reference reactions:

$\hspace{3.25cm}$

$[1]$ Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science: Sausalito, CA, 2006.


Why is copper(II) sulfate added drop by drop during the Biuret test?

The Biuret Test is a chemical assay that detects the presence of proteins in a sample. The test is not named after any famous scientist, but after an urea dimer called biuret ($\ce{H2NC(O)NHC(O)NH}$). However, funny thing is biuret is not a part of the test at all. The test is called Biuret Test because biuret gives a positive result to the test (See the image below).

The test relies on the characteristic color change (from blue to deep purple or violet; see the image), which confirms the presence of proteins. Biuret isn't a protein, but it gives a positive result to the test, because it has two amide bonds.

Your question (1): How does $\ce{NaOH}$ dissolve the protein exactly?

Proteins has a lot of amide (peptide) bonds, terminal amine and carboxylic acid groups, and along the $\alpha$-helix, a lot of amino acid side chains including amine groups (e.g., lysine), carboxylc acid groups (e.g., aspartic acid), thiol groups (e.g., cystine), etc. All of them can be reacted with $\ce{NaOH}$ to become ionic, which make it more soluble in water. Specifically, recall most of zwitterionic form of amino acids resist to dissolve in water. Regardless, $\ce{NaOH}$ is essential for the test, not only to make protein more soluble, but also to extract $\ce{H}$ from amide $\ce{N-H}$ bond to make $\ce{N}$ negative (Ref.1).

Your question (2): What is the significance of adding copper(II) sulfate drop by drop?

When you perform the Biuret Test in qualitative manner, you need to be careful to see color change clearly, I'd say picture perfectly. Suppose, you have very dilute protein sample. Adding the blue solution into a colorless solution is tricky, because if you are not slow enough, you won't realize there is a color change happens due to both (before and after) solutions been colored but intensities are low (see the near pipette tip of the picture). If you are doing the quantitative test, it doesn't matter how fast you add the solution. After color development, you use UV-Vis-Spectrometer to measure color intensity: at appropriate wavelength to purple color ($\pu{540 nm}$) where absorbance is zero for the $\ce{CuSO4}$ blue color, thus interference is minimal.

Your question (3): How would the reaction vary if copper(II) sulfate was added first and $\ce{NaOH}$ added subsequently, drop by drop?

When you search for the Biuret Test, you'd find out that biuret solution is already alkaline with $\ce{NaOH}$ (see the description of solution in the picture). Thus, I'd say, the order does not matter. Yet, if you want to do a crude test with aqueous $\ce{CuSO4}$ solution and $\ce{NaOH}$ solution, you should be careful, because initial $\ce{Cu(OH)2}$ precipitation might happens. If interested, you may read Ref.2 to see the effects of all solutions on the test, which gives very comprehensive review of the test.

References:

H. Sigel, R. B. Martin, “Coordinating properties of the amide bond. Stability and structure of metal ion complexes of peptides and related ligands,” Chem. Rev. 1982, 82(4), 385–426 (DOI: 10.1021/cr00050a003).
A. G. Gornall, C. J. Bardawill, M. M. David, “Determination of Serum Proteins by Means of the Biuret Reaction,” Journal of Biological Chemistry 1949, 177, 751-766 (http://www.jbc.org/content/177/2/751.short).


Software for chemical reaction simulation

If you mean finding saddle point, then you should consider Gamess and Quantum Espresso packages: free, stable, mature packages, useful for molecule simulations and estimating reaction barriers. They, however, have to be manually compiled and require some skill to setup, as they are meant to be used in POSIX environment (various UNIX clones), not windows, and does not have user friendly result viewer or job builder/scheduler. It is not a problem for experienced linux user, though...

I'm currently using quantum espresso and I found it useful for my cases.


What is pure oxygen (gas) density as a function of T or P?

A colleague showed me a website that can render almost any thermodynamic quantity as a function of another for various substances: http://webbook.nist.gov/chemistry/fluid/.

Using NIST's software allowed me to get an isobar graph of $\rho (T)$ from 55K to 1000K (at 1.0atm).

And $\rho (P)$ from 0 - 800atm (at 273.15K):

It even has an HTML data-point option (and downloadable tab-delimited form). Perfect! In the case of $\ce{O2}$ it uses data from papers in the '80s and '90s (experimental and (from the software) empirical)


How does conversion of a free radical to a carbonyl compound take place? Unsure about mechanism

You seem unsure about the homolysis between C1 and C2, but I speculate to consider the product's stability in this suggested reaction:

For one, two previously unpaired electrons form a $\pi$ bond in acetone.  For the other, you form a benzyl radical which may be described in more than one mesomeric form, for example for a radical bromination reaction of toluene.  Plus the neuter tropyl radical, the seven membered ring structure on the bottom right, result of a rearrangement.

Of course, the benzyl radical won't be persistent, eventually either looses a hydrogen, or combines with an other radical to form a closed shell molecule.


Is there a way to find the mixture type with just the molecular formulas and masses of the solute and solvent?

Formula alone does not give you the structure. But if you have the structure of the molecules, there are molecular/thermodynamic models which take into account group contributions for every section of a molecule. These can help calculate and predict phase behaviour and mixture type. See PC-SAFT as an example.


The toxicity of carbon monoxide in terms of its interaction with haemoglobin

Carbon Monoxide and dioxygen both reversibly bind to the iron atom in hemoglobin.

The two molecules compete for the same binding site.

So if both $\ce{O2}$ and $\ce{CO}$ are present, some of each will bind, but for a given amount of $\ce{O2}$, less $\ce{O2}$ will bind if $\ce{CO}$ is present. The affinity of the Fe for $\ce{CO}$ is much greater than for $\ce{O2}$, so a lower concentration of $\ce{CO}$ results in a given saturation level of hemoglobin compare to $\ce{O2}$.

The same would be true for cyanide which is isoelectronic with carbon monoxide.

A more complete explanation needs to go into the cooperation between the four hemoglobin subunits.  $\ce{CO}$ bound to two of the four subunits causes $\ce{O2}$ to being too tightly bound in the other subunits.  That the oxygen can not be released from the hemoglobin, is another aspect of toxicity.


What is meant by a 'hydrophilic' sphere?

The answer to your question depends on the size level (microscopic or not). But it sounds like you are referring to spheres at a size that you can hold, or larger.

A hydrophilic sphere could simply be a sphere of some solid material (like plastic), coated with a layer of hydrophilic material/substance.

This page on hydrophilic coatings has some information on what this material may be.


A lot of chemical nomenclature is historic and not very systematic.

Common names for chemicals are often determined by history and don't always completely describe the detailed chemical structure or formula. The theory of chemical bonding and structure was not fully developed until the 20th century but we did a lot of chemistry before that. People got used to common names for things and those names persisted even when more systematic names were developed to introduce more clarity about the relationship between names and formula/structure.

For example, acetone is the commonest name for propan-2-one but the common name gives no indication at all about the chemical structure. Many organic compound have very unwieldy systematic names (reflecting the gargantuan complexity of the chemistry of carbon compounds). Fluoxetine (the active ingredient in prozac) is N-methyl-3-phenyl-3-[4-(trifluoromethyl)phenoxy]propan-1-amine and that is too big even to fit on the pill bottle and even the formula, C17H18F3NO, is a bit of a mouthful despite not giving away a lot of structural information about how the atoms are connected.

In the case of simple inorganic compounds like simple hydroxides the situation is similar. The names arose because they described the key chemical parts making up the compound, in this case a metal (as an ion) and a hydroxide ion (OH-). Rather than the unwieldy but more precise names describing the number of ether part of the compounds, chemists apply a little chemical knowledge and keep simpler names for many compounds. Most chemists know that alkali metals (Na, K etc.) form ions with a single positive charge and, therefore, if they form hydroxides they will only contain one hydroxide. Alkaline earth metals (Mg, Ca etc.) form ions with two positive charges and their compounds will contain two hydroxides. This is because all atoms want to reach the lowest possible energy state, which usually is the elecron structure of noble gases. Since most chemists know this, there is no point in using a more complex and longer name. This isn't always true: transition metals can form ions with several different "oxidation states" (which basically means the charge on the ion in simple ionic compounds: iron for example is commonly found both as Fe2+ and Fe3+) and, in these cases the name needs to contain extra information to help the chemist tell them apart.


What chemical reactions could ancient people use to propel projectiles?

The answer depends on what do you mean for ancient people. There are a lot of reactions that could produce gas for ejecting projectiles. The most common was the mixture of sulphur, potassium nitrate $\ce{KNO3}$ and carbon that is known as gun powder and has been reported to be used since the IX century in China.

If you want some less efficient and not used reaction,  you could think of using methane explosion extracted from some swamp. Or some reaction between calcium carbonate and acetic acid. Like this one between sodium hydrogen carbonate:

$$\ce{NaHCO3 + CH3COOH -> NaCH3COO  + CO2 + H2O}$$