Propan-1-ol Heats Objects Faster than Propan-2-ol. Is this an anomaly?

The enthalpy of combustion $\Delta_\mathrm cH$ of a substance can be calculated from the enthalpy of formation $\Delta_\mathrm fH$. For a compound containing only carbon, hydrogen, and oxygen, the general combustion reaction is

$$\ce{C_$a$H_$b$O_$c$ + ($a$ + 1/4 $b$ – 1/2 $c$) O2 -> $a$ CO2(g) + 1/2 $b$ H2O(l)}$$

The corresponding standard enthalpy of combustion is

$$\Delta_\mathrm cH^\circ=-a\Delta_\mathrm fH^\circ(\ce{CO2,g})-\frac12b\Delta_\mathrm fH^\circ(\ce{H2O,l})+\Delta_\mathrm fH^\circ(\ce{C_$a$H_$b$O_$c$})$$

This equation applies if the reactants start in their standard states and the products return to the same conditions. The standard state pressure is $p^\circ=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ (note that most data published before 1982 used a standard pressure of one ‘standard atmosphere’, i.e. $p=1\ \mathrm{atm}=101325\ \mathrm{Pa}$). The definition of standard state makes no reference to fixed temperature; thus, the standard enthalpy of formation and the standard enthalpy of combustion remain functions of temperature. The most widely used reference temperature is $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$.

The following values for the standard molar enthalpy of formation at $p=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ and $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$ are taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

$$\begin{align}
\Delta_\mathrm fH^\circ(\text{propan-1-ol},\ \mathrm l)&=-302.6\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\text{propan-2-ol},\ \mathrm l)&=-318.1\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\ce{H2O, l})&=-285.8\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\ce{CO2, g})&=-393.5\ \mathrm{kJ\ mol^{-1}}
\end{align}$$

During your experiment, the products certainly did not completely return to the initial temperature. Anyway, for the difference in enthalpy of combustion of propan-1-ol and propan-2-ol, the exact state of the products is not relevant (provided that the conditions are reproduced) since the produced amounts of $\ce{CO2}$ and $\ce{H2O}$ are identical. The remaining difference is caused by the difference in enthalpy of formation of propan-1-ol and propan-2-ol.


Henry's law is:

  At a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.

If temperature changes, the Henry's law constant itself changes.


There is no sharp line between ionic, metallic or covalent bonds.

Most of transition metal oxides and sulfates have strong covalent characteristics. For example, the whole reason of using ligand field theory instead of crystal field theory is that the ionic description (CFT) breaks down, and the covalent effects are important in all but the simplest cases.

If you want to take even more covalent molecules, look for organometallic compounds, metal carbonyls, metallocenes, etc.


Spraying another compound to neutralize styrene (which is difficult) or whatever it was is probably not helpful, as it takes too long to spread and other potentially harmful materials are introduced, as one would need to apply the substance in excess.

Having readily installed outlets for water curtains at the plant might be helpful. The quickest emergency measure would have been if every household in the area was equipped with sufficient amounts of respirators and suitable absorbers ("filters") and training the people how to use them correctly (and teaching them that the absorbers have a limited capacity) might give enough time to escape. This would be a good and rather cheap provision for the future. Otherwise ventilation to dilute the materials is the only thing one can do, I'm afraid.


Assuming by thin, you mean a narrow line (not depth). It all depends on how thin you want to go. For a fine width line, a relatively cheap $\ce{CO2}$ laser cutter will get you down to 100–200 µm, but depending on the plastic and how thick it is, the heat produced may warp the plastic near the cut. I think most phone screen protectors are PET film. I've personally made ~200 µm features with one, and if the plastic is thin enough, you can keep the cut width pretty narrow. Things like pulsed UV lasers will give you much finer lines on plastics without thermal effects (as they tend to ablate rather than burn plastics), but they're much more expensive. If you don't have access to either, many companies will contract for this kind of work, even for small jobs.

Another option for coarser cuts that don't have too complex of shapes is a vinyl cutter—basically a plotter with a cutting blade instead of a pen. These are quite cheap, but can't cut all shapes and resolution is on the order of a hundred µm or so, though they don't cause distortions along the edges in most materials.

If you're really serious about it, it's certainly possible to coat photoresist on most plastics and make a mask for reactive ion etching, which will get you down to 1 or 2 µm easily, but it sounds like overkill for you.


Writing pressure equilibrium constant expression for a reaction with a single solid product

The expression $\mathrm{K_p}$ is an expression of the equilibrium of partial pressures. So, it just includes the the partial pressure of the oxygen because that's the only gas.

$$\mathrm{K_P} = \frac{1}{\ce{P_{O2}}}$$

More information about the equilibrium constant $\mathrm{K_p}$ can be found here.


Should I use specific heat to calculate how much fuel is required?

$$\Delta H = mc_p\Delta T$$

Where $\Delta H$ is either energy absorbed or released at constant pressure.
Assuming this specific problem refers to $3\ \mathrm{kg}$ of water:
$$
\Delta H = \left(3000\ \mathrm g\right)\left(4.184\ \mathrm{J\cdot g^{-1}\cdot K^{-1}}\right)\left(80\ \mathrm K\right)
$$

That means that the energy required is: $1\,004\,160$ joules.

Now, simply divide those 1 004 160 joules by the 36 000 joules your fuel releases per gram.
$$
m_\text{fuel}= \frac{1\,004\,160\ \mathrm J}{36\,000\ \mathrm{J\cdot g^{-1}}} = 27.89\ \mathrm g
$$


Are pKa values of acid and conjugate base the same?

Let consider the case of the phosphoric acid.

The phosphoric acid $\ce{H3PO4}$ has $pK_\mathrm{a1}$.
Its conjugate base $\ce{H2PO4-}$ has $pK_\mathrm{b1} = 14- pK_\mathrm{a1}$.

But as $\ce{H2PO4-}$  is at the same time a ( weaker ) acid as well,
it has its own acidity constant $pK_\mathrm{a2}$ independent  on $pK_\mathrm{a1}$ and $pK_\mathrm{b1}$.

Generally, acids like $\ce{HA-}$ have $pK_\mathrm{a2}$ and $pK_\mathrm{b1}$ values.
The former associated with $\ce{HA- <=> H+ + A^2-}$,
the latter associated with $\ce{HA- + H2O <=> H2A + OH-}$


Medical saline has a $\mathrm{pH}$ of 5.5 do to the dissolved $\ce{CO2}$ within the solution as well as factoring in the degradation of the PVC packaging.
I reference the attached article from the International Journal of Medical Sciences from 2013:
Benjamin AJ Reddi, “Why Is Saline So Acidic (and Does It Really Matter?),” Int. J. Med. Sci. 2013, 10(6), 747-750 (doi:10.7150/ijms.5868)(PDF).


will mercury vaporize in boiling water in a flask?

Mercury evaporates to some extent even at room temperature: Vapor pressure: 0.002 mm Hg at 250C (~300 K). "Although the vapor pressure of elemental mercury is low [at room temperature], an atmosphere that is fully saturated with mercury vapor contains approximately 18 mg/m . The levels attainable in
indoor airs at room temperature can therefore greatly exceed
safe levels and result in poisoning." Mercury metal stored under water in an open container will still release Hg vapor into the air, as it is insoluble in water. Since mercury vapor, and worse yet, mercury compounds that might be formed are so toxic, I would not suggest experimenting with it except with proper guidance and with proper equipment, such as a negative-pressure hood feeding into a scrubber and a containment area for possible spills.

At one time, mercury was used to treat STD's and was used by hat-makers to preserve furs, with serious neurological damage to the users. It accumulates through the food chain to cause some fish to be toxic. If feasible, avoid using mercury in the lab and substitute a somewhat safer chemical.


The perspective about ethyl acetate should be revisited.  EtOAc is slightly soluble in water; not infinite (as e.g. ethanol), nor this little as hexane(s).  It is greener than halogenated solvents such as dichloromethane, and the boiling point higher than the for dichloromethane or diethyl ether*) is an advantage during summer.  Thus, EtOAc is an often used polar solvent of extraction.
Equally recall the partition coefficient -- multiple extractions with a small volume of organic are more efficient than one with a large volume as long as after shaking the organic and the aqueous layer separate well from each other. Typically, for the isolation of your raw-product later to be purified by crystallization or chromatography, three extractions just «good enough» may suffice for you.  Except you intend a Craig-Extraction to separate similar compounds from each other (as here), a precursor of the countercurrent chromatography.
The same logic of using multiple small volumes instead of one large is applied when washing the organic phase, e.g. by brine, and back extraction.  Indeed, the presence of salts in the aqueous phase may break an emulsion during an extraction; prior to application, it should be tested on small (e.g., test tube) scale ahead though.
*) Compared to diethyl ether, consider MTBE ether as alternative: it is much less prone to the formation of peroxydes, has a higher boiling point, and is cheaper (per L) than the former, too.


Is the index 1 in the substituent name "1-methylethyl" a redundant index?

I'd not subtract a point for your "mistake," because there is no point to prove you are wrong. However, suppose the group is different than methyl. Then the position matter at that point. Say its bromo-group. So, 1-bromoethyl- and 2-bromoethyl- are two different things. That's why that rule is for. Can't give any favors just for methyl! :-)


Coefficient in front of molecules in a chemical equation

There's the trick: it's all of them.

The coefficients are a ratio. For every two unit of hydrogen molecule, you need one unit of oxygen molecule, and the result is two unit of water molecule.

That unit is often the particle itself, whether it's an atom, a metal ion, a more substantial molecule or even a protein or enzyme. It can also be the volume (for ideal gases under the same conditions), the moles used in reaction, or molarity in solution.

Simply remember that the coefficient applies to the item following it as a whole. As long as the ratio between coefficients is the same, you can use any unit that maintains it.


Why is picric acid more acidic than carbonic acid?

So you want to compare the acidity of different Brønsted–Lowry acids. This means ultimately you want to compare the equilibrium constants for reactions of the type $\ce{HX <=> H+ + X-}$. To do so, you'll have to analyse what stabilisation/destabilisation effects are present in each acid $\ce{HX}$ and each conjugate base $\ce{X-}$. This is easier said than done, especially quantitatively. There are many sorts of possible effects, often acting in a compounded fashion. Fortunately, there are some basic principles to help guide us.

Your question specifically asks to contrast carbonic acid, $\ce{H2CO3}$ and picric acid, more descriptively called 2,4,6-trinitrophenol. This already starts off as being a bit tricky, because they are not very similar substances; of course, the more similar the two substances are when being compared, the easier it will be to pick out the differences in stabilisation/destabilisation effects and the safer it will be to extrapolate from qualitative reasoning. With this in mind, it is best if we build up to the answer by making comparisons between several molecules, making small changes at each step.

First off, let us start with alcohols. These are substances containing hydroxyl groups ($-\ce{OH}$) linked to carbon atoms which are saturated (carbon atoms with four single bonds, i.e. $\rm{sp^3}$-hybridised carbon) and whose three other bonds are only to carbon or hydrogen atoms (in other words, linked to no heteroatoms other than the hydroxyl oxygen). This class represents neither of your substances, but is simple and is related to phenols, so it will be a useful benchmark. A simple alcohol is methanol, $\ce{CH3OH}$, which can act as an acid by the equilibrium:

$$\ce{H3C-OH_{(aq)} <=> H+_{(aq)} + H3C-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=3\times10^{-16}$$

The small equilibrium constant establishes that methanol is a very weak acid in water, and exists almost completely as the unionised reactant. Values of $\mathrm pK_\mathrm a$ around 15 are typical for simple alcohols.

Now let us take a step towards the structures we want, by examining at a hydroxyl linked to an unsaturated carbon atom, specifically a $\rm{sp^2}$-hybridised carbon atom linked to no heteroatoms other than the hydroxyl oxygen. What we have here is an enol. Simple enols are not very stable molecules, but for simplicity we shall disregard that. For the simple molecule propen-2-ol, the equilibrium is:

$$\ce{H2C=C(OH)-CH3_{(aq)} <=> H+_{(aq)} + H2C=C(O^{-})-CH3_{(aq)}}\ \ \ \ \ K_\mathrm a=1.3\times 10^{-11}$$

Evidently, enols are quite a bit more acidic than alcohols. There are two new effects at play here. First, $\rm{sp^2}$-hybridised carbon atoms show an electron-withdrawing effect compared to $\rm{sp^3}$-hybridised carbon. This ability to "suck away" electron density helps stabilise molecules when close to electron-rich regions such as the oxygen atom. This effect stabilises the reactant, but it stabilises the enolate product more (as it has extra electron density to be pulled), which drives the equilibrium to the right. A second effect is resonance. Some of the electrons from the oxygen atom are said to conjugate with the carbon-carbon double bond, creating resonance or canonical structures, which contribute to the true description of bonding in the molecule. Once again, spreading the electrons out contributes to decrease the energy of a molecule. The resonance structures for the reactant and the product are similar, but one of the resonance structures in the reactant contains a formal positive charge on oxygen (a very electronegative atom) and a formal negative charge on carbon (a not very electronegative atom), so this form of conjugation is not quite as favourable. In the product, though there is still a formal negative charge on carbon in one of the canonical structures, there is no formal positive charge on oxygen, making conjugation more favourable. Thus the resonance argument is also a part of the explanation for the acidity.

Next up is phenol, the parent compound of 2,4,6-trinitrophenol. The effects present here are much the same as in the enol above, except that conjugation happens in a larger extent in both the reactant and the product. Experimentally:

$$\ce{C6H5-OH_{(aq)} <=> H+_{(aq)} + C6H5-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=1.1\times10^{-10}$$

This represents only a small increase in acidity compared to propen-2-ol, which suggests conjugation is not the main factor in phenol acidity, but rather the inductive effect of the $\rm{sp^2}$ carbon.

Finally, we've built up to our first structure, 2,4,6-trinitrophenol. The new bit here is the addition of nitro groups ($\ce{-NO2}$), which are strongly electron-withdrawing both by induction and mesomerism (conjugation). Again, these effects decrease the energy of both the reactant and the product, but stabilize the product more because there is extra density to be spread around. As a quick interesting aside, the acidities of 2-nitrophenol and 4-nitrophenol are similar, and both are somewhat larger than 3-nitrophenol, which suggests most of the increase in acidity comes from electron-withdrawing by mesomerism rather than induction, represented by canonical structures similar to these. You also make a good point in mentioning the intramolecular hydrogen bond present only in 2,4,6-trinitrophenol and not its ionised form. Indeed, this helps stabilize the reactant a slight amount, and decreases its acidity. Let's see how all these effects add up:

$$\ce{(O2N)3C6H2-OH_{(aq)} <=> H+_{(aq)} + (O2N)3C6H2-O^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=0.4$$

The acidity increase from the electron-withdrawing effects of the three nitro groups far overwhelms the acidity decrease due to the formation of the intramolecular hydrogen bond. How could you have known this beforehand? Strictly speaking, you couldn't have deduced this from thin air (not without a computer, at any rate!), but generally you are expected to know the situation in similar cases and work out the answer based on previous knowledge. Intramolecular hydrogen bonding usually doesn't play a huge role in increasing or decreasing acidity, at least in simple molecules, as they are relatively weak interactions and so don't contribute massively to increasing/decreasing the energy of a structure. Furthermore, nitro groups are among the most electron-withdrawing there are, so if pulling electrons is going to have any role, it'll show some of its strongest effects with nitro groups.

To prove this answer comes to an end, we reach our final structure, carbonic acid. First, we notice it is related to carboxylic acids, and its acidic group is the $\ce{-COOH}$ in the structure. If you think about it, the carboxyl group looks like an enol where a $\ce{=CH2}$ group is replaced by an oxygen atom. Thus the analysis is similar to the enol. However, there is an increase electron-withdrawing effect due to induction (oxygen being more electronegative than carbon) and mesomerism (in the canonical structures, the negative formal charges are now on an oxygen atom, rather than a carbon atom as was the case for the enol). The instability of carbonic acid interferes, but leaving that aside, its dissociation would be something like:

$$\ce{H2CO3_{(aq)} <=> H+_{(aq)} + HCO3^{-}_{(aq)}}\ \ \ \ \ K_\mathrm a=2.5\times 10^{-4}$$

Comparing everything, the acidity increases in the order:

Simple alcohols < Simple enols and phenols < Carbonic acid and simple carboxylic acids < Picric acid

How would you know carbonic acid is a weaker acid than picric acid? Once again, it's not something you can deduce by pure logic; it requires some benchmarking and comparison with previous data and knowledge. The analysis above would have been helpful, but has its limits. In Chemistry, it is often far easier to explain than predict. Whether you should have been expected to know the answer to this question, I'm not sure. It would depend on how your chemistry course is conducted.


Is ingesting water during swimming harmful?

Could this be potentially harmful depending on the chemical content in the pool?

Probably not, unless you managed to drink quite a bit of pool water. Worldwide, millions (hundreds of millions?) of people have collectively ingested millions (billions?) of liters of pool water - with varying chemistries - over many decades (myself included), and nobody seems to have suffered too much. I believe this falls into the "generally regarded as safe" area.

  Are there studies which advise on the extent of permissible chemicals that can be added to a pool?

Numerous ones, Google away. These guidelines, from the State of Indiana, are representative of the recommended levels of various chemicals present in common swimming pools.


IUPAC advises against using the term racemic mixture because of its ambiguity.

  racemic mixture (obsolete) The term is confusing since it has been used as a synonym for both racemate and racemic conglomerate. Usage strongly discouraged.

The term that applies in this case is racemate.

  racemate An equimolar mixture of a pair of enantiomers. It does not exhibit optical activity.

There are three qualitatively different initial conditions to consider.

1. A racemate of (3S)-methylpentan-2-one and (3R)-methylpentan-2-one, nucleophile is not $\ce{OH-}$

Both enantiomers behave similarly, each yielding a stereocenter at the carboxylic carbon.

Verdict: not optically active but contains two pairs of equimolar enantiomers: $\mathrm{a), d)}$ and $\mathrm{b), c)}$. Even three if you count the reagents. Is a racemate if one relaxes the single pair criterium.

2. A racemate of (3S)-methylpentan-2-one and (3R)-methylpentan-2-one, nucleophile is $\ce{OH-}$

Verdict: not optically active, equimolarity is ensured, a pair of enantiomers. Is a racemate.

3. Either (3S)-methylpentan-2-one or (3R)-methylpentan-2-one, nucleophile is / is not $\ce{OH-}$

Verdict: optically active, does not contain enantiomers. Not a racemate.


Yes, process lactose could still contain protein from the milk. It's equally likely that the casein could contain lactose. Whey is much more likely to contain lactose since the whey and lactose are in the water when the cheese is made.

It would be extremely rare for a baby to be lactose intolerant since lactose is the major sugar they consume. As the child ages, the ability to digest lactose decreases. I think I heard that 5% of Caucasians adults can still digest lactose. Only a few people develop lactose intolerance.


What is 'Retention period' in chromatography and what is its importance?

To answer your three questions in series:

1) The time is a measure of how strong the interaction of a compound is with the column used. The stronger the interaction the longer the compound will stay on the column and this is therefore a way to separate two compounds. For example one molecule with a weak interaction spending a minute on the column, the other spending 5. Collecting fractions of solvent from a liquid chromatography column will yield pure compounds. Note that the time is also a measure for the amount of 'solute' flushed over the column. Many chromatograms also report retention in column volumes of solute. I will not go into detail about that here. It might be worth a separate question.

2) If no peak is found, no retention time can be reported. In chromatography it never (actually almost always) good when a compound fragments/degrades. That means the technique is not suited for purification or analysis. In unique cases, the degradation can be used to analyse whether or not a compound will degrade when interacting with certain materials.

3) In mass spectrometry a retention time in the machine is, as far as I am aware a 'Time of Flight'. Compounds are quickly brought into a gaseous and charged state and are pulled through a magnetic field. Depending on the weight of a compound, the time it flies, is shorter or longer. That is a way of calculating the weight of a molecule. In some cases, for example small proteins, the molecules are purposefully fragmented. The fragments can then be used to reconstruct of what amino acids the protein is made of.

Hope this answers your question! If any clarification is needed, please, do ask :)


Why oxidation number of oxygen in OF2(oxygen difluoride) is +2?

Oxygen is an electronegative element -True.

But oxygen is not the most electronegative element. It has an electro negativity of 3.5 on Pauling scale. That means if oxygen combines with an element which is more electronegative than it will surely possess a positive oxidation state.

Fluorine being the most electronegative element (electro negativity of 4.0 on Pauling scale) will in any case (except in fluorine gas) have an oxidation state of -1. So that's the reason why oxygen has a +2 oxidation state in OF2.

Dioxygen difluoride (O2F2) is another fluoride of oxygen in which oxygen has an oxidation state of +1.

EDIT: Fluorine in its gaseous form (F2) has zero oxidation state.


Is “amount of substance” the same thing as “number of moles”?

"Amount of substance" is one of the seven base quantities of the SI [1]. Its dimension is noted with the symbol N, and its base unit is the mole (mol). Similarly "length" is another base quantity with dimension noted L and with base unit the meter (m).

"Number of moles of [substance X]" rigorously means the numerical value (thus dimensionless) associated with the corresponding "amount of [substance X]" expressed in moles, but from my experience it is often misused for "amount of [substance X]" (thus of the dimension of an "amount of substance"). - Think about how you would respond to the question: "How many meters are there between your house and the nearest bakery?" You could answer "35" or "35 meters", both would be grammatically correct; but strictly speaking the question expects just a numerical answer.

Note that according to the SI, a number of entities (molecules, atoms,..) is dimensionless and the unit of a dimensionless quantity is simply the number one, 1 [1]. My way of reconciling these two views (an "amount of substance" of dimension N and of unit the mole vs. a number of entities, dimensionless and of unit 1) is that the "amount of substance" refers to a combination of an amount (a number) and a given type of entity of a given substance, whereas a number of entity simply refers to the number itself. Why not consider an amount of substance as a dimensionless quantity? I think it is a reminiscence of history, at the time when chemists had to measure amount of substance without knowing how this substance was structured (but I can be wrong).

Regarding the equations, I agree with this other answer by @desc (equation 1 is correct, equation 2 isn't). Importantly: "atomic weight" is dimensionless (it is a ratio between a mass and a reference mass) and should not to be mistaken with "atomic mass". [2]

Finally, regarding the units "gram-atom" or "gram-molecule", these are outdated units which, according to SI were connected with "atomic weights" and "molecular weights" [2]:

  Following the discovery of the fundamental laws of chemistry, units
  called, for example, "gram-atom" and "gram-molecule", were used to
  specify amounts of chemical elements or compounds. These units had a
  direct connection with "atomic weights" and "molecular weights", which
  are in fact relative masses. "Atomic weights" were originally referred
  to the atomic weight of oxygen, by general agreement taken as 16. But
  whereas physicists separated the isotopes in a mass spectrometer and
  attributed the value 16 to one of the isotopes of oxygen, chemists
  attributed the same value to the (slightly variable) mixture of
  isotopes 16, 17 and 18, which was for them the naturally occurring
  element oxygen. Finally an agreement between the International Union
  of Pure and Applied Physics (IUPAP) and the International Union of
  Pure and Applied Chemistry (IUPAC) brought this duality to an end in
  1959/60. Physicists and chemists have ever since agreed to assign the
  value 12, exactly, to the so-called atomic weight of the isotope of
  carbon with mass number 12 (carbon 12, $^{12}$C), correctly called the
  relative atomic mass $A_r$($^{12}$C). The unified scale thus obtained gives
  the relative atomic and molecular masses, also known as the atomic and
  molecular weights, respectively.
  [2]

https://www.bipm.org/en/publications/si-brochure/chapter1.html
https://www.bipm.org/en/publications/si-brochure/mole.html


Electrophilic addition on cyclopropane using sunlight

It should say "in addition to" rather than "instead of".
Cyclopropane suffers badly from ring strain. The bond angles in the ring are 60° rather than the normal ~109.5° when carbon makes four single bonds. In the absence of light, when in contact with $Cl_2$ or $Br_2$ the ring will break and addition occur.
In sunlight, both the expected substitution reactions and the addition reaction will occur.
Can't find the reference I thought I remembered, but here's a site or two discussing it:
https://www.chemguide.co.uk/organicprops/alkanes/halogenation.html
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.07%3A_Reactions_of_Alkanes


Calculating Final Temperature and Volume of Adiabatic Expansion

The final volume V and final temperature T can be determined exclusively from the information for the irreversible path if we assume that the gas continues to expand until it equilibrates at a final uniform pressure of $P=P_{ext}=100$kPa and a final uniform temperature of T.  Let $T_0$ be the initial temperature (300 K), $P_0$ be the initial pressure (800 kPa), and $V_0$ be the initial volume (5 L).  Then the number of moles is $$n=\frac{P_0V_0}{RT_0}$$
The work done on the surroundingsis given by:$$W=P_{ext}(V-V_0)$$
The change in internal energy is $$\Delta U=nC_v(T-T_0)$$
So, from the first law,$$nC_v(T-T_0)=-P_{ext}(V-V_0)$$
The final volume is given by: $$V=\frac{nRT}{P_{ext}}$$
These equations are sufficient to solve uniquely for the final temperature T and the final volume V exclusively in terms of the input data values.


Formation of a bubble which is not a sphere

Surface tension will relax most or all bubbles into a
spherical shape around internal pressure.  So, one solution is to create a situation where surface tension is nonuniform:a bubble adhering to the wall of a glass. Because the glass is not part of the liquid's surface, that
face of the bubble needn't be spherical.

Another is to modulate the pressure the film contains.  Two bubbles in collision is the
easiest way: if the bubbles are the same internal pressure, they
will form a flat wall where they meet.  If they do NOT have the same
internal pressure, the wall curvature is concave toward the
center of the smaller bubble.


Alkyl chain substituent has higher preference than benzyl substituent?

The longest carbon chain is the base name. Benzyl has a "one carbon chain". Compare the two structures below. In the top example, the butane chain, even though it bears a phenyl group, is the main chain. On the bottom example (yours), the phenyl is attached to a "one carbon" chain. therefore, propane is the main chain.


Why is silicon oxide giant covalent structure?

Carbon dioxide is a molecule with double bonds.
Elements of 3+ rows are not happy with double bonds, they prefer single bonds, even in compounds like $\ce{H2SO4}$ 'double bonds' are actually dative single bonds.
The reason for it is that pi-orbitals for 3+ rows are very diffuse, making overall overlapping with p-orbitals of 2-rows extremely ineffective
For this reason, $\ce{SiO2}$ forms structure with only singular Si-O bonds. It happens so that for it the only way to form one is to make 3-d infinite structure.
For similar reason $\ce{N2O5}$ is exactly like this in gas phase while $\ce{P2O5}$ forms tetrahedral $\ce{P4O10}$  and $\ce{SO3}$ loves to form trimers and chains.


Extrapolating from a calorimetry lab to find a new delta T with different volumes and grams of substance

The ratio strategy will definitely work, but I think going back to $q=m\cdot C\cdot \Delta T$ is easier to understand.

You've already figured out a bunch of these constants: $m = 50$, $C = 4.184$ (we assume that the heat capacity of the water doesn't change too much when adding the solid), and $\Delta T = 13.33$.

$$\begin{align}
q & =m\cdot C\cdot \Delta T \\
& = 50\cdot4.184\cdot13.33 \\
& = \ce{2789 J}
\end{align}$$

The heat produced by mixing $\ce{5 g}$ of $\ce{CaCl_2}$ is $\ce{2789 J}$, therefore,
the heat of reaction ($\ce{\Delta H_{rxn}}$) in $\ce{J/g}$ is $\ce{557.7 J/g}$.

When we mix $\ce{10 g}$ of $\ce{CaCl_2}$, the total heat produced will be $\ce{10 g \cdot 557.7 J/g = 5577 J}$.

Now we need to determine the heat change had we mixed in $\ce{10 g}$ of $\ce{CaCl_2}$. We can rearrange the specific heat formula to get $\ce{\Delta T = \frac{q}{m\cdot C}}$. We know $q = 5577$, $m=50$, and $C=4.184$.

$$\begin{align}
\Delta T &= \frac{q}{m\cdot C} \\ & \\
&=\frac{5577}{50\cdot4.184} \\ & \\
&= 26.66\, ^\circ \ce{C}
\end{align}$$

The reason the ratio works is the same logic as the gas laws:

$P_1\ V_1 = C$, $P_2\ V_2 = C$, therefore $P_1\ V_1 = P_2\ V_2$.

$q_1 / T_1 = m\cdot C$ and $q_2 / T_2 = m\cdot C$.

Therefore, $q_1 / T_1 = q_2 / T_2$ by the transitive property of equality


How can the electronic structure of pentaaquanitrosyliron be explained?

As explained in Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of
Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction
Revisited Inorg. Chem. 2002, 41, 4-10, the $\ce{Fe}$ is really 3+ oxidation state.

The $\ce{Fe^3+}$ is spin 5/2 and antiferromagnetically couples to spin 1 $\ce{NO-}$ to yield a S=3/2 state.


Rearrangement of Grignard reagent formed from exocycloproyl vinyl bromide

Grignard reagents are capable of rearranging, especially in this situation where there is significant strain. The specific vinyl bromide you drew here has been studied: J. Org. Chem., 1974, 39 (10), pp 1411–1416. The table of data shown below is hard to read without including more of the text, but it was found that at 66 C in THF, after 5-10 minutes, 41% of the initial Grignard had undergone rearrangement, so this is a fairly likely process.

A potential mechanism for the rearrangement is:


Numbering in IUPAC name of a trisubstituted cyclohexane

You have already correctly identified the principal characteristic group of the compound as carbaldehyde. You have also identified the senior parent structure as cyclohexane, which is correct since the compound has only one possible parent structure. The remaining substituents are correctly expressed as prefixes. The last problem is the correct numbering of locants.
The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The lowest locant is assigned first to the principal characteristic group according to rule (c), i.e. 1-carbaldehyde.
Next, low locants are assigned to the other substituents as a set accoding to Rule (f). However, this could be ‘2-bromo-6-methylcyclohexane-1-carbaldehyde’ as well as ‘6-bromo-2-methylcyclohexane-1-carbaldehyde’ since both names correspond to the locant set ‘2,6’.
According to Rule (g), the lower locant is assigned to the substituent that is cited first as a prefix in the name, i.e. ‘bromo’. Therefore, the correct name is 2-bromo-6-methylcyclohexane-1-carbaldehyde.


Crystal field splitting pattern with strong field and weak field ligands

Well, the split would depend on:
The identity of the metal ion as well as the identity of the ligands themselves. You'd have to consult a spectrochemical series to determine the magnitude of the split caused by the ligands.

Alternatively, you could beam white light through the complex and using the light that appears, determine the wavelength of the light absorbed (the complementary colour) to promote electron(s) from the lower split d orbital to a higher one... This would give you more information as to the magnitude of the split (sometimes, the experimental value will provide an answer that theoretically cannot be determined).


What will be the major enol form of methyl acetoacetate?

The enol forms of beta-diketones and beta -ketoesters are stabilized
by intramolecular hydrogen bonds in its enolic form. Enol structure with greater conjugation is more stable and amajor enol.

update :

In organic chemistry terms, conjugation is used to describe the situation that occurs when π systems (e.g. double bonds) are "linked together".
A conjugated system requires that there is a continuous array of "p" orbitals that can align to produce a π bonding overlap along the whole system.If a position in the chain does not provide a "p" orbital or if geometry prevents the correct alignment, then the conjugation is interrupted (broken) and therefore lost at that point. Due to extra π bonding interactions between the adjacent π systems there is an overall stabilization of the system .This is referred to as the delocalisation energy or the resonance energy or conjugation energy

source:

http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch10/ch10-1-1.html

According to MO theory,1,3-butadiene ,comprised of four overlapping p orbitals are mathematically combined to produce four molecular orbitals.As seen in figure , the four π electrons of a conjugated diene occupy the two lowest energy MOs (the bonding MOs).Specifically, the two electrons that occupy ψ1 are delocalized over four carbon atoms. This delocalization accounts for the observed stabilization energy .

source :
Organic chemistry ,David R. Klein.page no.773

ref :Advanced Organic ,Chemistry FIFTH EDITION ,Part A: Structure and Mechanisms FRANCIS A. CAREY
and RICHARD J. SUNDBERG


Can you synthesize phenol from salicyladehyde?

It is possible to convert salicylaldehyde into phenol in a two-step process. Firstly, salicylaldehyde is oxidized to salicylic acid with a standard reagent like $\ce{KMnO4}$ (see the answers to this question).

In the second step, salicylic acid is thermally decarboxylated to phenol in a simple distillation apparatus. Solid salicylic acid is strongly heated in a round-bottom flask equipped with a short path distillation head with thermometer. Heating is maintained to keep a reasonable distillation rate of the desired product (phenol passes over at a temperature range from 145 to 165 °C). Most by-products of the reaction, like phenyl salicylate, have higher boiling points and remain in the residue. The crude product can finally be purified by vacuum distillation, for example, at a pressure of 30 Torr and a temperature range of 85 - 90 °C (reference).


Difference between valency and oxidation state

Valency is a term used to describe the number of bonds a given atom undergoes in a certain environment. For example, the carbon atom in all of the following substances is tetravalent:

methane
methanol
difluoromethane
formic acid
carbon dioxide

An example for a formally trivalent carbon would be carbon monoxide. Note how valency only counts how many bonds and how many lone pairs an atom has.

The oxidation state is used as a formal measurement on how formally negative an atom should be under standard conditions. For this, each bond is split heterogenously and both electrons are counted as belonging to the electronegative partner only. Then, the electrons on each atom are counted, the difference to a neutral atom of the same kind is established which constitutes the oxidation number. The compound above — while all being tetravalent — have different oxidation numbers:

methane $\mathrm{-IV}$
methanol $\mathrm{-II}$
difluoromethane $\mathrm{\pm 0}$
formic acid $\mathrm{+II}$
carbon dioxide $\mathrm{+IV}$


Generally, inner electrons shield outer electrons. In order to shield you have to be closer to the nucleus, but it's harder for an electron in a higher energy orbital to be closer to the nucleus than an electron in a lower energy orbital. Therefore the shielding effect from inner on outer is a lot more pronounced than outer on inner.

EDIT (after better understanding of OP's question):
One thing to remember is that the inner electron can be on the same side of the nucleus as the outer electron, or it can be on the other side of the nucleus as the outer electron. In one case, the inner electron is being pushed towards the nucleus but the other case results in the inner electron being pushed away from the nucleus. A hand-wavy application of Gauss' Theorem suggests that these effects might cancel out over time. So, the net effect is that the outer electrons have minimal impact on the inner electrons. This argument fails if the outer electrons have greater nuclear penetration, but that doesn't happen by definition.


Significance of phase of atomic orbitals

You have quite a few questions indeed!

  What is the significance, if any, of the phase of the wavefunction for electrons in atoms when the atom is not interacting? Suppose you had a free hydrogen atom in space. Would it make a difference if you switched the phase of its 1s orbital?

No, see this question: Physical intuition behind negative values for wave function? and the answers within.

The phase is not related to any of the physical properties of the atomic orbital. For example the probability density of the electron in the orbital is given by $|\psi|^2$. If we write $\psi_\mathrm{ps} = \mathrm{e}^{\mathrm ix}\psi(r,\theta,\phi)$ ("ps" for phase shifted) it is easy to see that the probability density is independent of the phase chosen $x$:

$$|\psi_\mathrm{ps}|^2 = \psi_\mathrm{ps}^*\psi_\mathrm{ps} = \mathrm{e}^{-\mathrm ix}\psi^*(r,\theta,\phi)\mathrm{e}^{\mathrm ix}\psi(r,\theta,\phi) = |\psi^2|$$

so the phase shift has not done anything. Likewise, the energies etc. are independent of any constant factor that is placed in front of the wavefunction. This is easy to show using

$$E = \frac{\langle \psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}.$$

  What about if you had a multi-electron atom? Would there be a consequence of switching the phase of the 1s orbital but not of the other orbitals? My guess is that, if this is at all possible, then yes this would make a difference due to the penetration of higher orbitals- i.e. the 2s orbital has some penetration into the 1s orbital. Thinking about it in this way, I think atoms are most stable when all the s orbitals are in the same phase and the p orbital phases are aligned. Otherwise you would get cancellation of the wavefunction.

In hydrogen the 1s and 2s orbitals are orthogonal to each other (in general all AOs are orthogonal to all other AOs). This is because the 2s orbital has an inner radial node and two regions which are phased differently (conventionally, positive + negative; however you can draw it negative + positive and the overlap would still be zero).

  How do we form the bonding and antibonding MOs? Is it becaue each orbital is actually in a superposition of being 'phase 0' and 'phase pi' which is why you get both bonding and antibonding MOs?

The absolute phase of the orbital has nothing to do with it, but rather the relative phase. See the linked question above.

  So here we have two opposite phases existing together. Is there a significance of this? Or is it perhaps again only important in the interaction of p orbitals?

I'd be inclined to say that there's no real significance, except that that's how the maths works out. More fundamentally the presence of a node (in this case an angular node) means that there must be regions of different phases. The same applies to the 2s orbital which has a radial node instead of an angular node. This is analogous to the solutions to the particle in a box: the ground state solution has no nodes and is therefore positive throughout, but the first excited state has one node and therefore has one positive and one negative region.


Proper disposal of calcium hydride/calcium hydroxide

The safest way to get rid of calcium hydride is to pour over ice. The water quenches the reagent itself, with the ice helping to control the exotherm (if you just add water, a lot of heat and gas is liberated, having a tendency to spray hot chemicals over the whole hood).

Once everythings quenched, it's helpful to add some hydrochloric acid before disposing of down the sink to prevent it clogging the drains.

Note that whilst clean calcium hydride is safe (once quenched) to be disposed of via the normal drainage, you must ensure any chemicals are washed off first (hexane works well), especially odorous things such as amines which you may have been drying with the calcium hydride.


Your idea is in principle correct. The reaction you mentioned works well, you break one H-H bond and create 1 C-C bond (triple bond, in fact) and two C-H bonds. The energy is released. And in most crude approximation it also says where the activation energy barrier comes from - you have to break a H-H bond.

As for the paradox - that is the principle of chemistry, to make bonds we wish for. If it means increasing the energy level, thus being less favored by nature, you have to pay for it. By sacrificing other bonds. Those are the highly reactive chemicals humbly stated above reaction arrow, which quite often do all the magic.

As the most revealing example in this I consider the Nucleophilic Acyl Substitution reaction, where you have a series of derivatives of increased stability. You can always and simply make the more stable out of the less stable, having wide choice of reactions. And there are just a few to climb up, using very nasty chemicals.


Gases produced by bacterial decomposition of human urine

Carbon dioxide and ammonia when treated with bacteria along with water vapour


How do I determine if a process is reversible without numeric information?

Any process is irreversible if it involves:

Friction
Heat transfer across a finite temperature difference
Expansion when there is a finite pressure difference
Mixing of difference substances under any real circumstances
Anything else where work is not being maximized

So air friction against the balloon rising at a non-zero velocity makes the process irreversible.

Heat transfer from the balloon when the inside temperature is different from the outside temperature makes the process irreversible.

For more information see: http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node34.html


Molecular orbital theory & predicting the stability of a molecule?

You need to mix the orbitals, populate them with the electrons and see if you have net bonding.

Eg:  H + H two 1s orbitals mix to form sigma and sigma*.  Two electrons total, both occupy the sigma orbital, two more electrons in bonding than antibonding orbitals, the compound is stable.

Eg:  He + He; same mixing as above.  Four electrons, two in the sigma, two in the sigma*.  Since there are as many bonding electrons as as antibonding, there is no net bond.  He2 is not possible.

Eg:  He + H; same mixing as above.  Three electrons, two in sigma, one in sigma*.  One more electron in bonding than antibonding.  He-H forms a very weak bond.  Please note the diagram is for He2+ but the He-H is very similar

Eg: Li + H; Li has 1s + 2s, while H has 1s.  This mix to form a sigma orbital from H1s+Li2s, a sigma* orbital and H1s-Li2s, and a non bonding orbital from Li1s (lower in energy than the sigma).  Four total electrons.  Nonbonding sigma is occupied, and then the sigma orbital is occupied.  Net effect: Li-H forms a stable bond.

You might have noticed that the first three diagrams had identical structure, even if their energy levels were different.  This is a property of quantum mechanical symmetry, and solving the problem only requires knowing a few diagram types and filling the electrons in.  Consider the second row.  The diagram is as :

You simply count the valence electrons and fill in the diagram.  Then you subtract the number of bonding orbital electrons from antibonding, and if you have more, you will have a bond.

A couple of webpages are very good for further explanation:

Source of the diagrams:
http://www.chem1.com/acad/webtext/chembond/cb08.html

and
http://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html


Why is the molecular geometry of a molecule with 5 bonds not uniform?

You may want to look into VSEPR theory. To a good first order approximation you can think of electron pairs and external atoms on a central atom as taking up about the same space.
This explains why water's bond angle (104 degrees) is very close to the 109.5 degrees exact equidistant angle associated with methane, or $\ce{AlCl4+}$.

Your example with $\ce{SbCl5}$ is a good example where the theory gets a bit more complicated than simply maximizing distance between steric centers. The bonds to the chlorines are not equal. And yes this is how the chemical would look: see the Wikipedia article on orbital hybridisation.

Now, as your example has 5 identical groups, inter-conversion can occur, where the axial group "switches" with an equatorial group. But at any moment there will be two distinct chemical environments around the compound.


How can the ground state electron configuration for a d3 complex correspond to an F term symbol?

Free ion

The ground-state term symbol is only $\mathrm{^4F}$ in the case of a free ion. If you take a closer look at the Tanabe-Sugano diagram, the $\mathrm{^4F}$ term only appears at the far left-hand side of the diagram, where $\Delta = 0$. $\Delta$ refers to the ligand-field splitting parameter, and $\Delta = 0$ indicates that there is no ligand field, i.e. a free ion.

The quantum number $L$ (total orbital angular momentum) of the ground state can be obtained by coupling the individual orbital angular momenta of the d electrons using a Clebsch-Gordan series. The way to do this is described in most physical chemistry textbooks under the Russell-Saunders coupling scheme. For example in Atkins 10th ed. it is in page 386 under the chapter on "Atomic structure and spectra".

(Note that the symbol $\Lambda$ is used for diatomic molecules, not atoms.)

$L$ is said to be a "good" quantum number, in that the operator $\hat{L}^2$ (almost - this neglects spin-orbit coupling) commutes with the Hamiltonian $\hat{H}$. Quantum mechanically this means that the $\hat{H}$ and $\hat{L}^2$ (almost) share a set of eigenstates, and therefore for every state of the Hamiltonian (which correspond to the electronic configurations we are familiar with) one can (almost) calculate the corresponding value of $L$.

$$\hat{L}^2|\psi\rangle = L(L+1)\hbar^2|\psi\rangle$$

Octahedral complex

The ground state term symbol of a $(\mathrm{t_{2g}})^3$ ion is $\mathrm{^4\!A_2}$, not $\mathrm{^4F}$!

The $\mathrm{t_{2g}}$ set comprises the $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ and $\mathrm{d}_{xy}$ orbitals. These three d orbitals are what we call "real" spherical harmonics, which are linear combinations of the complex spherical harmonics which you have quoted. As such it is not possible to assign $m_l$ values like you have to the $\mathrm{t_{2g}}$ orbitals.

It's not correct to say that $\mathrm{d}_{xy}$ can have "either $m_l = +2$ or $-2$". That would mean that $\mathrm{d}_{xy}$, at any one point in time, is either equal to $Y_2^{+2}$ or equal to $Y_2^{-2}$, which makes no sense. It is not flip-flopping between the two spherical harmonics, it is its own thing: a linear combination of the two spherical harmonics, or a superposition, if you prefer that word. Furthermore, the spherical harmonics only have significance under spherical symmetry, where they act as simultaneous eigenstates of $\hat{H}$, $\hat{L}^2$, and $\hat{L}_z$. In octahedral symmetry, the spherical harmonics aren't of any significance at all and trying to "resolve" the $\mathrm{t_{2g}}$ orbitals into their components as spherical harmonics is physically unmeaningful (it helps with the maths, but that's all).

Under octahedral symmetry, the total orbital angular momentum $L$ is no longer a good quantum number (i.e. $\hat{L}$ no longer commutes with the Hamiltonian) and therefore the term symbol doesn't say anything about it!


Rationalising an apparent inconsistency in the determination of reaction spontaneity

I recommend the following article by Juan Quílez if you have access: DOI: 10.1021/ed100477x. The main point is summarized in their figure:

The lower curve shows your first approach, with $\Delta_r G$ as the slope of the curve. The upper curve (i.e. straight line) shows your second approach, where the slope is non-zero and constant. What the second approach leaves out (and it might be appropriate for certain cases, but usually not) is the entropy of mixing (the $R T \ln(Q)$ term in calculating Gibbs energy for given concentrations).

For some more interpretation of the graph and the different definitions, see Orthocresol's excellent answer here: https://chemistry.stackexchange.com/a/41864/72973


I'm looking for something as strong as Acetone

Mineral oil is often used to remove nail polish. Give it a try (J&J has a great one for babies with a soft smell).


Does it change the compound if the element letters in its molecular formula are switched?

Apart from affecting organic compounds, it also affects coordination compounds.

$\ce{Hg[Co(SCN)4]}$ is a significantly different compound from $\ce{Co[Hg(SCN)4]}$. The central metal atom and the counter-ions in both cases differ, as does the complex anion.
In fact, in the above compound itself, $\ce{Hg[Co(SCN)4]}$ differs from $\ce{Hg[Co(CNS)4]}$ or $\ce{Hg[Co(NCS)4]}$: this is called linkage isomerism. The order in which ligand atoms are written denotes the atom of attachment to the central metal atom, which happens to differ in this case.
The complex salts $\ce{[Co(NH4)6]^3+[Cr(CN)6]^3-}$ and $\ce{[Co(CN)6]^3+[Cr(NH4)6]^3-}$ - formed by swapping the two radicals - are coordination isomers, and not identical.


Degree of association, van't Hoff factor and dissociation constant

There is absolutely no difference in your two equations but you forgot the units which makes it seem like there is a difference.

Indeed it is in both cases $$k_a=\frac{c~\alpha^2}{1-\alpha}$$
since what happens if your concentration would be $c=1~\mathrm{mol~L^{-1}}\approx1~\color{\red}{\text{forgotten unit}}$?

   $$k_a=\frac{\alpha^2}{1-\alpha}~\ce{mol~L^{-1}}$$

How do I know that the c is the concentration?

Using the acid dissociation constant $$k_a=\frac{c(\ce{A-})~c(\ce{H3O+})}{c(\ce{HA})}$$ we have a good basis to obtain what we search. Now by including the degree of dissociation we can define equations for the undissolved acid molecules $c(\ce{HA})$ and the concentration of the conjugate base in solution $c(\ce{A-})$ $$\begin{align}c(\ce{HA})=&~(1-\alpha)~c_0(\ce{HA})\\c(\ce{H3O+})=c(\ce{A-})=&~\alpha~c_0(\ce{HA})\end{align}$$ and combine it with the equation for the acid dissociation constant to obtain $$k_a=\frac{(\alpha~c_0(\ce{HA}))^2}{(1-\alpha)~c_0(\ce{HA})}=\frac{c_0(\ce{HA})~\alpha^2}{1-\alpha}$$
and this is exactly your equation which tells us that the $c$ is indeed a concentration ... namely the concentration of your solution.


Is 6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol a valid name?

The suggested name
“6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol”
is not correct; the closest valid name is
6-ethenyl-4-(prop-2-en-1-yl)oct-6-en-3-ol.
However, this name does not describe the compound that is shown in the picture. Note that the parent chain actually is non-7-en-3-ol instead of oct-6-en-3-ol, and the longest side chain actually is but-3-en-1-yl instead of prop-2-en-1-yl.
Therefore, the correct name for the compound that is shown in the picture is
(7E)-4-(but-3-en-1-yl)-6-ethenylnon-7-en-3-ol.


Please refer me to some paper or book if you know any. A graph showing the adiabatic flame temperature vs equivalence ratio would be ideal.

In 2014 it appears that Glaude, Sirjean, Fournet et al. published a paper that gives exactly the data you are looking for.  See Figure 4 from their paper, which I have also included below.

Possibly relevant text from the paper:

  the
  maximum flame temperature is obtained for slightly rich
  mixtures, at an equivalence ratio around 1.05-1.1, because of
  the lower impact of dilution by nitrogen when the amount of
  air decreases. At the strict stoichiometric conditions, the rate
  of the endothermal molecular dissociations is also at its
  maximum which tends to depress the temperature.

Also this next bit:

  The results show that alcohols have flame temperatures
  comparable to those of conventional fuels, contrary to DME
  that develops significantly higher temperatures. This is
  consistent with the respective values of the heating values of
  these fuels and thermochemical equilibrium data.

  I couldn't find anything on google.

I found the paper by Googling dme adiabatic flame temperature.


How to determine an unknown metal based on the reaction products with oxygen and fluorine?

Your answer is Magnesium. How you get there requires you to first realize that the number of moles of Fluorine is double the number of moles of Oxygen. Since Oxygen will accept 2 electrons and Fluorine will accept one electron, you can deduce based off oxidation states and the number of moles that this metal has a 2+ charge when it is a cation. From this, you can write your chemical equations out X+1/2O2-> XO and X+F2->XF2 . Moving on, you can do some basic  stoichiometry to determine you have .412 moles of X present and use that fact that X's mass is 10g to determine the molar mass of 24.272 g/mols. Then look at your Periodic Table to find what corresponds to 24.272 g/mol and Magnesium does. Hope this helps


How easy it is for MoS2 to become oxidized

It seems that the problem has been examined even 50 years ago, probably in order to determine if/how oxidation would alter the the properties of $\ce{MoS2}$ as a lubricant.

Here is an old NASA document from 1949, and in Surface Oxidation of Molybdenum Disulfide from 1955, the authors state in the abstract, that the oxidation is limited to the outmost surface layer, which then passivates and protects the rest of the batch.


Flexibility of long-chain aliphatic compounds

PubChem 3D has limits on the number of 'rotatable bonds'

I recently answered a related question about PubChem 3D:

  Beyond the limits of MMFF94 and MMFF94s themselves, PubChem3D had several limits indicated in the accompanying manuscript: Bolton et. al. "PubChem3D: a new resource for scientists" J Cheminf. (2011) v. 3, art. 32)

  Not too large (with ≤ 50 non-hydrogen atoms).
  Not too flexible (with ≤ 15 rotatable bonds).
  Consists of only supported elements (H, C, N, O, F, Si, P, S, Cl, Br, and I).
  Has only a single covalent unit (i.e., not a salt or a mixture).
  Contains only atom types recognized by the MMFF94s force field.
  Has fewer than six undefined atom or bond stereo centers.

Your molecule falls into the "more than 15 rotatable bonds" category. I count about 18 $\ce{sp^3}-\ce{sp^3}$ single bonds.

Basically, the problem is that the search space for conformers with that many rotatable bonds is too big. In principal, it's $\approx3^n$ where $n$ is the number of rotatable bonds.


Detailed kinetic explanation for vapor pressure reduction by dissolved solute?

I'd like to specifically commment on this:
According to the NCERT for Class XII, Part I, [pg. 46, para 3][1],

In  a  pure  liquid  the  entire  surface  is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to give  a  solution  [Fig.  2.4.(b)],  the  vapor pressure  of  the  solution  is  solely  from  the solvent  alone.  This  vapor  pressure  of  the solution at a given temperature is found to be  lower  than  the  vapor  pressure  of  the pure  solvent  at  the  same  temperature.  In the solution, the surface has both solute and solvent molecules; thereby the fraction of the surface covered by the solvent molecules gets reduced.  Consequently,  the  number  of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapor pressure is also reduced.
[emphasis mine]

NCERT's explanation for why vapor pressure is lowered due to the presence of a dissolved solute is incorrect, because vapor pressure is independent of the surface area accesible to the solvent (assuming it's non-zero).  Consider these two examples:

You have two sealed containers, both of which contain an open beaker of identical solvent.  In container A, the beaker is low and wide.  In container B, the beaker is tall and narrow.  The surface area of solvent in container B is thus smaller.  Yet, assuming the conditions in the respective containers are identical, the equilibrium vapor pressures will be the same.

You have two sealed containers, both of which contain an open beaker of identical solvent.  The beakers are identical, so the surface area of liquid is the same. The conditions in the containers are also identical.  Glucose is dissolved into the solvent in container A, while sucrose is dissolved into the solvent in container B.  The final concentrations are the same.  Sucrose is larger than glucose.  Consequently, the fraction of the surface area covered by solvent in container A is larger than in container B.  Yet, ignoring non-ideality, the vapor pressures are the same; they depend only on the concentration of solute, not its nature.

The surface area only matters for kinetics, i.e., for how quickly the solvent can escape the container to reach equilibrium.  Clearly, with a larger surface area, the equilibrium vapor pressure will be reached more rapidly.
Feel free to send an email to NCERT with a link to this comment.
Here's the thermodynamic explanation: The addition of solute lowers the chemical potential of the solvent, due to the entropy of mixing.  [There can also be energetic effects, in either direction, but with a soluble solute the entropy term is dominant.]
At equilibrium, the chemical potential of the solvent in the liquid and vapor phases must be equal.  Thus, since the chemical potential of the solvent in the liquid phase has lowered, that of the the solvent in the vapor phase must lower as well.
And since the chemical potential of a gas increases with its partial pressure, the vapor pressure of the solvent in the gas phase will decrease until its chemical potential reaches the new (lowered) chemical potential of the solvent in the liquid phase.
As far as the question itself goes, I'm uncertain, because the question is asking for a kinetic, rather than a thermodynamic, explanation.  But I can offer two alternative pictures:
Let's represent the movement between the liquid and vapor phases of the solvent ("X"), as follows:
$$\ce{X_{(l)}<=>X_{(g)}}$$
Picture (I): Both "B" and "C" are correct.
Let's start with the vapor in equilibrium with pure solvent.  At equilibrium, the rates of the forward and backward reactions are equal.
Now suppose we add solute to the liquid phase.  This shifts the reaction to the left, which means that the rate of the forward reaction initially decreases (which is answer "B") and the rate of the backward reaction initially increases (which is answer "C").  Eventually the system reaches its new equilibrium point, at which point the rates of the forward and reverse reactions are again equal.
Picture (II): "B" is correct.
Again, let's start with the vapor in equilibrium with pure solvent.  At equilibrium, the rates of the forward and backward reactions are equal.
Now suppose we add solute to the liquid phase. The rate at which gas enters the liquid phase depends only on the concentration of gas, so the backwards reaction will be unaffected by the presence of the solute in the liquid.  However, the rate of the forward reaction will be reduced, since it is now more favorable to for the solvent to remain in the liquid state.  Hence the answer is "B".
As I mention in the Comments:
In Picture (I), the change in relative chemical potentials would raise the energetic barrier in the forward direction and lower it in the reverse direction, which would in turn increase the % of collisions that are succesful in the rev. direction, thus increasing the rate of the rev. reaction even though the collision frequency of the vapor with the liquid woudn't be changed by the presence of the dissolved solute.
But: If there isn't any barrier in the rev. direction, such that 100% of vapor molecules that collide with the liquid phase become liquid, then Picture (I) would not apply.
I'm afraid I don't know enough about the microscopic kinetics to have certainty on this.  Perhaps someone that has done simulation work on this could provide an answer.  [But to get this, you would want to vote to reopen the question.]


Why would substitution of H with D alter chemical shift?

OP's main question: Why would substitution of $\ce{H}$ with $\ce{D}$ alter chemical shift?

OP has correctly indicated that the $\ce{Sn-D}$ bond will be shorter/stronger than $\ce{Sn-H}$ due to the lower ZPE, which may cause the chemical shift difference. That is the major reason for isotope effect in chemical kinetics mainly due to bond strength (shorter the bond distance, stronger the bond). However, here, how change of isotope effected the chemical shift is caused by increasing the electron density of the adjacent atom ($\ce{Sn}$). Since $\ce{Sn-D}$ distance is shorter than $\ce{Sn-H}$, the sharing electron pair in $\ce{Sn-D}$ bond is closer to $\ce{Sn}$ tan that in $\ce{Sn-H}$ bond. Thus, it is safe to say that $\ce{Sn}$ of $\ce{Sn-D}$ is shielded more than that of $\ce{Sn-H}$. Therefore, we can expect up-field shift (red-shift) in relevant $\ce{Sn}$ chemical shift. That's what you have experienced $(\pu{650.6 ppm} - \pu{657.9 ppm} = -\pu{7.3 ppm})$ when you have changed the $\ce{H}$ with $\ce{D}$ of the dimer. This phenomena is common among deuterated organic compounds (Ref.1):
$$
\begin{array}{llrrr}
\hline
\text{Hydrated cpd} & \text{deuterated cpd}  & \delta \ \left(\ce{^{13}C-H}\right)/\pu{ppm} &  \delta \ \left(\ce{^{13}C-D}\right)/\pu{ppm} & \Delta \delta/\pu{ppm}\\
\hline
\ce{CHCl3} & \ce{CDCl3} & 77.36  & 77.16 \pm 0.06 & -0.20  \\
\ce{CH3C#N} & \ce{CD3C#N} & 1.79  & 1.32 \pm 0.02 & -0.47  \\
\ce{(CH3)2C=O} & \ce{(CD3)2C=O} & 30.60 & 29.84 \pm 0.01 & -0.76 \\
\ce{C6H6} & \ce{C6D6} & 128.62 & 128.06 \pm 0.02 & -0.56  \\
\ce{CH3-OH} & \ce{CD3-OH} & 49.86 & 49.00 \pm 0.01 & -0.86 \\
\hline
\end{array}
$$
These chemical shifts are also depend on the other factors such as temperature and solvent. For comparison purposes, all chemical shifts of hydrated compounds have been recorded in their deuterated versions (e.g., $\ce{CHCl3}$ in $\ce{CDCl3}$) at a constant temperature (Ref.1).
Note: I also realized that $^1\!J_{\ce{Sn-D}}$ constant must be zero or negligible according to your observation. I tried to find the value of $^1\!J_{\ce{Sn-D}}$, but failed. Note that in deuterated organic solvents, $^1\!J_{\ce{C-D}}$ varies between $\pu{20 Hz}$ and $\pu{30 Hz}$ (Ref.1).
References:

Hugo E. Gottlieb, Vadim Kotlyar, Abraham Nudelman, "NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities," J. Org. Chem. 1997, 62(21), 7512-7515 (DOI: https://doi.org/10.1021/jo971176v).


Is there a difference between dipole and dipole moment?

The (electric) dipole moment is a mathematically clearly defined quantity, the product of charge difference and distance

A dipole is anything (e.g. a molecule, or a part of a molecule), that has a non-zero dipole moment.

And a polar molecule is one that has at least local dipoles. CO2 is polar if you get close (i.e. in a condensed phase). From a slightly larger distance, it's unpolar, because the centers of the positive and negative charge distributions in it are identical, the two local C=O dipoles cancel each other out.


Molecular Orbitals (only pi system) of Allyl system using group theory

The calculation below is not exactly the same as yours but I had it to hand and it is very similar. The bit you specifically want is towards the end. In your case you add will vectors along the $\pi$ orbitals.

What projection operators  operators do is to extract the symmetry adapted functions from the basis functions used to form them and so produce a linear combination L of the basis functions. Many basis functions can be chosen, but the easiest to use are usually vectors pointing along bonds or those representing a p$\pi$ orbital. To form a linear combination, a 'victim' vector is chosen and operated on with each symmetry operation in the point group, $C_2\,, \sigma(xz)$, etc. to find what vector this turns into. This vector's name (the basis function) is then multiplied with the character from the point group corresponding to that symmetry operation and symmetry species. Finally, the names are added together by moving from one symmetry operation to the next along the top row of the point group. We expect the result to be the sum or difference of the displacement vectors for a molecular vibration, such as $2v_1 + v_2 - v_3 \cdots$ and so forth. The same formula or algorithm is used if the combinations of atomic orbitals that make up molecular orbitals are sought.
The equation to make a linear combination is

$$ L_M=\frac{d}{h}\sum_{j=1\cdots h}c_jS_j(v)$$

where $M$ is the symmetry species label, $A_g, B_{3g}$, and so forth, $h$ is the order of the group, $d$ the dimension of the irreducible representation, and the sum is over all the classes. Because the resulting vector will be normalized d/h can be ignored. One further point is important. In the $C_{3v}$ point group, for example, there is a heading $2C_3$ as there are two members in this class, and this must be split in the summation into $C_3^+$ and $C_3^-$. This is because moving a vector $120$ degrees to the right, say, will turn it into a different vector than turning 120 degrees to the left, similarly, any mirror planes must be separated out. The character for symmetry species $M$ is $c_j$ and $S_j(t)$ is the effect that operator $S$ in column $j$ of the point group has on our victim vector $v$. The result of this operation is to produce a vector whose name is recorded. For example, if $S$ is the identity then the result is $E_1(v) = v$, other operations may leave $v$ unchanged or change it into another vector. Finally, note that if the reduced representation of the vibrations or molecular orbitals contains two or more symmetry species of the same type, e.g. $2A_1$, then two or more different victim vectors will have to be chosen to obtain all the linear combinations.

In H$_2$O, SO$_2$%, and other triatomics with $C_{2v}$ symmetry, it is clear that two of the vibrational modes stretch the bonds and one changes the bond angle. To work out the normal mode vectors, some representative vectors are placed along the bonds as in the figure. You can choose where to put these but physically realistic choices will usually make the calculation simpler.

In $C_{2v}$, the normal mode vibrations have been found to comprise the symmetry species $2A_1$ and $B_1$. The $L_M$ formula is shown as a table with $v_1$ as the victim vector. The table also shows the characters for each symmetry species.

$$\begin {array}{l|llll|l}
C_{2v}  & E & C_2 & \sigma(x,z) & \sigma'(y,z) & L_m \;(d=4,h=4)\\
\hline
A_1& 1&1&1&1\\
L_{A1} & v_1 & v_2& v_1& v_2&  2(v_1+v_2)\\
\hline
A_2& 1&1&-1&-1\\
L_{A2} & v_1 & v_2& -v_1& -v_2&  0\\
\hline
B_1& 1&-1&1&-1\\
L_{B1} & v_1 & -v_2& v_1& -v_2&  2(v_1-v_2)\\
\hline
B_2& 1&-1&-1&1\\
L_{B2} & v_1 & -v_2& -v_1& v_2&  0\\
\hline
\end{array}$$

Notice that what the vector changes into is (obviously) the same for each symmetry species as this is determined by the symmetry operations. Only the value of the character changes in front of each term The final $L_M$ is not multiplied by $d/h$ because the resulting vectors are instead normalized giving, $L_{A1} = (v_1 + v_2)/ \sqrt{2}$ and $L_{B1} = (v_1 - v_2)/\sqrt{2}$. Note also that $v_3$ and $v_4$ do not enter into this table, they are at $90$ degrees  to $v_1$ and $v_2$ and no operation in this point group can inter-convert them.

There are two $A_1$ species produced from the reducible representation but only one has been found so far. This is because the vectors $v_1$ and $v_2$ cannot produce a bend, which is the other normal mode. Using vector $v_3$ as the victim this species appears. As this mode has $A_1$ symmetry all the characters $c_j$ in the $L_M$ formula are $1$, making the calculation easy and the combination is shown below.

$$\begin {array}{l|llll|l}
C_{2v}  & E & C_2 & \sigma(x,z) & \sigma'(y,z) & L_m \\
\hline
L_{A1} & v_3 & v_4 & v_3 & v_4 &  2(v_3+v_4) \\
\hline
\end{array}$$

which describes the bending mode shown in the figure in terms of its unit vectors. When normalized this mode is $(v_3 + v_4)/\sqrt{2}$. (As the centre of mass cannot move during a vibration, there being no force to make it do so, therefore the O atom has to move a small distance away from the H atoms. Vectors
could be added onto the O atom to show this and $L_M$ terms re-calculated.

In the case of degenerate symmetry species as occurs in the $C_{3v}$ and most other point groups the results of the summation $L_M$ is to produce two or more vector sums that are not orthogonal. Either these have to be made orthogonal, by choosing another victim for the degenerate term and then using the Gram-Schmidt method to make the two vectors orthogonal. R. Carter 'Molecular Symmetry and Group Theory' and A. Vincent 'Molecular Symmetry and Group Theory' both give method by which to do this.


Nomenclature and bonding of the complexes ions

The bonding of the $N$ end of the thiocyanato group to the Metal centre $\ce{Cr^3+}$ can be explained by both HSAB Theory and Symbiotic effect.According to HSAB Theory, Hard acids prefer to bind with  hard bases, while soft acids prefer to bind with soft bases. Here the metal centre is $\ce{Cr^3+}$, which has $3d^3$ electronic configuration. So,  $\ce{Cr^3+}$ satisfies the criterions for a Hard acid centre, as it has very less no. of $d $ electrons, high positive charge density on it, smaller radius and greater polarisation. So, the metal centre in this complex acts as Hard acid centre. And, now if you consider the properties of a hard base, you will find that $N$ is the harder basic centre, whereas, $S$ is the softer base centre. $N$ has higher electron charge density on it and also smaller in size than $S$ which makes it harder base. Thus, due to the presence of the harder metal acid centre, the ligand tries to bind with the harder basic centre, which in this case is $N$. The attachment of the $N$ site can also be explained by the Symbiotic effect. Observe that, in the complex, you already have two $\ce{NH_3}$ ligands donating through the $N$ site, which actually compells the thiocyanate ligands also to coordinate through the $N$ site, which is the main idea of this effect. Actually, donation through $N$ sites of ammonia makes the metal centre even more hard, and therefore it prefers also the $N$ site of other ligands.  But a remark which is necessary is that the HSAB theory doesn't always predict the correct coordinating site for an ambidentate ligand. If even stronger effects like Competitive $\pi$-Bonding effect start to interfere due to presence of $\pi$-accepting and $\pi$-donating ligands, HSAB theory may not give a correct answer at all. So, in the absence of those effects, HSAB theory  works well.


Understanding reason behind existence of chemical equilibrium

It is best to think of reversible reactions statistically

A good understanding about equilibrium can be had by thinking about the detailed molecular level processes involved at the molecule level.

Take, for example, the simple equilibrium between liquid water and water vapour in a closed vessel. At room temperature, the equilibrium clearly favours liquid water. But there is always some water vapour present (that's what we call humidity in the wider atmosphere). Why is there any vapour if the equilibrium favours liquid? The reason is that the water molecules don't all have the same amount of energy. some are travelling fast and some some are slow. A small percentage of the fast ones can escape the liquid and become vapour (ie gas-phase water molecules not liquid phase water molecules). In the gas above the liquid, the same is true. Not all the gas molecules have the same speed. Some are travelling slowly and, if they hit the liquid surface, stay in the liquid. Both processes are dynamic and are happening fairly fast. So molecules are constantly exchanging between the vapour and the liquid.

Equilibrium is not reached when everything stops: at room temperature there is a lot of kinetic energy shared out among the molecules of the system. Equilibrium happens when the rates of evaporation and liquefaction are the same. So the number of molecules per second going into the vapour from the liquid equals the number going the other way.

The same is true for other, more complex, reversible reactions. Equilibrium is not stasis but an average state resulting from dynamic reactions in both directions.

This point of view makes it clear why, even when one product is far more stable, the reverse reaction can still happen.


Diastereoselectivity in synthesis problem (4 stereocenters)

If we get rid of the two methyl groups and just consider what the 3D conformation of the molecule is likely to be (I don't have a pen on me right now, but the model below was minimised using very low level molecular mechanics).

What we can hopefully see is that there is very much a convex face and a concave face of this molecule.

In the absence of all other factors, the hydroboration should be faster from the convex face (the lower face as drawn).

This would be simple facial selectivity, as is often observed on ring systems.

However, we have additional chiral centres that are going to influence the reaction (diastereoselective reactions can be matched/mismatched such that the methyl groups present might help us or hinder us.

The problem in this question is that it isn't specified whether the methyl groups are up or down etc. So presumable what we're actually seeing is a resolution (i.e. we're only forming a small amount of product, but its optically pure because certain combinations just don't react at a sufficient rate).    [I did try and find the paper on Reaxys but had no luck, there is a chance he just made the question up].

So if we now consider the methyl group next to the acetal functionality, it would help us if it were pointing up (as I've drawn it, you could flip the rings to get it pointing down as shown in your product above) to leave the convex face free.

If it were pointing down (as I've drawn it..) then the approach of H-B would be hindered, making attack from convex/concave both unfavourable relative to the approach in which the methyl group is orientated into the concave face.


Wikipedia reports biological half-lives for several recreational drugs I just tried (cocaine, THC, MDMA). In each case, I went to the page for the particular substance, and Wikipedia keeps an information box at the top of the article containing the pharmacokinetic data. THC's half-life has a citation to the scientific literature, while the other two I checked do not. I think it might be slightly harder in general to get good data for the purely recreational drugs. I would expect drugs without an accepted medical benefit to be less well-studied. To find what's there, though, it looks like you will be able to look up each active ingrediate individually. Depending on what you need, Wikipedia might be okay. If it's not, follow the Wikipedia citation back to the literature if it's there, or try PubMed.


How can I tell which parts of this example drug are necessary parts of its pharmacophore?

From this? You can't. You need more information. Examples:

Do you know other drugs of this family? Which chemical features do they have in common? The interactions mediated by these features are likely important then.
Do you know how the drug is supposed to function (you mention competitive inhibition)? What does the natural substrate look like, and what are the similarities and dissimilarities to your example drug?

If you know only the figure shown above, you're really asked to identify electronic and steric features of the drug, that can possibly interact with the amino acid residues in the active site shown. In this case, all amino acid residues are capable of favourable interaction with a part of the example drug, so I would mention which interactions are present.

There are many good ionic interactions/H-bond interactions, which are qualitatively the strongest.

One possibility is that it is sterically unfavourable to jam something into the hydrophobic pocket, but the payoff is, that this is what stabilises a certain (active or inactive) conformation of the protein, and the ionic/H-bond interactions simply serve to increase binding affinity / to keep the drug in place.

Another possibility is that the interaction with the hydrophobic pocket only increases binding affinity. The drug then might share the same ionic interactions as the real substrate, but placed differently in space, so they don't induce the same conformation change as the natural substrate.

The point is, without knowing what the action of the drug is, and some knowledge about other drugs of the same family or the natural ligand/substrate, anything beyond a listing of the interactions you can spot between the amino acid residues shown, and the chemical features of the drug will really just be guess work :-).


How to draw the structure and find the oxidation number of Phosphorous in H2P2O7

If you didn't understand what  Poutnik has sain in his comment, this is how it should be done:
You have realized your parent compound is $\ce{H4P2O7}$ (adding two $\ce{H+}$). Correct? If so, concentrate on following hypothetical reaction:
$$\ce{2H3PO4 -> H4P2O7 + H2O} \tag1$$
I believe you know what $\ce{H3PO4}$ is looks like. Now, draw two $\ce{H3PO4}$ side by side. Choose two closest $\ce{OH}$ groups (that would be $\ce{-P-\color{red}{OH} \space \space \color{tequish}{HO}-P -}$). Then, remove one $\ce{\color{red}{-OH}}$ and the $\ce{\color{tequish}{-H}}$ from the second $\ce{\color{tequish}{-OH}}$ you chose to eliminate a $\ce{H2O}$ molecule. Finally combine two remaining molecules together such that final structure forms $\ce{-P\color{red}{-}\color{tequish}{O}-P -}$ bond. This is your $\ce{H4P2O7}$. Eliminate any two hydrogens to get $\ce{H2P2O7^2-}$ structure.
Note: Since above hypothetical equation $(1)$ does not require oxidation or reduction to form $\ce{H4P2O7}$, it is safe to assume that oxidation state of $\ce{P}$ in both $\ce{H3PO4}$ and $\ce{H4P2O7}$ is same.


Will fluoride evaporate from tap water at the same rate as chlorine

Short answer: no, fluoride does not evaporate in the same way/rate as chlorine.

Longer answer: chlorine will be present as covalent, molecular $\ce{Cl2}$. This is, when not in solution, a gas. Chlorine molecules leaving the surface of the water will (on average) not come back (at least unless there is a very high concentration of $\ce{Cl2}$ in the room's air, in which case you have other problems).

Fluoride, on the other hand, is present as $\ce{F-}$ ions (or other, complex ions) along with corresponding positively charged ions (usually $\ce{Na+}$, I believe). Even if the $\ce{NaF}$ were to be removed from solution, $\ce{NaF}$ is a non-volatile salt, and so would not evaporate away.


Determine compound acidity based on structure

Panagiotis, in order to solve the exercise, one needs to identify the relevant functional groups and explore their environment a bit.

Supposed that we're talking about "standard conditions", i.e., what is likely to happen if the dissolve/suspend the compounds in water, the situation is as follows:

In the structure of Moclobemide, we spot an amide, and - in the morpholin moiety - an ether and a tertiary amine. Amides do have a lone pair on the nitrogen atom, but due to the conjugation with the carbonyl group, the electron density on the N atom is strongly reduced, as compared to an amine. Ethers are rather lousy bases, in fact, the $pK_a$ of protonated tetrahydrofuran is as low as -2.05.

The remaining tertiary amine thus is the functional group that influences the pH: tertiary amines show a $pK_b$ around 3.3.

In Paracetamol we identify an amide (that we've already excluded as irrelevant) and a phenol group. The latter, once deprotonated, will stabilize the negative charge by resonance with the $\pi$-system (G M has already mentioned this). Therefore, phenols are far more acidic than alkanols. Consequently, the compound definitely acts as an acid.

I'm sure that you've already noticed two imide units in the case of Hexobarbital and in fact, it is possible to abstract the proton from the nitrogen atom. However, you'd need a base such as a carbonate under anhydrous conditions to deprotonate here. As a result, Hexobarbital will hardly do anything at all in water.


IUPAC Naming of CH3C(p-ClC6H4)2CH(Br)CH3

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical units that are linked by di- or polyvalent groups can be named in two ways:

by ordinary substitutive nomenclature, in which one of the parent structures is chosen as the senior parent structure and the remainder of the structure is expressed by substituent prefixes
by multiplicative nomenclature, in which two or more parent structures are connected by symmetrical or unsymmetrical single or concatenated substituent groups

Both methods yield correct names, which may be used in general nomenclature. However, the IUPAC recommendations define a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations.
For the compound that is given in the question, the preferred IUPAC name is actually generated using multiplicative nomenclature (see also What is the rationale of the IUPAC name for DDT?), which yields the name 1,1'-(3-bromobutane-2,2-diyl)bis(4-chlorobenzene).
In general nomenclature, nevertheless, you may choose the butane chain as the senior parent structure and generate the complete name using substitutive nomenclature.
The most important criteria for the numbering in substitutive nomenclature according to the current version of the IUPAC Recommendations are:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Since the compound that is given in the question has no principal characteristic group and the butane chain, which is taken as the parent structure, has no multiple bonds, the rules (c) and (e) are not relevant in this case.
Thus, all substituents are considered together according to rule (f). Therefore, the correct name is 3-bromo-2,2-bis(4-chlorophenyl)butane and not 2-bromo-3,3-bis(4-chlorophenyl)butane since the locant set ‘2,2,3’ is lower than ‘2,3,3’.


Titrating iodine starch solution with sodium thiosulphate - Colour change

I don't think your memory is serving you right. That is why we write everything in the notebook, especially color changes.

I think you are doing distribution experiments where iodine is distributed between aqueous layer and an organic layer. When we add indicator for titration, it is not a solid starch but starch which is boiled in water. So when you added starch $solution$ to heptane which contained iodine, I would not be surprised if the starch solution turned blue.

Remember that iodine is strong oxidizing agent as well. A very small fraction of it can easily convert into iodide. You really really need a trace of the triiodide ion to form a dark blue iodine complex.


Flame Test of Alkali Metals and Alkaline Earth Metals

Yes, the metals, when sent in a flame produce the same colors as their salts. In fact, the metallic ions do not produce any color when sent in a flame. It is only the neutral atom that produce the famous yellow double line of the sodium. This means that a tiny amount of the $\ce{Na+}$ and $\ce{Cl-}$ ions are first transformed into neutral $\ce{Na}$ and $\ce{Cl}$ atoms. And the color of the flame is due to the deexcitation of these rare atoms being first electronically excited in the flame. So it means that the following processes are occurring successively in the flame:$$\ce{NaCl(s) -> Na(g) + Cl(g)}$$ $$\ce{Na + heat -> Na^*}$$ $$\ce{Na^* -> Na + h\nu }$$ The famous yellow light of the $\ce{Na}$ atom is due to the return of the outer electron from the level $4s$ to $3p$ of the sodium atom, if my memory is good. Nothing to do with the $\ce{Na+}$ ion.


When DNA degrades, does any chemical environment preserve sequence information?

A mineralized form might be possible (maybe DNA can form a pattern of missing atoms in stone) but reading it will be next to impossible. We have many techniques to amplify and read tiny bits of DNA, doing this for imprints in stone would be a huge challenge, especially because many (all?) techniques for reading DNA depend on the amplification of DNA.

As for conservation, as far as I know most of the damage making the DNA unreadable is hydrolysis of the backbone. If the DNA is in an environment without water this will not happen. Of course you need an extremely dry environment for DNA to last millions of years.


Good quality quartz crystals for use in resonators can be found in nature, such as in Minas Gerais, Brazil. They can also be produced by crystallization in a hydrothermal process, simulating in the autoclave the same process that occurred over geologic epochs.

You can easily grow your own Rochelle salt (potassium sodium tartrate, $\ce{KNaC4H4O6·4H2O}$) piezoelectric crystals. Rochelle salt was commonly used in inexpensive crystal headphones and phonograph cartridges. The raw materials are household chemicals and the directions are straightforward. Again, these crystals are grown, not hot pressed.

Tuning forks watches such as the Bulova Accutron did not use crystals, but rather metal tuning forks, made from NiSpan C, with a low coefficient of thermal expansion, similar to invar. These did not rely on the piezoelectric effect, but on magnetic induction.


How are SN2 transition states stabilised by adjacent double bonds and carbonyl groups?

Unsaturated groups conjugated to the site of attack (aromatic rings, alkenes, carbonyl groups) all tend to have the effect of increasing the rate of an SN2 reaction at that centre.

As an example, displacement by nucleophilic iodide in acetone as a solvent is ca. 30x faster for allyl chloride than ethyl chloride, with benzyl chloride being faster still.[1]

The effect was initially thought to be steric, given that unsaturated groups are sp2 hybridised (and therefore planar) and allow the nucleophile a less hindered approach, but many empirical and computational studies now dispute that, with a stereoelectronic effect[*] widely accepted (though sterics may play a small role).

Empirical Evidence for a Stereoelectronic Effect

One of the earliest indications that the effect was stereoelectronic was via a series of experiments conducted by Payne (shown below) [2].The open (acyclic) compound reacts ca. 8000x faster than the closed (cyclic) compound. To understand why, we have to consider the orbitals involved...

The Orbital Explanation

If we accept a stereoelectronic effect (coming back to Payne’s findings later),  we then have to decide whether the effect is a ground state effect (weakening of the C-X bond, raising the energy) or a transition state effect (stabilisation of the TS, lowering the energy). To quote Fleming (emphasis my own):[3]

  [This conjugation] may partly operate by raising the energy of the starting material, but most probably [the conjugating groups] are primarily operating to lower the energy of the transition structures.

In practice, computational studies [4] prove this to be the case, with the pi bond interacting with the principal axis of the SN2 transition state (sigma-pi conjugation):

Calculation of bond lengths also agrees with this mode of conjugation, with the C-C bond length decreasing upon stabilisation of the TS:

Hopefully, by this point, Payne's experimental results are obvious. In the cyclic structure, there is no way of allowing the pi-orbitals on benzene to correctly align with the SN2 transition state (given that we know the TS is linear, not bent).

One final thing to notice is that in both the alkene and carbonyl case (as in your question) the dominant effect is thought to be as above (sigma-pi conjugation).

—

[1]: March, J.; Smith, M. B. Advanced Organic Chemistry Wiley: Hoboken, New Jersey, 2007

[2]: King, J. F.; Tsang, G. T. Y.; Abdel-Malik, M. M.; Payne, N.C. J. Am. Chem. Soc. 1985, 107, 3224

[3]: Fleming, I. Molecular Orbitals & Organic Chemical Reactions:Reference Edition Wiley: Hoboken, New Jersey, 2010

[4]: Bach, R. D.; Soddens, B. A.; Wolber, G. J. J. Org. Chem. 1986, 51, 1030

[*]: Sorry Jan, couldn’t help myself with the bold, I’m not shouting, I promise…


Geometries of SCl2(OCH3)2 and SF2(OCH3)2

Check the bond angles. The larger the angles around a bond, the more S character.
Electronegative elements prefer higher P character(Therefore less the bond angles).
Axial position has lesser bond angles.Hence higher electronegative element takes axial position.( F>O>Cl)

This concept however may not be applicable to every compound.


Why does increasing the concentration of oxidizing agent decrease the E° in a galvanic cell?

If the Zn2+ ions aren't involved in the redox reaction, then why does the extra concentration affect the cell potential?

The overall reaction is
$$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$

If this reaction were happening in a common reaction mixture (maybe with Zn and Cu powder on the bottom of a solution) we would write it as

$$\ce{Zn(s) + Cu^2+(aq) <=> Zn^2+(aq) + Cu(s)},$$

and would realize that the rate of the reverse reaction depends on the zinc ion concentration, thus affecting the equilibrium. However, the reaction at hand relies on electrons going through a single wire, and they would not go in both directions at the same time.

MaxW's comment is the key to the conundrum. There are reversible half reactions in either cell, and they determine the cell potential.

$$\ce{Zn^2+ + 2e- <=> Zn}$$

$$\ce{Cu <=> Cu^2+ + 2e-}$$

At each electrode, the half reactions are reversible, delivering electrons to the electrode and picking them up again. This back and forth establishes the potential difference that we can measure between the electrodes.

  I wanted a conceptual explanation

For some, a kinetics-based answer (like the one above) counts as conceptual (or particular as opposed to macroscopic?). Here is a second way of thinking about it, using a different but related system. Let's consider a concentration cell with zinc electrodes and zinc ions in both cells. With higher concentration of zinc ions in the right cell than the left cell, the reaction would be:

$$\ce{Zn(left) + Zn^2+(right) -> Zn^2+(left) + Zn(right)}$$

Using your logic, the potential would not depend on the zinc ion concentration in the left cell because that is one of the products. Now we add some zinc ions (in the form of zinc chloride or such) to the left cell, making its concentration higher than that of the right cell. The reaction now would be:

$$\ce{Zn(left) + Zn^2+(right) <- Zn^2+(left) + Zn(right)}$$

Again using your logic, now the potential would not depend on the zinc ion concentration in the right cell because that is now the product.

The question is how the half reaction on the right would "know" to reverse direction, and to stop influencing the potential via its zinc ion concentration. I'm hoping this reductio ad absurdum (I had to look it up) is a convincing argument that zinc ion concentrations in both cells are relevant.


Calculating amount of ice required for heat loss

HINT:

$\pu{4.18 J/g K}$ is the specific heat of water. Also question assumes that due to addition of tea leaves the specific heat doesn't change much and remains equal to that of water.

So if you mix ice and tea and want the final temperature at $\pu{10^\circ C}$ then what do you expect of ice. Shouldn't it must melt and reach to $\pu{10^\circ C}$ and that will require a further energy to raise it temperature from $\pu{0^\circ C}$ to $\pu{10^\circ C}$. Also the mass of ice will remain same even if it get converted to water


For alkali and lanthanide metals, is there a better liquid barrier to oxidation than mineral oil?

Not really as all liquids will, to some extent, dissolve a small amount of oxygen or water. The only liquid that comes to mind that will prevent either is mercury, but you probably don't want your stuff contaminated with mercury. If you really wanted to prevent water contamination, adding molecular sieves might help, but other than that, no not really.