What does the phase diagram mean by "liquid" at NTP? (The liquid and vapor coexist even when the situation is not on the curve)

The phase equilibrium diagram is for a system consisting of just water.

The diagram pressure is the total system pressure.

It may be equal to the saturated vapor pressure, if no external pressure is imposed on the system.

Or, it can be ( much ) higher, if some external pressure higher than the saturated vapour pressure exists.  In such a case all vapor is at equilibrium transformed to liquid or solid phase.

Really high pressures can be seen in the water phase diagram for very high pressures, displaying rare ice phases.


First I only wanted to comment, but then it got too long... But I wanted to pass on this info... this isn't a real answer...

Every simulator I know uses atomic units. I would check out the NAMD and CHARMM manual for their units, I imagine they use nanometer for distance and ultimately kJ/mol for energy. You will need a conversion for the electrostatic since you always input +/- charge, where charge is usually between 0 and 1, but units of charge are actually Coulomb. The conversion factor for this is ~138.9

This is from my own code... qq_convert is the electrostatic conversion I mentioned above. I use nm for all distances. I try to follow the same procedure as Gromacs.

const na = 6.02214129e23      # mol^-1
const R = 8.3144598e-3        # kJ mol^-1 K^-1
const kb = R / na             # kJ K^-1
const e = 1.6021765653e-19     # C
ϵ₀ = 8.854187817e-12 # C²/(J ⋅ m)
ϵ₀ *= 1e-9  # C²/(J ⋅ nm)
ϵ₀ *= 1000   # C²/(kJ ⋅ nm)
const qq_convert = e^2 / ϵ₀ * na / 4.0 / pi # 138.935458  # kJ mol^-1 nm e^-2

after each electrostatic interaction I multiply the energy by qq_convert

For instance the real part contribution of the EWALD summation is

@inline function ERealSingle(diff::MVector{3,Float64},qqCut²::Float64, coord1::SVector{3,Float64},
        coord2::SVector{3,Float64},q1::Float64, q2::Float64, kappa::Float64,
        overflow²::Float64, L::SVector{3,Float64})

    @inbounds for i=1:3
        diff[i] = vector1D(coord1[i],coord2[i],L[i]) # mirror image separation
    end

    rij_sq = diff[1]*diff[1] + diff[2]*diff[2] + diff[3]*diff[3]
    rij = sqrt(rij_sq)

    if rij_sq < qqCut²
        return q1 * q2 * erfc( kappa * rij ) / rij
    else
        return 0.0
    end
end

This is for a single charge interacting with another charge... I loop through all charges etc and then I multiply the summed energy by qq_convert.

I highly recommend checking your code against these SPC/E water benchmarks offered by NIST... really saved my bacon having a test set to duplicate... NIST Test Set. This set allows you to test your Lennard-Jones, all parts of the EWALD Summation and also tail corrections.

Lastly, I recommend using the Julia programming language... it is much faster than Python... Python can almost keep up if you use the Numba JIT decorators, but Julia has other perks.

Edit:

It looks to me like charmm uses Angstrom as you have said, so that is the only units of length you should be using. It would be dangerous for them to deviate.

In my experience with Gromacs, the parameters of each potential energy function will have units such that everything cancels out leaving you with $\ce{kJ/mol}$ (or $\ce{kcal/mol}$ in CHARMM I assume). In Gromacs all I need to do is make sure all units of length are nm. I imagine in CHARMM it is the same, but for Angstrom. Look at the angle parameter units, they do have units and it will work out that the energy is $\ce{kcal/mol}$ or else I say leave CHARMM.


Does a calibration curve of absorbance vs concentration need to pass exactly through the origin?

Good answers, but it is almost never a good idea to force the intercept to zero. Unless there is a clearly justifiable and compelling reason to do so, it is better to let the data set 'vote' for itself and not impose a preference/bias/mandate. Among other things, forcing the intercept to zero changes the standard error of the slope, which is an important statistic.

Below is a screenshot of the result of using Excel's Data Analysis tool on a set of linear calibration data I collected in the past:

As shown, the intercept is -0.0541 and its standard error is 0.0352, so the t test statistic of the intercept, defined as intercept/(standard error of the intercept) = -1.537. The negative sign is irrelevant for present purposes, but the P-value is 0.264.

The null hypothesis for the intercept is that the true intercept, which is unknown, is zero. The alternative hypothesis (aka research hypothesis) is that the true intercept is either 1) not equal to zero or 2) not equal to zero, but in a known direction. The former is the 2-tailed option and the latter is the 1-tailed option. For intercepts, the 2-tailed option is most common and this would almost certainly be the case for a Beer's law plot.

The P-value of 0.264 is much higher than the usual 0.05 value and this means that it would be quite risky to reject the null hypothesis based on the OLS analysis. In fact, the risk of mistakenly rejecting the null hypothesis, even though it was true, would be 26.4%. The confidence in the alternative hypothesis would be 100% - 26.4% = 73.6%, which is too little confidence for most people. Typically, you want P ≤ 0.05. Otherwise, it is too risky to 'jump ship' for the alternative hypothesis.

What does it mean? Well, the data resulted in a P-value of 26.4%, so the null hypothesis cannot prudently be rejected. But it is also not accepted: it still might be wrong, but the data did not show that. Replacing the intercept by zero would mean that a decision was made that the true intercept was zero, even though the analysis proved no such thing. It is not a good idea to pretend that the data showed what it did not show. Bottom line: the intercept should not be forced to zero unless there is a clear and compelling reasion to do so.


Most thermodynamic equations are derived assuming a perfect gas or ideal solution. However, for real a solution the chemical potential $\mu_i=\mu^0_i +RT\ln(x_i)$ may not be accurate for the solvent where $\mu^0_i$ is the chemical potential of the liquid solvent at 1 atm and $x_i$ is the mole fraction. The solute may, similarly, not follow ideal behaviour.  The activity is introduced to allow us to use the familiar equations but now as $\mu=\mu^0_{liq} +RT\ln(\gamma x_i)$ where $\gamma$ is the activity coefficient thus $\mu_i=\mu^0_{liq} +RT\ln(a_i)$ where $a=\gamma x$. Thus we can think of activity as effective concentration or effective pressure (now called fugasity) relative to its standard state.  The activity coefficient is unity in the ideal case. The activity has to be measured experimentally because $\gamma$ depends on exactly what intermolecular interactions exits as any temperature and pressure and cannot be easily calculated.  As an example, naphthalene dissolved in benzene behaves almost ideally and the activity coefficient $\gamma = 1$ but when dissolved in hexane this is not the case and $\gamma=2.5$. A positive deviation from Raoult's law indicates that the vapour pressure is larger than that expected from perfect behaviour and so the activity is larger than the mole fraction and $\gamma >1$.


You have to consider the $3$D structure and sterics when dealing with cyclic compounds.  If you look at the geometry of cyclohexene below you can that the ring is not symmetrical, thus the placement of the vinical methyl groups in regard to the double bond matters.

In (X) when the vinical methyl groups are both on doubly bonded carbons they are forced into a coplanar arrangement which is unfavorable and maximally strains the ring's double bond and the methyl group bonds.  In (Y) when they are opposite the double bond the methyl groups are optimally spaced apart and steric strain on the ring and methyl bonds is at a minimum.  When they are not located on the double bond they are neither at a maximum or minimum, just in between so (Z) just falls in the middle.  There are of course other conformers of cyclohexene but the relation will still hold.

The question could have also thrown in a fourth isomer where one methyl group was on a double bonded carbon, but it would not be easy to rank it above or below (Z) without computational methods.


How would you synthesise tetrahydrohyperforin and octahydrohyperforin from hyperforin?

That route won't work.

You need a hard base reduction that does not attack the enone, or you need to protect the enone to make Tetrahydrohyperforin.

Catalytic hydrogenation that is selective against carbonyl reduction is necessary to make Octahydrohyperforin.


Part A
-   Calculating  the initial $\pu{pH}$(before adding $\pu{0.02 mol}$ of $\ce{HCl})$

The equilibrium here is
$$\ce{CH3CH2NH2 + H2O <=> CH3CH2NH3+  + OH−}$$
For  $\ce{ CH3CH2NH2} , \mathrm{K_b} = 4.6\times{10^{−4} }$

$$\mathrm{K_b} = 4.6\times{10^{−4} }=\frac{[\ce{CH3CH2NH3+}][\ce{OH−}]}{ [\ce{CH3CH2NH2}]}$$
The nominal concentrations of the base and conjugate acid  are respectively $$\mathrm{C_b} =\pu{0.170 M}~\text{ and } \mathrm{ C_a } =\pu{0.150 M}$$.

The mass balance expressions are:
$$ [\ce{CH3CH2NH3+} ]+ [\ce{CH3CH2NH2}] = \mathrm{C_a} + \mathrm{C_b} = \pu{0.32M}~\text{ and}~ [\ce{Cl−}] = \mathrm{C_a}=\pu{0.150M}$$
and the charge balance is given by:
$$[\ce{Cl−}] + [\ce{OH−}] = [\ce{H+}] + [\ce{CH3CH2NH3+}] \approx{[\ce{ CH3CH2NH3+}]}$$
in which the approximation shown above is justified  that the solution will be weakly alkaline.

The equilibrium concentrations of the conjugate species are then:
$$[\ce{CH3CH2NH3+}] = 0.150 + [\ce{OH−}]$$

\begin{array}{  }
[\ce{CH3CH2NH2}] &= &(\mathrm{C_a} + \mathrm{C_b}) &-&([\ce{ CH3CH2NH3+} ])\\ &= &(0.32) &–&([\ce{ CH3CH2NH3+} ])\\ &= &0.320 &- &(0.150 + [\ce{OH−}])\\&=&0.170 &- &[\ce{OH−}]\end{array}
Because $\mathrm{C_a}$ and $\mathrm{C_b}$ are large compared to $[\ce{OH−}]$ (the solution is not expected to be strongly alkaline),
the $[\ce{OH−}]$ terms in the above expressions can be dropped and the equilibrium expression becomes
$$\mathrm{K_b}=5.6\times{10^{-4}} =\frac{ (0.150)[\ce{OH−}]}{(0.170)}$$
from which we find $[\ce{OH−}] = \pu{6.347\times{10^{-4}} M}$ , $\pu{pOH} = 3.2$ and $\pu{pH} = 14- 3.2 = 10.8$

Part B : Calculating  the final $ \pu{pH}$(after adding $\pu{0.020 mol}$ of $\ce{ HCl}$)

Calculate the initial amount of $\ce{ CH3CH2NH3+}$ and the initial amount of $\ce{CH3CH2NH2}$:
$$\begin{align}n_\ce{ CH3CH2NH3+} &=n_\ce{ CH3CH2NH3Cl}\\ &=V_\text{(solution)}\times{[\ce{ CH3CH2NH3Cl}]_\text{I}}\\ &=
\frac{\pu{550.0 ml}}{\pu{1000 ml/L}}\times{\pu{0.150 mol/L}}\\
&=\pu{0.083 mol}
 \end{align}$$
$$\begin{align}n_\ce{ CH3CH2NH2} &=V_\text{(solution)}\times{[\ce{ CH3CH2NH2}]_\mathrm{I}}\\ &=
\frac{\pu{550 ml}}{ \pu{1000 ml/L}}\times{\pu{0.170 mol/L}}\\
&=\pu{0.094 mol}
 \end{align}$$
Addition of the strong acid $\pu{ 0.020 mol}$ of $\ce{ HCl}$ , consumes this same amount of $\ce{ CH3CH2NH2}$ and produces $\pu{ 0.020 mol}$ of $\ce{ CH3CH2NH3+}$ ion as explained in the following table:

\begin{array}{c | c c c c c c c}
\text{rxn}&\ce{ CH3CH2NH2} &+ &\ce{HCl} &-> &\ce{CH3CH2NH3+} &+ &\ce{Cl-}\\\hline
 \text{I}&(0.094)&  &(0.020)&  &(0.083)& &(0.083)\\
\text{C} &(0.094-0.020)& &(0.020-0.020)&   &(0.083+ 0.020)&
   &(0.083+ 0.020)  \\
\text{F} &(\pu{0.074 mol})&  &(\pu{0.000 mol})&  &(\pu{0.103 mol})&  &(\pu{0.103 mol})
 \end{array}

We can continue using the approximations for the concentrations of the conjugate species
that we developed in the preceding part A.
The values of C$_\mathrm{b }$ and C$_\mathrm{a }$ are now

$$\mathrm{C}_\mathrm{b }= [\ce{CH3CH2NH2}]_\mathrm{I} =\frac{n_\ce{CH3CH2NH2}}{V}=\frac{\pu{0.074mol}}{\pu{0.550 L}}=\pu{0.133 M}\approx{[\ce{CH3CH2NH2}]_\mathrm{e}}$$
and
$$\mathrm{C}_\mathrm{a }= [\ce{CH3CH2NH3+}]_\text{I} =\frac{n_\ce{CH3CH2NH3+}}{V}=\frac{\pu{0.103 mol}}{\pu{0.55 L}}=\pu{0.187 M}\approx{[\ce{CH3CH2NH3+}]_\mathrm{e}}$$

Substituting these into the equilibrium constant expression yields
$$\begin{align}[\ce{OH−}] &=(\mathrm{K_b})\frac{[\ce{CH3CH2NH2}]} {[\ce{CH3CH2NH3+} ]}\\
&=(5.6\times{10^{-4}})\times{\frac{\pu{0.133 M}} {\pu{0.187M}}}\\&=\pu{3.983\times{10^{-4}} M }
\end{align}$$
the new $\pu{pOH}= 3.40$ ,and the new $\pu{pH}=14- 3.40=10.60$, so addition of the acid has changed the $\pu{pH}$, a decrease of only $0.20~~\pu{pH}$ unit.


Can sodium bicarbonate be considered an acid?

The statement you read that "baking soda is a base" comes from the fact that a solution of sodium bicarbonate (baking soda) and water has a pH of around $8.3$.

However, sodium bicarbonate is amphoteric with respect to Brønsted–Lowry acid/base theory, which means that it can act as either an acid or a base. More specifically, the bicarbonate ion is amphiprotic, meaning that it can either donate a proton, acting as an acid, or accept a proton, making it act as a base. The reason that a sodium bicarbonate solution is slightly basic is that it's ability to take a proton from a water molecule is greater than its ability to donate a proton to a water molecule. In other words, the chemical reaction causing the solution to be basic is:

$$\ce{HCO3- + H2O <--> H2CO3 + OH-}$$

The other equilibrium reaction at play, the reaction "trying" to make the system more acidic, is:

$$\ce{HCO3- + H2O <--> CO3^2- + H3O+}$$

Again, in an otherwise pure aqueous system, it is the first reaction that dominates, and thus giving the observed slightly basic solution.

Note that the $\ce{H2CO3}$ produced in the first reaction is also part of yet another equilibrium process:

$$\ce{H2CO3 <--> CO2 ^ + H2O}$$

Where the $\ce{CO2}$ is liberated as a gas. This is the cause of the well-known "fizzing" when baking soda is added to vinegar (acetic acid).


Safe, easily detectable gas or aerosol for testing fume hoods

Smoke generation is preferred over smell, as the visual test is a bit more objective. See this reference for more info, although $\ce{TiCl4}$is not available for home use. There are hobbyist alternatives such as smoke matches which will help you test the fume hood performance. (The above link is just an example, I did see better prices at less-known websites.)  An anemometer might be of help as well.


Which definition of van der Waals forces is correct?

van der Waals forces is not another name for London dispersion forces. In his answer, airhuff has given IUPAC definition for van der Waals forces.
London dispersion forces are the intermolecular forces that occur between atoms and between nonpolar molecules as a result of the motion of electrons in those atoms and molecules. They are induced dipole-induced dipole interaction, better described in following diagram:

These inter-atomic or inter-molecular forces are the weakest of van der Waals intermolecular forces. London dispersion forces got its name due to its discoverer, Fritz London (Ref.1). You can find some details of the distance dependence on London forces here.
References:

Fritz London, "The general theory of molecular forces," Trans. Faraday Soc. 1937, 33, 8b-26 (https://doi.org/10.1039/TF937330008B).


If concentration of K+ ions inside the cell is almost equal to concentration of Na+ ions outside of the cell why these ions cross the membrane?

In order to understand why this happens, you have to recall the permeability characteristics of the plasma membrane. It seems that the movement of ions to and fro the cystol is counterproductive however this is not so. Actually there is unequal distribution of ions in the extracellular fluid and cytosol

  Extracellular fluid is rich in Na+ and chloride ions (Cl–). In
  cytosol, however, the main cation is K+, and the two dominant anions
  are phosphates attached to molecules, such as the three phosphates in
  ATP, and amino acids in proteins.

Differences in intracellular and extracellular concentrations of ions result primarily from:

the Na +/K + pump and
the permeability characteristics of the plasma membrane.

Ions pass through the plasma membrane through ion channels. The two major types of ion channels are leak channels and gated ion channels.

i. Leak Channels

Leak channels, or non-gated ion channels, are always open and are responsible for the permeability of the plasma membrane to ions when the plasma membrane is unstimulated, or at rest. The plasma membrane is more permeable to K +and Cl- and much less permeable to Na+ because there are many more K + and Cl- leak channels than Na+ leak channels in the plasma membrane.
Because the plasma membrane typically has more K+ leak channels than Na+ leak channels, the number of potassium ions that diffuse down their concentration gradient out of the cell into the ECF extracellular fluid ECF is greater than the number of sodium ions that diffuse down their concentration gradient from the ECF into the cell. See below

The plasma membrane is relatively permeable to K + and much less permeable to Na + and to negatively charged molecules found inside of the cell. Consequently, positively charged K + tend to diffuse out of the cell through K + leak channels, leaving the negatively charged molecules behind, thus polarizing the membrane. The membrane is at equilibrium when the tendency for K + to diffuse out of the cell is resisted by the negative charge of the molecules inside the cell.

ii. Gated Ion Channels

These channels open and close in response to small voltage changes across the plasma membrane. The movement of ions into or out of the cell changes the charge difference across the plasma. membrane, which causes voltage-gated ion channels to open or close. Voltage-gated channels specific for Na+ and K + are most numerous in electrically excitable tissues

The Na+/K+ Pump

The differences in K + and Na + concentrations across the plasma membrane are maintained primarily by the action of the Na + –K + pump. Through active transport, the Na+/K + pump moves K+ and Na+ through the plasma membrane against their concentration gradients. Potassium ions are transported into the cell, increasing the concentration of K + inside the cell, and Na + are transported out of the cell, increasing the concentration of Na + outside the cell. Approximately three Na+ are transported out of the cell and two K + are transported into the cell for each ATP molecule used. These pumps help maintain the resting membrane potential by pumping out Na+ as fast as it leaks in. At the same time, the Na+/K+ ATPases bring in K+.

In summary it is important to note, there is uneven distribution of ions in and outside the cytosol.
Because the plasma membrane has more K+ leak channels than Na+ leak channels, the number of K+ ions that leave the cell is greater than the number of Na+ ions that enter the cell. As more and more K+ ions leave the cell, the inside of the membrane becomes increasingly negative and the outside of the membrane becomes increasingly positive. The inability of most anions to leave the cell also contributes to the negativity of the resting membrane potential. These trapped anions cannot follow K+ out of the cell because they are attached to non-diffusible molecules such as ATP and large proteins.

Nevertheless, sodium ions do slowly diffuse inward, down their concentration gradient. If not controlled, such inward leakage of Na+ would eventually destroy the resting membrane potential. The small inward Na+ leak and outward K+ leak are offset by the Na+/K+ ATPases (sodium pump)

This diffusion accross the plasma membrane is necessary because:

There is uneven distribution of ions which needs to be corrected.
The characteristics of the plasma membrane has lot to do with how transport of substances occur ( It is important to note that not only ions use this path, other substances vital for cell growth have to use this path as well, therefore strict control is necessary)

Hope this helps

References

Anatomy and Physiology (Seely)
Principles of Anatomy and Physiology (G. Tortora, B. Derrickson)


How do IR spectrometers produce light spanning the entire infrared spectrum?

This could be accomplished by a scanning instrument. A broad-band light source could output IR radiation over the whole range. This incident radiation is passed through a monochromator to individually select wavelengths to send to the sample. This monochromator can work many ways, using mirrors, gratings, prisms, etc.

Alternatively, you could expose the sample to the broad-band radiation and use a detector that captures all of the transmitted radiation at once. UV/Visible spectroscopy can be done with a photodiode array detector capable of this, but I know of no IR spectrometers that use this method.

Most modern IR spectrometers use Fourier transform. An interferometer, converts an IR laser beam into broad-band IR pulses through interference generated with moving mirrors or by other methods. This action converts the coherent beam into an interference pattern in time. The interferogram below shows what this might look like (where this mirror position is also the time axis). The image has been released to the public domain by Wikipedia user Sbyrnes321.

Absorption of IR radiation by the sample perturbs the interference pattern before it hits the detector. Electronics or software perform the Fourier transform, converting time data into frequency data.


I'll assume we're talking about aldol reactions, ie. the electrophile is a aldehyde, or less commonly, a ketone. Being a more strongly coordinating cation, lithium can actually coordinate to both the enolate oxygen atom AND the electrophile oxygen. This serves to "pre-organise" both reactants for a direct addition via a six-membered, cyclic transition state. The favoured transition states are often chair-like. This is known as the Zimmerman-Traxler model.

The model explains diastereselectivity when that is possible, ie. E enolates give predominantly anti products, while Z enolates give mainly syn products.

Lithium also stabilises the initial aldolate product, by chelation via both oxygens.

The more highly dissociated Na and K cations presumably don't greatly facilitate reaction via a cyclic transition state, nor form a stable chelate of the product. The reaction is also more likely to be reversible, and the products of conjugate addition more stable.

Sent from my Samsung Note 8


Relative acidity of p-chlorobenzoic acid and p-flurobenzoic acid

For acidity of para substituted benzoic acids, -I effect is helpful and +R effect decreases the acidity.

In 4-fluorobenzoic acid, the +R effect is prominent in $\ce{F}$ due to stronger $\ce{2p}\pi - \ce{2p}\pi$ overlap with nearest carbon. Thus, due to stronger +R effect of $\ce{F}$ than -I effect, it creates a negative charge on your starred carbon. This charge delocalises in the carbonyl carbon  and the electrophilicity is thus decreased of that carbon. Releasing hydrogen also doesn't help much as the negative charge created over $\ce{O}$ of $\ce{-OH}$ group can only delocalise a little.

Whereas in the $\ce{Cl}$, there is much weaker $\ce{3p}\pi-\ce{2p}\pi$ overlap. So, +R effect of $\ce{Cl}$ is much weaker and due to a relatively higher -I than its +R, there is more positive character on the starred carbon due to $\ce{Cl}$.

And, thus, in reality, acidity of 4-chlorobenzoic acid is a little higher than that of 4-fluorobenzoic acid.


Must compounds formed between solids be a solid as well?

No, they don't have to. In short, you have to memorize. The physical state of a compound has nothing to do with the states of its elements.

Examples:
$$\ce{Hg~(liquid) + Br2~(liquid) -> HgBr2~(solid)}$$
$$\ce{C~(solid) + S~(solid) -> CS2~(liquid)}$$

Now, there are some common patterns (an ionic compound is most certainly a solid; a compound with small non-polar molecules is probably a gas), but that's another story.


I see 4 possibilties to avoid EDTA sedimentation:

Mechanical ways like shaking, mixing or container turning - but you say it is not applicable.

Increasing viscosity by thickening glycerol to  decrease the rate of sedimentation. Gel stages could probably even stop the sedimentation, but may not be applicable. You may want to try

Carboxymethyl cellulose (CMC)
Polyvinyl alcohol(PVA)
Polyvinylpyrrolidone/polyvidone/povidone (PVP)
Polyethylene glycol(PEG)

Other way would have been adjusting the glycerol density to match EDTA density if EDTA had not had too low density $\pu{0.86 g/cm^3}$ ( while glycerol  $\pu{1.26 g/cm^3}$). OTOH, if there was a chance to replace glycerol by ethanol 73 w/w %, having the same density as EDTA has ... but I guess  there is not.

It may be worthy to try some emulsifiers, but EDTA particles may be too large and emulsifiers may not be applicable for the suspension purpose.

Soy lecithin is another emulsifier and thickener
Pickering stabilization – uses particles under certain circumstances
Sodium phosphates - not directly an emulsifier, but modifies behavior of other molecules, e.g. caesin
Mono- and diglycerides - a common emulsifier found in many food products (coffee creamers, ice-creams, spreads, breads, cakes)
Sodium stearoyl lactylate
DATEM (diacetyl tartaric acid esters of mono- and diglycerides) – an emulsifier used primarily in baking
Simple cellulose – a particulate emulsifier derived from plant material using only water


Reduction of 1,4-benzoquinone in the synthesis of 1,5-diazocane

Here is a relatively straightforward synthesis of 1,5 diazocane that uses the Beckmann rearrangement as a key step. The Beckmann rearrangement is a useful synthetic tool for converting ketones to amides.  If the carbonyl is in a ring, then conversion to the amide will enlarge the ring by one nitrogen atom.

The novel twist in the following synthesis is to perform a double Beckmann rearrangement in the penultimate step.

heating benzene with concentrated sulfuric acid containing excess sulfur trioxide (oleum) will lead to the disulfonated product, benzene-1,3-disulfonic acid
heating the disulfonic acid with potassium carbonate produces resorcinol (see here)
treating resorcinol with hydrogen gas at low pressure over a rhodium catalyst yields 1,3-cyclohexanedione (see here)
hydroxylamine can be prepared by oxidizing ammonia with hydrogen peroxide to produce ammonium nitrite which can then be converted by a number of methods to hydroxylamine
the hydroxylamine is used in standard fashion to afford the bis-oxime (various oxime stereoisomers are likely produced, but this does not matter, all oxime stereoisomers will undergo the Beckmann rearrangement in the next step)
the bis-oxime is converted to the bis-amide through a double Beckmann rearrangement (see here for this specific conversion; note that the double Beckmann could afford isomers but only the desired isomer is produced; see the preceding link for the authors explanation as to why this is so)
finally, standard reduction of the bis-amide with $\ce{LiAlH4}$ and aqueous workup gives rise to the target molecule

The double Beckmann could also be run starting with 1,4-cyclohexanedione (readily prepared from benzene, see here).  1,4-cyclohexanedione has the advantage of symmetry - the first Beckmann rearrangement can produce only one isomer.  Although the second Beckmann rearrangement could produce 2 isomers, only the desired 1,5-diazocane-2,4-dione was isolated (see here).


What happens when a solution of sodium chlorite is acidified?

When acid is added to sodium chlorite in solution (note: $\ce{ClO2-}$ is the chlorite ion), chlorous acid ($\ce{HClO2}$) is transiently formed which then goes on to decompose. However, you need hydrochloric acid, not just any acid, because chloride ion ($\ce{Cl-}$) itself acts as a catalyst for the decomposition. What happens is that chlorous acid disproportionates; that is, some of it gets oxidized while the rest gets reduced.

Specifically, out of five $\ce{Cl}$ atoms in chlorous acid (oxidation state +3), one will end up as chloride in the oxidation state $-1$, and four will end up as $\ce{ClO2}$ in the oxidation state +4:

$$\ce{5HClO2 -> 4ClO2 + Cl- + H+ + 2H2O}$$

Pretty much any concentration will do. Calculating the proportions is easy if you know a bit of stoichiometry: you need 5 moles of chlorite for 4 moles of hydrochloric acid. The reaction proceeds to completion, and you get 4 moles of $\ce{ClO2}$.

The $\ce{ClO2}$ remains in solution and is a good bleaching agent; see Wikipedia. It can undergo further disproportionation but if the solutions are kept in the dark, this process is slow.


Is butanoic acid more soluble in water than ethyl acetate?

Try to solve the question by analogy. Imagine that you perform a couple of experiments on different glasses of water:

Add vinegar to one glass with water. Does it mix? Yes, it does!
Add olive oil to another glass with water? Does it mix? No, it does not.

Vinegar is a carboxylic acid, whereas olive oil is made of esters. This should suggest to you that carboxylic acids in general are more soluble than esters.

Carboxylic acids, $\ce{R-COOH}$, are acids, that partly dissociate in water

$$\ce{R-COOH + H2O -> H3O+ + R-COO-}$$

to form ions. Supposing that the alkyl chain isn't extremely long, even the undissociated part of the acid added is solvated in water, a polar protic solvent, by hydrogen bonding interactions. In other words, the carboxylic acid is highly soluble in water.

The solubility of ethyl acetate in water is 8.3 g/100 ml (Wikipedia), whereas butanoic acid is completely miscible in water. That is to say, butanoic acid completely dissolves in water, regardless of the proportion of acid to water.


Why diethyl ether does not absorb in UV region?

We can measure the UV-Vis spectra for π → π* or n → π* electronic transitions (other transitions are forbidden according to selection rule) [1, pp. 577–578]. But in diethyl ether, being a saturated compound, it has no π-bonds. This successfully dismisses the idea of having π* orbital. And, as there is no π* orbital, we can not have a transition to the π* orbital resulting in no UV-Vis absorption spectra.
Reference

Pavia, D. L.; Lampman, G. M.; Kriz, G. S.; Vyvyan, J. R. Introduction to Spectroscopy, 5th ed.; Cengage Learning: Stamford, CT, 2015. ISBN 978-1-285-46012-3. (PDF)


Molecular chirality and optical rotation

since for every orientation of the molecule, we can reverse the orientation such that the light appears to be falling on the molecule from a direction other than the one for our original molecule.

This is false.

Let's take 2-butanol. For this stereoisomer, light is turning clockwise when viewed from the right side (I'm not sure of this, but we can assume).

Here, the disks are there to clarify the direction from which we are viewing the entire setup. Here, the "face" of the disks is to the right so we are viewing it from the right.

By the principle of reversibility of light, if light enters from the other side, we get the reverse path:

Rotate this diagram (by $180^\circ$ around an axis perpendicular to your screen):

Note that the disks are being viewed from the other side (the left) now. But we can flip them:

Take a close look at this. This is the same molecule as the first, just oriented backwards. And looking from the right, it still gives a clockwise twist to the angle.

So while rays of light incident on different orientations of a molecule may differ in the exact value of the angle of rotation, rays of light incident on a reversed (not stereoisomeric, just spatially rotated) molecule have the same direction and value of the angle of rotation and do not cancel.

[8]:


Hydrolysis of vinyl bromide vs alkyl chloride

Hydrolyzing the saturated alkyl chloride could occur by SN2 but I'd be inclined to expect SN1 in this situation. Chlorine would leave behind a (secondary) carbocation, which would be attacked by the nucleophile: hydroxide (basic conditions) or water (acidic conditions). The rate limiting step of SN1 is loss of the nucleophile, and the activation energy might be lowered in an acidic environment because the hydronium could coordinate to the chloride to stabilize it. The real issue is the resulting carbocation though. In basic conditions, hydroxide could backside-attack, kicking out the chlorine before it has the chance to leave. This concerted SN2 mechanism would avoid the carbocation entirely, but it is slowed by the steric hindrance of the secondary alpha carbon. So here acidic favors SN1, limited by formation of the secondary carbocation, and basic favors SN2, limited by secondary sterics.

Hydrolysis of the vinyl bromide I think must occur by SN1 in basic conditions or protonation-addition-elimination in acid. SN1 requires formation of a vinyl cation when the bromine leaves, which is quite unstable and will severely limit the rate of the reaction. The rate limiting step of the acidic P-A-E sequence is probably the protonation (slower than addition, and I would expect also slower than elimination because protonation is bimolecular). I would expect protonation of an alkene (PAE pathway) to occur much faster than loss of a nucleofuge (leaving group) to form a vinyl cation (SN1 pathway); forming a secondary carbocation would surely have a much lower activation energy barrier than forming a vinyl one. So I expect the acidic hydrolysis of the vinyl bromide via hydration-elimination (P-A-E) to occur much faster than the SN1 hydrolysis in basic conditions.

So the pH conditions may play a significant role here, with base severely slowing the hydrolysis of the vinyl bromide. In basic conditions, I would almost certainly bet the alkyl chloride will hydrolyze faster. Acidic conditions will speed up hydrolysis of the vinyl bromide, with the RLS being protonation to form secondary carbocation or possibly elimination of bromide to restore the double bond. Acidic conditions for the alkyl chloride promote SN1, whose RLS is loss of chloride to form the secondary carbocation. It's hard for me to predict with only theory whether vinyl bromide acidic RLS or alkyl chloride acidic RLS will occur faster.

An interesting note, hydrolyzing your vinyl bromide yields an enol, which will spontaneously tautomerize to a ketone. So your hydrolysis products are a ketone (from vinyl bromide) and an alcohol (from alkyl chloride). The difference of oxidation state in these products makes me wonder why you are asking this question.

Update: Here are my predicted mechanistic pathways. The details vary slightly from my analysis, especially how tautomerization factors in. I also included a couple of alternative pathways I was trying out (Acidic #2 and Basic #2 for vinyl bromide, and Basic #2 for alkyl chloride; these are probably negligible).


No. Per definition you need to have at least local supersaturation. You can archive it by different means (cooling, evaporating of solvent, mixing with other solvent...) but in the end you need supersaturation. See: http://xray.chem.ufl.edu/growing%20tips.htm or http://web.mit.edu/x-ray/cystallize.html

Of course you can use techniques like sublimation but your question was about solution (btw in case of sublimation you also have supersaturation, only difference is that your solvent is gas).


Can you extract specific compounds from infused oils?

THC is often extracted from cannabis by mashing up the plant with a solvent such as hexane. The slurry obtained may then be filtered off, and the solvent evaporate to give a crude extract from the cannabis which will contain the THC amongst other organic molecules.  Depending on the extraction solvent, the components of the cannabis plant are extracted in different concentrations, depending upon their relevant affinity for dissolving in the solvent.

Coconut and olive oil have both been used in extractions, as they are safe for human consumption and so can be used directly once isolate without the need to remove potentially toxic (hexane) or flammable (hexane) solvent.

In general, the solvents used are inert, that is, they do not react to degrade the extraction products as you mentioned.

With regard to homeopathy, this too involves the extraction of active components from plant/biological matter. However, in homeopathy, it is generally less of an exacting science (for instance specifically getting THC out of cannabis) and more a case of extracting ALL of the components of a plant, for instance by making a tea. If you are interested in this, you might like to look at the recent Nobel prize winner Youyou Tu (2015, Physiology or Medicine). She realised that boiling the plant in water to create a tea was destroying the functionality of the molecule that chinese medicine was trying to extract, limiting its potency. By exploring alternative extraction methods she was able to get the biologically active compound without degradation leading to a useful drug.


Composition of arcanite (potassium sulfate) agricultural grade. Potassium content of random drawn sample?

The deviation can be in range from negligible to huge, depending on the level of nonhomogeneity and locality of the sample.
Note that routine analysis of potentially non-homogenous sources involve sophisticated sampling and sample averaging schema, so the result is representing number.


Boiling chips for concentrating sulfuric acid

Clean porcelain or glass shouldn't react with concentrated sulfuric acid.
For boiling chips you want to use homogeneous material with a substantial/developed surface:

Porous unglazed ceramic.
Raschig rings, which are also widely used industrially for the similar distillation purposes, are probably the ideal choice.
An alternative for a consumer market would be porous ceramic rings for aquarium filters, which is basically the same thing (stick with the white ones).

The majority of glass types (borosilicate glass, quartz, …).
Broken pieces of chemical glassware or light bulbs would do.
For higher efficiency, use glass capillaries or sintered glass pieces.

Potential cause of cloudy solution when you used mug pieces might be the glaze on top of porcelain or the porcelain itself if it hasn't been properly thermally treated (remaining binder), and, of course, porcelain dust particles leftovers may float around forming a suspension.
Yellow color of sulfuric acid when you used glass pieces is likely an indicator that acid came in contact with organics (e.g. oil, grease).
Pure colorless glass won't cause this coloration.
On a side note, there is no need to use grease at all as sulfuric acid itself is quite viscous and can serve as a sealant for the joints (if you are using any).
Notes, advises

If the liquid has been cooled down, make sure to add at least one new chip before heating again to the boiling point (prevents superheating).
Avoid using boiling chips for vacuum distillation.
Instead, use mechanical stirring or bubble inert gas (argon, nitrogen) through a submerged glass capillary.
Boiling chips don't work under vacuum.
Never add boiling chips to a hot solution as it will result in rapid vaporization and the liquid is going to boil over or bump.
Keeping a substantial amount of sodium carbonate nearby might come in handy in neutralizing spilled acid, if there is any.


How to analyze a acid and salt mixture using titrimetry?

Prologue

The following will show you not only how to sketch a titration curve but how to produce an analytical form of a titration curve. So this might not be an easy solution even if I try to keep it as easy as possible. There is a wonderful yet german book that has (to my knowledge) never been translated to english which describes everything of the following in detail:

Bliefert, Claus; Linek, Alfred; Morawietz, Gerd (1978): PH-Wert-Berechnungen. Weinheim: Verlag Chemie.

There might also be some imprecise terms because I don't know many of the english counterparts to the german terms I am familiar with.

Answer

Let's think a bit about your problem first to simplify it from scratch:

As you mentioned, reaction (a) will take place right after all protons from hydrochloric acid have reacted and before reaction (b) and (c) would start.
The pK$_a$ of (b) and (c) are so close to another that they are unable to be separeted by titration and
the pK$_a$ of (b) and (c) are so high that they are unable to be determined by titration in diluted aqueous solutions.

If we take a look at the titration curve of phosphoric acid, we can see what is meant by "too high to be determined". You can clearly identify the two inflection points for the first two protons but there is no third inflection point that could be determined. What is likewise to be seen, is that one can neglect this third deprotonation step in titration curves of diluted solutions (yellow vs. blue titration curve).

Right now we can reduce your problem to a mixture of hydrochloric acid and M$^{3+}$. To continue we need to establish a proton condition (Protonen-Bedingung):

$$c(H_3O^+) = c(OH^-) + c(A^-) + c(X^-) - c(HB^+) \tag{1}$$

It means that protons come from the autoprotolysis of water (c(OH$^-$)), from hydrochloric acid (c(A$^-$)), from your metal ion (c(X$^-$)) and that the protons react with the used base (c(HB$^+$)).

To define an equation for the acid residue anions a small equation system (where small c's denote actual concentrations in solution and the big C denotes the origin concentration)
$$\begin{align}c(HA) \cdot k_a - c(H_3O^+) \cdot c(A^-) =&~0\\c(HA)+c(A^-) =&~C(HA)\end{align} \tag{2.1}$$
needs to be solved to give
$$c(A^-) = \frac{C(HA) \cdot k_a}{c(H_3O^+)+k_a}. \tag{2.2}$$

This can be done analogously for the other acid and also for the base.

What remains is the concentration of the hydroxide ions from the autoprotolysis of water that is simply determined by rearranging the equation for the autoprotolysis itself:

$$c(H_3O^+) \cdot c(OH^-) = k_W \Leftrightarrow c(OH^-) = \frac{k_W}{c(H_3O^+)} \tag{3}$$

Combining everything so far we get the following proton condition:
$$x = \frac{k_W}{x} + \frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}} - \frac{x \cdot C(B) \cdot k_b}{x \cdot k_b + k_W} \tag{4}$$

where $c(H_3O^+)$ has been replaced by $x$ for the sake of simplicity.

The plot of titration curves usually shows the pH over the base volume (see eq. 8) or the degree of neutralisation ($\tau$ or $\gamma$) that is defined as $$\tau = \frac{c(B)}{C(HA)} \tag{5.1}$$

In the case of your mixture of two acids - and as always neglecting the increase in volume - it can be written as
$$\tau = \frac{c(B)}{C(HA)+C(HX)} \tag{5.2}$$

To sketch the titration curve let's say that we know the two concentrations of the acids. To calculate (5.2) we also need to know c(B) that can easily be obtained by rearranging (4) to get:
$$c(B) = \left(1+\frac{k_w}{x \cdot k_b}\right)\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right) \tag{6}$$

If one decides to include the increasing volume the concentrations of the acids and the base have to be extended to give
$$\begin{align}C(HA) \rightarrow&~\frac{C(HA) \cdot V(HA)}{V(HA)+V(B)}\\C(B) \rightarrow&~\frac{C(B) \cdot V(B)}{V(HA)+V(B)}\end{align} \tag{7}$$

Inserting (7) in (4) and rearranging for V(B) we finally get a neat equation

$$V(B)=\frac{V(HA)~\left(\frac{k_w}{x}-x+\frac{C(HA) \cdot k_{a,HA}}{x+k_{a,HA}} + \frac{C(HX) \cdot k_{a,HX}}{x+k_{a,HX}}\right)}{\left(x-\frac{k_w}{x}+\frac{C(B) \cdot k_b}{\frac{k_w}{x}+k_b}\right)} \tag{8}$$

As the pH is defined as $pH = -log_{10}(c(H_3O^+))$ and $c(H_3O^+)=x$ you can replace every $x$ in equation 8 with $10^{-x}$.

Now you are able to calculate with e.g. Excel enough (V(B),pH)-pairs to draw a titration curve and get some information about your system of two acids during the titration with your desired base.

The following figure shows some examples of your mixture using equation 8.

Please let me know if someone is interested in the full equation including also the second and third deprotonation step with pK$_a$ values 13.3 and 13.5 and I will add it here.


Chemical kinetics of a reversible reaction

The simplest and most general way to solve this type of scheme is to calculate the effect of an amount x that reacts with initial amounts $A_0$ and $B_0$, thus
$$\ce{A~~~ <=>[k_a][k_b] ~~~B}  \\  A_0-x  ~~~~  B_0 +x $$
which produces the rate equation
$$ \frac{dx}{dt} = k_a(A_0-x)-k_b(B_0-x)$$

the variables can be split to give a straightforward integration instead of having to solve a complicated pair of differential equations.
$$\int _0^x \frac{dx}{k_aA_0-k_bB_0-(k_a+k_b)x} = \int _0^t dt$$

which, after some rearrangement, gives
$$\ln\left(1-\frac{(k_a+k_b)}{k_aA_0-k_bB_0}\right) = -(k_a+k_b)t$$
and this can be rearranged to
$$x= \left( 1-e^{-(k_a+k_b)t}  \right)\frac{(k_aA_0-k_bB_0)}{k_a+k_b}$$

with the initial condition that at $t=0$, $B_0=\ce{[B]=0 }$ this becomes
$$x= \left( 1-e^{-(k_a+k_b)t}  \right)\frac{k_aA_0}{k_a+k_b}$$
The concentration of A is $A_0-x$ which is

$$ \ce{[A]}_t =A_0 \frac{k_b+k_ae^{-(k_a+k_b)t}}{k_a+k_b}$$

This expression has the correct form at all times. $\ce{[B]}_t$ can be calculated similarly from $B_0+x$.


Methylnitrobenzene ring formation Synthetic Route

This is a challenging question as we have so few details.

One method would be to convert the orthonitrotoluene into the benzyl bromide with NBS or another free radical bromination reagent. Then you could react the dimethyl malonate with sodium methoxide in methanol. Next react the sodium salt of dimethyl malonate with 2-nitrobenzyl bromide.

Then filter off the sodium bromide thus formed, next stir it with the product of reacting Merifield's resin with tris-2-aminoethyl amine to scavenge anyunreacted nitrobenzyl bromide. Next subject the mixture to hydrogenation with a palladium on carbon cataylst. This will convert the nitro group into an amine.

Next evapourate it down to form a oil, heat this to make the amine react with the methyl ester to form methanol and an amide. When you are done with this step. Then react it with dilute sodium hydroxide to convert the remaining methyl ester into the sodium carboxylate. Make it acidic again and it will decarboxylate to give you a cyclic amide.

For the next step I would convert the amide to a thioamide with Lawesson's reagent and then treat the thioamide with rainey nickel to convert it into am amine. Between some of the steps it would be a good idea to purify the intermidates, but this is how I would (off the top of my head) try to make the conversion. I suspect that there will be cheaper ways of making the final product.


Why is acetyl cyanide not obtained from acetyl chloride?

Acetyl chloride on reacting with $\ce{KCN}$ in ethanol will give ethyl acetate. If acetyl cyanide forms at all, it will immediately react with the ethanol solvent as it is a more reactive species than acetyl chloride.

Acetyl cyanide is known and may be formed by catalytic reaction of hydrogen cyanide and ketene at elevated temperature. Details.

Aromatic acyl cyanides are less reactive than alkyl acyl cyanides much as aromatic acyl chlorides are less reactive. They may be formed by reaction of the neat aromatic acyl chloride with cuprous cyanide (details) and are stable to a wash with cold aq. sodium bicarbonate.


Anhydrous caffeine occurs in two different polymorphic crystal forms, with a transition between the two forms occurring at $\pu{136 °C}.$ The high temperature form has a melting point of $\pu{236 °C}.$
The vapor pressure of the high temperature solid form is described by the equation:
$$\log P[\pu{Pa}] = -\frac{5477}{T[\pu{K}]} + 14.395,\tag{1}$$
and for the liquid
$$\log P[\pu{Pa}] = -\frac{3918}{T[\pu{K}]} + 11.14.\tag{2}$$
So, $\pu{173 °C}$ is incorrect; the boiling point is more like $\pu{365 °C}.$
References

Griesser, U. J.; Szelagiewicz, M.; Hofmeier, U. Ch.; Pitt, C.; Cianferani, S. Vapor Pressure and Heat of Sublimation of Crystal Polymorphs. Journal of Thermal Analysis and Calorimetry 1999, 57 (1), 45–60. DOI: 10.1023/A:1010188923713.
Bothe, H.; Cammenga, H. K. Phase Transitions and Thermodynamic Properties of Anhydrous Caffeine. Journal of Thermal Analysis 1979, 16 (2), 267–275. DOI: 10.1007/BF01910688.


Are all monomers in a polymer covalently bonded?

I think your second quote is referring to some polymers such as sodium polyacrylate, $\ce{[-CH2-CH(COONa)$-$]_n}$. This polymer is basically just the sodium salt of polyacrylic acid (or polymerized sodium acrylate). You can see below the ionic bond between the anionic carboxylate group and the sodium cation:

However, the quote could be referring to so-called 'chain-link' polymers, in which the monomers are intertwined and looped/knotted together, like links in a chain, instead of just consisting of a bunch of long strings of monomers, like a rope. However, chain-link polymers are highly uncommon and rather difficult to synthesize. Note that these are not the same as cross-linked polymers.


Why would a solution of FeCl2 appear brown/yellow sort of like FeCl3?

The yellow/brown color you are seeing in the solution of $\ce{FeCl2}$ is due to formation of various types of hydrated iron oxide and not iron(III) chloride. They may include both the anhydrous($\ce{FeOOH}$)  or the monohydrate($\ce{FeOOH.H2O}$) often referred to as ferric hydroxide($\ce{Fe(OH)3}$). They are generally yellow to brown in color. The reaction is given here:
$$\ce{4FeCl2 + 6H2O + O2 → 4FeO(OH) + 8HCl}$$

Reaction between iron chloride (II), water and oxygen(moisture) to form
iron metahydroxide(Iron(III) oxide-hydroxide) and hydrogen chloride. The reaction takes place at
reflux.

Moreover, if you further heat iron chloride in oxygen at 450-480℃, iron oxide will form.
References(all pdf links)

http://www.jieas.com/fvolumes/vol081-5/3-5-11.pdf
http://www.gfredlee.com/SurfaceWQ/StummOxygenFerrous.pdf
https://repository.kulib.kyoto-u.ac.jp/dspace/bitstream/2433/77056/1/chd061_5-6_335.pdf


How were urolithins discovered and named?

The identification of urolithins as products of human metabolism of ellagic acid was reported by Cerda et al. in the European Journal of Nutrition in 2004. The full reference is
Cerda, B., Espin, J.C., Parra, S., Martinez, P. and Tomas-Barberan, F.A. (2004) The potent in vitro antioxidant ellagitannins from pomegranate juice are metabolised into bioavailable but poor antioxidant hydroxy-6H-dibenzopyran-6-one derivatives by the colonic microflora of healthy humans. Eur J Nut 43:205-220.
However, these researchers did not name the compounds as urolithins. Rather, they note that the compounds

have also been found as constituents of ‘clover stone’, a type
of renal calculus found in sheep which can also ingest large
amounts of ellagitannins[26]. In fact, the metabolite d
(3,8-dihydroxy-6H-dibenzo[b,d]pyran-6-one) has been previously named
as ‘urolithin  B’ [22, 26].

That explanation seems sufficient to understand the source of the name, but if you are interested in the specifics of the original discovery, reference 26 is to the Dictionary of Natural Products published in 1992 by Chapman & Hall, which is available only by subscription. Perhaps someone with access to that Dictionary can look up the entry on urolithin and provide more info.
UPDATE: Since this answer has received a decent amount of attention, I dug a bit deeper to find the references for the first isolation and naming of the compounds (thereby actually answering the question).
In 1963 at the 430th meeting of the Biochemical Society in Oxford, England, Nottle and Pope first reported isolating two compounds that they named "urolithin A" and "urolithin B". Their report was published among the proceedings of the meeting in the Biochemical Journal, which are available for free from the journal here. The full reference is
Nottle, M.C. and Pope, G.S. Two Constituents of 'CloverStone', a Type of Urinary Calculus Found in Sheep in Proceedings of The Biochemical Society. Biochem J (1963) 89 (1): 1P–67P.
The Nottle and Pope report is on p. 67P. In that report, however, the structures of the urolithins were not described, as they had not yet been determined. In a subsequent paper, Pope reported the structures of the urolithins, noting that they were actually previously known compounds (specifically urolithin A is identical to pigment I of castoreum, isolated from beaver scent glands), but despite the compounds already being known, Pope and others continued to use the new name, which seems to have stuck. That report is also freely available from the Biochemical Journal here. The full reference is
Pope, G.S. (1964) Isolation of Two Benzocoumarins from 'Clover Stone', a Type of Renal Calculus Found in Sheep. Biochem J 93:474.


Is it possible to write a redox reaction with two oxidations and one reduction together?

I wouldn't try to combine these reactions. You would want to consider both reactions and maybe look up the standard reduction potentials of all reactants and products involved to give yourself an idea of what is oxidizing what and what is reducing what. In general, you should let the SRP's be a guide for trying to figure out what actually happens in these reactions.

The reaction you wrote doesn't really describe anything chemically. We have to make sure not only the math works out, but also the chemistry. Also, your charge balance is wrong, so even if the chemistry did work out, the reaction would be impossible. You seem to have 24 (-) on the reactants side and net 2 (+) on the products side.

Remember to always check BOTH your mass (MB) and charge balance (CB) work out when you're doing REDOX problems.

The following two reactions are the acidic balanced REDOX reactions for the the species you have written. Of course, these may not actually be the real reactions that would occur. For that, you would have to look things up in tables in your book.

$\ce{6 H+ + 2MnO4- + 5SO3^2- -> 5SO4^2- + 2Mn^2+ + 3H2O} \quad\text{(CB: -6 : -6)}$

$\ce{16 H+ + 2 MnO4- + 5C2O4^2- -> 10 CO2 + 2Mn^2+ + 8 H2O} \quad\text{(CB +4 : +4)}$


why are amines more water soluble than alcohol for similiar molecular weight?

I think it may be due to van der waals force (v.w.f). For both alcohols and amines the v.w.f acting between them and water molecules is keesom force i.e. dipole dipole interaction. In such case v.w.f is directly proportional to molecular mass , but if the molecular mass is also same then v.w.f is directly proportional to surface area of molecule which is inversely proportional to branching. Hence since amines have more branching than alcohol they have less surface area and hence less van der waals force. Due to weaker v.w.f amines are less soluble than alcohol.
Please correct me if I am wrong.


Volume in decimetres cubed dimensional analysis

This touches a very powerful analytical technique called unit analysis (aka dimensional analysis).

Moles per decimetre cubed is $\dfrac{\text{moles}}{\text{dm}^3} = \text{moles}\cdot \text{dm}^{-3} \ne \text{moles}\cdot \text{dm}^3$

So the -3 indicates that the unit is in the denominator not the numerator.

PS - I learned chemistry when there were only four elements - earth, wind, water and air. I think of liters not $\text{dm}^3$.


How many times can we recycle one sheet of paper?

Well apart from the fact that the answer depends on the type of the paper that is used. Secondly how much is wasted and also the process used. Below is an article that might be helpful in answering the question

Some industry sources estimate that an ordinary sheet of paper made from cellulose fibers derived from wood can survive only four to six trips through the recycling process. The Environmental Protection Agency puts the figure at five to seven times.
It is not surprising that the rigors of remanufacturing take a toll on the fibers.
Ideally, paper for recycling is separated into types, because paper with long fibers, like white office paper, offers the most flexibility for recycling, while newsprint, with its shorter fibers, is usually reserved for making more newsprint and other low-quality papers.
The paper is shredded and chopped, then subjected to a mixture of chemicals and water and heated as it is repulped. It is centrifuged and screened to remove impurities; de-inked with more chemicals; then sprayed onto a wire screen, drained, dried and squeezed through heated rollers.
With each step, the fibers become shorter, coarser and stiffer, so that eventually, recycled fiber needs to be mixed with virgin fiber to make paper of the desired quality.
According to the American Forest and Paper Association, 63.4 percent of the paper consumed in the United States was recovered for recycling in 2009. C. CLAIBORNE RAY

SOURCES: New York Times - http://www.nytimes.com/2010/12/21/science/21qna.html?_r=0


A quick search in my university's library for Life cycle polymer brings me two reputable looking books:

Green composites: Polymer composites and the environment
Fundamentals of Materials for Energy and Environmental Sustainability

A Google Scholar search for life cycle analysis of plastic yields several good looking hits. A similar result for life cycle analysis of polymers.

This kind of information is out there, and there is a lot of it. My examples hopefully will inspire you to spend the effort (and it's not that much effort) to look for it.


Friedel–Crafts alkylation of benzene with tetrachloromethane

Fonken1 says that the reaction doesn't occur, but it is not due to steric hindrance or instability of tetraphenylmethane since the molecule is actually thermally stable. Instead, it argues that the formation of tetraphenylmethane due to alkylation of benzene using triphenylchloromethane is reversible in nature. The product readily dissociates into triphenylmethyl carbocation and benzene. Due to this unfavourable equilibrium, only insignificant amounts of tetraphenylmethane is formed.

Ben Norris also mentions that the reaction of benzene with tetrachloromethane is used to produce chlorotriphenylmethane, and provided an Organic Synthesis procedure2 that works on the kilogram scale.

References

Fonken, G. J. Alkylation of Benzene with Triphenylmethyl Chloride. J. Org. Chem. 1963, 28 (7), 1909–1909. DOI: 10.1021/jo01042a510.
Bachmann, W. E. Triphenylchloromethane. Org. Synth. 1943, 23, 100. DOI: 10.15227/orgsyn.023.0100.


What happens is that convex (outside) corners get severely undercut in wet etching. On a <100> cut wafer, the edges leading up to the corner reveal <111> planes which are resistant to etching, but right at the corner, where the two planes intersect, etching reveals a <110> plane, which is etched much more quickly. The difference in etch rates has to do with the number of "dangling" bonds—faces with fewer Si-Si bonds at the surface etch faster. The reason why concave corners don't undercut, despite having the same <111> plane intersection is that there are no dangling bonds in an inside corner so it doesn't etch any faster than the <111> planes.

It should also be noted that adding a small amount of surfactant to the etchant can dramatically decrease undercutting by adsorption to the corners.

This (fortunately open-access) article explains the phenomenon quite nicely.


Can I replace potassium nitrate with sodium nitrate in a smoke bomb?

In pyrotechnics, $\ce{NaNO3}$ is normally avoided because it is very hygroscopic.

With dried $\ce{NaNO3}$, the effect should be identical.

In any case, the usual safety rules apply!


Which one of the following substituents at para-position is least
  effective in stabilizing the phenoxide ion $\ce{PhO^{−}}$

Yes, the p-methoxy group is the least stabilizing of the choices. I'm not quite sure what you mean by

  its mesomeric effect which will counter the mesomeric effect of $\ce{O^{−}}$
  .

so let me add a few words of explanation for the choice.

In the figure below, I've drawn a few of the resonance structures possible for p-methoxyphenoxide (A-C).  The methoxy group donates electrons through resonance and resonance structure B illustrates this effect.  However, note that in resonance structure B we have placed two negative charges close to each other - this is very destabilizing.  Therefore, as the methoxy group donates electrons through resonance it destabilizes the system. In resonance structure C, the electron withdrawing inductive effect of the methoxy group is stabilizing.  However the resonance effect outweighs the inductive effect and overall the p-methoxy group is strongly destabilizing in the case of phenoxide.

Resonance structure D illustrates the p-methyl case.  Methyl is very mildly electron releasing both through resonance and inductively, so it also destabilizes the phenoxide anion, but since its effect is weak, it is not as destabilizing as the p-methoxy substituent.

Finally, notice in the p-acetyl case we can draw resonance structure E which actually stabilizes the phenoxide anion via resonance.  The p-acetyl group is also inductively stabilizing in the phenoxide example.


Why do some substances ionize instead of dissolving in water?

Dissolution is simply the mixing of two phases to produce a single homogeneous phase. Ionization is a process by which a substance with no net charge is converted into one or more ions. It includes the mundane example of dissolving an ionic compound in water that dissociates into its constituent ions as well as things like forming ions by knocking electrons off with an electron beam. They're both surprisingly complex topics, but I think the question can be answered conceptually by considering a few compounds: $\ce{NaCl}$ , acetic acid ($\ce{CH3COOH}$), and glucose:

(source)

All dissolve readily in water, but only two of them are electrolytes. As you note, $\ce{NaCl}$ is composed of ions and breaks apart into $\ce{Na+}$ and $\ce{Cl-}$ in solution. An electrolyte is any compound that ionizes when dissolved in a given solution, so $\ce{NaCl}$ is an electrolyte in water. Moreover, because there's is no way for a salt to be dissolved without ionizing in water (i.e. $\ce{NaCl_{(aq)}}$ isn't possible), $\ce{NaCl}$ is known as a strong electrolyte as when it dissolves, it ionizes completely.

$\ce{CH3COOH}$ also dissolves readily in water, but isn't an ionic compound. Because it's a weak acid, it exists in this equilibrium in solution:
$$\ce{CH3COOH_{(aq)} + H2O <=> CH3COO^{-}_{(aq)} + H3O+}$$

Because it can lose a proton to form ions, it does ionize in solution and is thus an electrolyte, but because it's a weak acid, the reaction doesn't go to completion and some $\ce{CH3COOH_{(aq)}}$ exists in solution making acetic acid a weak electrolyte as it doesn't completely ionize in solution.

Finally, glucose readily dissolves in acid, but none of its protons are readily lost so it doesn't ionize at all meaning that it isn't an electrolyte in water. Without getting too much into the thermodynamics of dissolution, the reason glucose is so soluble is because the many hydroxyl groups allow it to hydrogen bond extensively with water, where the equivalent compound without them, hexanal, is only slightly soluble due to the polar carbonyl group, and hexane is essentially insoluble.


What causes the anti-siccative effect of soot (amorphous carbon) as a pigment in oil paints?

According to the Wikipedia page on oil-drying agents (emphasis added):

  An oil drying agent, also known as siccative, is a coordination compound that accelerates (catalyzes) the hardening of drying oils, often as they are used in oil-based paints. This so-called "drying" (actually a chemical reaction that produces an organic plastic) occurs through free-radical chemical crosslinking of the oils. The catalysts promote this free-radical autoxidation of the oils with air.

The exposed surfaces of carbon black are highly functionalized, in many cases with oxygen-containing moieties.  The Birla Carbon website indicates that this surface chemistry "is a function of the manufacturing process and the heat history of a carbon black and generally refers to the oxygen-containing groups present on a carbon black’s surface." Per panel 3 of the figure at Asahi Carbon's page, "Carbon black properties," these functional groups can include at minimum carboxyl, hydroxyl, and carbonyl (aldehyde/ketone) functionality.  The section of the Wikipedia page on carbon black describing its surface chemistry indicates that quinonic, lactonic, and phenolic groups may also be present.

Most of the above-mentioned functional groups can participate in radical reactions, including reactions with the active species involved in paint hardening.  These side reactions of the carbon black siphon off the active radical species in the paint into pathways that do not contribute to the hardening/drying process.  The Lamp Black page on the Natural Pigments website provides support for this (emphasis added):

  However, [lamp black] is not generally recommended for use in oils, because carbon is an anti-oxidant that greatly slows the drying time of oil paint.


Anhydrous aluminium chloride ($\ce{AlCl3}$) exists; however, it cannot be obtained by heating of the hexahydrate ($\ce{AlCl3.6H2O}$).

$$\ce{AlCl3.6H2O \rlap{\ \ \not}->[\quad] AlCl3 + 6 H2O}$$

The crystalline hexahydrate $\ce{AlCl3.6H2O}$ actually is the complex salt $\ce{[Al(H2O)6]^3+ 3 Cl-}$.

The acidic hexaaquaaluminium(3+) ion is susceptible to hydrolysis.

$$\ce{[Al(H2O)6]^3+ <=> Al(OH)3 + 3 H3O+}$$

Therefore, the hexahydrate $\ce{AlCl3.6H2O}$ hydrolyses upon gentle heating and releases three equivalents of water and hydrogen chloride.

$$\ce{AlCl3.6H2O -> Al(OH)3 + 3 HCl + 3 H2O}$$


Which side of an aluminum foil should I put in contact with food?

After some thought, I guess I'll go with the first argument, that seems plausible (though I don't think it's accidental). As for coating one side with a layer, that seems like a bit too much.

Assuming the second argument is correct, then it's most probable that it doesn't matter which side you use. Because if one side is a 'coated', that 'coat' has got to be safe. Since well, aluminium foil is used pretty much only for packaging food, I don't think people could go about selling Al foil coated with some toxic film for the past couple of decades or so without someone pointing out something very wrong.

Now if it's the first argument, then it still doesn't matter which side you use. Aluminium oxide isn't really as dangerous as out make it out to be, especially if you're dealing with an oxide film that's still on the foil (Hogging a tablespoon of powdered aluminium oxide is another story altogether...). We sometimes call that process (aluminum developing an oxide coat) Passivation, because the foil is rendered 'passive' or 'inert'. Why else do you think it takes so long to get aluminium to react with acids or alkalies?

In a nutshell, regardless of which argument is correct, it'd be perfectly safe to use either side of the foil.

The reflective surface will reflect heat and the matte side will absorb heat. If you're baking or defrosting, the matte side will absorb more radiant heat and reflect less infrared heat while the shiny side will reflect more of both, so it makes more sense to bake and defrost with the matte side facing up. Conversely, during freezing, it makes more sense to reverse the sides as the shiny side face up would reflect heat away from the food.

I've always felt that one of the more practical reasons for creating two visually dissimilar sides on the foil, is to make it easier for you to identify which side you had initially used to wrap some food in. Say, you had just hurriedly unwrapped a cheeseburger and you don't feel like finishing it off. If you re-wrapped using the wrong side, you just get greasy fingers, which is quite annoying.

EDIT: I've read @Vedant's answer and he seems to have provided a good explanation as to why the two surfaces are dissimilar. But I wouldn't entirely dismiss the reason I've mentioned (identification), since, from my personal experiences...I'm pretty sure my keyboard would be covered in grease by now.

But all the same, I've included @Vedant's explanation as well, to give you the complete package.


Chloroform is soluble in water, though only slightly (~1g/100ml in cold water forming a slightly sweet liquid whose mild anaesthetic affects made it a recreational substance in Victorian times before its toxicity was fully recognised).

Moreover, its hydrogen will form (weak) hydrogen bonds which is probably a better explanation for the solubility in water than the polarity of the molecule.


Does boiling water alter the chemical properties of dishwashing liquid?

It depends rather strongly on your brand.

Soap, by itself, is unlikely to be affected by the boiling water. They're relatively stable, especially for an organic molecule. However, many of us use detergents that include enzymes that may denature in the boiling water, which is what your friend may have been referring to: A denatured enzyme is not very effective at cleaning at all.

So, look on your bottle! If it mentions enzymes, it probably won't go over well with boiling water. If it's just soap, and, say, bleach, boil away!


Why is the melting point of hydrogen iodide larger than that of hydrogen fluoride?

The increase in dispersive attraction between HI molecules is greater than the loss of dipole interactions. Only in $\ce{H2O}$ do dipole forces dominate (Intermolecular and Surface Forces, Jacob N. Israelachvili). Contribution of dispersion to intermolecular bonding for various molecules:

$\ce{CH4-CH4}$ - 100%
$\ce{HI-HI}$ - 99%
$\ce{HBr-HBr}$ - 96%
$\ce{HCl-HI}$ - 96%
$\ce{H2O-CH4}$ - 87%
$\ce{HCl-HCl}$ - 86%
$\ce{CH3Cl-CH3Cl}$ - 68%
$\ce{NH3-NH3}$ - 57%
$\ce{H2O-H2O}$ - 24%

What makes the contribution of dispersion so great is that it's non-directional and always attractive. Think of atoms as sticky spheres.


Oxidation of formate ion by peroxydisulfate in water solution: Why are there "strange" kinetics?

The rate of the third reaction affects the rate of the first reaction, i.e. they are competitive reactions and in that case the rds method can't be applied.

The rate of the first reaction is $r=-\mathrm{d}\ce{[S_2O_8^{2-}]}/\mathrm{d}t$, but the change of the concentration of the peroxydisulfate depends on the third reaction, too:
$$-\mathrm{d}\ce{[S_2O_8^{2-}]}/\mathrm{d}t=k_1\ce{[S_2O_8^{2-}]} + k_3 \ce{[CO_2^-]}\ce{[S_2O_8^{2-}]}$$


NMR deals with the nucleus of an atom (not just protons, but all nuclear isotopes with spin>0). Electrons also have a magnetic moment, and the technique used to probe this phenomenon is EPR/ESR. While NMR uses frequency in the MHz range (MHz), ESR uses higher energies in GHz range.

Lots of common references available on the internet to learn more, but you might be interested in:

https://physics.stackexchange.com/questions/171256/whats-the-difference-between-nmr-and-epr
https://en.wikipedia.org/wiki/Electron_paramagnetic_resonance
https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance


How does the volume of a mole of ideal gas relate to pressure and temperature?

Your starting point of reasoning is good, but you misread the equation. The question is about the volume of 1 mole of gas so let's rewrite the ideal gas law to give this relation:
$$P V = n R T \rightarrow \frac{V}{n}=\frac{R T}{P} $$
Now we have the volume of a mole of gas $\frac{V}{n}$ on the left hand side so we can see the proportionalities on the right hand side.

The volume of a mole of gas is inversely proportional to the pressure and directly proportional to the temperature, i.e. 4 in your list of answers


Why can't a reverse fission reaction be automatically classified as a nuclear fusion reaction?

The Wikipedia article on Nuclear Fusion starts off:

  In nuclear physics, nuclear fusion is a reaction in which two or more atomic nuclei come close enough to form one or more different atomic nuclei and subatomic particles (neutrons or protons). The difference in mass between the reactants and products is manifested as the release of large amounts of energy.  ...

$\ce{^4He}$ and $\ce{^{27}Al}$ are certainly different than $\ce{^{30}P}$.

The masses of the products however are more than the masses of the reactants

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\text{4-He} & 4.&\hspace{-1em}00260325415 \\
\text{27-Al} & 26.&\hspace{-1em}98153863\\ \hline
\text{total} & 30.&\hspace{-1em}98414188\\
\end{array}$$

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\text{30-P} & 29.&\hspace{-1em}9783138 \\
\text{n}   & 1.&\hspace{-1em}00866491588 \\ \hline
\text{total} & 30.&\hspace{-1em}9869787 \\
\end{array}$$

So it would seem that the definition used would require the products to have less mass than the reactants. This seems like a weird definition to me...


The question borders on a semantic argument. In the third mechanism (textbook), we would assume that the deprotonation event is very fast. The first mechanism (concerted) is basically saying that the deprotonation is so fast that it cannot be distinguished from the alkoxide leaving.

We know that acid-base reactions are fast. We also know that an anionic oxygen 'kicking out' an alkoxide is fast, at least relative to nucleophilic attack on a carbonyl (the rate determining step for any of these reactions AND the related trans-esterification is nucleophilic attack on a carbonyl to give the tetrahedral intermediate). What we're trying to answer is how fast is fast?

Here is my argument for the third mechanism. While acid-base reaction is fast, the top mechanism is essentially an E2 reaction, which does have a stereoelectronic preference for the hydrogen to be anti-periplanar to the leaving group. I have no doubt that conformation is readily accessible (the C-O bond will be rotating freely). The step is also bimolecular. In the third mechanism, an alkoxide kicking out a leaving group has no stereoelectronic preference. There are no alternate conformations to even consider. The C-O bond may be rotating, but it doesn't matter: there is always an electron pair arranged properly to kick out the alkoxide. Furthermore, the step is unimolecular.

The above is speculation, because I have never seen an alternative to the standard textbook saponification mechanism before. To be honest I never considered the concerted mechanism, and although I think it is a reasonable mechanism, I still prefer the textbook mechanism.


What is the 'electronic energy', in units of 'negative' hartrees, of a molecule?

The energy is the potential energy of all electrons in the molecule. Sometimes the term also includes the potential energy of the nuclei but in most cases it doesn't.
As reference or zero energy you can imagine the case where the electrons are non interacting or infinitely far apart from each other. In this sense it tells you the energy required to "remove" all electrons, i.e. pulling them away from each other so that they no longer interact with each other.
It can be interpreted as ionization energy where you remove all electrons in one go form the system but that is an theoretical process. The nuclei would push each other apart before you could ionize water to that degree. Normally the ionization energy refers to the removal of one electron. And the second ionization energy normally refers to the removal of a second electron after the system relaxed from the first ionization. And so on for higher ionization. This consecutive ionization process is different from removing all electrons in one go and would yield a different energy than the electronic energy given here.
This question/answer might also be helpuful :
https://chemistry.stackexchange.com/questions/38589/definition-of-electronic-energy#:~:text=Electronic%20energy%20is%20indeed%20an,nothing%20to%20do%20with%20electrons).


Do nitrogen and oxygen react to form nitric oxide or nitrous oxide?

Oxygen and nitrogen actually react to form nitric oxide.

  $$\ce{O2 + N2 ⇄ 2NO}$$

  Oxygen react with nitrogen to produce nitrogen(II) oxide. This
  reaction takes place at the high temperature[about 2000°C(without catalyst) ;
  1200-1300°C(with catalyst)), an overpressure and in the presence of a
    catalyst.
  In this reaction, the catalyst is can be platinum, manganese(IV)
  oxide.(Chemiday)

In wikipedia article of nitric oxide, it is written that:

  The uncatalyzed endothermic reaction of oxygen ($\ce{O2}$) and nitrogen ($\ce{N2}$), which is performed at high temperature (>2000 °C) by lightning
  has not been developed into a practical commercial synthesis (see
  Birkeland–Eyde process):

  $$\ce{N2 + O2 → 2 ·NO}$$

  From a thermodynamic perspective, $\ce{·NO}$ is unstable with respect to $\ce{O2}$ and $\ce{N2}$, although this conversion is very slow at ambient temperatures
  in the absence of a catalyst. Because the heat of formation of $\ce{·NO}$ is
  endothermic, its synthesis from molecular nitrogen and oxygen
  requires elevated temperatures above 1000 °C.

Why nitrous oxide is not formed can be explained through thermodynamics. From Quora:

  For any chemical reaction to take place spontaneously, we look at a
  factor known as free energy available for that reaction. Gibbs's Free
  Energy is given by the equation:

  $$\ce{\Delta G = \Delta H − T\Delta S}$$     This is one of the most
  important equations in all of chemistry
  because it provides information as to whether a particular reaction is
  spontaneously possible or not.

  The quantity $\ce{\Delta G}$ must be negative for a reaction to take
  place spontaneously. For the reaction $\ce{2N2 + O2 −>2N2O}$,
  calculating the values of change in enthalpy and temperature times
  change in entropy for the atmosphere, we see that this reaction can't
  take place.

Generally, reaction between nitrogen and oxygen produces $\ce{NO_x}$ which causes air pollution.

  $\ce{NO_x}$ gases are usually produced from the reaction among nitrogen and oxygen during combustion of fuels, such as
  hydrocarbons, in air; especially at high temperatures, such as occur
  in car engines.The term $\ce{NO_x}$ is chemistry shorthand for
  molecules containing one nitrogen and one or more oxygen atom. It is
  generally not meant to include nitrous oxide ($\ce{N2O}$).


Why is reduction potential of water ( 2H₂O + 2e⁻ → H₂ + 2OH⁻ ) is -0.83 insted of 0, as in the case of (2H⁺ → H₂)?

Note that the actual potential for a particular redox reaction is not a fixed value, but depends on concentrations ( more exactly activities ) of reagents. The standard redox potentials are potentials with activities equal to 1,

If we consider reactions

$$\begin{align}\ce{
2 H+ + 2e-    &<=> H2 \\
2 H2O + 2 e-     &<=>  2 OH- + H2 \\
}\end{align}$$

their standard redox potentials imply
$\rm{p}H=0$ for the former, $\rm{p}H=14$ for the latter.

For the former reaction and $\rm{p}H=14$, $E=-0.83\ \rm{V}$. As for $\ce{H+/H2}$ potential, at $\pu{p_{\ce{H_2}} = 1 atm}$

$$E_\ce{H2\ /H+}  = E^{\circ}_\ce{H2\ /H+} + 0.059 \log [\ce{H+}] = E^{\circ} - 0.059\  \rm{p}H$$

Thermodynamically, the equations they are equivalent, but kinetically, the latter reaction has in strongly alkaline environment the advantage of many concentration orders. And vice versa.

Elements can have multiple standard redox potentials, as such a potential can be expressed for any redox reaction (even if for some only formally from the thermodynamic data).

For example, some standard redox potentials for $\ce{Ag}$ are:

$$
\begin{array}{lcll}
\ce{Ag+ + e- &<=>& Ag(s)};\ & E^\circ_{red} = \pu{0.80 V}\\
\ce{AgCl(s) + e- &<=>& Ag(s) + Cl-(aq)};\ & E^\circ_{red} =\pu{ 0.22 V}
\end{array}
$$

(There are more; see the Standard electrode potential data table).

The latter can be calculated from the former using
$$E_{\ce{Ag/Ag+}} = E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log c_\ce{Ag+}$$

For the $\ce{Ag / AgCl / Cl-}$  system, we have to include into potential calculations the $\ce{AgCl}$ solubility constant.

$$\begin{align}
K_{\rm s, AgCl} &= c_{\ce{Ag+}}\cdot c_{\ce{Cl-}} \\[1em]
E_{\ce{Ag/AgCl/Cl-}} &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log c_\ce{Ag+} =\\
  &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log {K_{\rm s, AgCl}} - 0.059 \log {c_\ce{Cl-}}= \\
  &= E^{\circ}_{\ce{Ag/AgCl/Cl-}} - 0.059 \log {c_\ce{Cl-}} \\
E^{\circ}_{\ce{Ag/AgCl/Cl-}} &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log {K_{\rm s, AgCl}} =\\
 &= 0.80 + 0.059\log(1.77\cdot10^{−10}) = \\
 & = 0.22
\end{align}$$

Note that the $\ce{Ag/AgCl/Cl-}$ electrode is often used as the secondary standard of the electrode potential, aside of the calomel electrode $\ce{Hg/Hg2Cl2/Cl-}$, because maintaining of the primary potential standard electrode $\ce{Pt-H2/H+}$ is not convenient. The $\ce{Ag/AgCl/Cl-}$ electrode is used also as the electrode to measure $c_{\ce{Cl-}}$.


Proving RTln($a_i$) equal to non PV work done by system,where $a_i$ is activity

We know that chemical potential is defined as
  ${\mu}_i={\mu}_i^{standard}+RT\ln(a_i)$.

Be a bit careful with your wording. This equation is not part of the definition of the chemical potential but rather it is derived from it.
The derivation of this equation can be found in this answer of mine.
The definition of the chemical potential is

\begin{equation}
 \mu_{i} = \left( \frac{\partial G}{\partial n_{i}} \right)_{p,T} \ ,
\end{equation}

where $n_{i}$ is the amount of the $i^{\text{th}}$ component in the system.

  We know that Gibbs free energy of a system at constant temperature and
  pressure also represents Non-PV work done by system.

Again this is not quite correct: It is the change in Gibbs free energy of a system during some process, say a chemical reaction or a phase transition, at constant temperature and pressure that can be identified with non-PV work done by the system.

Now considering a general chemical reaction

\begin{equation}
 \ce{\nu_{1}A + \nu_{2}B + ... + \nu_{$x$}C <=> \nu_{$x+1$}D + \nu_{$x+2$}E + ... +\nu_{$n$}F}
\end{equation}

the differential change in Gibbs free energy is given by

\begin{equation}
 \text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \mu_{i} \text{d} n_{i}
\end{equation}

or introducing the extend of reaction, $\text{d} \xi = \frac{\text{d} n_{i}}{\nu_{i}}$

\begin{equation}
 \text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \nu_{i} \mu_{i} \text{d} \xi \ .
\end{equation}

From this it follows that the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an extend of reaction $\xi = 1 \, \text{mol}$) at constant pressure and temperature is given by

\begin{equation}
 \Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ .
\end{equation}

Here you can see that the non-PV work done by the system originates from the differences in the chemical potentials of the reactants.
Substituting the equation for the chemical potential

\begin{equation}
 \mu_{i} = \mu_{i}^{0} + RT \ln a_{i}
\end{equation}

the equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, leads to

\begin{equation}
 \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,
\end{equation}

where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced.
This is the relation between the non-PV work, represented by $\Delta G$, and the activity terms. Admittedly, it is not a very graphic explanation but I fear that it simply is a pretty mathematical matter.


Why doesn't the solute settle out of a solution, such as brine or sugar water?

You discuss two different phenomena, solubility and concentration gradients due to gravity.

Solubility can be explained using the concept of entropy. Solubility requires the the free energy of a solid be higher than the corresponding substance dispersed in solution. Surprinsingly perhaps, additional entropy associated with dispersion of the solute in the solvent suffices to drive the process (interactions can certainly be important, but are not at minimum necessary, although much depends on how much you are attempting to solubilize, and at what temperature).

Gravity can (and does) cause concentration gradients, and can therefore under the right circumstances cause precipitation. The problem is that gravity is a very weak force, so it's effects on light molecules are, well, weak. Gravitational effects require large scales or masses to be significant. Density differences are argued to be critical driving force behind water cycling in the oceans. The air atop Mt Everest is so thin people suffocate without access to an auxiliary supply of oxygen. So these effects do happen, and matter. And you can use a centrifuge to separate components in a mixture, which is analogous to applying gravitational forces, just scaled up in strength.


Why is the fluorescence spectrum of pyrene time-dependent?

I think that you have rediscovered the emission of pyrene exciplexes :)

As you have correctly assigned, the structured fluorescence spectrum shortly after the pulse is the emission of pyrene monomers.

Given sufficient time (100 ns), an excited molecule of pyrene diffuses towards a pyrene molecule in its ground state and forms an exciplex.

Fluorescence emission from these exciplexes is known to be broad and unstructured.


If you know LaTeX and you want to see what you are writing you can use lyx, and some TeX package like mhchem and chemfig. So you can write all the things you want and add comments. This is what I have done with my Mac just now:

Here is the code, you can make these lines as a default template:

\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\usepackage{chemfig}
\usepackage{tikz}
\usepackage[version=4]{mhchem}
\setlewis{4pt}{0.8em}{black}

From here you have to write! The date is automatically added!

\title{Introduction to chemistry notes}
\begin{document}
\maketitle
\section{mhchem}
You know that \ce{ CH4 + 2O2 -> CO2 + 2H2O}     this is a simple equation\ldots
Okay maybe you can do this with word, but if you are familiar with Chemistry S.E. you will be very fast with this notation.

\ce{Zn^2+
<=>[\ce{+ 2OH-}][\ce{+ 2H+}]
$\underset{\text{amphoteres Hydroxid}}{\ce{Zn(OH)2 v}}$
<=>C[+2OH-][{+ 2H+}]
$\underset{\text{Hydroxozikat}}{\ce{[Zn(OH)4]^2-}}$
}

\section{Chemfig}
\begin{figure}[h]
\setatomsep{1.8em}
\schemestart
\subscheme[90]{\chemfig{*6((-[@{a1}]H_2@{a2}\Lewis{26,N})=[@{r1}]-[@{r2}]=[@{b2}](-[@{b1}]N =[@{c1}] N -[@{c2}]*6(=[@{d1}]-[@{d2}] =[@{e1}](-[@{e2}] OH)-=-))-=-=)}\chemrel{<->}
\chemmove[-stealth,shorten <=1pt, shorten >=1pt]{%
    \draw(a2)..controls +(90:5mm) and +(135:5mm)..(a1);% first arrow
    \draw(r1)..controls +(225:12mm) and +(300:12mm)..(r2);% ring arrow
    \draw(b2)..controls +(10:3mm)..(b1);% ring arrow
    \draw(c1)..controls +(340:8mm)..(c2);% ring arrow
    \draw(d1)..controls +(225:12mm) and +(300:12mm)..(d2);% ring arrow
    \draw(e1)..controls +(20:4mm)..(e2);% ring arrow
    }
\chemfig{*6((=[@{a1}]H_2@{a2}\chemabove{N}{\oplus})-[@{r1}]=[@{r2}]-[@{b2}](=[@{b1}]N-[@{c1}] N =[@{c2}]*6(-[@{d1}]=[@{d2}] -[@{e1}](=[@{e2}] \chemabove{O}{\ominus}H)-=-))-=--)}
}
\schemestop
\caption{This is a coniguated system\ldots}
\label{fig:con}
\end{figure}
Okay the lesson will end and in one hour you'll draw this figure only, maybe you should use an external software\ldots %what the professor means whit this last phrase?!
\end{document}

Sincerely I think that is best to take handwrite notes because I think that writing and drawing chemical formula allow you to memorize better but of course this is only a thought.


How to run a B2PLYP-D3 calculation in Gaussian 09?

Search for B2PLYP on the Moller-Plesset page:

  B2PLYPD3 requests the same but with Grimme's D3BJ dispersion [Grimme11, Goerigk11].

The correct D3 parameters will be set for you automatically.

%nprocshared=4
%mem=4gb
%chk=fohsdft3
# opt freq b2plypd3/aug-cc-pvdz scf=tight

Input file taken from fohsdft

0 1
O                  -3.46278   2.54045   0.
H                  -2.50278   2.54045   0.
H                  -3.78324   3.44539   0.

Since B2PLYP-D3 is a method, it goes where the density functional is normally specified.


As you have known that metabolism of sucrose yields glucose and fructose.So formed glucose undergoes glycolysis producing pyruvate or lactate(under anaerobic condition). In case of fructose, it is converted into fructose-1-phosphate and which on action of aldolase enzyme produces dihydroxyacetone phosphate and glyceraldehyde-3-phosphate which are the intermediate products of glycolysis and finally converted to pyruvate.

The end product of glycolysis i.e. pyruvate is converted into acetyl coa in presence of coenzyme A , which is the precursor for the synthesis of cholesterol.

For the synthesis of cholesterol , acetyl coa undergoes different enzymatic reactions finally forming cholesterol at the end.


Your resonance structure doesn't do anything to stabilize the cation. You've created a singlet carbene at the top most carbon, so it has both the cation and the anion on the same carbon. These do not cancel out. As drawn, you have two hydrogens on the top carbon. You need to replace it with a single lone pair and an empty orbital.


What are glues/adhesives for cardboard boxes/crates made of?

Because the glue is not in direct contact with food you can use basically any glue. And cardboard is easy to glue so there is no limitation. Packaging should be cheap and therefore I would not expect a biodegradable glue.


Why is molarity independent of temperature?

First of all, molarity is temperature-dependent.

You might have meant molality, because molality is temperature-independent.

The reason is very simple, if you see molarity definition, molarity is the amount of substance of solute per litre of solvent and that mean it depends on volume.
You know that volume depends on temperature, therefore molarity is also temperature-dependent.

Molality on the other hand is the amount of substance of the solute per mass of the solvent, both are independent of the temperature.


I can provide some physical intuition, but it's still going to require math and previous work to make it happen (aside on this at the end).

Let's consider a particle in just one dimension. There are two quantities of interest: (1) the position of the particle along this coordinate, (2) the scalar momentum (related to the velocity) along this coordinate.

I'm going to represent this particle as a plot on a graph. The horizontal axis will measure position. The vertical axis represents some notion of how likely it is to find the particle at this position (formally, it would be squared and normalized). Finally, we also stipulate that momentum of the particle is going to be the frequency of what I draw. This may seem strange when we're talking about an arbitrary plot, but it's actually completely reasonable from the perspective of representing functions from as Fourier series. You're just going to have to assume that this is true because a formal analysis does require math (see other answers).

Now, let's make this more concrete with some plots.

Here's a standard sine function. It has a very well defined frequency, so we know the momentum exactly. Where is the particle? Well, the sine wave stretches out to infinity in both directions, so we have no idea where it "is."

Here's a different graph. Where is the particle? It's very close to $x=0$. What is it's momentum? It's kind of hard to pick out single frequency here, so you cannot easily establish what the momentum is.

And that's basically it. Identifying frequency (momentum) and position for a plot like this are traded-off against one another. You'll probably end up in the intermediate case where you know some information about both. This is wave packet.

This particle is roughly centered around $x = 0$, and we can roughly say what its momentum is.

The uncertainty principle simply formalizes this idea except using the formal mathematical structure of quantum mechanics.

Now, for the aside I promised. Your question was entirely reasonable until you got to the last line, and I quote: "No usage of earlier established principles and maths please."

This is impossible. Science does exist in a vacuum. It is based on the past observations and structure built by previous generations of scientists. If we were to start from scratch, would we arrive at the same science that we have today? Possibly, possibly not. But the understanding we are working with is built upon foundations that we are stuck with, for better or for worse.

You also can't really do this without math. Part of science is to describe relationships, and mathematics allows us to do so unambiguously. Words suck at conveying ideas because words mean different things.

Lastly, it's fine to try to build intuition on a concept by relating it to the macroscopic world. However, you should not assume that this is possible. It may be philosophically satisfying that objects behave in similar ways regardless of size, but that's not the way the world is.


Why are 3 equivalents of organolithium needed in synthesis of ketone from carboxylic acid?

The text box you are quoting from (p 219) refers to this specific reaction. The compound in question possesses an α-hydroxyl group, which needs to be deprotonated too.

Notice that three equivalents of organolithium are needed in this reaction: one to deprotonate the acid, one to deprotonate the hydroxyl group, and one to react with the lithium carboxylate. These chemists added a further 0.5 for good measure.

However, the scheme directly above this incorrectly indicates that 3 equivalents of MeLi are required for a generic carboxylic acid:

In the absence of any other acidic groups, only 2 equivalents of organolithium are needed. In practice, however, more than 2 may be added.
Note that the 3 protons are still needed to get to the protonated species shown directly afterwards. In practice the proton source (e.g. saturated aqueous $\ce{NH4Cl}$, or just plain water) is usually added in excess.


What does it mean that an electron quantum numbers are n1= 4 - l1?

$n_1 = 4 - l_1$ means the same as $n_1 + l_1 = 4$, so the $n$ and $l$ quantum numbers of the first electron added up should equal 4. This could refer to 3p or 4s.

For the second electron removed, the $n$ and $l$ add up to 5. This could refer to 3d, 4p or 5s.

For chromium, the first electron removed is the 4s, followed by the 3d. For copper, this is the case as well, but its mass is higher than that of iron, so it is not the answer.


The anode and cathode when corrosion happens

You are right! What you have written is an example of Galvanic corrosion (Galvanic corrosion occurs when two different metals have physical or electrical contact with each other and are immersed in a common electrolyte).
You have  successfully described Galvanic corrosion and given an example of it


What is the difference between IPA and Ethanol in terms of solubility?

[...] ethanol could be used in a solvent extraction of two solids, but it also says that isopropyl alcohol could also be used, but must be used in higher concentrations. Why? [...]

They hopefully didn't write about the "concentration of isopropanol" but mentioned that a larger amount of the solvent has to be used.

It is not uncommon that organic compounds with polar substituents are less soluble in higher alkanols.

Vanillin is a typical example, the solubilities are in the range of 4.15 mol/L in methanol, 2.5 mol/L in ethanol and 1.8 mol/L in 1-propanol. (Data are taken from a freely accessible collection of experiments performed by students: OpenNotebook Science Challenge). I couldn't find any data for 2-propanol.

  So would this suggest that if one were to use IPA as a solvent, one could use the same amount of IPA as they would Acetone?

No, not really. A ketone and a secondary alkanol, although looking similar on first sight, may have behave completely different as solvents. Polystyrene (foam), as an example is pretty much insoluble in isopropanol but highly soluble in acetone!


As the solvent evaporates the least soluble salts precipitate in succession.  This is used during seawater and salt lake evaporations to isolate values beyond NaCl. Overall, you will get a mixture of combinations of cations and anions.

Supersaturation is not uncommon.  Metastable solutions usually require nucleation sites to get over their initial activation energy for solids' crystallization.  High melting points (strong lattice forces) are difficult to supersaturate.  Benzil in xylenes remarkably supersaturates, and is stable for a week or more in silanized glassware and cooled very slowly within styrofoam and rags.


Why is Fluoride anion less stable than iodide anion?

Actually, the answer is a bit more nuanced. If we observe the pKb values (i.e. tendency to form HX from X-, where X is a B-L base) it is pretty clear that F- would form HF very quickly.

The explanation is that I- would have a much larger area to balance out the negative charge compared to F-, and the anion would be more stable as a result.


I'm really not getting why that post didn't answer OP's query.

Nevertheless, I'm posting the following argument as an answer.

  Why is the Gibbs free energy change equal to zero at equilibrium?

Well, it's not 'Gibbs free energy change' $\Delta G$ but rather Reaction Gibbs
Energy $\Delta _r G$ which is zero at equilibrium, if written accurately.

Reaction Gibbs Energy is defined as the rate of change of Gibbs Free Energy per unit change in composition i.e.,

$$\Delta_r G =\frac{\Delta G}{\Delta n} \tag 1$$

where $\Delta n$ is the change in composition of the system.

Now, OP must be aware of the first derivative test for finding local minima; that means for reaching the equilibrium position, the rate of change of Gibbs Free Energy per unit composition must be zero and the rate must change from negative in the neighbourhood of the extrema point of Gibbs Free Energy vs. Composition graph viz.

$$\Delta_r G= 0\;.$$

Now, question might arise as that of the physical implication of association of equilibrium position with $\Delta_r G= 0\;.$

The definition of equilibrium position sheds light on this:

A reaction reaches equilibrium position when it has no further tendency to change; that is, the reaction does remain 'spontaneous in neither direction' .

This happens when the reaction Gibbs energy becomes zero viz. $\Delta_rG=0.$


Reaction mechanism for reduction of alkyl halide to alkane

The action of HCl on Zn results in the formation of hydrogen.

Next, there is some surface chemistry where the diffusion of atomic hydrogen (or hydrogen radical) occurs with select metals. See, for example, this 2008 thesis "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena", by Saikat Adhikari, who cites the "ingress of atomic hydrogen into the metal".

Then, the action of a hydrogen radical on an alkyl halide:

$\ce{.H + CH3Cl -> HCl + .CH3}$  (Source on radical reactions)

$\ce{.H + .CH3 -> CH4}$

I suspect more reactions are possible resulting in minor products, for example:

$\ce{.CH3 + .CH3 -> C2H6}$

$\ce{.H + CH4 -> H2 + .CH3}$

......

The final products include HCl, CH4 and other minor products.