How to calculate Lennard-Jones potential with quantum mechanical methods

Yes, this is technically possible. A basic tutorial for this is in the excellent Psi4Numpy project, which I'll reproduce here with minor modifications. Their example fits the counterpoise-corrected MP2/aug-cc-pVDZ total interaction energy of the helium dimer.

from __future__ import print_function

import psi4

import numpy as np

import matplotlib as mpl
mpl.use('Agg')
import matplotlib.pyplot as plt

he_dimer = """
He
--
He 1 **R**
"""

distances = [2.875, 3.0, 3.125, 3.25, 3.375, 3.5, 3.75, 4.0, 4.5, 5.0, 6.0, 7.0]
energies = []
for d in distances:
    # Build a new molecule at each separation
    mol = psi4.geometry(he_dimer.replace('**R**', str(d)))

    # Compute the Counterpoise-Corrected interaction energy
    en = psi4.energy('MP2/aug-cc-pVDZ', molecule=mol, bsse_type='cp')

    # Place in a reasonable unit, Wavenumbers in this case
    en *= 219474.6

    # Append the value to our list
    energies.append(en)

print("Finished computing the potential!")

# Fit data in least-squares way to a -12, -6 polynomial
powers = [-12, -6]
x = np.power(np.array(distances).reshape(-1, 1), powers)
coeffs = np.linalg.lstsq(x, energies)[0]

# Build list of points
fpoints = np.linspace(2, 7, 50).reshape(-1, 1)
fdata = np.power(fpoints, powers)

fit_energies = np.dot(fdata, coeffs)

fig, ax = plt.subplots()
ax.set_xlim((2, 7))  # X limits
ax.set_ylim((-7, 2))  # Y limits
ax.scatter(distances, energies)  # Scatter plot of the distances/energies
ax.plot(fpoints, fit_energies)  # Fit data
ax.plot([0,10], [0,0], 'k-')  # Make a line at 0
ax.set_xlabel(r'intertomic separation ($\AA{}$)')
ax.set_ylabel(r'interaction energy ($\mathrm{cm^{-1}}$)')
fig.savefig('1b_molecule.pdf', bbox_inches='tight')

Once the set of energies is calculated, a least-squares polynomial fit is performed, giving

$$
f(x) = 6677721.45419193x^{-12} - 11394.79882998x^{-6}.
$$

This isn't a good (efficient or general) way of evaluating the fit function, but illustrates the placement of correct signs in the above code:

def f(z):
    return coeffs[0]*z**powers[0] + coeffs[1]*z**powers[1]

From the above we can see that the first and second coefficients in the polynomial fit correspond directly to $A$ and $B$, respectively. In this example, the energies are in wavenumbers ($\pu{cm^{-1}}$). Most quantum packages print in atomic units (Hartrees) and dynamics packages print in kJ/mol, which is fine, but be aware that in the above fit the units are incorporated into $A$ and $B$. Here is the resulting fit compared against the individual interaction energy calculations:

This was not in the question, but I think it's important to ask if this is the correct approach and investigate further. Notice how the fit is qualitatively ok near the minimum, but the shape in the dissociation region is qualitatively wrong, and the potential simply isn't attractive enough in the limit of infinite separation. This could arise in three ways:

The curve fitting isn't working properly, or there's a bug in the code.
The functional form of the chosen interatomic potential is incorrect.
The (Boys-Bernardi) counterpoise correction, which is known to overcompensate for BSSE, is giving strange results at large separation.

Assume that point 1 isn't an issue (if there's a bug anywhere, please comment). Point 2 can be investigated by also fitting against the Buckingham and shifted Morse potentials, which should display better short-range and long-range behavior:

\begin{align}
V_{\text{Buckingham}}(r;A,B,C) &= A e^{-Br} - \frac{C}{r^{6}} \\
V_{\text{Morse}}(r;A,B,C,D) &= A \left( 1 - e^{-B(r-C)} \right)^{2} + D
\end{align}

Note that for these energies, an unshifted Morse potential (no $D$ parameter) fails miserably. See here for more information about fitting Morse-type potentials. The Buckingham and Morse fits require a more general (arbitrary) curve fitting routine, and the script that made this plot can be found as an HTML comment in the post source code below.

I find it interesting that the Buckingham potential gives a worse fit than the Lennard-Jones one, even though it is supposed to reproduce the repulsive wall better with $e^{-r}$ rather than $\frac{1}{r^{6}}$. The Morse fit is remarkably good, which makes sense considering there are 4 free parameters rather than 3 (Buckingham) or 2 (Lennard-Jones). Before going any further, a check on point 3 by setting bsse_type=nocp:

This reveals that the counterpoise correction is definitely interfering with the quality of the fit, and the functional forms of the Lennard-Jones and Buckingham potential appear to not describe dissociation properly, though extended to infinite separation, the Morse potential is the one that is qualitatively incorrect. This can be attributed to using a single functional form to describe the total interaction or binding energy, and not the different components. I suspect that the apparent poor fit is due to a non-zero Coulomb interaction, rather than van der Waals-type interactions which at least the Lennard-Jones potential is meant to be used for.

To test this, here are interaction energy calculations at two levels of sophistication, both based on symmetry-adapted perturbation theory (SAPT). This is still the helium dimer, at 7 angstroms separation, with the aug-cc-pVDZ basis set and the monomer-centered basis approximation. All units are kcal/mol.

$$
\small
\begin{array}{lrr}
\hline
\text{Component}  & \text{SAPT0} & \text{SAPT2+3(CCD)}\delta_{\text{MP2}} \\
\hline
\text{Electrostatics} & -0.01796877 & -0.01777334 \\
\text{Exchange}       &  0.00000000 &  0.00000000 \\
\text{Induction}      &  0.01587019 &  0.01508578 \\
\text{Dispersion}     & -0.00013235 & -0.00016773 \\ \hline
\text{Electrostatics + Induction} & -0.00209858 & -0.00268756 \\ \hline
\text{Total}          & -0.00223093 & -0.00285529 \\
\hline
\end{array}
$$

At both levels of SAPT, the non-dispersive part of the interaction energy accounts for 94% of the total interaction! Is this true at 3 angstroms separation?

$$
\small
\begin{array}{lrr}
\hline
\text{Component}  & \text{SAPT0} & \text{SAPT2+3(CCD)}\delta_{\text{MP2}} \\
\hline
\text{Electrostatics} & -0.04852183 & -0.04836334 \\
\text{Exchange}       & 0.02045965 & 0.02219227 \\
\text{Induction}      & 0.04202349 & 0.04081412 \\
\text{Dispersion}     & -0.02228767 & -0.02843936 \\ \hline
\text{Electrostatics + Induction} & -0.00649834 & -0.00754922 \\ \hline
\text{Total}          & -0.00832637 & -0.01379630 \\
\hline
\end{array}
$$

The non-vdW interactions now account for 78% and 55% of the total interaction energy for each SAPT flavor, respectively. Although induction (also called polarization) is repulsive, I am including it in the Coulomb-type interaction since it is not purely quantum mechanical in nature like the exchange term. It is the exchange term, not induction, that is closer to the charge cloud penetration picture. Even so, the interaction between two helium atoms is not purely based on dispersion, and requires fitting another nonbonded term for electrostatics, the Coulomb term:

$$
V_{\text{Coulomb}}(\vec{r}_{i},\vec{r}_{j},q_{i},q_{j}) = \frac{1}{4\pi\epsilon_0} \frac{q_{i}q_{j}}{|\vec{r}_{i} - \vec{r}_{j}|}
$$

Our goal is to separate out the Coulomb-like terms from the interaction energy, so that $A$ will describe only the exchange contribution and $B$ will describe only the dispersion contribution; electrostatics and induction will be handled separately in the Coulomb term. This will be done by fitting only to the combination of exchange and dispersion, which are taken from SAPT calculations. MP2 binding energies are not CP-corrected.

As you can see, using SAPT energies provides a much better fit than naively using the total interaction or binding energies. I'm not sure why the total SAPT energy works so well, but it may be due to the absolute magnitudes at each point being so small.

The importance of only fitting exchange and dispersion is obvious for $\ce{Na^+---Cl^-}$ (using def2-SVP for the basis).

This is an unfair example, since the dispersion interaction is dwarfed by exchange, and it is an ionically-bound molecule. As a final example, consider $\ce{V(III)---Cl^-}$ (using def2-SV(P) for the basis, vanadium as a quintet).

I had some trouble converging many of the calculations, presumably due to a non-optimal spin state at long range. For this reason (spin-state crossing), Lennard-Jones type models are poor for metal-x interactions. The problem appears to be similar to the sodium chloride, but the exchange plus dispersion fit looks good.

If you want to reproduce any of the work, rather than fish through the source for inputs and scripts, everything is in a GitHub repository. More references for SAPT can be found here, along with the Psi4 manual. Note that I didn't do any literature searching here, but I expect people have used energy decomposition approaches in the past as part of force field design. Hopefully this serves as a good starting point.


How do you find the pH of a solution with a known pKa and Molarity?

Do you know the formula:

$$K_\text{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$

?

You will need to find a concentration first, however, on the equation.

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

And solve for the dissociated hydrogen ion, say $x$.

$$K_\text{a} = \frac{[x][x]}{[\ce{HA}]-x} = \frac{x^2}{[\ce{HA}]-x}$$

The logic is: if $x$ of $\ce{HA}$ dissociated, then we would get $x$ of $\ce{H}$, $x$ of $\ce{A}$ and $[\ce{HA}]-x$ left of $[\ce{HA}]$.

Does this make sense?


Conversion of s-trans butadiene to s-cis butadiene

During the interconversion of s-cis and s-trans 1,3-butadiene, the molecule passes through a geometry where the two ethylenic arms are rotated 90 degrees one from the other.  In this conformation the two ethylenic arms are orthogonal to each other and there is no overlap between them.  In an MO sense they can be treated as two separate ethylenes.  Rotation beyond this conformation allows overlap to reoccur and the phase of the HOMO will be correctly reestablished with only one node occurring between C2 and C3.


Does the Clemmensen reduction reduce the C=C bond of α,β-unsaturated ketones?

Both are not quite correct. A quick literature search doesn't turn up any hits for the Clemmensen reduction of this particular aldehyde, but March's Advanced Organic Chemistry, 7th ed. (p 1541) writes about the Clemmensen and Wolff-Kishner reductions:

Neither method is suitable for α,β-unsaturated ketones [...] Under Clemmensen conditions, both groups of these molecules may be reduced or if only one group is reduced, it is the C=C bond.

Large structural rearrangements are common under the vigorous conditions: see, for example, Helv. Chim. Acta 1976, 59, 962. "In all cases but one the formation of cyclopropyl acetates was observed."
It seems that there are some situations in which the ketone is selectively reduced, but this is not general: see Tetrahedron 1986, 42, 6615.


Which base is stronger methylamine or propylamine?

It does not make much of a difference. Both are short chain primary alkyl amines, so chemically much alike ($\ce{pKa} \approx 10.7$ due to the $\ce{-NH2}$ group). In what context do you want to know the answer to this question? That could help me direct my answer better.


Basicity is how well an atom can donate electrons. More is the electron density more is the basicity of the compound.

In Aniline the lone pair of electrons present on the nitrogen are involved in resonance, so the electron density on N is less than expected.

In compound II (Diphenylamine), since two benzene ring is present so the electron density will be less as now lone pair are involved in resonance of both the ring.
Thus electron density on Nitrogen is less than as compare to Nitrogen atom of Aniline

Similarly, in III compound (Triphenylamine), since three benzene ring is present the lone pair are more localized, and, thus electron density will be much less (least) as compared to the other compounds

In 4th structure, there is no resonance so, lone pair are present with nitrogen itself, and the electron density will be high, also due to inductive effect of the alkyl groups the electron density is more, and thus it can easily donate the electrons as compared to all other compounds. So the basicity is the highest for compound IV.

Therefore, the order of basicity is IV>I>II>III.


Saliva has alpha-amylase (ptyalin), which hydrolyzes carbohydrates (so starch starts to taste sweet in just a few seconds), kallikrein, an enzyme that breaks down protein, and lingual lipase, which helps disintegrates fats (lipids).

So next time you clean off a splattered insect from your wind-screen (windshield, should you prefer), think of it as external digestion, as practiced by flies and by starfish. Or perhaps you might not want to consider that thought...


How to calculate the rate constant of a zero order reaction?

According to your table, for a zero order reaction, you make a graph of the measured concentrations vs the corresponding times.  It will be a straight line, and the slope and intercept will give you the rate constant and the concentration at time zero.

For a first order reaction, you make a graph of the natural log of the concentrations vs the corresponding times.  It will be a straight line, and the slope and intercept will give you the rate constant and the concentration at time zero.


How does Dalton's Law relate to scuba diving?

In scuba diving it is essential to consider the partial pressures of different gases for the following reasons:

The partial pressure of oxygen should be greather than 0.16 bar to avoid hypoxia (at the surface this partial pressure is obviously 0.21 bar).
Oxygen is toxic at high partial pressure. For the dive one want to stay as close as possible to 1.3-1.4 bar (maximum), while decompression is made with maximum 1.6 bar.
Nitrogen is narcotic at high partial pressure and we usually try not to go higher to partial pressure similar to what we have at 30-35m.

As an example consider a deep technical dive to 80m (9 bar). If you have air in your tank (21% oxygen, 79% nutrogen), the partial pressure of oxygen will be
$$pp_{\ce{O2}} = p_\text{tot} \times x_{\ce{O2}} =  9\times0.21=1.89 > 1.6,$$
thus you are in serious danger because of oxygen toxicity (visual changes, nausea, anxiety, irritability, confusion, dizziness, tonic-clonic seizure, ...) which can lead to death. If you want to have an optimal oxygen partial pressure of 1.4 you will need to reduce your oxygen content to 15% (less than air). But since the partial pressure of this new mixture at the surface is 0.15 < 0.16 you should not start your dive with this tank.

In conclusion, what it is really important for your body is the partial pressure of the gas you are breathing, not the absolute pressure. Dalton's law is necessary to compute the different partial pressures.

This argument is quite advanced for scuba diving an you will have to search for technical diving in order to find the necessary informations. You may also want to look at other interesting physical phenomena that are important for scuba diving (Boyle's law, Charle's law, ...).


Curious as to what CorA1 is and what it does?

You see, CorA and CorA1 are two different things. CorA is a protein, while CorA1 is the gene that codes for that protein.

It is the genes that when translated give rise to the protein.

This gene was isolated from the Rhizobium meliloti, a bacterium commonly known for forming leguminous nodules in certain plants.

As for it's function, I'm not really a biologist, but I know that magnesium is a component of chlorophyll (It's coordinated at the center of the porphyrin complexes).

So you could say magnesium is rather important for plants. The $\ce{Mg^2+}$ ion needs to be transported from the soil to the leaves, and this protein facilitates this transport.


Identify the type of structural isomerism

Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. Constitutional or structural isomers are compounds with the same molecular formula but their structural formulas are different. They also called chain isomers because of the possibility of branching in carbon chains (e.g., n-butane and 2-methylpropane). Note that just the twisted versions of the original molecule can be mistaken by novices as structural or chain isomers (e.g., conformers).

Positional isomers: Positional isomers are constitutional isomers that have the same carbon skeleton and the same functional groups. They differ from each other in the location of the functional groups on or in the carbon chain. In other words, in position isomerism, the basic carbon skeleton of isomers remains unchanged, but important groups on the skeleton are moved around on it (e.g., 1-bromopentane, 2-bromopentane, and 3-bromopentane).

Functional isomers: Functional isomers are constitutional isomers bearing different functional groups (e.g., propanoic acid, methyl ethanoate, and 3-hydroxypropanal).

Accordingly, the answer given is correct: These are functional isomers, since each have at least one different functional group changed (keto vs aldehyde).


Multiplication of the reaction and Equilibrium constant

Rate will not be square in most of the cases and who said so? The equilibrium constant scientifically changes because Gibb's free energy is extensive quantity and it is related to equilibrium constant through relation that you might know $\Delta G=\Delta G^\circ +\mathcal{R}T\ln Q $. Where $Q$ is reaction quotient which at equilibrium becomes equilibrium constant and then it gives $\Delta G^\circ = -\mathcal{R}T\ln K$


Is “lime” short for something?

In this case it is not a fruit Lime is a mixture of calcium hydroxide and calcium oxide produced from calcining limestone.

  What is the meaning of “the floors of which had been covered with lime”?...Was the floor covered with this intentionally, or did this “lime” accumulate naturally?

Not really the place to ask, but the lime likely was from previously treating bodies. As people were left to die, lime would be put on their corpse to keep the smell of decaying flesh down. There is also some belief that lime can speed up decay and reduce bacteria growth, but this is disputed. Any way not all of the lime would land on the body or it would fall off when moved. This process had happened so many times prior that the floor was covered in it. lime tends to build into layers as it is chemically similar to cement and will harden over time the same.

  What would be the consequences of this to the people in the railway cars?

Lime is very caustic and causes great respiratory distress when inhaled. It is used to capture the putrid scents of decaying flesh.

Today lime is still used at mass grave sites to capture the scent of decay and keep soil pH high. Low pH soil is an indicator for a mass grave as the decomposition products are acidic and lower the soil pH. Adding lime reduces this acidity masking (trying to mask) the presence of a mass grave.


What is the oxidation state of copper in Gilman's Reagent?

Don't count bonds.  Count electrons.

Copper has $11$ valence electrons as a neutral atom, as the $d$ orbitals are able to participate in reactions.  In the oxidation state formalism the alkyl groups, whose atoms are more electronegative than copper, are each considered to take an electron away from the copper to complete their carbon octets.  But then the lithium, bonded to copper (bonding lithium to the alkyl groups would give a separate organolithium molecule instead), adds an electron.  So the copper goes from $11$ valence electrons down to $10$ in the oxidation state formalism, losing one overall; the oxidation state of copper is thereby $+1$.


According to KMT, is the velocity of an ideal gas always sqrt(3RT/M)?

$E=\frac 12mv^2 \implies  v=\sqrt{\frac{2E}{m}} $ is valid for translational kinetic energy and the speed of the centre of mass.
Vibrational or rotational energy does not count. An object may vibrate or rotate even if it's centre of mass has zero speed.
As each available degree of freedom has the mean energy $E=\frac 12kT$,
and as there are 3 independent translational degrees of freedom,
the mean translational kinetic energy is $E_\mathrm{transl}=\frac 32kT \le E_\mathrm{kin}$
and the respective
$v_\mathrm{RMS}=\sqrt{\frac{2E_{transl}}{m}}=
\sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}$
even for ideal gas consisting of  multiatom molecules.


Why are reactions of acids with metals exothermic?

Yes there is a rigorous way of calculating the enthalpy change.

First write down an example reaction.  It might not be as easy as you think.  What species are really formed when zinc chloride (for example) dissolves in water?

The simple version below...

$$\ce{Zn(s) + 2 H+(aq) -> Zn^2+(aq) + H2(g)}$$

...won't do, solvated zinc ions in hydrochloric acid are present in a variety of tetrahedral chloro-aqua complexes.

So maybe something like

$$\ce{Zn(s) + 2 H+(aq) + 2 Cl-(aq) + 2H2O(aq) -> ZnCl2(H2O)2(aq) + H2(g)}$$

This at least shows which bonds are being broken and which are being formed:

Zn-Zn bonds (metallic) are broken
H-H bonds are being formed
Zn-Cl and Zn-aqua (actually zinc-oxygen) bonds are being formed

To figure out the enthalpy change, you can figure out the approximate enthalpy of formation of each bond type.  H-H bonds are pretty "strong", i.e. the bond enthalpy is relatively high at 436 kJ/mol.  So when those bonds are formed, that enthalpy is released.  Zinc-water (i.e. zinc-oxygen) and zinc-chlorine bonds are also pretty strong at about 284 and 229 kJ/mol, respectively.  Zinc-zinc bonds enthalpies are only ~29 kJ/mol.

Add it all up, and you get a lot of enthalpy being released!


Why does decreasing the pressure of the system increase the relative volatility of a binary solution?

Let's assume that the vapor pressure of both components can be modeled with the Antoine equation.  This gives us a convenient way to address the question algebraically.  The vapor pressure of most chemicals is excellently modeled by the correct Antoine equation for that chemical.  The equation is:
$$\log_{10}{p} = A-\frac{B}{C+T}$$

where the the parameters $A$, $B$, and $C$ are compound specific parameters.  The equation can be solved for temperature explicitly yielding $T = \frac{B}{A-\log_{10}\, p} - C$.

So at $\pu{1 atm}$, the equation for pure species $i$ and also a different pure species $j$ that is higher boiling by $\pu{50 ^\circ C}$ is:

\begin{align}
T_1 &= \frac{B_i}{A_i-\log_{10}\, p_{\pu{1 atm}}} - C_i\\
T_1 &= \frac{B_j}{A_j-\log_{10}\, p_{\pu{1 atm}}} - C_j - \pu{50 ^\circ C}
\end{align}

where $T_1$ is the boiling point of $i$ at $\pu{1 atm}$.  To make the math easier, let's suppose the equation is in units of atmospheres, in which case $\log_{10}\, p_{\pu{1 atm}} = 0$.

\begin{align}
T_1 &= \frac{B_i}{A_i} - C_i\\
T_1 &= \frac{B_j}{A_j} - C_j - \pu{50 ^\circ C}
\end{align}

The question is, what happens when $p$ changes, in this case, when it is lowered by ~760-fold, let's call that 1000-fold to keep the math easy.

\begin{align}
T_{0,i} &= \frac{B_i}{A_i-\log_{10}0.001} - C_i = \frac{B_i}{A_i+3}-C_i\\
T_{0,j} &= \frac{B_j}{A_j-\log_{10}0.001} - C_j = \frac{B_j}{A_j+3}-C_j
\end{align}

Now we have four equations but six unknown parameters, $A_i$, $A_j$, $B_i$, $B_j$, $C_i$, and $C_j$.  So in general, there is no generally valid constraint for the temperature difference $T_{0,i}-T_{0,j}$:  It could be higher, lower, or about the same as the difference at room temperature.

However, we could make some further assumptions.  A good one here could be that both species in question have a constant heat of vaporization, which means that $C_i=C_j=0$.  If that assumption is true, then the four equations become:

\begin{align}
T_1 &= \frac{B_i}{A_i}\\
T_1 &= \frac{B_j}{A_j} - \pu{50 ^\circ C}\\
T_{0,i} &= \frac{B_i}{A_i+3}
 =\frac{\frac{B_i}{A_i}}{1+\frac{3}{A_i}}
 =\frac{T_1}{1+\frac{3}{A_i}}\\
T_{0,j} &= \frac{B_j}{A_j+3}
 =\frac{T_1+50 °C}{1+\frac{3}{A_j}}
\end{align}

Now, the temperature difference of interest $T_{0,j}-T_{0,i}$ is

$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{1+\frac{3}{A_j}} - \frac{T_1}{1+\frac{3}{A_i}}$$

To keep the formulas easy let me define new parameters $a_i$ and $a_j$ such that $a_x=1+\frac{3}{A_x}$.  Then the temperature difference is

$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{a_j} - \frac{T_1}{a_i}= \frac{a_i(T_1+\pu{50 ^\circ C})-a_j T_1}{a_i a_j}=\frac{(a_i-a_j)T_1 + a_i(\pu{50 ^\circ C}) }{a_i a_j}$$

Now we must introduce further assumptions about the Antoine $A$ parameters.  They are positive and thus the $a$ parameters must also be positive, and further positive $A$ implies  $a>1$.  If we further assume $a_j=a_i$, then

$$T_{0,j}-T_{0,i} =\frac{(\pu{50 ^\circ C}) }{a_j}$$

Since we know already that $a_j>1$, then the temperature difference (i.e. the difference in boiling points) at the lower pressure is less than the temperature difference of $\pu{50 ^\circ C}$ at \pu{1 atm}.  Now, even if $a_i$ is not precisely equal to $a_j$, then the temperature difference will still be less than $\pu{50 ^\circ C}$ as long as the $a_i$ and $a_j$ parameters are similar enough to keep $(a_i-a_j)T_1 << a_i(\pu{50 ^\circ C})$.  It would take a very unusual choice of chemicals $i$ and $j$ for that to happen.  I don't think I could identify such a pair.

So, in summary, to get to the result that I think your instructor was asking for, we required many assumptions:

Both chemicals obey the Antoine equation in the temperature range of interest (probably an OK assumption).
Both chemicals have approximately constant enthalpy of vaporization in the T range of interest so that we can neglect the Antoine $C$ parameters.
The Antoine $A$ parameters for both chemicals must be positive.  (A good assumption but it still must be made.)
The chemicals have Antoine $A$ parameters (and thus also $a$ parameters as I defined them here) that are not too different from each other.

It might be possible to weaken some of the assumptions and still find that the lower-pressure boiling point difference must be lower than $\pu{50 ^\circ C}$, but I'm confident that the result is not necessarily generally true.  It isn't a thermodynamic law or anything.  It definitely depends on several key assumptions about how vapor pressure "usually" or "typically" behaves for most chemicals.


Which component in my tap water "scale" does not react with acid?

A likely candidate is calcium sulfate, which forms a relatively soft mineral (gypsum), hence removable with a knife, but it has limited solubility in water and does not react with common acids.

See here for more on water hardness.


Is it always the case that when a reaction's products are higher in energy than its reactant, then the reaction is not favored?

This question brings up the broader point of what it means for a reaction to be favored. Your question describes thermodynamic favorability and that is just dependent on the relative energy of the products and the reactants. An analogy to this is to think of energy levels as literal heights subjected to gravity. If you have a ball on the ground (reactants) and you want to put it on a shelf (products), you have to put in energy. But, once it is in the product state, gravity will want to drive the ball back to the reactant state because it is lower in energy.

Another type of favorability is kinetic favorability; this relates to the rate at which the forward and reverse reactions can occur. There are cases where a thermodynamically favorable reaction doesn't occur (or rather occurs at an incredibly slow rate) due it being kinetically unfavorable. An example of this is diamond and graphite; it would be thermodynamically favorable for diamond to convert to graphite, however the reaction to do so is so kinetically unfavorable that you are unlikely to see it occur.


How do I find out what benzene containing compound this is with IR and NMR?

This is another example of how important it is to work with good quality data, which can only come from preparing good quality samples. Sample preparation comes from good technique and lots of practice. Without it, you are in a very difficult starting position. Enough of my soapbox rant.

Let's start with IR (not my strong point)

The first and second IR spectra are so different that it is hard to be confident about which one is correct. Are they collected in different media (one a solution and one a mull perhaps)? The first looks very low concentration

~3420 is clearly an -OH stretch. Probably too high to consider a N-H group of any sort. So, it could be an alcohol or an acid.
1682 a C=O group. Probably in the acid region, likely aromatic acid, conjugated or intramolecular H-bonded. Possibly amide, but not convincing based on other peaks.
There are lots of other peaks, most of which could mean many things, but I do see 749 and 708 which support the NMR notion of a monosubstituted benzene. I usually think that IR is only useful to corroborate peaks of a known structure, but then I'm an NMR advocate.

NMR

Again, the first and second spectra are very different; the obvious concern is that they are of different batches of sample - were the first and second analyses taken from the same sample, or from a different batch. In any case, let us look at the second, which is only a partial spectrum, and omits some possibly very significant information, especially in the region above 10ppm, where I would like to see a broad acid peak.

The aromatic protons shown exhibit a classic mono-substituted benzene ring pattern, labelled above. There is a strong downfield shift of the ortho (H2) protons, consistent with an electron withdrawing group substitution (such as -(C=O)R). Ortho coupling for the H2 proton is difficult to determine accurately as there are 3 peaks picked. This may be a digitizer point average, but considering the average of each peak, and the coupling from the first spectrum, we get a ortho coupling of 7.2-7.9. Not overly useful, but less than 8, so we can exclude ethers and halides. Couplings are calculated by:

(8.1651-8.1507) x 500 = 7.2

Using additive analysis for monosubstituted benzene rings, we can determine the following shifts from benzene:

H2 (8.162) 7.34 + 0.82
H3 (7.514) 7.34 + 0.17
H4 (7.651) 7.34 + 0.31

Looking at the published tabulated data for substituent effects on monosubstituted benzene rings (I'm using Pretsch, Buhlmann and Badertscher, but lots of other sources are available), we find one possible candidate that matches closely:

Benzoic Acid H2(+0.79), H3(+0.14), H4(0.28)

The caveat that I would add to this is that the data presented is not the best quality, and the solution presented is consistent with the data only (not definitive). The key concern is the missing -OH peak in the NMR spectrum. Unless this sample was run in D2O, or washed with D2O, I would expect to see an acid peak around about 12 or so. Depending on the amount of water in the sample, this may be shifted somewhat and broadened, but should still be visible, even if a broadened hump in the baseline. Additional NMR experiments could be useful to confirm the structure.


Why does KCl come in amber coloured bottles?

The potassium chloride solutions in my lab are in transparent glass bottles, as was the potassium chloride solution in my inorganic students’ lab at the LMU in Munich. If I recall it correctly, the solid potassium chloride we had there was in a transparent bottle, too, but don’t quote me on that.

There should be nothing in potassium chloride that will react under irradiation. And these types of impurities are generally not a reason to use such a bottle. But maybe the supplier of your choice just happens to have too many amber bottles?


Entropy paradox and Confusion with Gibbs energy

The essence of what's going on is this:
Your equation, $dG = Vdp -SdT$, is correct, but has restrictions you don't realize. Specifically, it applies only to systems that are (a) closed (so no adding or subtracting material), (b) have only a single component* (so no mixing, phase changes, or chemical reactions) and (c) that can do $pV\ \hspace{-.45em}$-work only.
[*It also applies to multi-component systems where the composition is fixed.]
To understand why, under such conditions, $dT = 0 \text{ and } dp = 0 \Rightarrow dG = 0$, let's apply the Gibbs phase rule to such a system.  The phase rule says:
$$F = C − P + 2,$$
where C is the number of components, P is the number of phases at equilibrium (not to be confused with $p$, pressure), and F is the number of degrees of freedom.
Since $C = 1 \text{ and } P = 1$, we obtain $F = 2$.  That means we only have two degrees of freedom, i.e., two independent ways in which we can vary the intensive properites of the system. If the only type of work we can do is $pV\ \hspace{-.45em}$-work, the only ways we can adjust the system's intensive properties are by changing its temperature or its pressure.
Hence, if we apply the restrictions $\boldsymbol{dT=0 \textbf{ and } dp =0}$ to a closed single-component system where we only allow $\boldsymbol{pV}$-work, the system can't change!  And if the system can't change, then of course $\boldsymbol{dG = 0}$!
But, you might protest, $dG$ is not generally zero, even at fixed  $T\text{ and } p$.  So how do we reconcile this with what we wrote above?  Well, we need a more general expression for $dG$ that allows for non-$pV\ \hspace{-.45em}$-work, addition and subtraction of material, and changes in composition:
$$dU = \text{đ}q + \text{đ}w +\sum_i \mu_i dn_i
= \text{đ}q + \text{đ}w(pV) + \text{đ}w (non\text{-}pV) + \sum_i \mu_i dn_i$$
Since we can calculate $dU$ using any path, let's use a reversible path:
$$dU = TdS - pdV + \text{đ}w (non\text{-}pV, rev) + \sum_i \mu_i dn_i$$
And since:
$$G = U+pV-TS \Rightarrow dG = dU + pdV +Vdp - TdS - SdT$$
$$\Rightarrow  dG =  VdP -SdT+ \text{đ}w (non\text{-}pV, rev) + \sum_i \mu_i dn_i$$
Here, $\sum_i \mu_i dn_i$ is the sum of the chemical potential of each species $i$  $(\mu_i)$ times the change in the amount of species $i$ $(dn_i)$. This accounts for the change in $U$, and thus in $G$, as we change the composition
Hence, even if $dT = 0 \text{ and } dp = 0$, if we have non-$pV\ \hspace{-.45em}$-work and/or a change in composition (e.g., mixing, a change in phase, gain or loss of material, or a chemical reaction), it will not be the case that $dG$ is restricted to be zero.


What is the oxidizing agent for Fe in the following redox reaction?

To answer your initial question:

The oxidizing agent is the reagent that gets reduced. Conversely: The reducing agent is the reagent that gets oxidized.

Now, addressing your trouble with the concept:
You are right that hydrochloric acid is found in the ionic form when in aqueous solution, so:
$$\ce{HCl ->[][\ce{H2O}] H+ + Cl-}$$

Now we have the reagent that gets reduced, which is the proton. Together with another proton and two electrons, they form hydrogen gas that then evolves from the solution:
$$\ce{2H+ + 2e- -> H2 ^}$$
This is basically our electron sink, because that's where they go "to die", i.e. to vanish from our solution.

But where do these electrons come from? Well, we have two potential electron donors here: The chloride anion and the iron. Which one will give up his electrons more easily?

The answer we already know: Iron.

Why? Because as you correctly stated, the chloride anion has all the shells filled and is very happy because of that filled octet state. So the energy to get out an electron from it is higher than to rip it away from an unsuspecting iron atom.1)
$$\ce{Fe^0 -> Fe+ + e- -> Fe^{2+} + 2e-}$$

The electrons then vanish (because of the formation of hydrogen, remember?) which leaves us with some iron cations and chloride anions. They will then start to mix with each other. Don't judge them, they don't know any better!
$$\ce{Fe^{2+} + 2Cl- <=> FeCl2}$$

Since ferric chloride (II) is quite soluble in water, it will most probably not precipitate.

Footnotes:

Actually, you could also compare the standard reduction potentials of the reactions $\ce{Fe -> Fe^{2+} + 2e-}$ and $\ce{Cl- -> Cl^{.} + e-}$. The one that has the lower standard reduction potential is the winner (because the energy of a system is always minimized). This is also what you would usually do, but in this example it wasn't necessary because you could get to the solution via chemical concepts.


Introduction to modeling chemical reactions

The topic you are asking about is called reaction dynamics, and is closely related to chemical kinetics (the former is more focussed on atomistic events, reactive and nonreactive collisions, while the later usually describes the mathematical treatment of the kinetic laws governing species concentrations). It was the topic of the 1986 Nobel Prize in Chemistry.

It is covered by most physical chemistry textbooks, and more in depth by specialized textbooks on that matter. It is sometimes taught together with basics of statistical mechanics or statistical thermodynamics. Check your favorite (or local) university for a list of reading material on this topic!


In search of a safe and inexpensive solvent that can dissolve depolymerized polypropylene waste plastic

Short answer: Not really, unless you can do the cleaning still at high temperature. Solid hydrocarbons tend to only become soluble rather close to their melting point.

The only reasonable solvent is petrol ether, i.e. a mixture of alkanes. Polar solvents obviously don't work, chlorinated solvents are ecologically problematic and only have the advantage of an increased boiling point. Toluene or xylene might be alternatives.

So basically what you should do is to switch your feedstock from polyolefines to liquid alkane (petrol, basically) before shutdown. I would however be amazed if your process engineers don't do exactly that anyway. Maybe they could do that for a little longer, or maybe your process should be run (at least from time to time) at a slightly higher temperature so solidified (i.e. crystallised) residues in the reactor can melt and be washed out.

And whatever remains after shutdown and cooling, you will have to scrape off by hand, sorry.

(A sloppily designed or maintained reactor with bad flow profile, insufficient/broken insulation/heating devices of course might or might not be fixable.)


Reaction of nitrobenzene with potassium ferricyanide

I'd go more for an oxidation, but on the ring (hydroxylation).
My guess comes from this paper here : "The Hydroxylation of Aromatic Nitro Compounds by Alkalies" by O. C. Dermer.

This is very harsh (and somehow old) chemistry.

First, for m-dinitrobenzene and sym-trinitrobenzene, hydroxylation is reported in your conditions, that is to say alcaline conditions with the help of potassium ferricyanide as oxidizing agent (reactions A and B, see reference in the paper).

Second, the authors report the reaction C (with less than 4% yield though, according to my calculations), with a small amount of p-hydroxyazobenzene as a side-product. They did this reaction by mixing only nitrobenzene with potassium hydroxide, so I would expect the same product (o-nitrophenol) with maybe a better yield in your conditions.


What kind of stereoisomerism do cisplatin and transplatin exhibit?

The definition of diastereomer does not include "requires two stereocentres" anywhere. It is simply

  Diastereoisomers (or diastereomers) are stereoisomers not related as mirror images. (IUPAC Gold Book)

Therefore cis- and trans-$\ce{[PtCl2(NH3)2]}$ are diastereomers.

(E)- and (Z)-alkenes are also diastereomers of each other. Your "cis/trans stereomer" is in fact simply a subset of diastereomerism.


It depends on the context how he meant it. I could imagine two scenarios:

Certain materials can produce ions when they dissolved in water. A typical example is table salt, NaCl. Table salt is solid form has a lattice that contains Na+ and Cl+ ions, and when you put it in water, the water molecules can help releasing these ions into the solution, where you now can find the ions. Note, that in this case the ions are atomic entities but molecules with charge are also ions, like $\ce{MnO4-}$ or $\ce{SO}_4^{2-}$.
Another scenario when the charged entities are formed as a reaction with water by e.g. giving an electron (oxidation) to water molecules or to other dissolved components e.g. oxygen in water. As you can imagine, it happens generally with very reactive materials (materials that are very sensitive to moisture), but essentially the rusting of iron is a similar reaction.


What is the rule for the punctuation marks for a bicyclo compound?

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

P-16.2.2 Full stops
Full stops are used to separate numbers that denote lengths of bridges in polyalicyclic names constructed according to the von Baeyer system (…) and of chain lengths between spiro atoms in von Baeyer spiro names (…).

Examples:
bicyclo[3.2.1]octane
6-oxaspiro[4.5]decane


If propane with specific parameters were to combust in open air, what is actual volume of CO2 produced

Rephrased homework question? Anyway, you are going about this the wrong way, thinking much too complicated.

If you combust m moles of propane, how many moles y of CO2 do you get? Now remember that moles and liters of ideal gas are totally proportional, and you have the answer.


In Ion Chromatography instrument, Does the specific conductivity (peak area) of an analyte depends upon background conductivity?

The heart of an ion chromatograph is the suppressor, whose sole purpose is to minimize the background conductivity and convert analytes into highly conductive ions for example fluoride ion becomes HF, chloride becomes HCl and so on. If your background conductivity is high, it means that the conversion of F(-), Cl(-) into corresponding acids is also not 100% efficient-peak height can potentially decrease as a result. The peak areas will also change. Although the background is higher, one can still make a calibration curve (it will be less sensitive than the one with low background).
Remember that peak area in chromatography is not an absolute measure like mass; it is only true for one particular condition and set-up.


In that equation, the acid and base are a conjugate acid/base pair.  We assume HCl quantitatively dissociates and protonates benzoates.  The concentration of the acid increases, and the base decreases.

Empirically, the pH should go down in this case.  Without seeing your specific calculation, we can't comment where you are going wrong, but the proper flow should look something like:

$$\mathrm{pH=pK_a+Log_{10}(\frac{[base]-[Added H]}{[acid]+[AddedH]})}$$

$$\mathrm{pH=pK_a+Log_{10}(\frac{0.39M - 0.036M}{0.10M + 0.036M})}$$

$$\mathrm{pH=pK_a+Log_{10}(\frac{0.354M}{0.136M})}$$
From your calculated $\mathrm{pKa = 4.20}$, you should get there pretty quickly.

Just applying the difference in the log values of the changed ratios, I get  $\mathrm{pH = 4.61}$


How can I increase the viscosity of dilute hydrochloric acid

If the solution is not too concentrated and you dont need it to be stable for a prolonged period, many viscosity modifiers can be used. I would start trying something simple, like CMC based wallpaper paste.


What causes the DNA fragments to stop moving in gel electrophoresis?

There are electrophoretic methods where the molecules stop automatically like Isoelectric Focusing, but in the common agarose gel electrophoresis of DNA you mention this is not the case.

The DNA is negatively charged, it will move along the electric field you set up until you stop applying any current. It will happily move out of your gel if you let it run for too long.

To see the progress of your gel some dyes are usually added to the sample like bromophenol blue or xylene cyanol. They don't interact with the DNA, they just provide you a reference on how far your gel has run. You still need to have a rough idea how your samples run compared to those dyes if you want to use the full length of your gel.


Which of these compounds does not liberate heat when water is added to it?

It is important to note that if a molecule or formula unit is capable of undergoing hydrogen bonding ($\ce{NaOH}$, $\ce{H2SO4}$), then the formation of this hydrogen bond will be extremely exothermic.

Also, if a substance can react to form a substance that will undergo hydrogen bonding ($\ce{CaO + H2O -> Ca(OH)2}$), this species will subsequently form these hydrogen bonds and liberate heat.

Lastly, as you have stated you know from personal experience, the reaction between a alkali metal ($\ce{Na}$) and water is extremely exothermic— so much so that many of these metals will melt from the heat liberated.

Otherwise, there is no fail-proof method for predicting the enthalpy of solution for any given substance. Generally, the enthalpies of solution are determined experimentally, so unless you are expected to perform all these reactions in a calorimeter yourself, I would recommend something like this: this.


Effect of dilution on the degree of dissociation of an electrolyte

If our dummy dissociation degree is $$\alpha=\frac{[\text{C}]}{[\text{A}]+[\text{C}]}=\frac{[\text{D}]}{[\text{A}]+[\text{D}]}$$ and the total concentration of dissociated and undissociated species is $$x=[\text{A}]+[\text{D}]=[\text{A}]+[\text{C}]$$ the equilibrium constant, on which le Chatelier's principle is (well, should be) based is $$K=\frac{[\text{C}][\text{D}]}{[\text{A}]}=\frac{\alpha^2x}{1-\alpha}$$ Now, we'd like to dilute (to decrease) the total concentration $x$. Since the equilibrium constant cannot change, $\alpha$ must increase to keep the constant constant (sorry). And all the concentrations decrease upon dilution, which you can easily prove by checking whether derivatives $\frac{\partial[\text{A}]}{\partial x}$, $\frac{\partial[\text{C}]}{\partial x}$ and $\frac{\partial[\text{D}]}{\partial x}$ are positive.

I really hate them easy "intuitive" principles. In my opinion, resorting to the discussion of the equilibrium constants gives you a clearer picture.

And yes, the reaction proceeds in both forward and reverse directions, if only because it is an equilibrium process.


Why does dissolving a salt in water get a high pH?

It's a little more complicated than that.. When you dissolved $\ce{Na2CO3}$ in water you had a solution of ions: $\ce{Na+}$, $\ce{CO3^2-}$, $\ce{H3O+}$ and $\ce{OH-}$ (the later are present in water even when there is no salt, since it dissociates too, however they are in equal quantity in pure water so the pH is 7).

Now when you mix all those ions, the equilibrium isn't only the equilibrium of water dissociation ($\ce{H2O + H2O <=> H3O+ + OH-}$). Now you have a whole series of equilibria occurring in the solution simultaneously:

1) Water dissociation:
$\ce{H2O + H2O <=> H3O+ + OH-}$

2) Salt dissociation:
$\ce{Na2CO3 <=> 2Na+ + CO3^2-}$

3) Base equilibrium*:
$\ce{NaOH <=> Na+ + OH-}$

4) Acid equilibrium*:
$\ce{CO3^2- + 2 H3O+ <=> H2CO3 + 2 H2O}$

The salt dissociation doesn't affect the pH directly. We have two equilibria though that do unbalance the proportion of $\ce{H3O+}$ and $\ce{OH-}$ ions, which are the "base equilibrium" and "acid equilibrium". We don't think about them at first sight because we didn't add any $\ce{NaOH}$ or $\ce{H2CO3}$ to the solution, but everything needed to form them is right there.

$\ce{NaOH}$ is a strong base, however $\ce{H2CO3}$ is a weak acid. So it's likely (and you witnessed it in experience) that more $\ce{H3O+}$ will be consumed in comparison with the $\ce{OH-}$, making the solution more alkaline.


Why do atomic orbitals have their unique shapes?

The pictures that we typically use to represent orbitals are really just graphs showing the 3-dimensional probability of finding an electron (that occupies the orbital being examined) in space.  $\ce{\Psi}$ is the wavefunction that describes the electron with a set of specific quantum numbers.  $\ce{\Psi^2}$, gives rise to the electron density distribution (the shape of an orbital) plotted in the graph.


How to remove white coating on the seashells after lemon acid and soap?

Seashells' outer layer (periostracum) mainly consists of conchiolin and polymorphic modifications of $\ce{CaCO3}$ (aragonite, calcite). To sum it up, I see two possible sources of the white coating after treating the seashells with citric acid:

Formation of calcium citrate, which has limited water solubility ($\pu{0.9 g/L}$ at $\pu{20^\circ C}$):
$$\ce{3CaCO3 + 2C6H8O7 -> Ca3(C6H5O7)2 v + 3CO2 ^ + 3H2O}$$
You can try to use acetic acid instead of citric acid in order to produce calcium acetate, which is very soluble in water ($\pu{347 g/L}$ at $\pu{20^\circ C}$). I don't think you can easily dissolve calcium citrate with acetic acid because calcium is complexated by citrate, and this is a very stable formation. The problem here is that the prolonged exposure to acidic medium can further damage the outer layer as more calcite will be washed away.
Conchiolin porous network remained after $\ce{CaCO3}$ has been eliminated with an acid might also appear white when dry and then, when the pores are filled with water, return to it's original appearance. This is rather a speculative theory, but if this is the case, you will need to perform mechanical surface polishing in order to get rid of the porous layer. This can potentially damage the seashells and should be done cautiously.


In Statistical Transition State Theory, why is the concentration of the transition state for the forward and reverse reaction identical?

In transition state theory, it is assumed that the forward and the reverse reaction occur via the same transition state. This is implied by assuming a quasi-equilibrium between the reactants and the products, which adheres to a common Boltzman statistics. Therefore the labelling $\ce{AB^\ddagger_r}$, $\ce{AB^\ddagger_f}$ is artificial, because it is the same transition state:
$$\ce{AB^\ddagger_r} = \ce{AB^\ddagger} = \ce{AB^\ddagger_f},$$
and therefore the concentration is identical, too:
$$[\ce{AB^\ddagger_r}] = [\ce{AB^\ddagger}] = [\ce{AB^\ddagger_f}].$$


Can water volume reduce with introduction of another chemical

Yes, at least if only the total volume should decrease:

Density of water: $\pu{1000 kg/m^3}$
Density of ethanol: $\pu{789 kg/m^3}$
Expected density of 52 Vol.% ethanol solution: $\pu{890 kg/m^3}$
True density of 52 Vol.% ethanol solution: $\pu{909 kg/m^3}$
All values are for the same conditions.

The volume of every ethanol-water solution is smaller than the sum of the original volumes of water and ethanol. For example if you add $\pu{52 mL}$ of ethanol to $\pu{48 mL}$ water you will get a solution with an volume of approximately $\pu{97.9 mL}$.
This effect is known as volume contraction and depends on the mixing ratio of both components in a nonlinear way. Its reason is the formation of additional bonding forces (e.g., hydrogen bridges) between the molecules, which results in a higher density.

Note: I don´t think that there is something that can increase density of water in such a way that the resulting total volume is actually smaller than the volume of water used.


Possible approximations in complex equilibrium

There is some chemistry behind this which you must understand to recognize the approximation. First the copper ammine complex doesn't really form in the manner indicated rather:
$$\ce{Cu^2+ + 4H2O <=>T[acidic solution] Cu(H2O)4^2+}$$

But in basic solution copper will form a gelatinous precipitate which can perhaps be represented as:

$$\ce{Cu(H2O)4^2+ + 2OH^{-} <=>T[basic solution] Cu(H2O)2(OH)2}$$

With ammonia in the aqueous solution various ammine complexes will be formed such that the soluble tetra-ammine complex is formed in a series of reactions something like:

$$\ce{Cu(H2O)2(OH)2(s) + NH3 <=> Cu(H2O)(OH)2(NH3)} $$

$$\ce{Cu(H2O)(OH)2(NH3) + NH3 <=> Cu(OH)2(NH3)2} $$

$$\ce{Cu(OH)2(NH3)2 + NH3 <=> Cu(OH)(NH3)3^+} $$

$$\ce{Cu(OH)(NH3)3^+ + NH3 <=> Cu(NH3)4^2+(aq)} $$

Ammonia is typically present in excess so the overall reaction is written "in shorthand" as the following equation (I highly doubt that the exact sequence of chemical reactions between the gelatinous precipitate and ammonia is fully understood):

$$\ce{Cu^2+ + 4NH3 <=> Cu(NH3)4^2+} \quad \text{K}_\text{eq}$$

and the equilibrium for this reaction is greatly favored to the right (the ammine complex form).

If you let $[\ce{Cu2+}]_i$ be the initial concentration of $\ce{Cu^{2+}}$, let $[\ce{NH3}]_i$ be the initial concentration of $\ce{NH3}$ and the final concentration of the ammine complex be $[\ce{Cu(NH3)4^2+}]_f$ then in excess ammonia a simplfying assumption would be:
$$ [\ce{Cu(NH3)4^2+}]_f \approx [\ce{Cu2+}]_i$$
$$[\ce{NH3}]_f = [\ce{NH3}]_i - 4[\ce{Cu(NH3)4^2+}]_f \approx [\ce{NH3}]_i - 4[\ce{Cu2+}]_i$$

the equations could of course be solved exactly, but you end up with a quadratic expression which really isn't a difficult mathematical problem to solve. It just takes a wee bit more calculating.


In the days before chemistry, alchemists searched for the Philosopher's stone which would enable them to transform base metals into gold. They spent a lot of effort searching for it but we know now that their search was futile as no chemical process can cause elemental transformation. Only nuclear processes can do that but they cost far, far more than mining the element ever will so this doesn't really count as a solution.

But plenty of people faked a result or were conned by their lack of understanding. Fool's gold, for example looks a lot like gold, but doesn't share many other properties but, presumably, the name arises because some people are gullible. There are also many metal alloys that look like gold but are not pure (including many alloys that contain some, but not much, gold). It is also possible to use a variety of chemical or electrochemical processes to coat the surface of many metals with a thin film of gold.

But careful analysis can reveal most of these methods of fakery (including the alloying of gold with other metals which was very tempting when coins were made of gold). One of the most famous stories of Greek science is based on one of the most reliable methods. Archimedes was faced with the task of distinguishing debased coinage or other gold objects and, in the legend, realised the answer while having a bath, leapt out the bath and ran down the street naked yelling "eureka". What he realised was that an object immersed in water displaces a volume of water equal to its volume and this can be measured by weighing the displaced fluid. Combined with the weight of the object this can be used to estimate the density of the object. Since the only other metal with a density close to gold is tungsten (gold is 19.28 g/cm3 and tungsten is 19.25 g/cm3) knowing the density is a good way to test the purity of the metal. Common metals used in alloys, like silver (density 10.5 g/cm3) will substantially lower the density.

So, in short, you can make a lot of things that look like gold, but it is very hard to make something that shares enough properties to convince a skeptical observer who will test more than the appearance.


An acid that can donate one hydrogen ion is called monoprotic (i.e., $\ce{HCl}$). An acid that can donate two is called diprotic (i.e., $\ce{H_2SO_4}$). An acid that can donate 3 is called triprotic ($\ce{H_3PO_4}$). I don't know of any which can donate four.


Assign stereodiscriptor to 4th carbon of 3-ethyl-2,4-dimethly-5-(propan-2-yl)heptane-1,7-diol

Not the proper, somewhat complicated, IUPAC ‘digraphs’… but might help.

Bold/colored bonds are the left and right branches' resulting paths taken when advancing to next sphere (because the decision could not be yet done) during determining a higher priority branch by finding a first difference of the compared atom set rankings. Sphere numbers a Roman. Grey digits are semi-conveniently arbitrarily chosen numbering just for reference, not nomenclature numbering nor priorities(!).

Following list shows chosen highest priority atoms in comparison in each nth sphere and left/right branch (bold; also comparison with other sibling atoms in the same branch, if they are not just hydrogen), chosen according to the next bonded atoms from the (n+1)th sphere (in 1st braces; atom number in 2nd brace).

sphere № : [left] vs [right]
I: [ C(CCH) (1) ]   =   [ C(CCH) (4) ]
II: [ C(CCH) (2) > C(CHH) (3) ]   =   [ C(CCH) (5) > C(CHH) (6) ]
III: [ C(HHH) (7) = C(HHH) (8) ]   <   [ C(OHH) (9) > C(HHH) (10) ]

Please note that in the II. sphere, left, the carbon (3) leading to terminal -OH group could not be chosen, because it's C(CHH), compared to higher priority, chosen C(CCH) (2).

Also note that in sphere III, left, any of the two methyls' carbons could be chosen (equivalent).

In this III. sphere, the decision finally could be made. The right branch (R1) has higher priority – finally only hydrogens bonded at left: C(HHH) (7), vs oxygen and hydrogens bonded at right: C(OHH) (9).
(Next priority is left branch R2, then methyl, then H.)

So the compound configuration/name is
(4S)-3-ethyl-2,4-dimethyl-5-(propan-2-yl)heptane-1,7-diol.


How to decide what pathway a reaction will follow, elimination, addition to a carbonyl or nucleophilic substitution?

The nucleophilic substitution with another equivalent of the primary bromide will never be favourable as the nucleophilicity of the enolate generated is pretty low (the charge is delocalized!)
Even if E1cb was a possible route, you would end up with an extremely reactive vinyl aldehyde, which would likely participate in a Michael addition with an enolate in the reaction medium. So, isolating the product from this route would be very unlikely, even though I cannot back my claim up immediately.

If you ask me, both the Michael addition and the Aldol reaction should compete, given the nature of the base the problem has used. Had it been something like LDA (or any other directed Aldol conditions), the greater stability of the enolate would probably have precluded E1cb as a possible route.


Partial pressure of gases in a container with a piston

$B=\pu{0.6226 bar}$ as the water-water vapour system would be in equilibrium.
Now,
$A+B= \pu{1 bar}\\
A= \pu{0.3774 bar}$
For initial volume of the container,
$
P_{\ce{Ar}_i}V_i=n_{\ce{Ar}_i}RT\\
V_i=C=\frac{n_{\ce{Ar}_i}RT}{P_{\ce{Ar}_i}}\\
C=\pu{\frac{0.1 \times 0.08314 \times 360}{0.3774} L}\\
C=\pu{7.93 L}\\
$
For initial moles of water vapour,
$
P_{\ce{H2O_{(v)}}_i}V_i=n_{\ce{H2O_{(v)}_i}}RT\\
n_{\ce{H2O_{(v)}_i}}=\frac{P_{\ce{H2O_{(v)}_i}}V_i}{RT}\\
n_{\ce{H2O_{(v)}_i}}=\pu{\frac{0.6226 \times 7.93}{360 \times 0.08314} L}\\
n_{\ce{H2O_{(v)}_i}}=\pu{0.165 mol}\\
$
Now,
$
n_{\ce{H2O_{(v)}_i}} + n_{\ce{H2O_{(l)}_i}} =\pu{1 mol}\\
n_{\ce{H2O_{(l)}_i}} = D = \pu{ (1 - n_{\ce{H2O_{(v)}_i}}) mol}\\
D = \pu{ (1 - 0.165) mol}\\
D = \pu{ 9.835 mol}\\
$
Now as the volume is doubled by Boyle's law we can say that the partial pressure of argon and water vapour will decrease to half of original.
Thus,
$
E=\frac{A}{2}=\pu{0.1887 bar}\\
F=\frac{B}{2}=\pu{0.3113 bar}\\
$
Now as the partial pressure of water vapour is less than the vapour pressure of water at that temperature, water will evaporate to make it equal.
Now as the temperature is same thus the vapour pressure of water is also same as initial. Thus only change in system is it's volume which was doubled. So we can conclude that twice amount of water vapour as compared to initial conditions will be required to attain this pressure.
Thus,
$
n_{\ce{H2O_{(v)}_f}}=2 \times n_{\ce{H2O_{(v)}_f}}\\
n_{\ce{H2O_{(v)}_f}}=\pu{0.33 mol}\\
$
But as the total moles of water is conserved,
$
G=H=\pu{(1-n_{\ce{H2O_{(v)}_f}}) mol}\\
G=H=\pu{0.67 mol}\\
$


Phosphorite is a mineral containing calcium phosphate. This particular 1kg sample has 75% percent of it by weight i.e. 750 grams. All others reagents are assumed to be in excess as you have to find the maximum p4 obtained which would be at a point where all of our mineral is used.


Why is Thorium (At. no. 90) an f-block element at all?

The bulk metal actually does involve the $f$ orbitals, distinguishing it from what would be its congeners in Group 4 of the $d$ block.  From Wikipedia:

Despite the anomalous electron configuration for gaseous thorium atoms, metallic thorium shows significant 5f involvement. A hypothetical metallic state of thorium that had the [Rn]6d27s2 configuration with the 5f orbitals above the Fermi level should be hexagonal close packed like the group 4 elements titanium, zirconium, and hafnium, and not face-centred cubic as it actually is. The actual crystal structure can only be explained when the 5f states are invoked, proving that thorium, and not protactinium, acts as the first actinide metallurgically.[1]

Cited reference:
1.
Johansson, B.; Abuja, R.; Eriksson, O.; et al. (1995). "Anomalous fcc crystal structure of thorium metal". Physical Review Letters. 75 (2): 280–283.


Calculating the concentration of acetic acid in vinegar from pH

First off $$\ce{HC_2H_3O_2}\leftrightarrow \ce{H^+} + {\ce{C_2H_3O_2}}^-\\and\\ K_a = \frac{[\ce{H^+}][{\ce{C_2H_3O_2}}^-]}{[\ce{HC_2H_3O_2}]}$$ So let's do some simplifying. The amount of acetic acid left is the orignial molarity minus the amount that's dissociated. The amount of Hydrogen ion and conjugate base left is equal to the amount of that's been dissociated from the original amount of acid. So let's let HA equal the molarity of the acid and x equal to the amount dissociated. Soooooo $$K_a = \frac{[x][x]}{[HA-x]} = \frac{x^2}{HA-x}\\\text{Now let's solve}\\\rightarrow K_a(HA-x) = x^2\\\rightarrow HA -x= \frac{x^2}{K_a}\\\text{ switching the x interms of ph}\\ HA -10^{-ph} = \frac{(10^{-ph})^2}{K_a}\\\text{ plugging in}\\HA- 10^{-3} = \frac{(10^{-3})^2}{.0000174} = .0575$$ This is the molarity of the acid at equilibrium. Add 10^{-ph} for the concentration of the acid before dissociation.


Does the property of additivity for the amounts of different substances hold up for ideal gas law?

The question would be better phrased as which gas contributes most to the total pressure? The contribution is referred to as the partial pressure. The sum of the partial pressures is the total pressure.

The formula you would use to compute the partial pressure $p_i$ of gas component $i$ is either of the following:

$$p_i=\chi_i p = \frac{n_i}{n}p = \frac{n_i}{V}RT$$

where $n_i$ is the amount of component $i$ in the gas and $\chi_i$ its mole fraction. So the argument "the gas that takes up the least amount of volume with the greatest amount of particles will exert the greatest pressure" is not quite right. It suffices to say that the gas that contributes the greatest amount of particles will contribute (exert) the greatest pressure. If you mix identical masses of each gas, that will be the gas with lowest molecular mass.


What effect is on the rate and rate constant when concentration or temperature increases/decreases?

Assuming you are speaking about an elementary reaction, and by that I mean a reaction on the form $\ce{A -> B}$ the rate of the reaction is

$$r = \frac{\mathrm d[\ce{B}]}{\mathrm dt} = -\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]$$

Most of the time $k$ can be express following the Arrhenius law:

$$k = A\exp\left(\dfrac{-E_\mathrm{a}}{k_\mathrm{B}T}\right)$$

Where $A$ is the pre-exponential factor (the constant rate when them temperature is really high), and $E_\mathrm{a}$ is the activation energy of the reaction.

In this case the initial concentration has no affect on the rate constant, only the temperature. Looking at the first equation the initial concentration has effect on the rate, this effect will depend on the order of the reaction…

$$r = k[\ce{A}]^2$$

$$r = k[\ce{A}]^3$$

Here a second and third-order reaction. If you study a complex mechanism, it can happen that initial concentration will affect the global rate constant, it can also happen with some approximation:

$$\ce{A + B -> C} \qquad r = k[\ce{A}][\ce{B}]$$

If $\ce{A},$ for example, is in really large quantity compared to $\ce{B},$ you can admit that for every time the concentration will approximately stay the same $[\ce{A}] = [\ce{A}]_0$ and then put it in the constant

$$k' = k[\ce{A}]_0$$

I hope that helped! If not, you need to study more.


What properties of later elements are predicted by scientists?

Though elements past 92 are quite unstable, so that chemical properties are difficult to determine, heavier elements behave more like metals, since the outer electrons are well shielded from the attraction of the positive nucleus and will therefore be easily lost.

Another predictable property is specific heat, which decreases with increased atomic mass.


When are we justified in making assumptions in questions

You can make certain assumptions in these and similar problems because concentration of uniform substances (whether it's expressed as molarity, molality, mass density, or something else entirely) is an intensive property, i.e., it has no dependence on the total amount of material in the object or sample being measured. For example, if you have a homogeneous solution, the solute particles are distributed uniformly among the particles of solvent, and the two form a constant ratio. If I split the sample in half, the solute particles are halved, but so are the solvent particles, so the ratio remains the same.

As it happens, there's no need to assume any particular quantities to solve this problem; instead, you can use algebraic substitutions using the known values given and common definitions to solve for either molarity or molality directly. Let $M = \frac{m_{solute}}{n_{solute}}$ be the molar mass of the solute, $\rho = \frac{m_{solution}}{V_{solution}}$ be the mass density of the solution, $c = \frac{n_{solute}}{V_{solution}}$ be the molarity, and $b = \frac{n_{solute}}{m_{solvent}}$ be the molality, and note that $m_{solvent} = m_{solution} - m_{solute}$. Then, we have:

$$
\begin{align}
b\ & = \frac{n_{solute}}{m_{solution} - m_{solute}}\\
 & = \frac{n_{solute}}{\rho \cdot V_{solution} - M \cdot n_{solute}}\\
 & = \frac{n_{solute}}{\rho \cdot \frac{n_{solute}}{c} - M \cdot n_{solute}}\\
 & = \frac{c}{\rho - M \cdot c}\\
\end{align}
$$

You can rearrange this equation to solve for $c$ (molarity). Note, of course, that units need to match, but it's trivial to convert between grams and kilograms, mililiters and liters, etc.


What is the simplest explanation of the ASTM G71 procedures that can be done in a laboratory?

ASTM maintains a variety of publications. To the practitioner, usually the most useful are the test Methods. I have also found the guides (such as G71) and glossaries to be useful background. G71 is not a test method, and it lacks any "procedures", the only answer to your question is: "There aren't any procedures in ASTM G71." This may seem confusing since the Guide contains a diagram which seems to imply that a laboratory test involves two equally sized sample rectangles partially submersed in an electrolyte and connected via an external wire.

Since you are unfamiliar with ASTM publications, allow me to use an analogy which may be more intuitive. Take for example a "Guide for baking bread" which recommends that only a circulating air oven, with an air flow of at least 4 cu. meters per minute and a volume of less than 0.1 cu. meter capable of controlling internal temperature in the range of 50° to 250°C to within ± 2° and that the baker should wash his/her hands prior to starting while keeping the tools, ingredients and preparation surfaces at 20°C. You'll note that my guide (I made it up, so don't take it too literally) completely avoids telling you:

what the ingredients are,
How to mix and process those ingredients, and
how to bake bread.

That is, it is not a "method"; it is a series of best practices (or preferred, consistent practices). When is something like G71 appropriate? It is appropriate when the test needs to be customized to fit the problem. The guide recommends the metal be "as large as possible" (meaning up to full commercial size) while being practical in either lab or field test conditions.

There is also the need for the exposed surfaces to be in the same area ratio, and to avoid edge effects, especially those which will not or do not appear on the "real" assembly. The guide mentions 4 techniques to evaluate corrosion: visual (and photographic), mass loss, current, and potential (voltage). Which of those to use (if any) is up to the individual. It is vague because it can't be specific. It can't be specific because it is intended to apply to a great variety of different galvanic corrosion scenarios.


Precipitate Sodium Hypochlorite from tap water before distillation

Sodium hypochlorite is not carried over into water by distillation.  Instead, at high (~100°C) it decomposes, primarily to sodium chloride (i.e. table salt) and oxygen gas, but also possibly to trace amounts of chlorate.

You could neutralize the hypochlorite by mixture with stoichiometric amounts of hydrogen peroxide, which cause immediate release of oxygen gas, a reaction which occurs even at room temperature.


Interestingly, the British had such a solution in WWI: the Hypo Helmet. This was simply a bag soaked in sodium thiosulfate solution and worn over the head.  Thiosulfate reduces chlorine (and halogens generally) to chloride (halide) ions:

$\ce{2 S2O3^{2−} + Cl2 → S4O6^{2−} + 2 Cl^{−}}$

As the article notes, this was not the best solution, but I think it fits your criteria for rough and ready protection.


Iodine sublimes for the same reasons that all solids do: because it has some equilibrium vapor pressure an normal conditions. Now, the value of that pressure varies greatly in different solids. For many of them, it is so extremely low that for all practical reasons it can be considered non-existent. (Say, if you'd leave a sample of NaCl for a thousand years, I don't think you will be able to detect the weight loss due to sublimation.) For iodine, the vapor pressure is significant, so a sample left in the open will vanish completely in a matter of hours.

Some sources may specify that iodine goes directly from solid to gas when heated, never becoming a liquid. That's an urban legend; iodine does have the melting point and the boiling point at standard pressure, and between the two it will be liquid. It is not easy to demonstrate, though; if you heat it in the open, it will probably sublime away even before reaching the melting point, and if you do it in a closed container, the vapors get so thick you can't see a thing.

That being said, there are indeed some compounds that would never turn liquid at ambient pressure. That's because their triple point is above that ambient pressure. Carbon dioxide is one example; water on Mars is another.

As for chlorine and bromine, I don't really see the point of your question. At normal conditions they are not solid, so they can't undergo sublimation simply by the definition of it. If you cool them down enough to be solid, they will.


How do I evaluate the PV term when calculating a change in enthalpy?

The derivation for $\Delta H$ is
$$H = U + PV$$
$$\Delta H = \Delta U + \Delta (PV)$$

The $\Delta (PV)$ term expands to your $P_2 V_2 - P_1 V_1$ term. However, when the pressure is constant $P_2 = P_1 = P$, and $\Delta (PV)$ term becomes $P\Delta V$. The most common situation when the latter is true is when the system is open to the atmosphere and $P$ is the external pressure. In the former case, $P$ is more likely the internal pressure, which changes during the reaction.


Section 4.1.1 (p 237) of Szabo and Ostlund's Modern Quantum Chemistry also describes this issue.

Using intermediate normalisation the correlation energy can be written as

$$\begin{align}
\langle \Psi_0|\hat{H}-E_0| \Phi_0\rangle &= \langle \Psi_0|\hat{H}-E_0 \left(| \Psi_0\rangle + \sum_{ar}c_a^r |\Psi_a^r \rangle + \sum_{abrs}c_{ab}^{rs}|\Psi_{ab}^{rs} \rangle +.... \right) \\
&= \sum_{abrs}c_{ab}^{rs}\langle \Psi_0|\hat{H}| \Psi_{ab}^{rs}\rangle
\end{align}$$

$\langle \Psi_0|\hat{H}| \Psi_{a}^{r}\rangle = 0$  according to Brillouin's theorem.
$$\langle \Psi_0|\hat{H}| \Psi_{abc}^{rst}\rangle = \langle \Psi_0|\hat{H}| \Psi_{ab}^{rs}\rangle \langle \phi_c | \phi_t \rangle = 0$$

and similarly for higher excitations, because $\hat{H}$ is a 2-electron operator.


The Evans auxiliary: rotational freedom of the benzyl group - stereoselectivity still guaranteed?

The comment left by CHM is very helpful.  You need to consider what this molecule actually looks like in 3D.  That may help your understanding greatly, but to be honest there is considerably more going on here than you would expect.

A couple pieces of information would be very helpful for clarifying this question.  The first is what type of a metal are you using for this synthesis step.  The second is what is the outcome.  Believe it or not, based on those two pieces of information the mechanisms for this reaction can be explained entirely differently.

Since the answer basically breaks down into two possibilities I will attempt to explain both.  One of the most helpful articles I came across in my search was entitled "Transition State Models for Probing Stereoinduction in Evans Chiral Auxiliary-Based Asymmetric Aldol Reactions" J. Am. Chem. Soc., 2010, 132 (35), pp 12319–12330.

In this article they go into depth about how the electronic and steric effects balance out to produce the almost complete stereocontrol throughout the course of this reaction.  The scheme below is what I find the most helpful:

Note how the selectivity is varied based on whether the auxiliary is chelating to the titanium or not.  This is why the metal lewis acid used in these reactions is important.

One final note not shown here is the rational behind the non chelating setup.  The opposing dipoles on "y" and aldehyde oxygen in the Zimmerman-Traxler model help to hold the six member ring in place while the aldol reaction occurs.

The scheme shown is incredibly useful for visualizing this transformation and I would like to compliment the authors for their efforts in this publication.

Hope this helps!


There's a couple of newer methods, using

zinc + N-Methyl-N-(trimethylsilyl)trifluoroacetamide (MSTFA)
copper + N-Methyl-N-(trimethylsilyl)trifluoroacetamide (MSTFA)
diethyl chlorophosphate $\ce{(EtO)2POCl}$
$\ce{PdCl2}$ in $\ce{CH3CN/H2O}$
$\ce{(COCl)2}$ with DMSO at -78 °C (Swern-type conditions)

Thionyl chloride, $\ce{SOCl2}$ is almost a classic here.

Under these conditions, the side products of the amide dehydration are $\ce{SO2}$ and $\ce{HCl}$.

$\ce{R-CONH2 + SOCl2 -> R-C#N + SO2 ^ + 2 HCl ^}$

On the first view, $\ce{PCl5}$ looks like leave quite a mess. Thinking in the workup, that's probably the last thing I would try.


The structure of sodium cuprous thiosulphate

For knowing these three dimensional complex structures, which are hard to predict theoretically, we have to use X-Ray diffraction or spectroscopic techniques (IR, NMR, or, mass). I don't have any such facilities to verify the complex structure, but I may logically predict a  logical (possibly incorrect) structure.

In the complex ion $\ce{[Cu_6(S_2O_3)_5]^4-}$, all the $\ce{Cu}$  is in +1 oxidation state, and thus can form a tetrahedral geometry around it with the help of other $\ce{Cu^+}$ ions and $\ce{S_2O_3^2-}$ ions. On the other hand,  thiosulfate can act as both a bidentate and a monodentate ligand.

So, in the predicted structure, all the $\ce{Cu^+}$ are in tetrahedral geometry. Three $\ce{S_2O_3^2-}$ act as bidentate ligands, whereas the other two $\ce{S_2O_3^2-}$ ions act as monodentate ligands.

Note that the donor atom of the thiosulfate anion can be either all $\ce{S}$, all $\ce{O}$, or a combination of $\ce{S}$ and $\ce{O}$. It is indeed hard to predict (because in some complexes, $\ce{Cu}$ binds with $\ce{O}$ regardless of being soft acid to preferably bind with $\ce{S}$ - as in $\ce{CuCl_2.4dmso}$) and thus is best to determine through IR spectroscopy by looking at $\ce{S-S}$ stretching frequency. Thus, the donor atoms may be different from what is shown in the possible structure.

The possible structure is:


Is it still possible for an amateur, a hobbyist, or a science buff to make relevant discoveries in chemistry today?

It's conceivable that somewhere out there, someone with a completely fresh point of view could have an annus mirabilis in an area like theoretical chemistry, working with just a paper and pencil (or a laptop) and home-brewed equipment and resources.  Yes, for most of us, all of the low-hanging fruit is gone---but for a real giant, the whole tree might be full of low-hanging fruit.

Most of us aren't giants, but as Newton said, we can stand on the shoulders those who are. Doing that requires some ability to search the extant literature, even for "small to moderate contributions".

In addition to the obstacles you've listed above, there is a third impediment: paywalls.

There was a time when you could walk into a university library and sit down with the Science Citation Index and do a good job of finding research related to yours. Nowadays, you need a faculty or student university ID to do such a search and retrieve the papers you need. Without an ID at a good university, our amateur, hobbyist, or science buff will  hit paywall after paywall. Amateurs might have to pay half a day's wages or more for the pdf of a single paper!

Paywalls are not only a major obstacle for amateurs---they discourage professionals in the third world, and they shut out young people who are beginning their careers outside of university.

Consider Jack Andraka's story, here: http://blogs.plos.org/thestudentblog/2013/02/18/why-science-journal-paywalls-have-to-go/


I think an excellent book for understanding molecular orbital theory from a qualitative basis is Orbital Interactions in Chemistry by Albright, Burdett, and Whangbo. It covers a wide range of topics from molecules to solids.

The book doesn't go into particular methods, but instead illustrates how various MO treatments can understand a great deal of chemistry.


While considering colligative properties we always assume solute to be non-volatile. So, it doesn't contribute to the vapor. The amount of vapor will only be due to solvent particles.
So, in the problem we can use the following relation:
$$ \Delta T = K_\mathrm b\cdot m $$
Initially $ \Delta T  = 0.53\ \mathrm{^\circ C}$ so plugin the value of $K_\mathrm b$ and you end up with the value of initial molality of solvent as $1$. From this you can find the initial moles of solute which remain constant throughout.

Next you put $ \Delta T  = 2.12\ \mathrm{^\circ C}$ and plug in the known values of $K_\mathrm b$ and 'moles of solute' in the above formula to get the final value of solvent amount that remains in the solution.
Subtracting the final amount of solvent left from the initial amount will let you know the amount of water vaporized which is $750\ \mathrm g$.


How can I interpret this solubility data?

Yes, the website is a little bit ambiguous. However, there is a clue in the colour legend. The solubilities are indicated in units of grams per 100 grams of water. (This is also a very common unit for expressing solubilities.) Therefore, a solubility of "$3.7$" means that $3.7~\mathrm{g}$ of $\ce{KNO3}$ dissolves in $100~\mathrm{g}$ of whatever solvent it is listed under. This interpretation is readily confirmed by checking the values against this Wikipedia page.

As for "$50\%$", my educated guess is that it refers to a mixture of $50\%$ ethanol and $50\%$ water.

And lastly "$20^\circ$" refers to $20~^\circ\mathrm{C}$.


How to compute the residence time for given flow rate in standard cubic meters?

The N stands for "norm" and should be avoided (in fact, the German Wiki page lists the Nm³ as wrong abbreviation) as it can create confusion.(qed)

So your flow should be written as "3 m³/h under standard condtions".
If you can guess German, have a look at the German Wiki page. It is much better than the English one.

Standard conditions vary a bit between different norms

T = 0 °C or 15 °C or 20 °C
p = 1.01325 bar (sometimes 1 bar)

The gas is supposed to be dry.


Acidity of thiamine - how comes hydroxide anion with pK about 15 ionizes thiamine with pK about 18?

In the ionized structure shown, which looks like an ylid, the carbanion is formed on an $\mathrm{sp^2}$ orbital and no resonance stabilization of this charge is possible.
Regarding your question about the ionization, remember that biochemical reactions don't always happen in water, they may happen in the active site of enzymes, which have a completely different environment: the dielectric constant is different, the $\mathrm{p}K_\mathrm{a}$ of substances change and the proximity of certain groups stabilize reactive intermediates that would not be formed in water.


Why does precipitate always form so naturally?

The ions combine while in solution, and since the resulting compound is not soluble, it precipitates. This happens so quickly because as the precipitate forms, it takes those individual ions out of solution, thus pulling the reaction toward the products side, according to Le Chatelier's principle.


It's easy to find a slew of test papers and strips online. For example, a quick search of Amazon for "test papers" turned up chlorine test strips for pools and sanitation, glucose test strips for diabetes, phenylthiourea genetic-linked taste tests, UV-C dose test cards, lead acetate paper test for $\ce{H2S}$, WBC and nitrite test strips for UTI... You could write a paper on all the test papers available.
Increasingly, medicine is making use of paper tests because of ease of distribution, storage and use, particularly important in rural location and during emergencies.
BTW, you can make a test paper for polarity testing, and for detection of peroxides, by dipping filter paper into a suspension of starch and potassium iodide, $\ce{KI}$, rather than buy it. Dampen and use it as you would a potato to determine the polarity of an electric cell.


I have clarified with my chemistry tutor and it appears that performing the substitution and the addition as two separate steps with two separate sets of conditions is absolutely necessary. The most important reason would be the temperature, which is also the most stark difference between the two different set of conditions.

Due to entropy considerations, addition should ideally be carried out at low temperatures. Thus, if we would like to have the addition of $\ce {HCN}$ to the substrate, it should be done at much lower temperatures as the reaction has a negative entropy change. However, the substition of $\ce {Br}$ by $\ce {CN}$ does not need such low temperlatures and should be done at higher temperatures to provide enough energy of activitation to break the carbon-bromide bond.


Mathematical definition of a symmetry operator

That's not entirely correct. You're making at least an extra assumption here. You're assuming that $\hat{R}$ is a function from $D\mapsto D$ where $D$ is the domain of $f$, but there's no reason why $f$ must operate as a function on $D\mapsto D$. It might just as well be $f: D\mapsto \mathbb{R}$.
I think what you're really asking for is that a symmetry element $\hat{R}$ such that $\hat{R}: D\mapsto D$, and $\forall x\in D$, $f(x) = f(\hat{R}x) = (f\circ \hat{R})x$.
Then for example, for $f(x) = (\cos x, \cos x)$, the transformation $\hat{R}$ mapping $x \rightarrow -x$ would would a valid symmetry element.


Are we capable of deliberatly creating DNA from inorganic compounds?

Yes we could.
It wouldn't be worth the effort since all the building blocks are available from biological sources, but if you really wanted to you could have a chemist prepare nucleotides from inorganic substances.
Commercial labs routinely build long strands of DNA from individual nucleotides. You can tell them what sequence you need and they will make it for you. There's a Wikipedia article on how it works.


All resonance structures of the hydrogen molecule

This is a reference kick-starter contestant solution.

There are 6 resonance structures:

$\ce{H:H}$ (the usual one)
$\ce{:H- H+}$
$\ce{H+ H:^-}$
$\ce{*H*H}$
$\ce{H*H*}$
$\ce{*H H*}$


What are the g-block's predicted properties?

First of all, from the perspective of chemistry, it is the group (column) that matters more than the period (row), and subshells act at the period level.  Second, the elements beyond the actinides have virtually no significant chemical research owing to their extreme radioactivity.  So there may not be many predicted properties for the g-block beyond the simplistic theoretical assumptions about groups, which are much less relevant for a hyper transitional period like the g-block.

As for you question about potential stability, Glenn Seaborg made some predictions here, which you can read that this Wikipedia page: Island of Stability.  In general, though, these elements are not predicted to be stable.


Obtain chemical insight from neural network results

Since you are interested in Artificial Neural Networks that also explain their decisions/predictions you may be interested in transparent neural networks, although they have not been used to my knowledge on chemical information.

Other interesting approaches include inductive logic programming which have among other things been applied to predicting binding of proteins to hexose


Does increasing the entropy of a system cause it to absorb heat?

I consider watching any video a waste of time, so I'll be judging from your words alone. (Anyway, your question is essentially self-consistent, which is good.)

Yes, entropy is a measure of disorder (sort of).
No, $dS={\delta Q\over T}$ only for reversible processes.
And no, entropy is not the Q-T ratio for two reasons. First, because the above formula deals with infinitesimal increments and not with finite quantities. Second, because like I just said, that formula is not universally applicable.

All in all, the increase of entropy alone is not necessarily accompanied by the heat flow to the system. It may be so, but it may be otherwise. You may imagine pretty much any kind of possible outcome:

You dissolve some $\ce{NH4NO3}$ in water (which surely increases entropy), and the beaker gets palpably cold. The system starts sucking heat from the surroundings, just like you anticipated.
You mix your $\rm^3He$ with $\rm^4He$, and... nothing happens (except the increase in entropy, of course, but that's a thing you can't see or touch). There are no thermal effects to speak about, and hence no heat flow.
You burn a candle in a closed vessel (supposedly, containing enough air), and guess what? Entropy increased, but heat is flowing outside.

As for your last remark: true, it takes some energy to decrease entropy locally (say, in your fridge). But in doing so, we inevitably increase entropy elsewhere, so the overall balance is positive.

So it goes.


Why are dendrites formed during dilute CuSO4 electrolysis black?

To quote a source as you also noted:

  Copper hydroxide Cu(OH)2 is metastable. It easily transforms into copper oxide CuO more stable, either in the solid state by a thermal dehydration or at room temperature, in aqueous basic solutions. In the solid state, the transformation is performed at a relatively low temperature, 423 K.

In the current case, a basic solution develops and copper hydroxide breaks down into CuO and water.

However, there may also be heat from applying the overpotential, and this could speculatively be the source of CuO observed.

More precisely related to the cathode, start by noting per this source (Lumen Learning), for example, to quote:

  Electrolysis of a solution of sulfuric acid or of a salt, such as NaNO3, results in the decomposition of water at both electrodes. Hydrogen will appear at the cathode and oxygen will appear at the anode.

Assuming on the cathode the formation of an intermediate species (like $\ce{H*}$ mentioned on StackExchange in an advanced discussion of the underlying mechanic on the electrolysis of water) upon self-reaction, to quote:

  In an alkaline electrolyte they propose the following mechanism, which only slightly deviates from the above. The intermediate species are the same, that's why they did not need to treat them separately. On the cathode they just summarised as $$\ce{2H2O + 2e- -> 2{}^{-}OH + H2}$$ as the mechanism is bound to be the same.

Next, the action at the cathode of created OH- on Cu++ forms Cu(OH)2 which, in the basic conditions, results in CuO.


Are lithium bicarbonate and magnesium bicarbonate both soluble in water?

According to this Wikipedia solubility table, lithium bicarbonate is soluble in water to $\pu{5.7 g/100mL}$ at $\pu{20^oC}$, while this Wikipedia page gives the solubility of magnesium bicarbonate as $\pu{0.077 g/100mL}$.

However, another source (found by Oscar Lanzi in his good answer), "The Solubility of Magnesium Carbonate in Aqueous Solutions of Certain electrolytes, The Journal of Physical Chemistry, Volume 7, pub. 1903, reports making two measurements of the solubility of magnesium bicarbonate and finding $\pu{3.1 g/100mL}$ and $\pu{2.0 g/100 mL}$.  Note that this source is from 1903, but also that the the Wikipedia link above does not give a source for their value at all. At least one reason for the dearth of information regarding magnesium bicarbonate is that it easily decomposes and it cannot readily be isolated outside of aqueous solutions. My feeling is that the Wikipedia value seems low, and that the values reported in the Journal of Physical Chemistry seem more reasonable.


What is the correct name for 2,4-diethyl-4-ethoxyhexane?

In short, your answer is correct. The steps that you have taken are correct, so it's likely the online homework tool has a key mistake.


When a salt is fully dissociated, than the atoms in a solution exist separately:
$$\ce{CuCl2<=>Cu^{2+}}+2\ce{Cl-}$$
Amount of substance is 0.074 mol. The molecule is made of 1 atom of Cu and 2 atoms of Cl.

$$ n=\frac{m}{M}$$
$$ n \cdot M =m$$
Amount of copper in the mixture then is: $$0.074~\mathrm{mol} \cdot 63.5~\mathrm{g/mol} = 4.7~\mathrm{g} $$ and the rest is formed by chlorine. $$ 10~\mathrm{g}- 4.7~\mathrm{g} = 5.3~\mathrm{g}$$


The pyridine/benzene stability 'paradox'?

When you think about stability, think of it as a relative value, meaning in comparison to atoms of other elements. Therefore, pyridine is more stable than the benzene but less stable than other elements, in general.

You are correct when you say stability is inversely proportional to reactivity. But because stability is relative, it is very difficult to comment on the relative reactivity.

EDIT

Heat of hydrogenation of an alkene in which the double bonds are conjugated is not additive because such an alkene is resonance stabilized, and, therefore, its Lewis structure, which the calculation would be based on, is a poor representation of the molecule.

I have found 2 aspects :

Pyridine is less reactive, than benzene toward electrophilic aromatic
substitution, because nitrogen is more electronegative, than carbon and acts like an
electron withdrawing group. As a result, the meta hydrogen is substituted.
Pyridine is more reactive than benzene because the presence of nitrogen
enables pyridine to react with nucleophiles. Nitrogen is an electron withdrawing
substituents enables the aromatic ring to participate in such reactions and the
substitution happens in the ortho and meta positions.


The key to this synthesis should be — in my opinion — how to generate the two hydroxy functions with anti stereoselectivity starting from at least one achiral reagent. Most of the time you would attempt building up chirality de novo using chiral auxiliaries, catalysts or reagents as these methods are often more cost-efficient than acquiring a specific chiral reactant — the obvious exception being if a compound exists in the natural chiral pool that you can easily use (which is not really the case here). Having identified our most important problem, we can turn to the other ones:

we need to generate a δ-lactone
we need to form a $\ce{C-C}$ bond, preferably in a chiral manner, to a $\ce{C6}$ body (the bit of the final product that is missing in the reactant).

Scheme 1: General retrosynthesis idea.

In my opinion, the δ-lactone is a non-issue. If you have an open-chain ester such as 3, you can probably form the six-membered lactone simply by $\ce{K2CO3}$-mediated transesterification. Practically all literature methods to synthesise esters should also predominantly give the δ-lactone especially since the unfavourable ζ-lactone is the alternative. (In the unlikely case that it is formed, it can easily be transesterificated as noted.) I would put this step last.

Scheme 2: Transesterification step.

We now have 3, an anti-configured 1,3-diol. When I see this, my mind immediately jumps to the Evans-Tishchenko asymmetric reduction. Thus, we can arrive there from either of the two possible β-hydroxyketones 4 or 5. For this to work best, I would want the carboxy group to be masked as a protected alcohol here, cf. 6.

Scheme 3: Evans-Tishchenko step. $\ce{X}$ would be $\ce{H}$ for the newly-formed hydroxy group and $\ce{CH3CH2CO-{}}$ for the inducting one.

Both of these β-hydroxyketones could be generated from aldol reactions; however, only ketone 4 suggests that it can easily be reached from the desired reactant 2. If we wanted to synthesise 5, we would need an extra carbon atom on 2 which we would need to introduce in an additional reaction. However, the route to 4 only requires selective oxidation to the aldehyde — my personal favourite is Dess-Martin–oxidation, but you should probably check all the others that suggest themselves (Swern and relatives, TEMPO, Ley-Griffith, etc.) as one could be superior over the others due to your lab spirits liking it.

Finally, this also tells us that the aldol addition to generate 4 from 7 and 8 should proceed in a stereoselective manner giving the (S)-configured product predominantly. Again, many methods are known and I would suggest a Paterson aldol reaction. Analysis of the relevant literature (unfortunately, I did not find a website, so a citation it is) reveals that $\ce{(-){-}Ipc2BOTf}$ and Hünig base should give the (S) product preferentially.[1]

Scheme 4: Final aldol disconnection.

This leads us to the following forward synthesis:

$$\ce{\mathbf{2} ->[DMP] \mathbf{7} ->[(-){-}Ipc2BOTf][ (iPr)2NEt, \mathbf{8}] \mathbf{4} ->[SmI2, EtCHO] \mathbf{6} ->[1) TBSOTf; 2) PPTS][3) DMP; 4) {Pinnick}] \mathbf{3} (R $=$ H) ->[DCC, DMAP][K2CO3] \mathbf{1}}$$

Reference:

[1]: A. S. Franklin, I. Paterson, Contemp. Org. Synth. 1994, 1, 317–338. DOI: 10.1039/CO9940100317.


Effect of pKa value of acid component in Mitsunobu Reaction?

I figured out the answe

The desired Mitsunobu reaction proceeds through path (a). On the contrary, in case of the reaction with less acidic HA (pKa>15) , the hydrazo anion (4) attacks the alkoxyphosphonium (5) directly to afford alkylated the hydrazine derivative (6) as a by-product path(b).

Since the anion (4) is not efficient in deprotonating the weakly acidic HA. In order to overcome these drawbacks, the acidic component should have pKa<15 to get desire product.


Why are electrons added in the ionic counting method?

In the ionic method of counting, you are not counting neutral compounds. This is reasonable on a molecular basis: the ligand is, in fact, not cyclopentadiene or the cyclopentadienyl radical (with four or five electrons to be donated) but the cyclopentadienyl anion $\ce{C5H5-}$. This species contains six electrons in its π system, meaning that six are donated.


Do enzyme-substrate reactions show reversibility(to a considerable amount)?

There are many enzyme catalyzed reactions that are reversible. In such cases, buildup of product may slow and eventually stop the net reaction as you reach equilibrium conditions. Product binding to the active site or to a secondary allosteric binding site may also inhibit the catalytic reaction. This can occur even in the case of irreversible enzymatic reactions.