Rust forming and being washed off by water

Rust appears because iron is oxidized in presence of oxygen (it is a oxydo-reduction reaction, meaning an electron transfer occurs between species), and water helps this phenomenon occur. Iron loses 3 electrons to the favour of the oxygen.

In presence of water, oxygen can oxidize iron by the mean of hydroxyl group. It leads to:

$$\ce{2Fe(OH)_3 \leftrightharpoons Fe_2O_3\cdot 3H_2O}$$

where water is not the oxidizer, rather it is a mediator for oxidation.

Hydroxides have low solubility, so water can only dissolve a small fraction of it.

On the other hand, poured water can abrade rust (via mechanical stress) and remove it of from the bulk metal, leaving the surface clean for producing more rust. This is how huge rods of metal can vanish in a few months in corrosive environmental conditions (such as wet air or water).

Preventing rust requires the use of specific alloys and surface finishing. You can also add a galvanic electrode (generally made of $\ce{Zn}$, a so-called sacrificial anode) to protect submerged metallic structures such as the hull of a boat.

Therefore, in your example water is both a mediator of oxidation and mechanical abrasion agent.


Why is the more electronegative atom lower in energy in an MO diagram?

I am summarizing from the excellent comments while adding some glue:

The convention is that lower electron energy means electrons that are more tightly bound. An energy of zero would mean a free electron (not bound in the electrostatic field of a nucleus).

As you go across from left to right in a period of the periodic table, the number of protons in the nucleus increases and the number of electrons bound (in an atom) increases likewise. In the simplest explanation of electron energies, you conceptually separate the electrons into inner electrons and valence (or outer shell) electrons, and treat the inner electrons and the nucleus together as an entity with some charge (the effective nuclear charge) that the outer electrons "see".

Comparing oxygen to nitrogen, the effective nuclear charge increases by one (same number of inner electrons, one more proton in the nucleus), so valence electrons of oxygen are more tightly bound than valence electrons of nitrogen. Thus, they appear lower in an MO diagram.

Not the OP's question, but for completeness: If you go from one period to the next (compare F with Na, say), the effective nuclear charge decreases (because we now also count the n = 2 electrons as inner electrons), and valence electrons of Na are less tightly bound than those of F. Most atomic properties change in a periodic pattern when you plot them against atomic number for the same reason (e.g. ionization energies, atomic radii)


Many organic molecules, including food dyes, are colored because they contain chromophores. A chromophore is a region in the molecule where the energy difference between two different molecular orbitals falls within the range of the visible spectrum. Light that hits the molecule can thus excite the electrons in the chromophore, resulting in the emittance of a particular wavelength of light.

$\hspace{1cm}$

In urobilin (seen above) specifically, the tetrapyrrole moiety, as a result of its conjugated pi-bond system, is the chromophore that gives this compound its characteristic yellow color.


Bond strength (bond dissociation energy, BDE) refers to homolytic cleavage of a bond ($\ce{C-H}$ in this case). Heterolytic cleavage is measured by $\mathrm{p}K_\mathrm{a}$. The two processes are the numerical inverse of one another. In the case of acetylene with $\mathrm{sp}$-hybridization (50% $\mathrm{s}$-character) of carbon, the electron density of the $\ce{C-H}$ bond lies closer to carbon when compared to ethylene (33% $\mathrm{s}$-character) and ethane (25% $\mathrm{s}$-character). Thus in acetylene, it is more difficult to break the $\ce{C-H}$ bond homolytically because the carbon is "unwilling" to give up an electron but is happy to keep both electrons and give up a proton. Thus, acetylene has a high BDE ($\pu{133 kcal/mol}$) and lower $\mathrm{p}K_\mathrm{a}$ ($=25$) than ethylene ($\pu{110 kcal/mol}$, $\mathrm{p}K_\mathrm{a} = 44$) and, ultimately, ethane ($\pu{101 kcal/mol}$; $\mathrm{p}K_\mathrm{a} \approx 50$).


Stretching frequency of nitro and nitrito

First of all, in the webpage, I think there is some shuffling between Nitro and Nitrito first in the headline. But, according to the contents, Nitro has closer stretching frequency, but Nitrito has distinguished stretching frequency.Let's take that one as the correct data.

The stretching frequency of nitro ($\ce{<-NO_2}$ )corresponds to the stretching of the bonds between $\ce{N}$ and $\ce{O}$, which are in between double and single bond, due to resonance, and the bond order between these $\ce{N}$ and $\ce{O}$ becomes almost equivalent in both the two  bonds.

Thus the stretching frequency of $\ce{-NO_2 }$ corresponds to partial double bond character between N and O and almost equivalence of two bonds.  On, the other hand, the two streching frequency of nitrito ($\ce{<-O^ -N=O)}$ corresponds to one $\ce{N -O}$ pure single bond and  one $\ce{N=O}$ pure double bond as there is no delocalisation in this case. So, due to that pure double bond stretching frequency of $\ce{N=O}$, one streching frequency of nitrito is higher than Nitro, which is consisting of only partial double bonds.


Which half cell equation do you switch when calculating E° cell?

Now, lets first clear this up. $E_\mathrm{cell}^\circ$ can be negative, it just means that the reaction won't spontaneously occur. This means that an external current is required to be applied to push the electrons in that direction for the reaction to occur. When $E_\mathrm{cell}^\circ > 0$, that just means that reaction occurs spontaneously. So in voltaic or galvanic cells, $E_\mathrm{cell}^\circ > 0$.

Now to calculate $E_\mathrm{cell}^\circ$, the formula is:
$$E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ \mathrm{(cathode)} + E_\mathrm{ox}^\circ \mathrm{(anode)}$$
As you correctly stated, if you switch the direction of the reaction (so it goes from an oxidising to reducing half equation), the sign of $E^\circ$ changes. Hence the equation for $E_\mathrm{cell}^\circ$ can also be written as:
$$E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ \mathrm{(cathode)} - E_\mathrm{ox}^\circ \mathrm{(anode)}$$
This is the formula that you have used and hence this is why you get the correct answer when you do this.


What is responsible for making the bonds identical?

The hydrogens are all equivalent. When thinking about MOT, you have to throw almost everything you thought you knew about bonding out the window. The electrons are not localized between any two pair of atoms—they are truly molecular orbitals. This graphic nicely shows the phasing of the molecular orbitals of methane:

$\hspace{4cm}$

[source]

The lowest energy orbital is a linear addition of the carbon $\mathrm{2s}$ orbital and all four of the hydrogen $\mathrm{1s}$ orbitals. The pair of electrons occupying this orbital is not specific to any one hydrogen, and bonds all the atoms in methane together (even the hydrogens to the other hydrogens). The other bonding orbitals are of degenerate energy and result from linear combinations of the carbon $\mathrm{2p}$ orbitals and the hydrogen $\mathrm{1s}$ orbitals.

Naturally a molecular orbital that results from a combination of a $\mathrm{s}$ orbital will be of lower energy than one than one that results from a $\mathrm{p}$ orbital.


Can the nitrogen of an amide displace a primary chloride in a SN2 reaction in the presence of a strong base?

Well it is difficult to compare an intra-molecular reaction with an extra-molecular one because it really depends on the conditions. However, unless you have a very high concentration of one reactant, the intramolecular way will always be favored.

In your question, you consider that "N" is not a very good "attacking group". Well this is wrong. "N" is always happy to attack any nucleophilic group, even if it becomes "N+". However, a "CON-" is much more nucleophilic than just "CONH".

Sodium methoxide here just serves at deprotonating CONH to activate it, but MeONa (sodium methoxide) is not nucleophilic enough to interfere.


Examples of non-volatile, but flammable liquids?

The element caesium melts at 28.4°C which is slightly above "room temperature" but below human body temperature, so I think we can consider it a liquid in "normal" conditions. It is so flammable it readily self-ignites in the air, but if you put it in inert atmosphere you will see that it isn't particularly volatile.


Why is work and heat energy always viewed from the perspective of the system?

The problem is in the word "always". The statement "Work and heat energy are always viewed from the perspective of the system?" is wrong. Please inform your teacher that is true in the chemist's world only- and chemists did not invent thermodynamics.
Chemists have adopted a different convention than physicists. All we have to do is to be aware of the context.
Recall the First Law of Thermodynamics, which has 2 different equivalent statements.
In some books it is
$$∆U=q-w$$
and in some it is
$$∆U=q+w$$
Why, because the word "always" is not correct.


Writing a electrochemical cell representation in correct way

The chemical reaction is:

$$\ce{2AgCl(s) + H2(g) -> 2 HCl(aq) + 2 Ag(s)}$$

If we split this into half reactions, we get an oxidation half reaction at the anode of:

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

And a reduction half reaction at the cathode of:

$$\ce{2AgCl(s) + 2e- -> 2 Ag(s) + 2 Cl-(aq)}$$

If you want all the species at standard state, you need a partial pressure of $\pu{1 bar}$ of hydrogen gas, a $\mathrm{pH}$ of about 0, and a chloride concentration of $\pu{1 mol/L}$.

In cells where the reagents can't react directly, you don't need a salt bridge. In the reaction we are looking at, $\ce{AgCl}$ is confined to the cathode because it sticks to it, and hydrogen gas is confined to the anode because that's where we are releasing it, and it has low solubility in water.

The cell notation starts with the conductor of the anode half reaction, and ends with the conductor of the cathode half reaction. Different phases are separated by vertical lines. If there is more than one species in a phase, they are separated by commas.

$$\ce{Pt(s) | H2(g, \pu{1 bar}) |  H+(aq, \pu{1 mol/L}) , Cl- (aq, \pu{1 mol/L}) | AgCl(s) | Ag(s)}$$

  The books has plainly told 3. to be correct.

The answer is (5) none of the above. Using the format of the answer key, the best answer would be:

$$\ce{Pt(s) | H2(g), \pu{1 bar} |  1 M HCl(aq) | AgCl(s) | Ag(s)}$$

You get this answer from answer 3 by changing the salt-bridge boundary ("||") to a phase boundary ("|"). Having a salt bridge between the hydrogen gas and the aqueous hydrogen ions would mean that you are generating hydrogen ions at the electrode where they are not in contact with $\pu{1 M}$ hydrogen ions.


XRD: Can stacking faults cause pseudo-centring?

The outcome depends on the typical length between stacking faults and the coherence length of the X-ray beam.

If the length between stacking faults is large compared to the coherence length, you will get a phenomenon similar to that of twinning (unit cells in same orientation) or of broadened diffraction spots similar to the model of high mosaicity (unit cells in slightly different orientations).

If the length between stacking faults is small compared to the coherence length, you will get a phenomenon similar to partial occupancy of sites in an otherwise perfect crystal (i.e. static disorder - same diffraction pattern with different intensities compared to absence of partially occupied sites).

I found a statement about coherence lengths of synchroton radiation (https://www.researchgate.net/post/What_are_typical_requirements_on_coherence_lengths_for_coherence_experiments_using_3rd_gen_Synchrotron_Light_Sources):

  [Keoki Seu, March 21, 2012] I work at a coherent scattering beamline at the Advanced Light Source. It is an undulator source optimized for coherent flux. At 500 eV, which is usually where we work, coherence lengths are on the order of 6 um vertical and 3-4 um horizontal. We'd love to have larger coherence lengths so we can coherently illuminate larger sample areas, but for the soft x-ray, the values we see here are typical values.

The question was about power patterns rather than single crystal diffraction.

  [OP] Will the situation described cause the powder X-ray reflection pattern to resemble space group symmetry Fm-3 m, e.g. by integral extinctions of reflections where h + k, h + l or k + l is odd?

I don't know what the typical dimensions of the solids in the powder would be, and I don't know what the coherence length of the Xray radiation used in the experiments are, but if you consider all of these factors, you should be able to say in which regime (i.e. twinning-like or static disorder-like) the experiment at hand is in.


Your final result is almost correct, although your derivation is somewhat questionable, and, in particular, this equation is wrong: $$\Delta G_m = V_m\Delta P = RT\ln\left(\frac{P_f}{P_i}\right).$$ For one, as you've mentioned, $V_m$ isn't constant, so we can't use the finite difference implied by $\Delta$, and I don't understand how you would have gotten the second equality without having integrated.

Instead, we should have $$\Delta G_m = \int_{G_{m,i}}^{G_{m,f}}\mathrm{d}G_m = \int_{P_i}^{P_f}V_m\,\mathrm{d}P = RT\int_{P_i}^{P_f}\frac{\mathrm{d}P}{P} = RT\ln\left(\frac{P_f}{P_i}\right) = RT\ln\left(\frac{V_i}{V_f}\right).$$ The third equality follows from $V_m = RT/P$, as you note, and the final equality follows from substituting $P = nRT/V$, the ideal gas law, and cancelling out terms.


Sodium metal burning off when conducting electrolysis and I don't know why

First, sodium burns very nicely in air. However, it also burns in chlorine, returning back to the $\ce{NaCl}$ with which you started.

Most welders provide alternating current, meaning the the electrode becomes alternately positive and then negative (50 times per second [50 Hz] in much of the world, and 60 Hz in North America). Therefore, both $\ce{Cl2}$ gas and $\ce{Na}$ metal are produced at each electrode and recombining.

To produce sodium metal you need a direct current supply, such as from a battery, and some means to separate the chlorine bubbling out at the positive (anode) electrode from the sodium metal at the negative (cathode) electrode. A diagram of the Down's cell should help you understand this, and you can view a video of sodium production.

As a demonstration, I've made (for a very short time) small amounts of sodium in an open crucible using a laboratory 10 volt, 6 ampere supply... One sees a small metallic bead forming under the molten salt at one wire and immediately burning when it hits the air, with sodium's characteristic yellow glare.


Well an acid can act as a base when there is a stronger acid present, But if you look at the molecular level acidity depends on how well the negative charge is balanced. so out of the two molecules present the one which can disperse(stabilize) the negative charge to a greater extent will act as acid.


After a unitary transformation, is Koopmans' theorem still valid?

Disclaimer: I like Tyberius' answer, but I'd like to go a little further.
$%
\newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle}
\newcommand{\lb}{\left|}\newcommand{\rb}{\right|}
\newcommand{\op}[1]{\mathbf{#1}}$

The beauty of Koopmanns' theorem is its simplicity. Quoting from the gold book:

  Koopmans' theorem

  Directly relates experimental ionization potentials with energy levels of molecular orbitals. The theorem states that the ionization potential required to remove an electron from the orbital $\Psi_i$ is given by the negative value of the energy of the orbital, $−\varepsilon_i$, as calculated within the Hartree–Fock approximation. The theorem is not applied to localized molecular orbitals, which are not eigenfunctions of the effective hamiltonian.

While Koopmans' text (in German: Physica 1934, 1, 104-113.) is a bit hard to comprehend, due to the ancient language I never learned, it basically boils down to two important facts.

There is a unitary transformation of the Lagrange multipliers which diagonalises them. These form a characteristic set of elementary wave functions, matrix, which in turn have a characteristic eigenvalue each. If all eigenvalues are different, then all elementary wave functions are specified through the Hamiltonian. (If there are equal eigenvalues, a freely chosen unitary transformation can be performed.)

In other words: There is one set of canonical orbitals that diagonalises the Fock matrix; each canonical orbital has an eigenvalue. For non-degenerate systems those orbitals are specified through the Hamiltonian.
This choice of functions has physical meaning, as the eigenvalue (disregarding a small correction) can be equated to the ionisation energies of the corresponding electron.

The second statement implies that there is no relaxation of the orbitals when removing an electron. This is oftentimes referred to as "frozen MO" approximation. Koopmanns himself appreciates that there should be a contraction of orbitals when removing an electron, however, he doesn't go into detail.

Nowadays we explain Koopmans' theorem in some simpler terms, where we remove the $k$th electron:

\begin{align}
&& E_N &= \sum_{i=1}^N H_{ii}
        + \frac12\sum_{i=1}^N\sum_{j=1}^N (J_{ij} - K_{ij})
        + V_{\mathrm{nuc}} \tag1\\
&& E_{N-1}^k &=\sum_{i=1}^{N-1} H_{ii}
        + \frac12\sum_{i=1}^{N-1}\sum_{j=1}^{N-1} (J_{ij} - K_{ij})
        + V_{\mathrm{nuc}} \tag2\\\hline
&& E_N - E_{N-1}^k &=  H_{kk}
        + \frac12\sum_{i=1}^N (J_{ik}-K_{ik})
        + \frac12\sum_{i=1}^N (J_{kj}-K_{kj}) \tag{$1-2$}\\
\therefore&&
E_N - E_{N-1}^k &=  H_{kk} + \sum_{i=1}^N (J_{ki}-K_{ki})\\
\therefore&&
E_N - E_{N-1}^k &= \varepsilon_k\\[2ex]
\text{with}&&
H_{ii} &= \langle \phi_i(\mathbf{x}_1)|
          \mathbf{H}^\mathrm{c} | \phi_i(\mathbf{x}_1)\rangle\\
&&
J_{ij} &= \langle \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) | r_{12}^{-1}
          | \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) \rangle\\
&&
K_{ij} &= \langle \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) | r_{12}^{-1}
          | \phi_j(\mathbf{x}_1) \phi_i(\mathbf{x}_2) \rangle\\
\end{align}

As you can see, Koopmans' theorem is not limited to the HOMO, but can be used for any occupied MO.

There is a very important point to be always be kept in mind when working with any Hartree-Fock based approaches:
Even though the equation
$$\op{F}_i\phi_i = \varepsilon_i\phi_i
     \tag{3}\label{fock-pseudo}$$
suggests an eigenvalue problem, it is not. Remember the definition of the Fock operator and the operators contained
\begin{align}
  && \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\
\text{with}&&
     \op{J}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\
\text{and}&&
     \op{K}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr.
\end{align}
As you can see, the "one-electron" Fock operator depends on the solution of all "one-electron" Fock operators (ref. Szabó-Ostlund p. 115). The Hamilton operator is not the sum of all Fock operators, and the total HF energy is not the sum of all orbital energies. As a result of that, the canonical orbitals are actually unique solutions.
The Fock operator is associated with the whole wave function, and while unitary transformations will keep the wave function and its energy equivalent, everything else will fall apart.
Another important consideration is the fact that the Fock operator is only well defined for occupied MO. You sometimes find the statement
$$E_{N+1}^{l>N} - E_N = \varepsilon_l,$$
which can not be applied in the same way; it is very basis set dependent.

From all that above it is obvious, that Koopmans' theorem only works for systems where a single determinant approximation is reasonable. It also explains, why it only works for HF; although there are generalisations for DFT.

Is Koopmans' theorem still valid after a unitary transformation?

No. As stated above, the Fock operator is associated with the $N$-electron wave function. After a unitary transformation and then removal of an electron, the energy of the $N-1$-electron wave function is not conserved.
As an illustrative example:
\begin{align}
&& \sum_{i=1}^{N} H_{ii} &= \sum_{i=1}^{N} H_{ii}'\\[2ex]
\text{with}&&
H'_{ii} &= \langle \phi'_i(\mathbf{x}_1)|
          \mathbf{H}^\mathrm{c} | \phi'_i(\mathbf{x}_1)\rangle\\
\text{and}&&
|\phi_i\rangle &\color{red}{\neq} |\phi'_i\rangle\\[2ex]
&& \sum_{i=1,i\neq k}^{N-1} H_{ii} + H_{kk}&=
   \sum_{i=1,i\neq k}^{N-1} H_{ii}' + H_{kk}'\\
\text{for } H_{kk} = H_{kk}':
&& \sum_{i=1,i\neq k}^{N-1} H_{ii} &=
   \sum_{i=1,i\neq k}^{N-1} H_{ii}'&&
\implies|\phi_i\rangle \color{red}{=} |\phi'_i\rangle\\
\end{align}

I have indicated the contradiction in red. Therefore it follows:
$$|\phi_i\rangle \neq |\phi'_i\rangle
  \implies H_{kk} \neq H_{kk}'
  \implies
  \sum_{i=1,i\neq k}^{N-1} H_{ii} \neq \sum_{i=1,i\neq k}^{N-1} H_{ii}'
$$

You can follow that trough for the other terms and will see that the expectation value of energy of the $N-1$-electron wave function needs to be different from the expectation value of energy of the $N-1$-electron wave function after unitary transformation.


What is the difference between graphite oxide and graphene oxide?

In short, graphite is several graphene sheets piled one above another.

Graphene is made up of one single sheet of carbon atoms arranged in hexagonal pattern (like a honeycomb), and graphite are several such sheets, each sheet linked to another by weak intermolecular forces, which gives the graphite its lubricative properties.

EDIT:
Similarly, graphene oxide is a layer of graphene with oxygen atoms, $OH$ groups and $COOH$ groups attached to it, while graphite oxide are layers of graphene oxide.

Because of the presence of $COOH$ groups, graphite oxide can disperse in basic solutions (or even water!) to give graphene oxide.

$CO_2$ is simply carbon dioxide, one carbon atom doubly bonded two two oxygen atoms.

By contrast, graphite oxide is like a huge field of carbon with the groups I mentioned earlier while carbon dioxide is like a small ball.


The density of $\pu{2.17 g/cm³}$ refers to the bulk density, i.e. within a crystal of NaCl.  In chemical engineering, the terms of powder density, tapped powder density and settled apparent density take into account for the air between the grains of a solid. Especially the later recognizes that there may be a difference between the solid simply poured into a container, and after light compression (still with air gaps between the grains) after applying a little pressure e.g. if you shake and knock the tin filling with freshly ground coffee powder.
References like this, this, this, or this .pdf state powder densities of $\pu{1.378 g/cm³}$ (fine table salt), $\pu{1.282 g/cm^3}$ (granulated salt, again from here), and $\pu{1.089 g/cm^3}$ (rock salt). From these values, your estimate of $\pu{1.40 g/cm^3}$ seems plausible.
However, these data lack to state the typical size of the grains (think about the diameter), as well as the dispersion of the grain sizes (presence of larger and smaller grains, equally known as particle-size distribution) of the samples characterized. Both influence the packing of the grains and thus the recorded density.  In this perspective, the softer / more airy harvest of fleur de sel possibly packs much less dense.


How to derive the relation between gibbs energy and equilibrium constant?

Firstly, if you know the relation $G= H-TS$, you can reach at the differential form of it just by taking differential at both sides, i.e.$$dG= dH-SdT-TdS$$ Now, recall the definition of enthalpy as $H= U +PV$, So, we can write, $dH= dU + PdV + VdP $, and also recall the heat supplied to the system as $dQ= TdS $ and by the first law of Thermodynamics, $dQ= dU+ PdV$ .Thus combining we have $$dG= dU+PdV+VdP-SdT-dQ = VdP -SdT$$ Thus you derive the required relationship.

Now coming to your next question, it's not about taking path where Entropy and volume remains constant. A state in a thermodynamic system can be defined in terms of three parametrs which are interrelated i.e Pressure, Temperature and Volume. In fact, you can only independently vary any two out of these three parameters and third one will automatically get adjusted. Here the independent variables are Temperature and Pressure. By changing the Pressure and Temperature independently, the change in the Gibbs' Free Energy is calculated. The coefficients are volume and Entropy only but that doesn't mean they are kept constant. So, the interpretation of the equation $dG= VdP- SdT$ needs to be done correctly.


Do we lose significant figures with digital instruments?

As far as I know the visual interpolation of analogous measurements is bad scientific practice anyway.
We assume that rounding errors work in both directions and compensate each other in the long run. On the other hand visual interpolation by a human can introduce systematic errors.

But putting this discussion away one has to note that even if you visually interpolate, you still digitalise your data (just with a finer grid). By digitalising I do not mean that you write them into a computer, but the abstract mapping of continous data points of $\mathbb{R}$ into a finite subset.

In the end it just is a question of how fine the grid is you are measuring with. And yes, if you don't introduce systematic errors a visual interpolation makes your grid a little bit finer.
On the other hand switching to electronic devices usually gives you orders of magnitude better results.

Some examples:

Let's assume that someone wants to measure lengths of $\mu$m with a normal ruler that has 1 mm spacing.
It is intuitively clear that this won't work.
(Even if you can interpolate.)

Now let's look at another example that was brought up in the comments by @Zhe. The problem that every number in your computer is an element of a finite set. This introduces rounding errors, when saving real numbers (real as in $\mathbb{R}$).
If you execute the following python code or something similar in your language of choice, you will see this rounding error.

import sys
print(sys.float_info.epsilon)
# for python2:
# print sys.float_info.epsilon

On my 64bit machine this is in the order of $10^{-16}$. The standard deviation of the meter is in the order of $10^{-11}$ (wikipedia).
Floating point noise will rarely be a problem, when dealing with experimental lengths.

This kind of analysation has to be done for every experimental measurement device you are using. Regardless of electronics vs. non electronic measurements. The advantage of "old analogous" measurements is that you can intuitively see from the scale how fine the grid is, that you are working with.


Is it possible to have a diatomic molecule of sodium in gaseous state?

According to molecular orbital theory, disodium should be stable in the gas phase, with a bond order of one. The molecular orbital diagram is the same for all the alkali metals since they all have one valence electron in an $s$ orbital.

(image source)

NIST chemistry webbook has a small page on disodium and quotes an enthalpy change of formation of $142.07~\mathrm{kJ~mol^{-1}}$ which is moderately endothermic.

This page has a brief discussion of alkali metal dimers and claims that they are all stable in the gas phase, in agreement with the prediction from MO theory.


In order to be in the "critical state", both temperature (x-axis) and pressure (y-axis) must be above the critical point. So answers to your questions become quite obvious:

Very hot but below the critical pressure, it is a gas. You don't need any more information than the temperature and the pressure.
Yes if its pressure is less than its critical pressure
Correct.


What makes dimethyl compounds so extreme?

This is because of two trivial facts, one from chemistry and another from math: (1) methyl is the smallest organic radical, and (2) it is not uncommon for a function to reach an extremum at the border of its domain.

Arguably, this is not even true in any significant way. Not sure about DMSO, but I firmly believe that organic compounds of mercury are all about as deadly as dimethylmercury. Well, those with bigger organic radicals might be slightly less dangerous because they are not quite as good at penetrating the cell barriers, or lower solubility, or because they contain less mercury per gram. The maximum is probably reached at methyl; but the maximum is not sharp.

All that applies to pyrophoric metalorganics as well.

So it goes.


Can we picture metallic bonding as an equilibrium between electrons and cations?

In metals, electrons are non-localized, forming a "sea" of electrons, rather than having them localized, as in the $\ce{Na+Cl-}$ lattice of crystalline salt. See Metallic bonding for a more complete description.

It is, of course, a matter of degree, as covalent, ionic and metallic bonding can "blend" from one to the other. A bond can be considered partially ionic and covalent, for example; see these helpful graphics


Here Cu plate is cathode and Zn plate is anode, it's written wrong in it. In the first case (Zn in  $\ce{ZnSO4}$ and Cu in $\ce{CuSO4}$) the electrodes are connected through an external circuit (You see that bulb glowing?). But what this article is trying to say is that in the cell (Zn, Cu in $\ce{H2SO4}$), if you leave the solution as it is without connecting externally, Zn having higher tendency to lose electrons, will lose electrons and these will appear on its plate surface making it negatively charged, the $\ce{Zn^2+}$ thus formed will combine with $\ce{ SO4^2-}$ and form $\ce{ZnSO4}$. The remaining two $\ce{H^+}$ ions will more towards copper plate where they will combine with the electrons on its plate and make it positive charged, releasing $\ce{H2}$. Now if a wire is connected across this setup then the electrons on Zn surface will leave it and go towards positive charged Cu plate.


The main reason ionic substances like salts are not used is because they tend to be corrosive to metals.

Polar organic liquids tend to be the best because many are very soluble in water, and they are fairly inert under the conditions antifreeze is used.


Can I add citric acid to KCl for iron removal in a water softener?

Your ion-exchange column removes $\ce{Fe^{3+}}$ ions from your water by exchanging them for $\ce{Na^+}$.

When you back-flush you use concentrated solutions of $\ce{Na^+}$ ions and citric acid to remove the iron out of your ion-exchange column. The flush water is waste of course which must be disposed.

The exact chemistry mechanism of all of this depends on a number of factors:

NaCl has a solubility of 36 g/L at 20 C which is 0.62 molar but
KCl has a solubility of 34.2 g/L at 20 C which is 0.46 molar.
I'd expect potassium citrate to be less soluble than sodium citrate but
I didn't find any data to support this notion in a quick search.
I'd expect that you column has a better attraction to $\ce{Na^+}$ ions
than to $\ce{K^+}$ ions.
If you have some sort of automatic monitor then it may not
function properly with $\ce{K^+}$ ions. I'd guess that the monitor
just looks to determine if the water is yellow, so I'd guess that
$\ce{K^+}$ ions would work fine in this regard.
Not sure what pH you'd want. I'd guess maybe half trialkali citrate with
half dialkali hydrogen citrate to keep the back-flush solution
slightly acidic.

All in all my guess would be that potassium/potassium citrate flush solution would work. I'd expect the column to need flushing more often. Maybe doing a flush once in a while with the sodium/citric acid mixture would give the ion-exchange column a good cleaning.

The other consideration here is cost. Since the sodium/citric acid mixture is the common back-flushing salt, I'd expect a potassium/citric mixture to be more expensive.


Is gravity an obstacle for homogenous solvation due to diffusion?

At normal Earth gravity, for all practical purposes homogeneous solvation is not hindered by gravitational force.
However: For some solutes such as sucrose and $\ce{CsCl}$, though, the "artificial large gravity" of centrifugation is capable of establishing concentration gradients in their solutions. One example in common laboratory use is buoyant density centrifugation:

One of the most common methods of fractionating DNA is to exploit differences in density between different types of DNA in concentrated salt solutions. Caesium chloride is the most commonly used. The method involves prolonged high speed centrifugation of purified DNA in concentrated caesium chloride. The centrifugation produces a gradient of salt concentration in the centrifuge tube and the DNA migrates to occupy a position in the gradient which corresponds to its own buoyant density.


Salt concentration and electrical permittivity of water

The relative electric permittivity of water at $20^oC$ under the effect of a static or low frequency is $\epsilon_r\approx80$. Hasted (1948) conducted an experimental study on the dielectric properties of salt-water solutions and observed a dielectric decrement with salt concentration. That is, the addition of sodium chloride to water results in a drop in electrical permittivity.

Source: Hasted, Ritson and Collie (1948).

This can be understood by examining the following diagram.

The sodium and chloride ions dissociate in solution, resulting in an an electric field between them (shown in bold line). The water molecules, which are polarised, get oriented under the effect of the electric field so that their oxygen atoms (carrying a partial negative charge) face toward the sodium ion and their hydrogens (carrying partial positive charges) face toward the chloride ion. This orientation of the polar water molecules causes its own electric field (dashed lines), which cancels out most of the electric field that would exist if the ions were in a vacuum. The end result is that the sodium and chloride ions are effectively 'sheilded' by a 'hydration shell' of polarized water molecules which effectively lowers their to the applied external field and hence lowering the dielectric constant (electric permittivity) as the ionic concentration increases ().

In dilute solution (less than 2.0M) the dielectric decrement is linear. At higher salt concentrations, the dielectric decrement is observed to saturate at about $\epsilon_r=45$. The effect of sodium chloride concentration on electric permittivity is shown in the graph below at various temperatures.

Source: Gavish and Promislow (2012).

The above graph indicates that for a given salt concentration below about 3M, the electric permittivity decreases with temperature.

REFERENCES:

Hasted, J.B., Ritson, D.M., Collie, C.H., "Dielectric Properties of Aqueous Ionic Solutions. Parts I and II." Journal of Chemical Physics 16, 1 (1948). http://dx.doi.org/10.1063/1.1746645

Gavish, N., Promislow, K, "Dependence of the dielectric constant of electrolyte solutions on ionic concentration", http://arxiv.org/pdf/1208.5169.pdf


It pretty much has to be a supercooling phenomenon.  This page from the University of Toronto Physics department presents an explanation that I think fits the scenario in the question nicely.  The description and photos used suggest that this is likely the case for this question:

  The short explanation is this: as the ice freezes fast under
  supercooled conditions, the surface can get covered except for a small
  hole. Water expands when it freezes. As freezing continues, the
  expanding ice under the surface forces the remaining water up through
  the hole and it freezes around the edge forming a hollow spike.
  Eventually, the whole thing freezes and the spike is left.

  A slightly longer explanation: the form of the ice crystals depends on
  the cooling rate and hence on the degree of supercooling. Large
  supercooling favors sheets which rapidly cover the surface, with some
  sheets hanging down into the water like curtains. These crystallites
  tend to join at 60 degrees and leave triangular holes in the surface.
  Hence, spikes often have a triangular base. The sides of the spike are
  sometimes a continuation of pre-existing subsurface crystallites, and
  can extend from the surface at steep angles.

Notice the triangular shape at the base of the ice spike in the photos in the question, as is also mentioned in the description above.  The spike also appears to possibly have a hollow region through the center.  Additionally, the flexible plastic lid would allow for some pressure buildup before the release through some weak spot in the ice sheet.


Why can't all electrophiles act as Lewis acids?

The definition of Lewis acids states:

  A molecular entity (and the corresponding chemical species ) that is an electron-pair acceptor and therefore able to react with a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis base. For example:

  (source: iupac.org)

In principle, since all electrophiles can be electron acceptors, they can usually act as Lewis bases, too. However, sometimes sterically demanding moieties stabilise the electrophile, so that it is impossible to form a Lewis adduct. These compound are then one part of a frustrated Lewis pair.


How much water is required to dilute glacial acetic acid to given pH?

The question is flawed; the $\mathrm{pH}$ of 100 % glacial acetic acid is not $\mathrm{pH=2.4}$. (Note that 100 % glacial acetic acid contains no water; therefore you cannot apply the usual equations for dilute aqueous solutions to calculate the $\mathrm{pH}$ of 100 % glacial acetic acid.)

Also note, since acid dissociation is not a redox reaction, you cannot apply the Nernst equation here.

The quantity $\mathrm{pH}$ is actually defined in terms of the activity of $\ce{H+}$ ions; however, for simplicity’s sake, we may use the old concept of $\mathrm{pH}$ with the definition in terms of the concentration of $\ce{H+}$ ions
$$\mathrm{pH}=-\log_{10}[\ce{H+}]$$
or
$$[\ce{H+}]=10^{-\mathrm{pH}}$$

The equilibrium of acid dissociation is usually written as:
$$\ce{HA <=> A- + H+}$$
where $\ce{HA}$ is a generic acid that dissociates into the conjugate base $\ce{A-}$.
The equilibrium constant $K_\mathrm a$ for this reaction is defined as
$$K_\mathrm a=\frac{[\ce{H+}]\cdot [\ce{A-}]}{[\ce{HA}]}$$
Usually, values are given for the logarithmic constant $\mathrm pK_\mathrm a$, which is defined as
$$\mathrm pK_\mathrm a=-\log_{10}K_\mathrm a$$
or
$$K_\mathrm a=10^{-\mathrm pK_\mathrm a}$$

The corresponding equilibrium for acetic acid is
$$\ce{CH3COOH <=> CH3COO- + H+}$$
The value of the logarithmic constant for acetic acid at 25 °C is
$$\mathrm pK_\mathrm a=4.756$$
Thus, the value for the acid dissociation constant is
$$K_\mathrm a=\frac{[\ce{H+}]\cdot [\ce{CH3COO-}]}{[\ce{CH3COOH}]}=10^{-\mathrm pK_\mathrm a}=10^{-4.756}\approx1.75\times10^{-5}$$

The chemical equation shows that, one $\ce{CH3COO-}$ ion and one $\ce{H+}$ ion are produced for every $\ce{CH3COOH}$ molecule that dissociates. This can be represented by subtracting a value $x$ from the original acetic acid concentration $[\ce{CH3COO-}]_0$, and adding this value $x$ to the concentrations of $\ce{CH3COO-}$ and $\ce{H+}$, i.e. when neglecting the $\ce{H+}$ ions from the autoprotolysis of water:

$$\begin{align}
[\ce{CH3COOH}]&=[\ce{CH3COOH}]_0-x\\[6pt]
[\ce{CH3COO-}]&=x\\[6pt]
[\ce{H+}]&=x
\end{align}$$

Thus, the equation for the acid dissociation equilibrium can be rewritten and solved for the original acetic acid concentration $[\ce{CH3COOH}]_0$ as follows:

$$\begin{align}
K_\mathrm a&=\frac{[\ce{H+}]\cdot [\ce{CH3COO-}]} {[\ce{CH3COOH}]}\\[6pt]
&=\frac{x\cdot x}{[\ce{CH3COOH}]_0-x}\\[6pt]
&=\frac{x^2}{[\ce{CH3COOH}]_0-x}\\[6pt]
K_\mathrm a\cdot\left([\ce{CH3COOH}]_0-x\right)&=x^2\\[6pt]
[\ce{CH3COOH}]_0-x&=\frac{x^2}{K_\mathrm a}\\[6pt]
[\ce{CH3COOH}]_0&=\frac{x^2}{K_\mathrm a}+x
\end{align}$$

Since $[\ce{H+}]=x$, $[\ce{H+}]=10^{-\mathrm{pH}}$, and $K_\mathrm a=10^{-\mathrm pK_\mathrm a}$, the required acetic acid concentration $[\ce{CH3COOH}]_0$ can be given in terms of the desired $\mathrm{pH}$:

$$\begin{align}
[\ce{CH3COOH}]_0&=\frac{x^2}{K_\mathrm a}+x\\[6pt]
&=\frac{[\ce{H+}]^2}{K_\mathrm a}+[\ce{H+}]\\[6pt]
&=\frac{\left(10^{-\mathrm{pH}}\right)^2}{K_\mathrm a}+10^{-\mathrm{pH}}\\[6pt]
&=\frac{\left(10^{-\mathrm{pH}}\right)^2}{10^{-\mathrm pK_\mathrm a}}+10^{-\mathrm{pH}}\\[6pt]
&=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}
\end{align}$$

For example, for the given acetic acid solution with $\mathrm{pH}=2.4$:

$$\begin{align}
[\ce{CH3COOH}]_0&=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}\\[6pt]
[\ce{CH3COOH}]_0&=10^{4.756-2\times2.4}+10^{-2.4}\\[6pt]
&=0.907630545\\[6pt]
&\approx0.9
\end{align}$$

and for the desired acetic acid solution with $\mathrm{pH}=4.3$:

$$\begin{align}
[\ce{CH3COOH}]_0&=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}\\[6pt]
[\ce{CH3COOH}]_0&=10^{4.756-2\times4.3}+10^{-4.3}\\[6pt]
&=0.000193338\\[6pt]
&\approx0.0002
\end{align}$$

Finally, since
$$c_1\cdot V_1=c_2\cdot V_2$$
you may use the concentrations to calculate how much the given acetic acid solution with $\mathrm{pH}=2.4$ should be diluted in order to obtain the desired acetic acid solution with $\mathrm{pH}=4.3$.


I strongly suspect that a hydride shift is not observed in this example.

The reason is that it seems much more likely that you will have an alkyl shift instead, accompanied by ring contraction and formation of a carbonyl. This is essentially a pinacol rearrangement.

EDIT:
Adding more details per @OscarLanzi comment...

I would think that this is the expected product after proton transfer.

It looks like the pinacol arrangement of 4,8-decalindiol has been known for quite some time. Hückel, Danneel, Schwartz, Gercke, Ann. 1929, 474, 121. and Meiser. Chem. Ber. 1899, 32, 2055.


How to separate lead condensate from aluminum

You could try hard anodizing Al and Ti (e.g., Calphalon cookware) to give them a non-delicate surface, and go on from there.  Black is a good gun component color.  CVD deposition of titanium boride, ion plating of titanium nitride or carbonitride gives exceptionally survivable hard surfaces.

http://www.brycoat.com/coating-services/pvd-coatings/titanium-nitride-tin.html
http://www.brycoat.com/coating-services/pvd-coatings/ticn.html

Among mostly cheap and harmless diddles, I'd try 1:2 mole ratio of choline chloride  with urea or ethylene glycol ("all natural," etc.).  Mix, the deep eutectic liquefies, soak.  Both eutectics conduct electricity, so you could make lead contamination the anode and plate it out on a closely spaced but not touching copper foil cathode. Test on contaminated waste Al or Ti.  Careful with the voltage - too high and you mobilize Al and Ti.  Avoid contact with skin pro forma, do not ingest, keep away from eyes.

Try vacuum thermal deposition of a micron or so thickness of the Parylene family on the newly fabricated and cleaned metal - intrinsic, especially C, or F.  Parylene polymer surface coats survive the insides of excimer lasers, protecting the excitation electrodes.  If the lead still sticks, chemical cleaning should stop at the polyparylene film.  Or not.  Gotta test it.  You are worried about vapor deposition not abrasion, so an inert barrier film is reasonable.

Google, parylene coating services

For aluminum or titanium vs. lead in chemistry, I'd pull one out of left field to try on test coupons:  Rauchfuss' elixir: elemental sulfur dissolved in N-methylimdiazole (pyridine less so, very smelly, toxic, and don't get it on your skin or breathe the awful vapors).  It goes through polysulfide-forming metals like a buzz saw, giving soluble products.  I would expect aluminum to be inert; titanium is a toss-up.  It should hit lead hard, probably.  Test it small scale.

Inorg. Chem. 30 (11): 2514 (1991)
DOI:10.1021/ic00011a011
http://dx.doi.org/ (resolve a DOI)

Last gasp:  Weapon Shield to clean before and after firing.

Concentrated nitric acid (already a bad idea) passivates aluminum and eats lead.  Toxic and corrosive NOx fumes come off.  However...aluminum plus lead in galvanic contact make aluminum the sacrificial anode.  Homeland Severity does not like people possessing concentrated nitric acid.

Wear gloves and goggles, have ventilation.  None of this stuff is good for you.


How to find molecular charge of fragments using quantum calculations?

Your question boils down to "where are the electrons" in a quantum chemical system?

The problem, is if there's any level of interaction between the two fragments, you don't know. The charge will be delocalized across the entire A+B complex.

Put it another way.. Let's say you have a bunch of atoms and you run a calculation and the overall charge on the molecule is +1. Where's the extra positive charge?

Well, you might make some guesses, but it's probably delocalized across multiple atoms. The same thing happens with intermolecular interactions, but to a lesser degree.

There are two approaches:

Constrained DFT: This is a method where you force the charge on a particular fragment to be the way you want. You can then compare this semi-localized charge with the distributed (unconstrained) calculation and compare.
Fragment MO: You can diagonalize the wavefunction onto multiple fragments that you define.

I don't use GAMESS much, but a quick search implies that FMO is supported in GAMESS.


Propose a reaction mechanism that explains the rate law for the formation of dinitrogen pentoxide

First step - the rate determining step or the slow step (activation energy is high).
$$\ce{NO(g) + NO2(g) -> N2O3(g)}$$

2nd step - the fast step.
$$\ce{N2O3(g) + O2(g) -> N2O5(g)}$$

Since the rate determining step does not include $\ce{O2(g)}$, the rate is independent of the initial $\ce{O2(g)}$ concentration.


Protonation of the carbonyl group of a ketone allow amines to react via nucleophilic addition. This is because the added proton pulls electrons further away from the reaction site, thereby allowing the lone-pairs from the amine to attack the carbonyl carbon.

In the case of an amide, the C=O beside the nitrogen causes electrons to be pulled away because of the positive charge of the carbon. This means that despite the possibility of having a protonated ketone, there will still not be a reaction because the nitrogen is not nucleophilic.


Deriving The Triple Point Temperature and Pressure as the following expression

On a phase diagram there are three 1-dimensional curves, the melting curve, the liquid vaporization curve, and the sublimation curve.

It is sufficient to consider just two of the three curve, and determine the point at which they intersect.

Given the limited information availible to you, (no information about how volume changes upon going from the solid to liquid phase), it is wise to choose the liquid vaporization curve and sublimation curve, rather than the melting curve, as $\Delta V$ can be approximated as the volume of gas, the volume of solid or liquid being approximately zero compare to the much larger volume of gas.

From the Clapeyron equation in the question, you need to change variables from  $\Delta S$ which you don't know to $\Delta H$ which you do know.

When two phases are in equilibrium, such as along any of the three curves discussed above, $\Delta G$ = 0.

$G = H-TS$

$0 = dG = dH-TdS -SdT$

Considering that $T$ is constant during a phase change:

$\Delta H = T \Delta S$

So substituting:

$\frac{dP}{dT}= \frac{ΔH}{TΔv}$

$\frac{dP}{dT}= \frac{ΔH_{vap}}{TΔv}$ and $\frac{dP}{dT}= \frac{ΔH_{sub}}{TΔv}$

Then, as mentioned above the approximation $\Delta V = V_{gas}$ can be made and the ideal gas law (or other equation of state) can be used to express $V$ as a function of $T$ and $P$.  Then (after intergrating) you have two equations with two unknowns, $T$ an $P$, the temperature and pressure at the triple point.


Can I use the mono ammonium phosphate found in fire extinguishers to make crystals?

If you google for MAP, you'll find out that it is quite soluble in water, ranging from 370 to 401 g/litre of H2O at temperatures around 25 C. Thus, what you report contradicts literature data on MAP's solubility. So, maybe the powder you found is not MAP, it's something else (CaCO3 comes to my mind, because it is insoluble in water and yields CO2 when heated, but I may be wrong).


As this video clearly confirms, the hydrolysis of SiCl4 is highly exothermic from breaking of chlorine bonds. The solution boils and solid insoluble Si(OH)4 is produced.
Also, per Wikipedia on the hydrolysis, to quote:

The reaction can be noticed on exposure of the liquid to air, the vapour produces fumes as it reacts with moisture to give a cloud-like aerosol of hydrochloric acid.[5]

As such, the balanced reaction is likely best expressed as:
$\ce{ SiCl4 (l) + 4 H2O (l) ⟶ 4 HCl (g) + Si(OH)4 (s)}$
where the gaseous and solid precipitates formed can push the reaction to the right.
There is an interesting comment in Atomistry, to quote:

Consider, for example, the two reactions:

$\ce{Ca(OH)2 + 2HCl → CaCl2 + 2H2O}$

$\ce{Si(OH)4 + 4HCl ← SiCl4 +4H2O}$

The difference between the metal and the non-metal is signalised by the direction of the reaction. This difference, however, is seldom absolute, and with elements of intermediate character it disappears. In the case of stannic tin, for example, the reaction is reversible:

$\ce{Sn(OH)4 + 4HCl ⇔ SnCl4 + 4H2O;}$

and stannic chloride can be prepared either by dissolving stannic hydroxide in hydrochloric acid, or by passing chlorine over heated tin or a mixture of stannic oxide and carbon.

So, part of the answer also relates to the non-metal characteristics of Silicon as well.


What is the yellow precipitate formed from reacting diamminsilver fluoride and potassium iodide?

The yellow precipitate that forms is silver(I) iodide, $\ce{AgI}$. It is very, very insoluble and hence will precipitate immediately if silver ions and iodide ions meet in solution.

In fact, all three common halides other than fluoride form insoluble salts with silver: silver(I) chloride is a white precipitate, silver(I) bromide is somewhere between white and yellow and silver(I) iodide is yellow. These precipitates are so characteristic that they were historically (and still in qualitative inorganic analysis lab courses) used to identify the different halide ions in solution. To further improve the distinction, silver(I) chloride can be dissolved in dilute solutions of ammonia while silver(I) bromide needs concentrated ones. Silver(I) iodide does not dissolve in ammonia.

$$\begin{align}\ce{Ag+ (aq) + Cl- (aq) -> AgCl v &->[NH3 (dil)] [Ag(NH3)2]+ (aq) + Cl- (aq)}\tag{1}\\
\ce{Ag+ (aq) + Br- (aq) -> AgBr v &->[NH3 (conc)] [Ag(NH3)2]+ (aq) + Br- (aq)}\tag{2}\\
\ce{Ag+ (aq) + I- (aq) -> AgI v &->[NH3] \text{no reaction}}\tag{3}\end{align}$$

I assume that the original silver solution was the diammine complex to prevent silver chloride from precipitating as soon as chloride ions are added (due to not perfectly deionised water being used).


Contactless non-ferrous metals hardening?

From this link I could find a few non-ferrous systems which show quench-induced or stress-induced martensite . Generally non-ferrous systems dont show this behaviour and they are hardened/strengthened by Precipitation hardening, work hardening, Solid Solution Strengthening.

Also work hardening is basically working/deforming the material plastically to strengthen it. So you can't work harden a material without contact. Induction hardening is used mainly for Case Hardening which is different from hardening(done to bulk of the material). So as long as the non-ferrous alloy is conductive and resistive , Induction hardening is enough for case-hardening.

And if it is hardening , the above mentioned methods are used for non-ferrous. Only the quenching/martensitic transformation is limited to limited systems.


Why don't Gringard Reagents react with carbonyls at the alpha position?

It actually happens sometimes and you end up with the enolate being formed in appreciable amounts. But it depends on other adjacent substituents that can withdraw or donate electron density to the alpha carbon.

If you draw the mechanism for the alpha attack you can see that the enolate can react further with the aldehyde and produce aldol condensation adducts. Furthermore, the carbon atom of the aldehyde is way more electrophilic than the hydrogen atom bound to the alpha carbon, that is the main reason why the enolate end up being formed in very small amounts and sometimes you don't even detect it.


Why does shower gel / shampoo foam a lot more on hairy than non-hairy body parts?

It's mainly due to the surface area and the spacial distribution of hair compared to that of hairless skin.

A typical human hair width is around one micron.  Given the various hair length and thickness (space between hairs) on the different hairy regions of the body, there are orders of magnitude more surface area on which soap bubbles (foam) can adhere as compared to smooth (or even very rough), hairless skin.

The second part of that answer, the spacial distribution, just means that the hair gives support to sustain the bubbles so they don't immediately collapse.

Short answer, but I think this pretty much covers it.


Some tools were made of "pure" copper. But tin was the best and only available addition to copper for strength. At the time tin was the only type of bronze. Today a variety of copper alloys are called "bronze"  that do not contain tin. Lead could possible have been available depending on time and place but it does not improve strength of copper and could actually weaken it. I see in alloy listings there are silicon bronzes ( today) that contain 1 to 2 % . Silicon was available, but it takes modern technology to alloy it into copper. Silver and gold could be added to copper ( they are all mutually soluble ) but make little increase in strength. There is a copper alloy today that contains about a quarter % of each phosphorus and arsenic ( C14200) , it is a wrought product so not available in the bronze age. The Copper Development Association  presumable has some information on the internet.


Having trouble understanding activity series of metals & electronegativity

There's electronegativity and then there's electropositivity. Electronegativity is the tendency of a substance to gain electrons and electropositivity on similar grounds is the tendency to donate electrons. What actually is a redox reaction? Exchange of electrons?

A substance with high electronegativity would like to snatch any electron it finds, while a substance with high electropositivity would like to force-fully hand over its electrons, however unwilling the substance on the receiving end may be. Metals are electropositive.

  how do metals replace hydrogen in a substance such as water. Isn't water happy with its configuration?

There was water, it was living happily. Now, water had two components. Oxygen and hydrogen. Let's pull up their electronegativities. Oxygen is at $3.44$ and hydrogen is at $2.2$. This means that the bond that oxygen and hydrogen shared in water was a polar bond. Oxygen being powerful (higher electronegativity) kept the bond in it's favor. The electrons of the bond were more towards oxygen. This gave oxygen a partial negative charge ($\delta -$) and hydrogen a partial positive charge ($\delta+$).

In came a metal which was highly electropositive (think alkali metals, as an example you can take up sodium whose electronegativity is at $0.93$). He gave oxygen a tempting offer:

"Oxygen, you seem to be caught up in a tussle, look at hydrogen, he doesn't want a positive charge because he's quite electronegative. You're constantly fighting with him to keep the electrons of the bond pair pulled towards yourself. Why don't you pair up with me? I'm low at electronegativity. This means I'm electropositive. As much as electropositive species like bearing a negative charge, we electropositive species like bearing a positive charge!"

And oxygen was greedy. A greater difference in electronegativities of elements forming the bond meant a bond which would be richer in ionic character, which appealed to oxygen.


Help with finding intermediates in the synthesis of Ritalin

In A and B, you should probably draw the C-C single bond in the nitrile longer, it's a bit difficult to see at the moment.
The nitrile is sequentially hydrolysed to an amide and then to an acid.
Treatment with $\ce{ROH/H+}$ is a rather obvious Fischer esterification.

You can arguably work all of this out from first principles: the nitrile carbon is electrophilic, and is therefore susceptible to nucleophilic attack by water. The reaction mechanism is given here. On top of this, the IR peaks strongly support the presence of amide and acid functional groups in D and E respectively.

With that said, I'd argue that there is always going to be an element of "I just know this because I read about it/was taught it before". The very fact that you said "I'm not familiar with the mechanisms/reactions of nitriles" indicates this - if you had happened to read a textbook chapter on nitriles, then you would probably have seen this, since it is one of the most fundamental reactions of nitriles. You would also have to read up about the IR wavenumbers - otherwise it's just gibberish.


How many stereogenic centres are there in discodermolide?

Summary

The confusion probably arises because of incompletely standardized definitions of at least some subset of the words "stereocenter", "stereogenic", "stereogenic center", "stereogenic unit".

IUPAC definitions

The place to turn for authoritative definitions of these terms is IUPAC's Gold Book, as noted by @jerepierre in the comments.  Here is what it has to say:

  stereogenic unit (stereogen/stereoelement)

  A grouping within a molecular entity that may be considered a focus of stereoisomerism.

  At least one of these must be present in every enantiomer (though the
  presence of stereogenic units does not conversely require the
  corresponding chemical species to be chiral). Three basic types are
  recognized for molecular entities involving atoms having not more than
  four substituents:

  A grouping of atoms consisting of a central atom and distinguishable
  ligands, such that the interchange of any two of the substituents
  leads to a stereoisomer. An asymmetric atom (chirality centre) is
  the traditional example of this stereogenic unit.
  A chain of four non-coplanar atoms (or rigid groups) in a stable conformation, such that an imaginary or real (restricted) rotation
  (with a change of sign of the torsion angle) about the central bond
  leads to a stereoisomer.
  A grouping of atoms consisting of a double bond with substituents which give rise to cis-trans isomerism.

Thus, the unequally substituted C=C moieties in discodermolide are "stereogenic units" according to IUPAC.

Stereocenter

Unfortunately, this term is not defined in IUPAC's Gold Book.

Stereogenic center

Unfortunately, this term is not defined in IUPAC's Gold Book.

Wikipedia

Wikipedia is not as authoritative as IUPAC but it is likely that chem.se readers will encounter wikipedia entries for these terms far more often than IUPAC's in the course of regular web searching.

Stereogenic unit

No Wikipedia entry

Stereogenic center

Redirects to "stereocenter", see below.

Stereocenter

  A stereocenter or stereogenic center is an *atom* bearing groups such that an interchanging of any two groups leads to a stereoisomer.

This definition does not apply to C=C double bonds, because only an atom can be a stereocenter, not a group of atoms.  Supporting this interpretation is a complete lack of reference to cis-trans isomerism on the rest of the page.

Usage by chemists

Discodermolide is referred to in many sources as having 13 "stereocenters" or "stereogenic" centers.  This conforms with the Wikipedia, atom-centric definition of stereocenter.  Here are some example sources:

Wikipedia's discodermolide article: (+)-discodermolide has
...13 stereogenic centers...
A presentation from 2004:
molecule possesses: 13 stereocenters
A paper from 2003: Paper title: Regulated-Stereoselective Construction of Thirteen
Stereogenic Centers...

All these uses comport with both Wikipedia definitions of "stereocenter"/"stereogenic center".  Chemists do not seem to use the IUPAC term "stereogenic unit" very much in actual practice.   I could not find any relevant pages using both the term "discodermolide" and the phrase "stereogenic unit", for example.

Bottom line

It is just possible to pass a thread of consistency through the eye of this needle of confusing terminology.

Discodermolide has thirteen (13) stereogenic centers.
Each of those stereogenic centers is by definition also a stereocenter, and is also a "stereogenic unit".
In addition to those 13 stereocenters, there are three additional "stereogenic units" in discodermolide corresponding to the unequally substituted C=C bonds.


Solid hydrophilic transparant materials that don't dissolve in water

Several commercially available titanium dioxide coatings are available, commonly used to kill bacteria or aid UV stability (I think). When exposed to light, the titanium dioxide has a contact angle of almost 0 degrees.

You should be able to deposit a thin layer on something like a PVA or polycarbonate sheet. The commercial coatings are somewhat expensive, though, as far as I can tell, and the common process for doing it yourself involves heating the surface in a mid-temperature flame for 10-15 minutes which excludes pretty much all plastics I know of that would otherwise be suitable.


Why do lipids in a cell hydrolyze into glycerol and fatty acids?

Organisms control rates of chemical reactions in two ways. First, by controlling the conditions, and second, by controlling the location of reagents.

Conditions

Hydrolysis of esters is very slow at normal physiological conditions: 37 oC and pH 7.4. A mixture of water and simple esters like ethyl acetate remains stable for years. At higher or lower pH, the reaction becomes faster. As you mention, the HCl in the stomach would definitely increase the rate of hydrolysis. The reaction is also accelerated by enzymes, and organisms use enzymes for all intentional ester hydrolysis.

Location

For a reaction to occur, the water molecule and the acid catalyst needs to get to the ester. In cell membranes, that ester functional group is not on the surface of the membrance; the ionic phosphate part is. The ester group is just below the surface, in the hydrophobic interior of the membrane. Neither water, nor acid can penetrate into this hydrophobic region.

In the stomach, there is an extra layer of protection: mucus, which prevents the stomach acid from reaching the cells.


Are molecular optimizations in quantum chemistry really in vacuum?

You raise an excellent question and the short answer is "yes, quantum chemistry calculations are intrinsically in vacuum".

Approximate methods like semiempirical (AM1, PM6, PM7) and hybrid DFT methods (e.g., B3LYP) use experimental data, which often comes from crystal structures.

For many years, single-crystal x-ray diffraction studies were the gold standard of "what is the geometry of a molecule." Performing gas-phase measurements of bond angles and bond lengths are extremely difficult, since they typically require a molecular beam of the intended target and then very high resolution IR,  rotational spectroscopy, or gas electron diffraction.

I once heard a talk by an extremely talented theorist, who asked the question "is an experimental crystal structure accurate enough?" He showed that in many cases, the answer is "no." (Consider that establishing the position of hydrogen atoms is difficult if not impossible with many single crystals.) The method he was proposing (geometry optimization with CCSD(T) and a basis set limit) isn't really appropriate for large molecules.

Until there's some huge database of gas-phase molecular geometry measurements and CCSD(T) calculations, most methods are compared with crystal structures.


Can you differentiate between Kc and Kp in this problem?

The equation yields Kp.

$$\Delta G^{o}=-RTln(K_p)$$

There is, however, a relationship betweek Kp and Kc.

$$K_p = K_c(RT)^{\Delta n} $$

From the balanced equation there are 3 moles of gas on the right (2+1) and 1 mole of gas on the left. In chem, delta means final minus initial and we arrive at $$\Delta n = 3-1=2$$

and you now can find Kc. Note that you should use 0.08206 L•atm/mol•K for R.

Sources:

Zumdhal example 13.5
https://www.quora.com/In-a-gas-equilibrium-which-K-constant-is-used-in-the-equation-%CE%94G-RTlnK-Kp-or-Kc


What is the true meaning of Gibbs energy and chemical potential?

Your first question:

Before the second law was understood it used to be thought that the maximum amount of work that could be extracted from a reaction was $-\Delta H$ but many experiments showed that this was not the case. It is true that, according to the first law, (the law of conservation of energy),  that external work done must be equal to the loss of energy of the system, unless some heat is given to or taken from the surroundings. This was first understood by Gibbs.  In an isothermal reaction working under  reversible conditions, the heat absorbed from the surroundings is $T\Delta S$, if this is positive then the work done will be even greater than the heat of reaction and so heat will be taken from the thermostat.

Second question:

The $\Delta G$ you mention is actually the gradient of the free energy with extent of reaction $(\partial G/\partial \xi)_{T,p}$ at constant temperature and pressure. This differential is sometimes written as $\Delta G'$. The equation in general is $\Delta G'=\Delta G^\text{o}+RT\ln(Q)$ where $Q$ is the ratio of partial pressures, for example, in a gas phase reaction.  At equilibrium a plot of $G$ vs $\xi$ reaches a minimum and the gradient is zero, i.e. $\Delta G'=0$ then $\Delta G^\text{o}=-RT\ln(K_p)$  where $K_p$ is the equilibrium constant and is used instead of $Q$ at equilibrium. This reaction tells us how the position of equilibrium can be determined in terms of standard state free energies of reactants and products at 1 atm pressure.

(The extent of reaction is zero when only reactants are present and is one when one mole of reactants have changed to product)


Can ferrocene type sandwich compounds be made with fluorene?

Yes there are, many in fact.  One of the examples is having a complex consisting of a fluorene molecule and of cp altogether with iron, described by Decken in Acta Cryst. E (doi: 10.1107/S1600536804028077):

(source)

You are not limited to complexation with iron, though.  Already by the title «Synthesis and characterisation of Ti, Cr, Mo and W bis(fluorene) complexes» by Cloke et al. in J. Organomet. Chem. (source, paywall) you may see that there is a lot of variation possible by either varying the metal, or the second molecule to build the complex.


Hybridization and excitation of electrons, how is it possible?

We don't really consider the excitation of electrons. If it appears like this, then this is probably due to an incorrect understanding of what an orbital actually is: The attempt to describe each single electron by its own wave function which is independent of all the other electrons (which it is not).

An orbital is defined as single electron wave function $\varphi(\vec r_i)$ where $\vec r_i$ are the coordinates of electron $i$. We then try to approximate the many electron wave function as an (anti-symmetrized) product of orbitals

\begin{equation}
   \Psi(\vec r_1, \vec r_2, \dots \vec r_N) \approx \prod_{i=1}^N \varphi(\vec r_i)
\end{equation}

But this is strictly speaking not true, because the electrons are correlated with each other, due to the Coulomb repulsion $V\propto \frac{1}{|r_i-r_j|}$. The product ansatz does not work for the kind of differential equation we need to solve. The Hartree-Fock method tries to do so anyway, and gives therefore wrong (or at least only approximate) result.

This means the whole concept of orbitals is an approximation, which may give a qualitative understanding, but are failing when trying to get quantitative results.

One approach to correct this approximation is expanding the many electron wave function in some basis and solve the resulting eigenvalue problem. This basis is not unique, but one possible choice is to use the Hartree-Fock determinant and the excited configurations you can generate from it. But these excited configurations are more of an abstract mathematical thing and have limited physical meaning, because they are based on the above approximation. The real physics/chemistry is in the solutions of such methods. And those solution cannot be describe by the above ansatz for $\Psi$.


The coldest (the most endothermic) reaction

There is an endothermic nuclear fusion for the elements heavier than iron. On Earth it's most likely a nuclear fission of light elements, like the one occurring in early "dry" thermonuclear bombs, or a fission processes preceding D-T fusion reaction:

$$\ce{^7_3Li + ^1_0n -> ^3_1H + ^4_2He + ^1_0n - \pu{2.467 MeV}}$$

It's probably not the answer you'd expect, but the absolute values of thermal effects of any chemical reaction would be much lower in comparison.

Special thanks to David Richerby for correcting me!


What is the geometry of an alkyl radical?

First off, it was very astute of you to recognize that if the radical formed at carbon 3 (the methylene carbon in the starting compound) is pyramidal, then the radical would be chiral ( the lone radical electron serving as the 4th different substituent on that carbon) and 4 different radicals would be possible.

In a typical carbocation, the carbon bearing the positive charge is $\ce{sp^2}$ hybridized, meaning that the p-orbital at the cationic center is empty and the carbocation is planar.  On the other hand, in a typical carbanion, the negatively charged carbon is pyramidal and approaches $\ce{sp^3}$ hybridization. This $\ce{sp^3}$-like orbital contains 2 electrons.  Therefore, it would be reasonable to expect that a typical hydrocarbon radical, where the orbital holds 1 electron would be somewhere in between these two extremes and be slightly pyramidal.

This is indeed the case and the degree of non-planarity is very sensitive to the substituents attached to the radical center.  In the case of hydrocarbon radicals, such as those created from your starting material, the barrier to inversion from one pyramidal form to the other is extremely low, on the order of a few kcal/mole or so.  Therefore, interconversion between the two enantiomeric forms of the radical would be extremely rapid at room temperature.

So yes, the $\ce{CH_3\dot{C}HC(CH3)3}$ radical is slightly pyramidal and it does transition rapidly between the two pyramidal enantiomers.  Your answer of 4 radicals is correct, as long as you understand that 2 of them rapidly interconvert at room temperature.


Jahn-Teller Distortions in Square Planar Complexes?

Introduction

Thank you for prompting me to look for square planar complexes that are not $\mathrm{d^8}$; I learnt some valuable stuff while researching the answer to this question! Do take this answer with a grain of salt, because I cannot access the paper I’m referencing here on holidays (I could when I’m back at work) but the supplementaries confirm what I’m trying to prove.

Dataset

Scott et al. reported the crystal structure of a square planar vanadium(II) complex in 1990. The central vanadium ion is surrounded by the oxygens of four substituted phenolates. A set of two oxygens is also coordinating a lithium ion to give an approximate trigonal planar geometry on the oxygens, and said lithium ions are further coordinated by a THF solvent molecule (again giving a trigonal planar geometry around lithium) such that there is an approximately straight line going through the following atoms: THF-oxygen, lithium, vanadium, lithium, THF-oxygen. I shall define this as the $x$ axis for further discussion; the square planar geometry being in the $x,y$ plane and $z$ being vertical.

Experimental section

I took the atom coordinates of vanadium, the six oxygens, the four oxygen-bound phenol carbons and the two lithii from the supplementary information. My first step was to move the coordinate system’s origin into the vanadium ion. Then I transformed into spherical coordinates and adjusted the axes so that when retransforming back into cartesian coordinates my definitions from above are valid.

Results

The vanadium-coordinating oxygens are slightly displaced from the $x,y$ plane — in sets of two such that one is about as much above the plane as the other is below the plane. This distortion carries through onto their corresponding carbons. The lithium ions are then again almost spot on $z=0$ — also coordinated by one oxygen with positive $z$ and one with negative $z$.

The oxygens do not form a perfect square. Two oxygens are slightly displaced outwards in $x$ direction; two in $y$ direction. The $y$ displacement is almost twice as strong as the $x$ displacement.

Discussion

The distortion of the square planar geometry resembles the one you predicted: Two opposite oxygens above the plane and two below the plane. This is coherent with a reported $\angle(\ce{O-V-O}) = 162^\circ$ (diagonally across the square).

The authors also reported a large $\angle(\ce{O-V-O}) = 103^\circ$ and a small $\angle(\ce{O-V-O}) = 79^\circ$ all between two neighbouring oxygens and a set of longer and shorter $\ce{V-O}$ bonds differing by $2~\mathrm{pm}$ all in accordance with the slight displacement in the crystal coordinates.

Seen as a whole, the entire thing is indeed en route towards tetrahedral, but also features additional slight distortion that further reduces symmetry from a $D$ point group to a $C$ point group.

Conclusion

The $\mathrm{d^3}$ ion vanadium(II) features a distorted square planar geometry. The distortion seems strong enough to consider a point group of $C_\mathrm{2v}$ around vanadium rather than $D_\mathrm{2d}$.

Reference

M. J. Scott, W. C. A. Willish, W. H. Armstrong, J. Am. Chem. Soc. 1990, 112, 2429. DOI: 10.1021/ja00162a059.


Rule of Thumb for separatory funnel volumes

If a general rule must be given than the volume of the separatory funnel is mostly dictated by that of the liquid to be treated.

Indeed is better to extract the product by repeating several steps each using small portions of extracting solvent. This ensures a more efficient extraction based on repartition, I.e. a more quantitative recovery of the product as well as a more convenient work-up (cost-effective, less time consuming, less waste).

I am not considering difficulties due to a slow phase separation which might prevent the use of a narrow funnel but in general use the smallest that allows an easy shaking and inter mixing. (When you get to know the actual separation you might even prefer a  specific funnel shape not just that V).

Example: 200 ml to be treated with three to four 25 or 50 ml portions : 500 ml funnel.


How do I solve this Ionic Equilibrium problem?

Chemistry
AgNO3 is a strong electrolyte it dissociates completely.
$\ce{AgNO3 -> Ag+ + NO3-}$
If it was just HCN in solution, then most of the $\ce{CN^-}$ species would be as HCN. But since $\ce{AgCN}$ precipitates, more HCN dissociates. So essentially the overall reaction is:
$\ce{HCN + Ag+ -> AgCN}\,\ce{v + H+}$
The $K_{sp}$ of $\ce{AgCN}$ is such that it is reasonable to assume that for all practical purposes that the precipitation is essentially complete.
Assumptions
(1) I assume that the correct answer should have two significant figures which are all that the equilibrium constants have. However I will carry 4 significant figures in the intermediate calculations to try to avoid significant rounding errors.
(2) $\ce{[H^+]} \approx 0.01000$ molar and $\approx 0.01000\ \mathrm{moles}\ \ce{AgCN}\ \text{ppt per liter of the mixture}$ (In other words about 1 ppt error for both)
Mass balance
If we consider 1 liter of the mixture of solution then we also have the following two mass balance equations:

The volume doesn't matter. Thinking of it as 1 liter just makes it simpler to get to the final result.

$0.01000 = \ce{[CN-] + [HCN]} +\ \mathrm{moles\ \ce{AgCN}\ce{v}\tag{1}}$
$0.01000 = \ce{[Ag+] +\ \mathrm{moles}\ \ce{AgCN}\ce{v}\tag{2}}$
$\therefore \ce{[Ag+] = [CN-] + [HCN]}\tag{3} $
Equilibrium equations
$K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}= \dfrac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]}\tag{4}$
$K_{sp}(\ce{AgCN})=\pu{2.2\times10^{-16}} = [\ce{Ag+}][\ce{CN-}]\tag{5}$
dividing (4) by (5) we can get:
$2.819\times10^6 = \dfrac{\ce{[H+]}}{\ce{[Ag+][HCN]}}\tag{6}$
Rearranging (6) and substituting $\ce{[H+] = 0.01000\ \mathrm{molar}}$
$\ce{[HCN] =}\dfrac{3.547\times10^{-9}  }{\ce{[Ag+]}}\tag{7}$
Rearranging (5) we get:
$\ce{[CN-]} = \dfrac{\pu{2.2\times10^{-16}}}{\ce{[Ag+]}}\tag{8}$
Substituting (7) and (8) into (3) we get:
$\ce{[Ag+] =} \dfrac{\pu{2.2\times10^{-16}}}{\ce{[Ag+]}} + \dfrac{3.547\times10^{-9} }{\ce{[Ag+]}}\tag{9} $
Solving:
$\ce{[Ag+]^2} \approx 3.547\times10^{-9}\tag{10} $
$\ce{[Ag+] =} 5.955 \times10^{-5}\tag{11}$
Rounding to two places... $\ce{[Ag+] =} 6.0 \times10^{-5}$
Solving for other species
Substituting for $\ce{[Ag+]}$ in (8) gives
$\ce{[CN-]} = \dfrac{\pu{2.2\times10^{-16}}}{5.955 \times10^{-5}} = 3.694\times 10^{-12}\tag{12}$
Substituting for $\ce{[Ag+]}$ in (7) gives
$\ce{[HCN] =}\dfrac{3.547\times10^{-9}   }{5.955 \times10^{-5}} = 5.956\times10^{-5}\tag{13}$
Argentocyanide Complex
The problem has ignored the argentocyanide complex. Is that reasonable?
$K = 5.6\times10^{18} =\dfrac{\ce{[Ag(CN)2^-]}}{\ce{[Ag+][CN-]^2}}\tag{15}$
rearranging
$\ce{[Ag(CN)2^-]} = \ce{[Ag+][CN-]^2}\times(5.6\times10^{18})=4.794\times10^{-9}\tag{16}$
So it is reasonable to ignore the additional complication.
Checking

You always check your assumptions on this sort of problem. It is easy to go down the wrong rabbit hole.

with $K_\mathrm{a}\ce{(HCN)}$
$\pu{6.2\times10^{-10}} = \dfrac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]} = \dfrac{(0.01)(3.694\times 10^{-12})}{5.956\times10^{-5}} = 6.202\times10^{-10}\tag{14}$
mass balance
It makes sense chemically that in the acidic solution $\ce{[HCN]} \gg \ce{[CN-]}$ so that $\ce{[Ag+]} \approx \ce{[HCN]}$. Making that assumption earlier using equation (6) would have reduced the work for this problem.
[H+] = 0.01000 molar and 0.01000 moles AgCN ppt per liter of the mixture
Also assuming that $\ce{[H^+] = 0.01000\ \mathrm{molar}}$ is certainly correct to two significant figures. However I tried to carry 4 significant figures in the intermediate calculations to avoid rounding errors. The assumption isn't that good.
!! PROBLEM !!
The assumptions that: $\quad\quad\ce{[H^+]} \approx 0.01000$ molar  $\quad\quad\approx 0.01000\ \mathrm{moles}\ \ce{AgCN}\ \text{ppt per liter of the mixture}$
are not good to 4 significant figures. It would have been better to use:
$\quad\quad\ce{[Ag+]} \approx \ce{[HCN]}$
$\quad\quad\ce{[H^+] \approx 0.01000 - [HCN] \approx 0.010000 -[Ag+]}$
and get a quadratic equation when simplifying equation (6). Then $\ce{[Ag+]} = 5.938\times10^{-5}$ and $\ce{[H+]} = 9.941\times10^{-3}$.
So the value $\ce{[Ag+]} = 5.956\times10^{-5}$ is off by +0.303%, or 3 ppt, which I think is acceptable, but it is marginal.


Copper plating, is sulphuric acid necessary? Alternative to H2SO4?

Copper acetate can be used, at least for a thin layer, and it can be made with chemicals found at home.

To get a bright copper coat, extra acid (vinegar, 5% acetic acid, in this case) or gelatin can be added.

That said, if the plating is for through-hole continuity, it might be simpler and cheaper to use a solderable conductive ink.


Does the first step in electrophilic aromatic substitution follow equilibrium? If so, can equilibrium constant be calculated?

Let me put it this way: Every compound exist in a 3N dimensional configuration space (may be simple XYZ coordinates of the system, may be internal coordinates, based on your choice), and based on the configuration, its potential energy is defined. The potential energy surface (PES) can be calculated quantum mechanically or using some force field classically. The probability of the system to be found in the given set of coordinates is given by the distribution function defined according to statistical mechanics. For a canonical ensemble, the formula is given as (Z is the partition function):

$$P(\vec{r}_{3N})= \frac{1}{Z}e^{-E(\vec{r}_{3N})/k_BT}$$

Now imagine this, a reaction between compounds containing M and N atoms is actually an exploration of the potential energy surface in 3(M+N) dimensions. Reactants, product and intermediates are given by local minima in the PES while transition state is given by maxima along one dimension and minima along all other dimensions. In other words, for a stable state (reactant, product, intermediate, etc.) all the frequencies (square root of the second derivate of the PES) is real, while for transition state one of the frequencies is imaginary.

Now coming to your question: You will require to calculate the energetics of each of the step of reaction considering the presence of solvent (this is probably happening in presence of solvent), plug in your values to the appropriate statistical ensemble (if it is inside solvent without release of gaseous products, canonical ensemble should be good enough) probability function and you will get an idea about the percentage of intermediates you get.

In most high school textbooks, the reaction marked with double arrows simple means the reactant, product (and in your case the intermediate) is not too much different in terms of potential energy. For a single headed arrow, it means that the difference between potential energies of the minimas (reactant and product) is significant and most of the compounds (in an ensemble) will be trapped in the lower minima.

In case you want to read more about it, I suggest the following books:

Statistical Mechanics by McQuarrie
Physical Chemistry by Atkins
Physical Chemistry by Levine

For more details about reaction dynamics, you may refer to:

Molecular Reaction Dynamics by R. D. Levine

Anyway, please reply in comments if you have more doubts, will be happy to help.

Greetings from India. जय हिन्द.  Jai Hind.


What is the procedure to create stock solutions?

Since it contains agar, you wouldn't be able to make a stock solution with the agar in it; but you can make a stock solution of the other items and add the agar when you need the working solution. Also, since there is a fair amount of sugar in the recipe, it would be best to add that to the working solution as well.

You should be able to make a 100 times concentrate (leaving out the sugar and agar) and then sterilize it in an autoclave. When you need to make a working solution, weigh 5 g of agar and 20 g of sugar, add 10 mL of stock solution, and about 900 mL of water. Boil the solution to dissolve the agar, cool (to about 50°C) and add water (before it sets) to get 1000 mL of solution.


Why would hydrogen not be generated at the cathode when electrolysing zinc sulfate?

In an electrolytic cell we are using the electrical energy from an outside power source to drive a chemical system against the direction it would spontaneously move.  In this case, we can take a solution of zinc ions and force them to become zinc metal.  As long as we put the zinc ions in an environment where they can be reduced, they will be.

Consider the half reaction that occurs when zinc metal is produced:
$\ce{Zn^{2+}(aq) + 2e- -> Zn(s)}$

This is a reduction reaction, and reductions always occur at the cathode, whether the cell is galvanic or electrolytic.

The other half reaction involves hydroxide being converted into water and oxygen.
$\ce{4 OH- -> 2 H2O + O2 + 4e-}$

This is an oxidation reaction, and oxidations always occur at the anode.

Platinum and carbon are nonreactive in this system, so they can be used for either electrode.  Zinc can be used as a cathode because that's where it is being deposited, so we're just adding additional metallic zinc onto our already existing zinc electrode.

With the idea that oxidation occurs at the anode in mind, what would happen if we used a zinc anode?


Why is the segment of the IR spectrum below 1500 wavenumbers called the fingerprint region?

I guess the naming "fingerprint region" results from the fact that while I can immediately identify amine, alcohol, carbonyl groups, double and triple bonds and so on in the "upper" and "middle" part of the IR spectrum, the peaks observed in the fingerprint region are often arising from vibrations that are very specific to the very compound you are looking at, thus a "fingerprint" of your compound.


Your professor obviously wants you to consider the $\ce{C-C}$ bonds of the $\ce{C=C}$ double bond plus any $\ce{C-H}$ bond to either of the alkene carbons.

The first step in double bond oxidation by permanganate will always be replacing a $\ce{C=C}$ by $\ce{C=O + O=C}$. If you consider the double bond two bonds, then for each of those bonds each carbon atom gets one $\ce{C-O}$ bond to again give a double bond.

After we have generated the inital carbonyl, we can turn our attention to the other bonds the former alkene carbon has. Remember that aldehydes are easily oxidised to carboxylic acids while ketones are rather resistant to oxidation. The difference is the presence of a $\ce{C-H}$ bond in the aldehyde; permanganate can oxidise this to another $\ce{C-O(-H)}$ bond. This is the second part of your rule.

If accidentally the carboxylic acid species generated is formic acid, which has another $\ce{C-H}$ bond, this can again be oxidised to give carbonic acid. Since carbon’s oxidation state is maximal in $\ce{H2CO3}$, this is truly the end of the oxidation.


Raoult's law & Dalton's law of partial pressures

In Raoult's Law, $p$ is the partial pressure of the vapor in equilibrium with some liquid which has vapor pressure of $P$. $x$ is the molar fraction of this liquid (it can be mixed with other liquid or solid.


Does a difference in IR signal size from 'normal' provide information on amount of material?

No. Frankly, in organic chemistry, IR is usually just used as support for what functional groups you think should be on your molecule. I only obtained IR spectra because you need to include them with supporting information when you want your paper reviewed.


Yes there are little cooling boxes, for camping, for example, where one of the walls (partially) is made of a sheet of a food-safe-metal, back-to-back to one side of a Peltier element as a cooling device.  This works fine, for example if attached to the car's battery.  Less efficient (by rate of heat transfer) may be an separation of the Peltier element viz. the inner of the cooling box by glass or food-safe plastic.
Of course, a complete immersion of the Peltier device into a cooler or cooling system won't work, as necessarily one of the flat sides of the element is going to become cool, but the other going to be warm.


What is the group number or name for elements between group 3 and 4 (F-block) on the periodic table?

I suppose IUPAC hasn't bothered with group numbers for the lanthanides/actinides quite yet because there would only be two elements for each group, and because there are few vertical similarities among the f-block elements. The entire row of lanthanides behaves very similarly due to the 4f orbitals being anomalously compact and not being available for bonding or ionization. The first half of the actinides is quite different to the lanthanides, as the 5f orbitals there can actually participate in chemical interactions, though the second half becomes more similar as the 5f orbitals are once again strongly trapped to the atom, this time by the very high effective nuclear charge.


Why do triple bonds have a low NMR value?

Chemical shifts arise due to differences in the local magnetic field in the different environments of a molecule. There are a number of contributing factors including the local diamagnetic contribution which is driven by the shielding effect of local electrons. Where there is functionality which generates electron circulation, a localised anisotropic magnetic field is created, and these can cause both downfield and upfield shifts, based on the position of the proton (nucleus) with respect to the functional group. These interactions, unsurprisingly, are called magnetic anisotropic effects.

The best example of this is the ring current effect arising from aromatic rings, with protons on the outside of an aromatic ring experiencing large downfield shifts, while protons either above the ring, or on the inside of the ring experiencing upfield shifts.

With double and triple bonds, the shielding cones are different in shape, and traditionally are drawn as shown below (you might find some better diagrams with a web search). Areas marked (+) increase the shielding (leading to decreased chemical shift) and areas marked (-) decrease the shielding (increasing chemical shift).

However, it is important to recognise that chemical shifts are not only influenced predominantly by electronics; sterics have a very significant contribution to chemical shifts also, often more so than electronics. A very good (recent) paper (Baranac-Stojanovic, M RSC Adv., 2014,4, 308-321) discusses traditional magnetic anisotropy models, and presents some very good examples that highlight the effects of steric influence on chemical shifts.


How to bond polymer to polymer (TPU on PMMA)

Sorry to say but you question is a bit vague because without knowing the project, it's pretty hard to help. Anyway, here is some general information.
Some of the sophisticated methods of joining these together include solvent bonding, vibration welding, friction welding and induction welding or pressure sensitive adhesives [ (Meth)acrylic copolymers are one of the most widely used polymer classes for the production of pressure-sensitive adhesives]. Acrylic requires an acrylic solvent adhesive. Cyanoacrylate (super glue) may work, depending on the project. UHU glue is very useful with all kinds of plastics, so you might wanna try that.  I myself have tried superglue and it works with both materials. But I think for your project, non adhesive methods woll be an optimal choice.
All the best.


According to Armarego, et al. (Purification of Laboratory Chemicals, 6th edition), they only provide a procedure to purify a solution of $\ce{KOH}$, likely owing to its very high solubility.  Further, the only impurity of note they cite is carbonate (from standing in contact with air), so if this is not your impurity then this procedure will not help.

If this is your impurity, they recommend adding a saturated solution of a barium salt, either $\ce{BaCl}$ or $\ce{Ba(OH)2}$, shaking, and then setting aside to allow the insoluble $\ce{BaCO3}$ to settle.  It sounds like you are working on large scale, so it is important to recall that the procedure just described should be done in several small batches to work best.

Not certain this is what you are looking for, but I hope it's a start.


The methoxy group actually interacts with the aromatic ring via two competing mechanisms.

It is indeed electron withdrawing via the sigma bond interaction between the O and C of the aromatic ring. However the lone pair on the oxygen conjugates through the delocalised pi network and increases the electron density at both the ortho and para positions. If you draw out the conjugation pathway you will see the electrons can sit on the ortho and para carbons. This is the more significant contributor to the overall electronics and, therefore, chemical shift. So, the protons ortho to methoxy are far more shielded than those ortho to the weakly electron donating methyl group.

This effect can be generalised across all aromatic directing groups, and the size of chemical shift difference from benzene (7.3ppm) is directly correlated to the strength of the directing group. So, aniline has ortho chemical shift of about 6.5ppm (-NH2 is a strong EDG), while the ortho protons in nitro benzene are up at about 8.5ppm (-NO2 is a strong EWG).


Why is it that the least reactive metals are the best electrical conductors?

One of the reasons why non reactive metals are good conductors is that they are good at staying as metals. Most metals react with the atmosphere to form oxides. And the majority of oxides are insulators or semiconductors. Of course there are few exceptions to this rule.

Also, just a note: calcium and iron have better conductivities than platinum.


Physical significance of magnitude of ΔG (change in Gibbs free energy) of a reaction

It is only a rule of thumb that a negative value for $\Delta G^0$ of a reaction means that a reaction is spontaneous.  If you start out with pure reactants (and no products present), all reactions are spontaneous, irrespective of the $\Delta G^0$.  A negative value for $\Delta G^0$ only signifies that the equilibrium constant is greater than 1.


What helps a sliced fruit or vegetables oxidize slower?

You can cover the cut pieces with aluminium foil which acts as sacrificial element and prevents oxidation of food.


What is the difference between chiral molecule and chiral carbon?

In addition to the excellent Mithoron link, heed Lord Kelvin.


First off, take a look at the Wikipedia article: Synthetic diamond. Basically, there are two ways to artificially generate the heat and pressure needed. First is the "conventional" high-temperature, high-pressure method, which uses a press to put a sample of pure carbon graphite under pressure of about 5 gigapascals (1GPa = about 10,000 atmospheres or about 145,000 psi) and about $\pu{1500 ^\circ C}$. This combination encourages the flat hexagonal lattice structure of the graphite to morph into the trihexagonal 3-D lattice of diamond. Second is the chemical vapor deposition method, in which a substrate of diamond crystal is exposed to sublimated carbon plasma (temperatures exceeding $\pu{3000 ^\circ C}$) in a hard vacuum (less than .0001 torr). The sublimated carbon attaches to the crystal substrate, "growing" it.

Both of these can be used to create so-called "memorial diamonds" from the cremated remains of a human. To do that, first the ash is chemically filtered to separate the carbon from "pollutants" naturally occurring in the body that don't combust into gaseous substances during cremation, such as calcium, phosphorous, iron etc. Then, this "amorphous carbon", much like any other carbon, can be used in HTHP or CVD processes, along with additional carbon as may be required to grow the diamond to the proper size. Usually, the HTHP process is used, as it makes the best use of the available carbon (CVD requires sublimating a lot of carbon, more than would be left behind in the ash, and most of which ends up on the chamber walls and not the substrate), and requires less energy (it's easier to maintain mechanical pressure once applied than to maintain temperature).


Most of the odor of cloves comes from volatile molecules, especially eugenol, found in the cloves.  Eugenol is a water-immiscible, oily liquid at ambient temperatures.

Plastic with resin identifiction code #6 is polystyrene.

Polystyrene is not compatible with most organic solvents, because it tends to be softened, swelled, or even completely dissolved by them.

Thus I suspect that the problem is that oil in the cloves is attacking the plastic that you use.  You could avoid the problem (mostly) by using a plastic with resin code #2 or #5 (i.e. high-density polyethylene or polypropylene) instead of #6.  But, you are probably right that glass would be better than any plastic for storing your cloves.


How do I determine the equilibrium concentration given Kc and the concentrations of component gases?

For a reaction in the form $\ce{ \mathit{a}\,A + \mathit{b}\,B <=> \mathit{c}\,C}$, the quilibrium constant $K_c$ is given by

$$K_c = \frac{[C]^c}{[A]^a[B]^b}$$

You might want to check yours.


How to find volume needed from a stock solution

This is calculated by equating the moles of the two solutions.

You are using $\pu{0.6402g}$ of sodium fumarate in $\pu{100mL}$ of solution.

$$ \text{No. of moles} = \frac{\text{Given weight}}{\text{Molecular weight}} = \frac{0.6402}{160.04} = \pu{0.0040 moles} $$

$$ \text{The concentration of the solution can be found easily} = \frac{0.0040}{0.1} = \pu{0.04 M}$$
If you require a solution of concentration $40$mM, the stock solution that you have prepared is exactly that.
If you need a more diluted solution, you may use the formula:
$$ M_1 V_1 = M_2V_2$$
where, $M_1$ and $M_2$ are the initial and final concentrations, $V_1$ is the volume of stock you would require for the dilution, and $V_2$ is the volume of final desired concentration.
Let us assume you require $\pu{50 mL}$ of $\pu{10 mM}$ solution, and you have a stock of $\pu{100 mL}$ of $\pu{40 mM}$.
Then
$$
M_1 = \pu{40 mM}\\
V_1 = ? \\
M_2 = \pu{10 mM} \\
V_2 = \pu{50 mL}
$$
Plugging in the values, you will find out that you require $\pu{12.5 mL}$ of stock, and dilute it to $\pu{50mL}$ using the solvent, i.e., add $\pu{37.5 mL}$ of water to $\pu{12.5 mL}$ of stock solution.


Compute minimum energy paths from arbitrary positions on the potential energy surface

There are some possibilities here:

The negative of the gradient of a function always points to the direction of optimal descent (for a discussion, see here).
A method that simply follows the gradient is called first order.
As permeakra pointed out, extrapolation techniques are commonly used for optimizations.
This is because we take finite steps in the PES and this introduces the risk of "jumping" over the minima.
Better approximations are either second order in nature (e.g., the Hessian matrix is calculated in each step) or quasi-second order (the Hessian matrix is approximated or approximately updated).

In order to simply follow the gradient you want the steepest descent-like, first order method with probably a very small maximum step and this can be accomplished by setting the Hessian or the Hessian approximation equal to the identity matrix.
In ORCA (I have never used Molcas, sorry) for example you may use (source):

! Opt
%geom
  inhess unit
  update noupdate
  step qn
end

permeakra also cited the Nudged Elastic Band, which is used for finding minimum energy paths connecting two structures.
This might be very useful and is available for instance in VASP and in NWChem (free).
You may also want to take a direction from one Hessian eigenvector.
This is commonly associated with a molecular vibrational degree of freedom.
This may be accomplished with NWChem.

If you just want to go down the PES from an arbitrary geometry without any jumping, solution 1. is probably the one you want.