You're right that Cu0 has one unpaired electron in the 4s orbital. If you were to observe a single Cu0 atom in the gas phase, you would observe some paramagnetic behavior.

However, in metallic copper, the valence orbitals are engaged in metal-to-metal bonding. The 3d and 4s orbitals interact in constructive overlap with neighboring copper atoms to produce bonding molecular orbitals.

When two 4s orbitals (each containing 1 unpaired electron) combine to form a sigma bonding MO, the electrons in that bonding MO become paired, per the image below.

Note: this MO diagram depicts the bonding in H2, but I'm just using it to illustrate the principle of partially filled atomic orbitals coming together to form a filled bonding MO

The process of deriving the MO diagram for a metallic solid is obviously more complicated than this, but the answer to your question is that a single Cu0 atom is paramagnetic, but Cu2 would be diamagnetic. By extension, metallic copper is effectively Cu$_\infty$ and also diamagnetic.

As you can tell, the process for determining whether an atom or ion is para/diamagnetic is a very different question than whether the bulk substance (metal) is para/diamagnetic. You can't do the latter by electron counting alone; you need to know more about bonding in the substance. As another example of this, you mention that lead is diamagnetic. Lead is another example of a paramagnetic atom (6p2; the 6p orbital contains two unpaired electrons) - but lead metal is diamagnetic because those electrons become paired in the bonding MOs.


Why are alkyl halides more soluble in alcohol water solution rather than pure water?

As others have mentioned, it generally comes down to "like dissolves like".  Water is highly polar, low molecular weight alcohols like ethanol a bit less so.  By increasing the ratio of alcohol to water in a solution you are effectively making the solution less polar as a whole.  There are effectively fewer hydrogen-bonding opportunities in an alcohol water solution than in pure water.  In general, alkyl halides tend to fall in the low-to-intermediate polarity range.  Methylene chloride for example is only very slightly soluble in water, which makes it an excellent candidate for performing extractions.  By adding alcohol to water, you are giving the solution as a whole a more non-polar character, and increasing the potential for favorable interactions between the alkyl halide and the solvent.


Equilibrium for the synthesis of methanol

Note that under the given conditions, methanol cannot exist in the liquid state, however since you are given in the problem that methanol is obtained in liquid state, you will only consider the equilibrium (for problem solving purpose)
$$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)}$$
And since $K_p$ is related to $K_c$ by the relation
$$K_p = K_c(RT)^{\Delta n_\mathrm{g}},$$ where $\Delta n_\mathrm{g}$ represents the change in the number of moles of gaseous species, so here $\Delta n_\mathrm{g} = -3,$ Therefore $K_\mathrm{c}=K_\mathrm{p}(RT)^3=9.28\times 10^{-3}\times (300 \times 0.0821)^3 \approx \pu{138.65 atm^3 L^3 mol^-3}$.


Sodium hypochlorite (the active ingredient in bleach) and acetone react to produce chloroform.

Chloroform can slowly decompose into phosgene in air with exposure to oxygen
$$\ce{2CHCl3 + O2 ->[h\nu] 2COCl2 + 2HCl}$$

but not in significant quantities under the conditions you describe.

You did the right thing by opening the window for ventilation and isolating the fumes to the bathroom and rinsing with copious water.

It is best not to mix bleach with anything except water and especially not household cleaning products.


Manipulate melting point of frozen water

One way of doing so would be to significantly increase the pressure. As you can see from this phase diagram, at a pressure of 1 GPa, water ice will remain frozen at temperatures up to about 30°C.


How does one actually get the energy from the Kohn–Sham equations?

A common mistake is the thought that the total energy is the sum of all orbital energies $\{\epsilon_i\}$.

From Step #6 of Daniel Crawford's SCF programming project (modified slightly in some places):

  The SCF electronic energy may be computed using the density matrix as:

  $$
E_{\text{elec}} = \sum_{\mu\nu}^{\text{AO}} D_{\mu\nu} (H_{\mu\nu}^{\text{core}} + F_{\mu\nu})
$$

  The total energy is the sum of the electronic energy and the nuclear repulsion energy:

  $$
E_{\text{total}} = E_{\text{elec}} + E_{\text{nuc}},
$$

where the density matrix is defined as (Step #8)

  $$
D_{\mu\nu} = \sum_{m}^{\text{occ. MO}} C_{\mu m} C_{\nu m},
$$

the Fock matrix as (Step #7)

  $$
\begin{align}
F_{\mu\nu} &= H_{\mu\nu}^{\text{core}} + \sum_{\lambda\sigma}^{\text{AO}} D_{\lambda\sigma} \left[ 2(\mu\nu|\lambda\sigma) - (\mu\lambda|\nu\sigma) \right] \\
&= H_{\mu\nu}^{\text{core}} + 2 J_{\mu\nu} - K_{\mu\nu}
,
\end{align}
$$

and the core Hamiltonian as (Step #2)

  $$
H_{\mu\nu}^{\text{core}} = T_{\mu\nu} + V_{\mu\nu}.
$$

I've also introduced the definitions of the Coulomb matrix $J$ and the exchange matrix $K$:

$$
\begin{align}
J_{\mu\nu} &= \sum_{\lambda\sigma}^{\text{AO}} D_{\lambda\sigma} (\mu\nu|\lambda\sigma) \\
K_{\mu\nu} &= \sum_{\lambda\sigma}^{\text{AO}} D_{\lambda\sigma} (\mu\lambda|\nu\sigma) \\
\end{align}
$$

Now, identify each of the terms in the Kohn-Sham equations with the terms from above.

$$
\begin{align}
\hat{T}_{e} &= -\frac{1}{2} \nabla^2 \rightarrow T_{\mu\nu} = \left< \chi_{\mu} \left| \hat{T} \right| \chi_{\nu} \right> \\
\hat{V}_{eN}(\vec{r}) &= \sum_{A}^{\text{nuclei}} \frac{Z_A}{|\vec{r} - \vec{R}_{A}|} \rightarrow V_{\mu\nu} = \left< \chi_{\mu} \left| \hat{V}_{eN} \right| \chi_{\nu} \right> \\
\hat{V}_{ee}(\vec{r}) &\stackrel{?}{\rightarrow} 2 \hat{J} \\
\hat{V}_{\text{XC}}(\vec{r}) &\stackrel{?}{\rightarrow} - \hat{K}
\end{align}
$$

This last part isn't quite correct though. Usually, when looking at the Kohn-Sham equations, one replaces the full electron-electron interaction $\hat{V}_{ee}$ with the sum of the Hartree potential $\hat{V}_{H}$, which gives the Coulomb energy, and the exchange-correlation potential $\hat{V}_{\text{XC}}$, which replaces the exact exchange $\hat{K}$ with a (currently approximate) expression for both the exchange term and the true electron-electron (correlated) interaction.

In terms of how the energy is actually calculated, all quantities from above are the same as in Hartree-Fock theory, except the calculation of the exact exchange integrals during the Fock build is replaced with calculating the exchange-correlation matrix $F^{\text{XC}}$, leading to

  $$
\begin{align}
F_{\mu\nu}^{\alpha} &= H_{\mu\nu}^{\text{core}} + J_{\mu\nu} + F_{\mu\nu}^{\text{XC}\alpha} \\
F_{\mu\nu}^{\beta} &= H_{\mu\nu}^{\text{core}} + J_{\mu\nu} + F_{\mu\nu}^{\text{XC}\beta}
\end{align}
$$

For a density functional approximation (DFA) based on the generalized gradient approximation (GGA), where the functional is dependent on both the density $\rho(\mathbf{r})$ and its gradient $\nabla \rho(\mathbf{r})$,

  $$
\begin{align}
E_{\text{XC}} &= \int f_{GGA}^{\text{DFA}}(\rho_{\alpha},\rho_{\beta},\gamma_{\alpha\alpha},\gamma_{\alpha\beta},\gamma_{\beta\beta}) \, \mathrm{d}\mathbf{r} \\
\gamma_{\alpha\alpha} &= |\nabla \rho_{\alpha}|^{2} \\
\gamma_{\beta\beta} &= |\nabla \rho_{\beta}|^{2} \\
\gamma_{\alpha\beta} &= \nabla \rho_{\alpha} \cdot \nabla \rho_{\beta} \\
\end{align}
$$

  The exchange-correlation parts of the Fock matrices are given by

  $$
F_{\mu\nu}^{\text{XC}\alpha} = \int \left[ \frac{\partial f}{\partial \rho_{\alpha}} \chi_{\mu}\chi_{\nu} + \left( 2\frac{\partial f}{\partial \gamma_{\alpha\alpha}} \nabla\rho_{\alpha} + \frac{\partial f}{\partial \gamma_{\alpha\beta}} \nabla\rho_{\beta} \right) \cdot \nabla(\chi_{\mu}\chi_{\nu}) \right] \mathrm{d}\mathbf{r}
$$

$f^{\text{DFA}}$, $\frac{\partial f^{\text{DFA}}}{\partial \rho}$, and $\frac{f^{\text{DFA}}}{\partial \gamma}$ are unique closed-form expressions for each DFA, and are usually evaluated numerically on an atom-centered grid (ACG) such as a Lebedev grid. This generally requires mapping the set of AOs/basis functions $\{\chi\}$ onto this grid.

References

John A. Pople; Peter M. W. Gill; Benny G. Johnson; Kohn–Sham density-functional theory within a finite basis set. Chem. Phys. Lett. 1992, 199 557-560

$\tiny{\text{As usual, sorry if I'm lazy with notation, being consistent is so difficult...}}$


Let's start with $\ce{HXO_3}$.

In $\ce{HXO_3}$, after removal of proton, an $\ce{XO_3-}$ ion is formed which is resonance stabilised. For example, if X was $\ce{Cl}$,

Now, this shows that the $X$ needs to have a tendency to pull the electrons towards itself which decreases in the order :$\ce{Cl>Br>I}$. So more resonance is present in $\ce{ClO_3^-}$ than in $\ce{BrO_3^-}$ making the conjugate base of $\ce{HClO_3}$ more stable. Hence it is more acidic.
So acidic strength decreases as: $\ce{HClO_3>HBrO_3>HIO_3}$

Now let's compare $\ce{HXO_3}$ and $\ce{HXO_2}$. You should now know that the acidic nature depends on the stability of the resulting conjugate base. Here, the number of resonance structure for $\ce{HXO_3}$ is more. Hence, the acid strength decreases in the order:   $\ce{HClO_3>HClO_2>HClO}$. So its actually the number of resonance structures that affects the acidic strength.

Finally, let's come to hydrides of 16th group. Let me represent any 16th group element as $Y$. Now, as size of $Y$ increases, the $\ce{H-Y}$ bond length increases. This means, the $\ce{H-Y}$ bond becomes weaker and removal of $\ce{H+}$ ion becomes easier. The size of atom decreases in the order :$\ce{O>S>SE>Te}$.

Hence, the acidic strength decreases as : $\ce{H_2Te>H_2Se>H_2S>H_2O}$

This shows that electronegative is not the only reason. For example, if you considered $\ce{HF}$ and $\ce{HCl}$, then you may think that $\ce{HF}$ is more acidic because $\ce{F}$ is more electronegative and hence will be happy to get a negative charge. But, charge on a small atom is unfavourable. As atomic size increases, the charge can spread over a larger area. Hence, $\ce{Cl-}$ is more stable.

The same trend is seen in case of hydrides of 16th group. The charge in $\ce{Se-}$ is spread over a larger area. Hence, its more stable.


How are the number of tetrahedral voids twice the number of octahedral voids in a CCP structure?

To be precise, if in a close packed structure (ccp or fcc) there are $n$ atoms or ions then the number of octahedral voids and tetrahedral voids will be $n$ and $2n$ respectively.

For example, there are 8 tetrahedral voids per unit cell of fcc structure $(Z_{\text{eff}}=4)$. If you divide the FCC unit cell into 8 small cubes then each small cube has 1 tetrahedral void located at its own body center. Thus, total number of TVs in a unit cell $= 8 = 2  \times  Z_\text{eff}$

Now, again consider a CCP or FCC unit cell. The body centre of the cube is not occupied but it is surrounded by 6 atoms (4 in same plane, 1 above and 1 below). On joining these face centers, an octahedral void is formed. Thus, number of OV at body centre of the cube is 1.

Besides the body center, there is one OV at the center of each of the 12 edges. It is surrounded by 6 atoms, three belonging to the same cell unit (two on the corners and once on face center), and the three belonging to two adjacent unit cells.

Since, each edge of the cube is shared between four adjacent unit cells only one fourth of each void belongs to a particular unit.

Therefore, OV at body centre = 1

OV at the 12 edges of the cube and shared between four unit cells  $= 12\times\frac{1}{4}=3$

Total number of OVs $= 1 + 3 = 4 = Z_\text{eff}$


Why does the two-electron exchange integral in Hartree-Fock theory have positive values?

The conclusion seems obvious but I just cannot find a proof.

No, I don't think it is obvious. At least I could not think about a simple algebraic proof that the exchange integral $\langle ab \vert ba \rangle$ defined as follows,
$$
\langle ab \vert ba \rangle :=
\iint \overline{\psi}_a(\vec{r}_{1}) \overline{\psi}_b(\vec{r}_{2}) r_{12}^{-1} \psi_b(\vec{r}_{1}) \psi_a(\vec{r}_{2}) \mathrm{d}\vec{r}_{1} \mathrm{d}\vec{r}_{2}
$$
is always positive. But I could think about relatively simple physical argument that proves that at least the sum of all the exchange integrals is positive. The argument is based on the very notion of a variation principle and goes as follows.

We know that exchange integrals arise only if we use an antisymmetric product of orbitals as our trial wave function. If we use just a simple product of them there will be just Coulomb integrals in the energy expression.
Now since the true wave function is antisymmetric, if we use an antisymmetric product of orbitals as our trial wave function instead of a simple product of them, we are guaranteed to get the lower energy since we use the qualitatively right wave function.
Exchange integrals enter the energy expression with the negative sign, thus, they have to be positive, otherwise we won't get the lower energy by using an antisymmetric product of orbitals instead of a simple product.

Again, it does not prove that each and every exchange integral is positive, only that their sum is.


Devitrification vs glass transition temperature

This graph comes from a so-called "differential thermal analysis". It is studying the effects of temperature on glass.

Differential thermal analysis (DTA) is a method that heats two sample pans, one with a sample and one empty as a reference so that the effects of the sample pan can be negated. The heating is performed at a constant heat flow and the temperature of the sample is compared to the reference. Any exo- or endothermic process will add or subtract additional heat to the material and cause a spike in temperature upward or downward.

  I dont understand what the difference between "glass transition" and "devitrification" is, and why one would be endothermic and the other would be exothermic.

Glass transition is the temperature where bonds of a glass can relax and stress can be anneal out of a sample. It is neither endothermic or exothermic but the specific heat capacity increases as the atoms now have additional degrees of freedom.
Devitrification is where enough is put into the bonds that the atoms may rearrange and form crystalline structures from their vitreous state (devitrification). Crystallizing is an exothermic process because the atoms become more ordered and minimally space compared to a vitreous state, this reduces the strain in the bonds, releasing energy.


Can we dissolve salts with higher lattice enthalpy by providing heat?

Heat itself it just indirect factor. It is temperature what determines the shift of solubility products. If the dissolution is endothermic process, the salt solubility product would increase with higher temperature.
But if the solubility product is still small even at high temperature, like e.g. $\pu{e-16}$ instead of $\pu{e-20}$, then only a tiny extra salt amount would dissolve when heated.
The effective result depends on the particular case. The typical cases where significant solubility increase is observed are  potassium salts  $\ce{KNO3, KClO3, KClO4}$. But they have limited solubility even at low temperature - see the solubility table.
Generally, the dependence of an equlibrium constant of a reaction, including dissolving/precipitation process, is described by van't Hoff equation

$$\frac{\mathrm{d}}{\mathrm{d}T} \ln K_\mathrm{eq} = \frac{\Delta H^\ominus}{RT^2}$$

For solubility products, being equlibrium constants, is valid the following reaction:
$$\Delta G^{\circ}_\mathrm{dis} = \Delta H^{\circ}_\mathrm{dis}  - T \Delta S^{\circ}_\mathrm{dis} = RT \ln K_\mathrm{sp}$$


If you want to demonstrate that the addition of ozone to an alkene is a 6-electron oxidation of the carbon framework, here are two methods. To assign a charge to the carbon atoms of a neutral compound (sum of all charges equals zero) treat all hydrogens as +1 and all oxygens as -2, then do the math. Each carbon in ethylene (1) has a calculated (in red) value of -2. Molozonide 2 has carbons with a charge of +1 but, for convenience, they are written as zero and +2. Already one can see that 1 $\rightarrow$ 2 is a 6-electron oxidation [-4e $\rightarrow$ +2e]. Collapse of the molozonide gives formaldehyde (3) of carbon charge zero and zwitterion 4 with carbon charge +2. These species recombine to form ozonide 7 that has the same carbon charge as the molozonide 2. That the zwitterion 4 is the equivalent of formic acid 6, follow the steps 4 $\rightarrow$ 5 $\rightarrow$ 6. Thus, both the molozonide 2 and ozonide 7 are equivalent to one formaldehyde and one formic acid.
Alternatively, oxidation of ethylene (1) to ethylene glycol (8) [-4e $\rightarrow$ -2e] is a 2-electron oxidation and a second 2-electron  oxidation [-2e $\rightarrow$ 0e] affords two moles of formaldehyde. Oxidation of one of the formaldehydes to formic acid [0e $\rightarrow$ +2e) is the third 2-electron oxidation. Both pathways demonstrate that the oxidation of ethylene to formaldehyde and formic acid are 6-electron oxidations.


A fish-tank bubbler with needle valve or regulator would not only provide bubbles but be controllable and would not need to be replenished, as would chemicals.

N.B. However you make bubbles in mineral spirits, it would help the highly-flammable liquid to evaporate, creating a potential explosion and fire hazard. You might get better images with less-miscible mineral or silicone oil, with less danger.


What are the steps to integrate the common rate law to find the integrated rate law for any order

The integrated rate laws can be derived by equating two expressions for the rate of the reaction. This gives us a first order differential equation which can be manipulated to get the integrated rate law. The steps that you need to follow are actually quite simple and structured. Here is an example for deriving the first order integrated rate law:

Step 1: Equate the two expressions for the rate of the reaction

For first order reactions, the two expressions for the rate of the reaction are: $$ r = -\frac{\mathrm{d}[A]}{\mathrm{d}t}$$ and $$r = k[A]$$ Equating them we get: $$-\frac{\mathrm{d}[A]}{\mathrm{d}t} = k[A]$$

Step 2: Rearrange so that all the same variables are on the same side to form an first order differential equation

So we rearrange this expression to get all of the same variables on the same side (note that k is a constant, not a variable, so it doesn't matter which side it is on) to get: $$\frac{\mathrm{d}[A]}{[A]} = -k\mathrm{d}t$$

Step 3: Integrate both sides

Now we integrate both sides: $$\int_{[A_0]}^{[A]}{\frac{\mathrm{d}[A]}{[A]}} = -\int_{t_i}^{t_f}k\mathrm{d}t$$
$$\mathrm{ln}[A] - \mathrm{ln}[A_0] = -kt$$

Step 4: Using some algebraic manipulation, rearrange expression to get an equation for [A]
$$\mathrm{ln}\frac{[A]}{[A_0]} = -kt$$
$$\frac{[A]}{[A_0]} = \mathrm{e}^{-kt}$$ Therefore we get as required:
$$[A] = [A_0]\cdot e^{-kt}$$

Using this guide, you should be able to derive the other integrated rate laws. If you have any trouble, feel free to ask.

Edit:

Since you have also mentioned that you are struggling with deriving the second order integrated rate law, I will show you how to derive it, using the steps that I have outlined above.

Step 1:

$$r = - \frac{\mathrm{d}[A]}{\mathrm{d}t} = k[A]^2$$

Step 2:

$$\frac{\mathrm{d}[A]}{[A]^2} = -k\mathrm{d}t$$

Step 3:

$$\int_{[A]_0}^{[A]}\frac{\mathrm{d}[A]}{[A]^2} = \int_{t_i}^{t_f}k\mathrm{d}t$$

$$\frac{-1}{[A]} + \frac{1}{[A]_0} = -kt$$

Step 4:

Okay now everything up to here has been exactly the same thing we did for deriving the first order integrated rate law. We have gotten an equation relating $[A], [A]_0$ and $t$. To make $[A]$ the subject, we multiply by $[A]$ and put all the terms containing $[A]$ on one side:
$$\frac{[A]}{[A]_0} + kt[A] = 1$$ Now we factorise $[A]$ out on the left hand side and divide both sides so that the left hand side equals to $[A]$:
$$[A]\left(\frac{1}{[A]_0} + kt\right) = 1$$
$$[A] = \frac{1}{\frac{1}{[A]_0} + kt}$$
Now we multiply the right hand side by $[A]_0$/$[A]_0$ and we get the required equation:
$$[A] = \frac{[A]_0}{1 + kt\dot~ [A]_0}$$


The voltage on the terminals is different by 12 V (± a few depending on the charge state), you can think of this as the positive terminal being at +12 V and the negative terminal at being 0 V. The polarity of the voltage does NOT change.

On the current, things are a bit different. During discharge, the current flows from the positive to the negative.

During charge, the current flows out of the positive terminal of the charger into the positive terminal of the battery (and out of the negative terminal of the battery into the negative terminal of the charger).


The collision theory aims to explain how reactions happen.

In the theory, when the particles of the suitable reactant collide with each other, the particles with the right conditions (orientation, velocity, etc.) will react and form new chemical elements.

It is quite difficult to form $\ce{NaCl}$ from sodium metal and chlorine gas alone. If you have to produce $\ce{NaCl}$ from sodium metal and chlorine gas (which you would not), you would do it in an aqeuous environment:

$$\ce{Na(s) + H2O(l) -> NaOH(aq) + H2(g) ^}$$
$$\ce{Cl2(g) + H2O(l) -> HCl(aq) + HOCl(aq)}$$
$$\ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O}$$

But in this method $\ce{NaOCl}$ is produced as a side product, so I do not recommend using this method.


Answering your question from the title: no, SCF does not necessarily mean HF. Essentially the same self-consistent field (SCF) procedure is used to solve both the Hartree-Fock (HF) equations and the Kohn-Sham (KS) one, so for the case of DFT, SCF energy means not HF energy, but rather KS energy.

Regarding discrepancies in results obtained with different programs, there might be few reasons out there: different definitions of the B3LYP functional, different harmonics used for 6-311G basis sets (spherical vs. cartesian), different integration grids, different SCF convergence criteria, etc.

I checked the TURBOMOLE documentation and indeed found that to use the very same B3LYP functional that is used in Gaussian (with the VWN3 correlation) you need to use b3-lyp_Gaussian in TURBOMOLE, not b3-lyp. So, first of all, I suggest you try b3-lyp_Gaussian in TURBOMOLE instead of b3-lyp.


What molecules can be eliminated in condensation polymerization?

As an abstract answer, to me it would be whatever is commercially viable to manufacture and sell. Commercial polymerisation processes eliminate ethylene glycol (in the manufacture of PET) which is probably recycled,  methanol in transesterification reactions, ethene or propene in metathesis type,  phenol (recycled) in the use of diphenylcarbonate with bisphenol A, etc. In home applications,  acetoxysilicone sealant gives off acetic acid. So, anecdotally, it seems "simple" molecules of MW around 100 or less might meet your criteria.


Can rays be considered a type of "element"?

I like your approach.  "Here's the way I see the categories of 'elements' and 'not elements.'  Why don't they draw the line this way?"  While the line you want to draw to include "rays" alongside the elements that appear in the air isn't the one science draws, it does point out something really important:  We draw the lines between categories because they are useful.

As it turns out, the differences in behaviors between "things" like hydrogen atoms and helium atoms, and "rays" like radio and X-rays is substantial as you dig further into science.  The "elements," for instance, are "massive," meaning they have mass.  Quantum scientists would say they "interact with the Higgs field."  However, you choose to phrase it, these atoms behave very differently than electromagnetic waves (which is the proper term for "rays").  Electromagnetic waves have no mass (other than some secondary effects that you'll learn when you study relativity).  Their propagation through space is well defined by Maxwell's Equations.  They don't form bonds (like most elements do).  They don't take up space (electromagnetic waves don't push each other away, but atoms can only get so close before the repulsive forces drive them apart).

Eventually, you will learn that the lines blur.  Relativity will start to explain why electromagnetic waves indeed do have mass.  Quantum Mechanics will point out that particles have wave-like properties.  However, as you start out, the differences between "massful things" and "electromagnetic waves" is pretty iron clad.  As you continue your studies, you'll become more comfortable with the distinctions drawn, and then we can slowly introduce all those strange line-blurring facets of our reality that makes science really interesting.


How do I find the geometry of a tetramer assembly?

In general, this is a very tricky problem. At its root, it's quite similar to solving the question of predicting a crystal structure -- you want to know the interaction between multiple molecules via hydrogen bonds and other electrostatic non-covalent interactions.

Here's the best general procedure:

Perform a conformer search on an individual molecule. As you guess, a "conformer search" (in this case in Spartan) is not going to sample different configurations of the supramolecular system. Worse, there's no guarantee that the lowest energy conformer is the one that's found in the tetramer. Still, it's better to start with a low energy conformer, since that's the most likely one.
Attempt to find a low-energy dimer configuration by copy/pasting your molecule. Consider how the hydrogen bonds might match. If you have some knowledge of the interactions great. I might consider, for example that the N-C=O combination you've circled in the bottom left corner probably matches up with the two amines above them (e.g., the dimer has one copy flipped "upside down" compared with this depiction).
Optimize the dimer geometry using hydrogen bond constraints and a reasonable computational method. Here, I would not use RM1 or PM3, since both methods do not include hydrogen bonding terms. I would use some flavor of PM6 or PM7 if you want to use semiempirical methods, or I'd try some type of dispersion-corrected DFT method, e.g., B3LYP-D3 or $\omega$-B97X-D.
Copy/paste the dimer to try to build up a tetramer structure.

I know some research groups have tools that somewhat automate this process, but I don't know of any that are currently widely-available and user-friendly. This is very much still an active area of research.

Moreover, as a human, you can often spot "how things fit" better than most automated strategies.

Good luck!


Expressions for constant volume and constant pressure heat capacities

The difference in internal energy ($\mathrm{d}U$) is a function of many different variables:

$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V \!$$

Therefore you must specify – via a partial derivative – with respect to which of the variables you are doing the derivative, otherwise you would need a total derivative, which is not what the equation requires.


Just to clear things up first (I might be saying something you already know, but I just want to be sure): the intent of the question is not to ask you what happens to the reaction quotient after equilibrium is re-established.

If a system is originally in equilibrium, with $Q = K_1$, and then you disturb it such that the equilibrium constant changes to $K_2$, then after the system re-reaches equilibrium you will have $Q = K_2$. If you interpret the question as asking whether $Q$ has changed after equilibrium is re-established, then the question basically amounts to asking whether the equilibrium constants $K_1$ and $K_2$ are different.

What the question is asking is whether $Q$ is affected instantaneously by the disturbance. Re-reading your question I think you got it, so...

Anyway, heat is not a product.

The reaction quotient for that reaction is defined as

$$Q = \frac{a_{\ce{CO2}}}{(a_{\ce{CO}})^2} \approx \frac{p_{\ce{CO2}}}{(p_{\ce{CO}})^2}$$

and $K$ is defined as the value of $Q$ at equilibrium.

Nowhere does heat appear in the equation. The idea of using heat as a "product" is nothing more than a mnemonic to help you remember the direction in which the equilibrium shifts. As long as the pressures are fixed, the temperature does not affect the instantaneous value of the reaction quotient.

When you increase the temperature, what happens is that the equilibrium constant $K$ drops. Let's say that before changing the temperature $Q = K = 0.01$ (just an example). If you increase the temperature, $K$ might fall to $0.001$ but $Q$ at that instant will still be $0.01$. As the system re-equilibrates, $Q$ will drop to $0.001$.

So, it is not that $K$ remains constant and $Q$ increases. It is more like: $K$ drops and $Q$ remains constant (at least instantaneously).


finding out fractions of ions(different valencies/charges) of an element in non-stoichiometric compounds

What you have done seems to be correct. But you might want to convert those numbers to percentiles. I can tell you another method based on charge conservation.
Let us take  100 oxygen atoms and 98 Nickel atoms into our consideration.
Total negative charge$= 200$
Total positive charge$= 2x+3y$ where $x$ and $y$ are what you have assumed to be in the question . Now Net charge =0.Equate the positive and negative charges.
Also $x+y=98$. Solve them and you will get the same number. This of course has an high degree of
similarity to what you have done


Volume expansion from phosphate buffer solution

There is an exact way of doing this and there is a practical way of doing this.

The exact way involves a volumetric $1~\mathrm{l}$ flask. You would first fill the salts into the volumetric flash, add some water to dissolve them and then carefully add more water until the flask contains exactly $1~\mathrm{l}$ of solution. That removes the error of the salts increasing the overall volume.

However, it is also part of the exact way to add a pH-monitoring device before you have finised to make sure the pH is exactly at $7.2$ — and to carefully add phosphoric acid or sodium hydrogen phosphate if the pH is off. (A better solution would be to make sure the pH is over the desired value so phosphoric acid can simply be added; as it is a liquid it can be added and dissolved easier.) For most applications outside of thermodynamics, this is an overkill.

The practical way involves one of the following:

using an Erlenmeyer flask, beaker or similar with a $1~\mathrm{l}$ mark; filling the salts first and then the solution up to the mark
simply measuring $1~\mathrm{l}$ of water and adding that.

Why are the practical solutions okay? In most cases, it does not matter whether the molar concentration of, say $\ce{NaCl}$ is exactly $0.15~\mathrm{M}$ or whether it is $0.14~\mathrm{M}$. This rather unimportant experimental error is simply accepted. It also does not really matter whether the pH is exactly $7.2$ or whether it is, say $7.5$. Therefore, one often decides to go the not-so-exact but easier and faster way.


What is the correct name of 3-isopropyl-4-methylhexane?

Both the names you have given are not the preferred name for the molecule above. You are correct that the preferred name maximises the number of substituents when there is a choice: see Alkane nomenclature - choosing the parent chain - maximize or minimize substituents? So, a name of the form a-ethyl-(b,c)-dimethylhexane (where a, b, and c are appropriately chosen locant numbers) is preferred over a name of the form a-isopropyl-b-methylhexane, as it has three substituents as opposed to two.

However, the numbering on the right is not correct either. If you number the carbon chain the other way around instead, you would obtain the name 3-ethyl-2,4-dimethylhexane:

This is preferred over your proposed name 4-ethyl-3,5-dimethylhexane, because the locant set (2,3,4) in 3-ethyl-2,4-dimethylhexane is lower than the locant set (3,4,5) in 4-ethyl-3,5-dimethylhexane.

Note also that according to current IUPAC nomenclature recommendations (P-57.1.1.1 of the 2013 Blue Book), the preferred prefix for $\ce{-CH(CH3)2}$ is "propan-2-yl", not "isopropyl".


Using your diagram, the boron-silicon bond with only the silicon electron has only half of the possible electron content.  A "hole" as you said. The hole is not confined to the one $\ce{B-Si}$ bond, but can be in any of the 4 positions. It's not like the boron sits there and the silicon pops its electron into the potential bond; the bond is there and it's only half filled.
Now this isn't a molecule, but you can focus on the boron part of the crystal and say that all four adjacent silicons are bonded to the boron, but you have only 7 electrons where the "molecular" orbitals could contain 8.
Electrons don't really "belong" to one atom or another; they comprise some sort of sea of electrons, and the sea above this boron is a little lower than elsewhere. I think of the sea of electrons in a copper-boron alloy as being very fluid, and having a level surface everywhere, but the sea in a p-type semiconductor is more like mud (quicksand?) and can have some dips that could catch and puddle some water (~electrons). The sea is still controlled mostly by the silicon, just modified by boron (submerged holes).


What is the chemical redox equation between hydrogen peroxide and melanin?

Your formula for melanin is incorrect. Melanin does not have a clear and concise chemical formula; it is a class of natural polymers that differ in their exact composition. The only thing they have in common is that they derive from the amino acid tyrosine and that they at one point polymerised.

Therefore, unfortunately there cannot be an exact description of what happened between melanin and hydrogen peroxide. You can draw a half-equation for the hydrogen peroxide part which would be:

$$\ce{H2O2 + 2 e- + 2 H+ -> 2 H2O}$$

But what exactly happens in melanin and which groups are oxidised is hard to say. What probably happens is a series of radical reactions that in some way oxidises away the aromatic system and thereby moves the absorpion wavelength out of the visible spectrum. But that’s probably about as much as you can say.


Why is the rate of reaction of polar EAS favored by inductive effects and the rate of less polar EAS reactions through hyperconjugation?

I may be oversimplifying things.

A look at electron density

Let's pick up the extreme cases of the substituted benzenes . $$\ce{A -> ArCH3}$$
$$\ce{D -> Ar(CH3)_3}$$

Case A shows the maximum hyperconjugation. Case D on the other has the maximum +I effect.

For a moment let's imagine that the electrons are coins.
Case D , has more coins, more methyl groups (let's say 3).
Case A, on the other hand has 1 coin , hyperconjugation would simply mean that I am  simply moving that coin in circles. It's simply redistribution of electrons.it doesn't actually increase the density.

So the first point which I'm going to use to answer the question is that -

Inductive effects increases  the electron density, hyperconjugation simply redistributes the electron density.

A look at Coulomb's law

$$\ce{ F \propto Q_1*Q_2}$$

More polar reactants will have a greater charge on them as compared with less polar reactants.
Since inductive effects concentrate  electron density at one area the negative charge concentration will be greater.

HOMO/LUMO

Unoccupied orbitals are high in energy and occupied orbitals are low in energy. For the best overlap the energy gap between the HOMO and LUMO must be small.

Rate Order for more polar reactions

In more polar reactions the electrostatic forces of attractions play the primary role as the polar reactants is highly charged. As established earlier ,inductive effects lead to a electron density thus a greater pull. Homo/Lumo interactions are secondary as the overlapping happens after a collision occurs.

The electrostatic forces are pulling the benzene and reactant. That would by itself increases the number of collisions.
A greater electron density leads to the better stabilization of the +tive charge on the reactant.

inductive effects support more polar reactions

Rate order for less polar reactions

The electrostatic forces of attraction are lower in the case as as the reactant is less charged.
So the control of the reaction is primarily through Homo/Lumo interactions.

Unoccupied orbitals of the reactant are high in energy.( They are Unoccupied).
To bridge the HOMO/LUMO gap the Lumo of benzene carbon must be high in energy. Non bonded electrons or the electrons which are delocalized through hyperconjugation are high in energy as they are held by only one atom.
So the HOMO/Lumo gap reduces in the case of hyperconjugation thus hyperconjugation supports less polar reactions.


What is the value for the reverse reaction?

${K_c}$ of reverse reaction should just be the reciprocal of forward ${K_c}$. So, ${K_c}$ of reverse reaction is $\frac{1}{1752}$.

Now, I am not sure, why you used 2nd equation. Suppose you have 1 unit conc of $\ce{NO}$ and $\ce{O_2}$, then after time $t$, if the conversion is $x$, then your equation should be
$1752 = \frac{[2x]^2}{([1-x])^3}$.

Can you please check your 2nd equation again?


If Mendelev hadn't put the periodic table in its current order, would it still have been made popular?

The periodic table is very different to things like the QWERTY keyboard layout which is essentially arbitrary.

The structure is determined by the fundamental chemical properties of the elements. When the table was first developed Mendeleev was wise enough to arrange the table with both gaps and with some elements in chemical property order rather than the order indicated by know physical properties such as atomic mass (which usually gives the same order as chemical properties but gets it wrong a small number of times).

These properties are fundamental to the elements and not a matter where human choice can have any impact at all. Sure, there are alternative ways to get the graphic design right, but none of these work if the violate the original principles of grouping similar elements together. Therefore, no alternative choices by Mendeleev could work since they would fail to achieve this.


What does the newest IUPAC nomenclature have against unsaturation?

I think it is important to note that these are recommendation and not rules (in the sense of which you must follow) and hence you can choose for yourself whether you like to implement them or not.

A generated unambiguous name still conforms to the general recommendations of the IUPAC, hence what you call outdated is still a valid name, it just isn't the preferred IUPAC name, which is a concept introduced in 2013. Hence names generated according to the 1993 recommendations will still be valid names according to the new recommendations; the new recommendations are backwards compatible.

The reasoning can be found in the Introduction of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) (It is also in the cited draft):

  P-10 INTRODUCTION
  [...]
  A major new principle is elaborated in these Recommendations; the concept of preferred IUPAC names' (PINs) is developed and systematically applied. Up to now, the nomenclature developed and recommended by IUPAC has emphasized the generation of unambiguous names in accord with the historical development of the subject. In 1993, due to the explosion in the circulation of information and the globalization of human activities, it was deemed necessary to have a common language that would prove important in legal situations, with manifestations in patents, export-import regulations, environmental health, and safety information, etc. However, rather than recommend only a single unique name for each structure, we have developed rules for assigning preferred IUPAC names, while continuing to allow alternative names in order to preserve the diversity and adaptability of the nomenclature to daily activities in chemistry and in science in general.

  Thus, the existence of preferred IUPAC names does not prevent the use of other names to take into account a specific context or to emphasize structural features common to a series of compounds. Preferred IUPAC names (PINs) belong to a 'preferred IUPAC nomenclature'. Any name other than a preferred IUPAC name, as long as it is unambiguous and follows the principles of the IUPAC recommendations herein, is acceptable as a general IUPAC name, in the context of a 'general IUPAC nomenclature'.
  [...]

If nothing more, the new guidelines provide a more robust nomenclature. In my very personal opinion that is a very good change. Let's look at a more practical approach.
Upon a chemical reaction which modifies a compound (in many if not most case) the principle chain will be retained according to the new change. Since unsaturation often provides the possibility of relatively easy modification, similar to a functional group (and some would even consider it one), a reaction at these bond will likely involve that the principle chain changes according to the old recommendations. Let's stick to the example of Loong's answer in the linked post; and let's just add $\ce{HBr}$ to the double bond:

According to the old recommendations the reader is likely to imply that there was a modification hexene to heptane, and without further inspection one could assume that there was an extension of the carbon skeleton, i.e. a carbon-carbon bond formation.
In the new guidelines it is much more obvious what the modification actually is, a simple addition. This obviously is of little concern for these simple compounds, but when it comes to natural product synthesis, it removes some of the ambiguity of what actually happens, highlighting the change of the functional group rather than implying a modification to the carbon skeleton.

There are probably instance where you deliberately want to deviate from PINs. A simple example would be the protection of an alcohol:

It might be a necessary step in your synthesis, it might even be an important one, for the chemistry involved and expressing what you are actually doing, PINs will be a nuisance. I personally would prefer the names in parentheses. However, when you are eventually writing your patent, you might want to stick to a PIN in order to avoid any ambiguity, after all that's why they were introduces in the first place.


Carbon has a valency of four which helps in bonding with other atom or groups easier and adds to catenation. When one talks of Carbon's ability to bond, then Catenation property should come to your mind. Catenation is the linkage of atoms of the same element into longer chains. Catenation occurs most readily in carbon, which forms covalent bonds with other carbon atoms to form longer chains and structures. This is the reason for the presence of the vast number of organic compounds in nature. One such marvel is DNA! So, while studying chemistry, Carbon compounds form a class of peculiar natured compounds and the vast range of molecules and development of their own domain in their own scale might have made people feel that a separate category should be made for Carbon compounds because of their vastness and availability, also predominately manifestation of them in almost every thing, even life! In fact life became possible only because of Carbon, such is the importance. They have got their own nomenclature system in their domain.. Its really vast. The study of everything begins with Organic compounds first.. be it isomerism, medicinal reaction or anything. So, you may see, how important Carbon is. If you are a biology student, you must have learned about biomolecules.. DNA and gene pool, replication, transcription, translation, protein coding, defective proteins and some viral components and its lifecycle. Molecular basis of genetics is based on DNA and other organic compounds. That gives you thrilling experience to see how stuffs work and wonder about their importance and manifestation. Carbon is a wonder.


You are acting under two false assumptions

Electrode potentials are measured relative to real zero. They are not, there is no practical procedure to measure absolute electron potential. All standard potentials are measured relative to standard hydrogen electrode (well, actually some more pragmatic standard electrode, but recalculated to be against standard hydrogen electrode)
That values got for gas has anything to do with values for condensed matter. Hydration energy can and does shift electrode potentials greatly, as well as energy of crystal lattice.


$$\ce{pH} = \mathrm{p}K_\mathrm {a} + \log \left(\frac{\alpha}{1-\alpha} \right)$$
Multiply the equation by $-1$:

$$-\ce{pH} = -\mathrm{p}K_\mathrm {a} + \log \left(\frac{1-\alpha} {\alpha}\right)$$
Rearrange the equation
$$-\ce{pH} +\mathrm{p}K_\mathrm {a} = \log \left(\frac{1-\alpha} {\alpha}\right)$$

Apply the definition of $\log$

$$10^{-\ce{pH} +\mathrm{p}K_\mathrm {a}} = \frac{1-\alpha} {\alpha}=  \frac{1}{\alpha} -1$$
Rearrange the equation
$$1+ 10^{-\ce{pH} +\mathrm{p}K_\mathrm {a}} =  \frac{1}{\alpha} $$
So,
$$\alpha= \frac{1}{ 1+ 10^{-\ce{pH} +\mathrm{p}K_\mathrm {a}}  } $$


Isopropyl and Butyl Groups - Relative Priorities

Comparing IsoPropyl and n-Butyl

On first position both have C.
On second position IsoPropyl has two C's whereas n-Butyl has only one C.
Thus IsoPropyl has higher priority and we do not see the further chain.


How do I write orders of magnitude in scientific notation when margin of error is involved?

Sometimes you need to go the extra mile to be entirely unambiguous. Your original formulation (3.78 ± 0.02 * 10^-3) could be interpreted in two ways, the way you intended and the way your TA suggested. The majority of people, I expect, would use context to interpret it correct as you intended, but some (as your TA demonstrates) will not.

There is no clear cut widely agreed standard that will resolve this. But you can apply standard mathematical notation (which will look superfluous to some) to make what you write entirely unambiguous. As Ivan suggested in a comment the simple addition of brackets giving (3.78 ± 0.02) * 10^-3 makes your statement unambiguous in its interpretation. This is more compact than 3.78 * 10^-3 ± 0.02 * 10^-3 but just as unambiguous.

If you were using properly typeset formulae, you might well be able to avoid some of the apparent redundancy of the last formulation, but that would take more effort. Better to accept that redundancy is sometimes required to make things clear.


Before answering the question, you might find it helpful to know that people tend not to go about making the dioxolane, but instead make use of natural products such as safrole to buy in a precursor with the ring already formed. If you work at a University or a research institution, this shouldn't be an issue. If you don't, I strongly advise you against trying this chemistry, as your desired product bears significant resemblance to multiple illegal drugs.

To answer specific parts of your question:

  1) Does the nitrogen or the carboxyl group react with the methanal? Do I have to use a protecting group?

The reactivity of the amino acid portion depends on the ionisations.

In both the protonated and deprotonated form, the carboxylic acid/carboxylate can't react with formaldehyde.

Amines do react with formaldehyde in the presence of an acid catalyst (think about the Mannich reaction). Whether or not your substrate (amino acid) as drawn would react with formaldehyde likely depends upon whether the amine is protonated or not (think about the amino acid and whether or not it would exist in its zwitterionic form under the conditions given).

  2) Is there a chance it reacts similar to a marquis reaction?

The mechanism of the Marquis reaction isn't fully elucidated, however it does give positive results for many alkaloid drugs

As already stated, given the similarity of your compound to several illicit drugs, I wouldn't be at all surprised if it did give a positive result.

Moving on to the actual chemistry you've proposed.

The standard method for making 1,3-benzodioxole (and derivatives) is to take the catechol and treat it with CH2Br2 under basic conditions.

This works better than formaldehyde (I'm not actually convinced the formaldehyde would work at all, but I'll presume you've researched the method) because rather than trying to expel water (which as you point out, you'd have to Dean-Stark to remove the water), you only expel halide ions, which don't interfere under the conditions.

(As another side note, experimentally you also need to add a phase-transfer catalyst as the CH2Br2 isn't miscible with the base/aqueous layer).


When to use internal energy and enthalpy in balances

Is it that ΔU is used for closed systems whilst ΔH is used for open systems?

That's not quite correct.

Consider the case of a simple system: a volume of gas. The problem is that when processes take place, the gas may expand or contract. Since the gas pushes back on the external pressure, some amount of work may be done the system ($W = p\Delta V$). The internal energy change $\Delta U$ does not automatically factor in this work.

Enthalpy, on the other hand, is a quantity which is defined ($U + pV$) such that it does account for the pressure-volume work of the system, so $\Delta H$ is the change the energy of the system and the pressure-volme work done by the system.

So, the summary is that if you have a constant volume process (where pressure volume work is by definition zero), use $U$. If you have a constant pressure system, use $H$.

Even the free energies are this way as well. There's Helmholtz free energy ($A$ or $F$) for constant volume processes and Gibbs free energy ($G$) for constant pressure processes.

In chemistry, most processes are done in open air (unless you're using a bomb calorimeter), in which case all processes are simply subject to atmospheric pressure, so these processes would all be best described using enthalpy and Gibbs free energy.


Use of calcium sulfate in food and pharmaceutical applications is widespread
and continues to expand. United States Gypsum Company offers two highly refined calcium sulfate products: USG Terra Alba and SNOW WHITE filler. Both fillers are food- and pharmaceutical-grade forms of calcium sulfate, a mineral that appears on the Food and Drug Administration’s GRAS (Generally Recognized as Safe) list of additives approved for nutritional and functional use in food products.

USG Terra Alba ($\ce{CaSO4 \cdot 2H2O}$), the dihydrate form of calcium sulfate, results from fine-grinding and air-separating a select, high-purity white gypsum that contains about $20$ percent water of crystallization.
SNOW WHITE filler ($\ce{CaSO4}$), the anhydrous form of calcium
sulfate, is produced by calcining and milling high-purity white
gypsum$^{[1]}$.

Gypsum is made suitable for commercial use by a process called calcination, which involves heating the mineral to remove approximately three-quarters of its water. Calcined gypsum can recrystallize into any shape with the simple addition of water.$^{[2]}$.

Presumably through repeated calcination, gypsum is made pure enough to meet FDA standards.

$^{[1]}$ USG, Calcium Sulfate Fillers
$^{[2]}$ Encyclopedia.com, USG Corporation


At the freezing point, why is the vapor pressure of the liquid phase equal to the vapor pressure of the solid phase?

If the vapor pressure of the substance in its liquid phase would be different from the vapor pressure of its solid phase then there would be a transfer of mass from the phase with the higher vapor pressure to the phase with the lower vapor pressure.

This would involve releasing or consuming energy (heat of fusion) even in a closed system in thermal equilibrium, which is obviously a contradiction.


The instability of gem-diols is due the fact that hydogen bonding between the 2 OH- groups increases the repulsion causing a strain on the C-O-H bonds.
In case of compounds of boron, a greater bond angle (due to sp2 hybridization of boron) prevents intramolecular hydrogen bonding.
Compounds of sulphur, phosphorous always contain a double bonded oxygen along with the OH- groups on the central atom. This oxygen atom forms hydrogen bonds with OH- groups preventing hydrogen bonding between the OH- groups, thus preventing any strain.


How do you find the average rate and instantaneous rate given a kinetics plot?

This is more of a question of calculus than chemistry, but in a first order reaction
$\ce{A -> B}$, the average rate is:

$$\frac{\Delta A}{\Delta t} = \frac{A_2 - A_1}{t_2 - t_1}$$

 Graciously borrowed from here

So, the average rate is the slope of a secant line between the two points,

$$ \frac {\Delta A}{\Delta t} = \frac{1 - 0.5}{400 - 200}$$

The instantaneous rate is the derivative of the plot:

$$\frac{dA(t)}{dt}$$ or $$\lim_{t\rightarrow 0} \frac{\Delta A}{\Delta t}$$

For the above first order rate law, the equation is in the form:

$$ A(t) = e^{-kt + ln(A_0)} $$ so the derivative is simply

$$ \frac{dA(t)}{dt} = -ke^{-kt + ln(A_0)} $$


Why is Omega-3 a polyunsaturated and not a monounsaturated fatty acid?

The term "omega-3" just denotes that the position of the first double bond in the fatty acid will be at position "3" (counting from the omega end of the chain).  It does not mean that there is only one double bond in the fatty acid.  The most common omega-3 fatty acids contain between 3 and 6 cis double bonds (therefore they are polyunsaturated).

Here are few examples, again note how they all have a double bond at the third carbon from the omega end of the chain.

α-Linolenic acid (ALA)

Eicosapentaenoic acid (EPA)

Docosahexaenoic acid (DHA)


Does the hydrolysis speed increase when more H+ ions are in the solution?

The glycosidic bond of a polysaccharide chain is essentially an acetal moiety, and hydrolysis of acetals proceeds under acid-catalyzed conditions. Hence, the addition of acid, within some ambit of concentration, will accelerate the hydrolysis reaction. The optimum proportion of acid, however, is probably something that needs to be determined experimentally on a case-by-case basis. It isn't necessarily true that the rate of the reaction will continuously increase, without bound, as the concentration of catalyst increases.

The mechanism proceeds through a series of proton transfers and SN1 reactions. As far as I'm aware, the rate-determining step is typically the actual cleavage of the glycosidic bond after an initial protonation of the oxygen of the leaving group. Given that, I would expect increases in acid concentration to have marginal effect beyond some threshold value. That's just a surmise on my part, however.

I did find one study that examined the kinetics of starch hydrolysis, which found the optimum pH to be 3 (with values of 2-5 being tested).


Total number of “hyperconjugable” hydrogens in 1-(3-ethyl-4-methylphenyl)-2-methylprop-1-ylium

There is one obvious $\alpha$-position, which is the tertiary carbon atom adjacent to the carbon carrying positive charge. This contributes one hypercongujation structure. It looks like there are no more apparent structures, but the methyl group that is para to the cation can also participate in hyperconjugation

There are three $\ce{H}$-atoms on the methyl group, which give us three more structures. Thus, there are a total of 4 hyperconjugable hydrogens in this structure.
Interesting to note is that the ethyl group at the meta position does not participate in hyperconjugation. Can you find a reason why?


Removing aluminum oxide from impure silica

You may try the famous Bayer process, which is used for extracting alumina form bauxite. When treated at $160°$C and high pressure with $\ce{NaOH}$, a mixture like yours dissolves $\ce{SiO2}$ and $\ce{Al2O3}$, and lets the other compounds insoluble. This is a first purification. $$\ce{SiO2 + 2 NaOH ->  Na2SiO3  +  H2O}$$ $$\ce{Al2O3  +  2 NaOH + 3 H2O -> 2 Na[Al(OH)4]}$$Later on, bubbling $\ce{CO2}$ in the solution makes $\ce{Al(OH)3}$ to precipitate. $$\ce{Na[Al(OH)4] + CO2  -> Al(OH)3(s) + NaHCO3}$$ Of course $\ce{NaHCO3}$ is immediately transformed into $\ce{Na2CO3}$ by $\ce{NaOH}$. $\ce{Al(OH)3}$ can be eliminated by filtration. And the final solution contains only sodium silicate, apart from the excess of $\ce{NaOH}$. Silica can then be obtained pure by adding  enough $\ce{HCl}$ to destroy $\ce{Na2SiO3}$ and the NaOH in excess.$$\ce{Na2SiO3 + 2 HCl + H2O ->  Si(OH)4 + 2 NaCl}$$ It is advisable to highly dilute the $\ce{NaOH + Na2SiO3}$ solution before adding $\ce{HCl}$. Because the precipitate of $\ce{Si(OH)4}$ is extremely voluminous, and fills up the entire solution.


Wikipedia already offers a decent summary of two theories of shear thickening behavior - namely the generation of transient 'defects' in a high-density suspension that either resist further growth or are incompressible in the direction of shear.

A potential macroscale analogue of shear thickening is the dramatic increase in viscosity of large particle-air mixtures upon pneumatic removal of the interparticle fluid, as is exploited in jamming skin enabled locomotion. (paper here) This should be roughly comparable to the transient decrease in particle spacing in a shear thickening fluid that is being compressed beyond its relaxation rate.


Does the density of a liquid and gas phase of a substance(nitrogen or water) converge when reaching critical point

Does the density of a liquid and gas phase of a substance converge when reaching critical point?

Yes. This is exactly what happens. Let's take a look at a qualitative $P-T$ (pressure–temperature) phase diagram of an arbitrary substance:

source: Cansell, F. et al. Supercritical fluid processing: a new route for materials synthesis. J. Mater. Chem. 9, 67–75 (1999). 10.1039/A804964E

As you can see, moving directly from gas to liquid involves crossing the phase boundary. However, if you're doing it in the supercritical, the change in density is continuous. Maybe you can get a better, more quantitative view with this example of $\ce{CO2}$:

source: http://what-when-how.com/nanoscience-and-nanotechnology/microweighing-in-supercritical-carbon-dioxide-part-1-nanotechnology/

This one is also a a $P-T$ phase diagram, with equal density $(\rho)$ lines. You can see that the lines radiating from the gas–liquid boundary lines are not equal on each side. For example, 0.1 and 0.9. However, as you get close to the critical point, $\rho$ on each side of the line converges to a single value: 0.2 and 0.7, 0.3 and 0.6 and so on.

YouTube has some terrific examples of this happening. For example: http://youtu.be/GEr3NxsPTOA
What you see is a function of density - you can see the phase boundary between liquid and gas because they have differing densities. Notice in the videos how the phase slowly and gradually disappears as the density of both phases converges, effectively turning the $\ce{CO2}$ into a supercritical fluid.


Water and Ethanol Vapor Phase diagram for specific range

The Dortmund Data Bank might be a resource to start searching.

They do seem to have a page for vapour-liquid equilibrium data for the ethanol/water mixture.

On a first view, however, the data seem to be for less than atmospheric pressures only and are compiled from Ind. Eng. Chem., 1942, 34, 1501–1504.


Derivation and visualization of cell potential dependency on temperature and entropy

What you are suggesting in your equation is that $\Delta G_m=-T\Delta S_m$ but you know that in fact $\Delta G_m=\Delta H_m-T\Delta S_m$. Your math is incorrect because you are missing a constant of integration.
The expression $\Delta G_m=\Delta H_m-T\Delta S_m$ can be rewritten using the Nernst equation as
$$-nFE = \Delta H_m -nFT\left(\frac{\partial E}{\partial T} \right)_p$$
by using the definition you provide for $\Delta S_m$ as the temperature-derivative of $\Delta G_m$, and the Nernst equation $\Delta G_m = -nFE$. In fact
$$\Delta S_m = nF\left(\frac{\partial E}{\partial T} \right)_p$$
Therefore if you plot $nFE$ as a function of T the slope of the curve at a given temperature will provide the value of $\Delta S_m$ at that T.


React these iron oxide, ammonium nitrate, and sucrose

Disclaimer: what I offer here is an educated guess.

$\ce{NH4NO3}$ is known to decompose quite easily, especially if proper catalyst is present. Thus, one of the reactions is

$\ce{NH_4NO_3 = N_2 + 2H_2O + [O]}$

generated oxygen may be used to oxidize sucrose or black iron oxide. I doubt that $\ce{N2O}$ is formed, since it is quite strong oxidizer and decomposes quite easily, especially if redox catalyst is present

$\ce{C_12H_22O_11 + 24[O] = 12 CO_2 + 11 H_2O}$

$\ce{2Fe3O4 +[O] = 3 Fe2O3}$

However, as you said, the snake is black, so the second reaction does not occur

the final process is dehydration of sugar, producing carbon (actually, the carbon may contain significant amount of hydrogen and oxygen)

$\ce{C_12H_22O_11 = 12 C + 11 H_2O}$

at the same time numerous other processes, producing gaseous reducing organics, happens, so I advise to use a fume hood.

When cooled, reducing atmosphere from sugar decomposition vanishes, and black iron oxide reacts with air

$\ce{4Fe3O4 +O2 = 6 Fe2O3}$

Since black iron oxide acts as a catalyst and does not melt on reaction conditions, it is good to use as fine one as you can get and maybe it may be used in smaller amounts. Since ammonium nitrate acts as main energy source here, it is likely that it can be used in larger amounts. Sucrose forms the body of the 'snake', so it may be used in smaller amount.

The 'snake' consists of (partially) carbonized/caramelized sucrose binding other ashes together and leftovers from iron oxide. The expansion of volume results from gases emitted in the processes of decomposition of ammonium nitrate and decomposition of sucrose, though decomposition of ammonium nitrate is probably the main source here.

BTW, thanks for recipe, I think it would come handy for me this year.


Does 1-ethylidene-2-(propan-2-ylidene)cyclobutane show geometrical isomerism?

The two isomers are distinct substances, differing only in the orientation of a double bond, so they must be geometrical isomers. However, as a practical matter, they may interconvert under mild conditions and/or one may be much less stable than the other.


Will magnesium undergo a redox reaction with zinc(II) or copper(II)?

Let’s assume for a second that magnesium reacts only with the zinc. The reaction would occur according the following equation:

$$\ce{Mg + Zn^2+ -> Mg^2+ + Zn}$$

You have already identified, if I read your question correctly that the redox potentials allow this and that the reaction is spontaneous etc. Assume that you have equimolar amounts of magnesium, zinc and copper for the rest of this thought experiment.

Now our new solution would contain magnesium(II), copper(II) and zinc(0) (okay, the zinc would deposit itself somewhere, but you get the picture). Notice anything? Yes, zinc and copper can react according to the well-known and often-displayed electrochemical reaction (depositing copper on a blank piece of zinc metal). The equation for that one is:

$$\ce{Zn + Cu^2+ -> Zn^2+ + Cu}$$

Again, check out the standard potentials to see that this reaction is possible and spontaneous.

If we add the two reactions together, we get an effective reaction:

$$\ce{Mg + Zn^2+ + Zn + Cu^2+ -> Mg^2+ + Zn + Zn^2+ + Cu}$$

$$\ce{Mg + Cu^2+ -> Mg^2+ + Cu}$$

You notice that the zinc is little more than a spectator ion.

Now how does this differ from the ethanol case? Well, permanganate can (and will) react with ethanol to form ethanal, and it can (and will) react with ethanal to form ethanoic acid — more rapidly. However, if permanganate (and therefore: electrons) is the limiting reagent, we end up with a mixture of ethanol, maybe ethanal and ethanoic acid. By a reaction analogous to the reverse Cannizarro reaction, these can comproportionate:

$$\ce{CH3CH2OH + CH3COOH <=>> 2 CH3CHO}$$

Therefore, the overall reaction seems to be one where ethanol only reacted with permanganate to only form ethanal, while in fact ethanoic acid was formed, but in the end the equilibrium favoured ethanal.


Why does adipic acid give a cyclic ketone on heating while glutaric acid and succinic acid give a cyclic anhydride?

According to Blanc Rule for Dibasic Carboxylic acids, A cyclic Ketone is formed unless a five or six membered cyclic acid anhydride can be formed. This rule is valid for 1,4; 1,5; 1,6. dicarboxylic acids


Double bond crossed lines representation

It is, for example, used by Marvin JS -- one of the editors in the GUI of Reaxys by Elsevier to query the database for structures, where you do not impose either (E)-, nor (Z)-configuration of the double bond.  Below an example if you were looking for both isomers of stilbene:

(demo version here).

Beside ChemDraw, ChemDoodle's sketcher (here) and PubChem's sketcher are other examples of "some software" equally allowing such a representation, too.  I speculate it has something to do how databases (like ISISBase, Beilstein/Reaxys, Scifinder, for example) traditionally store(d) chemical structures and reaction schemes in a machine-readable format.

One may, for example, draw (cis)-stilbene with ChemDoodle "in the standard way" and save it as *.mol format and set this file in comparsion with the result of an undefined double bond with the using the wavy bond towards one of the phenyl rings.  In the chemical table, there is no significant difference spot:

If the "standard (cis) *.mol" file is compared with the one where the double bond is crossed -- as in the OP figure -- there is a difference on line #26 (fourth column now reads a three, instead of a zero), while maintaining the other entries:

This picture, naturally, differs from the *.mol files of classically specified (cis) and (trans) configuration:

used structures:


There are a few things that seem to be wrong with the diagram you present, but since your focus is on the relative energies of the $\alpha$ and $\beta$ spin states I'll address only this. If you check a reliable source such as Cavanagh et al.s NMR textbook [ref 1] it is explained that

  the state with $m=+\frac{1}{2}$ is referred to as the $\alpha$ state, and the state with $m=-\frac{1}{2}$ is referred to as the $\beta$ state.  If $\gamma$ [the gyromagnetic ratio] is positive, then the $\alpha$ state has lower energy than the $\beta$ state.

Therefore for protons $\alpha$ is lower energy (since $\gamma>0$). If the nucleus is $\ce{^{15}N}$ ($\gamma<0$), $\beta$ becomes the lower energy state.

So while your concern is valid, it is a relatively minor flaw (in my opinion) with the diagram. Good for you for noticing it.

Reference

John Cavanagh, Nicholas Skelton, Wayne Fairbrother, Mark Rance, Arthur Palmer, III. Protein NMR Spectroscopy, Principles and Practice, 2nd Edition. Academic Press, 2006, ISBN: 9780121644918.


There are many different types of crystals and their fromation is far too difficult for me to understand. This answer simply deals with the thickness part and that too merely gives a rough approximation due to the lack of data.
 From CalTech

We have two formulae for calculating the width:

For Dendrite
$$\mathrm{h =a_2 D^f} $$
D: crystal diameter (cm) (major dimension)
h: crystal thickness (cm) (minor dimension)
For Needles
$$\mathrm{d = a_2 L^f} $$
D: crystal length (cm) (major dimension)
d: crystal thickness (cm) (minor dimension)

This is a table for $\mathrm{a_2, f}$

For d and l   (right most column )

I calculated the thickness of a $\pu{3 mm}$ Dendrite which is coming around $\pu{24  \mu m}$.

Source

https://web.archive.org/web/20180416083128/https://www.nsstc.uah.edu/data/uahsevere/owles/parsivel/references/Snowflake_Size_Distributions_Lecture.pdf


Being nonpolar does not require having the zero dipole moment.

Enough is being much closer to having a zero dipole moment than to having the a strong dipole moment. It applies also to parts of the molecule.

E.g. carbon dioxide has zero permanent dipole moment, yet liquid carbon dioxide is rather polar solvent due its polar bonds.


What is the difference between a ligand and a co-ligand?

I believe this refers to situations in tranistion metal complexes where there is more than one kind of ligand, and specifically, where one of the coligands could not ligate and endure by itself due to reactivity, such as a carbene being blocked from reactivity due to a coligands steric bulk


Does nitric acid release hydrogen only when reacting with Mg and Mn?

Actually, it usually doesn't. It produces all types of products ranging from $\ce{NO2}$ to $\ce{NH4NO3}$ with exact products depending on concentration and conditions.

The reason for it is high oxidation power of nitric acid, exceeding that of hydrogen cations, so nitric acid reduces first. As result, to produce some nitrogen, the oxidation power must be brought below that of hydrogen cations. I'm not aware if it is actually possible, but possibly high dilution may help.

As for Mg, it reacts with hot water, so it's unclear if the reaction of $\ce{Mg}$ with diluted nitric acid counts. Not aware about $\ce{Mn}$ though.


Spontaneity of Reactions- Gibbs Energy or Entropy change

Neither enthalpy nor entropy alone can decide the spontaneity of a process.

For example, evaporation of water from oceans is an endothermic process $(\Delta H \gt 0)$, which clearly violates the fact that a system aims for minimum energy, yet this process is spontaneous (or naturally occurring).

To determine the spontaneity of a process the Gibbs-Hemholtz equation was introduced. This combines the effects of both enthalpy and entropy for determining the spontaneity of a process.

$\Delta G=\Delta H-T\Delta S$

A system generally aims for two things; minimum energy and maximum probability to be found in the universe (or more entropy). Therefore, for a spontaneous process, $\Delta H \lt 0$ and $\Delta S \gt 0$, Which gives an overall negative change is Gibbs energy.

For a spontaneous process, $\Delta G$ is negative.


Why at constant pressure and temperature Gibbs energy change of a process can be negative?

The relation $dG = VdP - SdT$ implies that the Gibbs free energy of the system depends only on the two variables $T$ and $P$, e.g., as in the case of a single-component gas consisting of a fixed number of molecules. For such systems, the macrostate is completely determined if $T$ and $P$ are given; nothing can change further.

On the other hand, suppose that there is an additional parameter $X$ describing the system and that $X$ is not fixed. Then, according to the second law of thermodynamics, the system, being at constant $T$ and $P$, self-adjusts $X$ in such a way that $G$ is minimized.

In case there exist multiple species that can undergo a chemical reaction, $X$ could be a measure of how far the reaction has proceeded. The condition $\partial G/\partial X = 0$ gives the well-known law of chemical equilibrium.


Why is molar mass of a substance and atomic mass number usually the same?

The term mole is a short form of the German word Molekül which was coined by Ostwald in 1890s. At that time, the "atomic mass unit" did not exist. The common term at that time was gram atomic mass or gram-molecule. The original meaning of so-called was defined to be equal one gram molecule i.e, the mass of a molecule expressed in grams. Atomic mass unit came into use by the 1950s.

These values match by definition not by coincidence. Here is how Ostwald coined the concept of a "mole". In those times molecules weights were determined by colligative properties. This is a snapshot of Findlay "Principles of Inorganic Chemistry" (translation of Ostwald's work)


Is calcium chloride an acidic or basic salt?

Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH above 11, i.e. a strongly basic solution.

$\ce{CaCl2}$ solutions should be very slightly acidic if they were made from pure $\ce{CaCl2}$. This might not be the case. In industry calcium chloride is produces by reaction of calcium hydroxide with ammonium chloride, so industrial-grade calcium chloride is likely to be contaminated with calcium hydroxide.

Another potential reason for contamination is hot drying. Calcium chloride is quite hygroscopic, so it needs to be dried before use. When heated to hight enough temperature, moist calcium chloride hydrolyzes https://link.springer.com/article/10.1007/BF02654424


Can the degree of dissociation of a weak acid be more than 1?

The $\sqrt K/\sqrt C$ ratio is only an approximation.  Let's do something more exact.  (Why do I say "more exact" instead of "completely exact"?  Read on, nature has a nasty curveball in store for us!)
Let $C$ be the nominal concentration of the acid, assumed monoprotic ($\ce{HA}$) for simplicity, and $K$ be its dissociation constant.  So, recognizing all ions are of course solvated, we have
$[\ce{H^+}][\ce{A^-}]=K[\ce{HA}]$
$[\ce{HA}]=C-[\ce{H^+}]$
$[\ce{A^-}]=[\ce{H^+}]$
So
$[\ce{H^+}]^2=K(C-[\ce{H^+}])$
$[\ce{H^+}]^2+K[\ce{H^+}]-CK=0$
From the quadratic formula
$[\ce{H^+}]=\dfrac{-K+\sqrt{K^2+4CK}}{2}=\dfrac{2CK}{K+\sqrt{K^2+4CK}}$
And our dissociation is
$\dfrac{[\ce{H^+}]}{C}=\dfrac{2K}{K+\sqrt{K^2+4CK}}=\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{C+K/4}}$
Now, if $C$ is large compared with $K$ then the terms in the denominator with $K$ become small compared with $\sqrt C$ and we simplify to $\sqrt K/\sqrt C$.  But if we dilute the acid, then the terms with $K$ in the denominator become dominant and we find a  limiting value of
$\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{C+K/4}}<\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{K/4}}=1$.
Thereby the dissociation remains limited to no greater than 100%.
There is one little thing we forgot.  If we keep diluting the acid so its concentration approaches the ion concentrations from water autoionization, then we also have to count the hydrogen ions from the autoionization into the equilibrium.  This extra source of hydrogen ions cuts down on the acid dissociation, so the limiting amount of dissociation is less than 100%.  For acetic acid it's still about 99.4%, but if we pick a really weak acid like $\ce{HCN}$ with a $pK_a$ above $7$ at ambient temperature, we never get more than a small fraction to dissociate unless we force the issue by reacting with a base.


Determination of Urea and Hydrogen Peroxide in Cow Milk by ATR-FTIR(Attenuated total reflection-Fourier Transform Infrared Spectroscopy)

In contrast to a typical sample provided by a synthetic lab your sample consists of multiple components; milk -- already consisting of many chemicals for its own, plus urea added, plus hydrogen peroxide added (as aqueous solution).  Using ATR-FT-IR is indeed a good choice for characterizing such samples, both for qualitative, as well as quantitative analysis (even to the point to provide kinetic data) as non-distructive, and potentially on-line method for continous process control.

This being said, you may improve the situation in the way how the ATR data is recorded.  Most often ATR-spectrometers are designed in a single-reflection geometry, wich is fine for not too complex sample compositions.  Potentially better though here is to use a design with a crystal allowing multiple reflections.  (Do not forget to keep the surface of the crystal clean, as (bio)films and absorbed material in general may affect the analysis, esp. with ATR flow cells.)

(source)

As a second hint, after the accumulation of your spectra, you may facilitate recognizing the barycentre of the absorption peaks by looking at their first derivative.  This technique sometimes seen in the field of UV-Vis spectroscopy (ref) is equally helpful if absorption bands are broad, or -- for example because a multi-component system is scrutinized -- partially overlap (e.g., IR-imaging).

At third, data analysis.  Beer-Lambert's Law, likely known from UV-Vis spectroscopy, equally is applicable in IR spectroscopy and you may choose picking an absorption band belonging only to your component of interest.  Because of the complexity of your sample, however, I would suggest that you use Partial Least Squares regression analysis (PLS) as a multi-variate analysis instead.  Very much simplified, instead of performing a regression analysis at one wavelength of interest, you do so for all data points, across the complete spectrum recorded, which equally allows to account for multiple components, too.  Do not worry, there are programs around to assist in this process, either provided with the IR spectrometer (e.g., TQ analyst by ThermoScientific), or independented solutions (e.g., Unscrambler).  Often, PLS allows to lower the detection limit compared to the approach based on a single absorption band quite a lot.

An example of determining hydrogen peroxide in water with PLS by ATR-FT-IR may be found here, of urea in milk here, or milk adultered with Hydrogen peroxide (0.15 g/L) here:

(original source of the article.)


$\ce{SiO2}$ forms a tetrahedral covalently bonded network , while $\ce{CO2}$ is a strictly molecular compound that forms double bonds between carbon and oxygen. This is because Si is in the next period, so it has 3p orbitals instead of 2p orbitals. Orbitals of a higher shell (greater principal quantum number) are physically bigger, but oxygen still has smaller 2p orbitals, therefore there is less overlap between silicon and oxygen’s p orbitals than there is between oxygen and carbon’s p orbitals. This makes a Si-O π bond much higher in energy, so it does not form. Instead, only a lower energy tetrahedral network of σ bond forms, giving us silica as we know it. This explains why $\ce{SiO2}$ cannot form double-bonded molecules like $\ce{CO2}$. Also, single Si-O bonds are especially stable (466 kJ mol-1) and short. This is attributed to the overlap between oxygen’s p orbitals and silicon’s empty d orbitals. Additionally, silicon is a larger atom than carbon, so the bonding electron pair repulsion between the oxygens bonded to silicon in SiO2 is much lower than what would occur if such a tetrahedrally coordinated CO2 network could exist. Silica’s behaviour can, thus, be explained by enthalpy, MO theory and VSPER.

Crystal $\ce{CO2}$ can definitely be made in a lab, and it is known as dry ice. $\ce{CO2}$ freezes at -56 °C. It boils at -78 °C and you will often see this temperature written under arrows in synthesis because dry ice is the most common method of carrying out reactions at such a low temperature. At atmospheric pressure, the dry ice crystal structure does not, however, resemble network covalent silica, but some high-pressure solid $\ce{CO2}$ phases have network covalent structures. Here is a phase diagram:

Phase I $\ce{CO2}$, shown in the bottom-left corner, is the most common dry ice crystal structure, and it is molecular, exhibiting a face-centered cubic arrangement with respect to the carbon. The high-pressure Phase V $\ce{CO2}$ (top-right), on the other hand, has a "partially collapsed cristobalite" structure. Isostructural with one of silica's polymorphs, it is non-molecular.

As for gas-phase $\ce{SiO2}$, it would certainly be hard to make, as it boils at close to 3000 °C, presumably for the bond-stability reasons listed above. I haven't read amy literature about its gas-phase structure, but I assume it would be similar to that of $\ce{CO2}$ but extremely unstable.


Number of stereoisomers of tetramethylspirononane

The compound 2,3,7,8-tetramethylspiro[4.4]nonane has five potential stereogenic or pseudoasymmetric centres: the four C atoms with the methyl substituents and the spiro atom. In total there are $2^5=32$ possible configurations. However, many of these combinations are actually identical compounds.

A simple and obviously exhaustive (but also exhausting) approach is to draw all 32 configurations and name them. (This approach may also be used to check the results if someone finds a more elegant solution.) The resulting names for the 32 combinations are:

(2R,3R,7R,8R)-2,3,7,8-tetramethylspiro[4.4]nonane (2 times)

(2R,3R,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (8 times)

(2R,3R,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)

(2R,3S,5r,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)

(2R,3S,5s,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)

(2R,3S,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (8 times)

(2S,3S,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (2 times)

Therefore, there are seven different combinations.


Why highly unstable cyclic boron compounds should be stabilised by two electron reduction?

Your energy diagram is that of a linear moiety, like an allyl cation. For a cyclic structure, it would be somewhat different, but the general idea remains the same: there is one $\pi$-bonding orbital, and when two electrons go to it, everyone wins.

As for vibrational spectroscopy, well, it can help you tell the cyclic structure from linear, if that's what you want.


Accessible statement on the intuition and usefulness of group theory to chemistry

Individual molecules rely on point group symmetry - that is, the symmetry about a point. Crystal structures rely on space group symmetry including the translational symmetry in a crystal.

Many molecules, particularly important ones, are highly symmetric. The different symmetry operations (reflections, rotations, etc.) form a finite group.

There are a number of interesting group theory results, but a simple statement comes down to two things:

Understanding Orbitals, Bonding, Vibrations..: Using point groups, any function of the molecule corresponds to a linear combination of irreducible representations. This means I can quickly understand vibrations of a molecule, spectroscopy (as you mentioned) and orbital diagrams and bonding using just pencil and paper.
Faster Calculations: Using symmetry, you can greatly reduce the amount of time required for quantum calculations. For one, you only need to consider the symmetry-unique atoms and bonds. For another, you can show that certain types of integrals and calculations will always be zero and can be completely ignored. This leads to immense speedups.

For example, chemists label orbitals with symmetry labels based on the point group. Using the direct product we can realize that certain interactions will never occur.

Similarly, vibrations have symmetry labels, and we can determine what types of spectroscopy will interact with those vibrations, or whether they are "silent."


Connecting a computer to Genesys 10S Spectrophotometer

This is a long comment....
The webpage specification says that there are three USB connections.

USB Type A port for USB memory device (front panel)
USB Type B port for optional computer connectivity (rear panel)
USB Type A port for external printer (rear panel)

The printer uses the HP PCL 3.0 specification via a USB cable. You should be able to get the HP specification and emulate that to get the data. The spectrometer doesn't know what it is communicating with...
The type A port on the front is supposedly for a USB stick for data storage. I'd guess that the Thermo Scientific software formats the stick in some screwy manner. You'll need the Thermo Scientific  specification to duplicate that. You might be able to get someone with the software to format some USB sticks for you. No idea what format the data format would be. I'd also guess that the Thermo Scientific software manual doesn't detail the format of USB data.
The USB Type B port on the back must have some bidirectional communication protocol. Unknown if Thermo Scientific would release details of the communications specification even if you bought the program. If you know someone that has such a computer instrument setup then you could install a sniffer in the USB communications line and figure it out. A lot of work.


What is the difference between BET and Langmuir model for calculating the specific surface area?

Langmuir adsorption model considers that only a monolayer of adsorbate can be formed on top the surface.
In contrast, BET model considers multilayer formation.

Then, if (as generally occurs) multilayers are formed, BET is able to consider that they are not being formed by directly covering the surface, but by covering extra layers. Langmuir will consider that multilayers can only form as the first adsorbed layer, thus resulting in an exaggerate (unrealistic) surface area.

You can clearly see the differences in the Wikipedia page for Langmuir adsorption.


Difference between Superheated steam and Superheated water?

Boilers and similar steam generators usually produce wet steam, which is a mixture of saturated steam and liquid water droplets. Saturated steam is steam that is in equilibrium with liquid water. The temperature of the steam is equal to the temperature of the liquid water; it corresponds to the boiling point at the given pressure. At normal pressure ($p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$), the temperature is $T=99.974\ \mathrm{^\circ C}$. The corresponding specific enthalpy of the steam is $h = 2676\ \mathrm{kJ/kg}$; the specific enthalpy of the liquid water is $h = 419\ \mathrm{kJ/kg}$.

If heat is supplied to the wet steam, liquid water vaporizes. At constant pressure, the temperature remains constant until all liquid water has been vaporized, i.e. until the steam is dry.

If further heat is supplied to the dry steam, the temperature and the specific enthalpy of the steam increase. Such steam at a temperature higher than its boiling point is called superheated steam. For example, superheated steam at normal pressure ($p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$) and a temperature of $T=150\ \mathrm{^\circ C}$ has a specific enthalpy of $h = 2777\ \mathrm{kJ/kg}$.

Superheated water is liquid water under increased pressure at temperatures above the boiling point at normal pressure. For example, water at a temperature of $T=150\ \mathrm{^\circ C}$ and a pressure of $p=506\,625\ \mathrm{Pa}=5\ \mathrm{atm}$ is liquid. The specific enthalpy of the liquid water is $h = 632\ \mathrm{kJ/kg}$.

The energy content of superheated steam (e.g. $h = 2777\ \mathrm{kJ/kg}$ at $T=150\ \mathrm{^\circ C}$ and $p=101\,325\ \mathrm{Pa}=1\ \mathrm{atm}$) is much higher than the energy content of superheated water (e.g. $h = 632\ \mathrm{kJ/kg}$ at $T=150\ \mathrm{^\circ C}$ and $p=506\,625\ \mathrm{Pa}=5\ \mathrm{atm}$) because of the latent heat, i.e. the enthalpy of vaporization $\Delta H_\text{vap}$.


What safety measures should I take when operating a backyard aluminum furnace

Set this up in an area free of leaves and debris.

I recommend wearing safety gear: heat resistant goggles and fire-resistance clothing and gloves that will protect you from heat, especially when pouring.

Get a couple practice runs in pouring before using it so you get a feel for the best angle to hold in and learn what NOT to do. Even better, you could assemble something that would do the pouring for you. (I'm thinking of something along the lines of this, but it doesn't have to be as elaborate.)

You may also want to scrape the paint off the cans before putting them in. Otherwise, you'll  get a lot of crud from the paint in your molten aluminum.


Whenever you want a definition of a chemical concept, you should refer to the IUPAC's Gold Book which in this case states that antiaromatic compounds are

  compounds that contain  4 n(n ≠ 0) π-electrons in a cyclic planar, or nearly planar, system of alternating single and double bonds, e.g. cyclobuta-1,3-diene.

A corollary to this rule is that all the atoms must have an sp2-hybridization, so the orbitals can overlap. However, the carbon bearing the negative charge has an sp3-hybridization, i.e. they does not count as π-electrons but σ-electrons and that breaks the rule of alternating single and double bonds. So both anions are non-aromatic.

Edited after the OP opened my eyes on a big mistake I had made in answering his question. Sorry about that.


Are some alkali metals better conductors than others?

Li
Conductivity 1.08×10^7 S/m
Resistivity  9.26×10^(-8) mΩ
Na
Conductivity 2.18×10^7 S/m
Resistivity  4.59×10^(-8) mΩ
K
Conductivity 1.43×10^7 S/m
Resistivity  7.0×10^(-8) mΩ
Rb
Conductivity 8.0×10^6 S/m
Resistivity  1.25×10^(-7) mΩ
Cs
Conductivity 5.3×10^6 S/m
Resistivity  1.9×10^(-7) mΩ
Fr
Conductivity 1.4×10^6 S/m
Resistivity  7.1×10^(-7) mΩ


Attachment of disulfide (protected) thiolated oligonucleotide on gold surface

What actually happens in sulfur-gold bonding is not very well known. There have been many proposals and studies done, well-summarized here. There is some evidence, including computational studies, that suggest that dissociation into two thiolates is energetically favourable. Bridging across protruding bits of gold (we often draw gold surfaces as straight lines, but they're seldom that flat) may also be involved. This case, where you only have a chain on one of the two sulfurs, may be less interesting and I haven't found many studies of terminal disulfides so the exact process might be different.

To answer the practical question, Can I stick a disulfide-functionalized molecule onto gold?, yes you can, and they will stick very well.


Is there a substance that will eat through and completely dissolve carbon but not damage engine cylinder, piston nor rings?

If it's pure carbon, then the only way to remove it I can think of is by rigorous oxidation (e.g. by burning in oxygen flow or by reaction with strong oxidants (e.g. conc. $\ce{H2SO4}$, $\ce{HNO3}$)), but this is going to damage the metal parts either, as Ivan said.

However, if the clog is caused by soot and resinous residue from the incompletely burnt fuel, then most likely hot toluene or xylene can dissolve the residue and liberate the carbon particles. Both are widely available solvents which won't damage the metal parts (but avoid contact with rubber tubing) and are also used to remove clogs caused by paraffin wax.

P. S. Both are highly flammable liquids, so do not warm them up using an open fire (e.g. use an improvised water bath)!


Are all carbohydrates a subset of alcohols?

Whenever you see -OH is bound to a saturated carbon atom in organic chemistry it is Alcohol group.

So answer is YES. All carbohydrates are  subset of alcohols.

Though in linear forms they also have aldehyde or ketone group.


Calculate force on atoms given the energy of configuration, trajectory of simulation and position of atoms in each frame

tl,dr: Forces are equivalent to the negative of the gradient, use small, arbitrary $\mathrm{d}r$ for each atom

The problem, as you mention, with many neural network methods in chemistry at the moment, is that the $E = f(x,y,z)$ is a black box and thus analytical gradients (and thus forces) are often not available.

First, as you clearly know, $F = -\nabla E(x,y,z)$ so you need to solve $\frac{\mathrm{d}E}{\mathrm{d}r}$ for your neural network. (I indicate this for anyone reading who doesn't realize that the force will inherently be the negative of the gradient of the energy function.

So what you need is a finite difference method. This is, of course, a common problem (e.g., on Math Stack Exchange or Wikipedia):

$$ \frac{\partial E}{\partial x} \approx \frac{f(x+\epsilon)-f(x)}{\epsilon}$$

Picking a good step $\epsilon$ is a bit tricky, but I'd probably pick something like 0.01 Å or 0.001 Å, as a start. In your case, it sounds like you want Cartesian gradients, so you'll take your $f(x,y,z)$ and perform the partial derivatives for all three axes (i.e., the initial point $(x,y,z)$ plus 3 extra energy evaluations per atom.

That is, you'd loop through each atom, move it a bit in the x, y, and z directions and evaluate the system to get the gradient/force on that atom. You'll, therefore, have $3N$ energy evaluations to get all the forces on all the atoms.

You can perform better (and more computationally expensive) finite difference methods if you choose.


R, S Configuration and Branching - Which Path?

No,  you have a carbon figuratively attached to 3 oxygen atoms vs.  a carbon figuratively attached to 3 carbon atoms; the former has higher priority.


Why is twice as much hydrogen produced than chlorine in the electrolysis of brine?

It appears, as the others have said that the same number of moles of gas of chlorine and hydrogen is produced.

Then why are there different levels of solution?

Since the solution is taken out in the cathode chamber, the level of solution present is lower than the anode chamber where sodium chloride solution is being added in.


In H20 it is sp3 hybridisation, but in H2S it is purely p3 orbital , there is no hybridization in H2S, it cannot afford the cost of hybridization you may recall that all the p orbitals I.e. Px,Py, Pz are 90 degree to each other as along x y and z axis respectively and if u compare this with the Bond angle of H2S which is 92.12 degree( almost 90 degree) which clearly shows that it is formed by p orbitals only, hybridization is a costly process , when sulfur can easily bond with p orbitals itself , why will it go for expensive process of hybridization unnecessarily isn't it?


Why is incubation in dark place necessary?

Studies have not yet defined the exact mechanism of Lowry method. However, incubation in dark is suppose to promote the stability of copper tartrate complex with Folin-ciocalteu solution and thus their reaction/interaction with protein which ultimately leads to the oxidation of aromatic residues. You can read chemistry behind Lowry method for more details.


Dilution of whisky – the molecular perspective

What the paper says is that in that range of concentrations, that the excess of alcohol at the air-liquid interface is enough to extract some of the guiacol from the bulk solution. At higher bulk concentrations of ethanol, there's no particular benefit for the guiacol to separate to the interface (there's enough ethanol in the bulk to keep the guiacol satisfied, so to speak). I forget whether the paper determined if there was a concentration (proof) below which the guiacol did not segregate, but if there was, it would be because the slight increase in alcohol concentration at the interface wasn't attractive enough. In other words the guiacol only is preferentially concentrated at the interface when the difference between the concentrations of alcohol at the interface to the concentration in the bulk makes it worth-while for it. (Worthwhile here means the difference in energy is large enough). Here's an analogy (My family would now roll their eyes, my analogies are lousy). A little kid is deciding whether to get up from his video game and go trick-or-treating. There's no candy in his house, and maybe he'll get a half dozen or a dozen pieces if he hits 20 or 30 houses/doors. Nah, he says, not worth it. But what if he's in a rich neighborhood? If he hits 10 doors, he's sure to get at least 2 dozen, maybe 3 or 4 dozen pieces of candy. Well, if his home hasn't much candy then he'll probably go, but if his home is chock-full of candy (rich neighborhood, remember?) then he may decide it's not worth it either. The choice he makes depends not just on the difference in the absolute amount of candy he will end up with, but also depends on how much he values an extra couple of pieces. It's non-linear, or as Liberals might say "context sensitive". (added: just to be clear, going trick-or-treating is analogous to concentrating at the air-liquid interface, not going is analogous to staying in the bulk phase.)


Can you create a temporary bond in between an atom and a filter by electrically charging the filter?

In a sense, you can increase and decrease flow through an osmotic filter with electricity. The process is called electro-osmosis. This can even be used to create a pressure differential in the electroosmotic pump.


A “formula unit” would be closer to the answer than a “unit cell”. A unit cell is used in crystallography to define the smallest repeating unit of the crystal that has all the symmetries of the overall crystal structure. For NaCl, there are a variety of different ways to visualize the unit cell (2 interpenetrating FCC lattices, FCC with a 2-atom basis, etc. ), but regardless, there are four formula units of NaCl in the NaCl unit cell (4 Na and 4 Cl atoms).


Why do metallic bonds have delocalized electrons, whereas covalent bonds only share electrons?

To be somewhat flip this is like asking why are redbirds red and blue birds blue?

In order to pigeon-hole bonding ionic, covalent, aromatic, and metallic bonds are "artificial" constructs used to predict behavior. But there are always exceptions, and no bond is purely in any one of the categories.

So in general a sample which has metallic bonds has delocalized electrons and hence will conduct electricity. Such elements would be metals. If the sample only has covalent bonds then there would not be any delocalized electrons and the sample would be a nonmetal.

For some elements the nature of the bonding yields different results. For example graphite conducts electricity, but diamond won't.

To add another wrinkle in this we can think of gaseous ionic or covalent molecules but in general being metallic is a "bulk" property. So we have to consider not only the bonding in the "molecule", but how "bonding" is distributed through the solid matrix of the sample. For example there are discrete water molecules in ice, but in a diamond there are no discrete carbon "molecules."


Let's suppose a hypothetical irreversible reaction:
$$\ce{B + H -> D + F}$$
where H is a reactant with constant concentration during the reaction and that don't absorb light in the working frequencies. The absorbance of a species $i$ related with the concentration by the Lambert-Beer Law:
$$A_{i}=c_{i}\varepsilon_{i}l$$
where $c_i$ is the concentration of species $i$, $\varepsilon_{i}$ is the molar absorptivity and $l$ is the length of the cuvette.

At an arbitrary time $t$, the total absorbance of the solution is given by:

$$A_{t}=c_{B,t}\varepsilon_{B}l+c_{D,t}\varepsilon_{D}l+c_{F,t}\varepsilon_{F}l
\tag{1}$$

by the stoichiometry of the reaction, we have: $c_{D,t}=c_{F,t}=c_{B,0}-c_{B,t}$; where $c_{B,0}$ is the initial concentration of the reactant B, so replacing this in Eq. 1 and defining a auxiliary constant $\varepsilon_{i}l=k_{i}$, we have:

$$A_{t}=c_{B,t}k_{B}+k_{D}(c_{B,0}-c_{B,t})+k_{F}(c_{B,0}-c_{B,t})$$
$$A_{t}=c_{B,t}(k_{B}-k_{D}-k_{F})+c_{B,0}(k_{D}+k_{F})\tag{2}$$

At a time $t\rightarrow \infty$  (a time long enough that the reaction goes to completion), $c_{B,\infty}=0$ and $c_{D,\infty}=c_{F,\infty}=c_{B,0}$, so the absorbance is given by:

$$A_{\infty}=c_{D,\infty}k_{D}+c_{F,\infty}k_{F}$$
$$\boxed{A_{\infty}=c_{B,0}(k_{D}+k_{F})} \tag{3}$$

so,replacing Eq.3 in Eq.2 its follows that:

$$A_{t}=c_{B,t}(k_{B}-k_{D}-k_{F})+A_{\infty}$$
$$\boxed{A_{t}-A_{\infty}=c_{B,t}(k_{B}-k_{D}-k_{F})}$$

In conclusion: $A_{\infty}$ is important because is directly proportional to the initial concentration of reactant B. When you subtract $A_{\infty}$ from the absorbance at an arbitrary time $t$ $(A_{\infty})$ it allows you to know (indirectly) the concentration of the reactant at the aforementioned time $t$.