In fact you have three configurations to consider: t-shaped, parallel-displaced and sandwich. The dissociation energy depends on the configurations of the structures since they may interact differently. As the pi-pi interaction is directional, the different structures of the dimer interact differently.


Rotovapping Aqueous Solutions - Need a cold trap?

You will always want a cold trap not only to protect the pump but also to make sure it's running at peak efficiency at all times. Especially when you push the vacuum and heating to try to expedite the process, you never know when the whole thing might suddenly bubble up. A clean cold trap will also allow you to recover the material that has escaped the main chamber, whereas without one you'd probably be faced with loss of compound and having to repair the pump.


What is the effect of basis set on charge transfer?

Mulliken charges are not representative of charge transfer, I would argue, in any case. Mostly because of their enormous basis set dependency, as Geoff pointed out.

In any case, several points are missing here; for instance whether the geometries were optimized with each method/basis set combination. This is not trivial.

Assuming a single static geometry (or set of geometries) for all single point calculations, if you go with an MP2 approach, the larger basis set will give you the best result. This is so by definition. In fact, you should always use a large basis set with MP2 because this method converges slowly and requires a large amount of virtual orbitals. In DFT you never know, but I would try to use modern basis sets, such as Ahlrich's (RIP) "def2tzvp" (naming conventions change, but you will find it). Such basis sets are optimized for DFT, and are generally diffuse enough (as a rule of thumb, you should always include diffuse functions if you have an anion in your system).

For your goal, several methodological problems arise. First, if you are indeed optimizing the geometry, B3LYP might be very bad. Dispersion corrections or range separation might help, but you can check the B3LYP-MP2 difference to see if anything suspicious arises.

Secondly, defining charge transfer is not easy. It is not trivial in SAPT (if you want to use SAPT i'd recommend PSI4 dearly for speed, there's a paper by A. Misquitta about charge transfer in SAPT but its not trivial to implement), which is the most rigorous energy term decomposition scheme, and in principle free from intermonomer BSSE.

Defining charge-transfer is somewhat easier in some flavours of the many available EDAs, as in the Kitauma-Morokuma scheme which I think is implemented in GAMESS. Real space EDAs, as the Interacting Quantum Atoms method, might be useful here as well. I think real space charges are far more convenient than Mulliken's in any case. You see, charge-transfer is a nice term but a very diffuse quantity.

Otherwise, you might be better off by checking some more available, well defined observable. For example, the dipole moment of the dimer will capture the charge separation.

In any case, good luck with this project!


There are all elements in your exercise to solve it expect the heat capacities but this is easy to find them.

You are doing an exercise of thermodynamics and this is not mention that the equilibrium is not reached. So if your water heats up only by $10\ \mathrm{^\circ C}$ at the end even in one million years if your system is adiabatic, your solution will still be at $20\ \mathrm{^\circ C}$. And so at the equilibrium the copper must have the same temperature than the water.

So if you use $\Delta H=mc_p\Delta T$, for water $\Delta T=10\ \mathrm{^\circ C}$ and for copper $\Delta T=30\ \mathrm{^\circ C}$.

Are you able to finish now ? :-)


How to calculate hydrogen from SMILES string?

Other answers have done a good job of providing a description of the concepts and algorithms needed to answer the question.  Here I'll demonstrate the opposite: a black-box empirical approach.

In Python using rdkit:

from rdkit import Chem
from rdkit.Chem import rdMolDescriptors

my_smiles_string = 'C1=CC(=C(C=C1C(CN)O)O)O'
my_mol = Chem.MolFromSmiles(my_smiles_string)
print(rdMolDescriptors.CalcMolFormula(my_mol))

This shows the result C8H11NO3.  You could parse that further to get only the number of hydrogens, if you need to do so.  Alternately:

my_mol_with_explicit_h = Chem.AddHs(my_mol)
print(my_mol_with_explicit_h.GetNumAtoms() - my_mol_with_explicit_h.GetNumHeavyAtoms())

displays 11.  Slightly more cumbersome but it has the output you want.  If you take the latter approach you don't even need to import rdMolDescriptors.


Resonance only occurs when the result is beneficial, that is when a more-stabilized molecule will result. If resonance (delocalization) occurred in a compound such as cyclobutadiene to produce a square molecule with equivalent bond lengths, an antiaromatic molecule would be the result.  Therefore, resonance does not occur and a square molecule with equivalent bond lengths is not produced.


What does thermal stability of compounds depend on?

Lattice energy might be important if the decomposition took place in the solid phase.
But Li2CO3 melts at 723 C and boils (with decomposition) at 1310 C. Na2CO3 melts at 851 C and decomposes before boiling, according to an older CRC Handbook and Wikipedia, which seems to be pretty current.  According to those numbers, thermal stability of sodium and lithium carbonates seems fairly similar, and decomposition does not take place because of solid issues.

K2CO3 melts at 891 C, decomposes before boiling. Rb2CO3 melts at 837 C, decomposes at 900 C. Cs2CO3 melts/decomposes at 610 C. Lattice energy might be important with Rb and Cs, but thermal stability seems to become less stable as you go down the group, not more so.

The first three seem fairly stable for several degrees above the melting point. We could consider the melt to be a mix of CO3-- anions with the metal cations, all oxygen-bonded (similar to hydrogen bonding!), but much looser than in a crystal. The thermal stability of the CO3-- anion does not seem to be very dependent on the ionic size of these three metals (Li+, Na+, K+).

So, I would ask the textbook author to substantiate his claims.


How do enzymes affect a reaction's equilibrium?

I believe you have misread the first referenced paper. What it says is that the $V_{max}$ and $K_m$ values for the forward and reverse reactions can be varied to optimize a property such as the maximum velocity (represented by $V_{max}$) in one direction, but are always constrained by the fixed value of $K_{eq}$.
The reason that the $K_m$ values (which are pseudo equilibrium constants) can be adjusted is that they represent a reaction in which the enzyme is a reactant. For the substrate, this reaction is approximately $\ce{E + S <=> ES}$.
Likewise, the second paper also describes how the $V_{max}$ for one direction or the other can be maximized, but again subject to the constraint that the respective $K_m$ values must vary such that the overall $K_{eq}$ maintains its fixed value.


Does any element have nutritional (caloric) value?

Interesting question! You may get a better answer, but I believe the answer is no.

Calories are units of heat and represent energy available to the body. The form of fuels that can be (biochemically) "burned" by the body are hydrocarbons; i.e. molecules of carbon and water, like sugars. Carbon alone I do not believe can work.

The only other single element fuel is hydrogen, and again I do not think this is biologically available to the body. By the way, oxygen is required as the oxidizer in any combustion reaction, and we definitely need oxygen (from breathing) to convert the food we eat to energy. But oxygen is not a fuel: it is the oxidizer that is reduced during combustion.


Luckily for us, a really smart guy named Josiah Willard Gibbs figured out how these two tendencies work together back in the late 1800's. As a result, in chemical thermodynamics we now have a state function called the Gibbs Free Energy that describes what is called the thermodynamic potential of a system at constant pressure, temperature, and number of molecules.

Since most of the systems we work with under lab conditions are at constant pressure and temperature and are typically closed to mass transfer (abbreviated as NPT), Gibbs free energy is a very convenient way of determining what will happen to a given chemical system under normal lab conditions.

The reason it works is that it is a potential energy function - it describes how a system's potential energy will change as different state variables change, and therefore lets us predict where the system will "try" to go. An analogy that works pretty well is to imagine it like the surface of a planet, where gravity is the potential energy. If the derivative or slope is negative in a given direction, then things will tend to "roll downhill" at that point. On the other hand, to make things go uphill, you need to add energy.

The other thing we can get from this analogy is that given the opportunity, things will eventually move from a state of higher potential energy to a state of lower potential energy. So we can compare the energy at two different states (the reactants and products, for example), and decide based on the difference in potential energy whether the reaction would be spontaneous under those conditions.

The equation for Gibbs free energy is:

$G = H - TS$

Where $H$ is enthalpy (a.k.a heat of reaction), $T$ is absolute temperature (usually measured in Kelvin) and $S$ is entropy.

If you take the derivative at constant temperature and pressure, you get:

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

(This is for large changes. for small changes replace $\Delta$ with $\delta$). What this tells us is that change in Gibbs Free Energy is a function of change in enthalpy, change in entropy, and the absolute temperature.

To illustrate how this works, take a look at the following diagram.

Here the line represents the Gibbs Free energy "surface" - the thermodynamic potential as a function of some variable (a reaction coordinate, for example). The ball represents the system at some point along that coordinate axis. If this were regular potential energy, it would be easy to see what happens - the ball will roll downhill if it gets the chance, and it will stop at the lower energy state. To get back up the hill, someone would have to put in that amount of energy. For thermodynamic potentials, it works the same way - the system will move along the coordinate axis in the direction of decreasing thermodynamic potential energy ($G$ in this case).

This means that under constant NPT conditions, any process that involves an overall decrease in Gibbs Free energy will be spontaneous. It will also tend to move in the "direction" that has the most negative slope in G at any given time. In mathematical terms,

$\Delta G < 0$ - overall process is spontaneous

$dG < 0$ - process will be moving in that direction

$dG = 0$ - process is at equilibrium

So how does this all fit in with what you described about systems trying to reach minimum energy, and maximum entropy? The answer is: those are both thermodynamic potentials for different types of systems. For a system with constant entropy, volume, and number of particles, (NSV) the internal energy (or total energy) is the thermodynamic potential. For a system at constant entropy and pressure (NSP), enthalpy is the thermodynamic potential. And for a system at constant volume and internal energy (NVE), the negative of entropy is the thermodynamic potential.

In other words, all of these thermodynamic variables are interconnected, and a change in one affects all of the others. You can hold three constant at any given time and still allow the system to "move" through phase-space. Which thermodynamic potential you need to describe how the system will move depends on which variables you choose.

Let's look at the equation for $\Delta G$ and see how changes in enthalpy and entropy affect it.

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

If $\Delta H$ decreases, it will make $\Delta G$ decrease as well. This makes sense, since we know that exothermic processes tend to be spontaneous, because they are releasing energy and therefore the final system energy is lower than the initial.

$\downarrow \Delta G = \Delta H - T \space \uparrow \Delta S$

On the other hand, $\Delta G$ tends to decrease as $\Delta S$ increases - this is because the change in entropy is subtracted in the equation. This also matches up with what you know - an increase in entropy indicates a spontaneous process.

For your last question:

  I am really having a hard time in figuring out a reaction where energy decreases and entropy increases. Can this happen?

Yes, it can! In fact, under these conditions, the process is guaranteed to be spontaneous - if $\Delta H$ is negative, and $\Delta S$ is positive, then $\Delta G$ has to be negative - the reaction would be spontaneous at any temperature under these conditions.

Let's look at the other possibilities:

$\Delta H > 0; \Delta S < 0$

In this case, since $\Delta H$ is always positive, and we are subtracting a negative $\Delta S$, the reaction can never be spontaneous.

$\Delta H > 0; \Delta S > 0$

$\Delta H < 0; \Delta S < 0$

In these two cases, the reaction could be spontaneous or it could be non-spontaneous - it depends on the relative magnitudes of the enthalpy and energy terms as well as the temperature at which it occurs.

To summarize - all thermodynamic variables are related to each other in fairly complicated ways. When you hold three of them constant (two if you don't count number of molecules), you can derive thermodynamic potential energy functions that describe the behavior of the system in terms of a single quantity. For most cases in chemistry, Gibbs Free Energy is the thermodynamic potential that we use. It gives us the relationship between enthalpy, entropy and temperature under constant pressure conditions. Since it's the thermodynamic potential, it also lets us predict how the system will behave - how it will move through phase space.


Oxidation - Reduction Ionic Half Reaction

Credits to my awesome teacher for this method.
(I had to add a potassium ion to the right hand side, there was no other way)


How will the pressure of a gas in manometer change if it goes to a place with higher pressure?

Assuming an ideal gas with T constant, $$ PV = nRT $$ the entire right-hand term is constant. Therefore$$ P \propto \frac{1}{V} $$ which is Boyle's law. Since the atmospheric pressure goes up, the liquid portion of the manometer will go down on the side open to the atmosphere, and it will go up on the side open to the gas. Since the volume of the gas goes down, its pressure must increase by Boyle's law.


The best way to make magnesium carbonate is to opt the laboratory route (mix magnesium chloride and sodium bicarbonate) as described in the Wikipedia article of magnesium carbonate.

$$\ce{MgCl2(aq) + 2NaHCO3(aq) → MgCO3(s) + 2NaCl(aq) + H2O(l) + CO2(g)}$$

The advantage of this process is that the byproduct do not interfere with the main product (sodium chloride is soluble in water, carbon dioxide bubbles out) leaving behind magnesium carbonate as precipitate.

Alternatively, magnesium carbonate can be prepared from magnesium carbonate by bubbling carbon dioxide in it. The advantage of this process is that there are no byproducts.

Note: If you use sodium carbonate, you will get a  precipitate of basic magnesium carbonate—a hydrated complex of magnesium carbonate and magnesium hydroxide—rather than magnesium carbonate itself is formed.

$$\ce{5MgCl2(aq) + 5Na2CO3(aq) + 5H2O(l) → Mg(OH)2·3MgCO3·3H2O(s) + Mg(HCO3)2(aq) + 10NaCl(aq)}$$

You can see that a basic salt is formed and also magnesium bicarbonate is formed which add to the difficulty of separating them and decomposing them to get magnesium carbonate and thus this process is not feasible.


Calculating ΔH with non-integer coefficients in the balanced equation

There is not much to clarify: you determined dimolar enthalpy of combustion, whereas standard enthalpy of combustion is determined for a complete oxidation of one mole of the substance:

$$
\begin{align}
\ce{C2H6(g) + 7/2 O2(g) &-> 2 CO2(g) + 3 H2O(l)} &\quad &Δ_\mathrm{c}H^\circ\\
\ce{2 C2H6(g) + 7 O2(g) &-> 4 CO2(g) + 6 H2O(l)} &\quad &Δ_\mathrm{c}H^\circ ×2
\end{align}
$$

When composing the corresponding reaction equation, make sure to balance it in such a way that the stoichiometric number $ν$ of the substance the enthalpy is determined for is always $-1.$


IUPAC way of writing prefix of alkyl isomers of an ether

According to Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book) (P-63.2.2.2), the preferred prefix for the substituent $\ce{(CH3)2CH-O\bond{-}}$ is 'propan-2-yloxy'. Both 'isopropoxy' and '1-methylethoxy' remain permissible in general nomenclature.

Correspondingly, the PIN of the symmetrical ether $\ce{[(CH3)2CH]2O}$ is 2-[(propan-2-yl)oxy]propane.

I'm quite sure the double parentheses are required based on other similar examples given in the Blue Book, but am not 100% sure of the rationale for it and would be grateful if somebody could confirm that. I'm thinking one set comes from P-16.5.1.1 as 'propan-2-yloxy' is a compound substituent, and one more comes from P-16.5.1.2 to separate the locant '2' in 'propan-2-yl' from the remainder of the name, but don't quote me on that.


Difference between reversible and irreversible processes

In a reversible process, at each point along the process path, the system is only slightly removed from being in thermodynamic equilibrium with its surroundings.  So the path can be considered as a continuous sequence of thermodynamic equilibrium states. For an irreversible process, the system is not close to thermodynamic equilibrium with its surroundings at each point along the path.

At any point along the path of a reversible process, both the system and the surroundings can be returned to their original states without significantly affecting anything else.  This cannot be accomplished if the process is irreversible.


The "dep" probably stands for "depletion": read the paper you linked to for more context. Often times, subscripts on rate constants are used to refer to which reaction in a system of multiple reactions. You'll see $k_1$ and $k_2$ for rate constants of "reaction 1" and "reaction 2", etc.

I suspect in the paper, they were measuring the depletion of a substrate, and not what the substrate was converted to.  So they wouldn't know if they were measuring the rate of a single reaction $\ce{A->B}$, or the combined rates of multiple reactions such as $\ce{A->B}$ and $\ce{A->C}$.  Thus (probably) the desire to indicate $k_{depletion}$ instead of just $k$.


How does the inverted Marcus region explain chemiluminescence?

Marcus theory of electron transfer shows that it is possible for a reaction with a large negative free energy ($-\Delta G$ ) to be slower than one with a less negative free energy. The region where this happens is called the 'inverted' region and is shown in figure (c) in your diagram. You can see that the energy of the products is far below that of the reactants but that there is still a substantial activation energy, that is the energy from the minimum of the R curve to the crossing point of the R-P curve on the left.

In fig (A) this is called the 'normal' region and the free energy is close to zero and there is still an activation barrier, in (b) the reaction is slightly exothermic has an activation energy and is still in the 'normal' region. What is missing is the case when the reaction is activationless, this lies between figs (b) and (c) and occurs when the P curve crosses the R curve at its minimum. In this situation the reaction has its maximum rate constant.

The figure shows some data illustrating the rate constnats for electron transfer in the normal(left side) activationless (top) and inverted inverted (right) regions vs $-\Delta G$. The data is for electron transfer from bases in DNA to different intercalated dyes. The fitted Marcus curve is the solid line one from classical Marcus theory the other from a quantum calculation.

Hopefully this will description allow you to follow the paper you have been reading. (You may also want to  look at BBA v811, p265, 1985)

It was not until late 1980's that Marcus theory was confirmed experimentally, first in the solid state and then with chemically linked donor and acceptor molecules as this prevents diffusion, which had hitherto greatly confused interpreting data, and also meant that the inverted region could not be observed.

The energetics for an electron transfer reaction

$$\ce{D^* +A \rightarrow D^+ + A^-  }$$ is

$$\Delta G = E^o_{D^+/D} -E^o_{A/A^-}-E_{s_1} +\epsilon_r$$

for an excited state $\ce{D^*}$ with energy $E_{s_1}$ and electrode potentials $E^o$. Marcus in the paper also includes a solvation term in a solvent, of dielectric constant $\epsilon$ to account for change in redox. This term can often be neglected but is

$$\epsilon_r = \frac{\Delta e^2}{4\pi \epsilon_o}\left (\left(\frac{1}{r}-\frac{1}{2a}\right)\frac{1}{\epsilon}-\frac{1}{\epsilon_ar}\right)$$

where $\epsilon_a$ is the dielectric constant of acetonitrile, $2a$ is radius of donor + acceptor and r their separation.

The rate constant is calculated from
$$ k_\mathrm{et} = k^0\exp\left(- \frac{(\Delta G +\lambda)^2}{4\lambda k_\mathrm{B}T}\right)$$

where $\lambda$ is the reorganisation energy and is the vertical (positive) energy between the minimum of the acceptor potential and the donor potential. Its is determine experimentally from the maximum in the rate vs $-\Delta G$ plot as is $k^0$. In the plot $\lambda \approx 1~\mathrm{eV}$ and $k^0 \approx 10^{13} \pu{s^{-1}}$.

The chemiluminescent reaction can be considered as the reverse of the example given.


In what sense is the Inductive effect a permanent effect?

The effect is permanent because the group that does the 'induction' is always there in the molecule. For example, if the induction is caused by a t-butyl group, that group is held by a strong carbon to carbon ($\ce{C-C}$) bond that will not easily be broken. One could also consider the inductive effect of the Fluorine held by a very strong $\ce{C-F}$ bond.

A temporary effect might be caused by protontation of an oxygen atom where the protonation can catalyse a reaction. The protonation is reversible. The bond between the proton and oxygen would be much weaker, as $\ce{C=O^+-H}$ will be in equilibrium with $\ce{C=O  +  H^+}$.

Removal of the group that does the induction will remove the inductive effect.


Reduction potential and electron affinity

Oxidising power , i.e the ability to accept eelectrons for any element depends on its electron gain enthalpy as also 2 other factors:

1) Hydration energy - greater the better, which basically means that water molecules easily overcome attractive forces in the solute particles to break chemical bonds and that easy separation of solvent molecules to accommodate the solute occurs in a solution and that formation of new attractive interactions between solute and solvent particles is not that common.

Rem: sum of delta H 1, 2, and 3 is the overall enthalpy of the solution process. Rem also that H1, H2 are positive, H3 are negative...........read this up if you have forgotten.

Due to its small size Fluorine's hydration energy is very high.

2) Enthalpy of dissociation (i.e. bond dissociation energy) - lower the better

Fluoride has a greater electron-electron repulsion among the lone pairs in the small sized F2 molecule where they are much closer to each other than in case of Cl2, hence the enthalpy of dissociation of F2 is lower than Cl2.

These two results in fluorine to have a greater positive electrode potential than chlorine, which results in greater oxidising power for F2.


Which rate, the forward or reverse rate of acid dissociation, is more strongly affected when diluting acetic acid in aqueous solution?

Let's consider case 2. To make the math easier, I'm going to assume that acetic acid and any mixture of acetic acid/water has the same density as water. Thus, the molarity of pure acetic acid is approximated as 16.7 mol/L. That isn't exactly right, but conceptually that doesn't change anything. Also, since the mole fraction of water is a bit less than 1, let's use its actual molarity of approx 55.6 M.

So, to start, we have 0.1 M of acetic acid in water, which means only ~6 mL/L  water, and we can assume the water is still 55.6 M. And assume a total volume of 1 L. At equilibrium, we know that the forward and reverse rates of ionization are the same, so

$k_f[\ce{H2O}][\ce{AcOH}]=k_r[\ce{AcO-}][\ce{H3O+}]$

Based on the published $K_a$, we can calculate that [AcOH]~0.1 M and [AcO-]=[H3O+]=0.0013 M.

If we instantly add an additional 1 L water to double the volume, the AcOH, AcO- and H3O+ concentrations all drop by half, but the water concentration stays at about 55.6 M. Intuitively, it should be clear that the result is a much greater drop in the rate of reverse reaction than the rate of the forward reaction. More rigorously, we relative rate of the forward reaction is (55.6x0.05)/(55.6x0.1)=0.5, while that of the reverse reaction is (0.00065)^2/(0.0013)^2=0.25. Since the rates were equal before dilution, the forward rate is now approx twice that of the reverse, and in order to return to equilibrium, the AcOH concentration must get even lower and the ion concentrations must come back up.

Conceptually, this comes about because the chance of an acetate ion and a hydronium ion encountering each other diminishes as the solution gets more dilute. The dissociation of the acid in bulk water, on the other hand, is just a function of how many undissociated molecules are present.

With regard to the quotes you included, some brief responses are

  as there are more water molecules so the chances that say a CH3COO- ion attracting an H+ from it's surrounding water should increase

The concentration of water is not the same as H3O+, which is decreased by dliution. Water self-ionization will partially compensate, but not very much.

  since we are adding more water to it and since water has high dielectric constant so it'll ionise CH3COOH more

The dielectric constant of water does have an influence on the $K_a$, but it doesn't change when you add more water.

  since water is more so it'll hydrate the broken ion and surround it,make it harder for them to recombine

If we start with dilute acetic acid in water, we assume that the acetate ions are far enough apart from each other that they do not interfere with each other's solvation, so additional dilution doesn't factor in.

  What motivates or pushes [the acid] to dissociate more?

I think it's more appropriate to say that it reassociates less rather than dissociates more, but that's mostly semantics.


Here is a passage from Marschner P. (2012). Marschner’s mineral nutrition of higher plants. 3rd Edn. London: Academic Press:

In the context of the wikipedia table, it means that while you can have a high concentration of sulfate, you should consider the cation of the substance (e.g. calcium sulfate, potassium sulfate, ammonium sulfate) as well when trying to figure out the effects on a plant.


Why we don't talk about unit cells in hcp/fcc structures?

We do not remain silent about the unit cells in this context. The hexagonal close packed structure is called so because it has hexagonal unit cell, and the other one is called fcc precisely because its unit cell is face-centered cubic.

(source)


1.

Consider the phases for potassium hydroxide and benzyl chloride at room temperature.
Why might you not want to use ethanol as the solvent to form the benzyl alkoxide of vanillin in?
(As an aside, think about why you might not want to use water over, say, acetonitrile as a solvent for this reaction.)

2.

What governs the rate of the Williamson ether synthesis?
Why would you want to heat a chemical system you want to have a reaction in?

3.

What solvents would the product ether dissolve in?
Which by-product(s) would dissolve in water but not ethanol?
How would you remove the product ether from the solvent mixture?

I'm trying not to give away the answers directly but these bullet points should help you work out each point.


How to predict the decomposition products of for example barium carbonate?

Chemistry, despite being several hundred year old science, it is still very much empirical. In the beginning you will have to learn and accept some facts. Later you will start to see patterns (hence the periodic table). If you learn that the carbonate of barium decomposes into carbon dioxide and barium oxide, it will be safe to say that Mg, Ca, Ba, Sr, will do the same. Here you have learnt four more chemical reactions. If you recall the periodic functions in mathematics, which repeat themselves periodically, in the same way, elements repeat their properties in the periodic table.

When you strongly heat inorganic salts, usually you get smaller products (e.g., carbon dioxide and barium oxide) which are energetically more stable with respect to the starting material.


How do I determine the crystal field splitting for an arbitrary point group?

If you already know the symmetry of your site then it is quite easy. In a lot of books (e.g. this one) and on this web site you can find the character tables of the point groups supplemented with two additional columns which show the transformation properties of the basis vectors (e.g. $\ce{p}$ orbitals), their rotations and their quadratic combinations (e.g. $\ce{d}$ orbitals). In the following picture I highlighted the important parts in the character table of the $\ce{D_{3\mathrm{h}}}$ group.
So in the $\ce{D_{3\mathrm{h}}}$ group the $\ce{d_{z^2}}$ orbital transforms as the irreducible representation $A_{1}^{'}$ and the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals transform as the irreducible representation $\ce{E^{'}}$.

After you have identified the irreducible representations of the $\ce{d}$ orbitals you can tell their degeneracy by identifying which $\ce{d}$ orbitals belong to the same irreduzible representation. If two or three $\ce{d}$ orbitals belong to the same irreducible representation they are degenerate. In the case of the $\ce{D_{3\mathrm{h}}}$ group shown below this means that $\ce{d_{z^2}}$ forms an energy level of its own, while $\ce{d_{x^2-y^2}}$ and $\ce{d_{xy}}$ (both belonging to $\ce{E^{'}}$) are degenerate and $\ce{d_{xz}}$ and $\ce{d_{yz}}$ (both belonging to $\ce{E^{''}}$) are degenerate. So, your $\ce{d}$ orbitals split into three energy levels, two of which are doubly degenerate, when your site has $\ce{D_{3\mathrm{h}}}$ symmetry.


Is there a way to calculate how fast water will cool?

Check out Fourier's law. The rate of heat flux (transfer of thermal energy per unit time per unit surface area)  is proportional to the difference in temperature:

Since a change in temperature is directly proportional to the change in energy, you can form this into a fairly simple first-order differential equation. If we take one side (air in a room), at a temperature we'll call $T_{0}$ as a cold reservoir, we can say:

$$
\frac{d(cT_{1})}{dt}=c\frac{dT_{1}}{dt}=-kA\frac{T_{1}-T_{0}}{w}
$$

Where $T_{1}$ is the temperature of the water, $c$ is the water's specific heat, $A$ is the surface area, $k$ is the thermal conductivity of the material between them (i.e. styrofoam cup, say it has a styrofoam lid too so it surrounds the water completely), and $w$ is the thickness of the container.

You will notice that this equation leads to an exponential decay in the temperature difference.

Simply plug in the appropriate constants and solve this equation for any given $\Delta T$ and you can find how long it takes to get that close (it will never quite reach zero).

I'm a bit rusty, so please forgive any notational errors


(not yet a fully grown answer: experiment still pending)

I heard that $\ce{CO2}$/$\ce{H2CO3}$/$\ce{HCO3-}$ is responsible for "fresh" taste of water.

@NickT: Chloramine has a very characteristic smell, so Manishearth could most probably tell us whether this was the smell/taste.

So, here are first questions for @Manishearth:

does water that has been boiled and then cooled to the same temperature taste similar to the taste you're asking for?
What about de-ionized water?

Of course, there's also that average-of-all-that-has-ever-been-in-the-freezer taste...


When are all cannabinoids infused into grape seed oil using non-decarboxylated, dried plant matter?

The time taken (before there are diminishing returns) of extracting the herb is totally dependent on the solvent used, or in your case, the oil. If you were to use water, the answer might be days. If you used heptane, a couple of hours.

The decarboxylation and the extraction - a catch all chemical term for removal of components or a component from a particular solid liquid or gas phase, as opposed to infusion which seems to be specific to botanical extractions (so both are correct in this instance) - should be decoupled. If you start with the cannabinoidal acid and heat in solvent, you'll start extracting components at different rates, while also decarboxylating and extracting the cannabinoid. This reduces the selectivity of the extraction.

Again, depending on your chosen oil, the rate at which your desired components and undesired ones are pulled into your oil will differ. You seem to have a handle on  which of your two oils is better. I would try two experiments:

Heat the dry herb to decarboxylate; less than two hours should be ample and may even be too much. Then extract/infuse your oil. Bear in mind, the more oil you use also leaves you open to extracting more rubbish you don't want, as does extending the time. You're better cutting your time down first and monitoring potency (as extracting more undesired components will lower your potency in the final extract, so potency should increase as you bring your time down) and then experimenting on reducing the oil content until your potency decreases. A decarboxylation should take about one hour at around 110C, but it depends on the amount of herb you have and if it's whole leaf,  shredded,  milled etc. I'm talking fine material at kilo scale. Extraction under ideal conditions is done in six hours, depending on solvent.
The second option is to extract at cooler temperatures (cool temp to 50-60C absolute maximum) first to get the cannabinoidal acids. If you have the facility, add base (diluted sodium hydroxide is ideal and would take minutes . Sodium carbonate dissolved in water may work but would be slow (days?) and you'd have to monitor carbon dioxide evolution). The acids you want will form salts and extract into the water.  The neutral stuff you don't want (terpenes, waxes - long chain alkanes) stay in the oil. Separate the oil and water. You have a choice here.

Add acid to your water layer - this may cause your cannabinoid acid to precipitate, so filter it off.
Otherwise add fresh oil to the water,  then acid and your cannabinoid acid will now be in the oil after mixing.

Whatever you choose, heat your cannabinoid acid to over 100 degrees Celsius and you should be done in an hour or two. Over 120 degrees Celsius may be too much. As I mention, some experimenting will be needed but try to ensure your herbs are consistent and not mixtures of different varieties each time.

As for your oils. One is more like a liquid "fat" - the coconut oil. The grapeseed oil is still a mixture, but is primarily "polyunsaturated" ( it's actually di-unsaturated) and about 50% bigger than the main coconut oil component. These oil mixtures are not usually used in chemical extractions as they're not pure. I assume you use digestible, non-toxic solvent oils. Otherwise I'd suggest limonene.


https://en.wikipedia.org/wiki/Copper%28II%29_chloride

  Aqueous solution prepared from copper(II) chloride contain a range of copper(II) complexes depending on concentration, temperature, and the presence of additional chloride ions. These species include blue color of $\ce{[Cu(H2O)6]^2+}$ and yellow or red color of the halide complexes of the formula $\ce{[CuCl_{2+x}]^{x−}}$

On the Wikipedia page, you can see a photo of 3 solutions with different colour.


Difference between oxygens in a carboxylic acid?

First of all I would like to put my usual disclaimer about hybridisation: It is just a mathematical concept, nothing too real and it should always taken with a grain of salt. In the linked post I have already discussed certain aspects of that. Furthermore it has been discussed on this site in various occasions.
Going a bit further, especially for delocalised systems, the hybridisation picture starts to break down. In acetic acid (or any other carbon acids) this is not yet a big problem and we can still argue a lot with the involved concept. But I hope the following explanations will demonstrate some of this break down.

Now let's get back to the problem at hand. Here is the structure of acetic acid in the gas phase at 0 K, calculated with Gaussian09 at the DF-BP86/def2-SVP level of theory.

As you can see the carbon oxygen bonds are not equal at all. In the Lewis picture, we would say there is a single bond between carbon 2 and oxygen 1, which is a somewhat valid approximation. On the other hand we would say that between carbon 2 and oxygen 2 there is a double bond, which is also a somewhat valid approximation. In the molecular orbital picture we have three π-type orbitals (10, 13, 15). Therefore we have partial double bond character in the carboxyl unit due to delocalistion. Note that there is also an hyperconjugation effect involved. The main difference, however lies in the the molecular orbital 15, which is bonding with respect to C2-O2 and antibonding with respect to C2-O1. This orbital is basically the key, why the Lewis structure is a valid approximation.

Now we can unitary transform these orbitals to localised orbitals, therefore representing the bonding structure with hybrid orbitals. I am using the Natural Bond Orbital theory for this. The following summary gives the bonds of the carboxyl subunit, the numbers in brackets refer to the orbital number in the picture. Also note, that since these orbitals are no eigenfunctions of the Hamiltonian and they do not have an energy eigenvalue, hence there is no canonical order.

-----------------------------------------------------------
#  (Occupancy)  (Type)    (orbital)  || Contributions below
-----------------------------------------------------------
1. (1.99578)    BD ( 1)   C2 - O1   [6]
            ( 30.48%) C(2) s( 26.60%)  p 2.75( 73.10%)
            ( 69.52%) O(1) s( 35.03%)  p 1.85( 64.91%)

2. (1.99728)    BD ( 1)   C2 - O2   [5]
            ( 33.43%) C2   s( 33.28%)  p 2.00( 66.51%)
            ( 66.57%) O2   s( 42.68%)  p 1.34( 57.24%)

3. (1.99005)    BD ( 2)   C2 - O2   [14]
            ( 31.97%) C2   s(  0.00%)  p 1.00( 99.64%)
            ( 68.03%) O2   s(  0.00%)  p 1.00( 99.87%)

4. (1.98518)    BD ( 1)   O1 - H4   [7]
            ( 74.58%) O1   s( 19.52%)  p 4.12( 80.41%)
            ( 25.42%) H4   s( 99.60%)

5. (1.80107)    BD ( 2)   O1 - H4   [15]
            ( 99.93%) O1   s(  0.00%)  p 1.00( 99.93%)
            (  0.07%) H4   s(  0.00%)  p 1.00(100.00%)

6. (1.97613)    LP ( 1)   O1        [10]
                           s( 45.45%)  p 1.20( 54.52%)

7. (1.97395)    LP ( 1)   O2        [8]
                           s( 57.37%)p 0.74( 42.62%)

8. (1.83220)    LP ( 2)   O2        [16]
                           s(  0.00%)p 1.00( 99.91%)
------------------------------------------------------------

The total compliance with the Lewis bonding picture is 98.3%. Note the artefact in line 5, which should be a lone pair of oxygen.

From here we can see, that O1 is best described with an sp1.85 contribution towards carbon, an sp4.12 contribution towards hydrogen,
an sp1.20 and a p lone pair.
The O2 is best described with an sp2 contribution towards carbon, an π orbital towards carbon, an sp0.74 and a p lone pair.

Concerning the intramolecular hydrogen bond, one can analyse the electron density with the Quantum Theory of Atoms in Molecules (QTAIM). This shows that there is no bonding interaction between H4 and O2.

(Laplacian distribution, solid blue lines indicate charge depletion $\nabla^2\rho<0$, dashed blue lines indicate charge accumulation $\nabla^2\rho>0$, red spheres are bond critical points, black lines are bond paths, dark red lines are zero flux surfaces. Hydrogens H2 and H3 omitted for clarity.)

One could argue that there is an electrostatic interaction, causing the slightly smaller (105°) CHO angle, but since the H4-O2 distance is only slightly shorter than the sum of the van der Waals radii (H = 1.20 A, O = 1.52 A) in vacuum this might already be a stretch.

In solution there are many more factors to be considered, i.e. dimerisation, protonation/ deprotonation, intermolecular hydrogen bonds, and possibly many more.


What is the relation between molarity and molality?

$$m = \frac{M}{\rho - (M \times wt. of solute)}$$

$m=$molality

$M=$Molarity

$\rho=$density

all quantities in S.I. units


Free energy of solvation and Henry's law constant

The CRC Handbook of Chemistry and Physics [1] provides data for the solubility of $CO_2$ in $H_2O$ originally from reference [2]. The reported data are mole ratio solubilities $\chi$ of the gas at a partial pressure of one atmosphere (101.325 kPa), but can readily be converted into Henry's law solubility in other units:

$$H_{CP} = 1000 \delta_{H2O} \chi /M_{H2O} = 997 \times 6.15 \times 10^{-4}/18 = 0.034 mol/L atm$$

or in molal and bar units

$$H_{mP} = 1000\chi /1.013 (bar/atm)M_{H2O}=1000\times 6.15 \times 10^{-4}/(18 \times 1.013) = 0.034 mol/kg bar$$

Reference [2] is one of the original sources of the data tabulated in the Wikipedia, compiled by Rolf Sander in reference [3]. Other references in [3] provide similar values of $H_{CP} \approx 0.034$ mol/Latm. The value of $H_{CP}$ in the wikipedia is therefore consistent with the solubility reported in the CRC Handbook.

The article by Prasetyo and Hofer [4], referred to in the OP, cites the CRC Handbook as the source of the following reported experimental value (see Table 4): $$\Delta_{solv} G^o = 0.24 kcal/mol$$

However, from $H_{mP} = 0.034$ mol/kgbar (or  $H_{CP} =  0.034$ mol/Latm) we obtain $$\Delta_{solv} G^o = -RTlog(H_{CP} \frac{P^o}{m^o})= 2.0 kcal/mol$$ which is far larger than the value reported in the article.

To attempt further progress, note that comparing standard free energies of solvation requires awareness of the associated reference states.

Prasetyo and Hofer indicate (pg 6474) that their choice of standard state is P° = 1 bar, m° = 1 mol/kg, and T = 298.15 K. The experimental values Prasetyo and Hofer compare to simulation results (see Table 4) refer presumably to this standard state. But this is approximately the same as the molar standard state associated with the data reported in the Wikipedia (P° = 1 atm, c° = 1 M, 298.15 K). Certainly the energies computed in the two scales don't differ significantly, as shown above. I therefore suspect that a different standard state was used in their calculation, or free energies for different (chemical) processes are being reported.

It is worth adding that the value $\Delta_{solv} G^o = 0.24 kcal/mol$ is encountered in other articles that also reference the 75th Edition of the CRC Handbook.

References

[1] The CRC Handbook of Chemistry and Physics, 85th Edition, section 8-87, SOLUBILITY OF SELECTED GASES IN WATER, L. H. Gevantman.

[2] R. Crovetto, Evaluation of Solubility Data for the System CO2-H2O, J. Phys. Chem. Ref. Data, 20, 575, 1991.

[3] R. Sander: Compilation of Henry's law constants (version 4.0) for water as solvent, Atmos. Chem. Phys., 15, 4399-4981 (2015).

[4] N. Prasetyo and T.S. Hofer J. Chem. Theory Comput. 2018,  14, 12, 6472-6483


A little bit about me - I manage a busy research NMR Facility at a leading international tertiary institution. I deal with questions like this not infrequently, and there are some very important things you need to understand about implants and magnetic fields. I have all users of our Facility sign a waiver prior to access to indicate whether they have medical or surgical implants, and for those that do - there is probably 3-4 a year - we make a specific case assessment to determine whether it is safe for them to use the instruments. For some people, we simply arrange an alternative pathway to have samples run. This may be having others run them altogether, or load up samples into an autosampler for them.

A bit about you - that you 'think' it is a titanium implant is not sufficient to allow a proper assessment in your case. Unless you can get this is writing from your doctor, then I would have to assume a surgical grade of stainless steel. And it needs to be made perfectly clear that most surgeons don't know this information. There is no such thing as completely non-magnetic stainless steels - even the highest grade austenitic steels contain minute amounts of ferrite materials. I have had many examples over the years of people believing one thing about their implant and then discovering the truth later. In the most interesting case, I had a user who had surgery and believed they had a titanium plate in their skull. Their doctor put this in writing. Only after a follow-up MRI scan some years later did they discover that they did NOT have any implant. Nothing at all. The doctor checked his records.....ah yes - they didn't need to use one after all! Some implants have additional attachments and electronics that meed to be considered above and beyond the basic 'rod' and these can both be affected by, and affect magnetic fields.

Typically, implants will rarely affect the magnet of the spectrometer to the point of creating problems. In any case, problems only arise through moving through the field - a static object is generally no problem. With MRI, this 'movement' comes from the application of field gradients. Just as passing a current through a coil generates a magnetic field, moving a metal conductor through a magnetic field creates an electric current.

Even surgical grade metals are affected by the stray fields of magnets, and depending on the age, design and size of magnet, this may be of concern to you as a spectrometer operator. Movement of metals through the field will create local eddy currents which can result in localised heating. In MRIs, a common incident is serious burns from the embedded metal in bra straps. For biomedical implants, this risk is still considerable, and consideration should be given to:

how long ago the surgery was done. Recent surgery may not have allowed adequate time for bone growth to sufficiently embed the implant, and heating can create swelling and cause problems that may need to rectified. I am aware of one such case with an implant in the wrist that required follow-up surgery to rectify intense swelling that resulted from exposure to a strong magnetic field.
where the implant is. Some areas will be more susceptible to problems, or least symptoms. Not everyone that has an implant can detect any changes. Some people do. I'd say that most people who say they experience a symptom have implants associated with joints - elbows and wrists are the most common that come to mind. Symptoms have ranged from tingling in the joint to feeling warm. How much of this is genuine, and how much psychosomatic, I can't say. I once had an older tradesman working within the stray field for a day who commented that the tattoos on his forearm were itchy by the end of the day.
what the implant is. Some implants are just not worth taking a risk with. Aneurysm clips, for instance, are simply not worth taking a risk with. Finding an alternative way for a user to access the spectrometer is much easier than cleaning up an exploding head (yes, an exaggeration, I know) and doing the paperwork. Some pacemakers are MRI approved, and most are actually perfectly ok within the small stray fields of a spectrometer. But some aren't ok, so I wouldn't take this risk without plenty of written evidence from doctors and medical suppliers.
the extent of stray field form the magnet. All magnets are different through age, design and magnitude. An older unshielded 600 poses a very high dose of exposure; a newer shielded 300 has the 5G line contained within the magnet case, and has an almost neglible exposure. However, magnetic fields are invisible, and unless you know the details about the specific magnet you are using, it is best to assume the worst case. We base all of our stray field limits on our unshielded 400, which has a much greater stray field than our 500 for instance.

Regardless of the final assessment of your condition, I would still advise against undertaking high field procedures, such as changing probes, or doing low temperature connections etc.

Ferromagnetic materials can obviously be a problem because of the large attractive forces created. Most incidents with these arise from large external metal objects being brought within the stray field; gas cylinders and trolleys etc. MRIs are usually more problematic with stray fields due to the bore size, but small objects can also pose a serious concern in strong fields. These include handheld tools, but for NMR spectrometers the biggest problems actually come from things like staples and paperclips and even iron filings that migrate across the floor over time and can cause serious problems if they get near the magnet bore.

The take-home message for you is not to take advice from a bunch of stackexchange users beyond understanding that you need to discuss your individual case with your NMR Facility Manager, your doctor, and quite possibly your supervisor if alternative arrangements need to be considered for your access. A great resource for you to be aware of is Frank Shellock's MRI Safety site.


why is potassium the most free element in most of plants

There are many functions related to potassium some of which are shown below

Potassium regulates the opening and closing of the stomata by a potassium ion pump. Since stomata are important in water regulation, potassium reduces water loss from the leaves and increases drought tolerance. Potassium deficiency may cause necrosis or interveinal chlorosis. K+ is highly mobile and can aid in balancing the anion charges within the plant. Potassium helps in fruit colouration, shape and also increases its brix. Hence, quality fruits are produced in Potassium rich soils. It also has high solubility in water and leaches out of rocky or sandy soils. This water solubility can result in potassium deficiency. Potassium serves as an activator of enzymes used in photosynthesis and respiration[4] Potassium is used to build cellulose and aids in photosynthesis by the formation of a chlorophyll precursor. Potassium deficiency may result in higher risk of pathogens, wilting, chlorosis, brown spotting, and higher chances of damage from frost and heat.

and the answer to your question is the small size of potassium when compared to all the other cations in plants

another reason for this is its association with the chloride ion .
chloride and potassium ions will be playing follow the master game in which if any cation takes chloride ion near it then potassium will also follow chlorine and the same is the case with potassium whenever any other anion attracts it this makes potassium highly mobile.  source wikipedia

hope you got your answer. if any doubts left feel free to comment


You got it reversed: it is not a half-metal properties that make band structure different but it is the band structure that gives rise to half-metal properties.

That said, the proper question would be: why band structure may be different for electrons with different spin orientation?

The answer is that electrons do not exist in vacuum and unpaired electrons with different spin interact with each other through exchange which lead to having unpaired electrons having lower energy when their spins are parallel (please, note: what 'direction' we use for 'up' and 'down' when performing this type of calculation is completely arbitrary). So, when we have a half-filled band, we may expect electrons on it to have parallel spins. In reality it is usually not the case, because exchange interactions are fairly weak.

For more details you probably should repost on Physics.SE or go for specialized literature. You need at least basic understanding of quantum mechanics to understand the details. Some insight may be obtained here.


If a gas always occupies the volume of its container, will its volume always be 22.4L at STP?

According to Avogadro's law, equal volumes of gases at the same temperature and pressure have an equal number of particles. If you had one mole of particles at STP it would occupy $\pu{22.4dm3}$

However according to Boyle's law, pressure is inversely proportional to volume. Hence, by increasing the volume like you did, the vessel's pressure will decrease, thereby not making the measurement to be at STP. Hence, it will not be $\pu{22.4dm3}$.


Is there safe chemicals for making voice pitch or bass?

Assuming that helium and sulfur hexafluoride (as shown in the video) will give you the correct sounding "pitch" and "base", it would be fine if done correctly and safely.

Helium is incredibly inert and so will not cause any sort of toxicity. It is also a very light gas, so it will exit your lungs quickly. Just do not breath only helium, or for a long time or you could suffocate. Make sure to get lots of normal air.

Sulfur Hexafluoride is also rather unreactive, but more dangerous. This is because it is heavy and will tend to stay in your lungs. Only use it for a short period of time, or you can asphyxiate. While the same goes for helium, it tends to exit your lungs more easily.

The actual mechanism that causes them to work is based of how dense they are. It does not change the pitch, however. Instead, it changes something called timbre. This is basically the fact the helium allows sound to travel much faster through it, so it is a similar affect to playing a video at a faster speed.

Sulfur hexafluoride does the exact same thing, except it is much denser and does not allow sound to travel through it very quickly.

If you want to do this, be careful! You can suffocate, especially with sulfur hexafluoride.


Stoichiometric calculation of composition of a compound

This is an example of a classical method of determining chemical formulas by elemental analysis.
You are not supposed to know the chemical formula of copper oxide neither they are asking you to write a balanced equation. All they are telling you is that
(i) Mass of pure oxide of copper = 3.45 g
(ii) Mass of copper obtained after reduction (we do not need to know the chemical equation) = 2.76 g
Use mass balance, how many grams of oxygen atoms must be there in that oxide?
Now you know the masses of Copper and Oxygen.
Two options
(a) Use mol and find out molar ratios between the two elements
(b) Alternatively, if you do not need to invoke the mole concept ... old school way. Just for trying
Think about the fact that 16 (exactly 16) is the mass unit of 1 oxygen atom. What mass of copper atoms would be associated with 16 units of O if copper's atomic mass is 63.54 units. Use mass balance and some arithmetic.


Do commuting operators always share the same eigenfunctions?

If two operators commute, then there exists a basis for the space that is simultaneously an eigenbasis for both operators.

However, if one of the operators has two eigenvectors with the same eigenvalue, any linear combination of those two eigenvectors is also an eigenvector of that operator, but that linear combination might not be an eigenvector of the second operator.

As an example consider the states $|l,m\rangle = |1,1\rangle$ and $|l,m\rangle = |1,-1\rangle$. These are both eigenvectors of $\hat{L}^2$ and $\hat{L}_z$ (the angular momentum operators).

The eigenvalues of $\hat{L}_z$ are $\hbar$ and $-\hbar$, respectively, but the state
$$|1,1\rangle + |1,-1\rangle$$
is clearly not an eigenvector of $\hat{L}_z$, but it is still an eigenvector of $\hat{L}^2$. This is probably why we use the term "mutual" rather than "same". Finally, we form linear combinations of the $|1,m\rangle$ states to get the $l=1$ states that are eigenvectors of, say, $\hat{L}_y$.

As a straight-forward example, consider the following two matrices:
$$L = \left[ {\begin{array}{ccc}
0 & 0 & 0 \\    0 & 1 & 0 \\    0 & 0 & 1
\end{array}}
\right]$$
and
$$Z = \left[ {\begin{array}{ccc}
0 & 0 & 0 \\    0 & 1 & 0 \\    0 & 0 & -1
\end{array}}
\right].$$
The vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ are eigenvectors of both operators. The latter two are eigenvectors of $L$ with the same eigenvalue (namely $1$), but they are eigenvectors of $Z$ with different eigenvalues (namely, $1$ and $-1$). If we instead use the vectors $(0,1,1)$ and $(0,1,-1)$, these are still eigenvectors of $L$ with eigenvalue $1$, but they are no longer eigenvectors of $Z$, as you can verify.


Local resolution of a crystal structure in PDB File (protein data bank)

The question has false statements:

  Usually, the resolution of the PDB is assigned the resolution of the highest resolved part of the protein/nucleic acid.

Usually (in ~90% of the PDB entries) the experimental method used is X-ray crystallography. In such case the resolution refers to a distance between crystal lattice planes: it is assigned the value of the smallest distance between lattice planes that was resolved in the diffraction pattern.

Crystallographic experiment measures reflection. Each reflection corresponds to a set of lattice planes in the crystal. The spacing between these planes, d, is the measure of resolution. "Resolution range 3 - 20 Angstroms" means that the reflections used in the refinement had d between 3 and 20 Angstroms.

The resolution is simply dmin - the smallest distance between crystal lattice planes that resulted in a reflection usable in the refinement. Each reflection contributes to each point in the electron density map, so while we can talk about local quality of the structure, the resolution cannot be local.

Unless the data comes from a Cryo-EM experiment, where the resolution is defined differently and people can talk about local resolution.


Will toluene vapors damage a mirror made from PETG?

PETG is also not resistant against toluene.
A month-long contact under ambient conditions results a complete swollen/disintegrated PETG material (PDL rating 0), and an year-long exposure even causes whitening [1, p. 2660]:

References

Chemical Resistance of Thermoplastics; Woishnis, W. A., Ebnesajjad, S., Eds.; PDL handbook series; William Andrew: Norwich, N.Y, 2012. ISBN 978-1-4557-7896-6.


Must lactose-free milk be ultra-pasteurized?

You can now buy (at least in Switzerland where I live) lactose-free milk which is not ultra high-temperature pasteurized. They are kept chilled in the supermarket alongside "normal" milk. So I would say based on this, no lactose free milk does not have to be ultra pasteurized. Though I will note, they hardly tasted sweet at all and they are quite expensive, nearly double the price of ordinary milk.


Is there any rule that says that oxygen should have at least one free pair of electrons in Lewis dot structure?

$\ce{H4O^{2+}}$ is a counter example to such a rule.

http://pubs.acs.org/doi/abs/10.1021/ja00265a031


Does simmering sparkling water change/decrease its mineral composition significantly?

No, simmering sparkling water should have negligible effect on its mineral composition.

You're exactly right that simmering it will drive off the $\ce{CO2}$. This $\ce{CO2}$ leaves as only/exactly $\ce{CO2}$ molecules, and any minerals that might have been associated with the carbonate/bicarbonate $(\ce{CO3^{2-}}/\ce{HCO3^-})$ will remain behind in the water.  Most such minerals will be present in dilute enough of concentrations that they will remain dissolved in the no-longer-sparkling water, as long as only a relatively small amount of the water itself is boiled away.


Hybridization of orbitals and forming of bonds in the nitrogen dioxide molecule

In your simple model, you start off with:

nitrogen: $\mathrm{3(sp^2)^1 + p^2}$
left oxygen: $\mathrm{(sp)^1 + sp^2 + p^1 + p^2}$
right oxygen: $\mathrm{(sp)^1 + sp^2 + p^1 + p^2}$

The σ bonds are easy. Use $\mathrm{sp(O) + sp^2(N)}$. That leaves you with:

left oxygen: $\mathrm{sp^2 + p^2 + p^1 + (\sigma_\ce{O-N})^2}$
nitrogen: $\mathrm{p^2 + (sp^2)^1 + (\sigma_\ce{N-O(1)})^2 + (\sigma_\ce{N-O(2)})^2}$
right oxygen: $\mathrm{sp^2 + p^2 + p^1 + (\sigma_\ce{N-O})^2}$

Now the question is how to do a π system (obviously there can only be one σ bond between two atoms). The orbitals used for the π system must be orthogonal to the bond axis, since that axis must be a node. Both oxygens have two p-type orbitals which fulfill this condition. Nitrogen also must use its p-type orbital to form the π bond. The $\mathrm{sp^2}$ orbital cannot because it is in the same plane as the other two $\mathrm{sp^2}$ orbitals, angled at $120^\circ$ rather than $90^\circ$. This gives us a π system composed of two oxygen p orbitals (one on each oxygen) and one nitrogen p orbital; choosing our orbitals well will result in this π system having four electrons. Essentially, this is theme and variation of Your Average 4-electron-3-centre bonding; but you can also consider it akin to the allyl anion.
The one remaining unpaired orbital would be in $\mathrm{sp^2}$ on nitrogen, pointing away from both $\ce{N-O}$ bonds but being in the same plane.

The above are the considerations and reasonings that you should use from the starting point you had. It is not the only way. You could assume that nitrogens five electrons are actually distributed as $\mathrm{2(sp^2)^1 + (sp^2)^2 + p^1}$. In this case, the argument outlined above is essentially the same except that the π system now contains three electrons rather than four; it is an allyl radical system, if you wish. Since this ab initio construction will lead to all orbitals with s contribution ($\mathrm{sp^2}$ on nitrogen, $\mathrm{sp}$ on oxygen) being fully occupied, it is more consistent with Bent’s rule and likely to be a better proposition.
If you do it like this, you can – in the Lewis 2-electron-2-centre formalism – choose either oxygen and have the radical form an $\ce{N-O}$ π bond on that side; the other will remain a radical. By resonance, you can draw the mirrored molecule.
$$\ce{O=N-\overset{.}{O} <-> \overset{.}{O}-N=O}$$
In these structures, the radical is never centred on nitrogen although we find that dimerisation at low temperatures leads to $\ce{O2N-NO2}$.

Actual measurement shows, that the radical is centred in an orbital that transforms as $\mathrm{a_1}$. As $\mathrm{a_1}$ is symmetric with regards to all symmetry transformations, it cannot be a π type orbital which can transform either as $\mathrm{a_2}$ (node through nitrogen) or $\mathrm{b_1/b_2}$ (no node through nitrogen; depending on your choice of coordinates). You see, it gets more and more complicated the more you rely on experimental data.


Alkali metal carbonates, their thermal stability and solubility

Lattice energy is also decreasing down the group. hence thermal stability is decreasing. But since lattice energy is decreasing squarely and hydration energy is decreasing linearly, the effect of decrease in lattice energy is higher than the effect of decrease in hydration energy. So if energy required to break lattice is very less, then the lattice of the bottom elements in alkali group will break easily compared to the lighter ones. Thus if not lattice form, then they prefer the aqueous state and hence they love to stay dissolved and not form lattice. So solubility increases.


Amine group - electron donating or withdrawing group?

When the atome of nitrogen participates through its free electron pair in resonance, the amine group plays the role of a donating group by conjugation. When it cannot participate in conjugation through resonance, the amine groupe plays the role of inductive electron-withdrawer group, due to the higher electronegativity of nitrogen atom compared to carbon atom.

When the amine group plays the role of electron donating group through conjugation, its inductive withdrawing effect (still existing) is weaker and masked by the first effect.


How ‘lively’ would this ‘health’ potion be?

Explosives: It is true that sulfuric acid and sodium/potassium nitrate are used to make various explosives (by creating nitric acid), but the key is something to nitrate. The worst this mixture could generate is nitrated ethanol, and I was unable to find any source suggesting that could be done with nitric and sulfuric acid alone. Nitrating a molecule that small with that simple of a structure seems unlikely without extreme conditions or reagents. If it were possible, the most likely product does seem to be something unpleasant.

Poisons: What definitely would be produced is nitric acid, and the production of nitric acid via sulfuric acid and a nitrate salt produces nitrogen dioxide, a poisonous and caustic gas. This is some seriously bad stuff you do not want to come in contact with.

Based on that, I'm going to conclude this mixture is probably not explosive, but definitely could become so if an easily-nitrated compound was a contaminant. Those are more common than you might imagine. It definitely would create a nasty gas.


How do I know which is the "correct" way to write the reaction?

There is no single correct way.  Picking a different way to write the reaction affects both $\Delta H_{R_j}$ and $Rr_j$ in opposite ways that cancel out.

  Does it mean that the reaction rate should be specific to R1 or R2?

Yes.  The unit of your reaction rate $Rr_j$ is $\frac{\textrm{kmol}}{\textrm{kg}_{cat}}$.  But the key question is kmol of what?  A good way to think about the answer is "kmol of things that have a stoichiometric coefficient of 1 in the reaction".

So if the reaction is R2, then the rate of the reaction will be the rate of disappearance of $\ce{O2}$.  The rate of formation of water will be twice the reaction rate $Rr_2$.  But if the reaction is R1, then the reaction rate is equal to the rate of formation of water (or the rate of depletion of hydrogen).

But you don't directly measure "reaction" rates, you can only measure the formation or disappearance of products or reactants.  So say you measured the formation rate of water.  If you decide the reaction is R2, then $Rr_2 = \frac{1}{2}R_{\ce{H2O}}$.  If you decide the reaction is R1, then $Rr_1 = R_{\ce{H2O}}$.  The factor of $\frac{1}{2}$ will cancel out the factor of two in $\Delta H_R$.

So in the end it shouldn't matter what choice you make.

For the exact same idea in a totally different chemical context, see this se.chem question on electrochemistry.


Reaction of acyl chlorides with diazomethane and silver(I) oxide

Your first reaction is correct. You do get an acid chloride. The reaction would proceed with the first step being what you have proposed:
$$\ce{R-COOH ->[SOCl2] R-COCl}$$
Now, the next step is a reaction known as Arndt-Eistert Reaction:
$$\ce{R-COCl ->[CH2N2][Ag2O/CH3OH] R-CH2-COOCH3}$$
The mechanism for this reaction is as follows:

Note: Mechanism is from Organic-Chemistry.org


Enol form of acetone is treated with D2O

The oxygen of the water will attack the carbon of the C=O forming the hydrate, but this is in equilibrium and completely reversible and a dead end so of no consequence when considering the enol reaction.


Mathematical justification for Le Chatelier's principle

I think that I might have an answer to my own question:

$$K=exp \left( \frac{TΔS^o-ΔH^o}{RT} \right) $$

implies that K is dependent on T, but not on total pressure of system, p (nor concentrations).

For
$$ A_{(g)}⇌2B_{(g)} $$
$$ K= \frac{{p_{B}}^2}{p_{A}} $$
Partial pressure of a gas = mole fraction (mf) x total pressure of system

If p is increased then K initially increases:
$$ K=\frac{(mf_B.p)^2}{mf_A.p} $$
$$ ⇒ K=\frac{{mf_B}^2.p}{mf_A} $$

But p doesn't affect K, yet K has increased. So the mole fraction of A must be increased by shifting the position of equilibrium to the left (thus decreasing the mole fraction of B), in order to re-establish the value of K to before p was increased.


I assume that the answer is that they are all C6H6 compounds (sorry for the terrible image!):

As others have pointed out in the comments, it doesn't really make sense to talk about the valency of a molecule. You could say they all have the same elemental formula, but then so do around 200 other molecules (in theory).


Peak Wavelength of Emission Spectrum of a Flame vs. Actual Flame Temperature

There seems to be a number of problems here.

spectrometers need to be calibrated in intensity on a wavelength-by-wavelength basis. Each part of the system has an efficiency which can vary with wavelength, but the grating response tends to be one of the strongest. Here is a random example of a grating efficiency curve from a Thorlabs catalog :

You can see right away that a grating efficiency curve can impose it's shape on the measured spectrum, generating a shape superficially similar to a blackbody, but totally instrumental. Without calibration for efficiency, you just have ADC counts vs wavelength raw data!

Fitting of a Plank shape should be done to the spectrum without ignoring the peaks. In other words, a good fitting program will fit the shape you are looking for (Plank) PLUS fit the peaks at the same time. And don't forget to calibrate intensity first (see 1.)!

" Is fitting the emission spectrum to a blackbody curve even appropriate for a non-blackbody emitter such as a flame?" Probably not unless you really know what you are doing. Flames are beautiful and complicated. There's thermal emission from gas in addition to peaks, but stronger is the blackbody-like emission from hot soot particles. Since there will be a distribution in temperatures within the 3D volume of a flame, you have a spectrum recorded from a significant fraction of the flame volume will include a variety of Plank spectra corresponding to different temperatures.

So the answer is: you have to do fitting, remove peaks by fitting them simultaneously, and more importantly, you must calibrate the spectrometer intensity, because grating efficiency curves look strangely Plank-like, althouth they fall to zero much faster above and below their useful wavelength range. Also have to deal with considering the sample volume of the flame and whether or not it's true black-body emission or not.

Those "peak" values, are closer to peak efficiency wavelengths of the spectrometer, and not really related to Wein's displacement law at all!


What you say is correct.  The  [Ar] configuration we are left with does have 8 valence electrons.  But I think it is just semantics.  Elemental potassium has an [Ar] 4s1 electron configuration.  One would say it has one valence electron.  If we take that one valence electron away, it makes sense to say that it now has zero valence electrons because "1 - 1 = 0".


As you said, the molar percentage is just the molar fraction multiplied by 100. As to the L and V above the molar fractions, they refer to the molar fraction of the component in the liquid and vapor phase, respectively. They are needed since (in general) the molar fraction of a component in a mixture are different in the liquid and vapor phase.

So, from the data in your problem, you know that the molar fraction of carbon disulfide in the gas phase is 0.855, and let's call it $x_{CS_2}^V$. The total pressure is P=263 torr, and since the vapor exhibits ideal behavior, the partial pressure of CS2 is: $P_{CS_2} =x_{CS_2}^V P$. Using Raoult's law (since the liquid exhibits ideal behavior), this partial pressure must be equal to the molar fraction of CS2 in the liquid times the vapor pressure of pure CS2: $P_{CS_2} =x_{CS_2}^L P_{CS_2}^o$. And I am sure you can follow from here.


Your drawing is correct and yes, "I read that I can simply assign a priority with the fourth priority group on a wedge and then take the opposite of whatever I assigned" this is true too.  In your drawing 1->2->3 is a clockwise rotation (therefore "R" but), but the lowest priority item (4) is coming out towards you.  Therefor, the R/S assignment needs to be reversed (from "R") to "S".


Why does Sodium Hydrogen Sulfate (Sodium Bisulfate) produce moisture?

Does the Sodium Hydrogen Sulfate (Sodium Bisulfate) react in any way to produce this moisture?

This process is called deliquescence (mouthful word). This process is not specific to your salt, even table salt, NaCl, would do the same in a humid environment e.g. on a rainy day. This has nothing to do with the state of hydrated or anhydrous salt. It is the nature of the compound.

The opposite of this phenomenon is efflorescence. The hydrated salt becomes dry when exposed to air.


Why are hydrides of transition metals called hydrogen-deficient?

As you note, many hydride phases have the hydrogen occupying interstitial sites within a metal phase, not unlike many carbide/nitride/oxide phases as well. In theory this could range from no hydrogen at all up to full occupancy (and perhaps even double occupancy).
Since you mentioned lanthanum hydride, I pulled up a Calphad analysis for the H-La-Ni system, focusing on the H-La part. Hydrogen has almost no solubility in dhcp-La, yet shows a wide solubility range in both fcc-La and bcc-La phases.
Now for the hydride phase, where I'll quote from section 3.4 in the paper above.

The binary LaH$_{x}$ hydride has an fcc structure... Usually, La atoms occupy the 4a sites, while hydrogen atoms occupy the tetrahedral 8c and octahedral 4b sites. Experimental determinations using using several different techniques ... show that hydrogen atoms occupy tetrahedral sites first, for concentration up to $H/M = 1.95$, and only above this concentration value octahedral sites begin to be occupied.

Their calculated diagram is:

So, hydrogen happily fills in to interstitial spaces, but prefers some more than others. Eventually (perhaps strain, perhaps getting too close to each other) other sites become more desirable for the additional hydrogen. In any event, often a quite broad range of compositions are stable, and perhaps (depending on the metal) you never make it to the ideal stoichiometric composition.


Three general questions in silanization of a glass surface

Is this method correct or I should necessarily deep coat the glass plate into a container of silan solution (because in the 2nd method, there are larger number of available agents than can react with glass or have better accessibility to the glass surface)?

Your method is fine, anything method that covers the surface with solution will work.

  How durable is the silanized glass? i.e. after silanization, when I want to clean the glass surface from reaction products, can I rub something onto the surface to help clean it better or the treated layer of glass might be damaged? (rubbing with cleanex, etc)

In my experience hydrophobic coatings are durable, but will wear due the the small surface thickness.  My recommendation is to use rinsing only, if you can and use abrasive cleaning when you have to.

  For getting maximum hydrophobicity, can I use Octadecyltrichlorosilane>90% directly on glass surface (for how much time?) or I should dilute it?

I find that hydrophobic solutions typically make  very effective coatings nearly instantaneously.  You can use it directly which should only require a few seconds maybe a whole minute to work.  If you want to conserve octadecyltrichlorosilane you can dilute it. I have had 5% solutions also work nearly instantaneously, but if you want the most hydrophobic surface you should wait 1-5 minutes.


Taking a procedure for ICP and adapting it to use with AAS

In short - it depends. For most metals no problem in using the same digestion protocol for ICP (OES or MS) or AAS. Are you using flame or graphite furnace?  If you are using flame AAS -just prepare a blank digest (same ratio of acids but no sample) use that as you blank.  If you are looking to quantify Na, K or Ca there can be significant background issues from soil samples - most trace metals are not a problem - but you need a good blank to qualify.  The LOD (limit of detection) will vary depending upon the matrix S/N (signal to noise).
Hope that helps.  If using graphite AAS you are typically limited as to the amount (volume wise) you can add to  the furnace.  You can try multiple dry downs before atomization.  Again a good plank is paramount.


What happens to the hydroxide anion during the electroreduction of water?

You only show one a half-reaction, reduction at the cathode.  Write the other half-reaction, oxidation at the anode to give oxygen.  A real world water electrolysis cell requires a salt electrolyte for conductivity.  $\ce{NaCl}$ could be electrolyzed to $\ce{Cl2}$ and $\ce{Na}$, the latter reacting with water to give $\ce{H2}$ and $\ce{NaOH}$.

http://www.ineris.fr/ippc/sites/default/interactive/brefca/image16.gif

If you want $\ce{H2}$ and $\ce{O2}$, make the electrolyte $\ce{NaOH}$.  $\ce{Na2CO3}$ can be interesting.


Representation of chemical formulas as strings?

If I understand your question correctly, the simplified molecular-input line-entry system (SMILES) and the IUPAC International Chemical Identifier (InChI) are two examples of "human-readable" string notations for chemical formulas your are looking for.


1,2-dibromopropane reaction with OH and NaNH2

Substitution, meaning displacement reactions are Sn1 or Sn2. These reactions are dependent on the leaving group, nucleophilicity vs base of attacking group, and solvent. Currently you have two good leaving groups (Br) and strong bases. Strong bases (have localized electron density, no resonance etc.) favor proton attack due to partial + charge on H, no spatial constraints. So the protons are attacked preferentially. It would be much easier sterically and kinetically compared to either Sn2 (recall l, these also requires specific orientation of attack) or Sn1. These reactions will occur but at much much small amounts, such that on time of reaction they are minimal.
The reason for the -yne product is because there are two adjacent leaving groups. Once the first comes off (which is aided by partial resonance by the other Br), the electron densities around the double bond, the large radii of Br, along with the positive charge character on Br due to resonance, it easier to remove the next Br, but this requires a strong base. The amine is definitely strong enough, but so is the OH ( less so due to higher electronegativity of Oxygen which doesn't want to share its electrons so to speak )
Because you end up with a -yne it doesn't matter which Br is removed first, regarding the final regiostereochemistry.


Why would permeability for krypton be higher than argon?

As noted in some of the comments above, a paper that directly compares noble gas permeation through several materials is S.J. Schowalter et al., Permeability of noble gases through Kapton, butyl, nylon, and "Silver Shield", in Nucl. Instr. Methods A 615 267-271 (2010) . They note that $K = Db$ where $K$ is the permeability, $D$ is the diffusivity, and $b$ is the solubility of the noble gas in the material. For the noble gases, the diffusivity monotonically decreases from He to Xe in all materials. However, the solubility has an odder behavior, with Ne being the oddball with lower solubilities than He or Ar on either side of it.

The net result is that the permeation can end up not being monotonic in nature across the noble gases. In particular, for butyl, $K$ is highest for He, drops for Ne, increases for Ar, drops a bit for Kr, then continues lower for Xe.

Lots of interesting data in the paper, a nice direct comparison of the noble gases and several materials in the same aparatus under the same conditions .


what voltage is needed to reduce Silver (I) Oxide in alkaline medium against an inert electrode

You need to look up the standard electrode potentials of the materials involved.  While wikipedia provides an answer for $\ce{AgO}$ in acidic media, the abstract of this primary publication mentions a value of $(0.599 \pm 0.001)$ V of the $\ce{AgO / Ag2O / OH-}$ electrode, as determined against the H electrode at 25 C in KOH solution.  Note the authors find the potential dependent on the grid material their AgO-electrode was made; beside the expected-to-happen changes in function of temperature of the solution and concentration of $\ce{KOH}$ present.

A generally useful secondary literature reference for physico-chemical data you (still) may carry with you is the CRC Handbook of Chemistry and Physics by CRC Press, including tables like this, too.


Isn't the mechanism of aqueous sodium hydroxide and 1-bromobutane involves both heterolytic bond fission and attacking by a nucleophile?

Well your thinking is absolutely right but only for tertiary halogenoalkane as this is a prinary halogenoalkane it will follow SN2 mechamism. In which necleophile attacks on a partial positive carbon


Melting point and boiling point in isomers

Melting and boiling point in relation to cis and trans isomers

The factor that affects melting point is, for the lack of a better term- packability. Molecules that are symmetrical can fit into each other more snugly and are difficult to melt. Therefore trans isomers, being symmetrical, always have a higher melting melting point.

As for boiling, molecules that are held together by any sort of intermolecular force are harder to boil. Cis isomers being unsymmetrical will almost always have a net dipole moment, as opposed to trans isomers where the symmetric nature cancels out the dipole moment. Therefore they are slightly polar and attract each other. Therefore cis isomers have a higher boiling point

Edit: I don't know why isobutene has a higher melting point.


Why are esters meta directors? Can't the carbonyl donate electrons into the ring?

As your concept oxygen gets positive charge and that leads to least stability which is as follows:


What types of energies can be determined by molecular dynamics in gas phase or explicit solvents?

First, whether you use DFT or molecular mechanics to get your energies and forces doesn't really matter. If you have equations of motion, forces, and energies then you have the dynamics necessary to sample an ensemble and from there you should be able to get any physical observable you desire.

It is often very difficult to determine free energy changes using straight up molecular dynamics. The reason for this is that even very fast reactions are rare events when compared to typical molecular motions such as vibrations and rotations. For this reason, many methods have been developed to determine free energies without having to run simulations long enough to get good sampling of reactions. Usually, this is not possible anyways.

First, you may wish to read about umbrella sampling. Umbrella sampling, as a general idea, does not sample a trajectory on the potential energy surface determined the potential energy of the system (your output from DFT if you like), but rather, a harmonic potential energy is added to the system as well. This has the effect of lifting the molecule out of the potential well it may be stuck in so that sampling of rare events like reactions becomes much more common.

It is not obvious at all that any physically meaningful quantities should be able to be derived from this new system where the potential energy is constantly being modified. Yet, surprisingly and rather beautifully, it is possible. You should probably see ref. [1] and the citations therein.

The second way to do this I will mention is so-called meta-dynamics. The conceptual idea here is to first define the so-called "collective variables" of the system. These are literally any variables which can be used to describe the properties of a system. Typically one would choose some set of internal coordinates, particularly for a gas-phase system as you describe. However, my understanding is that it is also possible to choose various other coordinates such as intermolecular distances. One could imagine this being quite useful if you are interested in something like radial distribution functions in some hard-to-sample region.

Using these collective variables, again, a biasing potential is applied to the system, but this potential is said to be "history-dependent" in the sense that the form of the biasing potential might change based on what parts of the free energy landscape have already been seen. This differs from umbrella sampling where the potential is always harmonic and the rate at which the potential is biased and how extreme the bias is, is chosen by the user. You should see ref. [2] and references therein for more details on metadynamics.

So, yes, it is possible and even fairly routine to calculate free energies, enthalpies, entropies and pretty much anything else from molecular dynamics simulations. However, just because it is routing does not mean these methods are perfect. It can be very hard to measure error with these sorts of methods because one does not generally know the correct answer particularly for dynamical processes. That is, if your aim is to study details of dynamics which can't be observed experimentally, then it is very hard to know if the dynamics under your biased potential mean much beyond the fact they should give you pretty good ensemble averages. Also, there is the inherent subjectivity in metadynamics involved in choosing collective variables which has been known to cause problems. How many variables are enough and which ones?

Hopefully that is about what you were looking for.

I don't address the differences between explicit solvent and gas-phase much because these methods will work in any context (even implicit solvent) for which forces and energies can be supplied.

[1]: Kästner, J. (2011). Umbrella sampling. Wiley Interdisciplinary Reviews: Computational Molecular Science, 1(6), 932-942.

[2]: Barducci, A., Bonomi, M., & Parrinello, M. (2011). Metadynamics. Wiley Interdisciplinary Reviews: Computational Molecular Science, 1(5), 826-843.


Whether something rings when struck depends on the mechanical properties of the bulk material. Ringing depends on the ability of the bulk material to recover from a mechanical impact elastically with little loss of energy. That way the block of material can vibrate for a significant period of time and we hear this because of the interaction of these vibrations with air, creating sound.

The bulk properties of a material depend on both the nature of the material and the crystal structure of the bulk solid. Ringing doesn't happen if the material has ways of dissipating mechanical energy quickly rather than by vibrating elastically. This can be because of a low elasticity but also because the crystal boundaries may create ways to dissipate energy.

So some metals don't ring. Lead makes very bad bells (as does mercury when solid as demonstrated in this amusing video). Many polymers don't ring as they have many internal ways to dissipate vibrational energy quickly. But many elastic materials like brass, iron or glass do ring because they don't have any easy internal ways to absorb vibrational energy.


Finding mass percent through molality of potassium nitrate solution

I am not sure you even need the density in this case.
Say you have $\ce{1kg}$ or $\ce{1000g}$ water (the solvent). That would mean that there is $\ce{4.16 mol}$ of $\ce{KNO_3}$.
Since we can figure out the molar mass of $\ce{KNO_3}$ = $\pu{101.11g mol-1}$, the mass of $\ce{KNO_3}$ in the sample must be $(4.16)(101.11)= \pu{420.62 g}$.
We already assumed that we had $\ce{1000g}$ of water, the mass percent of $\ce{KNO_3}$ is then equal to: $$\frac{420.62}{420.62+1000
} \approx 29.6\%$$


Why does copper(II) carbonate decompose into copper(II) oxide rather than copper(I) oxide when heated?

It comes to stability of products, in this case interaction of metal ion with ligands (here oxygen). In water solutions $\ce{Cu^2+}$ is usually more stable than $\ce{Cu^1+}$, it's quite different story in non-aqueous solvents like acetonitrile. In solid: $\ce{CuCl2}$ is more stable than $\ce{CuCl}$, but iodide $\ce{CuI}$ is more stable than $\ce{CuI2}$.

The simplest explanation would be hard-soft acid-base theory: $\ce{Cu^2+}$ is a hard acid, and $\ce{Cu+}$ is a soft one.  Oxygen in copper oxide or $\ce{Cl-}$ in chloride are hard bases and $\ce{I-}$ is a soft one. The same to the same: soft acids goes to soft bases, and hard acids likes the hard bases.


Is there a theoretical way to predict the coordination number a metal can have in a complex?

Summary of molecular orbital theory for coordination compounds-Ligand Field Theory

Molecular Orbital Diagram can explain what you are looking for. First, you choose the geometries you expected to have with corresponding metal cation and ligands. You need to do this since different complex structures will have different more and less energetic d-orbitals.

Second, do MO diagrams for all geometries, and compare the energies, or more properly, stabilities of compounds. Looking at antibonding interactions with ligands you can even predict which bonds of the ligands are becoming more vulnerable to attack.

Additionally, there are such lists in inorganic chemistry textbooks in general. Ferric ion; for instance, can make stable 7-ligand-complex. However, ligands are also very crucial on this aspect and there is ongoing research, like this one:

8 coordinate ferric ion


Balanced reactions at the anode and cathode

The reaction at cathode seems OK, but at the anode another reaction will occur.

Hint:

At the cathode, 4 e$^-$ are consumed. The electrons have to be "generated" in the system.

Remark:
The value of current applied is irrelevant in this kind of question. More relevant will be the applied potential.


Why do we use glacial acetic acid in lidocaine synthesis?

I think acetic acid is used because it allows you to play funny games with the varying solubility of reactants, products, and by-products. Note that the final product is not soluble in acetic acid/water mixture, but the reactants and salt by-products are. This means that instead of performing an extract in a separatory funnel (extra step, plus you'll need to evaporate solvent and purify), you can just filter the solid product directly from the rest of liquid.

This is the kind of simplifying "magic" you can get when the process chemists get a stab at designing how to run the reaction.


What is the difference between an Adiabatic System and an Isolated System?

There are three ways a system can interact with its surroundings:

heat flow across the system boundary

work flow across the system boundary

a flow of matter across the system boundary

An adiabatic system is one for which there is no heat flow into or out of the system.
A closed system is one for which there is no flow of matter into or out of the system.
An isolated system does not interact with its surroundings at all, so there is no heat flow into or out of the system, no work done on or by the system, and no flow of matter into or out of the system.
Thus all isolated systems are adiabatic, but not all adibatic systems are isolated.  To put it another way, being adiabatic is a necessary but not sufficient condition for being isolated.
Also, your textbook is not quite correct in its language when it says:  "This would not allow exchange of heat between the system and surrounding through its boundary and we call this type of system as adiabatic." [emphasis mine]   If you're a high school student using a high school textbook, this imprecision is understandable (they don't want to overly burden you with subtleties), so don't worry about it.
But, if you are interested in exploring the topic more deeply, read on:
Heat is not something that is exchanged between the system and surroundings.  Rather, it is thermal energy that is exchanged, and we call this exchange of thermal energy "heat".
More specifically: If systems actually contained "heat", then they could exchange heat with their surroundings.  But they don't.  Instead, what they contain (assuming $T > 0\,\mathrm{K}$) is thermal energy.  It is this thermal energy that can (if the system is not adiabatic) be exchanged with its surroundings.  And we call this exchange of thermal energy, which manifests itself only during a change in the system, and appears only at the system boundary, "heat".  I.e., heat is not a property of the system.  It is a property of the path through which a system undergoes change.
To give an analogy:  Thermal energy is like cars, and heat is like traffic.  Suppose cars are being moved from one parking lot to another along a connecting road.  The flow of cars from one parking lot to another is called traffic. The parking lots don't themselves contain traffic, nor are they exchanging traffic; they're exchanging cars.
This distinction between system properties (like internal energy) and path properties (heat and work) will become increasingly important if and when you continue your studies in thermodynamics.


Predicting major product of a reaction with nitrous acid

I think (2) is correct, but either your or your book erred in writing it. This is what the mechanism (and major product) should be:


Comb and hair, is that an example for ionic boding?

No, not at all. It is an example for triboelectric charging. A mechanical example for this type of electrostatic generators would be a Van de Graaff generator.

But all this has nothing to do with ionic bonds.


Why does K have a greater electron affinity than Ca?

Neither K nor Ca "wants" to ionize.  Both ionization energies are endothermic.  This question refers to eelctron affinity, which is gaining an electron.

K has a more exothermic electron affinity because the electron gained fills the 4s orbital.  In Ca the electron goes to 3d.  The energy gap between 4s and 3d is enough to make the process barely exothermic.

Note that the numbers given in tables are positive but that the process is exothermic.


Why is the rate of a reaction always positive and why?

No mystery. Chemists write equations so that reactants (on the left) yield products (on the right). So:
$$\ce{H2O -> H^+ + OH^-}$$
and
$$\ce{H^+ + OH^- -> H2O}$$

The reactions are read from left to right and reactivity is defined as positive from left to right - this means the reactivity for the reaction we want to look at is always positive. (great clarification from user Stian Yttervik.