Relation between chemical kinetics and chemical equilibrium

The derivation that you cite only holds true if both forward and backward reactions are elementary steps. In this case, the rate of the forward reaction is indeed given by

$$r = k[\ce{A}]^a[\ce{B}]^b$$

and so on, and hence the derivation is logically sound. If they are not elementary steps, then even though the conclusion is correct, the method of arriving there is not, for the reasons you have already pointed out. See also: hBy2Py's answer and Ben Norris's answer where the assumption of an elementary step is explicitly mentioned.

In general the equilibrium constant does not need to be derived in this manner; a more conventional method of arriving at it is by using (pressure here is taken to equal an arbitrary but specified standard pressure $p^\circ$)

$$\mu_i = \mu_i^\circ + RT\ln a_i$$

where $\mu_i$ is the chemical potential of species $i$, $\mu_i^\circ$ is the standard chemical potential of species $i$, and $a_i$ is the chemical activity of species $i$. We now have, for a reaction

$$\ce{a A + b B -> c C + d D},$$

the following equation:

$$\begin{align}
\Delta_\mathrm r G^\circ &= c\mu_\ce{C}^\circ + d\mu_\ce{D}^\circ - a\mu_\ce{A}^\circ - b\mu_\ce{B}^\circ \\
&= c(\mu_\ce{C} - RT\ln a_\ce{C}) + d(\mu_\ce{D} - RT\ln a_\ce{D}) - a\mu_\ce{A} - RT\ln a_\ce{A} - b(\mu_\ce{B} - RT\ln a_\ce{B}) \\
&= c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B} - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right] \\
&= \Delta_\mathrm r G - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right]
\end{align}$$

However, at equilibrium, $\Delta_\mathrm r G = 0$ and hence

$$\Delta_\mathrm r G^\circ = - RT \ln\left[\frac{(a_\ce{C})^c (a_\ce{D})^d}{(a_\ce{A})^a (a_\ce{B})^b}\right]$$

The term in square brackets can then be identified as the equilibrium constant, $K$. As you can see there is no need for any recourse to kinetic rate laws.

Why is $K$ a constant? Well, the standard chemical potentials $\mu_i^\circ$ are thermodynamic constants, just like standard enthalpies of formation, standard entropies, etc. So $\Delta_\mathrm r G^\circ$ - a sum of constants - is itself a constant, and it follows that $K = \exp(-\Delta_\mathrm r G^\circ/RT)$ must be a constant.


Is there any relation between resonance and smell of a hydrocarbon?

I wish I could convey the spirit of Professor Louis Fieser on the development of organic chemistry. I couldn't find it in his textbooks, but in class, it went something like this:
Coal distillation led to a large number of different unsaturated compounds, most of which had strong odors, so they were classified as "aromatic". When more quantitative measurements (heats of combustion and heats of hydrogenation) were available, double bonds were investigated, and some compounds were found to be unusually stable. A single $C-C$ bond should have a certain energy, and so would a $C-H$ bond, and so also should a $C=C$ double bond, but some compounds gave surprising results.
Some molecules (like benzene) could be written with two positions for the double bonds - they could "resonate" between the two arrangements. Most of the compounds that had this sort of possibility were odorous (aromatic), so any compound with this sort of resonance possibility was called "aromatic", even if it had no appreciable odor. And they all had some sort of extra stability ("resonance energy") due to the double bond shifting.
Now while "resonance energy" ought to have some quantitative significance, it had some considerable variability, depending on what values you used to calculate the expected value for heat of hydrogenation or combustion. Another term popped into use: "aromaticity", which was never quantitatively defined, but was really a measure of the surprise at the difference between the calculated value and the observed value of the heats.
But I expect that odor is not a result of extra stability, but more likely a result of some reactivity of the multiple double bond system. And volatility.


Isothermal vs. adiabatic compression of gas in terms of required energy

To solve this, try to use what I call the "graphical apparatus". For an isothermal process:
$$
\begin{align}
PV&=\text{constant}\\
P\mathrm{d}V&=-V\mathrm{d}P\\
\frac{\mathrm{d}P}{\mathrm{d}V}&=-\frac{P}{V}\\
\end{align}$$
for adiabatic process:
$$
\begin{align}
PV^\gamma&=\text{constant}\\
\frac{\mathrm{d}P}{\mathrm{d}V}&=-\gamma\frac PV
\end{align}$$
Therefore, starting at the same point on a P-V graph, the curves for an adiabatic and isothermal processes will diverge and the adiabatic curve will have a steeper slope. For the same reduction in volume (the graph in the picture is for expansion, not for contraction. In case of contraction, the curves will be reversed, i.e. adiabatic curve will be above the isothermal curve, and will enclose greater area under it for the same reduction in pressure), more area will be enclosed by adiabatic, and since the area $\int P\mathrm{d}V$ gives the work required, isothermal work is smaller than adiabatic for the same reduction in volume.

Your argument is correct. To provide more mathematical support to it, you can observe the fact that it is both increase in temperature and reduction in volume which increases the pressure in adiabatic process and only reduction in volume increases pressure in isothermal process. The exponent of volume in adiabatic equation ($PV^\gamma=K$) is $\gamma>1$, as compared to 1 in the isothermal equation. Hence, pressure change is more sensitive to volume change and is larger in magnitude.


Fragmentation of phenyl radical cations in mass spectrometry

What about $\ce{phenyl^{*+} -> Dewar\ benzene^{*+} -> cyclobutadiene^{+*} + C2H2}$?


Negative charges in acid-catalyzed reactions

The acidic environment would not allow creation of anions but anions of strong acids, that are too weak bases to be significantly protonated.


Which areas of chemistry require calculus?

For first year college chemistry the only calculus, if there is any at a particular school, would be reaction rate equations.

$\mathrm{d}[A] = -k[A]\mathrm{d}t$ ;  $\mathrm{d}[A] =-k[A^2]\mathrm{d}t$ for example

Then second year you take organic chemistry; no calculus.

In the third year, thermodynamics has some multivariable calculus (partial derivatives) and quantum mechanics has partial differential equations.


Why is polonium tetrafluoride non-existent?

Before going into details, let me state Wikipedia:

$\ce{^{208}PoF6}$ was probably successfully synthesised via the same
reaction (passing fluorine gas over $\ce{^{208}Po}$) in 1960 where a
volatile polonium fluoride was produced, but it was not fully
characterized before it underwent radiolysis and decomposed to
polonium tetrafluoride.

Even attempts were made to dissolve polonium metal in hydrofluoric acid but no fluoride salts were isolated as a free salt as they radiolysed and decomposed. According to this 1961 paper:

Although no fluoride salts of $\ce{Po}$ have been isolated, $\ce{Po}$
is quite soluble in $\ce{HF}$. Solutions containing 2  curies of
$\ce{Po}$ per ml of $\ce{1 N HF}$ are reported to be unsaturated.
Attempts to prepare a volatile fluoride by reaction with $\ce{F2}$ gas or
with liquid $\ce{BF3}$ have been unsuccessful. However recent work with $\pu{1
 mg}$ of $\ce{^{208}Po}$ indicated the formation of a volatile $\ce{Po}$ compound,
which decomposed before it could be isolated.

Polonium tetrafluoride is very difficult to isolate as it is seen to form either a complex or oxidize/decompose radiolytically. Also, bringing analytical data on such compounds is hard as they are produced in minute amounts possible in the range of micrograms. Below is a text from this e-book 1

A white solid possibly polonium tetrafluoride is obtained by treating
polonium hydroxide/tetrachloride with aqueous dilute hydrofluoric
acid. Treating this solid suspension in sulfur dioxide yields a bluish
grey product (possibly $\ce{PoF2}$) which quickly reverts back to the
solid on standing owing to radiolytic oxidation. The solubility of
$\ce{Po(IV)}$ increases with the concentration of hydrofluoric acid
indicating a complex formation
There has been unsuccessful attempts to make volatile hexafluoride by
passing fluorine to $\ce{^{210}Po}$ but recently a fluoride has been
made with $\ce{^{208}Po}$ plated on platinum. This product appers to
be quite stable in vapor phase but on cooling, a nonvolatile compound
is formed possibly polonium tetrafluoride resulting from radioactive
decomposition. Analytical data are not recorded for any polonium
fluorides owing to difficulty of determining fluoride ion at microgram
level.

Reference

Advances in Inorganic Chemistry and Radiochemistry, Academic Press, 1962
https://www.osti.gov/servlets/purl/4367751/4367751.PDF


Product of the addition of 1 eq. bromine to 1-methylcyclohexa-1,4-diene

You misunderstood what hyperconjugation is. It occurs when the electrons in a $\ce{C-H}$ $\sigma$-bond are shared with an adjacent atom's empty or partially filled $\mathrm{p}$-orbital, giving the species an extended molecular orbital and thereby stabilizing it. This means that hyperconjugation doesn't really come into play until bromine is added across one of the double bonds. Let's consider the two possible intermediates:

$\hspace{7.cm}$

$\hspace{6.5cm}$

The first one is a tertiary carbocation, and the second is a secondary carbocation. The first intermediate is more stable due to the hyperconjugative effects of the methyl group, and will require less energy to form than the second. This means the reaction will occur much faster at the more substituted double bond, and is why that is the major product.


That picture is showing relative stereochemistry. Essentially, it is just saying that, in the shown confirmation, the hydrogens will be on one side and the bromines will be on the other, but half of the mixture will have the bromines coming out and the other half will have the bromines going back.

If the S or R labels had been shown in the molecule (or they had used just a d- or l- prefix) the picture would have been showing absolute stereochemistry. By saying dl-stilbene dibromide, they are saying that it is a racemic mixture of the compound shown and its enantiomer, but not any diastereomers.


Are the canonical orbitals of Hartree-Fock also the natural orbitals?

There is a bit of a terminology problem in the field that makes things very confusing and I will try to clarify some of this here. Part of the problem arises from the fact that sometimes only one kind of density matrix is relevant and so people drop the adjectives when talking about specific approximations.
The Density Matrix
We're talking electronic structure here (density matrices are used in other areas of quantum and physical chemistry. Honestly, it's the same density matrix but people use different language and representations and normalizations to discuss them and everyone gets confused when non-EST and EST people talk to each other about density matrices) so I will restrict myself to that topic.
The ($n$-particle) density matrix is given by:
$$\rho=|\Psi(x_1,x_2,\cdots x_n)\rangle\langle\Psi(x_1,x_2,\cdots x_n)|$$
where $x_i$ is the coordinates (space and spin) of the $i$th electron. That is, it gives the probability that the system is in a state with a given set of coordinates (space and spin) for all electrons in the system. The density matrix has many great properties, including being Hermitian and having the property that $\text{tr}{\rho A}=\langle A\rangle$ for any operator $A$. In other words, the $n$-particle density specifies everything about the wavefunction of your system. If you have the full wavefunction, you have the full $n$-particle density and vice versa.
Now, most operators are not $n$-electron operators. For example, the kinetic energy operator is a 1-electron operator (only the coordinates of a single electron are needed to calculate a kinetic energy), the standard electronic Hamiltonian is a 2-electron operator (because the Coulomb potential is only dependent on the positions of two electrons at a time). Hence, we don't need the full $n$-particle density matrix to get energies and properties.
The 1(or 2 or 3...)-particle (reduced) density matrix
Since we don't need all of those other coordinates we can integrate them out. Indeed, the trace operation that we described above would do just that. Say we want the expectation value of a 1-particle operator (like kinetic energy). We can define a 1-particle reduced density matrix (sometimes the word "particle" or "reduced" is omitted), often abbreviated as 1PDM, 1RDM, $P^{(1)}$, or $\gamma^{(1)}$ (also others).
$$P(x_1)=N\int \rho(x_1,x_2,\cdots x_n)dx_2\cdots dx_n$$
where $N$ is some normalization (commonly the number of electrons in the system ($n$) because, typically, people want the trace of the 1PDM to be the number of electrons in the system).
This is the matrix which is diagonalized to form the natural orbitals.
Relatedly, we can also define a 2-particle reduced density matrix (aka 2PDM, 2RDM, $\Gamma$, $P^{(2)}$, others) as
$$\Gamma(x_1,x_2)=N\int \rho(x_1,x_2,\cdots x_n)dx_3\cdots dx_n$$
where $N$ is a different normalization (usually, $n(n-1)$). Most operators that we care about (like the energy) are 2-particle operators so the problem of electronic structure is determining the 1 and 2PDMs.
The Hartree-Fock density
There are many ways to think about the Hartree-Fock approximation, but one way that is related to density matrices is that the HF approximation is that the 2-particle, 3-particle, etc. density matrices can be expressed as direct products of the 1PDM. Intuitively, since HF ignores electron correlation, it makes sense that the probability of finding electron 1 at position $x_1$ and electron 2 at position $x_2$ is related to just the product of the those individual probabilities (with appropriate fermionic antisymmetrization). In this way, the HF 1PDM specifies all density matrices in the HF approximation (which is why in HF you can compute $J$ and $K$ which are two-electron operators by contracting twice with the HF density).
$$P(x,x') = \sum_i^\text{occ} \phi_i(x)\phi_i^*(x')$$
where $\phi_i$ are the spin-orbitals. Note that this means for HF, the canonical orbitals are the natural orbitals. For a correlated method though, the 1PDM will generally be different. For example, CI methods like CAS express the 1PDM as the weighted sum of several HF-like 1PDMs, with coefficients determined by the CAS-CI optimization. Also, note that for Kohn-Sham DFT, the noninteracting reference system also looks like a system without electron correlation and so the 1PDM specifies terms that look like higher order density matrices. However, one does not obtain the 2PDM with KS DFT methods. One obtains the energy from the 1-particle density and an exchange-correlation functional although $i$-particle operators value can be extracted from this non-interacting $i$PDM.

That was a really long answer, but I sense there is a lot of confusion about these different density matrices, so I figured I would take the opportunity to give a very clear answer. In short, the answer to your question is that the HF (1-particle) density specifies all of the HF $i$-particle densities. Generally though, the $i$-particle densities are determined somewhat separately, as each are impacted by electron correlation differently. The orbitals that diagonalize the 1-particle density give the natural orbitals.


What is the difference between a configuration state function and a Slater determinant?

Slater determinants are not eigenfunctions of the $\hat S^2$ operator, but CSFs are.

The Hamiltonian commutes with the operators for total and projected spin
\begin{align}
  [\hat H, \hat S^2] &= 0 \\
  [\hat H, \hat S_z] &= 0
\end{align}
Therefore, a set of common eigenfunctions to all three operators exists. As Slater determinants are eigenfunctions of $\hat S_z$, but not $\hat S^2$, using them as a basis for the electronic wave function (=eigenstate of $\hat H$), will not guarantee the (approximate) solution to be an eigenfunction of $\hat S^2$. The spin multiplicity of the found solution may not be a pure Singlet (or Doublet, Triplet, etc.).

As an example, consider a 2 electron system, with both electrons occupying different spatial orbitals. The situation with same spin for both electrons can be represented as the Slater determinant $|\alpha\alpha\rangle=\frac{1}{\sqrt{2}}[\alpha(1)\beta(2)-\alpha(2)\beta(1)]$. The spatial part is omitted here, to keep the notation short. The corresponding eigenvalue equations are then
\begin{align}
    \hat S_z |\alpha\alpha\rangle &= 1 |\alpha\alpha\rangle \\
    \hat S^2 |\alpha\alpha\rangle &= 2 |\alpha\alpha\rangle
\end{align}
Thus we have $S=1$, from $S(S+1)=2$, and the spin multiplicity is $2S+1=3$, i.e. a Triplet. Here the Slater determinant directly corresponds to a CSF.

For the case of opposite spin, we have
\begin{align}
    \hat S_z |\alpha\beta\rangle &= 0 |\alpha\beta\rangle \\
    \hat S^2 |\alpha\beta\rangle &= |\alpha\beta\rangle + |\beta\alpha\rangle
\end{align}
and therefore $|\alpha\beta\rangle$ is not an eigenfunction of $\hat S^2$. The expectation value for total spin would yield $\langle\hat S^2\rangle=1$, which is neither a Singlet nor a Triplet (not even a Doublet where $S(S+1)=0.75$).

This can be fixed by taking suitable linear combinations of Slater determinants. In this example we have 2 options:
\begin{align}
    |^1\Psi\rangle &= \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle - |\beta\alpha\rangle \right) \\
    |^3\Psi\rangle &= \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle + |\beta\alpha\rangle \right)
\end{align}
This yields the eigenvalue equations
\begin{align}
    \hat S^2 |^1\Psi\rangle &= 0 |^1\Psi\rangle \\
    \hat S^2 |^3\Psi\rangle &= 2 |^3\Psi\rangle
\end{align}
which correspond to a Singlet and a Triplet state respectively.

Overall, this yields 1 Singlet component $|^1\Psi\rangle = \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle - |\beta\alpha\rangle \right)$, and the 3 Triplet components $|^3\Psi\rangle = \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle + |\beta\alpha\rangle \right)$, $|\alpha\alpha\rangle$ and $|\beta\beta\rangle$.

For the arithmetic on how to apply the $\hat S_z$ and $\hat S^2$ operators to multi-electron systems, see Chapter 2.5 in Modern Quantum Chemistry by A. Szabo and N. Ostlund.


Why does silicon not react with dioxygen?

Elemental sulfur commonly occurs in form of $\ce{S8}$, with relatively low melting point and, more importantly, relatively low bond energies. So, it easily breaks.

Carbon commonly occurs as graphite or structures composed of graphite-like fragments, but usually something we call carbon contains up to 10% of other elements, meaning it can produce some combustible gases on heating.

However, more importantly, both elements mentioned above produce gaseous oxides on oxidation, that leave the surface of the reacting compound, exposing it to further reaction.

Silicon, on the other hand, produces solid oxide on heating, that prevents further reaction. It is still possible to make a very fine dust of silicon to react with oxygen, but it is not an easy feat. For reaction to occur to bulk silicon, the products must be removed from the surface. Given that, bulk silicon will not give silicon oxide, but it can react with chlorine in reasonable conditions, giving volatile silicon tetrachloride.

Why silicon forms solid oxide with high melting point is another matter, with answer considerably more obscure. The simplest approach to rationalize it would be to consider atomic radius of carbon, sulfur and silicon and consider bonding in alternative structures for this compounds.

Carbon dioxide is an oxide with small central atom, that can form $\pi$-bonds effectively with oxide, so it does not form polymeric structure.

Sulfur dioxide is an oxide of larger atom, but it has a lone pair on central atom, so polymeric form would be considerably hindered. (on contrary, sulfur trioxide forms polymeric forms, but still has gaseous form  which is only marginally less stable).

Silicon dioxide in monomeric form has two $\pi$-bonds, and this is uncharacteristic for third row elements, and silicon is significantly larger than carbon, having enough space for four oxygens around it. So, it easily and effectively forms 3-d network of $\ce{SiO4}$ units with shared oxygens, that is quite hard to melt and vaporize, and, once formed on the surface of silicon, it prevents further reaction.

Still, formation of silicon dioxide films on surface of bulk monocrystalline silicon is essential for microelectronics.


What's the difference between alpha-glucose and beta-glucose?

$\alpha$-D-glucose and $\beta$-D-glucose are stereoisomers - they differ in the 3-dimensional configuration of atoms/groups at one or more positions.

  $\alpha$-D-glucose

  $\beta$-D-glucose

Note that the structures are almost identical, except that in the $\alpha$ form, the $\ce{OH}$ group on the far right is down, and, in the $\beta$ form, the $\ce{OH}$ group on the far right is up.

More specifically, they are a class of stereoisomer called an anomer. Anomers are capable of interconverting in solution. All cyclic structures of monosaccharides exhibit anomeric $\alpha$ (down) and $\beta$ (up) versions. These differences occur at the anomeric acetal carbon (the only carbon with two $\ce{C-O}$ bonds.

These two forms exist because all monosaccharides also have an open-chain form with one fewer stereocenter. When the chain closes to the cyclic structure, the aldehyde or ketone carbon becomes a stereocenter, and it can do so in either configuration. One configuration is preferred ($\beta$), but both exist.

  Open chain form of glucose:

In the presence of acid or base (although water can fulfill this role if need be), the two anomers interconvert through the open form until dynamic equilibrium is established. The mechanism below starts with $\alpha$ in the upper left and finishes with $\beta$ in the lower right. The open-chain form is in the middle.


Why does hyperconjugation help for ring cleavage?

This is a bit late, but my instinct tells me you form the 1,3 diradical. The fusion to a cyclopropane seems pretty reasonable. The alkene is formed from a 1,2-hydrogen atom shift (I admit, this is a strange abstraction), but the formation of the alkene (and the fact that this is flash vacuum pyrolysis) make me think this is not completely unreasonable. Since the mechanism involves breaking the H/D bond, you would expect some selectivity 1:1 versus 2:1. The fact that there is essentially no primary KIE observed is consistent with diradical formation as the r.d.s.

The other reaction, I suspect, forms the cyclopentene with the deuterium substitution on at the 3- position. The secondary KIE would be consistent with diradical formation.


Calculating concentration of dissolved gas from its mole fraction

The gist is that as long as you have a specified mole fraction, the concentration of oxygen does not depend on the amount of water present. This is what is called an "intensive property". Intensive properties don't depend on how much stuff there is, and therefore, you can arbitrarily specify any volume you like. In order to make the calculations simpler, this specified volume is usually some unit quantity, like $\pu{1 L}$.

It is somewhat analogous to determination of an empirical formula, in that it doesn't matter how many grams of the compound is present, you will always end up with the same empirical formula. A common tactic is to assume $\pu{100 g}$ of compound is present, such that if there is $x\%$ of carbon by mass, there is $x\ \pu{g}$ and therefore $x/\pu{12 mol}$ of carbon.

Here, if you have $\pu{2 L}$ of water, then $n_{\ce{H2O}}$ is double of that if you only have $\pu{1 L}$. But since you have twice as much water, to maintain the mole fraction (that you determined via Henry's law), you have to dissolve twice as much oxygen - otherwise the mole fraction won't be the same. So $n_{\ce{O2}}$ is double of that if you only have $\pu{1 L}$. The factor of 2 will cancel out in the equation you gave:

$$\begin{align}
x_{\ce{O2}} &= \frac{2n_{\ce{O2}}}{2n_{\ce{O2}} + 2n_{\ce{H2O}}} \\
&= \frac{n_{\ce{O2}}}{n_{\ce{O2}} + n_{\ce{H2O}}}
\end{align}$$

If you are really uncomfortable with assuming the volume, then simply assign an unknown value: $k$ litres, for example. You should find that $k$ cancels out and you are left with the exact same value for the molar concentration of $\ce{O2}$.

There is, perhaps, a more "elegant" way to prove this. Note that

$$\begin{align}
\frac{1}{x_{\ce{O2}}} &= \frac{n_{\ce{O2}} + n_{\ce{H2O}}}{n_{\ce{O2}}} \\
&= 1 + \frac{n_{\ce{H2O}}}{n_{\ce{O2}}} \\
&= 1 + \frac{V_{\ce{H2O}} \times [(\pu{1000 g L-1})/(\pu{18 g mol-1})]}{n_{\ce{O2}}} \\
&= 1 + \frac{\pu{55.56 mol L-1}}{c_{\ce{O2}}}
\end{align}$$

where $V$ is the volume and $c = n/V$ the molar concentration. You can see that the molar concentration is uniquely determined by the mass fraction $x$ and the exact value of $V$ doesn't matter at all.


What is the most stable conformation (Newman Projection) of CF3CH3?

For $\ce{H3C-CF3}$, the stable conformation is staggered. I would expect this to be case for any similar, freely rotating molecule (the exception being conformations imposed by rings etc.), with the reason being the repulsion between the $\ce{C-F}$ and $\ce{C-H}$ bonds.

To confirm, I have optimized the molecular geometry on different levels of quantum chemistry, starting from a conformer halfway between staggered and eclipsed. In all cases, I have observed convergence to the staggered conformer. The levels are

HF/def2-TZVP
PBE-D3/def2-TZVP
B3LYP-D3/def2-TZVP
PW6B95-D3/def2-TZVP
PW6B95-D3/def2-QZVP

In response to comment: I have performed PW6B95-D3/def2-TZVP optimizations for propene, acetaldehyde, and propanaldehyde (the latter starting from two different conformers). The result is that in all cases, the double bond is eclipsed by a single bond (and two non-equivalent conformers exist for propanaldehyde).


Lucas test in the presence of EWG like -CN

Lucas reagent follows SN1 pathway. $\ce{Zn}$ is a lewis acid that tends to bond with $\ce{-OH}$ and form a partial positive charge leading to the formation of a carbocation.
According to Hammond Postulate, the stability of the transition state should allow successive formation of product or intermediate. The stability of the positive charge induced on the carbon if stabilised in the transition state should therefore stabilise the carbocation. The rate determining step is the formation of the carbocation.
Since tertiary carbocations are the most stable the carbocation is formed quite easily and the product formation happens quickly.
Similarly, the primary and secondary alcohols, need to form a primary and secondary carbocation, which is not stable. Hence the reaction is slow. Zinc forms a weaker bond with $\ce{-OH}$ and induces a partial charge on them, thus not forming the whole carbocation.
So I think that in such cases where the carbocation is not stable, the reaction is slow, and may not have a a very well defined SN1 or SN2 classification i.e. truly borderline reaction may not occur, due to accompanying facets like solvation and ion pairs.


What's the difference between Carvacrol and Thymol?

On a reverse phase C18 column  (which itself is hydrophobic, it is an octadecyl hydrocarbon chain bound to the support material) the more hydrophobic the molecule, the more strongly it will bind to the stationary phase [like dissolves (or in this case binds to) like], and the longer it will take to elute with an organic solvent. Below are the structures of carvacrol and thymol.  The more exposed the hydroxyl group, the less hydrophobic the molecule.  The hydroxyl group in carvacrol (ortho to a methyl group) is more exposed than the hydroxyl group in thymol (ortho to an isopropyl group) because of the relative steric sizes of a methyl group vs. an isopropyl group. Therefor thymol is more hydrophobic than carvacrol.  Consequently we would expect thymol to bind to the column more strongly than carvacrol, and therfor we would expect the carvacrol to elute first which matches your observation.


How is the time for a test tube in a centrifuge determined?

The short answer: check out equation 5 below

I will explain the whole story in terms of sedimenting particles, but you can easily read this as sedimenting droplets as well.

Necessary condition for sedimentation

If you want particles to sediment the first thing to check is whether their density is higher than the density of the liquid, if not: don't even bother. This can be easily shown on the basis of a force balance with buoyancy and gravity acting on the particles:
$$\tag{1} F_{net}=(\rho_{p}-\rho_{l})V_{p}g $$
where subscripts $p$ and $l$ stand for particle and liquid respectively.
What I should also mention here is that I neglect Brownian motion which can, for very small $V_p$ become a significant effect that avoids sedimentation even when the density of the particles is above that of the liquid.

For now, let's assume the particles are big enough to ignore Brownian motion. If we have a centrifuge it is typical to specify the centrifugal acceleration in terms of $a=Ng$ where $N$ is a multiplication factor dependent on the rotation speed of your centrifuge and the distance of the test tube to the center of rotation ($a=\frac{v^2}{r}$, see the wiki for details). The net force acting on the particles in a centrifuge is thus:
$$\tag{2} F_{net}=(\rho_{p}-\rho_{l})V_{p}Ng $$

Settling velocity

After the initial acceleration of the particles there will quickly be a force balance between the net force introduced by the centrifuge and the drag force on the particle. For a perfectly spherical particle that doesn't settle to fast this is simply the Stokes drag:
$$\tag{3}F_{drag}=6\pi\mu_l R_p v_s$$
where $\mu_l$ is the liquid viscosity, $R_p$ is the particle radius and $v_s$ is the settling velocity of the particle.

Balancing the drag with the centripetal force we obtain:
$$\tag{4} v_s=\frac{2(\rho_p-\rho_l)R_p^2Ng}{9\mu} $$.

Finally, to determine the settling time we need the liquid height in the test tube $h$ so we find $$\tag{5} t_s=\frac{h}{v_s} =  \frac{9\mu h}{2(\rho_p-\rho_l)R_p^2Ng } $$
Note that with respect to the settling height we take the worst case scenario here because we want all particles (also those close to the surface of the liquid) to settle.

Equation 5 nicely shows the influence of various parameters on the time a test tube has to stay in the centrifuge. Smaller particles means longer times, lower density difference also and obviously higher centrifugal acceleration means shorter times. Plugging in some typical values for $1 \mu$m particles in water $(\Delta\rho=100$kg/m$^3)$ and a strong centrifuge ($N=10^4$) we find that we need $O(10 s)$.

Discussion of the simplifications

Obviously there are quite some assumptions in the derivation above.

Brownian motion might not be negligible therefore either increasing the settling time, or avoiding settling altogether.
Particles need to accelerate to their settling velocity first, this will also increase the settling time a bit.
In general particles will not be spherical. If, for example, the particles are more rod-shaped then the orientation will determine whether the settling time increases or decreases with respect to the spherical particle prediction.

For many types of particles a Svedberg coefficient is tabulated in literature which will allow you to calculate the settling time without having to assume sphericity of the particles. In this document they indicate the use of Svedberg coefficients.

Clearly, these simplifications can mean that the prediction of equation 5 is of by a factor 10 (or maybe even 100 if you're unlucky) therefore in most practical situations you would use this equation as a starting point for some trial-and-error experiments. You can also use it to determine what will happen to the settling time when you e.g. run a new experiment with a different liquid or particle type.


The effect of dilution on the rate of a system in equilibrium

The rate forwards and back are the same at equilibrium  and differ before this is attained. Let the reaction be $A+B=C$. If each concentration is suddenly halved then the reaction is not necessarily any longer at equilibrium and in this case the rates forwards and back differ until a new equilibrium is reached. The figure shows this. The red curve is the forwards rate $k_1[A][B]$ and the green one $k_{-1}[C]$ the reverse. As you see the new equilibrium takes a short while to re-establish. (You can ignore the scales they depend on the particular rate constants and concentrations used but the behaviour is general.)
The figure you show is not fully supported by the calculation. The red curve (forwards rate)  agrees but the blue one does not.  If you do not halve the product concentration you get the  blue curve in the figure.


Does melting HDPE plastic produce toxic fumes? (melting, not burning)

This table indicates that polyethylene (HD/LD not specified) "melts at" 135°C, decomposes in the range 335-450°C, and produces vapors that will ignite between 341-357°C.

I'm sure melting/softening temperature is more complex than that, but I'm not sure that you should expect much decomposition into toxic gases at temperatures below 260°C.

However, if you're melting it in an oven, it's possible to overshoot the setpoint temperature. Your oven's thermostat might be slow to kick on and off. Direct thermal radiation from the heating element can heat an object in the oven well beyond the air temperature. Your plastic can even spill/drip onto the heating element.

I wouldn't like the idea of disposing of canola oil contaminated with unknown HDPE breakdown products. I'd be happier using some sort of double-boiler arrangement, where the HDPE is in a container that's heated by a bath of some other fluid. I'd suggest paraffin as a heat-transfer fluid, but its flash point is a bit low (200-240°C?)


Analysis of cations using sodium thiosulphate

I do not think the question had sodium dithionite. This is a mistake. It must be sodium thiosulfate, because sodium dithionite is such a strong reducing agent that it will not solubilize these metals. The properties mentioned in the answer match with thiosulfate.
Sodium dithionite is such a strong reducing agent that it will reduce copper (II) to copper.
With this correction...
Interesting question but requires some knowledge of colors of sulfides. Sometimes you would have to use intuition in such exam questions. The hint is "...finally turn black on standing". Recall that silver tarnish is brown black, so are lead, copper and bismuth sulfides. Mercury also forms black sulfide. So in a state of exams, a safe bet would be all cations.
Thiosulfate has complex properties (a) it is a strong reducing agent (b) it forms complexes and (c) it decomposes under various conditions to pure sulfur and sulfite ion in acidic medium.
From Jander and Blasius, Textbook of Analytical and Preparative Inorganic Chemistry (in German)

Von den Salzen sind nur $\ce{BaS2O3}$, $\ce{Ag2S2O3}$ und $\ce{PbS2O3}$ schwer löslich. Sie werden jedoch durch Sodalösung quantitativ in das lösliche Natriumsalz überführt. Mit vielen Schwermetallionen (z.B. $\ce{Ag(I)}$, $\ce{Fe(II)}$ und $\ce{Cu (I)}$) bildet Thiosulfat im Überschuss lösliche Komplexe.

An english translation of this would be

Of the salts only $\ce{BaS2O3}$, $\ce{Ag2S2O3}$ and $\ce{PbS2O3}$ are poorly soluble. However, they are quantitatively converted into the soluble sodium salt by soda solution. With many heavy metal ions (e.g. $\ce{Ag(I)}$, $\ce{Fe(II)}$ and $\ce{Cu(I)}$) an excess of thiosulfate forms soluble complexes.

In the book the equations of conversion to sulfide are given as. Ask your teacher if this a typo in the exam.

4. Nachweis als $\ce{AgS2O3/Ag2S}$
\begin{align}
\ce{S2O3^2- +2Ag+ &-> Ag2S2O3\downarrow} \\
\ce{Ag2S2O3 + 3S2O3^2- &-> 2[Ag(S2O3)2]^3-} \\
\ce{Ag2S2O3 + H2O &-> Ag2S + H2SO4} \\
\end{align}

That is all you have to remember that thiosulfate will precipitate $\ce{Ba}$, $\ce{Ag}$ and $\ce{Pb}$ initially but excess would dissolve it.
I would be interested to know otherwise.


What are the limitations of the Hendersson-Hasselbalch equation?

The key approximation made in deriving the Henderson-Hasselbalch equation is that the equilibrium constant can be written as $$K=\frac{c_{H^+}c_{A^-}}{c_{HA}}$$
that is, we assume activity coefficients are unity. If you take the base-10 logarithm of this equation and rearrange terms you arrive at the HH equation:$$pH=pK_a+\log_{10}\left(\frac{c_{A^-}}{c_{HA}}\right)$$ Problems usually arise with approximations made beyond the HH equation, in the context of a polyprotic compound particularly  by ignoring one of the equilibria, including perhaps water autodissociation, when this shouldn't be done.

In the case of the problem you outline, you might

ignore the second dissociation equilibrium and
assume that added base neutralizes an equimolar amount of remaining acid $\ce{H2SO4}$ to form additional $\ce{HSO4-}$.

Then

$$\begin{align}c_{HA} = c_{HA,\,initial}-c_{added\,OH^-}&=\pu{0.0705 M} - \pu{0.100 M} \frac{\pu{5 mL}}{\pu{30 mL}}\\&=\pu{0.0538 M}\end{align}$$
$$\begin{align}c_{A^-} = c_{A^-,\,initial}+c_{added\,OH^-}&=\pu{0.0295 M} + \pu{0.100 M} \frac{\pu{5 mL}}{\pu{30 mL}}\\&=\pu{0.0462 M}\end{align}$$

Applying the HH equation we then have, as the approximate pH
$$\begin{align}pH &=1.91 + \log_{10}\left(\frac{0.0462}{0.0538}\right)\\&=1.84\end{align}$$ which is above the starting pH of 1.53.

A more accurate value (although still approximate) can be obtained starting from the condition of charge balance for the system, and by assuming that the $\ce{OH^-}$ concentration is negligible, from which one obtains pH=1.75.


Calculating the x, y and z distances between two molecules - AMBER (Molecular Dynamics)

I ended up realizing that if you run:

parm prmtop-file
trajin netcdf-file
vector v0 :1-4@heavy-atoms-from-molecule-1 :5-8@heavy-atoms-from-molecule-2 out v0.out magnitude
run
quit

You will get the total distance as well as x, y and z distances between your two molecules. The output file will contain 8 columns. The AMBER manual states that they are identified as follows:

Frame | x-Vector-Coordinate | y-Vector-Coordinate | z-Vector-Coordinate | x-Origin-Coordinate | y-Origin-Coordinate | z-Origin-Coordinate | Magnitude

The vector coordinates are your x, y and z distances. Magnitude is the total distance.

Decided to share this as someone might run into the same problem one day.


How to get the SMILES of all compounds on PubChem?

The best way to download bulk data from PubChem is actually FTP, as documented in their documentation.

For example, if you want the unfiltered SMILES of every CID in PubChem, the URL is ftp://ftp.ncbi.nlm.nih.gov/pubchem/Compound/Extras/CID-SMILES.gz

You can also download subsets using the PubChem Structure Download service

And as mentioned above, there are Python and other tools to access both PubChem and ChemSpider through their documented web APIs.


Why does the melting point get lower going down the Alkali Metal Group with increase in atomic number?

You have to understand that melting point is related to the bonds in a metal between the metal cation and the 'sea of electrons'. With an increase in atomic number, you have an increase in electron shells.

As the radius of atoms get larger down the group, you could say that the force holding them together is spread over the greater area and hence, the metal cations are more weakly bonded. And coming back to my first point, when the bonding is weaker, the metal's melting point will decrease.


First if all, I appreciate you and thank you for putting up such a good question.

One thing I will request you to note here is the definition of temperature. Temperature, in broad sense, is defined as the average kinetic energy (for equilibrium conditions, I hope you are speaking of that) of a system. The contribution mostly arises from translational kinetic energy, and the fluctuation in temperature is proportional to $\frac{1}{N-1}$. (I don't remember if it has some power or something.) In other words, a particle has infinite temperature fluctuation. So if you want to do finite temperature DFT calculations, you will need an ensemble of particles, and it will only add the cost.

In case you have infinite computational power (or at least enough to support no of molecules in the order of Avogadro number), you can, in principle, go for DFT at a finite temperature.

If you are worried about occupation of the higher states, you can calculate the higher states using Kohn-Sham Hamiltionian (these are just eigenfunctions of the Hamiltionian), and then calculate the occupation at a temperature using the formula for the concerned ensemble. Since the states of the KS hamiltionian are its eigenstates, these should not change if you have a static distribution of electrons. The eigenstates mean that these are decoupled by the same KS Hamiltonian, and in principle, you are not allowed to take an electon from one state to other without external purturbation that couples the states.

However, at a given temperature, everything is dynamic. This invokes that the states have finite lifetime, which from Uncertainty principle, imposes broadening in energy, i.e. it is not an eigenstate anymore. This imply there can be coupling between different states, and hence electrons can jump between states without external perturbation. You can externally add broadening to the states using broadning matrix, that mimics the temperature to some extent.

For TDDFT, you actually introduce an alternate electric field, and then this purturbation is used to calculate coupling between the concerned states. Thus, TDDFT calculates the excitation using light. It has nothing to do with the temperature.

Please let me know in comments if you have further doubts.

Greetings from India. जय हिन्द. Jai Hind.


Why are chlorides, sulfates, and bicarbonates of Ca and Mg solely responsible for the hardness of water?

If I remember correctly, hardness is defined by the Mg and Ca content of water. Temporary hardness is the mineral content which can be removed by boiling and permanent is the total Mg and Ca content - this is what counts when the precipitate with the fatty acid is being formed.

Now typical potable water will only contain sulphide, bicarbonate, halide and perhaps nitrate anions. Other anions either don't occur naturally or they are harmful so they are not considered in the context of drinking water.


Physical intuition behind negative values for wave function?

The wavefunction of a particle actually has no physical interpretation to it until an operator is applied to it such as the Hamiltonian operator, or if you square it which gives its probability of being at a certain place. So having a negative wavefunction doesn't mean anything physically. However, let's say for a particle in a box, if you solve the momentum operator for the $n^\text{th}$ stationary state, you would get two solutions: $+\hbar k$ and $-\hbar k$, where $k = n\pi/L$. The plus and minus signs mean that there is a $50$% chance the particle could be moving from the left to the right and a $50$% chance of the particle moving from right to left.

In the case of summing of two wavefunctions to get the wavefunction of a molecular orbital, a negative wavefunction doesn't mean anything at all. The reason why when a positive and negative wavefunction is added, an antibonding orbital is formed is simply because when you add a positive and negative number, you get $0$ or a really small number. This is known as destructive interference.  Therefore the new wavefunction will contain a region where the y values equal $0$. So when you square this new wavefunction of the molecular orbital you are going to get that there is very 0 probability of finding the electron in that position (known as a nodal plane), and hence it is called an antibonding orbital.However note that just because there is negative region in the wavefunction, doesn't mean that it will always add to form an anti-bonding MO. For example, if you add 2 negative wavefunction values (known as constructive interference), you get a large negative value. When you square becomes a very large positive value. Hence this means that there is a large probability of the electron being in that region and hence it will be a bonding MO.


Why does this flame travel (in a circular path, multiple times)

A flame is a dynamic phenomenon, using up fuel and oxygen in places and driven by convection and diffusion. Flames are sensitive to factors such as external sound, and can be used to measure frequency and waveform with a Rubens tube.

Under micro-gravity, flames burn quite differently, due to the lack of convection.

So though not a specific answer to your questions, I assume that oxygen is being used up rapidly in one spot, and the flame front moves along as convection brings in fresh oxygen. You could test this hypothesis by piping in just a little air in one spot on the "ring of fire" to see if it stabilizes the flame there.

Thanks for sharing that interesting video!


Are all ambidentate ligands monodentate?

When a ligand has multiple donor sites, but uses only say $x$ of them at a particular time, then it's called ambidentate ligand. However, if it uses different number of sites in different compounds (say $x$ in one and $y$ in the other where $x\ne y$), then flexidentate is a more appropriate term as said in the other answer. But this doesn't mean all ambidentate ligands are monodentate.

For example, dithiooxalate is bidentate and ambidentate:

This example is from the book Concise Inorganic Chemistry by J.D. Lee (Adapted by Sudarsan Guha).


How can this ketone show Cannizzaro reaction?

There are a number of discussions on this site regarding the Cannizzaro reaction. In strong base there are competing reactions (aldol self-condensation) that can occur with ketones that have at least two hydrogens vicinal (next to) the ketone. Even aldehydes suffer the same fate. That is why the aldehydes are usually aromatic. But I believe in this case reduction may well be facilitated by an intramolecular hydride transfer. Intramolecular reactions are, as a rule, faster than intermolecular reactions owing to the high concentration of the reacting species. The keto alcohol 1 forms alkoxide 2 which reversibly adds to formaldehyde. Intramolecular hydride transfer occurs in intermediate 4. The formate group is saponified (5 ---> 7) forming the diol 9 and sodium formate 8. The deprotonation 7 ---> 8 should not be taken literally. Base and water intercede. Just saving space. See transition state for reduction below reaction mechanism.


Colloidal solutions scatter light of different wavelenghts in different directions, and that's also dependent on the size(!) of the colloidal particles.
To see any scattering at all, there only has to be a difference in the refractive index of pure solution and dissolved particles.
The individual colour (i.e. absorption spectrum) of the particles is irrelevant to the scattering, but of course it additionally changes the colour that you see. Absorbed light does not get scattered, obviously.


Adding a bit of information for your question a). The reactin we are looking at is:

$$\ce{2 H2 + O2 -> H2O}$$

As you can clearly see, the final product water has one oxygen atom for 2 hydrogen atoms, while in the beginning, we only had diatomic molecules. This means that during the reaction, at least one fo the diatomics had to break to single atoms. In this case, this happens with the reaction:

$$\ce{H2 -> 2H^.}$$

From this step, we have a nice reaction scheme called a radical chain reaction. However, we first need to break the $\ce{H-H}$ bond somehow, and this is what flame is doing. If you break a significant amount, then you don't need to provide flame anymore, the reaction is self-supporting — the flame is ignited.

Now, a catalyst is doing exactly this. Many metals, including copper, can adsorb hydrogen molecules to the surface, and this also results in breaking the $\ce{H-H}$ bond — so the explosion can start this way too.


FMO analysis of the [3,3] Sigmatropic reaction regarding geometry. Is it accurate?

For chair-like transition states the angle at which you draw your chair can make a significant difference to how easy it is to visualise stereoelectronic interactions. Sadly, there is no real way to know which is the best way to draw it, except by either trial and error, or prior experience (or leveraging on other people's experience, i.e. copying it from somebody else!).

If you rotate the chair a bit so that the forming bond is in front (as opposed to at the side), then you can have something like this:

You don't need to overthink how the sigma bond rotates; this is a sigmatropic rearrangement, not a electrocyclic reaction where there are conrotatory/disrotatory modes. The geometry at the substituted carbon definitely changes (from tetrahedral to trigonal planar), so there is some kind of "flattening" as the reaction proceeds, but all you really need to pay attention to is the relative disposition of the R substituent and the rest of the carbon chain: here they are clearly trans to one another, which is an (E)-alkene. This is stereospecific, so if you start with the conformation shown in the left, you will end up with a trans alkene. There isn't a different "rotation mode" that leads to the cis alkene.

The main way of getting a cis, or (Z)-alkene, is not actually via a boat TS, but rather via a chair TS with the R substituent axial instead of equatorial. Note that this is a different conformation from the previous image, so again it's not about which direction the sigma bond rotates in, but rather the original location of the substituent (axial/equatorial). The orbital diagram is exactly the same, so there's no need to redraw it, but it should be easy enough to see that this leads to the reverse stereochemistry.

The boat TS is trickier to draw. This is my preferred representation:

but I am not entirely sure which TS is higher in energy (i.e. whether the substituent prefers to point up or point down), even after consulting several references. I don't think it's a matter of great significance; after all, the boat TS usually only plays a very minor role, as far as I know.


Does tetrachlorodibenzo-p-dioxin (2,3,7,8) have a nonzero dipole moment?

The best answer for this seems to be in Structure, Energy, Vibrational Frequencies, and Potential Energy Curve of
2,3,7,8-Tetrachlorinated Dibenzo-p-dioxin: Ab Initio MO Studies J. Phys. Chem., 1996, 100 (12), pp 4810–4814.

This research finds that though planar is the lowest energy conformation, the energy for bending with respect to the O-O axis from 180 degrees to 170 degrees is only 0.008 kcal/mol.  The molecule flaps like a butterfly.

The article says:

  At ambient temperatures, nonplanar conformations
  will be significantly populated for this potential. This prediction
  accounts for the experimental results that the electric dipole
  moment of the DD molecules is reported to be 0.64 D. [reference 24]

Where "DD" means the unsubstituted dioxin and where reference 24 is Higashi, K.; Uyeo, S. J. Chem. Soc. Jpn. 1941, 62, 396.


Which base is stronger? Ammonia of phosphite?

Your professor could be wrong, or rather could mean it in some subtly different context that would change everything. The way you put it here, you are right. Normally the constants would tell you the truth. I see no reason to believe this case is an exception.

Besides constants, there is actually a simple way of comparing two acids: let them fight each other to the death. Let us consider $\ce{NH_4^+}$ vs. $\ce{PO_4^{3-}}$ (I know $\ce{PO_4^{3-}}$ is not what you asked; this is just an illustration). Who would end up having that proton? It turns out that ammonium phosphate is unstable, which implies that the proton would go from $\ce{NH_4^+}$ to $\ce{PO_4^{3-}}$, meaning that $\ce{NH_4^+}$ is stronger acid than $\ce{HPO_4^{2-}}$ (in accordance with what the constants would predict), and hence ammonia is a weaker base than phosphate. Now, that was phosphate, the salt of phosphoric acid, while you ask about phosphite, the salt of phosphorous acid, which by all accounts should be weaker. Hence its anion must be an even stronger base.

Simple as it may seem, there is one more thing to mention. Phosphorous acid is actually diprotic, that is, capable of ionizing only two of its hydrogens. The third hydrogen is connected directly to the phosphorus atom. Organic esters $\ce{(RO)_3P}$ are known, but that's another story. The very existence of ion $\ce{PO}_3^{3-}$ seems somewhat dubious. By any means, even if you manage to get one in water solution, it will be instantly hydrolized. This puts it in the ranks of bases too strong to exist in water, which is of course way stronger than our lowly ammonia.


Does a Grignard reagent react with enones to give the 1,2- or 1,4- product?

The correct answer is the 1,2-addition product (i.e. the allylic alcohol 1). In general, Grignard reagents and organolithium reagents add directly to the carbonyl carbon, while organocuprates (organocopper reagents) add to the beta-position of an unsaturated ketone.

This exact transformation was reported by Akai and coworkers recently (Org. Lett. 2010, 12, 4900) and they obtained compound 1 in 95% yield (the supplementary information (PDF) is free).


Why is the H−C−H bond angle in сhloromethane larger than 109.5°?

Chlorine is more electronegative than carbon which is further more electronegative than hydrogen and thus chlorine pulls the bond pair towards itself, polarizing the bond. This decreases the electronic repulsion between the $\ce {C-Cl}$ bond pair and the $\ce {C-H}$ bond pair, decreasing the $\ce {H-C-Cl}$ bond angle. Consequently, the $\ce {H-C-H}$ bond angle increases to 110.5°.


Why does phosphoric acid mask the colour of iron(III) complex in water?

When phosphoric acid is added to an aqueous solution of iron (III), the ligands in the yellow complexes $\ce{[Fe(OH)(H2O)5]^2+}$ and $\ce{[Fe(OH)2(H2O)4]+}$ ( $\ce{[Fe(H2O)6]^3+}$ are weakly colored but fairly acidic), are successively substituted by $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$. This yields colorless hydrogenphosphatoiron(III) complexes, like $\ce{[Fe(H2PO4)(H2O)5]^2+}$ and the water-free $\ce{Fe(H2PO4)3}$ complex, depending on the phosphoric acid concentration (source). In less acidic solutions, where $\ce{PO4^3-}$ is present, the colorless bis(phosphato)ferrate(III) $\ce{[Fe(PO4)2]^3-}$ is formed (source).


The H stands for hydride, but most everyone just shortens nicotinamide adenine dinucleotide hydride to NADH.


From the comments:

  [OP] Okay Thank you all I finally got it. so K is just thrown out the window. and we are left with the equation (F- + H2O <-> FH + OH-) So i take the Ka to make Kb by (Kw/Ka) and my concentration of F- and OH- was (0.02/0.3) which I plugged into sqrt((Kb)*(M of OH- or F-)) took the -log of that got the pOH and then converted it back to pH (14 - pH) One final thing why did I K out the window does that mean its neutral. Oh and the answers 8.14

Paraphrased with paragraphs:

A strong base ($\ce{KOH}$) reacts with a weak acid ($\ce{HF}$) at stoichiometric ratio:

$$\ce{KOH + HF -> H2O + F- + K+}$$

The major species is fluoride, a weak base with pKb = 14 - pKa, where pKa is that of hydrofluoric acid. The potassium ion is a spectator. To find the pH, use your favorite strategy for a pure weak base. The result is pH = 8.14.

As MaxW pointed out in the comments, this relies on getting the stoichiometric ratio just right. If $\ce{KOH}$ is even in slight excess (let's say one tenth of a millimole per liter too much), the pH will be strongly basic (pH 10). If you err in the other direction, you will have an $\ce{HF, F-}$ buffer with an acidic pH.


Photoelectric effect with 0 Kinetic Energy?

If you irradiate a metal with the exact amount of energy to produce a free electron with KE = 0, it will just stay free, in the "sea of electrons" of the metal, while the nucleus, with its positive charge, also sits in the same sea - near "other" electrons, each of which has KE due to temperature. Then, after a while, an electron drops in around that nucleus, giving a bit of (thermal) shock. The observable effect will be that irradiation with this wavelength warms the metal as much as possible without exciting any electrons to the "outside" world.

Now if you attach a negative electrode to the metal and a positive electrode close to the surface, you can push/pull that electron (or one other from the sea of electrons) off - the application of a potential gives KE to a free electron. Actually, you don't even have to irradiate: if you apply enough potential, you can pull electrons out of a metal. If you heat the metal, the electrons will come out more easily.

This is the basis of the Edison effect, which makes vacuum diodes possible: a heated cathode "boils off" some electrons which would stay near the wire, but if you have a nearby anode (an electrode made positive by an outside battery), electron current will flow easily from the heated cathode to the (cold) anode. But not from the cold electrode to the hot electrode. So you can rectify AC current to pulsed DC.


Can halogens exist in their elemental state in nature?

I agree with Matt Black's comment on the question. The question is rather tricky and the answer can only be determined by the process of elimination.

Have to assume we're talking about STP.
"In nature" is ambiguous, so assume the question means "on earth."
It is also somewhat ambiguous what "elemental form" means. Assume that it means that element is only combined with itself since only the noble gases exist as monoatomic species.

The noble gases (helium and neon) and exist as monoatomic gasses at STP and can be found in the atmosphere.

Oxygen and nitrogen are diatomic gases and both exist in earth's atmosphere.

Sulfur is found in volcanic vents.

Carbon is found both as graphite and diamonds on the earth.

So, the list becomes:

neon, phosphorus, fluorine
helium, hydrogen, iodine
helium, carbon, fluorine
nitrogen, chlorine, sulfur
oxygen, chlorine, phosphorus

Oxygen is ubiquitous on earth and phosphorus is so reactive with oxygen that elemental phosphorous just won't be found on earth. However, elemental phosphorous might be found on the moon for example.

neon, phosphorus, fluorine
helium, hydrogen, iodine
helium, carbon, fluorine
nitrogen, chlorine, sulfur
oxygen, chlorine, phosphorus

So now we are left with the halogens -- fluorine, chlorine, and iodine. Iodine is the least reactive, so "elemental iodine" is the most probable to be found in nature.

Thus the answer must be #2. However this is really an "educated guess," not a "fact" since the question is so ambiguous.


How is an aqueous Ag+ the same as [Ag(H2O)2]+?

There is no way to absolutely predict the co-ordination number in any complex entity, without knowing say for example, magnetic or geometric properties of the complex. All we can do is give a probable configuration based on intelligent guesses. The co-ordination number depends on the characetristics of the ligands and the central ion.

Silver ion, $\ce{Ag+}$ usually forms complexes with Co-ordination Number two. This is partly because the silver ion is small sized, making it difficult to accommodate higher co-ordination numbers around itself owing to increased ligand-ligand repulsion, and partly due to the low charge on silver ion which makes it difficult to hold more than two ligands together. Therefore, in most of its complexes, silver ion co-ordinates with two ligands to form a linear entity. But to confirm this, you need to study solvated silver ions spectroscopically, magnetically or geometrically (diffraction), without which the above arguments can only give you, at best, a good guess.

Other examples include $\ce{[Ag(NH3)_2]+}$, and the polymeric $\ce{[Ag(en)]ClO4}$, both of which have co-ordination number of the silver ion 2.


The reason might be one of kinetic versus thermodynamic control.  The actual halogenation step is stage 2 in this scheme:

(reaction scheme from chem.libretexts.org)

The 2-radical might well be more stable, but that doesn't mean that it will react faster than the 1-radical.  If the reaction rate constant for the 1-radical with chlorine is much faster then the 1-chloro- product will predominate.

More likely, though is that it is that the 1-radical is formed in preference over the 2-radical in step 1. The halogen can attack at one of three positions, so statistically speaking you'd expect more 1-radical over 2-radical intermediates.  If this species then reacts immediately in step 2 without rearrangement then you'd get a predominance of the 1-product formed.
(Thanks to Karl for pointing this out).


Role of acetate ion in formation of hydrogen sulfide ion from thioacetamide

You are correct.

Adding sodium acetate to an acidic solution forms a buffer and thus decreases the acidity of the solution without making it basic.

$$\ce{CH3COO- + H3O+ <=> CH3COOH + H2O}$$

This helps to precipitate sulfides $\ce{S^{2-}}$ without forming basic precipitates like hydroxides that might disturb the analysis.

$$\ce{S^{2-} + H3O+ <=> HS- + H2O}$$
$$\ce{HS- + H3O+ <=> H2S + H2O}$$

Heating the test tube just speeds up the hydrolysis of thioacetamide. To keep $\ce{H2S}$ in the system this is sometimes done in a hermetically sealed glass tube.


Rationalizing the Planarity of Formamide

Most of the amides are planar (due to steric reasons the restriction may be lifted) and so is also formamide.

The carbon is obviously $\ce{sp^2}$ hybridised (as this concept is very well applicable here), hence organising it's ligands in one plane with roughly $120^\circ$ angles. Naturally one would assume the nitrogen to be $\ce{sp^3}$ hybridised, which is the case for most amines. However, the inversion barrier for these molecules is (depending on the substituents) very low. $$\ce{[NH3]^{pyr-top} <=> [NH3]^{TS-plan} <=> [NH3]^{pyr-bot}}$$ For the nitrogen that means going from $\ce{sp^3}$ to $\ce{sp^2}$ and back again. You can now stabilise the intermediated structure with conjugation, and that is exactley the case here. In your chart, that refers to entry 2. This causes the nitrogen to most likely be of $\ce{sp^2}$ hybridisation and the lone pair of entry one would be in a $\ce{p}$ orbital.

The conjugation happens as stated in you chart by the overlap of that orbital with the antibonding $\pi^*~\ce{C-O}$ orbital. This causes the $\ce{N-C}$ bond order to increase, while the $\ce{C-O}$ BO has to decrease.

All those resonance structures are only descriptions of extreme states, the truth lies between them. The following scheme considers the most common ones and adds a third, that might explain delocalisation (in a non-traditional Lewis way) up to a certain visual point.

In molecular orbital theory you can form 3-centre orbitals from all molecules perpendicular
to the molecular plain. If you choose this plane to be $xy$ then the contributing orbitals will be $\ce{p_{$z$}}$. The following scheme may support this claim, the dipicted orbitals were obtained by a BP86/cc-PVTZ calculation. (The HOMO is an in-plane lone pair of oxygen.)

While there is certainly no intra molecular hydrogen bond ($d(\ce{O-H}\approx 2.57$ same level), there will also most certainly be an attraction between the $\ce{C-O}$ and $\ce{N-H}$ bond helping to stabilize the planarity. But I will not go into detail about that, because it would involve breaking away from the very handy hybridisation concept.


What classifies as a solvent/base in SN1, SN2, E1, and E2 reactions?

Okay in the question i am only seeing CH30H that simply means that methanol is acting both as a base and a solvent.

Since methanol is acting both as a solvent and a reagent, it will undergo E1 reaction as you will know that it is a polar protic solvent and if the solvent was like **water(polar) or DMSO **  then it would follow SN1 and SN2 respectively and *CH30 (-) * would act as the nucleophile.  Thus solvents play an important role in understanding the type of reaction it would undergo.  So it is really important to understand the solvent. As per as i know there are no such conventions..( That  solvent is written below and reagent/base at the top).  If only one reagent  is given it is acting as a both solvent and a base. I would like to tell another thing that sometime we are mistaken a reagent to be a solvent :
DMSO is often used as a oxidizing agent in presence of phosphoric acid.  We mostly confuse it and hold phosphoric acid as the oxidizing agent.


Why can't carbon and nitrogen be detected in ICP-MS and ICP-OES techniques?

As you say, air is allowed to enter the plasma. Air consists primarily of $\ce{N2}$ and $\ce{O2}$, and would clearly interfere with any oxygen or nitrogen in the sample. The problem is not that ICP-MS or -OES cannot detect them, the problem is air.


Difference between flame test and ability to show photoelectric effect

I think you have a misconception. Observing an emission spectrum and photoelectric effects are different things. Photoelectric effect requires not only the electron is ejected with some kinetic energy, but it is ejected by light of a suitable wavelength. No heat or flame is involved.
In (Bunsen burner) flame emission spectroscopy, your temperature is not high enough to remove electrons from Mg atoms in the gas phase. Eventually, the flame temperatures can be made high enough that you do start to observe emission from Mg atoms, but it is in the UV range. Your eyes cannot see UV light.


Why can copper have multiple coordination numbers?

the coordination number of for example $\ce{Cu^2+}$ can be either 4 or 6.

Allow me to preface this answer by debunking that myth.

Balamurugan and Palaniandavar published a set of interesting sterically strained trigonal planar copper(I) complexes, among them 1. Under cyclic voltammetric conditions, they observe two distinctly different oxidation potentials, one of which they attribute to the oxidation to the corresponding trigonal copper(II) complex. Unfortunately, they do not report isolation.[1]

Tetracoordinated copper(II) species are not as common as hexacoordinated ones, but they are also rather well-known. For example, Kitajima et al. synthesised the structurally interesting complex 2, a formal copper(II) and boron(III) binuclear complex. Naturally and as expected for tetracoordinated copper(II) complexes, its structure is tetragonal around both the copper and the boron centres.[2]

Slightly more interesting is the second complex Kitajima et al. were able to crystallise: it is the formal addition of a DMSO solvent molecule to complex 2 to give complex 3. Here, copper(II) is pentacoordinated and equally interestingly the coordination sphere represents a square pyramidal structure, not the usual trigonal bipyramid.[2]

I’ll skip hexacoordinated complexes, because they are damn common and just note that everybody who witnessed complex chemistry at any educational institution has come in contact with $\ce{[Cu(H2O)6]^2+}$ and $\ce{[Cu(NH3)4(H2O)2]^2+}$.

Finally, Martínez-Bulit et al. were able to synthesise an interesting heptacoordinated copper(II) complex 4. Unfortunately, the complex did not crystallise like the corresponding cobalt(II) and zinc(II) counterparts so we are left with analogy to the other syntheses, MS and IR data for establishing it’s structure. In figure 2, you can see a depiction of the coordination sphere.[3]

A variant of heptacoordinated, albeit only characterised for copper(I) to the best of my knowledge, is $\ce{[CuCp(TMS-C#C-TMS)]}$, complex 5.[4] Cyclopentadienyl — even though it is a $\unicode[Times]{x3b7}^5$ complex — should be characterised as pentacoordinating because there are five coordinating atoms. Likewise, acetylene is dicoordinating for a total of 7. It should be possible to extend the concept of sandwhich complexes to $\ce{[CuCp2]}$ 6; however, that complex is currently unknown outside of calculations.[5]

Figure 1: Structures of the complexes mentioned above.

Figure 2: Three-dimensional representation of 4’s ligand sphere.

Now why does copper(II) assume these structures? Well, you can argue things like complex stability, Jahn-Teller distortion, steric strain, electronic effects and much more. In the end, it all boils down to: if the MO scheme of a complex provides sufficient stabilisation with respect to the individual components, the complex is viable to be formed. If removing a ligand will substantially destabilise the complex, it is likely to remain inert. Unfortunately, I was unable to find a linear copper(II) complex (or any linear copper complex for that matter); however, I would assume it to exist if a ligand provided a large cavity with one coordination site and only a single anion was present to coordinate the other. Or maybe not even anions but neutral donors; the anion being $\ce{F- . SbF5}$, a very non-coordinating ion. I’m also sure that the cuprocene sandwhich complex will be synthesised given sufficient time. That would extend copper(II)’s coordination count range from 2 to 10 — always given beneficial circumstances for a certian coordination number.

So tl;dr: There is no simple explanation for the choice of exactly that coordination number.

Note that in aquaeous solution the ligand sphere of the seemingly so stable complex $\ce{[Cu(H2O)6]^2+}$ is in constant fluctuation with dissociating and reassociating ligands. Otherwise, there would be no rapid exchange mechanism that would allow the formation of $\ce{[Cu(NH3)4(H2O)2]^2+}$ upon addition of concentrated ammonia. Compare iron(II): while the hexaammin complex $\ce{[Fe(NH3)6]^2+}$ is undoubtedly stable and characterised,[6] you won’t be able to get it by reaction $\ce{[Fe(H2O)6]^2+}$ with ammonia. Instead, $\ce{Fe(OH)3}$ will precipitate, demonstrating that ligand exchange in iron(III) complexes is slow with respect to deprotonation, while in copper(II) complexes it is reasonably fast.

References:

[1]: R. Balamurugan, M. Palaniandavar, Inorg. Chem. 2001, 40, 2246. DOI: 10.1021/ic0003372.

[2]: N. Kitajima, K. Fujisawa, Y. Moro-oka, J. Am. Chem. Soc. 1990, 112, 3210. DOI: 10.1021/ja00164a052.

[3]: P. Martínez-Bulit, A. Garza-Ortíz, E. Mijangos, L. Barrón-Sosa, F. Sánchez-Bartéz, I. Gracia-Mora, A. Flores-Parra, R. Contreras, J. Reedijk, N. Barba-Behrens, J. Inorg. Biochem. 2015, 142, 1. DOI: 10.1016/j.jinorgbio.2014.09.007.

[4]: D. W. Macomber, M. D. Rausch, J. Am. Chem. Soc. 1983, 105, 5325. DOI: 10.1021/ja00354a023.

[5]: D. W. Clack, K. D. Warren, Inorg. Chim. Acta 1978, 30, 251. DOI: 10.1016/S0020-1693(00)89045-3.

[6]: G. L. Schimek, J. W. Kolis, Chem. Mater. 1997, 9, 2776. DOI: 10.1021/cm970095c.


Thermal decomposition of calcium carbonate

Pump a hard vacuum and calcium carbonate will calcine at a lower temperature.  825°C is about one atmosphere of carbon dioxide at equilibrium.  Thermodynamics proposes, kinetics disposes - finely ground goes faster than bulk.


Is entropy change of the system for reversible process same as entropy change of the system for irreversible change?

Being a state function, entropy is independent of path, so the entropy change of the system is indeed the same for both a reversible and an irreversible process. The transferred heat, however, is not, and does depend on process.

You've probably heard mentioned that the entropy change through a reversible process is less than that through an irreversible process. At first glance this proposition seems suspicious, because we've just said that as a state function entropy is independent of path$-$the resolution is that the entropy change of the system remains the same regardless of process, but the entropy change of the environment is indeed different, such that the total entropy change is greater for an irreversible process than for a reversible process. [Interpreted differently: the environment is in a different state after the system undergoes a reversible process than after it undergoes an irreversible process.]

The internal energy does depend on the system's entropy$-$by the first law we may write $$\mathrm{d}U = T\,\mathrm{d}S + \mathbf{f\cdot\mathrm{d}X}$$ (where $\mathbf{f\cdot\mathrm{d}X}$ represents a generalized work term; commonly this is $-p\,\mathrm{d}V$), so we see that $U$ is a natural function of $S$ and $\mathbf{X}$; $U = U(S,\mathbf{X})$. To give a functional form for $U$ as a function of $S$ and $\mathbf{X}$, we'll need in general an equation of state, an equation for energy, and a type of process$-$let's restrict ourselves to the case of an ideal gas, for which the two equations are $$pV = nRT \quad \text{and} \quad U = \frac{3}{2}nRT$$ (assuming also a monatomic gas), and work out how $U$ varies over an isobaric process. We want to evaluate $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$; let's start by working out $-p\,\mathrm{d}V$: $$-p\,\mathrm{d}V = -p\,\mathrm{d}\left(\frac{nRT}{p}\right) = -nR\,\mathrm{d}T = -\frac{2}{3}\,\mathrm{d}U.$$ Now plug that into our starting equation, substitute, and rearrange... $$\frac{5}{3}\,\mathrm{d}U = T\,\mathrm{d}S = \frac{2U}{3nR}\,\mathrm{d}S \quad \Longleftrightarrow \quad \frac{5}{2}nR\,\frac{\mathrm{d}U}{U} = \mathrm{d}S,$$ which integrates to $$\Delta S = \frac{5}{2}nR\ln\left(\frac{U_2}{U_1}\right) = \frac{5}{2}nR\ln\left(\frac{T_2}{T_1}\right).$$ This is no different from working out $q_\text{rev}$ for such a process and then using $\Delta S = q_\text{rev}/T$; the role of energy is simply more explicit in this approach.


Does anyone know a database that shows the effect of electrolyte concentration on the production of hydrogen gas during electrolysis of water?

You need to look for the data in a different way. You can use electrical conductivities to get the current flow the solution. From the imagined current flow you calculate how much hydrogen would be generated.


Is bamboo based plastic less likely to leak chemicals?

Inspired by the other answer and the comment to my question, i have dug a little deeper. The tableware in the picture is sold by a company called Zuperzozial and made from bamboo and corn reinforced with melamine resin.

Many similar products are sold, almost always without a brand name and always without any specification of the actual material. So I will answer my question for this specific set of cups, there is no way to tell if the answer holds for 'bamboo' tableware in general.

Melamine from tableware may leak into the food, especially when heated, according to the FDA. It is ok to serve cold drinks in a melamine resin cup. The claim that these cups are micro-wave safe is not true.

As for the amount of BS of the rest of the marketing claims: bamboo and corn are renewable sources. But i'm not sure how degradable a resin is.


How can I find the ratio of molecular masses using the ideal gas equation?

The ideal gas equation is $PV = nRT$
But for this problem V, R, and T are constants. So we can collect terms as:
$\dfrac{P}{n} = \dfrac{RT}{V}$
So $\dfrac{P_A}{n_A} = \dfrac{P_B}{n_B}$
Now you correctly determined the number of moles of A and B.
$n_A = \dfrac{4}{\text{MW}_A}$
$n_B = \dfrac{8}{\text{MW}_B}$
But you need to consider the partial pressures of A and B.

The cylinder was empty before A was added, and after A was added the pressure was 1 atmosphere. So the partial pressure of A is 1 atmosphere.
Before B is added the pressure is 1 atmosphere and after B is added the pressure is 1.5 atmospheres.  So the partial pressure of B is 0.5 atmospheres.

$P_A = 1$
$P_B = 0.5$
Now substituting into our simplified equation
$\dfrac{1}{\frac{4}{\text{MW}_A}} = \dfrac{0.5}{\frac{8}{\text{MW}_B}}$
$\dfrac{8}{\text{MW}_B} = \dfrac{2}{\text{MW}_A}$
$\dfrac{\text{MW}_A}{\text{MW}_B} = \dfrac{2}{8} = \dfrac{1}{4}$


Soluble compounds to absorb radio-frequency electromagnetic radiation

Of course! Salt water is effective, because it is an ionic solution. Submarines and SCUBA divers cannot use radio-frequency (RF) communication over any distance because of that issue. Very high power, very-low-frequency (VLF) land-based transmitters can penetrate to perhaps 20 meters depth to subs, but salt water is essentially opaque to RF and microwaves.


How to calculate the rate of formation of a product when given the rate law?


I believe this question is both simple and difficult -- especially for those who are not native English speakers. I'm going to suggest a straightforward answer, but first let me explain the difficulty.

Unfortunately, terminology related to the nanoscale is often misused. But if we can agree on what the nanoscale is, the answer to the question becomes much simpler. So I'd like to respond to this question in two parts: 1) What is nanoporous? and 2) What is a nanoporous structure?

1) What is nanoporous?
This question relates to a more general question, "What is the nanoscale?" In my experience authoring articles in nanotechnology and serving as a referee for ACS Nano, I've struggled with this question.  It seems no accepted authority has defined the term. The closest I've found is this website which says clearly, "...the nanoscale, which is about 1 to 100 nanometers". I realize that good arguments can be made for extending this range to larger values… but surely it must stop short of 1000 nm = 1 micron. If someone knows of a more authoratitive body making a definition, I’d love to hear of it.

Unfortunately, I’ve found a significant number of scientific experts publishing statements suggesting that the nanoscale extends up to the 1000 nm range. This does not make sense to me: that would more properly be called the "microscale".  As a referee I've been surprised at the number of articles submitted for publication which claim nanoscale work when actually the objects of study are at the microscale. The motivation for this is obvious: it’s a lot more impressive to say you’re working at the “nano” than the "micro" scale these days.

So, my suggestion is that if we accept nanoscale means 1-100 nm, then nanoporous means the presence of pores in the 1-100 nm range.

2) What is a nanoporous structure?
Strictly speaking then, a nanoporous structure would be any structure having pores in the 1-100 nm range. The term itself implies nothing about the uniformity of the pores, and that is why we refer to sponges as porous despite very different sized pores. It is true that uniformity of pore size is of great interest to scientists for important reasons, but the term itself does not include that meaning. It is a case of denotation verses connotation. Though scientists have a reputation for precise terminology, they can be just as guilty as others in applying terms imprecisely – and for the same reasons: convenience and self-interest.


Determining Enthalpy Change of the Combustion of Ethanol

I see your concern, and it's absolutely valid. Any time you perform a calorimetry experiment, you need to be certain that you're not introducing extra energy into the system. Typically, the spark used to ignite the sample has a negligible energy input, but it is still there. However, any heat released by the flame (other than the initial energy put in to generate the spark) is generated by the combustion process. Without the combustion, the flame would cease to exist. If you were running this experiment, you could use an electric sparker and measure the power input for the sparker to know how much energy was inputted. Another option would be to run a control experiment with just an ignition (but no ignitable material) and see how much, if any, the water heats up.

Of course, if you used a lighter to start the flame, and then held the lighter open throughout the experiment, you would be introducing extra energy from the combustion of lighter fluid, and this would give you false results as you feared. That's why we don't run calorimetry experiments in that way.


Rule of mutual exclusion: is the converse true?

If there is a molecule of $O$ (as opposed to $O_h$) or $I$ (as opposed to $I_h$) symmetry, it would be such a counter example (no mode that is both IR and raman active, and not centrosymmetric).

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=903&option=4

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=905&option=4

Apparently a snub cube is an example of the $O$ point group and in Chemistry for the 21st Century it is stated:

  9.5.2.5 Snub Cube (6) We have recently demonstrated the ability of six resorcin[4]arenes and eight water molecules to assemble in apolar media to form a spherical molecular assembly which conforms to a snub cube (Fig. 9.3)


Soil conductivity: How do I get a near equivalent between meq/100 g to mS/cm?

The mineral ions (nutrients) in soil can be expressed as two ways: (i) Cation Exchange Capacity (CEC), which has units of $\pu{meq}$/$\pu{100 g}$; and (11)  Electrical Conductivity (EC), which has units of $\pu{mS/cm}$.

When the conductivity is measured in 1:1 mixtre of soil:water, it is called $\mathrm{EC_{1:1}}$ (This is a way of measuring electrical conductivity of soil water or $\mathrm{EC_{sw}}$).

According to USDA Publication, $\mathrm{EC_{1:1}}$ readings less than $\pu{1 dS/m}$ ($\pu{1 mS/cm}$), soil are considered non-saline and do not impact most crops. This publication also gives you a easy procedure for how to measure $\mathrm{EC_{1:1}}$ of your soil.

My best bet is you don't have to worry about your second soil sample, which has CEC of $\pu{45 meq}$/$\pu{100 g}$, because none of the cations (exchangeble cations in the soil) in concern is $\ce{Na+}$ (the given CEC is the sum of all exchangeable cations in equivalents). According to above publication, soils with a high concentration of sodium salts (sodic conditions) have given problems.

To my understanding, there is no direct relationship between $\pu{meq}$/$\pu{100 g}$ and $\pu{mS/cm}$. For more info on Cation Exchange Capacity (CEC) and units of $\pu{meq}$/$\pu{100 g}$, read:

J. P. Reganold, J. B. Harsh, “Expressing cation exchange capacity in milliequivalents per 1 100 grams and in SI units,” J. Agronomic Educ. 1985, 14(2), 84–90 (https://www.agronomy.org/publications/nse/archive/vol-14).


As @Ivan Neretin states, it depends. For example, potassium sulfate and aluminum sulfate form "alum", with a single crystal structure.

Tartaric acid (and its salts) occur in different forms, and even though they have the  same chemical makeup, these stereoisomers can be separated because they form differently shaped crystals. One of Louis Pasteur's first studies was separating the isomers from a mass of crystals.

A good source on growing crystals is Crystals and Crystal Growing by Alan Holden and Phylis Morrison


How do you know if electron is deexciting or exciting itself?

The 'longest wavelength of electron' does not make any sense. I assume that the question asks for the longest wavelength transition when a photon is absorbed or emitted (as fluorescence). One way to answer this is to look up the Balmer formula for hydrogenic like atoms and choose two principle quantum numbers for the transition you want. Alternatively you take the difference in energy between two levels using the formula given. As you start with n=3 the closest is either going to be n=4 (absorption) or n=2(emission). I leave it up to you to do the calculation but the answer must be positive as it corresponds to a photon's wavelength :) ( Note that the question gives the energy in Joules and you will want the wavelength in nanometres).


How does a d4 high-spin complex have a spin-allowed transition without violating the selection rule for total angular momentum?

The photon has one unit of angular momentum and as angular momentum is conserved this must be taken up by the molecule; in terms of symmetry species this means that the product of the symmetry species of the two states involved and of the transition dipole must be worked out. The electron spin in not affected by the photon angular momentum so the spin selection rule applies $\Delta S=0$ unless there is also some spin - orbit coupling then otherwise spin-forbidden transitions can occur but with generally low probability compared to a spin allowed one.

To work out the symmetry species involved in a transition the value of the integral

$$\int\psi_i\mu\psi_fd\tau$$

between initial $i$ and final state $f$ has to be evaluated.; $\mu$ is the transition dipole. This integral is very hard to evaluate but fortunately symmetry can help.

Using symmetry the integral can be determined to be zero (transition forbidden) or not zero. The equation to use is

$$\Gamma\psi_i \Gamma\mu \Gamma\psi_f = \;'A_1\,'$$

where $\Gamma$ means the symmetry species, $A,\; E$ etc and $'A_1\,'$ means the product must contain the totally symmetric representation in the point group used. This is the label given to the top row of characters in the table ($A_1$ or $A_{1g}$).

The transition dipole symmetry is found in the column of the point group just to the right of the characters and are the symmetry species with x, y, and z. In the $T_d$ and $O$ types of point groups the transition dipoles are either the $T_2$ or $T_1$ symmetry species respectively.

Thus in your $A_1\to T_1$ example the symmetry product is

$$A_1 \otimes T_1 \otimes T_1 = A_1$$

in an $O$ point group. The symbol $\otimes$ means multiply characters in each symmetry species column by column.  Hence you can work out the product  by multiplying the characters together or more easily by looking up the values in direct product tables.

In this case this is easy to do as anything multiplied by $A_1$ is unchanged (as its totally symmetric) and any symmetry species multiplied by itself always contains the totally symmetric representation $A_1$. So the transition
$A_1\to T_1$ is allowed in octadedral point groups.

In the tetrahedral point group $T_d$ where the transition dipoles each transform as $T_2$ the product is

$$A_1 \otimes T_2 \otimes T_1 =  T_2 \otimes T_1$$

and now the characters have to be multiplied and the the irreducible representation found (see link to answer below). Instead looking up the product in direct product tables gives

$$T_2 \otimes T_1 = A_2+E+T_1+T_2$$

and in this case as the totally symmetric representation $A_1$ is absent the transition would not be allowed. This means that any transition will be weak, i.e. has a very low but finite extinction coefficient. There are effects in molecules that 'break' the symmetry, such as vibrations distorting the molecules 'ideal' symmetry and these can make forbidden transitions slightly allowed.

This answer gives some details on point groups.
Understanding group theory easily and quickly


How to compare the rate of decarboxylation?

First thing to see is to see carbanion stability.

Now the question arises that between  carbonyl and COOH  groups which can stabilize better. For that you can look at the COOH 's already possible resonance,within its molecule, which decreases it electron withdrawal  tendency.
Hence order would be: I>IV>III>II.


Relationship between Hartree-Fock and Hückel coefficients

Let start by definition the Hartree-Fock-Roothaan equation:

$$\sum_\nu F_{\mu\nu}C_{\nu i}=\varepsilon_i\sum_\nu S_{\mu\nu}C_{\nu i}$$

Where the Fock matrix elements are giving as:

$$F_{\mu \nu} = \left< \phi_\mu \left| \hat{f} \right|\phi_\nu\right>$$

And the overlap matrix is given as:

$$S_{\mu \nu} = \left<  \left. \phi_\mu \right|\phi_\nu\right>$$

Here $\phi_\nu$ is our atomic orbitals (i.e. what you get from your basis set).

When we do a Hartree-Fock calculation we want to find a set of orthonormal orbitals, also known as molecular orbitals. The molecular orbitals are constructed as:

$$\psi_i=\sum_{\mu=1}^KC_{\nu i}\phi_\nu$$

For the molecular orbitals to be orthonormal we can write the following condition:

$$\left< \left. \psi_i \right|\psi_j\right>=\delta_{ij}$$

This means that the orbitals have zero overlap with one-another and an overlap of $1$ with them self. If we insert the expansion of the molecular orbitals in atomic orbitals in the above equation we get:

$$\sum_{\mu=1}^K\sum_{\nu=1}^LC_{\nu i}^\dagger C_{\mu j}\left< \left.\phi_\nu \right|\phi_\mu\right>=\delta_{ij}$$

I.e. we can see that when we do a Hartree-Fock calculation we have the constraint on our molecular orbital coefficients that have to transform our atomic orbital overlap to a matrix with ones on the diagonal and zero elsewhere. This can also be written more compact in matrix form:

$$\mathbb{C^\dagger SC=1}$$

Which is a condition equivalent to the one you know from Hückel theory. At last let me make a breif comment about basis set, since you also talked about those in your question. In general we use contracted Gaussian type basis sets, just like STO3G, that you mentioned. Our atomic orbitals can thus be written as:

$$\phi_\nu = \sum_{k=1}^M N_k d_k P(r)\cdot e^{-\alpha_k\cdot r^2}$$

Here $P(r)$ is a function related to the angular moment quantum number.The way the contraction coefficients $d_k$ are choose implies that we normalize our Gaussian functions, i.e:

$$N_k^2 \int_{-\infty}^\infty \left( P(r)\cdot e^{-\alpha_k\cdot r^2} \right)^2 dr = 1$$

We can now write $\left< \left.\phi_\nu \right|\phi_\nu\right> = 1$ as:

$$\overline{N}\sum_{k=1}^M N_k d_k\sum_{l=1}^T N_l d_l\left< \left.P(r)\cdot e^{-\alpha_k\cdot r^2} \right| P(r)\cdot e^{-\alpha_l\cdot r^2}\right> = 1$$

Where $\overline{N}$ is the normalization we need. As we can see there is no easy condition we can put on the contraction coefficients, in order to avoid evaluation $\overline{N}$. It can be shown (I will not show it here) that $\overline{N}$ only depends on the total angular moment and on the Gaussian exponents, thus it is possible to choose the contraction coefficients in such a way, that we do not need to evaluate $\overline{N}$.


Identity a metal based on its properties

Copper is much denser than water and it does not react with it. Magnesium is also denser than water and reacts only very slowly. Calcium will also sink in water, although it will react with it producing hydrogen, so small pieces of the element may float on the surface. However, it would definitely not melt. The correct answer is potassium - it is a very soft metal, less dense than water and reacts with it producing enough heat to melt it (potassium melts at only $\pu{63.5^\circ C}$).


What do you want?

If you're just looking for a source of potassium ions you could just buy $\ce{KCl}$ or some other compound, they're relatively cheap if not pro analysis. If you actually want to test for $\ce{K+}$:

What else do you know about the product?

Tests are completely different for different amounts of $\ce{K+}$. Does the vendor specifies the amount of each compound on the product's label?

How sure do you need to be?

Do you need to do a QUALITATIVE analysis or a QUANTITATIVE analysis? If you need to be absolutely sure about the concentration then you need to guess it's ridiculously low - you'll probably need at least AAS (Atomic Absorption Spectroscopy) equipment, which can't be accessed by anyone nor find anywhere, but you can order an analysis. Alternatively, you can use an electrode - keep reading .

Some ways of testing that I can think of:

(Of course you'll need to dissolve the powder in at least distilled water to do all the following tests, and do a blank with the same water - the more pure the water the more accurate your results.)

For big amounts of $\ce{K+}$

The flame test suggested by Paul J. Gans on the comments above could work, although I never actually did it.

Another way is doing the following titration:

$$\ce{KCl(aq) + AgNO3(aq) -> AgCl(s) + KNO3(aq)}$$

$\ce{AgNO3}$ can be found at any store that sells chemicals. We used 0.1 M for small amounts of $\ce{Cl-}$.

What you are actually testing for is $\ce{Cl-}$ ions, but if the amount of K = the amount of $\ce{Cl}$ the job is done. Now, since the powder contains potassium citrate this is just a qualitative test, although it can be very accurate for small concentrations if properly executed (see Mohr method). Silver chloride is a white solid that is barely visible in small amounts, and a similar  titration's end point is just a few drops (Mohr method), and there are a lot of possible interference so you should not expect precision unless you do some pro work.

For small amounts of $\ce{K+}$

I think the best way of doing this is using a potassium-selective electrode. I never used one of those, but other ion-selective electrodes work great on water analysis, detection levels were a few mg/L on our old ones!

http://en.wikipedia.org/wiki/Potassium_selective_electrode

And of course, you can use AAS.


Naming ester as a substituent to carboxylic acid

As you have correctly stated in the question, the $\ce{-COOH}$ group has higher priority than the $\ce{-COOR}$ group. In fact, the $\ce{-COOH}$ group has the highest priority (excepting cations) in IUPAC nomenclature.
The $\ce{-COOH}$ group should be included as the suffix of the parent chain, while the $\ce{-COOR}$ is indicated using a prefix.
In fact, Section P-65.6.3.2.2 of the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names, 2013 (aka the Blue Book) specifies the rules for this.

When, in an ester with the general structure R-CO-O-R' or R-S(O)XO-R', another group is present that has priority for citation as the principal group or when all ester groups cannot be described by the methods prescribed for naming esters, an ester group is indicated by prefixes as 'acyloxy' for the group R-CO-O—, and 'alkyloxy(alkany1)...oxo' or 'alkyl(allcanyl)oxycarbonyl' for the group —CO-OR'.

I quote the example provided in the text:

And another one (not from the Blue Book though)

3-Ethoxy-3-oxopropanoic acid

References: Favre, H. A., Powell, W. H. Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 R. Soc. Chem 2013 DOI: 10.1039/9781849733069


How to measure the percent weight of NaCl in a potato chip

I'm somewhat skeptical that atomic absorption spectroscopy (AAS) would work as you outlined the procedure. The "solution" would have starches and fats floating around in it.

(1) I'm not sure that you could get a homogeneous solution.

As far as the homogeneous solution, think of some parts of the solution settling. So what happens if the solutions sit for an hour before going into the AAS?

You could test this by homogenizing a sample and analyzing it immediately, then analyze it after an hour. Do you get the same result?

(2) Even if you did I'm not sure that all the gunk would burn correctly (uniformly) in the AA flame.

Using a plasma spectrometer would solve this problem.

(3) Depending on what sort of accuracy you want there is also the problem of moisture in the chip. How "dry" is a potato chip?

(4) There is a difference between salt added to the chip and $[\ce{Na^+}]$ and  $[\ce{Cl-}]$ in the chip. I'd guess that $[\ce{Na^+}] \approx [\ce{Cl-}]$ in a biological system, but I'm sure that it isn't exactly $[\ce{Na^+}] = [\ce{Cl-}]$.

Such are the joys of method development in analytical chemistry.


Galvanic corrosion: Copper, iron and aluminum in an automotive cooling system

If the copper components are not in direct contact with the aluminum or iron parts, then there should be no electrolytic corrosion. This would require that non-conductive rubber hoses are used for coolant lines (avoid those with steel spring cores to prevent collapsing on the vacuum side, lest the rubber crack through and allow metal-to-metal contact). Also mount the copper parts on rubber shock mounts, rather than directly to the chassis.

Actually, corrosion was not well prevented in the past. For example, some Renault engines had an aluminum block with steel cylinder liners (shrunk with liquid nitrogen and dropped in place to expand for a very tight metal-to-metal seal, if I remember correctly). This short-circuited cell corroded rapidly. Modern antifreeze contains corrosion inhibitors such as silicate/phosphate mixtures that form an insulating film on metal and should help prevent any issue with the copper parts. Use a good quality ethylene glycol antifreeze with inhibitors.


Determining identity of unknown compound with NMR data

Peaks of acidic protons in an NMR spectrum can change (in position, form and intensity) due to exchange with other volatile protons.

A very fast exchange makes the peaks coalesce, you get one singlet between the two "original" (expected) chemical shifts, the exact position depends on the respective amount of both.
A slower exchange, one "on the NMR timescale", broadens that peak considerably, typical case is residual water and a COH proton. The peak might in some cases be so broad it becomes invisible.
Even slower, this peak splits into two, which become sharper and further removed from each other.
A very slow exchange makes no change to the positions, but an exchange with deuterium (i.e. from the solvent) changes the relative intensity of the respective peaks over time.


To clarify Apeiron's answer. You have drawn 2-methyl butane, the same isomer that was given in the question. The rotation of a bond is probably the source of any confusion. A 3D model of your two structures would be the same.

To answer your question:

Neopentane is the third isomer of the pentane series.