How to calculate melting/boiling points at different pressures

I don't believe there is an equation that you can use for melting points of a general substance as a function of pressure (since the melting phase transition has a lot to do with the geometry of the molecule and the structure of the solid), but there is one for the boiling point of any pure substance when you are not near the critical point.

The liquid-vapor transition follows a very specific curve on the pressure-temperature plane of a PVT diagram. This is given by the Clausius-Clapeyron relation, which at temperatures and pressures that are not close to the critical point, can be approximated as:

$$
\ln\frac{P_2}{P_1} = \frac{-\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)
$$

This is the two-point form, which lets you predict the pressure of the vapor phase at a given temperature if you know the temperature and pressure at another point, and the enthalpy of vaporization.

You can use this to predict boiling points as a function of pressure by recognizing that:

  The boiling point is the temperature at which the vapor pressure at equilibrium exceeds atmospheric pressure

This means that if you know the boiling point of a substance at 1 atm (the normal boiling point), and you know $\Delta H_{vap}$ for that substance, you can predict the boiling point at another pressure by plugging in values for $P_1$ and $T_1$ (the normal boiling point), $P_2$ (the pressure you are interested in), and $\Delta H_{vap}$, the enthalpy of vaporization (also sometimes called the latent heat), then solving for $T_2$ (the new boiling point.)

Like I said, this only works when you aren't near the critical point, but for most applications that's not a concern.

For your specific case, especially since you are probably more interested in the liquid-solid transition at this temperature, you are better off looking at a phase diagram.

Here's one for gallium (source) (original source):

The solid-liquid line has a negative slope at low pressure (as ron pointed out) which means the melting temperature increases as pressure decreases. Depending on how high the vacuum is, and on what temperature "room temperature" is, you might be in the solid phase - but the slope is steep, so even at 0 atm the melting point is still right around 300 K.


How does overvoltage affect the products of electrolysis?

When you electrolyse an aqueous solution of NaCl, the product formed at cathode is H2 gas (preferential discharging), but the product formed at anode is Cl2 gas.

There is another complication. The book is missing the fact whether chlorine or oxygen is evolved depends on the concentration of the salt. At low salt concentrations, oxygen will be formed and at high NaCl concentrations $\ce{Cl2}$ will be produced at the anode.

Keep in mind the thermodynamic "competition" is between the oxidation of water and oxidation of chloride ions. Electrochemists have spent their lives and careers in understanding the oxidation reduction of water for hundreds of years. So if you are slightly confused and you have a lot of questions that is a sign of a healthy scientific mind. I am glad that you did not digest what the textbook was saying right away.
Wikipedia hints as how complex is the process of hydrogen and oxygen production at a electrode.

The numbers which you are quoting are like comparing apples and oranges.
This half cell
$$\ce{2H2O(l)} → \ce{O2(g) + 4 H+(aq) + 4e−}$$ $E^o$= +1.23 V is valid for pH of 0, i.e., 1 M H$^+$
ion. Your aqueous NaCl solution is not present in 1 M H$^+$.

You might need to use the Nernst equation to see what is the true potential at a neutral pH.


This is not about mere arbitrary notation.  It is about fundamental concepts in classical thermodynamics, and the notation flows from these concepts.
So: q and w are not state functions; rather, they are path-dependent. Hence their differential forms are inexact rather than exact. The dyet, $\text{đ}$, is used to indicate that (though typically it is an angled slash rather than a horizontal bar; I don't know how to make that with MathJax).
Since the differential forms of heat and work are inexact, any integration with respect to them must be a path integral. While not all physical chemistry/thermodynamics texts use this specific notation for inexact differentials, I've never seen a physical chemistry textbook that doesn't carefully use some sort of notation to distinguish exact from inexact differentials, because of the conceptual and pedagogical importance of distinguishing between state functions and path functions.
[After seeing the comment about Atkins, I went back and looked at my old Atkins pchem book and, sure enough, he doesn't use distinguishing notation for inexact differentials.  I didn't know this because Atkins' thermo treatment was considered poor, so we only used him for kinetics.  We used Castellan for thermo.]
Hence we have, for a closed system:
$d\text{U} = \text{đ}q + \text{đ}w$
And:
$\Delta \text{U} = q + w$
Or, alternately:
$\Delta \text{U} = \text{Q}  + \text{W}$
Heat and work are path functions that reveal themselves only during a change in state—during a process.  Hence, for a given process, there will be a certain amount of heat flow, q, and work flow, w.  One should not use "$\Delta q$"  or "$\Delta \text{Q}$", as this is conceptually incorrect: It implies that heat is a state function, and thus you could have a change in heat.  No.  There was (or could have been) a change in internal energy, along with some heat flow; there was no "change in heat".
More explicitly:
$$\int_{state 1}^{state2} dU = U_2 - U_2 = \Delta U$$
But:
$$\int_{state 1}^{state2} \text{đ}q \ne q_2 - q_1 = \Delta q$$
because there is no q for state 1 or state 2—q is associated with the path only, not with the state of the system.
Instead:
$$\int_\limits {\text{path from state 1 to state 2}} \text{đ}q = q$$
The only context in which it might make sense to use $\Delta q$  (or $\Delta \text{Q}$) would be if you were comparing two different processes, and calculating the difference in heat flow between those two processes.  This is acceptable, but perilous, since such usage risks people falling into the common misconception of thinking that it is acceptable to use $\Delta q$ to characterize individual processes, which it is not.
Adding something from one of my comments: A key property of exact differentials that both connects them with state functions, and distinguishes them from inexact differentials, is that their cyclic integrals must always be zero. Hence:
$$\oint dU = 0= \Delta U_{cyclic}$$
Conversely, it is not generally the case that:
$$\oint \text{đ}q = 0$$
Interestingly, if we carefully define the path to be reversible, and apply an integrating factor $(\frac{1}{T}) $, we can convert the inexact differential $\text{đ}q_{rev}$ into one of the most recognizable exact differentials in thermodynamics, as follows:
$$\oint \frac{\text{đ}q_{rev}}{T} = \oint dS =0$$


What's unique about the most commonly used hetero-cyclic elements?

A possible reason for this is that heterocycles with O, N or S are often easier to obtain and more stable than the corresponding compounds which contain other main group elements as heteroatoms. Phosphole, for example, is less stable than furan, pyrrole and thiophene because of its lower aromaticity. Borole, the structural analog with the lighter heteroatom B, has not yet been isolated in unsubstituted form. Ring strain and the stability of X-C and X-H bonds (X = heteroatom) also need to be considered. Bond lengths tend to increase with heavier heteroatoms. With decreasing electronegativity of X in relation to C, the X-C bond becomes less covalent and is more prone to be cleaved by hydrolysis, for example.


Mechanism for hydroxyethylation on nucleobases (e.g. Thymine)

The second part of the question is easy to answer: the imide attacks at C3 of propylene carbonate instead of C1 (the carbonyl carbon) because carbonates are stabilized (double ester-resonance), whereas carboxylate is a so-so leaving group.

The initial product from nuclephilic attack at C3 is a carbonate half-ester, which subsequently loses CO2 to form the observed N-hydroxyethyl imide.


What happens to the H electron during acid reaction?

The electron belonging to elemental hydrogen forms part of a covalent bond with $\ce{X}$. In this acid-base reaction the $\ce{HX}$ bond breaks heterolytically, meaning that both electrons constituting the bond go to $\ce{X-}$, hence the negative charge.

The base then donates electrons to the proton, forming the electron deficient species $\ce{HB+}$.


Generating the systematic names of organic compounds follows the rules for IUPAC nomenclature.

In this system, a name has several parts:

Prefix (meth-, eth-, prop-, but-, pent-, etc.) - which indicates the number of carbon atoms in the parent chain.
Infix (-an-, -en-, -yn-) - which indicates the presence or absence of carbon-carbon multiple bonds in the parent chain.
Suffix (-e, -ol, -amine, -oic acid, -al, etc.) - which indicates the highest priority functional group in the parent chain.
Substituents (bromo-, methyl-, hydroxy-) - indicate the presence of other things on the parent chain.
Locants (the numbers) - indicate where all of these things happen.

You are confusing the roles of infixes and suffixes.

  Infixes

The base infix is -an- indicating no carbon-carbon multiple bonds. The presence of alkenes or alkynes in the parent chain changes the infix to -en- or -yn-, respectively. No other functional group can change the infix, as there are no carbon-carbon bonds of higher order than 3 or lower order than 1. It is also possible to have a complex -enyn- infix if you have both, but let us not worry about that now. In your example compounds:

3-bromo-3,4-dimethylhexane - the infix is -an- because there are no carbon-carbon double or triple bonds.
but-3-en-1-amine - the infix is -en- because there is a carbon-carbon double bond.
3-aminopropan-1-ol - the infix is -an- again because there are no carbon-carbon multiple bonds. The amine and alcohol functional groups do not consist of carbon-carbon multiple bonds so they cannot change the infix.

  Suffixes

The base suffix is -e, signifying a hydrocarbon with some substituents. Many functional groups change the suffix of a molecule's name, but the name can have only one suffix. If two or more suffix-changing functional groups are found on the parent chain, then the one with the highest priority is chosen to be the suffix and all others become substituents. The common functional groups by decreasing priority are:

functional group     suffix         substituent name
carboxylate anion    -oate
carboxylic acid      -oic acid      carboxy
ester                alkyl -oate    alkoxycarbonyl or alkylcarboxy
acid halide          -oyl halide    halocarbonyl
amide                -amide         amido
nitrile              -enitrile      cyano
aldehyde             -al            oxo
ketone               -one           oxo
alcohol              -ol            hydroxy
thiol                -ethiol        mercapto
amine                -amine         amino
hydrocarbon          -e             varies by group

The name 3-aminopropan-1-ol is correct. Both the alcohol and the amine cannot change the suffix. Only the higher priority group (alcohol) changes the suffix. The other (amine) becomes a substituent. As mentioned before, neither group changes the infix.

  Substituents

In addition to the substituents listed above, several functional groups are always substituents (which are listed alphabetically in front of the parent name):

functional group     substituent name
alkyl                alkyl
alkyl halides        halo
ethers               alkoxy
nitro                nitro
sulfides             alkylthio

There are some other common substituents that deal with exception cases, like phenyl, aryl, vinyl, methylene, formyl, allyl, and acetyl.


Electronic density distribution in hydrogen bond

There are a few descriptors in 'conceptual'  DFT just to characterize hydrogen bonds. E.g.: non-covalent interaction (NCI): note that hydrogen bonds are often consider as NON-covalent bonds, Density Overlap Regions Indicator (DORI).

One can also look at the  deformation density contributions originating from natural orbitals for chemical valence within ETS-NOCV theory.


Why is Cr(2+) a stronger reducing agent than Fe(2+) in water?

The short answer is thermodynamics. Reduction with $\ce{Cr^2+}$ must be more exergonic than reduction with $\ce{Fe^2+}$, we'll get to some numbers in a bit, but let's deal with the concept.

It is tricky to compare the "stability" of two possible products that occur from different pathways. What is more important for the spontaneity of the reaction is the change in (free) energy that occurs from start to finish.

To put it another way, it may be that $\ce{Fe^3+}$ is more stable than $\ce{Cr^3+}$ on an absolute scale, but what we really care about is how much more stable $\ce{Cr^3+}$ is to $\ce{Cr^2+}$ compared to how much more stable $\ce{Fe^3+}$ is to $\ce{Fe^2+}$.

Let's examine all four using electron configuration as you have done:

$\ce{Fe^2+}$ is $\ce{d^6}$ or more probably $\ce{[Ar] 4s^1 3d^5}$ - two half-filled half shells!
$\ce{Fe^3+}$ is $\ce{d^5}$, which is $\ce{[Ar] 3d^5}$ - one filled half shell.

The difference between the two iron ions might not be that large.

$\ce{Cr^2+}$ is $\ce{d^4}$, which is $\ce{[Ar] 3d^4}$ or $\ce{[Ar] 4s^2 3d^2}$
$\ce{Cr^3+}$ is $\ce{d^3}$, which is $\ce{[Ar] 3d^3}$ or $\ce{[Ar] 4s^2 3d^1}$

The energy difference between chromium ions might be larger. Represented graphically, the reaction coordinate energy diagram for the two process might be:

  Now the number part.

We can go looking for some standard thermodynamic data to help make our case. The best data are for standard reduction potentials, because these data are for exactly what we want!

Taking the reduction potential data and writing the equations in the direction we care about, we have:

$$\begin{align}
&\ce{Fe^2+ -> Fe^3+} &&&E^\circ=\pu{-0.77 V}\\
&\ce{Cr^2+ -> Cr^3+} &&&E^\circ=\pu{+0.44 V}
\end{align}$$

Spontaneous reactions produce positive potential differences, so we can see right now that $\ce{Cr^2+}$ is a better reducing agent.

Let's go a step farther to free energy.

$$\Delta G^\circ =-nFE^\circ$$

However, to deal with free energy, we need a full reaction. Technically, comparing the two half-reactions in isolation is just as bad. However, the definition of the standard electrode potential and the standard free energy come with a common zero point in terms of half-reaction: $\ce{2H+ + 2e- -> H2}$.

The two full reactions (net ionic equations anyway) are:

$$\begin{align}
&\ce{2Fe^2+ + 2H+ -> 2Fe^3+ + H2} &&E^\circ=\pu{-0.77 V}&&&\Delta G^\circ=\pu{+74kJ/mol}\\
&\ce{2Cr^2+ + 2H+ -> 2Cr^3+ + H2} &&E^\circ=\pu{+0.44 V}&&&\Delta G^\circ = \pu{-42 kJ/mol}
\end{align}$$

And the winner is chromium, by a whopping $\pu{116 kJ/mol}$. In fact, the data might show that $\ce{Fe^2+}$ is more stable than $\ce{Fe^3+}$, while $\ce{Cr^2+}$ is less stable than $\ce{Cr^3+}$.


Relation between ionisation enthalpy and nature of bonds in boron

At first, let me give a picture of what ‘relatively high’ actually means. I present the sum of the first three ionisation enthalpies of some elements typically found as metal(III) ions:

Boron $6887.4~\mathrm{\frac{kJ}{mol}}$
Aluminium $5139~\mathrm{\frac{kJ}{mol}}$
Scandium $4256.7~\mathrm{\frac{kJ}{mol}}$
Iron $5281.4~\mathrm{\frac{kJ}{mol}}$
Cobalt $5640.4~\mathrm{\frac{kJ}{mol}}$
Ytterbium $4195.2~\mathrm{\frac{kJ}{mol}}$

We see that boron’s ionisation energy sum is $1.2~\mathrm{\frac{MJ}{mol}}$ larger than that of the next largest in my sample. Also keep in mind that boron is rather small per se so a $\ce{B^3+}$ ion must be even smaller.

But what does that actually mean? To generate an ionic compound from its elements, one usually considers an overall formation enthalpy composed of bringing both elements into the gas phase, turning them into ions and letting the crystal lattice form — this last step liberates most energy and is usually the driving force for salt creation. In boron’s case, the cations of such a salt would be $41~\mathrm{pm}$ — exceptionally small; way smaller than $\ce{Al^3+}$ ($68~\mathrm{pm}$) which is already tiny and smaller than $\ce{Be^2+}$ whose $59~\mathrm{pm}$ are typically cited as the smallest ionic radius known ($\ce{H+}$ doesn’t count).[1]

The structure of $\ce{BeF2}$ can already be considered as the greatest possible compromise in keeping the cations and — more importantly — the anions apart from themselves; hence its spacious structure. These problems would be multiplied if an even smaller cation were used here: It would be extremely hard to generate a crystal lattice that would allow for sufficient separation of the anions from each other since the cation is so tiny (and we need lots of anions: three negative charges per boron!).

So all things considered, ionic boron salts would have a very unstable crystal structure and would hence be hard to generate, especially since the lattice enthalpy cannot counteract the large ionisation enthalpy required for boron adequately. Luckily, covalent compounds are always a possibility and boron’s atomic orbitals are well shaped and well placed in energy to allow covalent bonding.

Note that boron’s electronegativity of $2.0$ on the Pauling scale also shows how badly boron would form ionic bonds: The tiny sizes of ions would draw electrons in in an even stronger way, potentially reducing boron before it would reach the ionic state.

The Polish chemist Kazimierz Fajans simplified this in his set of Fajans’ rules:

A compound will be covalent, if:

there is a high positive charge;
the anion is large; and/or
the cation is small.

A compound will be ionic, if:

there is a low positive charge;
the anion is small; and/or
the cation is large.

Note that in ambiguous cases such as $\ce{AlF3}$, (small cation, high positive charge but small anion) the rules fail to predict a mainly ionic compound, but correctly make us assume a rather high covalent character of said ionic bonds.

[1] Even beryllium forms rather covalent compounds most of the time. Due to the electronegativity difference of $2.5$ on the Pauling scale, we may just be able to consider solid $\ce{BeF2}$ as an ionic compound. All of these are borderline cases in one way or the other, though.


The way my teacher first explained this is due to the definition of atomic mass. According to Google:

  The mass of an atom of a chemical element expressed in atomic mass units. It is approximately equivalent to the number of protons and neutrons in the atom (the mass number) or to the average number allowing for the relative abundances of different isotopes.

Now, the key words are

  ...the average number allowing for the relative abundances of different isotopes.

What does that mean? We go back to the definition of an isotope. An isotope of an element is a species that has the same number of protons in the nucleus, yet a different number of neutrons. For example, $\ce{C^12}$ and $\ce{C^14}$ are isotopes of each other. Sidenote, isotopes are not to be confused with allotropes, which are different forms of the same element. For example, graphite and diamond are allotropes of carbon, since both, either, or neither might be made up of $\ce{C^12}$ ($6p$, $6n$)  or $\ce{C^14}$ ($6p$, $8n$). Now how do we represent this with the periodic table? Simply, we take the weighted average. Now, I do not have enough information or data to calculate the relative abundances of each isotope of carbon, and the periodic table does not either. We have an average. The atomic mass of $\ce{C}$ is given as $12.011$. That means a few different things. Firstly, there are more isotopes that are closer to having a sum of $p+n=12$, and less that have $p+n=14$ or $15$. Secondly, this means that although this number is not exact, it represents the most common $p+n$. Averaging all of those different $p+n$'s up, from each isotope, multiplying they by their relative abundances, and averaging should give our periodic table value of $12.011$.

Now it is important to remember that this number is not rounded. You might be asking, "Then why, in our calculations, do we sometimes round this number?" Well it depends on the problem, because for many problems, you do not round, but for the ones that you do, it is because it is more convenient to assume that the number that is given rounded is the most common isotope. Was that sentence confusing? Essentially, it is easy and generally correct to say that $12.011$ for carbon can be rounded to $12$, and $12$, meaning $\{p,n\}\in\{6,6\}$. We can do this by assuming that carbon in its most common form has $6p$, and $6n$.

In orthocresol's comment above, he is also correct. He has a perfectly valid point regarding common word usage in speaking. However, I do not think that this is the main reason.

Regarding Jon Custer's comment above, yes, that would be true, but only if we had a single isotope, which we do not.

Now, make sure you can differentiate between atomic mass and atomic weight! :)

Hope this helps!


Why is the spectroscopic entropy of carbon monooxide higher than that of nitrogen?

One contribution is the number of identical conformations of a molecule that can be generated by symmetry operations, which is described by a symmetry number $\sigma$. In the high temperature limit
$$S_\mathrm{rot} /R = 1 + \ln\left( \frac{8\pi^2IkT}{h^2\sigma} \right) $$
A larger symmetry number decreases the entropy. This number is 2 for nitrogen and 1 for carbon monoxide. That alone can account for $\log(2)=  0.69 \times R$. The larger symmetry number for nitrogen decreases the entropy by this amount.
In addition the moments of inertia (inversely proportional to the rotational temperature) of the two molecules are not really identical (it is ~5% larger for $\ce{CO}$), but the difference accounts for only ~$0.05 \times R$.


Clarification needed regarding description of molecules without hybridization using pure p orbitals

Yes, what I wrote on the topic in the previous question was very, very brief and while writing it I was not thinking of boron — an element that usually calls for exceptions due to it only having three valence electrons.

However nonetheless, the first thing you should do for any compound is to consider all atoms in ground state configuration and see whether you can form the required bonds from occupied orbitals in that state. Let’s take the example compounds $\ce{BF3, NF3, AsH3, C2H4}$ (tricoordinate) $\ce{CO2, H2O, H2Se, BeCl2, CH2}$ (dicoordinate) and $\ce{HCl, N2, CO}$ (monocoordinate) for the discussion. Any structural data referenced herein is taken from the corresponding Wikipedia articles.

Ideal cases

I consider the examples $\ce{AsH3, H2Se}$ and $\ce{HCl}$ to be ideal cases. Taking each atom in its ground state ($\ce{As}:{ [\ce{Ar}]}\,\mathrm{4s^2\,4p^3}; \ce{Se}{: [\ce{Ar}]}\,\mathrm{4s^2\,4p^4}; \ce{Cl}{: [\ce{Ne}]}\,\mathrm{3s^2\,3p^5}; \ce{H}{: \mathrm{1s^1}}$), arsenic has three half-filled p orbitas, selenium has two and chlorine has one. Each of these elements (except hydrogen) has a single lone pair in an s orbital and if there are further lone pairs, they are in p orbitals (arsenic has no more). Thus, we can construct $\sigma(\ce{s-p})$ bonds until all valences are satisfied.

Moving on to bond angles: well, $\ce{HCl}$ has no problems as there are only two atoms. But $\ce{H2Se}$ and $\ce{AsH3}$ also have hardly any problems since the central atoms are so large that the comparatively smaller ligands can be attached at almost perfect $90^\circ$ angles without significant steric repulsion. The bond angles repored ($\ce{AsH3}{: 91.8^\circ}; \ce{H2Se}{: 91^\circ}$) are in adequate agreement.

Ideal cases with steric distortion

The next set of examples along the route you should be going are $\ce{NF3}$ and $\ce{H2O}$. These compounds fulfil the general criteria outlined in the previous section. However, their central atoms are decidedly smaller. Thus, perfect $90^\circ$ angles are unobtainable as there would be a steric clash of the ligand atoms. The ligand atoms move apart to prevent steric repulsion; this is achieved by mixing in s-contribution into the bonding orbitals. $25~\%$ s-contribution would correspond to a tetrahedral bond angle of $109.5^\circ$ and $\mathrm{sp^3}$ hybrid orbials. The table below compares the bond angle and length data of the lighter homologues discussed here and the heavier ones discussed above (consider the fluorines in $\ce{NF3}$ to be pseudohydrogens).

\begin{array}{lcccc}\hline
\text{element} & \ce{N} & \ce{As} & \ce{O} & \ce{Se} \\ \hline
d(\ce{X-H})/\mathrm{pm} & 137 & 151.9 & 95.84 & 146 \\
\angle(\ce{H-X-H})/^\circ & 102.5 & 91.8 & 104.5 & 91 \\ \hline\end{array}

$\ce{CH2}$, carbene, is a dangerous case. As far as I know, no definite structures are reported (although Wikipedia mentions bond angles determined by EPR) since it is too reactive and since it is susceptible to dimerisation to form ethene. It also comes in three distinct forms: a singlet form, a linear triplet and a bent triplet. The most ideal structure of these three is the singlet carbene, which also fits nicely into this subsection.

For a singlet carbene, you would also start off from carbon’s ground state $[\ce{He}]\,\mathrm{2s^2\,2p^2}$. The two p orbitals each form a $\sigma(\ce{s-p})$ bond to the hydrogen atoms. The two hydrogens, however, need to move away from each other to ease steric repulsion; thus a little bit of s-contribution is mixed in for a final bond angle of $102^\circ$. A third p orbial remains unoccupied — note the similariy to $\ce{H2O}$ which has everything carbene has plus two electrons in a p-type lone pair.

Practically ideal cases which naturally lead to multiple bond

Herein, we shall discuss $\ce{N2}$ and $\ce{CO}$. For both, we can apply the method outlined above; I’ll start with the simpler $\ce{N2}$ case. We initially have one s lone pair and three, singly occupied p orbitals on each nitrogen. One of these forms a standard $\sigma(\ce{p-p})$ bond with the other nitrogen. Now, thankfully, the other p orbitals are nicely aligned and fit perfectly for generating two $\pi(\ce{p-p})$ bonds. Together, that gives us three bonds or a triple bond, composed of one σ and two π bonds.

The carbon monoxide case is essentially the same except that carbon would have one empty p orbital and oxygen would have one lone pair in a p orbital; only one π bond can be formed in the way outlined above. However, the unoccupied carbon orbital and the fully populated oxygen orbital can align likewise and form what could formally be described as a second, dative π bond. That renders nitrogen and carbon monoxide isoelectronic. Note, however, that the simplification introduced herein is not able to explain the big differences between $\ce{CO}$ and $\ce{N2}$ that arise from their actually rather different molecular orbital schemes.

Cases, in which an s type lone pair is impossible

I’ve cleared away all the simple cases. If you go through the list, the only ones that remain are $\ce{BF3, C2H4}$ and $\ce{BeCl2}$ — although we might as well include linear and bent triplet $\ce{CH2}$ in here. These all have in common that assuming an s type lone pair would not allow for sufficient electrons available for bonding.

If that is the case, the simple answer always suggests hybridisation (athough the MO description does not require it and nonlocalised molecular orbitals would not support it). At least one p orbital must be mixed in to allow for the generation of enough σ bonds; yet you should always try and make the p contribution of the hybrid orbitals as small as possible (in simplified terms) as that means the s lone pair ‘loses’ less energy en route. (Beware, the terms and images I used are inaccurate!) Any orbitals, populated or not, that are not required for σ bonds remain as p type ones. For $\ce{CO2}$ and linear triplet $\ce{CH2}$, that gives us an $\mathrm{sp}$ configuration of the central carbon with two unused p orbitals; in $\ce{C2H4}$, bent triplet $\ce{CH2}$ and $\ce{BF3}$, the central atoms are $\mathrm{sp^2}$ configured with one additional p orbital.

If possible — i.e. in $\ce{CO2}$ and $\ce{C2H4}$ — π bonds should then be formed. Each of the oxygens in carbon dioxide has one p-type semipopulated p orbital available for a π bond as does the second carbon of $\ce{C2H4}$. The carbenes cannot form π bonds due to the lack of partners, and thus they remain as diradicals. The linear one has two degenerate p-type radical orbitals, the bent one almost has an $\mathrm{sp^2}$ type radical orbital and one p type. (In fact, bent triplet carbene has a larger bond angle than the $\mathrm{sp^2}$ predicted $120^\circ$. This is in line with removing part of the s contribution of the radicals and supplying that to the bonding orbitals. Nobody expects you to correctly analyse the structures of carbenes, though.)

Beryllium

You may have noticed that that element had gone missing at the end of the previous section. This is because it is the most special of all cases. All other alkaline earth metals behave as metals (form dications and practically ionic structures) while beryllium still features notable non-metal properties.

I think that the best description of $\ce{BeCl2}$ is via a four-electron-three-centre bond akin to $\ce{XeF2}$. That again requires an unhybridised central atom. The lowest of the three relevant MOs of the 4e3c bond would be formed of the fully bonding $\mathrm{p}(\ce{Cl}) + \mathrm{s}(\ce{Be}) + \mathrm{p}(\ce{Cl})$ combination; the second bonding MO would then be $\mathrm{p}(\ce{Cl}) + \mathrm{p}(\ce{Cl})$ where the nodal plane traverses $\ce{Be}$ — mixing in its 2p orbital probably increases the bonding character — and a final antibonding $\mathrm{p}(\ce{Cl}) + \mathrm{s}(\ce{Be}) + \mathrm{p}(\ce{Cl})$ orbital wherein the phase of beryllium’s orbital is of opposite sign with respect to the chlorine orbitals’. If that is the case, it should possibly be discussed with the ‘hypervalent’ cases of the other question.

I could be terribly wrong, though. Thankfully, nobody will request you to predict the MO scheme of $\ce{BeCl2}$ until at least the final bachelor’s examination.


Nope.

While $\ce{NaOH}$ is a strong base, the mixture of $\ce{C6H5COOH}$ and $\ce{C6H5COONa}$ is a  buffer solution and tends to resist large changes in pH.

After adding the $\ce{NaOH}$, the extra $\ce{OH-}$ ions would react with the undissociated $\ce{C6H5COOH}$, effectively being removed from the solution. The solution would now be $1.1 \text{L}$ and contain a bunch of ions, like $\ce{Na+}$, $\ce{C6H5COO-}$, $\ce{H+}$, and $\ce{OH-}$. You can calculate the value of $\ce{[Na+]}$ now.

Use the Henderson's buffer equation:

$$\text{pH}=\text{p}K_a + \log{\left(\frac{\ce{[C6H5COO^-] }}{\ce{[C6H5COOH]}}\right)}$$

I think you can continue from here.


How could chlorine dioxide be stabilized in aqueous solution?

Your assumption is correct. The term “stabilized chlorine dioxide” is misleading since the solution does not contain relevant amounts of chlorine dioxide $\ce{ClO2}$.

The solution actually contains sodium chlorite $\ce{NaClO2}$, which is the salt of the unstable chlorous acid $\ce{HClO2}$. At low pH, chlorite solutions contain chlorous acid
$$\ce{ClO2- + H+ <=> HClO2}$$
which decomposes to form chlorine dioxide approximately according to
$$\ce{5HClO2 -> 4ClO2 + HCl + 2H2O}$$
or
$$\ce{4HClO2 -> 2ClO2 + HClO3 + HCl + H2O}$$

At higher pH, decomposition of sodium chlorite slows down. Therefore, sodium chlorite is mainly stabilized by using a buffered solution. In your product, the buffer consists of trisodium phosphate and citric acid. Thus, a better description would be “stabilized sodium chlorite” instead of “stabilized chlorine dioxide”.


Stereochemistry of cyclic hemiketal formation

No, the stereochemistry of the methyl group is not dependent on which face the $\ce{-OH}$ attacks from. This is apparent in the products of the cyclization of aldopyranoses; only two isomers (called anomers) are formed, which are classified as either $\alpha$ or $\beta$:

Depending on which position we are considering the stereochemistry of, a wedge in the straight chain could appear as either a wedge or a dash in the cyclic form. For the molecule you are considering:

$\hspace{.7cm}$

To visualize this, convert the skeletal structure into its Fischer projection:

$\hspace{4.5cm}$

$\hspace{2cm}$


Regarding part 1

Camphor will readily sublime and so, in proportion to the nitrogen gas, it is the minor constituent

For part 2,

  If nitrogen is solvent, then how can the solution be in gaseous state?

if the nitrogen is the solvent then of course it will be in the gaseous state as nitrogen is a gas.


You are right that intensive properties are not additive like the properties which depend on the extent of the system. Pressure is indeed an intensive property because it is force divided by an extensive quantity (area). Now partial pressure is an interesting property. It is a hypothetical construct which cannot be directly or "experimentally" measured in a mixture of gases yet it can be calculated for individual gases in that mixture.
The basis of calculation of partial pressure is the number of particles in the container of each gas (hence an partial pressure originates from the number of particles of gas-A and gas-B).  Greater the number of particles of each gas in a container of fixed volume, greater is the calculated partial pressure of each component. The number of particles in a container determine the partial pressure which is an extensive property and hence partial pressures are additive.
Wikipedia explains the limitations of classifying quantities as extensive or intensive.
Consider reading the Journal of Chemical Education Article (free download from Google Scholar)
Redlich, "Intensive and Extensive Properties" J. Chem. Educ., 1970, 47 (2), 154–156. Redlich says that this classification is an incredible source of confusion in thermodynamics. I feel that he is not wrong.

The general validity of the division of physical properties into extensive and intensive kinds has been addressed in the course of science. Redlich noted that, although physical properties and especially thermodynamic properties are most conveniently defined as either intensive or extensive, these two categories are not all-inclusive and some well-defined physical properties conform to neither definition. Redlich also provides examples of mathematical functions that alter the strict additivity relationship for extensive systems, such as the square or square root of volume, which may occur in some contexts, albeit rarely used.

> Other systems, for which standard definitions do not provide a simple
> answer, are systems in which the subsystems interact when combined.
> Redlich pointed out that the assignment of some properties as
> intensive or extensive may depend on the way subsystems are arranged.
> For example, if two identical galvanic cells are connected in
> parallel, the voltage of the system is equal to the voltage of each
> cell, while the electric charge transferred (or the electric current)
> is extensive. However, if the same cells are connected in series, the
> charge becomes intensive and the voltage extensive. The IUPAC
> definitions do not consider such cases.
>
> Some intensive properties do not apply at very small sizes. For
> example, viscosity is a macroscopic quantity and is not relevant for
> extremely small systems. Likewise, at a very small scale color is not
> independent of size, as shown by quantum dots, whose color depends on
> the size of the "dot".


The instability of hydrated carbon dioxide or "carbonic acid"

The situation is described by the following equilibrium
$$\ce{H2O + CO2 <=> HO-CO-OH}$$
In a closed system, using Le Chatelier’s principle you can manipulate the equilibrium in either direction (e.g. pump in carbon dioxide and increase the concentration of carbonic acid, etc.).  However in an open system, as carbon dioxide escapes, the amount of carbonic acid will decrease.

When all is said and done, water and carbon dioxide are two extremely stable molecules.  Since
$$\Delta G = -RT\ln K$$
Their stability is what drives the equilibrium to the left and why carbonic acid appears to be (relatively) so unstable.


Considering the d-d transition how, does tetracyanidonickelate(II) ion exist as a colored complex?

This is a diagram of the d orbitals of a generic d8 complex in a tetrahedral and square planar configuration. The tetrahedral complex would be expected for pi donor ligands (Cl-, OH-, etc.) where the pairing energy is greater than delta t, while the square planar complex would be expected for strong field pi* acceptor ligands (think CO, NO+, and CN-, which are effectively isoelectronic anyways).

As you can see the frontier orbitals of this complex are all mostly on the metal complex , and I'd expect the excitation of electrons from lower-energy orbitals into these anti bonding orbitals to be the reason for the colors.


How does one refer to the amino acids in a peptide?

A part of the peptide that was previously an amino acid molecule when free can be correctly referred to as an amino acid residue when bound in the peptide.

A source


Why does zinc react with sodium hydroxide?

Indeed, $\ce{Zn}$ is lower than $\ce{Na}$ in activity series of metals, so the following reaction won't take place

$$\require{cancel}\ce{Zn + 2NaOH \cancel{→} Zn(OH)2 + 2Na}$$

Remember, however, that $\ce{Zn}$ is amphoteric, so it can reacts with a strong base such as $\ce{NaOH}$ as an acid forming sodium zincate

$$\ce{Zn + 2 H2O + 2 NaOH -> Na2Zn(OH)4 + H2}$$

P.S. I forget a lot of good ol' chemistry, and I have just realized that usually we do not call metals themselves amphoteric, but rather use this term for their oxides/hydroxides. For metals we just say that some of them (such as $\ce{Zn}$) dissolve both in acids and in bases. The point is that $\ce{Zn}$ does not react as an acid in the reaction above since the reaction is not an acid-base one, but rather oxidation-reduction. So, the answer then is that $\ce{Zn}$ reacts with aqueous solution of sodium hydroxide because it can be oxidized in these condition in accordance with the reaction mentioned.


Are metals hydrogen bond acceptors or donors?

Hydrogen bonds are three centre 4 electron interactions that lower the energy of the molecular orbitals involved. We can hypothetically imagine two scenarios,

Metal donor (M$^{\delta ^{-}}-H^{\delta ^{+}}...C$)
Metal acceptor (M$^{\delta ^{-}}...H^{\delta ^{+}}-C$)

(where C is just something else). Metal-Hydrogen bonds where the metal is a donor are scarce since the metal hydride bond is the opposite polarity of the metal hydrogen bond donor, however they do exist.

Metal acceptors are likely to be low oxidation state late row metals with filled $d$ orbitals to donate via $\sigma$ symmetry interaction.

You get these types of interactions all the time in crystal chemistry where we find bridging hydrogens! So in short, it depends where the metal is in the period, if its $dz^2$ is available to donate etc.

If you had a specific example I would be happy to try and work it out, but it looks like you are aware of any literature I would consult!! :)


Ion/Counter ion layers in the colloid precipitate of silver chloride

It isn't absolutely clear from the wording, but it seems you're adding $\ce{Cl^{-}}$ to a $\ce{Ag^{+}}$ solution.

As some of the $\ce{Cl^{-}}$ is added, $\ce{Ag^{+}}$ is still in excess in the solution so virtually all of the $\ce{Cl^{-}}$ precipitates. $\ce{Ag^{+}}$ will be the primary adsorbent on the precipitate, and the colloidal particles will have a positive charge.

Since the colloidal particles have a positive charge, the ion in the counter-ion layer around the particle needs to be negatively charged. There isn't any  $\ce{Cl^{-}}$ in solution so it has to be $\ce{NO3^{-}}$ as the anion.

Thus the correct answer is (D).


Synthesis; effect of hydrogen gas and platinium

The selective debenzylation of benzyl ethers can be carried out under very mild conditions, leaving other functional groups intact. Commonly carried out under H2 using a heterogeneous catalyst such as Pd/C or Raney Ni, Pt can also be used. Solvent choice is critical, and reactions times are longer, but it is still effective.

I don't like red herrings.


Can the first ionization energy for a specific element be calculated?

I tried to find out empirically whether your formula makes sense. The data I used was effective nuclear charges from this Wikipedia page (always the minimal value, Pt was discarded), and ionization energies already gathered for the answer to this question on this site. The principal quantum number was also manually added to the data set.

I explored the data in many ways and tried to find a simple expression which could give an estimate for where the ionization energy should lie.

First, I looked at what your formula would give:

It is quite obvious that your formula describes two atoms perfectly and is absolutely terrible for the overwhelming majority of elements in the periodic table.

I present to you the best guess so far, based on simple manual adjustments and some exploration of the data:

$$ E_\text{ion} / \mathrm{eV} = 6 \left(\frac{Z_\text{eff}}{n^{1.5}} \right)^2 + 4$$

Below is the graphical representation:

The error will still be large for many atoms, I estimate the median error to be about 2 eV. But if you want to use a simple formula to calculate the first ionization barrier, this would be the one I would use.

The data exploration was done in Python and is recorded in this iPython notebook. Feel free to have a look at it and even come up with a better model.


Mechanism for reaction between hydrochloric acid and magnesium

Based on the three step mechanism you have provided, what if the slow step is your step 3 which is H + H --> H2  so rate law is now k3[H]^2. Since H is an intermediate, you can substitute for this intermediate from earlier steps and you will end up with
Rate = (k3.k1.k2.[Mg][H+]^2)/[Mg2+]   and if [Mg]/[Mg2+] is constant as the reaction of a solid always happens at the surface, you can get a rate law = k [HCl]^2


How to calculate the amount of substance of barium sulfate that is precipitated?

Yes, your answer is correct.

However, since you are still learning I recommend, that you take the time to write out all the equations, with proper units and every step. This should help you to gain a better understanding of the chemistry that is actually going on.

One of the key words of the exercise is excess of sodium sulfate solution, which indicates that the reaction will be complete, i.e. entirely on the product side.

Hence, you can (and you correctly did) determine from the reaction equation,
$$\ce{Ba(NO3)2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaNO3 (aq)},$$
that the stoichiometric ratio is one, $n(\ce{Ba(NO3)2})/n(\ce{BaSO4}) = 1$. Therefore you can also write $$n(\ce{Ba(NO3)2}) = n(\ce{BaSO4}).$$

Now you simply have to calculate the amount of substance of barium nitrate. You have been given the volume of your solution $V(\ce{Ba(NO3)2}) = 0.200~\mathrm{cm^3}$ and the concentration $c(\ce{Ba(NO3)2}) = 0.450~\mathrm{mol\,dm^{-3}}$. Using the relation
$$c = \frac{n}{V}$$
you can calculate the amount of substance of barium nitrate.

You have also correctly converted cubic centimetres to cubic decimetres. In any case, you should get familiar with metric prefixes since they will come up in every science class you will take.


Is the Haber Process here proceeding at positive Gibbs free energy change?

It's all about the pressure! The Haber process requires a pressure of about 200 atm. This is how the process becomes favorable, since LeChatelie's principle says that increasing pressure shifts toward the side with less moles of gas.

As long as Q is smaller than K, the reaction will go forward. For example, if nitrogen and hydrogen are supplied at stoichiometric ratio and 1% of a given reaction mixture is product, we would have the following partial pressures:

$$ P_\ce{N2} = 0.2475 \times \pu{200 atm}$$

$$ P_\ce{H2} = 0.7425 \times \pu{200 atm}$$

$$ P_\ce{NH3} = 0.01 \times \pu{200 atm}$$

Accordingly, Q would be

$$ Q = \frac{2^2}{50 \times 75^3} = \pu{1.9e-7} $$

In the Haber process, the concentration of product is kept small by continuously removing product (liquifying ammonia at low temperature) and feeding the remaining reactants back into the reaction mixture.


This is probably a more complex problem than your instructor intended. In addition to the various metals having different toxicities, different compounds of these metals have different toxicities.

For example, nickel metal has low toxicity, but the compound $\ce{Ni(CO)4}$ is a poisonous, flammable, volatile liquid.

Likewise alkyl mercury compounds, e.g. $\ce{(CH3)2Hg}$ are vastly more toxic than the low solubility mercury salts like $\ce{Hg2Cl2}$.

If you just want to know about the pure metals, then this can be easier. Look up a Safety Data Sheet for each metal. It would be better if you pull all your SDS from the same source, that way they are all using the same data sources.

For example, here is a link to the SDS for mercury from Fisher Scientific. Section 12 contains ecological information, including LC50 (concentration that kills half of the population in the time period given) data for various aquatic organisms. Section 11 contains toxicological information, including LD50 (dose that kills 50% of the population) for various organisms.

The best way to do this is to try to find one set of data for all the metals with the same organism for the best comparison. Then, devise your scale.


The oxygen is dissolved in the water, just as salt can be dissolved. It does not (to any appreciable degree) combine with water molecules to form hydrogen peroxide.

The reason that oxygenated water is not fizzy like soda water (CO2) is the solubility of oxygen in water is about 2% that of CO2. So the amount of bubbles formed when the cap is removed is only about 1/50 as much as for soft drinks. Or champagne.


Can hydrobromic acid perform substitution with propane?

I am afraid that you cannot support your argument. Here are two reasons why as I expand on my comment above. If this process were to be a radical chain reaction, the following process should be considered. Hydrogen bromide has a weaker bond then any of the C-H bonds in propane. An initiation step (heat, light?) would be expected to generate atoms of hydrogen and bromine. If you use a bromine atom as a reactant in propagation step 1 (P1), you will produce a hydrogen atom at the end of P2. So we start with a hydrogen and abstract a secondary hydrogen (primary hydrogen is higher in bond strength but it will only make the analysis less favorable) to produce a secondary propyl radical. The bond dissociation energies (BDE's in kcal/mol) are listed over the bonds of interest. The first step is good news. Exothermic by -5 kcal/mol. But P2 is endothermic by +14 kcal/mol. If you add P1 + P2 you get the overall reaction that has a heat (enthalpy) of reaction of +9 kcal/mol. The reaction is endothermic and uphill. The initiation step, the hydrogen atom and the propyl radical do not participate in the computation. The second method is to use heats of formation (in red). Using Hess's Law (ΔHorxn= ΔHfoproducts - ΔHforeactants) gives the same result, give or take a little. However, the reaction will be exothermic in the reverse direction given the right conditions. Sorry not to have better news.


As you have figured out, the zinc electrode is undergoing oxidation since zinc is a stronger reducing agent than copper.

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

Thus, the zinc electrode becomes thinner as shown in your image. It is oxidizing into soluble $\ce{Zn^2+}$ ions and thus its mass decreases.

In the right solution, $\ce{Cu^2+}$ ions are being reduced to metallic copper which deposits on the copper cathode

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

The $\ce{Cu^2+}$ ions are already there to begin with. The copper electrode isn't dissolving to form $\ce{Cu^2+}$ ions to begin with (the copper electrode does not react with the $\ce{Cu(NO3)2}$ solution), so there isn't any loss in mass of the copper electrode initially. Rather, during the operation of the cell, the reduction of copper(II) ions forms metallic copper, which causes the copper electrode to increase in mass and gets thicker, as seen in the image.


m-CPBA Selectivity between Baeyer Villiger Oxidation and Epoxidation?

The answer appears to be "it depends" (perhaps this should not surprise an experienced organic chemist).

According to the abstract of the paper by Mendelovici and Glotter [1] in steroid chemistry:

  Treatment of 3β- and 3α-hydroxy-(acetoxy-)cholest-4-en-6-one and of 6β-and 6α-hydroxy-(acetoxy-)cholest-4-en-3-one with MCPBA gives two types of product, depending on the initial site of the peroxy acid attack. Attack at the carbonyl group gives a Baeyer–Villiger rearrangement leading first to enol lactones and then by epoxidation of the latter to epoxy lactones. Alternatively, attack at the double bond gives epoxy ketones which can subsequently undergo a Baeyer–Villiger rearrangement leading to epoxy lactones. With one exception (3α-hydroxycholest-4-en-6-one), the Baeyer–Villiger oxidation of the enone is the dominant process. Epoxidation of the double bond is suppressed in the presence of an axial 3α- or 6β-acetoxy group.

However, Kazmeier and Horn [2] working with unsaturated aldehydes find

  Purified m CPBA is a useful reagent for the oxidation of several classes of aldehyde. Although linear unbranched aliphatic aldehydes are oxidized to the corresponding carboxylic acids, α‐branched ones undergo Baeyer–Villiger oxidation to formates. α‐Branched α,β‐unsaturated aldehydes provide enolformates and/or epoxides, which can be saponified to α‐hydroxy ketones with shortening of the carbon chain by 1 carbon. Unbranched α,β‐unsaturated aldehydes undergo an interesting Baeyer–Villiger oxidation/epoxidation/formate migration/BV oxidation cascade, which results in formyl‐protected hydrates with an overall loss of two carbon atoms.

Overall it appears that both processes can occur so Murphy's Law applies. It would be a good idea to explore alternative epoxidation conditions for an unsaturated ketone.

References

Mendelovici, M.; Glotter, E. Epoxidation and Baeyer–Villiger Oxidation of γ-Hydroxy-Αβ-Unsaturated Ketones on Exposure to m-Chloroperbenzoic Acid. J. Chem. Soc., Perkin Trans. 1 1992, No. 13, 1735–1740. DOI: 10.1039/P19920001735.
Horn, A.; Kazmaier, U. Purified MCPBA, a Useful Reagent for the Oxidation of Aldehydes. European Journal of Organic Chemistry 2018, 2018 (20–21), 2531–2536. DOI: 10.1002/ejoc.201701645.


Will the electrolyte surface become charged if one electrode is raised out of solution?

But, what about the air-water interface? Would the electric field induce a surface charge?

Yes.  Basically, by locating one of your electrodes out of the solution, you've assembled an air-gap capacitor, though a fairly inefficient/low-quality one.

Per that above link, typical metal-plate "[a]ir capacitors have a small capacitance which usually lies between $\pu{100pF}$ and $\pu{1nF}$". High-level, the definition of capacitance is:

$$
C = {q\over V} \quad\quad\Longrightarrow\quad\quad q = CV
$$

Assuming that this water/electrode air capacitor has similar properties to the above-referenced metal-plate air-gap capacitor, the charge accumulated at the electrolyte surface under a $\pu{24V}$ applied potential should be between $2.4$ and $\pu{24nC}$. A similar amount of charge, but of opposite sign, should of course accumulate at the surface of the elevated electrode.

  I don't expect anything to happen at the immersed electrode interface because the solution should remain equipotential.

This will be true at steady-state, as the entirety of the potential drop is going to be supported by the air gap.

However, you will have some minuscule amount of transient electrochemical activity at your submerged electrode right when you turn on your power supply, as the system adjusts from the $\pu{0V}$ to the $\pu{24V}$ applied voltage.


Hazards of Keeping Freshwater Fish in a Container that was Used for Adipic Acid

Adipic acid occurs naturally in beets and sugarcane, common foods, so it is not likely to be very toxic to fish if sufficiently diluted by washing. As @DGS and @ghost state, normal cleaning of the container followed by adjusting the pH for the fish should be OK. If you need to use an electric in-tank heater, though, check that it is safe to use in a plastic container.


Reaction mechanism in fluoride adsorption to aluminum oxide

Besides the presence of Al and O atoms on an Aluminum Oxide surface, there are also H atoms, which are bounded to the solid by OH bonds, - O --- H+ . Direct evidence for the presence of H atoms, and OH bonds, come from XPS or FTIR (grazing incidence) data. Indirect evidence comes from several sources, but especially from the reactivity of the surface to some compounds, like Silanes, where Si --- O --- R groups react readily with OH. Based on what is said above, I´d say the best explanation for the F– anion interaction with the Aluminum Oxide surface is a chemisorption process, through an indirect process,  bonding of the F– anion to the  H+ of the surface, forming an -O --- H+ --- F– , anion (F–) – to – dipole (+H  --- O– ) complex structure. A real solid – state Hydrogen bonding.


The answer to your question is no in general. Because amorphous metal is usually an alloy rather than a pure metal. These alloys are made of atoms of largely different sizes, which results in low free volume and higher viscosity (compared with other metal and alloys) in molten state. The viscosity hinders the atoms movement to form a crystal.

So, the significantly different atomic radii of the components that constitute the amorphous metal (about 12%) favors the high packing density and low free volume. This leads to a complex crystal units with higher potential energy and lower chance of formation.

Accordingly, I think that even if we heat an amorphous metal above a threshold temperature  and let it cool slowly, we won't have a conventional crystalline structure. At best, we get regions of some order.


How to delete everything but a specific ligand from a PDB file using Linux-compatible free software?

What about a straightforward

grep -w " AAA " file.pdb > ligand.pdb

? Where you substitute AAA with the name of the residue you want.


Consider a carbonyl group. We know that, under acidic conditions, $\ce{H+}$ can protonate on carbonyl $\ce{O}$ and hence activate the carbonyl group to nucleophilic attack at $\ce{C}$, and we call such an effect an activating effect. The metal ions $\ce{Li^+}$ and $\ce{Na^+}$ serve the same purpose as $\ce{H+}$, though $\ce{Li^+}$ is a stronger activator due to its high charge density and hence stronger attraction to $\ce{O}$.


How do I make 3d molecular graphics similar to those shown on Wikipedia?

You can generate the geometry for your molecule of choice by drawing it in Avogadro. More often than not, you want to refine the geometry through a force field calculation.

Avogadro supports multiple ways of visualization, which all can be configured according to your preferences. A ball-and-stick model, which you seem to prefer, is included.

If you want to generate an animation from the input file, QuteMol might be one option.

The interactive visualization on the wikipedia page of benzene is generated using JSmol, which is a JavaScript implementation of Jmol.

If you are only interested in online tools, MolView could be an option.


Usually, nucleophilic substitution reactions of anionic nucleophiles occur more rapidly in polar aprotic solvents. In such solvents (e.g. DMSO) the typical relative reactivity order is
$$ \ce{SCN-} < \ce{I-} < \ce{Br-} < \ce{Cl-} < \ce{F-}$$

In protic hydrogen-bonding solvents (e.g. ethanol), anions are solvated. Therefore, the reactivity of nucleophiles is decreased. Since hard nucleophiles are more strongly solvated than soft nucleophiles, the relative reactivity of soft anions is increased in such solvents. The typical relative reactivity order is
$$ \ce{F-} < \ce{Cl-} < \ce{Br-} < \ce{I-} < \ce{SCN-}$$

In the given example, $\ce{-Cl}$ shall be replaced by $\ce{-I}$; and in alcohols, the nucleophilicity of $\ce{I-}$ is higher than the nucleophilicity of $\ce{Cl-}$.


How can we separate carbon monoxide from wood gas?

This master's thesis by Dahiru Rufai Ahmed at the university of Oulu provides a comprehensive background into the removal of CO2 from wood gas.

On pages 36-37 it mentions the water-gas shift (WGS) reaction as a method of adjusting the hydrogen/carbon monoxide mixture or removing CO entirely to get pure H2.  "The WGS reaction is an exothermic reaction
(∆Hr = -41.1 kJ/mol) that converts excess carbon monoxide into hydrogen."

CO + H2O -->  CO2 + H2 + heat

"At low temperature the reaction can be catalyzed by catalysts based on copper, aluminum and zinc oxides. (Basu, P. (2010). Biomass gasification and pyrolysis: Practical Design and Theory. 1st ed., Elsevier Inc. 365pp. ISBN: 9780123749888.)

However, in recent times, interest has been shifted to a catalyst-membrane system that incorporates the WGS reaction and H2 separation via a hydrogen selective membrane (Mendes, D., Chibante, V., Zheng, J-M., Tosti, S., Borgognoni, F., Memdes, A. and Madeira, L. M. (2010). Enhancing the production of hydrogen via water-gas shift reaction using Pd-based membrane reactors. Inter J Hydrogen Energy. 35 (22), 12596–12608.)."

This article describes a membrane based catalyst for the WGS reaction.  The catalyst in the membrane uses Palladium alloys with Silver and Copper due to their higher selectivity and permeability to Hydrogen.

Since Palladium is an expensive noble metal and since catalytic converters for automobiles already use Palladium for CO oxidation, I am considering using an automobile catalytic converter (CAT) for this purpose.  The CAT uses oxidation rather than WGS to convert CO, so I am concerned that it would also oxidize the hydrogen.  However, I think the CAT would perform WGS if the input gas mixture is lacking oxygen which would be the case for wood gas.  Any useful comments and feedback appreciated.


No, it's not possible to have a negative activation energy in a simple reaction such as an isomerisation because there is no possible way to draw to potential energy curves to give a negative activation energy. The best you can do is to be activationless, and is commonly observed in electron transfer reactions (see for example J. Am. Chem. Soc. 2002, 124 (19), 5518–5527).

Experimentally, however, negative activation energies can be observed when the reaction comprises several steps some of which are not identified at the time of doing the experiment. For example if each reaction follows an Arrhenius type rate constant, the first may be $k_a = A_a\exp(-E_a/RT)$ and the second $k_b=A_b\exp(-E_b/RT)$, but because there are, say, equilibria involved in the mechanism, what may be observed without realising it is that the rate is a ratio of the two individual rates:

$$\begin{align}
k_\mathrm{obs} &\propto \frac{k_a}{k_b} \\
&\propto \frac{\exp(-E_a/RT)}{\exp(-E_b/RT)} \\
&= \exp\left(\frac{-(E_a-E_b)}{RT}\right) \\
&= \exp\left(-\frac{E_\mathrm{obs}}{RT}\right)
\end{align}$$

so it is possible for $E_\mathrm{obs} = E_a - E_b$ to be positive or negative.


Energy of electrons in the same subshell

It's important to remember (or to learn, if it hasn't been explained before) that the only atomic systems for which we can resolve the Schrödiger equation exactly are what we call hydrogenoid atoms - atoms with only one electron, such as $\ce{H}$, $\ce{He+}$, $\ce{Li^2+}$...

As soon as you add a second electron, atomic orbitals change and they aren't exactly the same as for the hydrogen atom (this is a form of the three-body problem that also comes up in other fields of physics; if you remember your Newtonian mechanics, that's why you cannot solve analytically a three-body gravitational problem). The source of this change is electron-electron interaction.

However, electron-electron interaction, in general, introduces only minor distortions in atomic orbitals - for the most part, the orbitals of poly-electronic atoms are very very similar to hydrogenoid atoms in terms of shape, size, etc, which is why we use the same labels for them as for hydrogenoid atoms ($\mathrm{1s}$, $\mathrm{2s}$, $\mathrm{2p}$, etc).

There is one property of hydrogenoid atoms that is very importantly changed by electron-electron interactions, though: in hydrogenoid atoms, the energy of an orbital is only a function of its main quantum number, $\mathrm{n}$, which means that all the orbitals of the same shell will have the same energy. In a single-electron atom, an electron in a $\mathrm{2s}$ orbital will have exactly the same energy as an electron in a $\mathrm{2p}$ orbital, because they only interact with the nucleus through a Coulombic interaction - which has spherical symmetry (it's the same in all the directions of space; it only changes with distance to the nucleus).

If there are other electrons in the same atom, occupying different orbitals, this breaks down the energy equivalence of same-shell orbitals - because, for instance, an electron in a $\mathrm{1s}$ orbital interacts differently with an electron in a $\mathrm{2s}$ orbital or in a $\mathrm{2p}$ orbital - due to symmetry, overlap, and a number of factors (these interactions are very complex, can't be solved analytically, and to some extent aren't even completely understood). The energy corrections are generally small, but the orbitals belonging to the same shell are no longer equal in energy - that's why we can order subshells, and we know that $\mathrm{2s}$ orbitals fill up before $\mathrm{2p}$ orbitals.


In E1 reactions, does carbocation stability have impact on the final product?

The answer above only shows one of the two possible isomers that could be formed by elimination from the tertiary carbocation intermediate.

I suspect that both isomeric cyclohexenes with tri-substituted double bonds would probably be produced in roughly equal amounts, along with a smaller amount of the exo-methylene compound.

Edit:  Thanks to Bon and Yolo123 for their comments

Elimination towards the isopropyl group is also a possibility, but two factors argue against this being the major elimination pathway.  First there is a statistical factor;  there is only proton on the central isopropyl carbon available for elimination, whereas the are 4 (2+2) protons available alpha to the carbocation center in the 6-membered ring.  Second, if you look at a model and consider that the proton to be eliminated must be lined up with the carbocation p-orbital you see that each methylene group adjacent to the carbocation has one hydrogen well lined up for elimination, whereas to properly line up the isopropyl proton results in some adverse steric interactions between the isopropyl methyl hydrogens and the ring methylene hydrogens.

My guess is that the gem-dimethyl group in the cyclohexane ring will not favor one in-ring elimination pathway over the other - my model doesn't show them exerting any adverse steric effect one way or the other.


This article on Wikipedia says your compound, ethyl sulphate, can be synthesized, by adding sulfuric acid drop-wise to boiling ethanol and temperature should be kept below 413K which agrees with your data. I found also an article according to which authors conducted esterification at about 300K, but with starting presence of anhydrous sodium sulfate.

Also proportion of reagents should be right, if you add more than equimolar amount of acid, removal of leftover is problematic, as sulfuric acid has high boiling point and there's no suitable selective extractant.


Decreasing order of electron density in benzene ring among the halobenzenes

Firstly, I would like to talk about the order of withdrawing effect among halogens. Since all the four halogens are electron-rich, they tend to donate electrons (+M effect). However, the high electronegativity of these elements hinders the donation by applying an even greater electron-withdrawing effect (-I effect) that overcomes the +M (resonance) effect.
In the case of halogens, the -I effect varies as -$\ce{F}$ > -$\ce{Cl}$ > -$\ce{Br}$ > -$\ce{I}$ due to the decreasing electronegative character as we move from fluorine to iodine.
The +M effect also varies as -$\ce{F}$ > -$\ce{Cl}$ > -$\ce{Br}$ > -$\ce{I}$ due to the decreasing overlap between the 2p orbitals of carbon and the 2p, 3p, 4p and 5p orbitals of the halogens respectively. Since the -I effect is stronger than the +M effect as mentioned earlier, the final order of electron-withdrawing capability is
-$\ce{F}$ > -$\ce{Cl}$ > -$\ce{Br}$ > -$\ce{I}$
Equipped with this knowledge, we come back to the question of comparing the electron densities of mono halogen benzene derivatives. From the above-mentioned trend, we can infer that the order of electron density is
$\ce{Ph-F}$ < $\ce{Ph-Cl}$ < $\ce{Ph-Br}$ < $\ce{Ph-I}$
This trend is validated by the order of reactivity of these compounds towards electrophilic substitution reactions.
Hope it helps!


Is there a gauche interaction between C-3 and C-6 in cyclohexane?

Yes, you're exactly right. Not only is the axial methyl gauche with respect to the methylene unit that you point out, but there are also gauche interactions of the two methylenes that are drawn in the "bridging" positions of your Newman projections.

So, why is the chair conformation so stable?

For one, gauche interactions are really not that destabilizing. You're talking about maybe a few kJ mol-1. Realistically, the axial interactions between that methyl group and other axial hydrogens is going to be more energetically unfavorable. But this is easily solved by a ring flip, which maintains the chair conformation.

Whoever mentioned that chair cyclohexane "acts like a long carbon chain" was probably referring to the fact that the bond angles are close to 109.5 degrees, which means that it is far less strained than rings that are fewer than 5 carbons or greater than 7 carbons. However, that's true for boat cyclohexane as well, so it's not that relevant to the question at hand.

For reference, boat cyclohexane is about 25 kJ mol-1 higher in energy than chair cyclohexane, so the gauche interactions you're talking about are a very minor contribution the the energy of the molecule.


Why is the addition of a singlet carbene to an alkene stereospecific?

Here is a diagram that might help explain things.

Carbenes have two electrons that, for the moment, are not being used for bonding.  These two electrons can exist in the same orbital with their spins paired (singlet carbene) or in separate orbitals with their spins unpaired (triplet carbene).  Normal hydrocarbons exist as singlets, spins are paired in all bonds, as they must be according to the Pauli exclusion principle.

If we look first at the pi electrons in ethylene, their spins are paired since ethylene is a singlet in the ground state.  When we add a singlet carbene to ethylene, a bond is formed by using one electron from the carbene and one from ethylene.  Those two electrons are selected such that their spins will be paired in the newly-formed bond (Pauli exclusion principle).  Either 1) a second bond is formed at the same time using the two remaining spin-paired electrons (one on ethylene and one on the carbene) to form cyclopropane in a spin-conserving, concerted addition or 2) a transient (short-lived), intermediate 1,3-biradical is formed that still contains two spin-paired electrons is formed, since the spins are paired on these two remaining electrons in the biradical and since they are held in relatively close proximity they can close quickly to form a three-membered ring without violating the Pauli exclusion principle.

Let's now contrast this to the triplet carbene case.  The spins are unpaired in the triplet carbene.  When it forms a bond with one of the ethylene pi electrons, in order to obey the Pauli exclusion principle we must select our electrons so that they are paired in the newly formed bond.  This requires that the two remaining electrons in our 1,3-biradical have their spins unpaired.  Before the biradical can form the final bond and close to cyclopropane, we must wait for one of the electron spins to flip so that we have a singlet biradical.  The triplet cannot close directly since this would violate the Pauli exclusion principle.

This "waiting" for the spins to pair in the triplet biradical means that the triplet biradical has an appreciable lifetime.  Had we run our reaction using cis-2-butene instead of ethylene.  We could see the effects of biradical lifetime by observing the stereochemistry in the resultant products.  The singlet biradical, which either reacts in one concerted step or passes through a short-lived, singlet biradical would produce only cis-dimethylcyclopropane.  In the case of the triplet carbene, the triplet biradical exists long enough (waiting for one electron spin to flip) that rotation about the various carbon-carbon bonds is now possible.  As a result, we obtain both cis- and trans-dimethylcyclopropane.


How long do the bonds in a molecule hold under nominally stable conditions?

Water with its fluctuating hydrogen bond networks is a special case. If you feel for it, have a look at Water clusters: Untangling the mysteries of the liquid, one molecule at a time, but it doesn't answer your question ;-)

It is more interesting to have a look at paleochemical/geobiological findings.

A complex organic molecule, such as 2$\alpha$-methylhopane was isolated from

  2700-million-year-old shales from the Pilbara Craton, Australia
  (Source)

Have look at the website of Roger Summons at MIT and his freely available publications, such as The Exceptional Preservation of Interesting and Informative Biomolecules.

In short, bonds can last a pretty long time!


What kind of math should I expect in first year chemistry?

Let's answer three ways:

Specific expected math  Important mathematical skills will include algebra (including understanding variables and constants conceptually), interpreting graphs, and logorithms. There is very little chance you will need to know calculus in first year chemistry. If you have specific concerns, like those described in the comments, your advisor or department assistant should be able to directly address those.
First year chem philosophy  A good goal for first year chemistry is to understand fundamental chemistry concepts so that you can apply them in future courses and work. To do this, you need to know enough math that it doesn't prohibit you from understanding the concepts.
Continuing Education This one is just my opinion and I'm sure many will disagree. If you want to work in chemistry, take more math and physics than the department requires. It will give you a more thorough and complete understanding of chemistry. It will give you an edge in the future in solving interesting problems. This can be done in parallel however- the department won't expect that you're doing this.

Bonus: As many people before me have suggested, go to 6-8 classes on syllabus day even if you're only signed up and intending to take 4. Its always better to know your options and theres no risk to going and seeing whats available. In this case, you'll learn if you've got the background the professor expects for this course and can compare it with other options.


Endo rule and stereoselectivity in the Diels-Alder reaction

Do you agree with the stereocenter with the CF3/CN in the product?

No, I would have expected the other isomer, the one with the $\ce{CN}$ and $\ce{NO2}$ groups pointing down in your drawing.

Rationale:

When exo and endo isomers can form in the Diels-Alder reaction, there is often a preference for formation of the endo isomer.  Although this endo preference is often referred to as the "endo rule", it is not a "rule", just a general, and rather weak, preference for endo product formation.  In fact, in most cases both isomers are formed, just more of the endo isomer (see figure below).

The explanation most often advanced for the endo preference is "secondary orbital interactions." The following diagram illustrates the two possible transition states (TS) in the reaction of cyclopentadiene and an $\alpha,\beta$-unsaturated carbonyl compound.  One TS leads to the exo product, the other TS to endo.  Note how the endo TS has the unsaturated carbonyl bond tucked under the double bonds in the cyclopentadiene ring; the exo TS does not.  It is speculated (1) that the overlap of the carbonyl unsaturation with the diene unsaturation somehow stabilizes the endo TS making endo product preferred.  The accompanying table illustrates how weak the endo preference can be.

image source

In your top example, this endo preference for a TS with overlapping unsaturation would lead me to suspect that the product with the unsaturated nitro and cyano groups (the $\ce{CF3}$ group is not unsaturated) oriented away from the 3-carbon bridge and tucked towards the double bond would be preferred.

Also, keep in mind that the endo preference applies to cases were the Diels-Alder reaction is run under kinetic control.  As the following example shows, the exo product is generally thermodynamically preferred (less steric crowding). So if the reaction is run under conditions (high temperature) where the initially formed kinetic (endo) product reverts to starting material, eventually the thermodynamically favored product (exo) will be formed eliminating the kinetic preference for the endo product.

(1) see here for an alternate explanation for the endo preference


Adding more solid to a solid/gas equilibrium

The concentrations of the solids change only negligibly with temperature or other reaction conditions and so are involved in the equilibrium only as constants. The amount of solids present does not change the concentration of each solid in its crystal structure. Therefore the equilibrium of the reaction is written as:

$\mathrm{K = constant}\times \ce{[CO2]}$

If one of the reactants ($\ce{CaCO3, CaO, CO2}$) is not present, there is no equilibrium.


Why is CuSO4 used as electrolyte while purifying copper?

During copper electrorefining, impure $\ce{Cu}$ from the anode is oxidized and dissolved, while $\ce{Cu^2+}$ is reduced at the cathode, forming the deposit of refined copper. Because anodic impurities which are less noble than copper – most notably $\ce{Ni}$, $\ce{As}$ and $\ce{Sb}$ – also dissolve in the electrolyte, the cell voltage is kept in the range of 0.2-0.4 V to avoid co-deposition of these metals at the cathode.

Copper dissolution is the main oxidation half-reaction at the anode, and the oxidation of anions from the electrolyte needs to be minimized, since it is an unwanted side reaction which reduces current efficiency of the process. In a $\ce{CuSO4}$ electrolyte, this would be oxidation of $\ce{OH-}$ (from water dissociation, so concentration will be low) to $\ce{O2}$, while in a $\ce{CuCl2}$ solution, $\ce{Cl-}$ will be oxidized to chlorine gas, which would present a greater hazard because of its toxicity.

The electric conductivity of the electrolyte is another important factor, which varies with different copper salts and their concentrations. $\ce{CuCl2}$ and $\ce{CuSO4}$ are both readily soluble in water; however, the relative molar conductivities of these solutions will be different. In this study, a copper chloride electrolyte proved to be inferior to copper sulfate and copper nitrate in terms of deposit quality and stability of cell potential.


How does different diffraction grating effect resolution and sharpness of lines in a line spectrum?

Sharpness is colloquial for how narrow a peak is. Resolution is how well we can see two features as they move closer together. You might do well to re-read your book to learn these definitions better. As already noted, skinnier and more clear have no meaning in spectroscopy.

A search on terms such as diffraction grating equation will give you further reading to support or invalidate the guess (hypothesis) that you propose.


Solubility of N-[(4-hydroxy-3-methoxyphenyl)methyl]acetamide at different pH

I would agree that it is most soluble in NaOH.

However, the molecule will only deprotonate the OH proton not the NH proton.  I wouldn't give your explanation much, if any, credit because you aren't saying which proton would deprotonate, and you are implying the molecule will lose all protons.

Also, at low enough pH the N will protonate, i.e. it would gain a second proton.


Does milk drinking prevent long-term chemical poisoning?

This post only concerns chronic poisoning, i.e. exposure to a harmful chemical in low amounts for an extended period of time. The answer is not applicable to acute poisoning; in that case a doctor should immediately be called.

Basically there are three types of chemicals that have a harmfulness threshold to which one can be exposed over extended periods of time.

Type one includes substances such as mercury that are primarily aggregated in the body rather than excreted. In rare cases, drinking milk may enhance excretion thereby decontaminating the body. I haven’t heard of any such examples, though, and I would assume it to be a very rare case so neglegible.

Type two includes those substances that the body can actually metabolise to harmless products. This would include most protein poisons. However, the metabolism shouldn’t be enhanced by drinking milk so it will likely have a non-effect.

Type three includes those that, while they aren’t metabolised, just don’t do enough damage at low concentrations. That would basically include your list of acidic and basic substances but also non-aggregating metals at low concentrations. For these, excretion is not significantly enhanced or reduced by drinking milk so again no effect is expected.

Comparing the acute exposure to acids to the chronic one, in the former giving a person milk to drink is actively encouraged. This is because milk contains a lot of proteins that can act as buffers, increasing the body’s buffering potential and reducing the effects of the acid. The body does have more than enough buffering capabilities to cope with chronic cases though, so even there regularly drinking milk would not help.

Tl;dr: Milk is a great drink but cannot act as a type of preventive antidote.


Orthogonality of wavefunctions for different normal modes

given the above, how can one show that vibrational wavefunctions
  corresponding to different normal modes are orthogonal?

The Hamiltonian for a system with two (or more) normal modes can be written as a tensor product of (Hermitian) Hamiltonians for each normal mode. For example for two modes (stretching and bending):

$$
H_{\rm{total}} = H_{\rm{stretching}}\otimes H_{\rm{bending}}
$$

The tensor product of two Hermitian matrices is also Hermitian.
Therefore the wavefunctions are orthogonal for the same reason that you figured out on your own for the single mode case.

More generally, all Hamiltonians (Hermitian operators) have orthogonal wavefunctions (eigenfunctions) for distinct energy levels (eigenvalues). So wavefunctions corresponding to different energy levels are orthogonal as a consequence of the Schroedinger equation which is based on a Hermitian object called the Hamiltonian, and is the basis for the existence of all distinct energy states (vibrational or not) in quantum chemistry.


Why does rice become sticky after cooking?

Starch is what makes rice sticky. Starch surfaces, when dry, are closely packed together and they do not stick to each other. If they get moist, especially if left to soak, or boiled, the starch will "unwind" as it absorbs water and become sticky. Depending on the source of starch and the type, you can have cleavage of starch to sugar and it will become even more sticky. You can read this pdf for more information

Funfact, you can make your own glue with flour and water as an example.

I do not take responsibility for any mess or any unwanted results if for instance you decide to glue the broken baby seat with home made glue. =)


Hyperconjugation does play a role in your example.  Acetic acid ($\mathrm{p}K_\mathrm{a}=4.75$) is a weaker acid than formic acid ($\mathrm{p}K_\mathrm{a}=3.75$).  To understand this we must consider the relative stabilities of both the ionized and unionized acids.

In acetic acid there is a partial positive charge on the carbonyl carbon.  Both the inductive effect and resonance effect (e.g. hyperconjugated resonance structures like B) of the methyl group act to stabilize this charge, thereby stabilizing the unionized form of acetic acid.  In formic acid, without a methyl group, similar stabilization of its unionized form is not possible.

Now let's look at the ionized acids.  The negative charge in the anion is delocalized as shown in resonance structure C. This negative charge on the two oxygens makes them much less electron withdrawing.  Consequently, there is much less positive charge on the "carbonyl" carbon in the ionized form.  In turn, there is less need for stabilization of positive charge by the methyl group.  The methyl group does not stabilize the ionized form as much as it stabilizes the unionized form.

Since the methyl group stabilizes the unionized form of acetic acid more than a hydrogen and since the methyl doesn't stabilize the ionized form much more than a hydrogen, there is a higher energy barrier for acetic acid to ionize compared to formic acid.  This means that acetic acid will be a weaker acid than formic acid.


“[M]ass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity.” (Wikipedia quoting [de Silva, G.M.S. (2002), Basic Metrology for ISO 9000 Certification, Butterworth-Heinemann, 214p.]
Well-known relationship:
$$
W = m \times g
$$
where W is weight, m is mass and g is the acceleration of gravity.
Because g is a constant that cancels out in a ratio, mass/mass and weight/weight are identical.
Likewise, the ratios % m/m and % W/W are identical.


This is a Steam explosion generally the flame is due to the hydrogen produced when water is superheated so thermal decomposed into $\ce{H2}$ and $\ce{O2}$.

Quoting wikipedia:

  Steam explosions are not normally chemical explosions, although a
  number of substances react chemically with steam (for example,
  zirconium and superheated graphite react with steam and air
  respectively to give off hydrogen, which burns violently in air) so
  that chemical explosions and fires may follow.


Will there be change in electronegativity difference in C-O and C=O?

I think that you didn't understand the answer to this question.

A dipole is created when two equal (only in magnitude) and opposite charges are separated by some distance. The dipole moment depends on the magnitude of the charges and the distance between them. Electronegativity is just a concept that helps us to get an idea of the magnitude of the charges. So, in order to find the dipole moment, you should focus on the magnitude of the two charges rather than the electronegativity difference.

Look at the hybridization of the atoms in the single and double bond. That will tell you that both the atoms should be more electronegative after forming a double bond, but won't help you comment on the electronegativity difference.


Cobalt electroplating over nickel-plated steel

in the past cobalt was used as a brighterner for Nickel plating electrolyites, now since Cobalt is about 5 times more expensive than Nickel, is hardly used but you can add about 30 grams per litter of CoSO4 to your regular nickel plating solution and you will end up with a Nickel/Cobalt alloy that is stronger and whiter than plain Nickel, also keep in mind that in these type of baths Cobalt plates out faster than Nickel, that means that a 10% Cobalt Sulfate(90% Nickel Sulfate) plating bath will produce about a 70% Nickel/ 30% Cobalt alloy plating.


How to calculate molecular dipole moment from a known wavefunction?

The necessary formal derivation has already been nicely done by AngusTheMan. I'll start from the last equation:

$$
\langle \mu_{z} \rangle = \langle \Psi | \hat{\mu}_{z} | \Psi \rangle
$$

where $\Psi$ is the variational wavefunction; it can be any molecular state. It's important that it's variational, otherwise the expectation value approach is not exact. So, this works for SCF, CI, and MCSCF wavefunctions, but extra derivatives need to be taken for Moller-Plesset and coupled cluster wavefunctions. More work needs to be done for multideterminental wavefunctions like CI and MCSCF, but the complexity is no different for a single state in each wavefunction. There may be some MO space partitioning I'm neglecting that's required for MCSCF, so I'll restrict my work to a single-determinental wavefunction.

Expand the wavefunction as a linear combination of molecular orbitals (MOs)

$$
\Psi = \sum_{i} \psi_{i},
$$

where each molecular orbital is a linear combination of atomic orbitals (AOs)

$$
\psi_{i} = \sum_{\mu} C_{\mu i} \phi_{\mu},
$$

where $C_{\mu i}$ is the MO coefficient matrix, so our expectation value now looks like this:

$$
\langle \mu_{z} \rangle = \sum_{i}^{\textrm{occ MOs}} \sum_{\mu\nu}^{\textrm{AOs}} C_{\mu i} C_{\nu i} \langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle.
$$

The indices $\mu,\nu$ run over all AOs, and the index $i$ runs over the occupied MOs. There's only one index because this is a one-electron operator. I'm also neglecting any complex values here, since we almost always work with real-valued AOs and MO coefficients.

We do one last rearrangement. Replace the MO coefficients with the density matrix

$$
P_{\mu\nu} = \sum_{i}^{\textrm{occ MOs}} C_{\mu i} C_{\nu i}
$$

to give the first explicit "working equation":

$$
\langle \mu_{z} \rangle = \sum_{\mu\nu}^{\textrm{AOs}} P_{\mu\nu} \langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle
$$

I say first for two reasons: we usually try and avoid explicit loops like this, and the expression can be broken down further, depending on what molecular properties are of interest; I'll be more clear about this later. Inside the sum there are two terms:

The density matrix $P_{\mu\nu}$, which comes from a converged SCF calculation.
The integral of the dipole operator over two basis functions, $\langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle$. Atomic orbitals are represented as atom-centered basis functions. These can be calculated once and at any time, since the quantities here don't change over the course of a calculation.

Each of these terms is represented as a matrix. Since the index $\mu$ runs along the rows and $\nu$ runs along the columns for each matrix, "contraction" involves either a matrix product followed by the trace, or an elementwise product followed by an accumulation sum over all matrix elements. There are other details one needs to be careful about, such as what units the result should be in, which changes the prefactor (programs work internally in atomic units), and what the origin for the dipole operator is, but that's really it.

Well, sort of. I'm actually treating some of the program internals as a black box. If you're familiar with Hartree-Fock, it should be clear where $P_{\mu\nu}$ comes from, but what about the integral? For a general expectation value $\langle A \rangle$ with its corresponding operator, where does $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ come from? If it's already available in the code, then you call a wrapper function that then calls the integral engine to do all the nasty work, and you get back a tidy matrix without having to worry about the details. If $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ isn't present, depending on the complexity of $\hat{A}$, there can be a non-trivial amount of derivation required for the working integral equation, followed by the implementation.

Ignoring any possible contraction of primitive basis functions, expand $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ using the definition of $\phi$ in Cartesian coordinates:

$$
\phi(\mathbf{r}; \mathbf{A}, \mathbf{a}, \zeta) = (x-A_x)^{a_x} (y-A_y)^{a_y} (z-A_z)^{a_z} e^{-\zeta |\mathbf{r} - \mathbf{A}|^2}
$$

where $\mathbf{r} = (x, y, z)$ is the electron position, $\mathbf{A} = (A_x, A_y, A_z)$ is the position of the basis function (almost always atom-centered), and $\mathbf{a} = (a_x, a_y, a_z)$ are the angular momenta for each coordinate, with $l_{\textrm{max}} = a_x + a_y + a_z$ total angular momentum of the basis function. $(0,0,0)$ is an s-function, $(1,1,0)$ and $(0,0,2)$ are d-functions, and so on.

Forming the integral more explicitly gives

$$
\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle = \int\int d\mathbf{r}_1 d\mathbf{r}_2 \left[ (x_1-A_x)^{a_x} (y_1-A_y)^{a_y} (z_1-A_z)^{a_z} e^{-\zeta_a |\mathbf{r}_1 - \mathbf{A}|^2} \right] \\ \times\left[ \hat{A} \right]\left[ (x_2-B_x)^{b_x} (y_2-B_y)^{b_y} (z_2-B_z)^{b_z} e^{-\zeta_b |\mathbf{r}_2 - \mathbf{B}|^2} \right]
$$

Before going any further, $\hat{A}$ must be defined. If $\hat{A} = 1$, this becomes an overlap integral. The dipole operator in the z-direction is given by $\hat{A} = \hat{\mu}_{z} = -ez = -e(z_3 - C_z)$, where $z_3$ is the integration coordinate and $C_z$ is the origin of the dipole in the z-direction, usually taken to be zero. Everything is kept in atomic units until after the integral/density contraction, so drop the prefactor $-e$. We can now generalize this to an arbitrary Cartesian multipole moment operator,

$$
\hat{A} = \mathfrak{M}(\mathbf{r}_3) = (x_3 - C_x)^{c_x} (y_3 - C_y)^{c_y} (z_3 - C_z)^{c_z}
$$

where $(c_x, c_y, c_z)$ determine the coordinate of each multipole, and their sum is the total multipole order; for example, $(1,0,0), (0,1,0), (0,0,1)$ are the x-, y-, and z-directions of the dipole operator. The Cartesian moment operator looks just like a Gaussian basis function where $\zeta = 0$.

Once the form of an operator has been derived, it needs to be implemented as part of an integral package, each of which implement one or more algorithms for computing integrals. Each algorithm is named after the authors of the paper in which they were introduced, and are usually abbreviated. For example, the first one I know of is the Taketa, Huzinaga, O-Ohata paper (THO, DOI: 10.1143/JPSJ.21.2313), where explicit working equations are given for 2-center overlap, 2-center kinetic energy, 2-center electron-nuclear attraction, and 4-center electron repulsion integrals. A working implementation can be found in the PyQuante package. I made an IPython notebook translation of the code snippets on the front page of the official documentation.

Other, more complex algorithms are from the Pople-Hehre (PH), McMurchie-Davidson (MD, DOI: 10.1016/0021-9991(78)90092-X), Obara-Saika (OS, DOI: 10.1063/1.450106), Dupuis-Rys-King (DRK), and Head-Gordon-Pople (HGP) papers. I'm sure I'm neglecting some, including the seminal paper by Boys which introduced the use of Gaussian functions as a substitute for Slater-type functions in basis sets. A good review of these algorithms is found in a Peter Gill paper (DOI: 10.1016/S0065-3276(08)60019-2); he is the original author of the integral code in both Gaussian and Q-Chem.

To bring these things full circle, I wrote some code a few months ago to calculate the dipole moment using pyquante2 and a wrapper that calls an implementation of the Obara-Saika recursive integral algorithm. You can find it here with some comparisons to "industrial" quantum programs.


The fictitious substance “ammonium hydroxide” $(\ce{NH4OH})$ was suggested in accordance with the Arrhenius theory. In 1882–1883, Svante Arrhenius had researched into the conductivity of electrolytes. His theory of electrolytic dissociation was discussed in his dissertation in 1884 and published in refined form in 1887. In 1903, the Nobel Prize in Chemistry was awarded to Svante Arrhenius “in recognition of the extraordinary services he has rendered to the advancement of chemistry by his electrolytic theory of dissociation”.

According to the Arrhenius definition, acids are compounds that dissociate and release hydrogen ions $(\ce{H+})$ into the solution:

$$\ce{HCl -> H+ + Cl-}$$

Bases are defined as compounds that dissociate and release hydroxide ions $(\ce{OH-})$ into the solution:

$$\ce{NaOH -> Na+ + OH-}$$

The need for a hydroxide group in bases according to the Arrhenius definition led to the proposal of $\ce{NH4OH}$ (i.e. “ammonium hydroxide”) as the formula for hydrated ammonia in water. (Note that this problem does not exist in the definition of bases according to the Brønsted–Lowry theory, which was proposed in 1923.)

However, ammonium hydroxide cannot be isolated; the fictitious solid compound does not exist. Nevertheless, the misleading traditional name “ammonium hydroxide” is still widely used for solutions of ammonia in water.

Actually, a solid compound (melting point: −79 °C) with the correct atomic composition (molecular formula) can be crystallized from aqueous solutions of ammonia. But even this compound is not ammonium hydroxide $(\ce{NH4OH})$. It actually is a real hydrate $(\ce{NH3.H2O})$ in which the molecules are linked by hydrogen bonding.


The enantiomers 1a and 1b can have their double bonds defined by CIP rules 1a and 5. Rule 1a dictates that CH3>H while Rule 5 has R>S. Thus, 1a has an E-double bond  and 1b is of the Z-configuration as exemplified by the red bonds. This issue has been addressed previously on this site.

Addendum: While the enantiomers 1a and 1b (vide supra) have their double bonds labeled in upper case E/Z (ChemDraw), JSmol takes into account the enantiomorphic nature of the double bonds by using lower case e/z (vide infra). See the answer of @orthocresol.


What is the easiest way to concentrate a solution without causing it to hydrolyze?

Try adding Na2SO4 to the mix and freeze forming Na2SO4.7H2O. Slow, but at least, no oxygen-related chemistry, discussed below.

As a sidebar, you likely have formed basic ferric acetate from boiling, which adds oxygen, to your acidic (vinegar) and aqueous ferrous acetate mix.

Likely reaction mechanics are complex, see, for example, Table S1 here . Per the source my take of a possible reaction sequence:

$\ce{ O2 (aq) + Fe^{2+} → Fe^{3+} + .O2− }$ ( moderately fast) )

$\ce{ . O2− + Fe^{2+} + 2 H+ → Fe^{3+} + H2O2}$  (fast )

$\ce{ H2O2 + Fe^{2+} → Fe^{3+}  + .OH + OH− }$ ( slow)

$\ce{  .OH + Fe^{2+} → Fe^{3+} + + OH− }$ ( fast )

Net radical reaction:

$\ce{ O2 + 4 Fe^{2+} + 2 H+ → 4 Fe^{3+} + 2 OH-}$

The creation of FeOOH is also possible in less acidic conditions per his source: 'Air Oxidation of Ferrous Iron in Water'.


In this simple case it is pretty obvious that a hydride ion is a much better base than it is a nucleophile. But that doesn’t answer everything; we could consider cases with much better nucleophiles ($\ce{Br-}$).

Another reason is that hydroxide is an even lousier leaving group than hydride is a nucleophile. It is literally impossible to do $\mathrm{S_N}$ displacements of uncharged oxygen atoms unless you are opening an epoxide (in which case you lose ring strain making the overall process favourable. It does not matter whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$; $\ce{RO-}$ will never leave. To remove an oxygen atom, you always need to use strong acids that can protonate it, giving the far better leaving group $\ce{ROH+-R}$ which gives the free alcohol/water when displaced. Thus, the following reaction works well due to the very strong hydroiodic acid (I tried it):

$$\ce{R-CH2-OH + HI -> R-CH2-OH2+ + I- -> R-CH2-I + H2O}$$


Removing titanium from tantalum crucible

Straight hydrochloric acid- Say 20% is virtually  inert to Tantalum- but significantly corrosive to titanium. If you heat it to boiling point of hydrochloric- it will eat through the titanium.

http://www.parrinst.com/wp-content/uploads/downloads/2011/07/Parr_Titanium-Corrosion-Info.pdf
http://www.tantaline.com/Tantalum-Corrosion-71.aspx


Acetic Acid % in vinegar? Given average molarity of acetic acid

$1000$ mL of vinegar will weigh $1106$ g and contain $1000 · 0.017078$ g of acetic acid (= $17.078$ g). Then $17.078/1106 = 0.0154 = 1.54$% acetic acid.
The stockroom manager must be diluting the vinegar, perhaps for a specific experiment. The vinegar I buy is $5$% acetic acid (rarely $4$%).