Nomenclature problem, when would ring be parent?

You have already found some relevant rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, for the preferred IUPAC name (PIN), the ring is generally selected as the parent structure according to P-52.2.8, which also corresponds to the method described in P-44.1.2.2 (1). However, this rule applies to structures with “the same number of characteristic groups cited as the principal characteristic group”. If the ring and the chain have different numbers of substituents corresponding to the principal characteristic group, another rule comes first, which stipulates that the senior parent structure is the one that has the maximum number of characteristic groups cited as the principal characteristic group.

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

For example, for compounds like 3-propylcyclohexan-1-ol or 3-cyclohexylpropan-1-ol, the senior parent structure is given according to P-44.1.1 by the principal characteristic group, which is expressed as suffix (here: ‘-ol’). The further rules P-44.1.2.2 and P-52.2.8 are not relevant in these cases.

The compound given in the question does not have any substituents that are cited as the principal characteristic group (by means of a suffix or a class name, or implied by a trivial name).* Thus, no decision can be made using P-44.1.1. According to P-52.2.8, the ring is selecetd as the parent structure to construct the preferred IUPAC name [(but-3-yn-2-yl)oxy]cyclopentane. In general nomenclature, however, the context may favour the the chain according to P-44.1.2.2 (2), which would lead to the name 3-(cyclopentyloxy)but-1-yne.

* Note that some sources try to simplify the nomenclature rules by combining different concepts into a single one. For example, the so-called “Table of Functional Group Priorities For Nomenclature” shown on masterorganicchemistry.com tries to combine the seniority order of suffixes with prefixes and endings. This can be misleading and result in wrong names. In particular, there is no suffix ‘-ane’, ‘-ene’ or ‘-yne’ (these are endings), and multiple bonds (and all groups that follow in that table) are actually not considered in the seniority order. Please do not use such sources.


What is the significance of the 2 and 4 positions in 2,4-DNP?

Probably any DNP could work if you could get it.  But, because of the directing  character of substitutents in electrophilic substitution you inherently favor the 2,4 isomer:

the hydrazine group (or more accurately the chlorine atom which is converted to a hydrazine group) is ortho/para directing.
the nitro groups are deactivating and meta directing.

So you get the nitro groups meta to each other and ortho/para to the hydrazine group.  Therefore 2,4 DNP.


Mechanism for condensation reactions of amino acids

Here is a proposed mechanism in a 2005 paper. One thing to point out is that this is a proposed mechanism but it is all I could find:

link to pdf: https://www.pnas.org/content/pnas/102/35/12395.full.pdf
Typically, these reactions involve the nitrogen attacking the partially positive carbonyl carbon as a form of nucleophilic substitution. However, OH is a bad leaving group so often, reagents such as DCC can be used to make the OH into a good leaving group before reacting. This is why it is easier to make nylon using a diacid chloride instead of a diacid.


From the IUPAC Gold Book:

  Carbocations may be named by adding the word 'cation' to the name of the corresponding radical. Such names do not imply structure (e.g. whether three-coordinated or five-coordinated carbon atoms are present).

So since this is the dication of hexamethylbenzene, it would probably  least incorrectly be called hexamethylbenzene dication. At the very least, this seems to be how a lot of people refer to it (or more verbosely, pentagonal-pyramidal hexamethylbenzene dication).


Reactivity of Alkanes, Alkenes, and Alkynes

Maybe it will help if you look at reactivity in terms of ability to attract electrons instead of bond enthalpies. Alkanes are $sp^3$ hybridized, and hence have $25$% $s$ orbital character and $75$% p character. $s$ orbitals are closer to the nucleus and thus have a 'contracting' effect on the hybrid orbital. Greater the $s$ character, 'smaller' the hybrid orbital. This means that electrons are more closely packed. Alkanes have only 25% $s$ character, hence the hybrid orbitals are comparatively larger, and the effective nuclear charge on outermost electrons is less. Alkenes ($sp^2$) and alkynes ($sp$) have 33.3% and 50% $s$ character respectively. Thus their hybrid orbitals are SMALLER; i.e. effective nuclear charge is more. This means that it is easier to accept electrons, as now the effective strength of nucleus is more. Hence, alkynes can easily accept electrons, followed by alkenes and then alkanes. Thus, alkynes are most reactive, followed by alkenes and alkanes. Even though increase in $s$ character increases bond enthalpy, but it also increases the electron withdrawing capacity; and it is the latter which predominates.

(In fact, in general, greater the $s$ character, more the electronegativity and more the acidic nature).


Why does HCN boil at a higher temperature than NH3?

The enthalpy of vaporization of $\ce{HCN}$ is higher than for $\ce{NH3}$, which suggests that $\ce{HCN}$ molecules interact more strongly than $\ce{NH3}$ molecules. $\ce{C-H}$ bonds are not usually considered good hydrogen bond donors, but $\ce{HCN}$ is unusual. For example $\ce{HCN}$ has a $\mathrm pK_\mathrm a$ value of 9.2, indicating that the $\ce{CN}$ group is electron withdrawing and that it is a reasonably good hydrogen (bond) donor. This is likely due to the electronegativity of nitrogen, and also the high "s-content" of the sp-hybridized $\ce{CH}$ bond, which keeps the electron pair close to the nucleus.


Only d orbital electrons for spin only magnetic moment

Any unpaired electron is taken into consideration while trying to quantify spin only magnetic moment.
valence electrons can be unpaired (not always), and inner shell electrons are paired so the latter don’t contribute (since they are fully filled). That’s why usually only the outer shell electrons are considered.
and no it’s not calculated for ions only. You can use it for individual atoms too!
In your example $\ce{Cr^0}$, you can certainly consider that one unpaired electron in the s orbital,

so giving a total of 6 unpaired electron it’s magnetic moment would be $\sqrt{48}$

By using the spin-only formula $M =\sqrt{n(n+2)}$  where  $n$ is the number of unpaired electron and $M$ has the units of magnetic moment.

Caution this formula is applicable for d block elements only.

Another example would be niobium, molybdenum it also has one unpaired electron in the 5s orbital and certainly we consider that electron while quantifying  spin only magnetic moment.

Furthermore, the spin only magnetic moment of molybdenum is very close to that of chromium.

I am telling very close but not same because theoretically calculated magnetic moment don’t usually coincide with experimentally calculated they differ ever so slightly

Sources  -

NCERT Chemistry for class 12 (pg no 221)
Descriptive Inorganic Chemistry by James E. House and Kathleen E. House by Elsevier. 2nd edition (pg no 329)
Shriver and Atkins Inorganic Chemistry, 5th edition (pg no 478)


What is the reason for the anomalous electronic configuration of palladium?

The Aufbau principle states that, hypothetically, electrons orbiting one or more atoms fill the lowest available energy levels before filling higher levels (e.g., 1s before 2s). In this way, the electrons of an atom, molecule, or ion harmonize into the most stable electron configuration possible.

Electron behavior is elaborated by other principles of atomic physics, such as Hund's rule and the Pauli exclusion principle. Hund's rule asserts that even if multiple orbitals of the same energy are available, electrons fill unoccupied orbitals first, before reusing orbitals occupied by other electrons. But, according to the Pauli exclusion principle, in order for electrons to occupy the same orbital, they must have different spins.

The order in which these orbitals are filled is given by n+l rule known as Madelung Rule or diagonal rule.Orbitals with a lower (n+l) value are filled before those with higher (n+l) values where n represents principal quantum number and l represents azimuthal quantum number.

The fact that most of the ground state configurations of neutral atoms fill orbitals following this n + ℓ, n pattern was obtained experimentally, by reference to the spectroscopic characteristics of the elements.

The Madelung energy ordering rule applies only to neutral atoms in their ground state,even in some cases there are examples which show different electronic configuration than those of experimentally determined.Copper,Chromium and Palladium are common examples of this rule.

For Palladium,according to Madelung's rule, 5s^2 has lower (n+l) value i.e. (5+2=7) and 4d^8 has higher (n+l) value i.e. (4+8=12),so the 5s orbital  fills up first before 4d orbital.

Hence, Palladium shows anomalous electronic configuration.


Why is the plateau on a heating curve in equilibrium?

You are right about the equilibrium existing because both freezing and melting occur at the same rate.
Imagine ice and water kept in a thermos flask under normal atmospheric pressure. You will observe that there is no change in the mass of both the phases.
But there is actually a lot of activity going on. Some molecules of ice enter the liquid phase and some molecules of water enter the ice phase. As both these processes occur at the same rate you don't see any change in the mass of both the phases.
In fact such equilibriums exist for all pairs of solids in their own liquids and gases above their own liquids at particular temperatures.
In your case I think there is a mechanism to extract heat from the system which shifts the equilibrium towards liquid changing to solid ie freezing.


Adding hydrogens to multiple pdb files using Openbabel GUI software

If you got an installation of openbabel's GUI, you equally have an installation of openbabel for the terminal (e.g., in Linuxes) / command line cmd.exe (in Windows).  There is nothing wrong using the GUI for an edit on one file or a few data -- your instructions set are right -- however using the CLI is closer to the engine, thus often more powerful.

Enter the directory containing the .pdb to work with from the command line.
The instruction babel *.pdb -opdb *_h.pdb -h will request babel to look for all *.pdb there and write new .pdb with hydrogens added.  Each file by name in pattern of example.pdb will have its edit recorded in a file named in pattern of example_h.pdb.  While the CLI will report the progress of this conversion, I recommend to slice the task into smaller batches than just one of 9k.  How many .pdb per batch depends on the computing power available on your machine.
The new files written are both discernable by name and time stamp from the retained input.  So you have two criteria to sort and discern the files.  Assuming you subsequently moved the new ones into their own folder, you may rename them en bloc back to the pattern of example.pdb.  If you work with Windows, a program like Ant Renamer may considerably accelerate this task.

Commands and grammar openbabel uses are documented in depth.  For commands accessible from the CLI, just a few are compiled here.  As perspective: note its use may be automated (scripted), for example with Python like the portable WinPython.


Why does butanoic acid have a higher boiling point than isobutyric acid?

In general, linear chains are more easily structured and packed than branched ones.  You need more energy to disturb the more relatively packed structure.This could explain why butanoic acid has a higher boiling point than isobutyric acid, and yet they are isomers.


Yes, calcium oxide can be used as an desiccant.

The following quote is an excerpt from this site:-

  Calcium oxide is calcinated or recalcinated lime having a moisture adsorptive   capacity of not less than 28.5% by weight. The
  distinguishing feature of calcium oxide is
  that it will adsorb a much greater amount of water vapor at a very low
  relative humidity than other materials. It is most effective where a
  low critical relative humidity is necessary, and were there is a high
  concentration of water vapor present.

  Calcium oxide removes water from a package very slowly, often taking
  days to reach its maximum capacity. As calcium oxide adsorbs moisture,
  it swells. Proper desiccant packaging is required for effective use.
  For these reasons, its use has been limited to primarily the packaging
  of dehydrated foods.

This following is a table which compares the properties of various dessicants taken from this site:-

And the following graph depicts the adsorption capacity of various dessicants:-

For more information click on the following sites:-

http://www.ohe-chem.co.jp/e_qa.html
http://www.agmcontainer.com/support/selecting-desiccant
https://www.usaemergencysupply.com/information-center/self-reliance/food-storage-frequently-asked-questions/types-of-desiccants


How is a reaction with more products than reactants spontaneous?

Note that whether the forward or reverse reaction is favorable depends entirely on the relative magnitudes of Q and K (the reaction quotient versus the equilibrium constant):

$$\begin{align} Q&>K \rightarrow \text{reactants favored}  \\ K&>Q \rightarrow \text{products favored}  \end{align}$$

Next, consider that when you declare a value for $\Delta G^\circ$ you can translate that into a value for K, using the following equation:

$$\Delta G^\circ = -RT\log K$$

For instance, if $\Delta G^\circ = 0$ then $K=1$.

For the example here, this means that

$$\begin{align} Q&>1 \rightarrow \text{reactants favored}  \\ 1&>Q \rightarrow \text{products favored}  \end{align}$$

Next, consider the standard state of all substances. For solutions this is typically $\pu{c^\circ = 1 M}$. Standard states are important because the reaction quotients and equilibrium constants, when described in terms of concentrations, contain ratios of the concentrations in solution to the standard concentrations, for instance

$$Q= \frac{\prod_{\text{products i}} (c_i/c^\circ)^{\nu_i}}{\prod_{\text{reactants i}} (c_i/c^\circ)^{\nu_i}}$$

If all substances are present at the standard concentration, $Q=1$. In the present example $K=1$ so the equilibrium state consists of all substances being at standard concentrations ($c_i=\pu{1 M}$).

Finally, consider the following way to relate Q, K and $\Delta G$:

$$\Delta G = RT\log \left( \frac{Q}{K}\right)$$

If $K=1$ as in the present example then

$$\Delta G = RT\log Q $$

This just reinforces what was stated earlier for the special example presented here: when $Q<1$ the free energy difference is negative and products will be favored, and vice-versa when $Q>1$.

Inserting the definition of the reaction quotient

$$\begin{align} \Delta G &= RT\log \left( \frac{\prod_{\text{products i}} (c_i/c^\circ)^{\nu_i}}{\prod_{\text{reactants i}} (c_i/c^\circ)^{\nu_i}}\right) \\ &= RT\log \left( \frac{c_B^2c_C^2}{c_Ac^{\circ\,3}}\right) \end{align}$$

If we assume all substances are present at the same concentration c, then

$$  \Delta G= 3RT\log \left( \frac{c}{c^{\circ}}\right) $$

Clearly if $c<\pu{1 M}$ then $\Delta G<0$ and products are favored.


Disulfide batteries use a redox shuttle between disulfide bonds ($\ce{R-SS-R}$) and two thiolates ($\ce{R-S-}$), e.g.:

Often the counter-ions are $\ce{Li+}$ but this is hardly a requirement. Since two electrons can be transferred in the charging / discharging cycle, the theoretical energy density is very high.

One downside (e.g., in my past research) is that you're breaking the polymer bonds during discharge, creating oligomers and monomers that might be more soluble in the electrolyte than the original polymer. This might cause problems with cyclability, since your electroactive material might leave the electrode.


Let us address the elephant in the room first and foremost.  All of these objects do not have any symmetry if you wanted to get real technical about it.  However, let us just assume certain things for this useful exercise to be relevant.

b.) The teacup has a vertical plane of symmetry splitting the teacup in half along the handle axis and intersecting with the centroid of the cup and saucer.  I would say that your guess is correct though it would be useful to have drawn in a plane rather than simply a line (for all examples).

c.) At first glance the pliers have a couple planes of symmetry BUT upon further thought it may not due to the way the rotating JOINT is constructed.  See the example below:

While it is difficult to see in your picture, below it is made very clear.  Therefore I will say that this object does not contain a plane of symmetry.  It would be POSSIBLE that it could have a plane of symmetry, one that lies in the plane of your monitor screen, cutting the object in half from tip to handle end... but it would be impossible to tell without seeing the other side. Many times pliers come with an inscription on one side of the joint, thereby breaking this symmetry.

e.) The brick does indeed have multiple planes of symmetry.  One vertical plane bisects the brick along its width and one bisects the brick along its length.  In addition, one plane of symmetry lies in the plane of the brick (the plane the brick is lying on), cutting the brick in half (think of it as coming in through the side of the brick).

f.) The hand has no plane of symmetry that I can see at all.  The back of the hand doesn't look like its front... the left and right side of the hand are clearly not mirror images of each other...

The brick has three planes of symmetry.


What is the chemical reaction in a home-made high-bouncer ball?

I have google it and got the link which explains the process but don't show the organic reactions(might be because they are complex)

  This activity demonstrates an interesting chemical reaction, primarily between the borax and the glue. The borax acts as a “cross-linker” to the polymer molecules in the glue – basically it creates chains of molecules that stay together when you pick them up. The cornstarch helps to bind the molecules together so that they hold their shape better. - See more at: Sciencebob


Why is iron pentacarbonyl known, but iron(III) hexacarbonyl is not?

There are several things which are needed to be known to be able to deal with this question.

Firstly there is the 18 Valence Electron (18VE) rule (description at libretexts.org). The iron pentacarbonyl is an 18VE complex. It is coordinatively saturated. A low valent transition metal complex with strong ligand field ligands is considered to be coordinatively saturated if the metal has 18 valence electrons (noble gas electron count).

An iron atom has eight valence electrons, each carbon monoxide (carbonyl) ligand adds two electrons to the valence count. Thus we have a total of 18 in total.

Each positive charge which the complex has causes it to have one less valence electron, so the cationic complex will have an iron which will have 17 VE, this alone would make it less stable.

But there is another issue which is more important, the bonding of carbon monoxide to transition metals is dominated by the withdrawal of electron density from the metal atom to the carbon monoxide. So the less electron-rich the metal the less able the carbon monoxides are to bind.

The fact that the iron in the +3 oxidation state will make it less able to bind to $\ce{CO}$ molecules. This is the main reason why it will not be possible to make the complex.

If you consider the series of complexes $\ce{[V(CO)6]-}$, $\ce{[Cr(CO)6]}$, and $\ce{[Mn(CO)6]+}$, the strength of the bonding between the $\ce{CO}$ and the metal decreases along the series $\ce{V, Cr}$ and $\ce{Mn}$ ($\ce{Mn}$ has the weakest bonds).

Secondly, we should only use the 18VE rule when the oxidation state of the metal is low and the ligands are strong ligand field ligands. I do not think that the iron(III) is sufficiently low in oxidation state for the chemistry to be dictated by the 18 VE rule. I would predict that if I was to bubble $\ce{CO}$ through a solution of iron(III) in dilute acid that nothing would happen.

However, if I was to make a solution of an iron(II) complex of a dialkyl glyoxime then it would be able to react with $\ce{CO}$. Here the iron is in a lower oxidation state and has more electron density on it.


Half-life and shelf-life of second-order reaction

Your mathematics is correct until you calculated for specific cases. Let's go back to your $n$th order version:

$$\frac{[\ce{A}]^{1-n}}{1-n} = -k_nt + \frac{[\ce{A}]_\circ^{1-n}}{1-n} \ \text{where } n \ne 1 \tag1$$

At this point, since you are working on second order kinetic, it is easy if you substitute $n = 2$ in equation $(1)$:

$$\frac{[\ce{A}]^{-1}}{-1} = -k_2t + \frac{[\ce{A}]_\circ^{-1}}{-1} $$

Once simplify, it becomes:

$$\frac{1}{[\ce{A}]} = k_2t + \frac{1}{[\ce{A}]_\circ}   \tag2$$

For self life, substitute $t = t_{10\%}$ and ${[\ce{A}]} = 0.9{[\ce{A}]_\circ}$ in equation $(2)$:

$$\frac{1}{0.9[\ce{A}]_\circ} = k_2t_{10\%} + \frac{1}{[\ce{A}]_\circ}  $$

$$\therefore \; t_{10\%} = \frac{1}{k_2}\left(\frac{1}{0.9[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.9} - 1 \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{9} \right) = \bbox[yellow]{\frac{0.11}{k_2[\ce{A}]_\circ}}$$

Similarly, for half life, substitute $t = t_{50\%}$ and ${[\ce{A}]} = 0.5{[\ce{A}]_\circ}$ in equation $(2)$:

$$\frac{1}{0.5[\ce{A}]_\circ} = k_2t_{50\%} + \frac{1}{[\ce{A}]_\circ}  $$

$$\therefore \; t_{50\%} = \frac{1}{k_2}\left(\frac{1}{0.5[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.5} - 1 \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{0.5}{0.5} \right) = \bbox[yellow]{\frac{1}{k_2[\ce{A}]_\circ}}$$


Determine the concentration of hydrogen cations to prevent precipitation of monocalcium phosphate?

I was looking for something else and stumbled across this old problem. I noticed Martin's comment about the source, and I looked for one. Noticing that this was sponsored by ACS made me wonder if my initial analysis had been correct.

The answer key, the OP, and I agree that the $\ce{H2PO4^-}$ concentration must be $8.2\times10^{-2}$ molar to cause $\ce{Ca(H2PO4)_2}$ to start to precipitate.

Problem 1 - The problem statement lists $K_{\alpha1} = 7.1\times 10^{-3}$ but the answer key uses $7.3\times10^{-3}$

My faith in the infallibility of the ACS is shaken...

Problem 2 - There are really two ways to solve the problem.

A. Neutralizing the acid

The solution starts off as 0.25 molar $\ce{H3PO4}$ and 0.15 molar $\ce{Ca^{2+}}$. You add $\ce{NaOH}$ until $\ce{Ca(H2PO4)_2}$ starts to precipitate. What is the pH?

Mass balance for all phosphate species is:

$0.25 = \ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$

The solution is very acidic so we can assume $0 = \ce{[HPO4^{2-}] = [PO4^{3-}]}$ so

$0.25 = \ce{[H3PO4] + [H2PO4^-]}$

$\ce{[H3PO4] = 0.25 - [H2PO4^-] = 0.25 - 0.082 = 0.168 }$

Now using the equilibrium expression for the first ionization of phosphoric acid

$K_{\alpha1} = \dfrac{\ce{[H^+][H2PO3^-]}}{\ce{[H3PO4]}}$

Rearranging and using $7.3\times10^{-3}$ from the answer key

$\ce{[H^+]} = \dfrac{K_{\alpha1}\ce{[H3PO4]}}{\ce{[H2PO4^-]}} = \dfrac{(7.3\times10^{-3})(0.168)}{0.082} = 0.014956 \ce{->[Rounding]} 1.5\times10^{-2}$

Fortuitously using 7.1 from the problem statement yields the same rounded value.

$\dfrac{(7.1\times10^{-3})(0.168)}{0.082} = 0.014546 \ce{->[Rounding]} 1.5\times10^{-2}$

B. Making a buffer

To the 0.25 molar solution of $\ce{H3PO4}$ you add $\ce{NaH2PO4}$ until $\ce{Ca(H2PO4)_2}$ starts to precipitate. What is the pH?

Let $x$ be the nominal molarity of the $\ce{NaH2PO4}$.

Again the solution is very acidic so we can assume $0 = \ce{[HPO4^{2-}] = [PO4^{3-}]}$ so

$0.25 +x = \ce{[H3PO4] + [H2PO4^-]}$

But $\ce{[H2PO4^-] = 0.082}$, so

$\ce{[H3PO4]} = 0.25 + x - \ce{[H2PO4^-]} = 0.25 + x - 0.082 = 0.168 + x$

The charge balance is given by

$\ce{[Na^+] + [H^+] = [H2PO4^-] + 2[HPO4^{2-}] + 3[HPO4^{3-}] + [OH^-]}$

Since the solution is very acidic we can neglect $\ce{[HPO4^{2-}]}$, $\ce{[HPO4^{3-}]}$, and $\ce{[OH^-]}$. So

$\ce{[Na^+] + [H^+] = [H2PO4^-]}$

$\ce{[H^+] = [H2PO4^-] - [Na^+]} = 0.082 - x$

Now substituting into the equilibrium expression for the first ionization

$K_{\alpha1} = \dfrac{\ce{[H^+][H2PO4^-]}}{\ce{[H3PO4]}} = \dfrac{(0.082-x)(0.082)}{0.168+x}$

$0.0073(0.168+x) = 0.082^2 - 0.082x$

$(0.0073 + 0.082)x = 0.0893x = 0.082^2 - 0.0071(0.168)$

$x= 0.06194$

So the buffer solution would be 0.25 molar $\ce{H3PO4}$ and 0.062 molar $\ce{NaH2PO4}$.

$\ce{[H^+] = 0.082 - 0.6194 = 0.02006 ->[Rounding] 2.0\times 10^{-2}}$


Why do no known atoms have electrons in the g or h subshells?

Well, the first available sub-shell for "g" electrons would be 5g (i.e., 1s, 2p, 3d, 4f, so 5g). Based on current trends, we'd need row 8 of the periodic table. We just haven't found those elements yet.

Let's look at it this way.

After row 1 (1s electrons only), there are 2 rows of the s and p-block elements before "d" opens up with $\ce{3d^1 4s^2}$
After two rows of s- p- and d-block elements, the "f" block opens up with cerium: $\ce{4f^1 5d^1 6s^2}$

So we'd need two more rows of the f-block (i.e., the lanthanides and actinides) and then complete the row.

Right now, we have found element 118, so if we can synthesize a few more, we can open up the "g" block.

Now, there are elements that have g electrons in excited states, so g and h-orbitals are relevant to some chemistry. But the ground-state elements haven't been discovered yet.


A is wrong. $\ce{AgBr}$ is one of the least soluble compounds of $\ce{Ag}$. Why don't you have a look in the table of solubility products to be convinced? Further more, $\ce{AgBr}$ will never react with $\ce{H_2O}$ to produce a strong acid like $\ce{HBr}$ and a base like $\ce{AgOH}$. The reaction goes the other way round. Strong acids react with hydroxydes to produce a salt and water.

C is wrong, because of the Le Chatelier's principle. Try to get the reason your self.

To solve D, you must know that $\ce{Zn(OH)_2}$ is soluble in an excess of $\ce{OH^-}$ ions


How to transform an aliphatic primary amine to the corresponding nitro compound?

Using silica gel, primary amines can be oxidized to their corresponding nitro compounds using ozone ($60$-$70 \%$ yield). In this paper, dry silica gel (dried for $24\ \mathrm{h}$ at $450^\circ \mathrm{C}$) was mixed with an amine. The silica gel (ca. $\mathrm{30\ \mathrm{g}}$) containing the amine ($0.1$-$0.2\ \mathrm{wt/wt \%}$) was cooled to $-78 ^\circ \mathrm{C}$, and a stream of $3 \%$ ozone in oxygen was passed through it.

Aliphatic amines and aromatic primary amines are rapidly and effectively  oxidized to nitro compounds by dimethyldioxirane ($80$-$90 \%$ yield). Dimethyldioxirane is prepared by reaction of OXONE ($\ce{2KHSO5 \cdot KHSO4 \cdot K2SO4}$) with buffered aqueous acetone. In a typical reaction, $0.7\ \mathrm{mmol}$ of amine in $5\ \mathrm{mL}$ acetone is treated with $95\ \mathrm{mL}$ of dimethyldioxirane in acetone ($0.05\ \mathrm{M}$). The solution is kept at room temperature for $30\ \mathrm{min}$ with the exclusion of light.

Oxidation of amines to nitro compounds in high yield has also been carried out using oxyacids like peracetic acid, peroxytrifluoroacetic acid, and meta-chlroperoxybenzoic acid, though the anhydrous forms of these acids are hazardous and difficult to use.

Ono, N. The Nitro Group In Organic Synthesis; Wiley-VCH: New York, 2001.


are there any stable four membered chelate rings?

Four-member rings can be formed with simple oxy-anion ligands such as nitrate.  An example with iron as the central ion is given here.

Reference

G. R. A Wyllie, O. Q. Munro, C. E. Schulz, W. R. Scheidt,
"Structural and Physical Characterization of (Nitrato)iron(III)
Porphyrinates [Fe(por)(NO3)] — Variable Coordination of Nitrate",
Polyhedron, Oct 10 2007, 26(16):  4664-4672.


What happens when heating water in a pressure vessel?

Look at the phase diagram of water to see what happens. Josiah Willard Gibbs invented these diagrams o help understand what happens as a substance is heated under pressure, and transitions from solid to liquid to gas and supercritical fluid (neither quite gas nor liquid).

Given sufficient heat, the vessel will burst. Above ~650 K, the water can no longer be liquefied by pressure, and behaves more like an ideal gas, so pressure continues to increase.


Nylon reaction:

$\ce{nH2N(CH2)6NH2 + nClOC(CH2)8COCl → H2N[(CH2)6NHCO(CH2)8]_{n}COCl + nHCl}$

Looking at the reagents, 1,6-hexanediamine is dissolved in water as it can form hydrogen bonds with the water molecules. Meanwhile the sebacoyl chloride is non-polar and thus dissolves best in non-polar solvents. The two solvents (polar and non-polar) cannot dissolve each other, just like oil and water.

The hexane is poured so it floats on top to minimize mixing -- the reaction only happens at the boundary between layers. Then we pull the thread causing more of the reagents to mix and react at the boundary. If we poured the hexane solution in quickly, we'd end up with a large lump of nylon instead of a thread.

To help with your other questions, some questions...

I've given some details about the solvents above. What kind of solvents would you choose to replace the hexane and water with?
Looking at the densities of hexane (0.659 g / mL) and water (1 g / mL), and looking at the procedure, why was the hexane poured? Why not pour the water onto the hexane?
Following on, do some research into the density of dichloromethane. That will give you your answer to (a)
You've mentioned "not enviromentally friendly" for dichloromethane. Are there any other issues? A good starting point are the CLEAPSS student safety sheets.


It seems like the calculation done or copied from the board by you has gone a little wrong.

In your first step when taking $-\log$ of the given concentration you have ignored the signs on the lower of the argument of the log function.
$$-\log 10^{-3} = \log 10^3 = 3 \log 10 = 3$$
since the base of log here is 10. And same correction for next steps.

And to your other question ... We add up the concentrations (adding up p values is irrevalent) because here NaCl is a strong electrolyte and HCl is a strong acid. They both dissociate completely and individually contribute to the concentration of chlorine anion.


Why are N-H stretching vibrations often sharp and not broad?

The sharpness or broadness of a stretch in IR spectra depends on extent of Hydrogen bonding present in the molecule. Basically, if it undergoes immense intermolecular hydrogen bonding, the peaks tend to be broader and the lesser the hydrogen bonding becomes, the sharper the peaks get in the spectra.

Now, if you recall the criteria needed for stronger hydrogen bonding, the more electronegative the atom attached to hydrogen is the better is the extent of hydrogen bonding as the more electronegative atom will be able to create more $\delta +$ve cahrge on hydrogen which is better for stronger hydrogen bonding. $\ce{O}$ is far more electronegative than $\ce{N}$. $(\chi_o = 3.44, \chi_N = 3.04)$. So, the extent of hydrogen bonding is much much lesser with $\ce{N}$ being the attached atom than $\ce{O}$. (If you want to get an idea how strong the hydrogen bonding is in case of $\ce{O}$ than $\ce{N}$, simply consider $\ce{H2O}$ and $\ce{NH3}$, due to hydrogen bonding the former has its boiling point at $100^o C$, while that for the latter is $-33.4^oC$.) But anyways Hydrogen bonding is there in amines, only the extent is lesser.

However if you can somehow increase the $\delta+$ charge on hydrogen, by attaching some electron withdrawing group to the nitrogen, like in the case of amides, there you will be again able to see broad $\ce{N-H}$ stretches as now it can undergo stronger hydrogen bond formation, as $\ce{N}$ pulls the electron density more towards itself due to resonance., and $\ce{H}$ becomes electron deficient, and peak becomes broad. Here is the IR spectra of $\ce{N-methylacetamide}$ from the book Introduction to Spectroscopy which shows broad $\ce{N-H}$ stretches due to the reason mentioned above.

In this spectra you can also see the $\ce{C=O}$ stretch is also relatively broader than observed generally. The reason is again extent of intermolecular hydrogen bonding, which is greater due to higher electronic density on the $\ce{O}$. But apart from that, the general reason for broadening the spectra is the extent of intermolecular hydrogen bonding, which is somewhat lesser for normal amines.


Is cobalt (II) chloride in silica gel dangerous when heated?

I have to assume things to answer this question. It would be great if you could include a picture of the actual stuff you have so that I may assess whether my assumption was correct or not.

The only aggregate state of desiccant silica gel, whether it be blue gel, orange gel or a colourless one, that I have seen to date is best described as a very coarse granulate. From these, unless you vividly shake the canisters repeatedly, there should be no dust evolution. Stuff like that is routinely stored and handled in chemical laboratories without anybody ever giving a damn because of dust — much unlike the very fine silica gel used for chromatography which can very much, as sixtytrees points out, cause silicosis.

The carcinogenic stuff inside blue gel is a cobalt compound. The salts of cobalt are only carcinogenic if you indigest them, and they do not permeate well through skin. They are even less prone to produce dust or carcinogenic volatiles even when heated to a hundred or so degrees. What they can do is stay on the skin after you touched the gel and then be transferred onwards, for example in your mouth on the crunchy apple you decided to eat — so wash your hands thoroughly if you accidentally handled the actual gel and did not use gloves.

Orange gel is slightly different, it uses organic moisture indicators. These have a much more significant vapour pressure than inorganic salts do. On the upside, these are also less carcinogenic. So the overall effect is probably similar.

Aside from that, you are at a very low risk. Sitting by a campfire is very likely more carcinogenic than putting blue gel desiccant into the oven for a few minutes.


How to calculate the concentration of the diluted solution given a source concentration in ppm?

Let us assume for a moment, that the parts-per-notation is actually unambiguous. We then know that we can convert the concentration given in ppm $c^\dagger$ into the concentration $c$ in SI units $\mathrm{mol\,L^{-1}}$, or $c^\dagger\propto c$. Then we know that the concentration is defined as the amount of substance per volume of solution.
$$c = \frac n V$$

When diluting a solution the amount of substance stays the same, hence we have two equations, that we can combine.
$$\begin{align}
  c_0 &= \frac{n}{V_0}  &&\text{source solution}\tag1\\
  c_\mathrm{dil} &=\frac{n}{V_\mathrm{dil}}  &&\text{diluted solution}\tag2\\
  c_\mathrm{dil} &=\frac{c_0V_0}{V_\mathrm{dil}}  &&\therefore\tag{1=2}
\end{align}$$

Which is exactly the formula you used (when I assume that $M$ stands for molarity).

Ergo your calculation result, as orthocresol already pointed out, of $c^\dagger_\mathrm{dil} = 18.7~\mathrm{ppm}$ is correct.

The error therefore has to be in the question, or in copying the same.


Why won't ethene and ethyne react with halogens?

Both substance classes have an inherently different reactivity. Alkanes are very unreactive — in fact, save radical reactions, there is not much that will adequately attack them. Alkenes and alkynes have $\unicode[Times]{x3c0}$-bonds which are much less energetically stable than the alkanes’ $\unicode[Times]{x3c3}$-bonds.

$\unicode[Times]{x3c0}$-bonds can be attacked by halogens directly, the mechanism is called electrophilic addition and the usual product is an anti-dihalide with the halogen atoms attached to the carbons previously involved in the double bond. An intermediate species of this reaction is the halonium ion, a three-membered ring consisting of the ex-double-bonded carbons and one positively charged halogen. The other halogen can then attack according to $\mathrm{S_N2}$ from behind.

However, if you try hard enough you will also manage to create halogen radicals in the presence of alkenes. The radical will now not attack a $\ce{C-H}$ bond, but rather the energetically less stable $\ce{C=C}$ bond forming a $\ce{C-Hal}$ bond and a new carbon-centred radical. This carbon radical will then most likely polymerise with further double bonds by the same mechanism, creating a carbon polymer rather than a haloalkene.

I’m not entirely sure how alkynes will react. Their triple bond is more stable than a double bond, so selective halogenation of double bonds is possible in the presence of triple bonds. However, in general there is no reason why they couldn’t react according to the same pathways.


Hydrogenation of 1-hexene with Raney nickel

To meet your brief of a) performing a hydrogenation and b) using Raney nickel, but performed in the safest manner, I would use aluminium-nickel alloy.

This is cheaper than sponge/Raney nickel and easier to handle i.e
 It's not pyrophoric until activated.

I would advise the use of nitrogen, but not a sweep as it can be detrimental to the procedure I'll outline.

I won't give molar details, but work on about 20% of alloy by mass vs your substrate to start with. I won't quote solvents either as I don't know if you need to isolate your hexane or not, nor how you intend to monitor the reaction.
If you just need to do the reaction, go for a less volatile solvent.

To perform the hydrogenation itself, slowly and in a controlled manner with constant monitoring of temperature,  add 4M NaOH.  This reacts with the aluminium in the alloy to make Raney nickel.  At the same time, hydrogen is generated which the nickel can adsorb and hydrogenate your substrate with.

Bear in mind that the NaOH solution introduces a lot of water,  so if your solvent is immiscible with water, your stirring needs to be all the better.

You could use solvents that mix with a lot of water e.g. 1-butanol can contain up to ~20% water, but from which the bulk aqueous layer can be separated if needs be.

So long as the NaOH solution is added slowly enough, you can control the temperature.  One way to manage is this is to only introduce 25% of your solution to be added at one time. That way if you accidentally add it all, it shouldn't go too crazy and boil/eject the contents uncontrollably.

Once complete, you can decant off the liquid but ensure you leave water behind to cover the catalyst.  Rinse and decant until the pH is 8 or below. Then you have some options:

Dispose of the wet catalyst into bags that go off site from your school for disposal,  using the appropriate school procedures or,
Use dilute (1M) HCl to gradually deactivate the nickel. Again, add very slowly over hours if necessary.  Then follow appropriate waste disposal guidelines for the nickel solution.


We call them "pairs" because there are two electrons in each grouping. Why are there two instead of three or just one?

Spin and orbital energies.

One electron up and one electron down, in order to have all electrons in the lowest energy orbital available. Having them separate is a vastly higher energy state for the particles as a whole.

If you have not yet studied orbital theory - your question will be answered in much more depth when you do.


How can geometric isomerism arise due to steric factors?

Let's take a biphenyl compound:

Due to steric factors, the rings cannot be planar. Instead, they have an approximate right angle between them:

(Image source)

Note that for the right angle to "flip" over to the other side, the $\ce{NO2}$ and $\ce{CH3}$ would have to pass. These groups are bulky enough to prevent that from happening1 unless sufficient energy is provided.

Now, try to look from it along the axis. You'll notice that for this particular molecule, the nitro group farther from your eyes is a clockwise step from the closer nitro group. This happens regardless of which side of the axis (left/right) you look at it from.

On the other hand, if we twisted the molecule so that the angle is again a right angle on the other side, there is now an anticlockwise step from the closer nitro to the farther nitro.

These two configurations of the compound are stereoisomers.

Since the steric barrier can be overcome with sufficient energy, such compounds usually racemize over time (though when the groups themselves  are as bulky as a phenyl, this doesn't happen at any appreciable timescale, IIRC).

1. I actually had some trouble using ChemDoodle to draw the molecule without any overlapping bits because of that. Doesn't exactly prove the bulkiness, but one gets the idea.


Don't put too much trust in the absolute signs of wavefunction, for they all are arbitrary anyway. Look at it this way: an s orbital has one sign (*). One lobe of p orbital has the same sign, so when they add up, it grows huge. Another lobe inevitably has the opposite sign, so when they interfere, it is reduced to a tiny pig tail.

(*) This is not quite true if we consider radial nodes. That's when things get really hairy.


How to set up equation for buffer reaction?

There is a general way to do these sorts of problems. The idea is to consider equilibria of both acid/base and of the water.

When there is solution of a weak acid and it salt, or just the weak acid or just the salt, i.e. pure HA, NaA + HA or pure NaA, then there is no distinction between these types of solution because of the equilibria involved. ( NaA just represents any salt ). The following calculations apply for each or any of these solutions.

Start by defining the equilibrium constants for the reactions. A molecule dissociates as $\mathrm{HA \leftrightharpoons H^+ + A^-}$, but to be general we let the base be [B] instead of $\mathrm{[A^-]}$ and so write $\mathrm{HA \leftrightharpoons H^+ + B}$

$$K_a=\mathrm{[H^+]_e[B]_e/[HA]_e} \tag{1}$$

where the concentrations are those at equilibrium.

There is also the equilibrium $\mathrm{H_2O \leftrightharpoons H^+ + OH^-}$ to consider and

$$K_w=\mathrm{[H^+]_e[OH^-]_e}$$

We know the amounts of acid $c_a$ and base $c_b$ added at the start of the reaction. To obtain the pH at equilibrium the amount $\mathrm{[H^+]}$ has to be calculated. To do this work out the concentration of acid and base in terms of the initial amount and the ionised species.

The total concentration of HA is

$$c_a=\mathrm{[HA]_e+[H^+]_e - [OH^-]_e}\tag{1a}$$

and for the base

$$c_b=\mathrm{[B]_e -[H^+]_e + [OH^-]_e\tag{1b}}$$

The e subscripts are now dropped for clarity. [Note that some authors use mass and charge balance instead to work out the concentrations.]

These values can be substituted into the equilibrium constant equation (1) and using $K_w=\mathrm{[H^+][OH^-]} $. Substitution leads to the general equation

$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - K_w/\mathrm{[H^+]} }{c_a - \mathrm{[H^+]} + K_w/\mathrm{[H^+]}  }\tag{2}$$

which is a cubic equation in $\mathrm{[H^+]}$ that is best solved numerically in the most general case. This equation can be simplified under different conditions as the examples below demonstrate.

Note that when $c_b=0$ equation (2) describes the case of pure HA, when $c_a=0$ it describes the hydrolysis of pure NaA solution.

The figure shows how the pH changes for a weak acid/conjugate base with the p$K_A$ shown and $c_a$ = 0.01 M over a range of $c_b$ concentrations. The curve was calculated by numerically solving eqn 2.  The straight line (coloured grey) is the Henserson-Hasselbalch equation described below in (A). It is clear where this approximate equation fails.

Figure 1. pH vs concentration of base $c_b$. Red line full calculation, straight line (grey) the approximate Henderson-Hasselbalch eqn. The dashed lines show the p$K_A$ and the acid concentration $c_a$ used. The region where the Henderson-Hasselbalch eq. is a good approximation is clear; $c_b$ should be no less than $\approx 0.1c_a$. The plot uses the values in example (B).

Example (A) The Henderson-Hasselbalch eqn is an approximation but limits should be tested.

When the concentrations $c_a$ and $c_b$ are far larger than $(\mathrm{[H^+]} + K_w/\mathrm{[H^+]})$ or $(\mathrm{[H^+]} - K_w/\mathrm{[H^+]})$ these terms  can be ignored without error. This produces

$$\mathrm{[H^+]} = K_Ac_a/c_b,  \qquad \mathrm{pH}=\mathrm{p}K_A - \log_{10}(c_a/c_b)  $$

which is the formula often used. The log form is called the Henderson-Hasselbalch eqn.  It is often the case that $c_a/c_b=1$ then $\mathrm{pH}=\mathrm{p}K_A$.

Example B. What is the pH of a buffer solution consisting of 0.01 M of a weak acid with $K_A$ = 1.5$\cdot$ 10-4, and 0.01 M of its conjugate base?

We could use eqn. 2 but this means solving a cubic which is not necessary. Realising that in acid solution $\mathrm{[H^+] \gg [OH^-]}$ then

$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - K_w/\mathrm{[H^+]} }{c_a - \mathrm{[H^+]} + K_w/\mathrm{[H^+]}  } \to \mathrm{[H^+]}\frac{ (c_b + \mathrm{[H^+]}) }{ (c_a - \mathrm{[H^+]} )}   $$

This equation has to be solved and not approximated to the Henderson-Hasselbalch eqn. since $\mathrm{[H^+]}$ may not be very different to $c_b$.

Letting $x =\mathrm{[H^+]}$ then $x^2+(c_b+K_A)x-K_Ac_a=0$ or

$$x=\frac{-(c_b+K_A)+\sqrt{(c_b+K_a)^2+4c_aK_A}}{2} $$

which gives $\mathrm{[H^+]}=1.45\cdot 10^{-4}$ or a pH = 3.84. This answer is close to that of the HH equation which gives $\mathrm{pH = p}K_A =  3.82$, so as it turns out the HH equation was good enough.


Well, you have to ask: how stable? and compared to what?

Epoxides, including ethylene oxide, are generally considered relatively unstable molecules, with a high chemical activity and involvement in numerous reactions, including polymerization and thermal decomposition. However, ethylene oxide does exist as a molecule, as does cyclopropane, which also contains a three-membered ring.

So, let's compare for example the ring strain in ethylene oxide and cyclopropane. I found some computation data on their relative stability from these lecture notes, which indicate that “ethylene oxide has less ring strain than cyclopropane”. I would attribute this to the fact that the C–O–C angle has less strain because its “relaxed” value would be 104.5°, compared to the “relaxed” C–C–C angle of 109.4°.


How do I calculate the amount of preservatives one would need for a cleaning solution?

Most producers/suppliers have recommendations for concentration ranges of their products. If you can´t get it directly, it is often possible to find the information on their websites (e.g. Schülke or BASF).
The ranges are typically around 0,1% - 0,5%, but different products are often mixtures of different preservatives and with different concentrations, so it can be difficult to compare products directly.
Citric acid has some biocidal effect but is not very efficient, so I would not recommend it.
Why do you need a chealant? You use demineralized water and a nonionic surfactant, so ions in the product should not be a big issue.
The CAS-number you refer for the surfactant covers a variety of chemicals, with different C-chains and degrees of ethoxylation. They are generally classified as mostly soluble in water, but it may be necessary to add a hydrotrope (e.g. Cumenesulfonate) or co-solubilizer (e.g. Butylglycol).


Logic behind this statement (On Transcription)

This textbook would never have made it into my classroom. The second "answer" is an arguably correct concept but is specious as a reason. The first "point" has a silly justification.

The real reason is an evolutionary one. In polypeptides, particular sequences of amino acids fold in ways that provide function. Tiny variations in this sequence can render the resultant polypeptide non-functional. With this sort of sensitivity what are the odds that the non-coding strand, whose sequence is dictated by the coding strand, could actually possess a viable gene? Furthermore, genes evolve over time. Any changes in the sequence of one would necessitate a change in the other. What are the odds of that being efficatious?

That said, I believe there are some cases where a gene on one strand has some overlap with a gene on the sister strand. I doubt these are lengthy or common, possibly occuring with the help of introns, portions of the DNA message that are "spliced out" of the final RNA transcript.


How does water evaporate completely when at room temperature?

As your small percentage of molecules with high enough kinetic energy evaporates, the remaining liquid water cools down. But in doing so, it drains heat from its surroundings and thus stays at room temperature (or close to it), so there is still some fraction of molecules that can evaporate, and they do so, and more heat is transferred from the surroundings, and so it continues until all water is gone.


Why do diamond and graphite have different structures, when they are both composed of carbon?

Chemical structures are a tradeoff of several factors, including the conditions on how they were formed.

The stability of any given chemical structure depends on the ease with which any specific reaction can turn it into something else. Both graphite and diamond are very stable structures which basically means they are hard to easily convert into something else. Another way to think about this is that to convert either into a different allotrope (different versions of the same element) of carbon requires a lot of bonds to be broken and rearranged and this requires a big input of energy. That's why both are stable.

But why is there more than one stable allotrope of carbon? Note that there are more allotropes than just graphite and diamond: buckminsterfullere is also a stable allotrope of carbon with a completely different structure that involves molecules of 60 carbons not a covalent network solid. The basic answer is that there are a lot of ways to arrange the atoms to give a stable structure (partially because there are a lot of options for stable C-C bonds given the chemistry of carbon). This is related to the same reason that organic chemistry is a big subject: C-C bonds are very "stable" and can be organised to give a huge variety of structures.

Not every arrangement of C-C bonds is equally stable: many will fall apart into other structures because there are mechanisms that make the rearrangement easy. But there are no easy mechanisms for rearranging the bonds in diamond or graphite into something more stable. Buckminsterfullerene is only formed from a plasma of carbon atoms when it is allowed to cool: you can't just make it by simple routes from other allotropes of carbon.

Both graphite and diamond have very similar energies of formation at normal conditions, though graphite is marginally more stable (the particular arrangement of bonds has a slightly lower energy than the arrangement in diamond under normal conditions). But this is not true at all conditions. Under extreme pressures, diamond is more stable. And this is how diamond is formed: it crystallised from solution in hot melted volcanic rocks deep in the earth containing lots of carbon and under high pressure. In these conditions, diamond is the more stable way to arrange carbon atoms and the conditions are such that the less stable form will covert to the more stable form because there is enough energy around to break the C-C bonds until the most stable structure is created.

Normally (by which I mean at atmospheric pressure) carbon rich compounds can be pyrolised to give pure carbon in the form of graphite. Here the heat drives off the non-carbon atoms and the carbon atoms rearrange to the form stable at lower pressures.

Once formed neither major allotrope is easy to convert to the other without extreme conditions. Which is why you can wear diamond jewellery without constant anxiety.

In short the stable allotropes of carbon are a function of the way they were made. And there are a variety of ways to join carbon atoms up to give different structures. Once formed they don't easily interconvert. Hence we find a variety of allotropes depending on the variety of conditions that led to their formation.

NB carbon isn't the only element with multiple allotropes. Tin, for example, changes crystal structure and properties depending on the temperature (and, unlike carbon, this interconversion is fast enough to have been a problem in the era when tin cans were actually made of tin and stored in very cold conditions).


Why does the SnAr reaction proceed by the Meisenheimer complex?

Basic level explanation:
SN2 occurs only at sp3 carbons, where you can get backside attack.

Intermediate:
Addition to the double bond generates a highly stabilized anion. This is a lower energy transition state than the concerted SN2.

Advanced:
The reaction occurs at the site of maximum frontier molecular orbital overlap. The HOMO (highest occupied molecular orbital, the hydroxide nucleophile) adds to the LUMO (lowest unoccupied molecular orbital,  π* of the arene). To do SN2, one would need to interact with σ*, which is not the LUMO, and has an unreachable trajectory (points towards the middle of the ring).


Help balancing reaction between Iron (II) chloride and potassium phosphate

This question was the continuation of your question and since both question are related,  I am answering the question by emphasizing the mistake you are doing and adding a relevant portion to explain it.

Mistake:

Iron(II) chloride and potassium phosphate react. Write and balance the molecular, ionic, and net ionic equations.

You need to consider potassium phosphate to be $\ce{K3PO4}$ not $\ce{KH2PO4}$ according to IUPAC's Red book (see sections IR-8.2 and IR-8.4 and the anions in question are explicitly named in table IR-8.1 therein.)

If $\ce{KH2PO4}$ is meant to be added, the correct term should be potassium dihydrogenphosphate.

Extra info.:

Iron hydrogen phosphate are new in this world and has been recently synthesised and as not as stable as iron phosphates. So, only iron (II and  III) phosphate and potassium chloride forms from the reaction.

$$\ce{3FeCl2 + 2K3PO4 ⟶ Fe3(PO4)2 + 6KCl}$$

$$\ce{Fe3(PO4)2 ->[air oxidn⋅,400C] FePO4}$$

or actually

$$\ce{4Fe3(PO4)2 + 3O2 ->[400 C] 8FePO4 + 2Fe2O3}$$

Iron hydrogen phosphate can be made by reacting iron(II) phosphate with dilute hydrochloric acid.

$$\ce{Fe3(PO4)2 + 2HCl → 2FeHPO4 + FeCl2}$$


Which is More Stable with Boranes: Amines or Phosphines?

OK, first off, let's comment on the stability question. When making a Lewis adduct (and rationalizing with the frontier orbital approach), you want the resulting complex to be stable. That requires two possibilities:

The HOMO of the Lewis Base ($\ce{NH3}$ or $\ce{PH3}$ here) is low (i.e., hard to remove an electron).
The coupling between the acid and base orbitals is large.

I want those electrons to be stabilized. Either (or both) approaches can work. So to correct your statement, you don't want a "high HOMO" or "low LUMO."

The catch, of course, is that a large splitting between the Lewis acid LUMO and Lewis base HOMO can lead to low coupling (poor orbital overlap).

Now, I went and ran some quick B3LYP/6-31G(d) calculations on both $\ce{N(CH3)3}$ and $\ce{P(CH3)3}$ because I couldn't find any quick comparisons in the textbooks I use.

The amine has a computed orbital energy of -5.64 eV, and the phosphine has a computed orbital energy of -5.98 eV. (Standard disclaimers about orbital energies being non-physical apply.)

This would suggest that the phosphine would form a stronger bond with a borane.

There's supporting evidence here: "Ab initio molecular orbital study of the substituent effect on ammonia and phosphine–borane complexes" Journal of Molecular Structure: THEOCHEM, (2004) 709(1–3), pp 103-107.

  The complexation energies of $\ce{H3BXH_{$3−n$}F_{$n$}}~(\ce{X~$=$~N, P}; n=0–3)$ and the proton affinities of $\ce{XH_{$3–n$}F_{$n$}}$ compounds have been investigated at the G2(MP2) level of theory. The G2(MP2) results show that the phosphine complexes are more stable than the corresponding ammonia ones. ... The NBO partitioning scheme shows that the stability of the phosphine complexes was related to the hyperconjugation effect.

Basically, the phosphine complexes seem more stable because there is some back-donation of electrons from the borane into the phosphine. It's not just a simple frontier orbital picture - more orbital interactions come into play.


Why won't water freeze if you put ice in it, despite ice being frozen?

Why won't water freeze if you put ice in it?

It will, even at room temperature. You just need a big enough, cold enough ice cube.

Don't believe it? Add a few drops of water to an ice cube in an ice cube tray (which is the same as adding an ice cube to a few drop of water). Wait a few seconds, turn the tray upside down. No water will fall, presumably because it has frozen. Repeat, but immediately turn the tray upside down. You'll get a wet floor. You can conclude that, given enough time, the water will be frozen by the ice. To the whys and hows:

From the second law of thermodynamics we know that:

  An isolated system, if not already in its state of thermodynamic equilibrium, spontaneously evolves towards it.

And

  Heat cannot spontaneously flow from a colder location to a hotter location.

So if you put ice and water together in a cup, heat will flow from the water to the ice until they reach thermal equilibrium, which will be a temperature between the water's initial temperature and the ice's initial temperature. If the temperature of the ice was low enough or if there was enough ice for the water to keep giving away heat, then the water will freeze. That's considering our cup is an isolated system.

Why don't you usually see that happening? Well, regular ice cubes are probably around −5 °C, and there usually is more water than ice in a cup, so before the water has a chance to reach 0 °C all the ice has already melted. Besides that, the ice is receiving heat from the room, unless you're in a place where's below 0 °C. In that case, the water WILL eventually freeze, but mostly because of the environment.


Well, one thing is as plastics. Nylon is a linear polyamide; it's traditionally produced from alkyl-diamines and a dicarboxylic acid (to donate the acyl group of the amides, as well as a second alkyl chain joined directly to the carboxyl), and the typical product is a poly-hexanamide, but nylon-8,x variants are also possible, as are reactions involving a more direct route of welding the monomers together and arriving at the same place.

Some Google-fu turns up a number of possible end products for which octanamide aka caprylamide would be a useful intermediate. By itself, it's an emulsifier and common solvent in organic synthesis. Other alkyl amides including derivatives of octanamide, often with a few additional alkyl or amino groups welded on, are used in drugs, inks/dyes, fragrances, and skin care items. N,N-dimethyloctanamide and N,N-diethyloctanamide appear to be common ingredients in deodorants and sunblocks, though I'm not sure if they're an actual fragrance or a carrier/stabilizer (Edit: they seem to be emulsifiers and stabilizers for essential oils of the fragrances in water-based products).


Does an "excited atom" lose energy gradually?

Yes, it certainly can. In fact, it can do either, traverse gradually across excitation states down to its ground state, or jump from its current state to its ground state, or to any lower state for that matter. So essentially, as long as the target state is equal or lower in energy than the current one the electron can jump to it.

Those jumps have different probabilities though. Determining which process is more likely to occur, and what the probabilities are is a matter of quantum mechanic calculations. Have a look here for example: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/fermi.html


Composition of a typical printed circuit board (PCB)?

This article should be useful for you...

  Material Recovery and Characterization of PCB from Electronic Waste
  Jessica Hanafi, Eric Jobiliong, Agustina Christiani, Dhamma C. Soenarta, Juwan Kurniawan, Januar Irawan, (doi:10.1016/j.sbspro.2012.09.1194)
  Abstract
  Printed Circuit Boards are commonly exist in computers, laptops and mobile phones. Since it contains of a variety of metals and semiconductors, it is difficult to recover the materials in PCB. This paper aims to characterize the composition of these materials and a method to properly recycle and recover the materials in them. This paper utilized simple mechanical and chemical procedures to recover PCBs. Collected PCBs were disassembled, pulverized and separated by using density and magnetic separation method. By using the chemical recycling method, 98.82% purity of CuSO4 hydrate and Al2(SO4)3 hydrate were recovered from the PCBs. It is found that the milling methods, size of the sample and the density of the separating solution determine the effectiveness of the materials recovered.


Carboxylic acid bond lengths in 3-carboxycyclobutane-1-carboxylate

It's most certainly true that proton transfer between the two acid groups can occur and is facile (particularly in an intermolecular manner). However, that only means that over a sufficiently long period of time, the four C–O equilibrium bond lengths will all average out to be the same. It does not mean that at any one instant in time, they are all the same.*

As you said, this is not the same thing as resonance. Resonance in a carboxylate group is a phenomenon whereby at all points in time, the two bonds are equal. There is no flip-flopping between two different structures.

Before we get to the answer for the actual question, I'll talk a little bit about your last question.

You bring up the idea of using spectroscopy to measure bond lengths, which is a very good point. The general answer is beyond the scope of this question, but it basically comes down to how "fast" your spectroscopic method is. To put it very simply, if it takes $x$ seconds (or microseconds, nanoseconds, whatever) to obtain the information that you are trying to get, then what you will observe is a time average over those $x$ seconds. Bear in mind that every method of spectroscopy takes a finite time, including the interaction of photons with the system. There is no such thing as an instant method; there are merely faster and slower ones. So it comes down to how fast the system is moving versus how fast your method is.

It's exactly analogous to taking a picture of a moving object, like a fast car.† If you try that with an phone camera with a long exposure time you will probably just get a blur. But a high-speed camera is good enough to take clear snapshots of Formula 1 cars during the race.

Of course, for photons you need to phrase that in a more quantum mechanical manner, using e.g. the energy–time uncertainty principle. But that's the general idea, at least. If you use a method that is "slower" than proton transfer between the acids (perhaps NMR), then what you may find is that all four oxygens (and hence C–O bonds) appear to be equivalent. If you use a method that is "fast" enough (maybe IR?), then you may well find that they are inequivalent.

Back to your question.

Generally, it is the instantaneous (static) picture which chemists deal with. That means that, at least for the purposes of this question, the dynamic motion of the system is not relevant, even though you are very correct to bring up the proton transfer. Equivalently, we could say that these questions are usually meant in the context of a (hypothetical) infinitely fast measurement method.

An infinitely fast method would be able to tell the difference between the $\ce{CO2H}$ and the $\ce{CO2-}$, because at any one point in time, one of them is protonated and one isn't. They can't be in a state where they're both half-protonated.

But an instantaneous method can't differentiate between the two oxygens in the carboxylate group. Those are entirely equivalent by resonance. And therefore you get three different bond lengths: Q > (R = S) > P.

* Of course, we are neglecting vibrational motion of the bonds, which technically mean that at any point in time it's far more likely that all the bonds are different. For example, even in the carboxylate group in which both oxygen atoms are equivalent (due to resonance), if an asymmetric stretch occurs then the instantaneous bond lengths will be different. Here, when we talk about "bond lengths", we are only referring to equilibrium bond lengths. We are also neglecting single bond rotations...

† I am not an expert on photography, so you may want to check out Photography Stack Exchange.


How can I model cellular membranes in terms of topology?

As Buck Thorn and MaxW already said in the comments, there is less to it than the textbook excerpt might suggest. The use of the term topology in this context is non-mathematical, similar to when someone says there has been a quantum leap in product development.

The actual topology of most proteins is the same as that of a sphere, and the topology of the typical cartoon channel is that of a donut. Having hydrophilic residues in the channel or outside of the channel is not an inversion because those parts of the surface are topologically equivalent.

  [OP] I can't seem to state this membrane chemistry in terms of the mathematical formalism.

That's because it is non-mathematical. There is some real topology associated with membranes, and that is the concept of inside and outside. Membrane proteins insert in one direction, and they don't "flip" directions over time, even though they are perfectly mobile to move around in the membrane (or have lipids move around them) while pointing in one direction or the other one. During endocytosis and exocytosis, the relative orientation of membrane proteins is conserved but inverted, so that parts that are inside organelles (such as glycosylation) end up on the outside of the plasma membrane.


MO diagram for pentaamminechlorocobalt(III)?

Let's start at the bottom. There will be three a1 bonding orbitals. First, we can have the metal $4s$ orbital interact with all six ligands in phase. Let's call this $1a_1$. Next, the metal $d_{z^2}$ can interact with all six ligands as well, but the axial ligands are now opposite phase from the equatorial. Let's call that $2a_1$. Both will of course have antibonding counterparts. The third $a_1$ orbital is an interaction of the metal $p_z$ with out of phase axial ligand orbitals. The equatorial ligands do not contribute to this orbital because they are on the nodal plane. Let's call this $3a_1$.

Mixed in with these a1 orbitals will be the $b_1$ orbital comprised of the equatorial ligand orbitals interacting with $d_{x^2-y^2}$. That will be $1b_1$. It is lower in energy than $3a_1$ but likely higher than $1a_1$ and $2a_1$. (This $1b_1$ and the $2a_1$ correspond to the $1e_g$ orbitals of an $O_h$ complex.)

Near $3a_1$ we have the two $e$ orbitals comprised of the metal $p_x$ and $p_y$, each interacting with a pair of equatorial ligands. Let's call these $1e$. (These along with $3a_1$ correspond to the $1t_{1u}$ orbitals of an $O_h$ complex).

If we are only considering sigma interactions, we next have the $e$ and $b_1$ orbitals representing metal $d_{xy}$, $d_{xz}$, and $d_{yz}$, all of which are nonbonding for sigma. However, the pi interactions with Cl mean that the $e$ orbitals are stabilized and have antibonding destabilized counterparts. Let's call those $2e$. The $b_1$ orbital ($d_{xy}$) remains nonbonding.

Getting the ordering exactly correct is tough in this sort of qualitative analysis, but that should get you a pretty good start.


What are the plastic connectors called (for glassware)?

Those are called Keck Clips. 24/40 is the size of the joint.


Ethene to ethanol: electrophillic additions involving ethyl sulfates

Yes, I believe you are correct. Water attacks at the sulfate containing carbon by an SN2 mechanism. You are also right that sulfuric acid is a stronger acid than hydronium (or protonated ethanol), which means that in solution the bisulfate anion will predominate. Often people (including textbook authors) will draw a mechanism that goes to neutral products. Most likely water will carry out the final deprotonation (especially given that water is the solvent).

Incidentally, most likely this reaction is self catalyzed, as one sulfate protonates another, activating the sulfate for SN2 reaction by a weak nucleophile.


Is scopolamine same as scopolamine hydrobromide?

If the extraction procedure does not involve HBr, then I would doubt it is the hydrobromide because:

hydrobromide salts don't dissolve well in acetone;
hydrobromide salts are rare in nature.

According to the work by Bracci et al. [1], the major component of MeOH extraction of Brugmansia is atropine.

Śramska et al. [2] used ethyl acetate extraction and isolated scopolamine and atropine.

No mention of hydrobromide salts in either.

References

Bracci, A.; Daza-Losada, M.; Aguilar, M.; De Feo, V.; Miñarro, J.; Rodríguez-Arias, M. A Methanol Extract of Brugmansia Arborea Affects the Reinforcing and Motor Effects of Morphine and Cocaine in Mice. Evidence-Based Complementary and Alternative Medicine 2013, 2013, 1–7. https://doi.org/10/f96zpc.
Śramska, P.; Maciejka, A.; Topolewska, A.; Stepnowski, P.; Haliński, Ł. P. Isolation of Atropine and Scopolamine from Plant Material Using Liquid-Liquid Extraction and EXtrelut® Columns. J. Chromatogr. B Analyt. Technol. Biomed. Life Sci. 2017, 1043, 202–208. https://doi.org/10/f9wvv8.


Why will a strong base neutralize as much acid as a weak base?

Well it's because the strength of an acid or base is a measure of it's dissociation in water. But for the the reaction of an acid with a base the equilibrium lies completely to the right as the products (salt and water) are much more stable than the acid and base.

You're trying to compare the dissociation of a species in water with the equilibrium constant of an actual reaction. I understand why, but think of them as two different situations.


Why are (3-methylbut-2-en-2-yl)benzene and (3-methylbut-1-en-2-yl)benzene not stereoisomeric?

I thought the only time you cannot assign E/Z is when the product is either mono or unsubstituted.

Well, you cannot assign E/Z when it doesn't make any sense.

tl;dr: When on at least one side of double bond two identical groups or atoms (e.g., Hydrogen), there's only one way to arrange substituents near double bond.

E/Z is used to explain how substituents on different sides of double bond located, which excludes 1,1-substituted ethylenes (which is the case with minor product on your figure).

When on one side of double bond is two identical groups, e.g., methyl (which is the case with major product) there's no E/Z isomerism too, because there's only one way to arrange substituents. You can actually think of first case as of the second, minor product has 2 identical atoms at one side of double bond.

Hope I haven't made more confusion than there was.

Also, I would suggest to play with some ball and stick models (They could be at the uni library), I think that building couple of examples and trying to rearrange them to stereoisomeric form will help with understanding stereochemistry in general.


Draw a simplified MO diagram for the pi system of Methyl vinyl ether

I am unsure how to use this knowledge to actually draw the MO diagram?

First, chemists used Lewis structures to explain molecular properties, but eventually this method was replaced with resonance structures (like you've drawn in part iv above).  Resonance structures focus primarily upon p orbitals in a molecule and usually do not consider the sigma system.  Molecular Orbital (MO) theory was a further improvement on resonance theory.  MO theory still focuses on p orbital connectivity and typically disregards the sigma system, but it lets you view the actual patterns that the molecular orbitals generate.  It is very similar to the orbital diagram you included above, but we leave out the sigma system and focus on the p orbitals and the electrons they contain.

In part ii) above you conclude that the ether oxygen is $\ce{sp2}$ hybridized.  This means that there is one p orbital on oxygen that contains a lone pair of electrons (there are also 3 $\ce{sp2}$ orbitals on the oxygen, 2 are used to form sigma bonds to carbon and the third holds the second lone pair of electrons on the oxygen).  This p orbital can interact with the 2 p orbitals involved in the carbon-carbon double bond.  These 3 p orbitals can interact to produce 3 molecular orbitals (see pictures below).  These 3 molecular orbitals are produced by combining (adding and subtracting) a p orbital and the 2 ethylenic molecular orbitals (the bonding and antibonding pi orbitals for the double bond).  All of the orbitals shown in the MO diagram below are p orbitals.  We will fill the MOs with 4 electrons, 2 from the pi double bond and 2 from the oxygen p orbital that contains one of oxygen's lone pair of electrons.

So, we have determined that we have 3 p orbitals (one on oxygen, 2 on the olefinic carbons) that can overlap and form 3 molecular orbitals.  The system we have created with 3 p orbitals from methyl vinyl ether is similar to the all carbon allyl analogue with 3 p orbitals, one p orbital on each of the 3 carbons.  Specifically your system is analogous to the ally anion with 4 pi electrons (2 from your double bond and 2 from the  oxygen lone pair).

Image Source

For more information on how to construct molecular orbitals in organic molecules see here and here.


Part 1. Let's consider first law of thermodynamics.

$\delta Q = dU + \delta W$ (classical sign convention)

or

$\delta Q = dU - \delta W$ (IUPAC sign convention)

In the case of adiabatic process, $\delta Q = 0$.

$0 = dU - \delta W$

$dU = \delta W$

$dU = C_VdT$ (by definition)

$C_VdT = \delta W$

After integration, assuming CV is constant.

$C_V\Delta T =  W$

Part 2. Why not Cp?

By definition, $\delta Q_p = dH = C_pdT$. Obviously, we don't have an isobaric process here.

P. S. Yes, those sign convention differences also make me feel bad.


You asked two very different questions.

  (1) How do I identify how many reactions are occurring when precipitating a mixture?

There is no information in way the reactants are shown in a chemical equation as to what the products will be. That is to say that you have to know the products of the reaction.

  (2) For this one, would the reaction be two equations?

Chemists use shorthand all the time. Reactions are typically netted out to the pertinent aspect under consideration. However if I were writing the reactions for the second problem I'd do it much differently that you have presented. I think I'd write:

\begin{align}
x\ce{CaCl2} + y\ce{RbCl &->} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-}\\
x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-} &\ce{->[excess AgNO3]} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{AgCl v}\\
\end{align}

From an analytical chemistry perspective there are two steps. First the chloride salts are dissolved and then the silver nitrate is added. You wouldn't want to add the salts to a silver nitrate solution since the silver chloride could coat the Ca and Rb salts preventing a complete reaction.

I even might write the second reaction as:

$$ (x+y)\ce{Cl-} \ce{->[excess AgNO3]} (x+y)\ce{AgCl v}$$

I'm not a fan of one line jobs like:

$$x\ce{CaCl2} + y\ce{RbCl ->} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-} \ce{->[excess AgNO3]} (x+y)\ce{AgCl v}$$


The terms are used in a different context to mean slightly different things.

The kinetic product of a reaction is the one which is formed the fastest. In the case shown above this is the 1,2 product which results from formation of the most stable cation, which has a lower activation energy for formation.

The thermodynamic product is the most stable product. In this case it is the 1,4 product because it has a more substituted double bond. However, it forms from a less stable carbocation and so the activation energy is higher and it is formed slower than the kinetic product. Raising the temperature favours the thermodynamic product because the activation energy becomes less important. Another way to get the thermodynamic product is to run the reaction under equilibrating conditions, allowing the product to convert back to reactants. This will eventually form more of the more stable product.

Nucleophilicity is often referred to as a kinetic property because it has to do with how well a species can attack an electrophile and is affected by factors such as steric hindrance.

Basicity is often referred to as a thermodynamic property because it is to do with the position of the acid-base equilibrium: $$\ce{B + H+ <=> BH+}$$


Molecular Orbitals of Hydrogen Fluoride (HF)

Bearing in mind the energy and symmetry conditions, in the case of $\ce{HF}$ one can construct MOs using the $1s$ AO on H and the $2s$ and $2p$ AOs on F.

In general, the contribution to MOs is determined by the coefficients in the linear combination.

Here, one observes that the $1s$ electrons are almost completely localized on the $\ce{F}$ atom. Also, $1 \pi$ electrons are completely localised on the $\ce{F}$ atom because the $2p_x$ and $2p_y$ orbitals on F have a zero net overlap with the $1s$ orbital on $\ce{H}$.

Electrons in MOs localized on a single atom are referred to as nonbonding electrons.

Also, I would note that the $3 \sigma$  MO has less bonding character and the $4 \sigma^*$ MO has less anti-bonding character.

Note that the total bond order is approximately one because the $3 \sigma$ MO is largely localized on the F atom, the $3 \sigma$ MO is not totally bonding, and the $1\pi$ MOs are completely localized on the F atom.

On account of fluorine being a more electronegative atom, in the $2 \sigma$ bonding orbital the electron density is much greater on the more electronegative fluorine than on the hydrogen. However, in the anti-bonding $4 \sigma^*$ orbital, this polarity is reversed.

Caveat: The preceding paragraph may support your intuition and may be right in a few simple cases, but I wouldn't rely on it too heavily.

Given below is a diagram showing $2 \sigma$ , $3 \sigma$ and $1 \pi$ MOs in HF


Why is the current draw dropping in water electrolysis

If you look at the copper wires after the electrolysis, the cathode (connected to the negative pole of the current source) will be white. This electrode is coated with Mg(OH)2.

The low activity of copper on the electrochemical scale makes it unlikely (impossible) for copper to displace magnesium from an aqueous salt.

However, the initial voltage is high enough to theoretically evolve metallic magnesium, which would instantly react with water to produce H2 and Mg(OH)2, which will precipitate because of its low solubility. The Mg(OH)2 adheres, at least to some degree, on the copper electrode and interferes with current flow. The resistance to the electrolysis at the same voltage leads to a lower resulting current.

Even at electrode potentials below that needed to evolve metallic Mg, current carried by the aquo Mg++ ion will result in an increased pH and deposition of Mg(OH)2 around the cathode. An interesting question is whether the deposit from a low voltage electrolysis has the same adhesion as from a high voltage experiment. Maybe low voltage electrolysis would deposit Mg(OH)2 farther out in the solution and not be so blocked off by adhesion to the electrode.


What is so special about positive hydrogen ions that they are used in pH calculation?

What we're really talking about with the acidity of a solution is the proportion of water which is dissociated into hydroxide $OH^-$ or hydronium ions $H_3O^+$ (hydronium is usually a more accurate representation than a free proton).

Other ions do affect the acidity of a solution, but by inducing water to dissociate. Check out the example on the wiki page. Pure water alone has a $pH$, because of spontaneous dissociation of some water molecules into ions. The $pH$ of pure water is 7, because the $pK_a$ of water as a weak acid is 14.

Basically, water is the special thing, and its constituent ions happen to be hydroxide and hydrogen / hydronium. Those ions are what we measure with a $pH$ electrode, so the scales are named after them. Water has a concentration around 55 M, vastly predominating in aqueous solutions.


Influence of pH on precipitate crystallite sizes?

At low pHs there is little "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ cations. Most of the carbonate species are dissolved $\ce{CO2}$, $\ce{H2CO3}$ and $\ce{HCO3-}$. Thus the precipitate forms slowly and you get relatively large crystals.

At high pHs there is a lot "free" $\ce{CO3^{2-}}$ to precipitate the $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ cations. Thus the precipitate forms relatively rapidly and you get relatively small crystals.

I have no idea what literature would lead you to believe otherwise.


Using Distribution Coefficient to Determine Mass and Percent of Acid Extracted

You got it almost right ;-)

The distribution coefficient gives the ratio of equilibrium concentrations in a biphasic mixture.

It is crucial to realize that a caffeine concentration of $2.0\,\mathrm{g}$ in $200\,\mathrm{mL}$ of water is the initial concentration, but not the concentration after a single extraction step! The latter will be lower, since a part of the caffeine went into the organic layer.

It doens't matter whether you express the concentrations in $\mathrm{g\cdot L^{-1}}$ or $\mathrm{mol\cdot L^{-1}}$ since $K_D$ is a dimensionless constant.

$$K_D = \frac{{[\mathrm{caffeine}]_{\ce{CH2Cl2}}}}{{[\mathrm{caffeine}]_{\ce{H2O}}}} = \frac{x}{10-x}\ \mathrm{g\cdot L^{-1}} = 7.8$$

$$x = 7.8\cdot (10-x)\,\mathrm{g\cdot L^{-1}} = \mathbf{8.86} \,\mathrm{g\cdot L^{-1}}$$

Note that $x$ is the concentration in $\mathrm{g\cdot L^{-1}}$. In order to calculate the absolute amount of caffeine in the organic layer or the remaining amount of caffeine in the water phase, you have to consider the volumes.

By the way, there's an article by G.S. Klebanov, L.N. Mednikova, and A.D. Ovcharova on this topic: Extraction of caffeine from aqueous solutions, published in
Pharmaceutical Chemistry Journal, 1967, 1, 221-223 (DOI).


How does an atom's electronegativity change as it becomes more negatively charged?

Electronegativity is a measure of the ability of an atom to attract electrons to itself (usually in a bonding framework; outside of a bonding framework the ability to attract electrons is electron affinity).

So from simple Coulombic considerations, we can surmise that as your carbon becomes more and more negatively charged, the less it will be able to attract electrons to itself.


Would an amalgam filling under or next to a gold crown cause any electrochemical reaction?

The answer is basically yes, if the dental procedures are done improperly.  Here is a good answer from a DDS:

  Galvanic current is a term that has been used in dentistry for over
  100 years. It is a condition created by the presence of dissimilar
  metals in the oral cavity of the teeth and gums, with saliva serving
  as the electrolyte. Have you ever felt a “shock” to your teeth caused
  by a piece of tin foil or a spoon that touches a sliver or mercury
  filling in your mouth? If you answer yes to this question, you’ve
  experienced a galvanic event.

  There are several different types of galvanism: 1) A silver/mercury
  filling is placed in opposition or adjacent to a tooth restored with
  gold. These dissimilar metals in conjunction with saliva and body
  fluids constitute an electric cell. When brought into contact, the
  circuit is shorted, the flow of electrical current passes through the
  pulp, and the patient experiences pain. 2) Dissimilar metals coming
  into contact when the upper and lower teeth come together and touch
  each other. 3) Two adjacent teeth are restored with dissimilar metals.
  The current flows from metal to metal through the dentine, bone and
  tissue fluids of both teeth resulting in discomfort and tooth
  sensitivity.

Additionally, from the same site:

  Metals will not necessarily always cause the galvanic effect in the
  mouth. It depends on the specific metal and alloy being used along
  with quantity and placement in the mouth.

In other words the amalgam filling and the gold cap may form a weak galvanic cell.  The proximity of the amalgam to tooth nerves creates a situation in which only a very weak cell current is required in order for the resulting sensation be perceived.  The rate of degradation could thus be very small, or if the filling and capping procedures are done correctly there will be no galvanic affect at all.


Rate of hydrolysis of silicon tetrahalides

We can justify the rate of hydrolysis of Silicon tetrahalides if we just think on the statement carefully. The statement claims that the rate of hydrolysis is in the order $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$. So, we are mainly concerned about the rate of the reaction, not on the thermodynamic stability or any energetic criterion for this reaction( I have dealt with it in the next paragraph). If we recall the characteristics for rate of   a reaction, it depends on the slowest step of the reaction. In the hydrolysis reaction of silicon tetrahalides, the first step, i.e the attack of a $\ce{H2O}$ molecule to form the trigonal bipyramidal intermidiate is the rate determining step. Also note that, no cleavage of $\ce{Si-X}$ bond happens in the rate determining step, but it actually happens in the next step, which is a fast step, so it doesn't contribute to the rate of the reaction. So, basically strength of $\ce{Si-X }$ bond doesn't affect the rate of hydrolysis reaction (the situation is exactly same as Nuleophilic aromatic substitution reactions ($S_NAr_2$) also). But if the $\ce{Si}$ atom becomes more and more electrophilic, the attack by a water molecule becomes more easier which subsequently results in the increase of the rate of the reaction, So the electrophilicity of $\ce{Si}$ in the tetrahalides increase in the order $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$, and so the rate of the hydrolysis also.

Here is the mechanism of the first out of the four subsequent substitutions in the hydrolysis reaction of Silicon tetrahalides.

Secondly, if we really want to judge the rate of the reaction by numerical calculations, observe that the ultimate product of hydrolysis reaction of all the tetrahalides are same. Also, all the reactions are exothermic in nature. So, $\Delta _r H^0$ (assuming the  hydrolysis reactions are done in standard conditions; it's not a strict assumption, it may not be at standard conditions also, but the changes in the heat of reaction will be almost similar for all the tetrahalides) will be more if $\Delta_f H^0$ is more for the reactants (note that, $\Delta _r H^0 = \Delta_f H^0_{products} -\Delta_f H^0_{reactants} $ and we are bothered about reactants. If $\Delta_f H^0_{reacants}$ are more $\Delta_r H^0$ will be more negative and hence energetically favourable). The values for $\Delta_f H^0$ have the order, $\ce{\Delta_fH^0_{SiF4} < \Delta_fH^0_{SiCl4} < \Delta_fH^0_{SiBr4} < \Delta_fH^0_{SiI4}}$. So, the hydrolysis reaction of $\ce{SiF4}$ will be the least exothermic and that of $\ce{SiI4}$ will be the most exothermic, which is expected also from the bond energy considerations. So, if you have less release of heat energy, i.e thermodynamically less stable product, it will have lesser activation energy and higher rate constant, and thus the rate of the reaction will be highest for $\ce{SiF4}$ and least for $\ce{SiI4}$ as we will get the product along with relatively more thermodynamic stability. So, by thermodynamic consideration also, rate of hydrolysis should be, $\ce{SiF4 > SiCl4 > SiBr4 > SiI4}$.

Lastly, the $\pi$-bonding is not important in case of large size difference between $\ce{Si}$ and $\ce{F}$ ($2p$ and $3d$ have course greater size difference though you may not admit). The fluorine atom makes the silicon more electrophilic by its electron pulling ability rather than  back-bonding.


C-H would be the weakest bond, then C-D, then C-T would be the strongest.

This is due to the zero point energy of the bond and the fact the zero point energy is proportional to:

$$\sqrt{\frac{m_C + m_H}{m_C m_H}}$$

Zero point energy is the minimum energy the quantum harmonic oscillator can have.

Because the greatest mass isotope has the smallest zero point energy, the energy of the bond is greatest.


Is sodium acetate a neutral or a basic salt?

Normal salts do not have to be neutral, and vice versa. There are two different features of a salt: (1) containing a replaceable hydrogen, and (2) making an acidic solution in water. They are not necessarily related. For example:

$\ce{NaHSO4}$ has both.
$\ce{NaHCO3}$ has the first and not the second (in fact, being a salt of a strong base and a weak acid, it has to be basic in nature, as you correctly recall).
On the other hand, $\ce{CuSO4}$ is partially hydrolyzed in water solution and hence has the second feature, but not the first.
To make the picture complete, $\ce{NaCl}$ has neither.

Likewise, the opposite features of (1) having a hydroxy group and (2) making a basic solution in water are even less related. For example:

$\ce{Cu2(OH)2CO3}$ has the first feature, but not the second (for it is not even soluble in water).
$\ce{CH3COONa}$ is partially hydrolyzed in water solution and hence has the second feature, but not the first.
$\ce{MgCl2}$ and $\ce{Mg(OH)2}$ form several basic salt phases with both features (J. Chem. Eng. Data 2017, 62 (4), 1384–1396).  They are only sparingly soluble but still form measurably alkaline solutions, especially when warmed.
$\ce{MgSO4}$ has neither.


Reaction of a compound with water is called hydrolysis.During hydrolysis, Si atoms of the chlorides make up of their empty d-orbitals in their valence shells to form coordinates bonds with water molecules.
However, C is in period 2 does not have d-orbitals in its valence shell to form coordinate bond with water and is not hydrolysed by water.

extra reading https://www.chemguide.co.uk/inorganic/group4/chlorides.html


Mechanism of retro aldol condensation of hydroxycyclopropylketone

You deprotonate the alcohol, and then flip bonds to create the enolate. There are two enolates possible, so swapping bonds gives you the terminal one, which can attack the aldehyde to form the 5-ring. Elimination of water gives you the product.


Why might a planar, conjugated molecule be higher in energy than a nonplanar system?

I reproduced your calculations with PBE/def2-SVP (using Orca). The energy "jump" is even smaller than with MP2, however the causes are probably the same. So here is my answer:

The planar structure is a Transition State, with one negative frequency.
The dihedral you are scanning is probably not the best to describe this twisting.

Now in more detail. Here you can see the Minimum Energy Path (calculated with pDynamo using the NEB method) between the two symmetrically related minima.
 You can see that, as expected, the path is smooth.

The main change between the structures is the pyramidalisation of the terminal NH$_2$. This can be described by the N-C-N-H dihedral, which also changes smoothly as you can see here:

I don't think the reason why the terminal NH$_2$ is not planar is steric problems as @Ivan Neretin suggests. I think its more a question of nitrogen hybridisation, which is between sp$^2$ and sp$^3$. It is the same reason why aniline is not planar. This is discussed in these two questions:

Is nitrogen of aniline sp² or sp³ hybridized?
How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?

Hope this helps!


How can you change the proportions of reactants and products at equilibrium if Le Chatelier's principle is occurring?

The Henderson Hasselbalc equation is a good way to calculate both the pH and the concentration of acid/conjugate base in an equilibrium solution. According to it:

pH = pKa + log([A-]/[HA])

Since you want to know when [A-] = [HA], you are looking for when the pH is equal to the pKa. This means that you will have equal concentrations of the acid/conjugate base whenever your pH is equal to your pKa, which you can find by monitoring the pH. While adding more conjugate base will shift the equilibrium towards acid formation, it will normally increase the acid concentration less than the amount of base that you added.


What temperature is required to burn pencil lead graphite?

You have two problems.
1.) pencil lead graphite is actually a graphite-clay composite.
2.) carbon does not sustain burning easily unless it is held at very high temperatures.

Pencil lead will not burn effectively for the first reason as clay is non-combustible and smothers any fire load. If you did have pure graphite to burn, a simple flame wouldn't be hot enough to combust it.  You need to sustain temperatures of around $1000\mathrm{-}2000~^\circ\mathrm{C}$ in order for pure carbon to burn.


pH implications of redox balancing half reaction method

If the reaction happens at neutral or mildly alkalic $\mathrm{pH}$:

$$\ce{6I- + 4 H2O + 2 MnO4- -> 3 I2 + 2 MnO2 (s) + 8 OH-}$$

If the reaction happens at mildly aciding $\mathrm{pH}$, it could be as well:

$$\ce{6I- + 8 H+ + 2 MnO4- -> 3 I2 + 2 MnO2 (s) + 4 H2O }$$

In any case, either spending  $\ce{H+}$ either production $\ce{OH-}$ increases $\mathrm{pH}$. Both iodide and permanganate carry a negative charge while their spent forms are neutral. This negative charge must be passed to other anions like $\ce{OH-}$ or neutralized by $\ce{H+}$.

If one considers further disproportionation of iodine to iodide and iodate in alkalic solutions, then the total reaction is:

$$\ce{ I-  +  H2O + 2 MnO4- ->  IO3- + 2 MnO2(s) + 2 OH-}$$

In strongly alkalic it could be also
$$\ce{ I-  + 6 OH- + 6 MnO4- ->  IO3- + 6 MnO4^2- + 3 H2O }$$

It may be obvious if the equation is written together with spectator ions:

Mildly acidic:
$$\ce{6KI + 8 HA + 2 KMnO4 -> 3 I2 + 2 MnO2 (s) + 4 H2O + 8 KA}$$
resp neutral/mildly alkalic
$$\ce{6KI + 4 H2O + 2 KMnO4 -> 3 I2 + 2 MnO2 (s) + 8 KOH}$$
resp alkalic
$$\ce{ KI  +  H2O + 2 KMnO4 ->  KIO3 + 2 MnO2(s) + 2 KOH}$$
resp. strongly alkalic
$$\ce{ KI  + 6 KOH + 6 KMnO4 ->  KIO3 + 6 K2MnO4 + 3 H2O }$$


A much more significant example of this is methyl bromide, which has historically been used as a soil fumigant. Its use has been slowed phased out due to its toxicity and ozone depletion abilities.


How to calculate the final concentration of calcium ions from known starting masses of carbonate and calcium ions?

The premise of your question is weird. $\pu{50 mg}$ of calcium ion and $\pu{50 mg}$ of carbonate ion are not equimolar. However, let's assume that they have appropriate counterions somewhere. This problem can be approached two ways.

Limiting reactant

We have a precipitation reaction between the calcium and the carboncate ions,  as calcium carbonate is nearly insoluble in water:
$$\ce{Ca^2+ (aq) + CO3^2- (aq) -> CaCO3 (s) v}$$

Convert your masses of ions to amount of substance and figure out which ion is in excess. If calcium ion is in excess, then use the amount of substance in excess to determine the remaining calcium concentration.
For example: Let's say you determined that your solution has $\pu{0.1663 mol}$ of $\ce{Ca^2+}$ and $\pu{0.1221 mol}$ of $\ce{CO3^2-}$. The maximum amount of $\ce{CaCO3}$ that could precipitate is $\pu{0.1221 mol}$, because we would run out of carbonate ion. Then there would be $\pu{0.0442 mol}$ of calcium ion left in $\pu{1 L}$ of solution for a concentration of $\pu{0.0442 M}$.

Solubility product

Calcium carbonate is not completely insoluble. It's solubility product constant is $4.8\times 10^{-9}$. Thus we can set this up as an ICE equilibrium problem. I'll use the same fake amount of substances from the previous example:

\begin{align}
\ce{CaCO3 (s) &<=> Ca^2+ (aq) + CO3^2- (aq)}\\
K_\mathrm{sp}&=[\ce{Ca^2+}][\ce{CO3^2-}]
\end{align}
\begin{array}{lcc}
\hline
                   & \text{calcium}/\pu{mol}  & \text{carbonate}/\pu{mol}\\
\hline
\text{Initial}     & 0.1663                   & 0.1221     \\
\text{Change}      & -x                       & -x         \\
\text{Equilibrium} & 0.1663 - x               & 0.1221 - x \\
\hline
\end{array}
\begin{align}
4.8\times 10^{-9}&=(0.1663-x)(0.1221-x)
\end{align}

Solving for $x$ gives you $x=0.1221,0.1663$ and only the first number makes sense. In this case the final calcium amount in solution would still be $\pu{0.0442 mol}$. The carbonate amount in solution is not actually zero. It is $\pu{1.09E-7 M}$, but that is for another day.