DNA-Sequencing (detection of the change of pH-value) after emulsion PCR

In molecular biology, it is a very important that DNA should be replicated in order to facilitate sequencing. Basically, this is a very important concept that DNA should be replicated, simply in order to amplify signals. No matter the method of sequencing, without a proper amount of amplification, it's near impossible to detect each base call (signal). All Next-Generation Sequencing (NGS) techniques replicate DNA before sequencing is done.

Emulsion PCR is a PCR variation that some NGS technologies (e.g., NGS flatforms: Roche/454’s GS FLX, Titanium; Life/APG’s SOLiD 3; and Polonator G.007) use to replicate DNA sequences. It is conducted on a bead surface within tiny water bubbles floating on an oil solution.

Procedure (Source: https://binf.snipcademy.com/lessons/ngs-techniques/emulsion-pcr):

Fragmentation of DNA: The library is first fragmented either by sonication (high sound energy) or nebulization (forces DNA through a small hole) to fragments ranging from 300 to 800 base pairs.
Adapters are ligated: Adapters are then ligated onto the DNA fragments. These allow the strands to bind to the emulsion beads.
Denature to single strands: The double stranded DNA's with adapters are then denatured by heating the DNA up to $\pu{95 °C}$. Denaturing DNA simply means to go from double stranded DNA (dsDNA) to two single strands (ssDNA) - the hydrogen bonds keeping the two together are broken.
Formation of clonal bead populations: Each bead coated with streptavidin (e.g., Streptavidin-coupled Dynabeads), which is resistant to organic solvents, denaturants, detergents, proteolytic enzymes and extremes of temperature and $\mathrm{pH}$. Over a billion beads are used with a primer that matches the adapters attached earlier. The ssDNA is then attached to these beads. Each bead is emulsified in a water-in-oil droplet with PCR reagents (DNA polymerase, primers, buffers, dNTPs):

Emulsion PCR (emPCR) amplifies DNA strands on beads: Within these droplets, PCR is conducted. This involves the steps Denaturation, Annealing, Elongation. Firstly, the strand is elongated with DNA polymerase and dNTPs. Then the double-strand is denatured, allowing for the strand to ligate to another site on the surface of the bead. Eventually, 1 million copies of the target is amplified on the surface of each bead. The water-in-oil droplet is approximately 1-um. Follow the figure to see how each bead is able to replicate DNA on its surface.

Emulsion Breaking: After the DNA strands are amplified, the emulsion from the preceding step is broken using isopropanol and detergent buffer. The solution is then vortexed, centrifuged, and magnetically separated. The resulting solution is a suspension of empty, clonal and non-clonal beads, which will be filtered in the next step.
Bead enrichment: After PCR is conducted, you are left with a mixture of some beads that have amplified DNA attached on its surface, and some that do not. We may take out the enriched beads by attaching streptavidin coated magnetic enrichment bead. With a magnet, we can then pull out the beads with amplified DNA. There are other methods of bead enrichment that include using larger beads that are able to bind to beads with amplified DNA. After centrifugation, the beads with amplified DNA and without can then be separated.

Bead Capping: Attach a capping oligonucleotide to the 3' end of both unextended forward EmPCR primers and the RDV segment of template DNA. This helps in cover-slip arraying, which is used to polony sequencing, and prevents fluorescent probes from ligating to the ends.
Result: The beads with amplified sequences are then placed on a slide and are sequenced. Due to their high density of the same DNA molecule, the signal is amplified, allowing computers to read the sequencing data.

For example, the method used by the Roche/454 sequencer to amplify single-stranded DNA copies from a fragment library on agarose beads. A mixture of DNA fragments with agarose beads containing complementary oligonucleotides to the adapters at the fragment ends are mixed in an approximately 1:1 ratio. The mixture is encapsulated by vigorous vortexing into aqueous micelles that contain PCR reactants surrounded by oil, and pipetted into a 96-well microtiter plate for PCR amplification. The resulting beads are decorated with approximately 1 million copies of the original single-stranded fragment, which provides sufficient signal strength during the pyrosequencing reaction that follows to detect and record nucleotide incorporation events. sstDNA, single-stranded template DNA (Ref.1):

Keep in mind that in pyrosequencing method, each incorporation of a nucleotide by DNA polymerase results in the release of pyrophosphate, which initiates a series of downstream reactions that ultimately produce light by the firefly enzyme luciferase. The amount of light produced is proportional to the number of nucleotides incorporated (up to the point of detector saturation). Above diagram shows an example of four-color mapping sequence. For additional info, should read all of Ref.1-3.

References:

Elaine R. Mardis, “Next-Generation DNA Sequencing Methods,” Annu. Rev. Genomics Hum. Genet. 2008, 9, 387–402 (doi: 10.1146/annurev.genom.9.081307.164359).
R. Kanagal-Shamanna, “Emulsion PCR: Techniques and Applications,” Methods in Molecular Biology, Volume 1392: Clinical Applications of PCR;  R. Luthra, R. Singh, K. Patel, Eds.; Humana Press: New York, NY, 2016, pp. 33–42 (https://doi.org/10.1007/978-1-4939-3360-0_4).
Michael L. Metzker, “Sequencing technologies —the next generation,” Nature Reviews Genetics 2010, 11, 31–46 (doi: 10.1146/annurev.genom.9.081307.164359).


NMR - coupling of chemically equivalent protons

The "n+1" rule refers to a situation where you have a proton of type A with $n$ protons of type B next to it. Proton A's signal will be split $n+1$ times by the B's. However, none of the B's will split each other because they are equivalent.


This statement is relevant for the initially [incorrectly] determined structure of $\ce{B2O3}$ by Berger [1, 2] who used powder x-ray diffraction data.
You can still access the structure and visualize it in 3D (ICSD 24047), but it was shown later by several researchers and summarized in paper by Gurr et al. [3] that this structure has been erroneously determined:

After failing to grow a useful single crystal by this process, Berger (1952, 1953) was forced to use data from powder photographs in an attempt to determine the structure. The basic units of Berger's structure were two different irregular tetrahedra of oxygen atoms each surrounding a boron atom. This structure has been considered unsatisfactory in several respects (e.g. Wells, 1962), particularly as it results in two different $\ce{BO4}$ tetrahedra with six different $\ce{B-O}$ bond lengths ranging from $1.31$ to $\pu{2.14 Å}.$ In this laboratory a least-squares refinement (Hamilton, 1961) on Berger's powder data was attempted from his positional parameters. The attempt was abandoned when the refinement diverged. The tetrahedrally coordinated structure proposed by Berger was also puzzling because in various borate systems there is a regular variation of the fraction of boron atoms in fourfold coordination as a function of increasing alkali content (both in glasses and crystals); yet in each instance there was an obvious trend towards only threefold coordinated boron as the composition approached $\ce{B2O3}.$
[…]
The n.m.r, absorption for $\ce{BO4}$ groups is sharper and stronger than that for $\ce{BO3}$ groups (Bray et al., 1961; Silver & Bray, 1958) and thus the absence of sharp components in the spectrum of $\ce{B2O3}$ I precludes the presence of tetrahedral $\ce{BO4}$ groups in this material.

A new crystal structure was proposed: same space group $P3_1,$ but lacking tetrahedral units (ICSD 16021).
This is the one that is shown on Wikipedia, hence the dissonance.

Figure 1. Comparison between the unit cell content of two $\ce{B2O3}$ structures (boron atoms and boron coordination polyhedra are depicted in yellow color). Left: crystal structure according to Berger (1952). Right: crystal structure according to Gurr et al. (1970).
References

Berger, S. V. The Crystal Structure of $\ce{B2O3}.$ Acta Cryst 1952, 5 (3), 389–389. DOI: 10.1107/S0365110X52001155.
Berger, S. V.; Boeri, E.; Bonnichsen, R. K.; Sörensen, J. S.; Sörensen, N. A. The Crystal Structure of Boron Oxide. Acta Chem. Scand. 1953, 7, 611–622. DOI: 10.3891/acta.chem.scand.07-0611.
Gurr, G. E.; Montgomery, P. W.; Knutson, C. D.; Gorres, B. T. The Crystal Structure of Trigonal Diboron Trioxide. Acta Crystallogr B Struct Crystallogr Cryst Chem 1970, 26 (7), 906–915. DOI: 10.1107/S0567740870003369.


No, acids can't have pH greater than 7. By definition, at 25 °C, solutions with a pH less than 7 are acidic and solutions with a pH greater than 7 are basic.

In your question you say thiopental and pH = 10.5. Actually this is sodium thiopental with pH range of 10.2-11.2 for a 8% solution SOURCE. The solution is alkaline because of hydrolysis.

The pH value however can be less than 0 or greater than 14 for very strong acids and bases respectively. (Lim, Kieran F. (2006). "Negative pH Does Exist". Journal of Chemical Education. 83 (10): 1465.)


The Oxford Dictionary does not have an entry for enclathrate, but does for clathrate:

  A compound in which molecules of one component are physically trapped within the crystal structure of another.

Using the word enclathrate should be redundant.


What would be the effect of the addition of an inert gas to a reaction at equilibrium?

Dissociation obviously increases the number of moles.
The addition of an inert gas can affect the equilbrium, but only if the volume is allowed to change.

There are two cases on which equilibrium depends. These are:

Addition of an inert gas at constant volume:
When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change.

Hence, when an inert gas is added to the system in equilibrium at constant volume there will be no effect on the equilibrium.

Addition of an inert gas at constant pressure:
When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase. Hence, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases.

Consider the following reaction in equilibrium:
$$\ce{2 NH_3(g) ⇌ N_2 (g) + 3 H2 (g)}$$
The addition of an inert gas at constant pressure to the above reaction will shift the equilibrium towards the forward direction(shift to the right) because the number of moles of products is more than the number of moles of the reactants.

Please read about the Effect of adding an Inert Gas


As discussed in your question and ensuing comments, there are ways to predict via theory whether a particular compound would be a polyprotic acid, but your last sentence asks how to prove that an acid is polyprotic. This pretty much limits us, as you proposed, to laboratory experiment, namely acid-base titration.

Although your suggestion "Acid base reaction with something like sodium hydroxide NaOH and then distillation to separate the product/products from the water that is produced" is on track, the experiment needn't be nearly so complex.

The experimental procedure is as simple as preparing a solution of the unkown acid, then titrating with a strong base solution while monitoring the pH.

The following is a titration curve for the diprotic oxalic acid, using sodium hydroxide as the strong base titrant. (Image taken from this boundless.com post):

Note that there are two equivalence points, at 25 and 50 ml of added titrant. These equivalence points are the features to look for when titrating the unknown acid. A single equivalence point indicates a monoprotic acid, two indicate a diprotic acid, etc.

There are a couple drawbacks to this experimental procedure. First, the equivalence points are not always as well defined as that shown in the figure. For instance, two acidic moieties may have similar Ka values and thus may almost appear as one.  Making a derivative plot of the titration curve can help to distinguish the equivalence points. In other words, you plot the change in pH vs volume of titrant added rather than just the pH of the titrant added.

The second drawback of this procedure is that it doesn't tell you which protons were titrated by the base. This would require a far more complex experiment, for instance by synthesizing the acid using deuterium in place of hydrogen at specific sites, then analyzing the compound after each equivalence point to see whether the deuterium was pulled off.

I hope I understood your question correctly and that my explanation of the titration procedure to determine polyprotic acids was clear. Don't hesitate to ask for clarification in the comments.


A problem in solubility of salts with multistep hydrolysis of an ion possible

Let's denote all processes of interest which are occuring or might occur in the system:

\begin{align}
\ce{MnS &<=>[$K_\mathrm{sp}$] Mn^2+ + S^2-} && K_\mathrm{sp} = [\ce{Mn^2+}][\ce{S^2-}] \tag{R1} \\
\ce{MnS + H2O &<=>[$K_\mathrm{h1}$] Mn^2+ + HS- + OH-} && K_\mathrm{h1} = [\ce{Mn^2+}] [\ce{HS-}] [\ce{OH-}] \tag{R2} \\
\ce{MnS + 2 H2O &<=>[$K_\mathrm{h2}$] Mn^2+ + H2S + 2 OH-} && K_\mathrm{h2} = [\ce{Mn^2+}][\ce{H2S}][\ce{OH-}]^2 \tag{R3} \\
\ce{H2S &<=>[$K_\mathrm{a1}$] HS- + H+} && K_\mathrm{a1} = \frac{[\ce{HS-}][\ce{H+}]}{[\ce{H2S}]} \tag {R4} \\
\ce{HS- &<=>[$K_\mathrm{a2}$] S^2- + H+} && K_\mathrm{a2} = \frac{[\ce{S^2-}][\ce{H+}]}{[\ce{HS-}]} \tag {R5} \\
\ce{H2S &<=>[K_\mathrm{d}] 2 H+ + S^2-} && K_\mathrm{d} = K_\mathrm{a1} K_\mathrm{a2} \tag{R6}
\end{align}

and also agree on labeling $C_\mathrm{A}$ total concentration of sulfide anions in solution in form of $\ce{S^2-}$, $\ce{HS-}$ and $\ce{H2S}$; $\alpha_2$ – fraction of $\ce{S^2-}$ among those; $s$ – solubility. Then from (R1)

$$K_\mathrm{sp} = [\ce{Mn^2+}] C_\mathrm{A} \alpha_2 \implies s = \sqrt{\frac{K_\mathrm{sp}}{\alpha_2}} \tag{1}$$

$\alpha_2$ can be expressed via known equilibrium constants and $\mathrm{pH}$:

\begin{align}
\alpha_2 &= \frac{[\ce{S^2-}]}{[\ce{S^2-}] + [\ce{HS-}] + [\ce{H2S}]} &&\Bigg| \cdot \frac{[\ce{H+}][\ce{HS-}]}{[\ce{H+}][\ce{HS-}]}\\
&= \frac{K_\mathrm{a2}}{K_\mathrm{a2} + [\ce{H+}] + \frac{[\ce{H2A}][\ce{H+}]}{[\ce{HA-}]}} &&\Bigg| \cdot \frac{K_\mathrm{a1}}{K_\mathrm{a1}} \\
&= \frac{K_\mathrm{a1}K_\mathrm{a2}}{K_\mathrm{a1}K_\mathrm{a2} + K_\mathrm{a1}[\ce{H+}] + [\ce{H+}]^2} \tag{2}
\end{align}

Since hydrolysis might occur partially (R2) or completely (R3), in order to find$[\ce{H+}]$ from

$$[\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{OH-}]} \tag{3}$$

let's estimate whether the equilibrium in (R3) is going to be shifted left or right. Solubility of sulfide anions in the absence of hydrolysis is

$$[\ce{S^2-}] = \sqrt{K_\mathrm{sp}} = \sqrt{\pu{2.5e-10}} = \pu{1.6e-5 M} \tag{4}$$

whereas solubility when $\ce{H2S}$ is forming (from (R6)) would be

$$[\ce{S^2-}] = \sqrt[3]{\frac{K_\mathrm{a1} K_\mathrm{a1}}{4}} = \sqrt[3]{\frac{10^{-7} \cdot 10^{-10}}{4}} = \pu{1.4e-6 M} \tag{5}$$

so that formation of less soluble product would indeed favor (R3). Hence, in order to estimate the increase in $\mathrm{pH}$, it's better to use $K_\mathrm{h2}$:

\begin{align}
K_\mathrm{h2} &= [\ce{Mn^2+}][\ce{OH-}]^2[\ce{H2S}] && \Big| \cdot \frac{[\ce{H+}]^2[\ce{S^2-}]}{[\ce{H+}]^2[\ce{S^2-}]} \\
&= \frac{K_\mathrm{sp}K_\mathrm{w}^2}{K_\mathrm{a1}K_\mathrm{a2}} \tag{6}
\end{align}

On the other hand,

$$K_\mathrm{h2} = [\ce{Mn^2+}][\ce{OH-}]^2[\ce{H2S}] = 4[\ce{OH-}]^4 \tag{7}$$

Equating (4) and (5):

$$[\ce{OH-}] = \sqrt[4]{\frac{K_\mathrm{sp}K_\mathrm{w}^2}{4 K_\mathrm{a1}K_\mathrm{a2}}} = \sqrt[4]{\frac{\pu{2.5e-10} \cdot (10^{-14})^2}{4 \cdot 10^{-7} \cdot 10^{-10}}} = \pu{5e-6 M} \tag{8}$$

Now back to (3):

$$[\ce{H+}] = \frac{10^{-14}}{\pu{5e-6}} = \pu{2e-9 M} \tag{9}$$

Knowing how the $\mathrm{pH}$ of the system has been changed, let's find what fraction of sulfide anions are in solution:

$$\alpha_2 = \frac{10^{-7} \cdot 10^{-10}}{10^{-7} \cdot 10^{-10} + 10^{-7} \cdot \pu{2e-9} + (\pu{2e-9})^2} = 0.0467 \tag{8}$$

Finally, solubility can be determined:

$$s = \sqrt{\frac{\pu{2.5e-10}}{0.0467}} = \pu{7.3e-5 M} \tag{10}$$

e.g. taking hydrolysis into account, solubility of $\ce{MnS}$ is approx. $4.5$ times higher (see equations (4) and (10)).


Why is the bond dissociation energy of C-H bond higher than that of a N-H bond?

If you look at an orbital diagram of nitrogen you will see that each p-orbital has one electron in it, if you want to add a second electron to any one of them, it makes the orbital higher energy. Atoms would generally like to be in the lowest possible energy state, and half or completely filled shells are far more low energy, and thus, more stable, this is most clearly seen with the electron configuration of copper and chromium.

Cu: [Ar] 4s1 3d10

Cr: [Ar] 4s1 3d5

Nitrogen has a half filled 2p sub level, so it has to move to a substantially higher energy state to add a fourth electron. If it gets one in a bond with Hydrogen, Nitrogen finds it energetically cheaper to dissolve the bond and return to its base configuration. The reverse applies to Carbon. Because it now is able to get another electron from Hydrogen, it now has a half filled 2p sub-level and is in a fairly stable configuration that it would like to keep.

Most of this comes from Chemistry by Maciel, Traficante, and Lavallee. The book was published in 1978, so I am sorry if anything is dated.


Why is use of calcium carbide as an artificial fruit ripening agent banned?

According to Anwar et al. the use of $\ce{CaC2}$ is discouraged due to

  dangers of explosion and carryover of toxic materials like arsenic and phosphorus to consumers, thus making the healthy fruit poisonous. Since no technical knowledge is considered necessary for its anomalous use, higher quantity of calcium carbide needed to ripen immature fruit, makes them
  tasteless.

For references and more information, see: Anwar, R., A.U. Malik, M. Amin, A. Jabbar and B.A. Saleem, Int. J. Agri. Biol. 2008, 10, 35–41. Available at researchgate.net.


What is the electrical conductivity of pool water?

The pH should be around 8.4 and the conductivity in the range of 2.5 - 3 mS/cm. The lower value is appropriate for freshly prepared water. The conductivity is used for the determination of TDS (total dissolved particles).


Not a chemist, needs help with a calculation regarding addition of HCl to sodium formate solution

At $\mathrm{pH}=\mathrm{p}K_\mathrm{a}=3.75$ , there will be 50% of non-dissociated formic acid and 50% of formiate.

You need 50 % of stoichiometric amount of hydrochloric acid to convert 50% of the formiate to the acid.

If we consider the percentage as (w/w) %,  then $\pu{1 kg}$ of 75% formic acid ($M=\pu{46
g/mol}$) contains $\pu{16.3 mol}$ of formic acid.

For usage of $\pu{5 kg}$ to $\pu{1000 L}$, $\mathrm{pH}=\frac 12 \left(\mathrm{p}K_\mathrm{a}-\log(c)\right)\\
=\frac12\cdot \left(3.75-\log{\frac{5\cdot 16.3}{1000}}\right)\\
=\frac12\cdot \left(3.75-\log{0.0815}\right)=2.42$

For $\pu{10 kg}$, it is 2.27.

$\pu{1 kg}$ of the above formic acid  is equivalent to $\pu{ 16.3 mol}$ of sodium formiate, i.e. $\pu{ 1108 g}$

For 50% conversion, you need $\pu{ 8.15 mol}$ of $\ce{HCl}$, i.e. $\pu{ 8.15\cdot 36.5 = 297 g}$ $\ce{HCl}$, what is equivalent $\pu{ 804 g}$ of 37% $\ce{HCl}$


Monosaccharides configurations (alpha beta, D L) identification

It’s easy to understand your frustrations, however you need to understand the basic stereochemistry and conformations of sugars.

A monosaccharide is designated D if the hydroxyl group on the highest numbered asymmetric carbon is drawn to the right in a Fischer projection, as in D-glyceraldehyde:

Note that the designation D or L merely relates the configuration of a given molecule to that of glyceraldehyde and does not specify the sign of rotation of plane-polarized light.

As I covered this explanation already on one of my posts:Are glucose and galactose cis-trans isomers of each other?

  D sugars have the same absolute configuration at the asymmetric
  center farthest removed from their carbonyl group as does
  D-glyceraldehyde. The L sugars, in accordance with this convention,
  are mirror images of their D counterparts

Cyclic forms

The designation α indicates that the hydroxyl group at the anomeric center (hemiacetal or hemiketal carbon)  is, in a Fischer projection, on the same side as the hydroxyl attached at the farthest chiral center, whereas  β indicates that these hydroxyl groups are on opposite sides:

In the α anomer, the $\ce{OH}$ substituent to the anomeric carbon is on the opposite side of the sugar ring from the $\ce{CH2OH}$ group at the chiral center that designates the D or L configuration ($\ce{C5}$ in hexoses). The other anomer is known as the β form.

The Haworth projections represent the stereochemistry of sugars more realistically than do the Fischer projections, which is why they are preferred:

For a D sugar, any group that is written to the right of the carbon in a Fischer projection has a downward direction in a Haworth projection; any group that is written to the left in a Fischer projection has an upward direction in a Haworth projection.

The terminal $\ce{-CH2OH}$ group, which contains the carbon atom with the highest number in the numbering scheme, is shown in an upward direction.

  Similarly  in the  α-anomer, the hydroxyl on the anomeric carbon is on the opposite side of the ring from the terminal $\ce{-CH2OH}$ group (i.e., pointing down). In the β-anomer, it is on the same side of the ring (pointing up).

References

Lehninger Principles of Biochemistry
Voet and Voet biochemistry
Biochemistry Grisham
Biochemistry Campbell and Farrel


It sounds like you're needing help with chemical graph theory.

I'm no expert in this topic, but it is one that has interested me in the past. Hence I can recommend 'Topological Methods in Chemistry' by R. E. Merrifield and H. E. Simmons—a book which explicitly deals with chemical graph theory in chapters 3 and 4; various topological indices feature throughout.

As a complement to this book, I would recommend the book on Topology by Munkres. The above book isn't rigorous enough to approach this topic perhaps, and so a mathematical treatise is also needed. I'm currently working through these topics myself for my own research.

It doesn't quite answer your question, but I hope it helps in your lofty quest!


The Smith-Hieftje background correction is basically another means of "chopping" the signal. The amplifier locks onto the chopper frequency to cancel out the background noise. "Locking on" essentially means that the amplifier will measure the difference between the two signals.

So one way to chop the "light" from the emission tube would be to insert a "fan blade" to mechanically chop the light on and off.

The Smith-Hieftje background correction is a bit more complicated. Basically what happens is that this emission tube has two parts. A heater to vaporize the source of the atoms in the tube and an electron beam to stimulate an excited state. In this type of emission tube the heater is pulsed high and low. When the heater is low it produces enough metals atoms so that the atomic line emission is created. But when the heater is on high heat then it vaporizes so many atoms that the vapor cloud of metal atoms self-quenches the emitted atomic line emission. The high density cloud of atoms also means that the atoms bump into one another. So instead of the atoms in the cloud all having about the same kinetic energy they have a range of kinetic energies which results in line broadening.


Do hydrogen halides (HX) form strong or weak acids?

No, it can’t be generalised. Amongst the haloacids, only $\ce{HF}$ is weak.

Acid strength can be determined by the stability of the conjugate base (for most acids, the conjugate base is what you get when you remove $\ce{H+}$ from the acid). In this case the conjugate bases are the halides.

As the size of the halide ion increases, the charge is dispersed over a larger volume, and thus the ion is more stable. In this manner, one can say that the order of acid strengths is $\ce{HF<HCl<HBr<HI}$ (I’m not considering the radioactive ones here; they fit in the same trend though).

$\ce{HCl, HBr, HI}$ are all strong acids. $\ce{HCl}$ has a $\mathrm{p}K_\text{a}$ of −7 (the smaller the $\mathrm{p}K_\text{a}$ is, the more acidic it is), $\ce{HBr}$ is approximately −9, and $\ce{HI}$ is also approx −9.

On the other hand, $\ce{HF}$ is 3.2. The $\ce{H-F}$ bond is pretty strong, so it has a smaller tendency to dissociate—making it a weak acid (not to say that it isn’t dangerous).


How does changing the concentrations of intermediate products/reactants change the rate law?

The steady state approximation allows us to assume that the concentration of $\ce{B}$ does not change.

$$\dfrac{d[\ce{B}]}{dt}=0$$

Thus the rates of production and consumption of $\ce{CH(CN)2-}$ must be equal.

  $\ce{CH(CN)2-}$ can be produced by one pathway:

$$\ce{CH2(CN)2 ->[k_1] CH(CN)2- + H+} \\ \mathrm{rate}=k_1[\ce{A}]$$

  $\ce{CH(CN)2-}$ can be consumed by two pathways:

$$\ce{CH(CN)2- + Br2 ->[k_2] BrCH(CN)2 + Br-} \\
\ce{CH(CN)2- + H+ ->[k_{-1}] CH2(CN)2} \\
\mathrm{rate}=k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}]$$

  Thus:

$$\dfrac{d[\ce{CH(CN)2-}]}{dt}=k_1[\ce{CH2(CN)2}] -\left( k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}] \right)= 0$$

Now you can can solve for $[\ce{CH(CN)2-}]$ and substitute into the rate law equation for the second step:

$$\dfrac{d[\ce{BrCH(CN)2}]}{dt}=k_2[\ce{CH(CN)2-}][\ce{Br2}]$$

You should get a final rate law that contains $[\ce{CH2(CN)2}],\ [\ce{Br2}],\ \&\ [\ce{H+}]$, and then you can answer your questions.

  Update based on revised question.

Your new rate late is correct:

$$\dfrac{d[\ce{BrCH(CN)2}]}{dt}=\dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_2[\ce{Br2}]+k_{-1}[\ce{H+}]}$$

Your interpretation of the simplification you can do is correct for one case and almost correct for the other.

  $[\ce{H+}] \ll [\ce{Br}]$

In this case, we assume that $k_{-1}[\ce{H+}]\ll k_2[\ce{Br}]$ so

$$k_2[\ce{Br2}]+k_{-1}[\ce{H+}] \approx k_2[\ce{Br2}] \\
\therefore \
\dfrac{d[\ce{BrCH(CN)2}]}{dt}\approx \dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_2[\ce{Br2}]}\approx k_1[\ce{CH2(CN)2}$$

which is as you figured.

    $[\ce{H+}] \gg [\ce{Br}]$

In this case, we assume that $k_{-1}[\ce{H+}]\gg k_2[\ce{Br}]$ so

$$k_2[\ce{Br2}]+k_{-1}[\ce{H+}] \approx k_{-1}[\ce{H+}] \\
\therefore \
\dfrac{d[\ce{BrCH(CN)2}]}{dt}\approx \dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_{-1}[\ce{H+}]}$$

Note that this approximation does not remove the $k_2[\ce{Br2}]$ term in the numerator.


Smelly matters in urine - do they have lower or higher boiling point than water?

There are several reasons that a smell vanishes.

One obvious reason is that a smelly component has a very low boiling point and simply evaporates. Another reason might be a chemical reaction of the "stinker" when exposed to air, which converts it to something untraceable.

On the other hand, other reactions might result in quite the opposite but I rather doubt that you want to perform a test run with open bowls of urine over the next week at home ;)

Here is an older article from 1971 on the gaschromatographic analysis of volatile organic compounds in human urine after extraction with diethylether.

On a first view, however, the ketones that they found usually have a rather nice smell.

In order to get a hand on the components in the vapour phase above the urine, others have performed a headspace GC-MS analysis, see this application note
and this article.

Again, the ketones usually have a pleasant smell, but they also identified some of the highly volatile bad boys, such as trimethylamine, methanthiol, and dimethylsulfide. These have boiling points of ~5, ~6 and ~40 °C, respectively.


What's the most complete basis set for calculating chemical shifts with DFT?

Using pcS-4 only tells you how much the error from your basis set is. This doesn't necessarily have any link to the total error, which comes from basis set, DFT functional, molecular geometries, treatment of solvation, and various other aspects of your chosen method. So, if all you want to do is to investigate the effect of basis set while keeping everything else the same, then sure, go ahead and use the pcS-n basis sets. They were designed with convergence to the complete basis set limit in mind, so pcS-4 is probably as close to the CBS limit as you can reasonably get. For a recent (and very thorough) example of benchmarking using pcS-4 (technically the segmented version pcSseg-4), see J. Chem. Theory Comput. 2018, 14 (2), 619–637.

If your aim is just to get the computed shifts as accurate as possible, then CCSD(T) with large pcS-n or cc-pVnZ (or even extrapolation to the CBS limit) is probably what you want. For benchmarking you're not likely to be using any excessively big molecules anyway, so for a one-off calculation I imagine the computer time needed is quite tolerable.

As far as DFT is concerned, as has already been pointed out, if you don't use a good functional then no matter what basis set you use, the shifts will not be great anyway. Hybrid functionals like PBE0 or mPW1PW91 are popular in the literature (and B3LYP of course), but there are some other interesting ones like M06-L (see J. Phys. Chem. A 2008, 112 (30), 6794–6799), and recently double-hybrid functionals have been suggested (J. Chem. Theory Comput. 2018, 14 (9), 4756–4771).

Aside from that, there are also empirical ways of increasing the accuracy of DFT-predicted shifts, which you may or may not want to take into account when benchmarking. A popular one is linear scaling of shifts (the slope and intercept being determined from a given test set). Tantillo has a nice collection of these scaling factors at http://cheshirenmr.info/index.htm. Although some of the work cited can be a bit outdated in such a rapidly-moving field, the methodology itself is still highly relevant.


In general the cooling water is attached according to the countercurrent exchange mechanism to maximize the efficiency. Of course this doesn't work well for the Liebig condenser you pictured, but you generally wouldn't use this kind of condenser for reflux anyway. The most common condenser for reflux is the Dimroth condenser which can be used for countercurrent exchange without any problems as the entry and exit of the water are both at the top of the condenser.

Here is an image from Arbeitsmethoden in der organischen Chemie (page 43) that shows where to attach the cooling water for different condenser types.


First, these Coulombic interaction integrals can be simplified to $ \langle \mu\nu|\frac{1}{r_{12}}|\lambda \sigma \rangle $, which is the common two electron integral in the AO basis (your atom-centered gaussian basis functions).
You are correct that energies are only well defined for eigenfunctions of the Hamiltonian, and thus basis set functions (the choice of which is arbitrary) do not have a well-defined energy. Therefore, the most straightforward workaround for this is to calculate the two electron integrals in the AO basis, forming the rank 4 tensor of two electron integrals, $g(\mu,\nu,\rho,\sigma) = \langle \mu\nu|\frac{1}{r_{12}}|\lambda \sigma \rangle $. Then, you must (approximately) solve the Schrödinger equation with some level of electronic structure theory (e.g. CCSD, MP2, etc.) to solve for the molecular orbital coefficient matrix, C. The a$^{th}$ column of the matrix C is the weightings of the AO basis (atom-centered gaussian basis functions) to give the a$^{th}$ molecular orbital with well-defined energy $\epsilon_a$.
Finally, you should rotate $g(\mu,\nu,\rho,\sigma)$ with the coefficient matrix C to yield $g(a,b,c,d)$:
$$g(a,b,c,d)=C_{\mu a}C_{\nu b}g(\mu,\nu,\rho,\sigma)C_{\rho c}C_{\sigma d}$$
Note that, in this form, this change of basis scales as $N^8$, where N is the number of basis functions used, but can be implemented efficiently (by doing one rotation at a time) to scale as $N^5$. See the Crawdad programming tutorial on MP2 (site seems to be down right now) for more discussion on tensor rotations. Although it was done for MP2, it works identically for you.


Why does the solubility of Group II hydroxides increase and the solubility of sulphates decrease down the group?

Okay, for this scenario, the two concepts that play a key role are the lattice enthalpy and the hydration enthalpy.
Lattice enthalpy is the energy released when one mole of a compound is formed from its constituent gaseous ions under standard conditions. Lattice enthalpy is dependent on the length of the ionic bond between the cation and the anion in the compound. If we look at Coulomb's law, we find that the attractive force between two bodies with different charges is inversely proportional to the square of the distance apart between their centres of mass, i.e. $F\propto 1/r^{2}$. A shorter bond would require more heat energy to be broken. Therefore, the shorter the bond, the higher the lattice enthalpy.

Hydration enthalpy is the energy released when one mole of gaseous ions is completely hydrated by water molecules under standard conditions. If two ions of different sizes were surrounded by an equal number of water molecules each, the smaller ion would be more hydrated compared to the larger one due to the size differential. Therefore, the smaller the ion, the higher the hydration enthalpy.

Let's deal with $\ce{OH-}$ first. While progressing down Group 2, the size of the cation increases since the number of shells increases to accommodate to the extra electrons. Due to this, the distance between the centres of mass of the cation and the anion increases and by definition, the lattice enthalpy decreases. The hydration enthalpy also decreases since the size of the cation increases. However, due to the square factor, the lattice enthalpy decreases faster than the hydration enthalpy. This is why the solubility of Group 2 hydroxides increases while progressing down the group.

Now let's look at $\ce{SO4^2-}$. The same thing applies to the cation while progressing down the group. However, the $\ce{SO4^2-}$ ion is shaped in such a way that the change in the distance between the centres of mass of the ions is far less significant compared to that in the case of Group 2 compounds containing $\ce{OH-}$ ions. Therefore, the lattice enthalpy decreases much slower when progressing down the group compared to the case of Group 2 $\ce{OH-}$ compounds. Again, the hydration enthalpy decreases the same way as it does in the case of Group 2 cations bonded to OH⁻ ions. Since the hydration enthalpy decreases faster than the lattice enthalpy in the case of Group 2 sulphates, the solubility of Group 2 sulphates decreases while progressing down the group.


Why is the para carbon in Benzaldehyde NMR more deshielded than the ortho and meta carbons?

Resonance effects come into play along with the inductive effects.  I've drawn a few of the resonance structures for benzaldehyde below, notice how resonance effects place some positive charge on carbons 2, 3 and 5 which results in their deshielding.  I might have guessed that the combination of resonance and inductive effects would place C2 below C3, but that's not the case.  At least C5 is at lowest field and C2 and C3 are close and both are below C1. The interplay of inductive and resonance effects is complex and estimating the magnitudes of these effects can get tricky.


Why does ozone have such a large solubility in water?

If you look at the molecular structure of ozone below, you can see that it is polar.  Diatomic oxygen is non-polar.  This mild polarity of ozone greatly enhances the solubility of the molecule in a polar solvent like water.


The effects of adding neutrons to an atom

Even if such a phenomenon were to occur, and the resulting nuclei were sufficiently stable to participate in chemistry (they would not be sufficiently stable, in reality), the "inflation" of the charge of the nucleus would almost certainly have a negligible effect on the chemistry.  The nucleus is really really really tiny with respect to the length scales of bonds and the electron cloud.

To give an idea of the scales involved: the largest stable element, uranium, has a nuclear radius of about fifteen femtometers $\left(15~\mathrm{fm}\right)$, per Wikipedia, or $1.5\times 10^{-14}~\mathrm m$.  Bonds and electron clouds are measured in Angstroms $\left(10^{-10}~\mathrm m\right)$.  Thus, the nuclear radius would have to expand by well over $\sim\!\!100$-fold in order for the assumption of a point-charge nucleus to begin to fail.

It is well accepted that the effect of isotopic substitutions on chemistry primarily is due to the change in mass of the nucleus, leading to the kinetic isotope effect.  Another potential isotope effect on reactivity involves the change in nuclear spin, leading to a "magnetic isotope effect" only relevant for certain classes of reactions between species with unpaired electrons (e.g., organic radicals and/or various inorganic species).

Isotopic substitution does affect the behavior of nuclei in certain spectroscopies, however, leading to isotopic shifts.  The isotopic mass change can have noticeable effects on the vibrational spectrum, due to the effect on the local distribution of mass. As described in the link to ADF's documentation provided by Aesin in a comment to the question, these effects can also be seen in the hyperfine structure of atomic spectra.  As well, the NMR isotopic shift of light elements, especially hydrogen, can sometimes be observed.


To identify hydrogen bond donors: Look for a polarized hydrogen (typically bonded to O, N, S)
Hydrogen bonds form from the interaction of a partially positively-charged hydrogen and a (partially) negatively-charged atom. Thus, to "donate" a hydrogen for hydrogen bonding, you need a polarized hydrogen. Hydrogens are polarized by being bonded to electronegative elements: in organic molecules that will typically be O, N, or S.

To identify hydrogen bond acceptors: Look for atoms with electron lone pairs, or otherwise partially negatively-charged sites on the molecule. (Typically O, N, S, or halides).
To form an interaction with a polarized hydrogen, you need a source of (partial) negative charge, which typically can come from electron lone pairs. The most common species you'll encounter in an organic molecule include O, N, S, or halides (F, Br, Cl, I).


If the helium is expanding isothermally against a constant external pressure $P_{ext}$, the work is calculated as
$$w = -P_{ext}(V_2 - V_1)$$
This process is irreversible. The first expression you gave for work is (sort of) the expression for work for a reversible isothermal ideal gas expansion.

If helium is assumed to be an ideal gas, then $\Delta U = 0$ because the process is isothermal. You have $q = -w$ (and not zero) for this process. $\Delta H$ will be zero too, since $$\Delta H = \Delta U + \Delta(PV) = \Delta U + nR\Delta T = 0$$.


Is Geologic Mineral Trapping of Chlorine Plausible?

One of the problems of geological sequestration of chlorine is that most chloride minerals are very soluble.

The number of non-soluble (or very little) minerals is limited, but there are several options. The obvious ones are sodalite ($\ce{Na8(Al6Si6O24)Cl2}$) and chlorapatite ($\ce{Ca5(PO4)3Cl}$). There are other options though.

Webmineral lists a lot of chlorine-bearing minerals, and most of those are soluble. However, there could be several minerals with high percentages of chlorine that are insoluble enough under the conditions of your reservoir. But in general, your best bet are anhydrous silicates and phosphates, because those are the most likely to be stable over a long time.

That said, this stuff could be dumped in the ocean. The $\ce{HCl}$ could be neutralised with something like feldspars, turning it into clay and a solution of harmless sodium chloride.

Alternatively, you can oxidise the HCl and make chlorine gas, a useful product.


With molecular formula, how do I know if it linear or one element surrounded by other elements?

The formula is insufficient. Consider pentane, $\ce{C5H12}$, which occurs in straight and branched isomers, the extreme case being neopentane, in which the central carbon is linked only to four methyl groups.


Molecular dynamics after modeling a protein loop. Necessary?

To expand on Mithoron's comment, it's absolutely necessary to equilibrate (or at least minimize the energy) of your system once you've introduced an appreciable perturbation of any sort (in this case, you've appreciably changed the system itself!). To give you an idea, some lesser changes that would still justify equilibration are: changes in solvation, temperature, pressure, protonation of amino acids, oxidation state of cofactors, etc.

Note: by equilibration, I mean letting your system relax in the presence of surroundings. By minimization, I am referring to minimizing the total energy of the protein (explicitly solvated, unless you're doing some odd gas-phase or implicit solvation) in the absence of temperature and pressure (0 K, no thermostat or barostat).


Possible methods for copper (II) reduction to copper (I)

There are several ways to make copper(I) oxide from copper:-

Heat it

$$\ce{4Cu + O2 ->[\Delta] 2Cu2O}$$

Copper react with oxygen to produce copper(I) oxide. This reaction takes place at a temperature of over 200°C.

Heat it with nitrous or nitric  oxide

$$\ce{2Cu + N2O->[\Delta] Cu2O + N2}$$
$$\ce{4Cu + 2NO->[\Delta] 2Cu2O + N2}$$

Copper react with nitrous and nitric oxide to produce copper(I) oxide and nitrogen. Both reaction takes place at a temperature of 500-600°C.

Heat it with copper(II) oxide

$$\ce{Cu + CuO->[\Delta] Cu2O}$$

Copper react with copper oxide to produce copper(I) oxide. This reaction takes place at a temperature of 1000-1200°C.

If you made copper (II) oxide, just heat it

$$\ce{4CuO ->[\Delta] 2Cu2O + O2}$$

The thermal decomposition of copper oxide to produce copper(I) oxide and oxygen. This reaction takes place at a temperature of 1026-1100°C.

Since you are doing this at home, method 2,3,4 cannot be used as it requires laboratory grade heater but method 1 can be used since it require 200°C which is achievable.

But the best method(according to the comment) I would recommend is the Fehling's test. All you need is Fehling's solution and glucose which can be purchased from sigma-aldrich at affordable price.

$$\ce{RCHO + (2Cu^2+ + 5OH^{-})->Cu2O + RCOO^- + 3H2O}$$


Is this analogy explaining the difference between enantiomers and diastereomers correct?

Enantiomers are non-superposable mirror images (all the signs of one spatial coordinate are reversed).  Diastereomers are  formed by inverting some but not all stereocenters.

http://osf1.gmu.edu/~bbishop1/CHEM%20814-579%20Stereochemistry%20Lecture%20slides.pdf
 pp. 11-13.  Excellent overall.

Have some fun with your prof!  An acyclic, aliphatic, unstrained, tetrahedral carbon atom is $\ce{sp^3}$-hybridized.  It is an uncharged, valence-saturated singlet.  It bears four freely rotating, rigorously identical substituents, e.g., $\ce{CR4}$, with identical bond lengths to that central carbon.  Said carbon is a chiral center.  It has no ${S_n}$ improper axes of symmetry.

1) Show how this can be true.
2) Use any system of nomenclature to name left-handed and right-handed stereoisomers.

Acyclic, unstrained, etc., trigonal planar $\ce{sp^2}$-hybridized carbons ${R2>C=C<R2}$ are each identical stereocenters.  Each carbon and the olefin overall have no ${S_n}$ improper axes of symmetry.  The olefin is resolved chiral.  (1) and (2) as above.


Does 6-methyl-2-(methylamino)pyrimidin-4-ol show aromaticity?

Each Nitrogen has 5 valence electrons. They are using 2 of those in sigma bonds to their neighbouring atoms, 2 in an sp2 lone pair in the plane of the ring, leaving only 1 electron to contribute to the delocalised pi system.

Each carbon is also contributing 1 electron to the delocalised pi system, and has 3 involved in sigma bonds.

This means each member of your 6 membered ring is contributing 1 electron, 6 in total. This means it is aromatic, as all 3 bonding orbitals are fully occupied.


The thermodynamic equilibrium constant $K$ (different from both $K_p$ and $K_c$) and the reaction quotient $Q$ are defined with reference to a standard state. When all reactants and products are at standard state, $Q = 1$. $K$ and $Q$ are related to the Gibbs energy of reaction via:
$$\Delta_\mathrm{r}G = RT \ln\frac{Q}{K}$$ No matter which standard state you use, this relationship gives the same Gibbs energy of reaction (because $K$ and $Q$ are equally affected). However, depending on which standard state is used, $K$ and $Q$ will have different numeric values.

According to wikipedia,

  In chemistry, the standard state of a material (pure substance, mixture or solution) is a reference point used to calculate its properties under different conditions. In principle, the choice of standard state is arbitrary, although the International Union of Pure and Applied Chemistry (IUPAC) recommends a conventional set of standard states for general use.

The recommended standard state for a gas is, again according to wikipedia,

  the hypothetical state it would have as a pure substance obeying the ideal gas equation at standard pressure ($\pu{e5 Pa}$, or 1 bar). No real gas has perfectly ideal behavior, but this definition of the standard state allows corrections for non-ideality to be made consistently for all the different gases.

Because neither the temperature nor the pressure is given in the problem the OP asked about, you can't use this standard state. Instead, you can use the standard state for a solute (still wikipedia):

  For a substance in solution (solute), the standard state is the hypothetical state it would have at the standard state molality or amount concentration but exhibiting infinite-dilution behavior. The reason for this unusual definition is that the behavior of a solute at the limit of infinite dilution is described by equations which are very similar to the equations for ideal gases. Hence taking infinite-dilution behavior to be the standard state allows corrections for non-ideality to be made consistently for all the different solutes. Standard state molality is 1 mol kg−1, while standard state amount concentration is 1 mol dm−3.

What is the problem with this problem?

  Am I wrong in the way I am approaching this problem, or is the problem itself wrong?

The problem is incomplete. If they use a funky standard state, they should say it explicitly.


Is there a catalyst that will reduce an alcohol to an alkane?

Here is a method for converting secondary and tertiary alcohols into the corresponding alkane using $\ce{InCl3}$ as a catalyst.  Chlorodiphenylsilane is the only reagent used in addition to the catalyst. The method works in the presence of common functional groups leaving them untouched.

Since primary alcohols are unreactive, this reaction can be used to selectively reduce secondary and tertiary alcohols in their presence.  An abstract showing a number of examples of the reaction can be found here.


Reimer–Tiemann-like reaction of catechol with diiodomethane

This is not really an answer to your question, but I do not think that it is possible to answer it. From one of the many comments we learn that the reference is actually the "preparation of piperonal". It uses 3-4-dihydroxy-benzenecarbaldehyde instead of catechol, but the reaction is same, with $\ce{CH2I2}$ and $\ce{NaOH}$.
I was able to find this reaction described in Chemistry of Natural Products
by Sujata V. Bhat, B.A. Nagasampagi, Meenakshi Sivakumar on pages 307f.

4.22.B. Synthesis of Piperine
Piperine was synthesized (Ladenburg, 1894) by the reaction of the piperic acid chloride with piperidine, which confirmed the structure of the molecule (Figure 4.57). The synthesis of piperic acid was achieved starting from piperonal 4.288, which was obtained from catechol using Reimer-Tiemann reaction followed by the condensation with diiodomethane in the presence of a base. Piperonal was condensed with acetaldehyde in the presence of sodium hydroxide and the product obtained was then heated with acetic anhydride and sodium acetate to yield piperic acid.

I find this a lot more in accord from what you would expect. I cannot imagine any way in which a Reimer-Tiemann (RT) like reaction would occur given the starting reagents. Instead the first step from catechol to 2,4-dihydroxy-benzenecarbaldehyde occurs via an RT reaction. This is very well discussed by H. Wynberg (Chem. Rev. 1960, 60 (2), pp 169–184, DOI: 10.1021/cr60204a003). The mechanism can also be found on Wikipedia.
The second step then involves diiodomethane and forms the acetal. I suppose after deprotonation of the phenolic hydrogen, it can attack at the diiodomethane in nucleophilic substitution fashion, further following another deprotonation and an intramolecular SN step to close the ring.

Unfortunately I could not find anything to further back up my suspicion, since the original synthesis only starts at 4.288. See here (in German): A. Ladenburg,  and M. Scholtz, Ber. Dtsch. Chem. Ges. 1894, 27, 2958–2960. DOI: 10.1002/cber.18940270356.
However, since there is proof that the reaction with 2,4-dihydroxybenzenecarbaldehyde forms the acetal, it would be highly suspicious if it were not doing a similar reaction with catechol.


Boiling point elevation for a mixture of ethanol and water

Are you sure that is the right equation to answer the question? It only holds true for non-volatile compounds. You should be able to use Raoult's Law. Perhaps this helps http://www.chemguide.co.uk/physical/phaseeqia/idealpd.html


Absorption Spectrum Using a Spectrophotometer

This is simple mathematical conversion. Firstly, you find the fraction of each absorbance value.

Let, the absorption spectrum has a three absorbance values say $A_1, A_2$ and $A_3$.

Now, find the fraction of each absorbance as per following:

fraction of $A_1 = A_1/(A_1+A_2+A_3)$

fraction of $A_2 = A_2/(A_1+A_2+A_3)$

fraction of $A_3 = A_3/(A_1+A_2+A_3)$

Now convert all the respective fraction values into percentage as per:

percentage of $A_1 = (\text{fraction of }A_1 \cdot 100)\%$

percentage of $A_2 = (\text{fraction of }A_2 \cdot 100)\%$

percentage of $A_3 = (\text{fraction of }A_3 \cdot 100)\%$


What white powdery compound turn black immediately after heating?

The answer to this question partially answers yours.Let me speculate. The compound could be some crystallized organic compound, making it appear white. On heating, it degrades and becomes solid carbon, which is black, giving of some carbon dioxide and water vapour. It wouldn't have any sulfurous groups, as it would give off sulfur dioxide.


Why do carboxylic acid derivatives show a more upfield shift than carbonyls in 13C-NMR?

Shouldn't the carbon in the carboxylic acid be more deshielded than the analogous carbonyl, since there are two oxygens instead of one?

Well, no. I love this misconception and I'm always having fun when doing oral exams about reduction reactions and grignards, which is basically the same problem.

If you look at reactivity of acids/esters and ketones/aldehydes against nucleophilic attack (grignard reagents or hydride reagents) you will see that ketones and aldehydes are much more reactive towards nucleophiles than acids/esters are. So why are ketones and aldeyhdes more electron deficient on the carbonyl carbon?

Yes, in case of acid/ester you got two oxygens both with electron withdrawing -I properties, but the additional oxygen also got lone pairs so a +M effect is possible. This actually pushes more electron density towards the carbon than the -I is pulling away leaving the carbonyl carbon in acids and esters overall more electron rich than in aldehydes and ketones.


Is it possible to remove a proton from the atom via some technique
  such as bombardment with another particle or applying an appropriate
  amount of energy to the system?

Yes, this occurs in nature and can also be done in the laboratory.  In nature the process is known as radioactive decay.  Listed below are two decay processes that will remove a proton(s)

alpha decay: an atom ejects a helium nucleus (2 protons and 2 neutrons)
$$\ce{^239_94Pu -> ^4_2He + ^235_92U}$$
beta-plus decay: a positron is emitted and a proton is converted to a neutron.

In the (nuclear) lab, a process known as phototransmutation can be employed.  If a nucleus is irradiated with high-energy gamma rays (high-energy photons) it can absorb the energy and change to another element by ejecting a proton and\or neutron
$$\ce{^25_12Mg ->C[{h\nu}]\   ^1_1H + ^24_11Na}$$

You can make a rough estimate of the energy required as follows:
$$\mathrm{24.98583692\,amu~ \longrightarrow 1.00727647\,amu + 23.99096278\,amu}$$
comparing the starting and final mass, we find that 0.01240233 amu has been gained in the process.  Since 1 amu = 931.5 MeV, the gamma ray photon would have to have an energy of at least 11.6 MeV.


Are molecules chiral if a rotation leads back to the same compound?

The property that you are describing for D-idaric acid is called C2 symmetry. This means that when you rotate a molecule (or any object) by 180 degrees, it appears identically to the original orientation. In D-idaric acid, the C2 symmetry axis is halfway along the bond between C3 and C4.

Having a C2 symmetry axis doesn't tell us if the compound is chiral or not. Achiral compounds have an (internal) plane of symmetry or a point of symmetry. D-Idaric acid does not meet either of these requirements and is a chiral molecule.

Galactaric acid does not have a C2 symmetry axis but does have a plane of symmetry. Galactaric acid is achiral. Since galactaric acid has stereocenters but is not chiral, we describe it as meso.


L-arginine is the most basic amino acid since it carries guanidyl group in addition to the $\alpha$-amino group. The arginine molecule is a zwitterion with the guanidyl group, rather than the primary $\alpha$-amino group, accepting extra proton from carboxylic acid group (Ref.1). Protonated or not, when consider guanidine as a molecule, there shouldn't be any $\mathrm{sp^3}$ nitrogen in the molecule because it is resonance-stabilized molecule. As a guanidyl group, this principle stays the same and it stays flat. There are few excellent answers to similar question you could find in here. In one answer it states that:

  Guanidine is isolobal to urea, where the carbonyl oxygen has been replaced by an imine $\ce{NH}$. However, in principle it is still the same flat, resonance-stabilised molecule. The main difference is that there is no ‘preferred’ site for the double bond — it could point towards any of the three nitrogens in theory; you could say the diamide resonance is enhanced (Author: Jan).

And, accepted answer showed following diagram to depict the concept of resonance stabilization (Author: Ron):

Further, I'll attached recent crystal structure of L-arginine, which shows clear equivalent three nitrogen atoms with hydrohen bonding network (Ref.2):

  Two planes characterize the arginine molecule: one through the acid group and the other through the extended side chain, which contains the guanidyl group. The dihedral angle between these two planes is $74^\circ$ (Ref.1).

References:

I. L. Karle, J. Karle, “An application of the symbolic addition method to the structure of L-arginine dehydrate,” Acta Cryst. 1964, 17, 835–841 (https://doi.org/10.1107/S0365110X64002250).
Emilie Courvoisier, P. Andrew Williams, Gin K. Lim, Colan E. Hughes,  Kenneth D. M. Harris, “The crystal structure of L-arginine,” Chem. Commun. 2012, 48, 2761–2763 (DOI: 10.1039/C2CC17203H).


Combustion of non-hydrocarbon organic compounds

Simultaneous elemental analysis of organic carbon, hydrogen, and nitrogen is classically accomplished by combusting the material under a stream of $\ce{O2}$, usually in the presence of a catalyst to ensure complete combustion, at a temperature of 950-1050$^o$C.  Carbon, hydrogen, and nitrogen are quantitatively converted to $\ce{CO2}$, $\ce{H2O}$ and $\ce{N2}$, respectively.  This is a common method in particular for determining nitrogen concentration in amino acids.


Naming rules and acidity of Cl2CHCH2COOH

1- 3,3-dichloropropanoic acid. Please see http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/

2- The first one  is more acidic, as the inductive effect, due to chlorine atom decrease fast with the distance. (So, your answer is correct!).


It depends. if your pH is really high, silicate $\ce{SiO3^2-}$ or $\ce{SiO4^4-}$ ions will form and will become a stable dissolved solute in the solution. if it is moderately high the rate of hydrolysis will increase greatly and the silicon will crash out in the form of orthosilicate hydrates [$\ce{Si(OH)4}$], which may agglomerate to form silica particles. These results are a bit of a spectrum and there is no pH limit where the reaction gives a binary change in result, as it is also sensitive to extrinsic properties of the solution such as the amount of base.


After asking more people who already wrote reports with similar structures and a paper I found on the net the most convincing answer I could find was the term 'tetracapped'.

Source: 10.1002/zaac.201400143


Which free radical is more stable, allyl free radical or benzyl free radical?

Yes, generally more resonance structures mean more stability but this case is an exception.

In three of the four resonance structures of the benzyl radical, the aromaticity is being broken which makes then less stable whereas in the allyl radical both the canonical forms are stable.

Therefore the allyl radical is more stable than the benzyl radical.


The average atomic mass is useful because its numerical value is equal to the molar mass of the element. This in turn is useful to know how much of a solid to take when you wish to react it with some known quantity of another reagent, because weighing is typically the easiest way to quantify a substance.


Why does a weak acid and a salt containing its conjugate base form an acidic buffer?

The statement "a weak acid plus a salt containing the conjugate base forms an acidic buffer." is not correct. More properly, a mixture of equal amounts of a weak acid (but not too weak: with a pKa < 7) with its conjugate base (salt) does indeed give an acidic buffer. To get a basic buffer you would use equal amounts of an even weaker acid (with a pKa > 7) and its conjugate base (salt). Similar considerations apply to formation of buffers from a weak base plus a conjugate acid salt.


Why is the first energy of ionization of oxygen lesser than that of nitrogen?

Two electrons of the oxygen molecule are located on a $\pi $* level. They are relatively easy to ionize. It is not the case in $N_2$ : all electrons are located in binding orbitals.


Termolecular reactions almost never happen.  The chance of three molecules "colliding" at the same instant is so small as to be ignorable.

On the other hand there are "almost" termolecular reactions when one molecule hits another and "sticks" to it, and then a third molecule hits the two that are stuck together.  This most often happens when one of the molecules is part of a solid.  But this is really two sequential bimolecular reactions.


How is the pKa of extremely weak acids determined?

Standard pKa measurements

pKa is, as you mentioned, commonly measured in water. This worked well initially, but as chemists got more ambitious in their measurements, they began to come up against the levelling effect, whereby the acidity/basicity of the compound being measured was limited by the acid/base properties of the solvent being used to measure it.

To a certain extent, this was mitigated by moving from water to a solvent that is unchanged over a wider range of pKa (the range of pKa is known as the discrimination window), DMSO being the example chosen. This change to DMSO allowed the pKa of far more compounds to be measured. DMSO has been used to measure the pKa of LDA, the example you gave.

Even in cases where DMSO is unsuitable pKa may be measured in other solvents. There is however an issue with using different solvents to measure pKa in that the values are not directly comparable since different solvents affect the ease of protonation/deprotonation. The discrimination windows of several solvents are given below (Source: Wikipedia).

Standard pKa measurement is carried out by dissolving the compound in the solvent and titrating in acid/base slowly, whilst measuring the pH change, which can be done in many ways. This gives a curve from which the pKa may be read off (pKa is the point at which the compound is 50% ionised).

Measuring pKa of weak acids

Whilst pKa has been determined for many common chemicals, there are instances where pKa will be extremely high/low. In these situations, we can abandon the direct measurement as described above, and look for some other way of doing it.

-A brief interlude- To give an analogy that may (or may not) be more familiar. In medicinal chemistry, logP is an incredibly useful property. It was traditionally measured by partitioning the drug molecule in-between octanol and water and seeing how much of the compound ended up in each layer. This was laborious, and so medicinal chemists moved from directly measuring logP to carrying out other analysis from which logP could be derived. The common way of doing this is to use a column (similar to chromatography, but specially adapted). The chemist first runs a set of samples with known logP through the column, followed by the compound being studied. By looking at retention times and then comparing and doing some maths, a calculated logP can be obtained which is generally fairly reliable.

With pKa measurement, the situation is a similar in that we can indirectly measure the pKa and extrapolate.

There are, however, two separate issues here. A first problem is that pKa measurements at the extreme are just difficult (i.e. very small or very large) due to the titration curve becoming very difficult to measure accuratly, it isn't impossible however. This is really the main issue.

The other problem is related to levelling, but as you can see from the table above , we do have solvents available to measure over pretty much as wide a range as you could ever want (note:  if you're considering measuring a pKa in hydrofluoric acid, you should probably reconsider why you need to measure it in the first place).

The following paper (free to view) details the most common direct and indirect methods of determination.

From my experience, there are three methods I think are worth discussing here as alternatives to direct pH change measurement. . .

HPLC measurement in which the pH of the mobile phase is varied. As the pH changes, the compound may ionise, which changes its retention time on the column. This may be used to extrapolate pKa by comparison to samples of known pKa. The HPLC columns used now are able to tolerate the use of highly acidic/basic mobile phases for pKa determination.
NMR measurement in which we 'titrate' in the NMR tube, adding acid/base, and watching as the chemical shift changes. This has allowed some super accurate pKa's to be measured.
Comparative measurement in which a compound with a known pKa is reacted with our compound for analysis, and we see the extent to which they protonate/deprotonate one another. By looking at this equilibrium constant we essentially can derive the pKa.

Last, but not least...

Computational determination of pKa

Although not a 'measurement' of pKa, in silico computational methods are becoming increasingly accurate. This is especially true within a normal pKa range, but has also found applications with highly acidic/basic molecules. Some common packages for this are ACDpKa, Epik and Marvin.


I'd like to propose a dark, non-radical pathway :)

Pyrroles, supposed that they are not fully substituted, undergo electrophilic aromatic substitutions with halogens.

For the given starting compound, a complete bromination isn't possible due to the lack of a proton that would have to be eliminated in the final step.

However, the addition of an electrophile is conceivable. Addition of "$\ce{Br+}$" at C-2 of the starting material, usually the prefered position in the reaction of unsubstituted pyrroles, is a dead end here. Addition at C-3 yields an $\alpha$-enamine cation intermediate from which a proton might be eliminated. The resulting species is both

a double enamine and
an allylic halide

This intermediate might rearrange upon elimination and addition of a nucleophile $\ce{Br-}$ in a way that the aromaticity of the pyrrole is reestablished and bromine is introduced in the side chain.

Edit 1

Supposed that you are mainly interested in the hydroxyalkyl-substituted pyrrole as the target molecule, you might want to consider an alternative strategy. Let's have a second look:

The target compound (1) has an $\ce{OH}$ group in a side chain. Alcohols may be prepared by different routes, such as

reduction of the corresponding ketone
reaction an aldehyde with carbon nucleophile

Let's focus on the second idea. This would leave us with the 2-formyl pyrrole (2), that could be transformed into (1) by a Grignard reaction with $\ce{EtMgBr}$.
Again, several methods are conceivable for the synthesis of the aldehyde (2), such as

selective reduction of a $\ce{-CO2R}$ substituent (or reduction to $\ce{-CH2OH}$ and subsequent oxidation to $\ce{-CHO}$)
direct introduction of a $\ce{-CHO}$ substituent at carbon 2 of  pyrrole (3)

The latter seems promising and you will probably realize that this an application of the Vilsmeier-Haack reaction, where the reactive intermediate, the $\ce{CHO}$ synthon, is prepared in situ from N,N-dimethylformamide and $\ce{POCl3}$.

This would leave you with a strategy for the synthesis of (3), but that's another question.


Can sodium hypochlorite soak all the way through an unglazed ceramic cup and bleach blue paper wrapped around the outside of the cup?

It is not the ionic properties that control the diffusion. Diffusion is a fundamental property of molecules dissolved in a liquid. A bleach solution stored in an unglazed container can potentially diffuse to the wall. The pore size of the material will mainly control the rate of movement of the fluid (viscosity, surface tension etc.) If the pore size and the connectivity channels are very small it will take ages to wet the wall under atmospheric pressure.


What will be produced when sucrose and potassium bitartrate react with each other?

Rather than "cream of tarter" I'm going to stick to the pure chemical potassium bitartrate and tartaric acid, and I'm going to speak from the point of view of a chemist about the difference between potassium bitartrate and tartaric acid (rather than a cook or winemaker).

First a simple answer to your questions and then the details. Simplistically speaking, when you add potassium bitartrate to an aqueous solution of sucrose you will get some hydrolysis (producing glucose and fructose) and some ester formation. The key word here, however, is some -- and the amount depends strongly on the reaction conditions. At room temperature, for example, you're likely to get essentially no reaction within, say a day. More details below.

Tartaric acid is a dicarboxylic acid (a weak acid). Each of its carboxyl groups (-COOH) can donate a proton (H+ ion), and when this happens it's "acting as an acid". This is important to your question because it is the H+ ion that acts as a catalyst to sucrose hydrolysis and not the tartaric acid itself. For this situation, chemists call it a specific acid catalyzed reaction. By contrast, if a specific molecule such as tartaric acid were involved in the catalytic step chemists would call that "general acid catalysis".

Acidity of molecules are gauged by assigning a pKa value. If there are 2 acidic groups in the molecule (as with Tartaric acid) then it has 2 pKa values. Adding KOH to a solution of Tartaric acid will give you potassium bitartrate. At 25C, the pKa for removing Tartaric acid's first H+ is about 3.0 and the pKa for removing potassium bitartrate's H+ is 4.3. This makes both of them what chemist's call "weak acids" because they don't release anywhere near as much H+ into a solution as a strong acid (like HCl).

The other reaction, ester formation, is also specific acid catalyzed and so would also occur as I mentioned, but to a small extent. Esters are formed in acidic solutions whenever carboxyl (-COOH) and alcohol (-OH) groups are present, as is the case here. Note that there are many -OH groups here, on both molecules.

To understand the extent of each of these reactions it's important to consider the role of water. Water is a reactant in the hydrolysis reaction and this tends to drive the reaction forward. By contrast, water is a product of the ester formation reaction. This is especially important for the ester reaction since it is highly reversible. In other words, whenever molecules collide to form an ester, it's much more likely that that ester will collide with the far more abundant water molecules and revert back to the starting materials, resulting in no net reaction.

Usually one thinks of these reactions happening somewhere near room temperature. Under these conditions neither of the above reactions happens to a great extent; they are both generally very slow. In order to push them forward you have to do things like increasing the temperature, adding more concentrated reaction materials or catalysts, or removing reaction products. Of course, in this case increasing the concentration of the reaction materials also increase the amount of catalyst (H+), but since they are such weak acids, adding HCl (or another strong acid) would be much more effective.


Free radicals are usually very reactive.  If we bring two free radicals or a free radical and a molecule with electrons available for bonding together a bond will usually form.  Yes, it does cost energy to bring those electrons close together.  But look at the payoff, a new chemical bond is formed and this is usually quite exothermic (stabilizing), usually more than enough energy to offset the energy cost of bringing those electrons close enough to form a new bond.  Because of the exothermicity of bond formation the reaction profile usually has a small activation energy.  Translation: a small activation energy means a  fast reaction.  A caveat, at the outset I said that "free radicals are usually reactive".  There are many known free radicals that are quite stable.  This is usually due to steric factors (the reactive free radical center is surrounded by bulky groups and just not sterically accessible for reaction) or electronic factors (some free radicals exist in very large delocalized systems, hence the spin density at any one atom in the system is so small that reaction is unlikely).  One further point, most of what I've said relates to free radicals, which was the title of your question, however you did mention "unpaired electrons" in the beginning of your post.  Molecules with unpaired electrons (note the plural, electrons) are different than free radicals with (usually) just one unpaired electron.  Some molecules like $\ce{O2}$ are quite stable even though they have two unpaired electrons.


Symmetry of an in-phase and an out-of-phase wavefunction "approaching" each other in an MO

Considering this diatomic molecule, the total wavefunction $\Psi$ is formed via a linear combination of the atomic orbitals $\psi_1$ and $\psi_2$ (which is an approximation, but not a too bad one).
$$ \Psi^+ = c_1 \psi_1 + c_2 \psi_2 $$
$$ \Psi^- = c_1 \psi_1 - c_2 \psi_2 $$

The (electronic) energy of the system can be described using the following (stationary) Schrödinger equation:
$$ H_\text{el} \Psi_\text{el} = E_\text{el} \Psi_\text{el}$$

So somehow this all factors into each other and in the end gives you the curves (for the potential energy of the system vs. internuclear distance) that you posted. Where is the symmetry? Well, it is still there, with respect to the center of one atom. If you imagine rotating the functions around the $y$ axis, you will get a symmetric image. And you're allowed to do this because space expands in all three directions. So in theory (if you follow through and expand the function to everywhere) you get a 3D image of the energy at every point around one atom.

But: The wavefunction in itself is not all that interesting to us, because there is no "real" physical information that can be directly extracted from it. But fret not, we can at least get out the electron probability density by complex-squaring it:
$$ \rho = |\Psi|^2 = \Psi \Psi^* $$

And this is where those nice small bubbles come into play that we see in your second figure. And we have again found the symmetry that you seek, because both linear combinations have $D_{\infty h}$ symmetry (I think... I'm not all too good with point groups and the such).


Hydrogen cyanide is an acid with a pKa of 9.2, meaning that at pH 9.2 in water you have about a 50-50 mix of cyanide anion and hydrogen cyanide. If you used sodium hydroxide to adjust the pH of a NaCN solution to, say, 12.2, you would inhibit the decomposition of NaCN by shifting the equilibrium to 99.9 percent cyanide anion and 0.1% hydrogen cyanide. Each pH unit higher gives you an additional factor of 10 lower concentration of HCN.

Another factor is temperature: since HCN boils at 28 C, it tends to evaporate from a room-temperature solution of NaCN, driving the reaction to the right. The solution would be more stable if kept cold.

Finally, it would be better to keep the solution in a tightly closed bottle, since carbon dioxide could be absorbed from the air, acidifying the solution and releasing hydrogen cyanide.


Why does Gibbs free energy correspond solely to non-expansion work?

For a closed system in contact with a constant temperature reservoir held at the initial system temperature T, and in contact with a surroundings at an external pressure held at the initial system pressure P, and experiencing a change between two thermodynamic equilibrium states, the change in internal energy is given by:  $$\Delta U=Q-W$$If we express the work W as the sum of expansion work and non-expansion work (the latter, say, obtained by generating electricity or turning a shaft), then $$W=P\Delta V+W_{NE}$$And, because there is a single constant temperature bath comprising the surroundings, the heat added to the system is related to the change in entropy between the initial and final thermodynamic equilibrium states of the system by:
$$\Delta S=\frac{Q}{T}+S_{gen}$$where $S_{gen}$ is the entropy generated as a result of any irreversibility's that may be present.  So, if we combine these equations, we obtain:$$W_{NE}=-\Delta G-TS_{gen}$$where $\Delta G=\Delta U-T\Delta S+P\Delta V$.  So, for any process that goes between the same two thermodynamic equilibrium  states, the maximum non-expansion work that the system can do on its surroundings is $-\Delta G$ (since $S_{gen}$ can only be positive).  The maximum corresponds to a reversible path between the two end states.


Covalent bond nature in electronegativity gap

Strength of covalent bond depends on many factors, one of the key factors is "extent of overlapping of orbitals". For e.g., if you consider $\ce{H2}$ molecule, there is a head on overlapping of $\mathrm{1s-1s}$ orbitals involving two hydrogen atoms, but in the case of, say, $\ce{HCl}$, there is a head on overlapping of $\mathrm{1s-3p_z}$ orbital ($\mathrm{1s}$ of hydrogen, and $\mathrm{3p_z}$ of chlorine), and we all know that $\mathrm{p-s>s-s}$, that is, overlapping between a $\mathrm{p}$-orbital and a $\mathrm{s}$-orbital is stronger than $\mathrm{s}$ and  $\mathrm{s}$ orbital, and stronger the overlapping more is the covalent bond energy.

So, it largely depends on the type of orbitals involved and the nature of overlapping.

Check this to learn more about overlapping.


What's the difference between Zinc Ascorbate and Zinc Gluconate?

As I understand your question, “the two” refers to Zinc Gluconate and Zinc Ascorbate.
Zinc gluconate and zinc ascorbate are distinctly different compounds.  They are both adducts of Zn with one Zn and two bidentate oxygen-based ligands of organic acids.  Some basic information from PubChem for both zinc gluconate and zinc ascorbate is given below.
Zinc gluconate: $\ce{C12H22O14Zn}$
Zinc gluconate [Gluconic Acid Zinc (II) Salt] is a zinc salt of gluconic acid comprised of two gluconic acid molecules for each zinc cation ($\ce{Zn^2+}$). Zinc gluconate is a generally recognized as safe (GRAS) substance by FDA.

Of interest, as related to periodontic issues note that a phase 4 clinical trial was completed 2014-12-16 “The effect of Zinc-D-gluconate on bleeding propensity of the periodontium in patients with chronic parodontitis - a randomised, placebo-controlled, double-blind interventional study”.
Zinc ascorbate: $\ce{C12H14O12Zn}$
Zinc ascorbate is a zinc salt of ascorbic acid[Vitamin C] comprised of two ascorbic acid molecules for each zinc cation ($\ce{Zn^2+}$). Zinc ascorbate goes by a variety of different names, L-Ascorbic acid, zinc salt (2:1), L-Ascorbic acid, zinc salt, (-)-Ascorbic acid, zinc salt to name a few.

A ResearchGate reference discusses zinc-gluconate, taurine in reducing halitosis and gingivitis.  A portion of the abstract follows:

Objectives: to evaluate the efficacy in reducing halitosis and gingivitis insurgence in fixed orthodontic patients of a
zinc-gluconate, taurine and cetylpyridinium-chloride (CPC) oral-rinse,
sole or combined with a toothpaste with the same active principles.

References:

https://pubchem.ncbi.nlm.nih.gov/compound/443445
https://pubchem.ncbi.nlm.nih.gov/compound/Zinc-ascorbate#section=Structures
Dalessandri, Michela & Dalessandri, Domenico. (2011). Halitosis And Gingivitis Reduction Using A Zinc-gluconate-taurine-cetylpyridinium-chloride Oral-rinse And Toothpaste. (link)


What is the correct equation for calculating the electric potential of an electrochemical cell?

I am afraid your equations are other way around.

  Ecell = Eoxidation - Ereduction

This is not the accepted convention among electrochemists. Either the reference is outdated or there is a typo. All electrochemists in the world now follow the convention of writing

Ecell= Ereduction-Eoxidation or Ecell = Ecathode-Eanode  ---Eq(1)

The reason is historical and I have written details elsewhere, but briefly the main reason is that Americans used one sign convention and the Europeans used another. It took more than 50 years to come to a consistent sign convention, which was originally used by a German chemist, Ostwald, 100 years ago. Given that all electrode potentials are now written as reduction potentials the above Eq. (1) is valid. If you look up the tables, the signs of the electrode potentials correspond to the actual electrostatic sign (+) or (-) of the electrode.

Now this simplistic equation (1) ignores any "connections" among the half cells. In reality we always need an electrically conducting connection to the half cells to keep them separated yet electrically connected. This "connection" could be a salt bridge, or a porous ceramic frit (used in pH probes and so on).

There is a general interesting electrochemical fact whenever there is an interface (the location where two surfaces meet) one can build up difference of electrical potential. This is a fundamental material property. Recall the simple voltaic cell, which has copper and zinc pieces separated by a piece of wet paper containing some common salt.

Liquid junction potential (Ej) is another type of potential which originates at the interface of two electrolytes with different concentration of ions (=charges). In that case we add the liquid junction potential to this Ecell as follows

Ecell= Ecathode-Eanode+ Ej

I have not seen your other equation.  Ecell = Eoxidation + Ejunction - Ereduction  - Eelectrode. Which book is using this expression and what is its year of publication?


Electrophilic aromatic substitution with fluorine

Fluorine reacts so rapidly with benzene that aromatic fluorination requires special conditions
and special types of apparatus. Even then, it is difficult to limit the reaction to monofluorination


How many d-electrons are in the outer shell of the metals in following compounds:

In zinc sulfide, two s-electrons have been removed from $\ce{Zn(0)}$ to make $\ce{Zn^{2+}}$. Therefore, yes, it does have 10 d-electrons.

In its lattice, zinc is bound to four sulfur atoms which fulfills the 18-electron rule (10 from $\ce{Zn^{2+}}$ and 2 each from four separate $\ce{S^{2-}}$; the 18-electron rule for transition metals is the equivalent of the octet rule for main-block elements). In this case, however, the metal (zinc) still only has 10 d-electrons. The additional electrons are in hybrid, non-d orbitals.


Random conformer search generates random conformers

Open Babel, like many conformer sampling tools, has a set of common dihedral angles to sample. This minimizes the search space, e.g., only 3-4 likely cases for each rotatable bond.

The RandomConformerSearch method indeed generates conformers using a random sampling of dihedral angles. It then (optionally) runs a force field for some number of steps to minimize any steric clashes. No energies are considered.

If you're finding only one conformer output, that's because you're not using the --writeconformers flag:
http://open-babel.readthedocs.io/en/latest/3DStructureGen/multipleconformers.html


I suspect the answer is that AlGaAs is an alloy with a bandgap between that of AlAs - 2.16 (which is indirect) and GaAs - 1.42 eV (which is direct). If there is enough aluminum, the resulting semiconductor is an indirect bandgap semiconductor, which means that to emit a photon, it must be coupled with a lattice phonon, greatly reducing the probability of photon emission, which would make a AlGaAs LED of short enough wavelength rather inefficient at emitting light, instead producing a lot of heat rather than light.


Oxidation state of hydrogen in the LaNi5H intermetallic

At an intro chem level, the oxidation state of all elements in an alloy or intermetallic is zero.
At a deeper level of analysis, fractional, integer, and/or decimal oxidation states may be more appropriate, depending on the system in question.

Per Wikipedia:

An intermetallic, also called an intermetallic compound, intermetallic alloy, ordered intermetallic alloy, and a long-range-ordered alloy, is a solid-state compound exhibiting metallic bonding, defined stoichiometry and ordered crystal structure. Many intermetallic compounds are often simply called alloys.

However (emphasis added):

Schulze in 1967, defined intermetallic compounds as solid phases containing two or more metallic elements, with optionally one or more non-metallic elements, whose crystal structure differs from that of the other constituents.

So, in a high-level description, an intermetallic containing hydrogen such as the $\ce{LaNi5H}$ of the question can be considered essentially to be an alloy with hydrogen as a constituent.  Based on this, assigning a $0$ oxidation state to all elements is warranted as a first approximation:

For intermetallic compounds, the ultimate choice of the oxidation state zero at all atoms is best if needed in redox chemistry.  (source | doi)

A more detailed investigation might turn up evidence supporting assignment of nonzero fractional, integer, and/or decimal oxidation states to various atoms:

Subtler estimates and round offs are required for compounds with electrons delocalized over non-equivalent atoms, as expressed by several resonance formulas with weights in arbitrarily long decimal numbers. Without round offs of bond orders in Lewis formulas, decimal values of oxidation states would be obtained for certain bonding connectivities. [...] [U]nambiguous and reasonable fractions of small integers are obtained for oxidation states in compounds such as dithiolate and catecholate (see Section 11) or in (car)boranes such as $\ce{B6H10}$ (see Section 6) and $\ce{B10C2H12}$ (p1041), or when vicinal oxidation states are indistinguishably mixed, such as in $\ce{YBaFe2O5}$ (p1058). Reasonable fractional oxidation states appear also in ions where the charge is distributed over several equivalent atoms such as $\ce{C7H7+}$, $\ce{B6H6^2−}$ (p1040), $\ce{I3−}$, and $\ce{N3−}$ (p1037). (source | doi)


Why does chromium(III) have the highest tendency to form complexes, comparatively, among Mn(III), Cr(III), V(III) and Ti(III)?

The premise of the statement is skewed. All four of those have the same (rather strong) tendency to form complexes since naked metal cations in aquaeous solution simply do not exist. However, a point may be made that the complexes of chromium(III) are the least labile (or most inert) ones of the series.

Th reason for this lies in the typical octahedral arrangement observed in transition metal aqua complexes. This arrangement causes splitting of the $\mathrm d$ orbitals (which are degenerate in ground state) into two sets of $\mathrm d$ orbitals: $\mathrm{t_{2g}}$ and $\mathrm{e_g}$. In very basic crystal field theory, this can be understood as the $\mathrm{e_g}$ orbitals being those pointing towards the ligands approaching along the coordinate axes and bearing a point-shaped negative charge; thus, these orbitals are destabilised (raised in energy). This affects the orbitals $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2 - y^2}$. On the contrary, the other three $\mathrm{d}$ orbitals — $\mathrm{d}_{xy}, \mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$ — are considered stabilised since the entire transformation should preserve the total energy. (This concept is not physico-chemically accurate in the way it is presented here but close enough.)

The energy difference between the two energy levels is defined as (the relative value of) $\pu{10Dq}$. Again, using the ‘conservation of total energy’ principle, the higher-up orbitals are defined as having an energy of $\pu{+6Dq}$ each while the stabilised orbitals are defined as having $\pu{-4Dq}$ each. A ligand field stabilisation enthalpy or LFSE can be calculated by taking the electrons in each orbital and adding up the corresponding $\pu{Dq}$ value to give a total energy; the higher this is, the more stable the configuration is relatively.

All complexes in question are high-spin complexes, meaning that even though the $\mathrm d$ orbitals are no longer degenerate they are still filled according to Hund’s rule. Since the order of the elements in the periodic table is $\ce{Ti, V, Cr, Mn}$ and since each is in the same oxidation state ($\mathrm{+III}$), we can instantly see that they differ by the number of $\mathrm d$ electrons only: they have the configuration $\mathrm{d^1, d^2, d^3}$ and $\mathrm{d^4}$, respectively.

If we add between one and four electrons into the five $\mathrm d$ orbitals according to Hund’s rule and then calculate the LFSE, we will get the following result:

\begin{array}{cr}\hline
\text{conf.} & \text{LFSE}\\\hline
\mathrm{d^1} & \pu{-4Dq}\\
\mathrm{d^2} & \pu{-8Dq}\\
\color{red}{\mathrm{d^3}} & \color{red}{\pu{-12Dq}}\\
\mathrm{d^4} & \pu{-6Dq}\\
\hline\end{array}

The highlighted value is the configuration with the lowest energy, i.e. the one which is most stabilised during complex formation. Therefore, chromium(III) complexes are the most inert complexes of the series.


To answer your question, if you haven't figured it out by now, RRSS is the same as RSRS and it is the only one of the five that is optically inactive (Compare with nonactin). Structure A numbers the positions of chiral chlorine-bearing carbons. Structure B is RSRS and C is RRSS. Clearly, they are superimposable and, therefore, identical. A similar exercise with RRRS and RSSS will give a similar result.


Comparing basic strength order of amines

You don't have problem with assigning compound IV (1, 4, 5, 6- tetrahydropyrimidine) as the most basic compound among given four amines. I agree with you, but not the same reasons. According to Ref.1, the 1,4,5,6-tetrahydropyrimidines (THPs) are extremely basic compounds, as might be expected from their amidine structure (e.g., 1,8-Diazabicyclo[5.4.0]undec-7-ene or DBU). The parent THP, 1,4,5,6-tetrahydropyrimidine (IV) has a $\mathrm{p}K_\mathrm{a}$ of about $13$ at $\pu{20 ^\circ C}$, which is even reacted with water (Ref.1). This highly basic nature of IV can be attributed to its stabilization by resonance between two equivalent structures, V and V', when protonated:

Also, keep in mind that it can also get protonated on 3-nitrogen atom (VI; inside the blue box) as well (Ref.1).

Among the rest of other 3 amines, two are aminopyridines (I and II), and the third one is aniline (III). Since the lone pair of nitrogen in pyridines does not contributed to aromatic system (hence it is readily available for proton) while that of aniline is involved in aromatic system by resonance (not readily available), generally pyridine is more basic than aniline. Even though I and II both have extra amino group, they are expected to behave like pyridines. By this analysis, we can conclude that the least basic amine among given four amines is aniline (III).

Now, let's look at what happens when I (4-aminopyridine) is protonated:

From above illustration, you can clearly see conjugate acid of I is well stabilized by resonance including full contribution of the lone pair of 4-amino-nitrogen. If you visualize the same with II (3-aminopyridine), it is not stabilized by the extra amino grop as depicted in following diagram:

Thus, by these facts, we can conclude that II is less basic than I. Therefore, the order of basicity strength of four given amines is: IV > I > II > III.

References:

D. J. Brown, R. F. Evans, “100. Hydropyrimidines. Part I. 1, 4, 5, 6- Tetrahydropyrimidine and its derivatives,” Journal of the Chemical Society 1962, 527–533 (https://doi.org/10.1039/JR9620000527).


Why are polysaccharides not sweet in taste?

Well, while indeed polysacharides are metabolised to simpler units, even in mouth (saliva in the mouth can account for 30% of initial starch digestion), it usually doesn't happen fast enough, so full digestion follows later in you digestive system.

You can`t sense the sweetness of polysaccharides as they do not fit the receptors.


Why is the boiling point of pentane (36.1 °C) slightly higher than that of diethyl ether (34.6 °C)?

This is perhaps not a very scientific answer, but consider how branched alkanes have a lower boiling point than straight chain alkanes.  This is due to the inability of the individual molecules to "stack" (if you want to get technical it's to do with surface area).  Now imagine stacking diethyl ether molecules.  The problem arises that two oxygen molecules (2 lone pairs each) on two separate diethyl ether molecules would not want to stack directly on top of eachother as the oxygen atoms have decently high electron densities.  The repulsive force would likely be minimal, as demonstrated by the insignificant difference in boiling points between diethyl ether and n-pentane.  However, this is only an educated guess.

To strengthen my argument, check out Diethyl sulfide, which has a much higher b.p. than both n-pentane and diethyl ether.  This could be due to the larger atomic radius (and therefore larger surface area) of Sulfur in comparison to Oxygen.  The charge has a larger area to be distributed over, resulting in less repulsion between "stacked" sulfur atoms.


Solubility: Is the common ion effect proportional to individual solubility values?

There are basically 3 effects.

The solubility is controlled by the solubility product, the constant being the multiplication of ion activities

$$K_\mathrm{sp}={a_\mathrm{M}}^m\cdot  {a_\mathrm{X}}^n$$
for a salt $\ce{M_{m}X_{n}}$

For low concentration of ions, it can be approximated by concentrations, related to activities via activity coefficients $\gamma$, that are supposed to be equal to 1.

$$K_\mathrm{sp}={c_\mathrm{M}}^m\cdot  {c_\mathrm{X}}^n$$

where $a_\mathrm{M}=\gamma_\mathrm{M} \cdot c_\mathrm{M}$, the same for $\text{X}$

Therefore the presence of other salt with the common ion decreases the salt solubility, as the common ion contributes to the rate of precipitation.

The second effect is general effect of dissolved non common ions, given by ionic strength

$$I=\sum_i {c_\mathrm{i}\cdot {z_\mathrm{i}}^2}$$

and its effect on the ion activity, given for diluted solutions approximately by Debye-Hueckel equation

$$\log \gamma=-0.5 \cdot z_+ \cdot z_- \cdot \sqrt I$$

where $\gamma$ is the mean activity coefficient for given salt and $z$ are respective ion charges. The coefficient $0.5$ is rounded up.

For a particular ion, it is often written as

$$\log \gamma=-0.5 \cdot {z}^2 \cdot \sqrt I$$

For less diluted solutions, it is often modified as

$$\log \gamma=-0.5 \cdot {z}^2 \cdot \frac {\sqrt I}{c + \sqrt I}$$

where $c$ is ion dependent constant close to 1.

Generally, the higher is overall ion concentration, the higher is salt solubility. But the effect is smaller than effect of common ions.

For high concentration of general salts, ion activity coefficients revert their decreasing trend and start raising. For very high concentration to values much higher than 1.

As result, the solubility decreases.


Looks like a classic to me. But since this is homework, I will not provide a full reaction scheme and only give some hints.

Your starting material is a $\beta$-ketoester (1,3-dicarbonyl compound).

Where is the most CH-acidic position in this molecule?
Benzylchloride is an alkylating agent. What will happen when it reacts with the anion of your starting material?
For extra credit: Why is sodium ethoxide in ethanol a reasonable choice when you want to isolate, purify and characterize the product?

Your starting material did have an ester group. Does it still have it after the first step? If so, what will happen to it in aqueous alkaline solution?
(This step would not work with the saponification product of a simple ester.)
But you started with $\beta$-ketoester! Heating the product of step 2 in acidic media will let it loose some atoms. The mechanism may be drawn using a cyclic intermediate, in which the ketone has an assisting role.

EDIT

Deprotonation and subsequent benzylation of the $\beta$-ketoester occur at the most CH-acidic position, i.e beween the keto and the ester group. As a side note, benzylation at the methyl group adjacent to the carbonyl group is possible when using 2 equivalents of a strong base.

Using ethoxide rather than methoxide as a base makes sure that transesterification does not lead to a mixture of products.

The resulting benzylated $\beta$-ketoester is cleaved under alkaline conditions to the corresponding carboxylate. Under acidic conditions, this from a $\beta$-ketoacid.

Upon heating, $\beta$-ketoacids undergo decarboxylation (= they loose $\ce{CO2}$) to form an enolate which tautomerizes to the final methyl ketone.


Why does plastic tear easier when you pull it quickly?

Polymers have a special property called viscoelasticity, which means they may strain elastically (elasticity) then flow (visco). The models below are mechanical representations of this phenomenon using springs and dashpots (dampers).

$\qquad\qquad\qquad$ Maxwell Model   $\qquad\qquad\qquad\qquad\qquad\qquad$ Kelvin–Voigt Model

$\qquad\qquad\qquad$Time Stress Deformation Plot

This property of viscoelasticity allows plastics, when pulled slowly (low strain rate), time for the polymer chains to rearrange and the plastic to flow which means it is less rigid but more elastic (think stretching silly putty). When the plastic is pulled quickly the plastic is more rigid and breaks (like breaking silly putty). AS I have alluded you can observe this with silly putty which flows when strain is low and will stretch to long lengths while snap into two pieces when pulled rapidly.

So far I have only explained why plastics will stretch or snap, but not why they are hard to tear after stretching. This is caused because as the plastic is stretched the polymer chains align as shown below. As it turns out polymers are hardest to tear perpendicular to the direction of the molecules because you have to break the bonds of the chain rather than some bond breaking and some chains separating. As a result, the plastic becomes hard to tear across the stretched part, but it is now easier to tear along the stretch (parallel) since in is mostly chain separating to do so, and indeed a tear in a stretched bag will follow the direction of a stretch.


You are right, both abstracting $\ce{Cl-}$ from the alkyl chloride with Lewis acid as well as dehydration of isopropanol with $\ce{H2SO4}$ are suitable ways to generate the 2-propyl carbocation, your electrophile in this reaction. Option (c) will not work because $\ce{AlCl3}$ reacts with alcohols to form aluminium alkoxides.

Source: Master Organic Chemisry


2 modes of radioactive decay for hydrogen-5?

Sorry, you haven't.  I don't know where your information is from but I think there is an error in your decay chain.  For the 0.9% of $\ce{^8_2He}$ decays that proceed via spontaneous fission your equation should look more like this:
$$\ce{^8_2He ->[fission] ~^3_1T + ^5_1He + \beta ^{-}}$$
$$\ce{ ^5_1He  ->[neutron~emission]  n + ^4_2He} $$


Could you estimate the rate of reaction from the chemical structure of the reactants? If so, how?

As Paul J. Gans correctly pointed out in his comment, in general it is impossible.

In practice you can calculate reaction rates only for quite restrictive class of reactions - the unimolecular gas-phase reactions - using RRKM theory.
The whole potential energy surface is actually not needed, as far as I know, you just need to identify reactant, product, and transition state.
RRKM is based on few assumptions, with the first one being quite obvious: since we are talking about potential energy surfaces, the electronic motion and the nuclear motion is separated by the Born-Oppenheimer approximation.
The nuclear motion is also assumed to be well separated into rotational, vibrational, and translational one. The internal degrees of freedom are treated classically with quantum mechanical corrections.

RRKM is quite useful for predicting gas phase dissociation and isomerization rates  under low pressure (low enough so that reaction can be safely assumed to be unimolecular). For that reason, this theory is ofthen used in mass-spectrometry, since the above mentioned conditions are quite often met here, but it is of no use for reactions in the condensed phase.