How to Experimentally Quantify the Response of a Receptor to a Ligand?

What you're asking about is a wide topic.

You have the basic ideas of agonist and antagonist, but there are also molecules called inverse agonists. Every receptor has some base level of activity, even without an agonist to activate them, or in the presence of an antagonist. Inverse agonists push this activity even lower. There are even "partial agonists" which are molecules that activate receptors, but not as well as the endogenous ligand, so act like competitive inhibitors because they occupy some of the receptors the endogenous ligand would have. In addition, the binding site is important. The endogenous ligand binds to a location called the orthosteric binding site, but other molecules may bind to other locations on the receptor. These are allosteric binding sites. Allosteric binders may activate or inhibit a receptor, but they may also be modulators. Modulators don't directly activate a receptor, but allow a stronger response when another ligand binds at the orthosteric site. Here's a schematic to help show how different types of ligand affect receptor activity.

Anyway, ligands have two important properties, affinity and efficacy. Affinity is how strongly a ligand binds to a receptor, and efficacy is how the ligand affects the receptor's activity.

Before we can dig too deep into agonists or antagonists, we should think about receptor theory. Receptors, like all proteins, have complex 3 dimensional structure, or conformation. This conformation is affected by solvent, pH, temperature, and other factors, but always has some fluctuation. Classical receptor theory used a lock and key model, where an inactive receptor bound a ligand and became active. However, this did not explain the base activity and activity of inverse agonists, so an equilibrium based model was developed. Here, the population of receptors has an ensemble of conformation states, some of these are active. When an agonist binds, it drives the equilibrium of the population toward the active state by stabilizing the active conformation. This diagram shows a simplified example of how drugs affect the equilibrium. Ri is inactive receptor, Ra is active. D is drug. Ra leads to some activity, while Ra-D leads to strong activity.

Affinity is relatively straightforward to measure becuase the same idea can be applied to almost any receptor-ligand system. Usually, a radiolabled version of the drug is added to a known amount of receptors in a binding assay. The goal is to determine a dissociation constant, Kd. Kd is used to quantify how well a ligand binds to a receptor, and smaller numbers are tighter binders.

Look at this reaction: D + R <--> DR with forward rate constant k1, reverse rate constant k2. ( how to get the right arrows here? )

At equilibrium, k1[D][R] = k2[DR]
and $K_d =\frac{[D][R]}{[DR]}$ = $\frac{k_2}{k_1}$

So incubate the radioligand and receptor until some DR complex is formed, then seperate the bound and unbound ligand so you can measure D and DR. The seperation method is important because if it's too slow the DR complex will revert to D and R, creating inaccurate data. But this is commonly done using dialysis, centrifugation, and vacuum filtration. Dialysis works well when dissociation rate is slow. You put receptor in one side and drug in both sides. It's slow, but it's theoritically most accurate becuase you can measure the true steady state by taking an equal volume of solution from each side of the membrane without affecting equilibrium. But you also need a large supply of receptor and drug.
Centrifugation works when you can immobilize the receptor on a large particle, usually you do this by using whole cells. You add an excess of of drug, then spin it down and take off the supernatent. Timing is still an issue and free drug can get caught in the pellet.
Vacuum filtration is usually considered the best because it can seperate the components very quickly, by pulling the solution through a filter and trapping the receptor on the surface. However this requires specialized and expensive equipment.
Once you measure bound and free drug at several concentrations, you can do a double reciprocal plot

Measuring efficacy is different because you need to choose an activity to measure. Idealy it will be objective and easy to quantify. If the receptor is an enzyme, you can measure its enzyme activity. If it activates a gene transcription factor, you can use cells modified to express green fluorescent protien or luciferase when that transcription factor is activated. If it's an ion channel you can measure a cell's membrane potential. If you cannot measure anything else, you can use behavioral assays. There are several assays used to measure mouse activity and try to get at certain aspects of their brain and are often used to study depression and other mental illnesses. These are subjective. Efficacy is quantified by EC50, the concentration at which response is half of maximum response. Smaller EC50s are more efficacious.

This field is much much larger than my answer. For more you should read this article : http://www.ncbi.nlm.nih.gov/pubmed/15063082


Does 95.8% silver tarnish as fast 92.5% silver?

If you look here: https://blog.centimegift.com/does-sterling-silver-tarnish/, it says that pure silver is very tarnish resistant. Sterling silver on the other hand being mixed with some Cu to enchance the metal's mechanical properties tarnishes as a result of the Cu in it. Therefore, considering that the amount of Cu in Britannia silver is smaller, it should tarnish more slowly than sterling silver. In practice i dont know if there will be a significant difference as the increse in silver purity is not that much.


How to debug a failed Gaussian IRC calculation?

According to the output, it seems like OP is using Gaussian 09 software package. And according to the error message, the problem is that at 9th step of IRC procedure, Gaussian could not optimize the geometry in 20 steps (which is the default number). I suggest first to increase the number of geometry optimization iterations for each and every IRC step by using the MaxCycle=N option as follows:

IRC=(...,MaxCycle=100)


Why are there different chemical structures for epinephrine (adrenaline)?

There are often shortcuts taken in drawing complicated molecules, particularly if there's not any interesting chemistry going on at those functional groups.

The $\ce{-CH3}$ is implied at the end of the "stick," just like how the carbon atoms aren't explicitly labeled in the benzene ring.

Hydrogens are rarely explicitly drawn for the same reasons, given that you can infer the hybridization by inspection.

See Skeletal formula for more details.  Briefly:

  Carbon atoms are usually depicted as line ends or vertices with the assumption that all carbons have a valence of 4 and carbon-hydrogen bonds, usually not shown explicitly, are assumed to complete each C valence. A skeletal formula shows the skeletal structure or skeleton of a molecule, which is composed of the skeletal atoms that make up the molecule.


Ways to change the spontaneity of a reaction

Natural variables of $\Delta G$ are $p$, $T$, and $N_i$.

$p$ = pressure

$T$ = Temperature

$N_i$ = number of particles  (or number of moles)

The derivation showing this is listed on the Wikipedia.


The crystal structure of NaCl forms at the growing surface of the crystals, where there is plenty of time for the ions to sort themselves into a regular lattice. Further, each ionic species is very evenly distributed also in the solution, because of the repellent electrostatic forces between the members of each. Whenever there is a local excess of positive or negative charges, it will level itself out very quickly.

If you have something larger than sodium and chloride ions and try to crystallise very fast, you will indeed get an increasing amount of errors in the structure. Freeze-drying also tends to leave behind rather unordered structures. In both cases, this is, of course, more likely with noncharged or even nonpolar substances.


In equilibrium, if bonds are constantly being created and broken, where does the energy that does so come from?

The energy comes from the molecules themselves. See, they have certain energy of thermal motion. You might have heard of the ${3\over2}kT$ thing, but that's only an average value. Molecules have different energies, described by the so-called Maxwell distribution. Those with higher energy might take part in some reactions which are not possible for the entire bulk.

As for your second question, it's option 1: the equilibrium typically will shift. (Though the reaction might slow down a bit, too.)


How can I deduce the linearity of XeF2 from the IR spectrum?

Let's start with the rule of mutual exclusion, it will come in handy.  It states that,  for centrosymmetric molecules (molecules with a center of symmetry), vibrations that are IR active are Raman inactive, and vice versa.

A linear $\ce{AB2}$ molecule will have $(3n-5) = 4$ vibrational modes, but because it is linear 2 of the modes are degenerate.  $\ce{CO2}$ is an example of a linear $\ce{AB2}$ molecule, look at the following diagram showing the 4 vibrations for $\ce{CO2}$ and notice that the bending vibration ($v_2$) is degenerate.  That is, there is no difference between those 2 bending modes, they are identical.

Notice too that $v_1$ is the only vibrational mode that does not change the dipole moment of the molecule, therefore it is the Raman active mode. Since the linear molecule is centrosymmetric, the Raman band does not appear in the IR, nor do any of the IR bands appear in the Raman spectrum. Hence, for a linear $\ce{AB2}$ molecule we would expect 2 IR active bands and 1 Raman band. This is consistent with the spectral data you reported.

If the $\ce{AB2}$ molecule is bent rather than linear (water for example), then there are $(3n-6) = 3$ vibrational modes as pictured below.  All of these vibrations result in a change in the dipole moment.  Therefore, for a non-linear $\ce{AB2}$ molecule, all 3 of the bands will be IR active. Since the molecule is not centrosymmetric all 3 bands may also appear in the Raman spectrum (note: while a change in dipole moment is required for an IR-active band to appear, a change in polarizability is required for a Raman band to appear; looking at the water vibrations pictured, they should alter both the dipole moment and the polarizability of the molecule). Hence, for a non-linear $\ce{AB2}$ molecule we would expect 3 bands in the IR and 3 in the Raman spectrum. This is not consistent with the spectral data you reported above.

In summary, your spectral data is consistent with $\ce{XeF2}$ having a linear geometry.

Note: Thanks to @porphyrin for pointing out that bent $\ce{AB2}$ molecules are indeed not centrosymmetric.


Is reforming and isomerization the same thing?

TL;DR: Isomerization is a subpart of reforming process. There are four major reforming reactions. Isomerization is one of them.

Reforming is a processing technique by which the molecular structure of a hydrocarbon is rearranged to alter its properties. The process is frequently applied to low-quality gasoline stocks to improve their octane number and thereby improving combustion.
There are four major types of reforming process:

Dehydrogenation of naphthenes

Dehydrogenation and aromatization of paraffins to aromatics (dehydrocyclization)

Hydrocracking of paraffins

Isomerization of normal paraffins to isoparaffins

You can see that during all the process except isomerization, there is a loss of atoms either carbon or hydrogen. However, there is no change of atoms during isomerization process.


It depends on what you mean by "spin".

If you mean "have intrinsic internal angular momentum, independent of its trajectory through space", then yes, electrons spin, and that's what the quantum number is measuring.

Though if by "spin" you mean "undergoes rotation" ("there's a little billiard ball, and if I were to put a mark on it and watch it, the mark would be rotationally displaced around the center in a periodic fashion"), then no, electrons don't spin. They're point particles(*): there's no billiard ball to mark, and even if you were somehow to mark a "side" of a point particle (you can't), a point particle can't undergo internal rotational displacement - there's no internal to displace! Likewise, the "frequency" of the spin is also meaningless - attempt to find out how many revolutions per minute an electron "spins" and your calculator will spit garbage at you. It's not something that has any physically realizable meaning.

So the spin on an electron is a very real thing and not just a convenient bookkeeping label. You can do experiments where you couple the spin angular momentum of an electron to "macro-scale" (or at least non-sub-atomic) angular momentum, and you find that the total sum of angular momentum - both "macro-scale" and electron spin - is conserved. Electron "spin" is as real and functional as a gyroscope's spin is, but not in the same way. The electron has "spin" (angular momentum) without actually rotating.

Bizarre, but that's Quantum Mechanics for you.

*) I'm assuming the Standard Model of Quantum Physics here. If you start to get into String Theory things get all sorts of complicated, but there's no universally accepted model of String Theory, so I'm just going to ignore it exists. (As most Chemists do.)

Also note that the Standard Model doesn't even attempt to predict where this intrinsic internal angular momentum comes from. It's just a property of electrons, taken as a given. Asking "why do electrons have spin?" is like asking "why are electrons charged?": They just are.


Step P is std base-driven alpha alkylation of ketone. Step Q Wacker details here takes the terminal alkene to methyl ketone then step R is base-catalysed internal aldol mechanism here gives the product shown. So the answer is (b).


Is precision measured by significant digits in chemistry?

The single measured values provided by 3 different tools gives us just a hint about possible  accuracy or precision of the tools. The tool providing the value 1.39 is probably, but not necessarily more precise and accurate than the other two.
What these values really say is just the tool resolution.

There are 4 terms:

resolution
precision
accuracy(1), also defined by ISO as trueness
accuracy(2), defined by ISO as combined trueness and precision

Resolution relates to the least distinguishable difference of values provided by a tool. A tool able to tell ..1.38, 1.39, 1.40.... has resolution 0.01. It is not the same as the tick mark distance, as even the decimal fraction between the marks can be estimated for analogue output ( not digital one ).

Precision relates to the random errors during repeated measuring of the same value by the same tool or measurement method at the same conditions. It is usually expressed as a estimation of the standard deviation or confidence interval. It can be absolute, or relative wrt the measured value. It is usually worse than resolution, unless the latter is not fine enough. E.g. the resolution may be 0.001, but precission +- 0.02

Accuracy(trueness) relates to the systematic error of the tool or method aka the bias. It is the absolute or relative error of the mean measured value wrt the true or known value. The acccuracy(trueness) error is usually, but not necesserily bigger than precission error, e.g. the precission can be +- 0.05, but the accuracy(trueness) can have bias +0.1.

As formulated by Wikipedia in above links:

In measurement of a set, accuracy is closeness of the measurements to a specific value, while precision is the closeness of the measurements to each other.

Accuracy has two definitions:

More commonly, it is a description of systematic errors, a measure of statistical bias; low accuracy causes a difference between a result and a "true" value. ISO calls this trueness.
Alternatively, ISO defines accuracy as describing a combination of both types of observational error above (random and systematic), so high accuracy requires both high precision and high trueness.

Precision and accuracy(trueness) are mutually independent. The former is related to random errors, while the latter is related to the bias of the measuring tool a/o method. The accuracy(ISO) can be expressed as error $+0.1 \pm 0.05$.


Why does silicon tetrafluoride have a higher melting point than sulfur tetrafluoride?

High symmetry molecules fit into crystal lattices especially well (higher m.p.), but are volatile for having fewer van der Waals interactions (lower b.p.).

$$\begin{array}{lrr}
\hline
\text{Compound} & \text{m.p.}/\pu{°C} & \text{b.p.}/\pu{°C} \\
\hline
\text{pentane} & −130 & 36.1 \\
\text{isopentane} & −160 & 27.2 \\
\text{neopentane} & −18 & 9.5 \\
\hline
\text{octane} & −57 & 125 \\
\text{2-methylheptane} & −110 & 117 \\
\text{2,2,4-trimethylpentane} & −107 & 99 \\
\text{2,2,3,3-tetramethylbutane} & 101 & 107 \\
\text{cubane} & 131 & 133 \\
  & & \text{(sub. at r.t.)*} \\
\hline
\text{decane} & −30  & 174 \\
\text{isodecane} & −75 & 167 \\
\text{adamantane} & 270 & \text{sub.} \\
\hline
\end{array}
$$

* sub. — sublimes; r.t. — room temperature.


It is not. $\ce{NaOH}$ is more soluble ($>100~\mathrm{g}/100~\mathrm{ml}$) then $\ce{NaCl}$ ($<30 ~\mathrm{g}/100~\mathrm{ml}$). $\ce{NaOH}$ when dissolved produces $\ce{OH^-}$ anion, that has relatively small size and forms specific hydrogen bonds (where it is bonded to relatively small positively charged hydrogen) very well. In solid $\ce{NaOH}$ it has to form hydrogen bonds with quite big $\ce{Na^+}$ cation.


Meaning of mass concentration "units" % m/v

Since the question raised a discussion in the comments, I guess it won't hurt putting up a brief summarizing answer.

Notation % m/v, as you might've guessed, refers to the ratio between the mass and the volume, or, more precisely, mass concentration $ρ_i$ (sometimes denoted as $γ_i$ [1, p. 48] to distinguish from density, see How to distinguish mass concentration and density?):

$$ρ_i = \frac{m_i}{V}$$

where $m_i$ is the mass of solute and $V$ is the volume of solution, suggesting the SI unit $\pu{kg m-3}.$
Where does percent in % m/v come from?
Quoting Wikipedia:

  In biology, the "%" symbol is sometimes incorrectly used to denote mass concentration, also called "mass/volume percentage." A solution with 1 g of solute dissolved in a final volume of 100 mL of solution would be labeled as "1%" or "1% m/v" (mass/volume). The notation is mathematically flawed because the unit "%" can only be used for dimensionless quantities.

So, in order to account for this percentage “unit”, one should normalize the mass concentration to the mass of solute in grams in 100 mL of solution.
This can be done in one's head in most cases, but you must watch out for the units.

In this particular example, only option b satisfy the definition of the % m/v quantity we are expressing since only b lists volume of KI solution.
Let's check the math.
Since $\pu{575.0 mg} = \pu{0.5750 g},$ there is $\pu{0.5750 g}$ $\ce{KI}$ per $\pu{10.00 mL}$ of $\ce{KI}$ solution, and normalizing to $\pu{100 mL}$ we indeed obtain the aforementioned concentration:

$$ρ_\ce{KI} = \frac{\pu{0.5750 g}}{\pu{10.00 mL}} = \frac{\pu{5.750 g}}{\pu{100.0 mL}} = 5.750\%~(\text{m/v})$$

The exercise really just requires to pay more attention and follow the analytical chemistry definitions.

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.


Deriving fluorescence intensity equations

Concentrating on fluorescence intensity for the moment the equation becomes $I_O = I_D $ where $I_O$ is that for the free dye and $I_D$ for that bound to the DNA. If $\alpha$ dissociates the equilibrium is then

$$\begin{align}  &I_O \quad \rightleftharpoons  &I_D \\ &1-\alpha &\alpha \end{align}$$

and the equilibrium constant $K=I_D/I_O=\alpha/(1-\alpha)$.

The total emission is $\displaystyle F=I_O(1-\alpha)+I_D\alpha=I_O\frac{1}{1+K}+I_D\frac{K}{1+K}$

Rearranging this gives $\displaystyle \frac{1}{I_O+KI_D}+\frac{1}{I_D}=\frac{1}{F}$

Replacing intensity with product of fluorescence yield and concentration and using the rather odd definitions given  should give the equation you seek. Note that $F-\phi_{AO}C_{AO}$ is the intensity of the bound dye.

The first question states that the intensity due to the bound dye $F-\phi_{AO}C_{AO}$ is equal to the amount bound times the difference in yields, this because the dye fluoresces at the same wavelength whether bound or not.


[...] it would be great if you could support your answer with a real world example or application.

Free radical polymerizations of monomers with $\ce{C=C}$ double bonds are not completely useless ;)

Let me give some examples:

Polymethyl methacrylate (PMMA) is obtained by polymerization of methyl methacrylate (MMA). Think acrylic glass.
Styrene can be polymerized to polystyrene (PS). Think Styrodur and food containers.
Vinyl chloride is polymerized to polyvinyl chloride (PVC). Think pipes, insulation for electric cables, flooring.


I think long story short: the difference between (lim) and not is that the dominant step drives the limiting case more strongly than the non-limiting case.

There are two aspects to a substitution reaction:

Stoichiometric mechanism: the sequence of elementary steps that take us from reactants to product

Associative (A): the bond formation between the metal and the entering ligand i
Dissociative (D): the bond breaking between the metal and the exiting ligand is the important step, i.e. the intermediate has a decreased coordination number
Interchange (I): there is no evidence of an intermediate

Intimate mechanism: what affects the rate constants of particular steps?

The reaction rate is affected by the entering group (a)
The reaction rate is affected by the leaving group (d)

You can see that SN1 and SN1 (lim) both have dissociative intimate mechanism, which means that the breaking of the bond between the metal and the leaving group is the rate determining step. In SN1 (lim), the leaving ligand bond breaks first forming an intermediate with a lower coordination number; the entering ligand then forms a bond with the metal. However, SN1 is an interchange stiochiometric mechanism which means that there isn't a true intermediate where the leaving group has totally left. In this case you could imagine that the leaving metal-ligand bond merely weakens or 'stretches' before the incoming ligand bonds to the metal.

Similar principle applies to SN2 (lim) and SN2. In SN2 (lim), an intermediate with higher coordination number is formed when the entering ligand bonds with the metal. For SN2, no such intermediate step is observed. You could say that the incoming ligand simply begins to form a bond to the metal before the leaving group bond starts breaking.

(http://nptel.ac.in/courses/104106063/Module%205/Lectures%208-10/Lectures%208-10.pdf for image above)


What can I say about (A) and (D)?

When a carbonyl compound and water are present together, an equilibrium is set up between the carbonyl compound and the corresponding gem-diol. Like any equilibrium, this equilibrium can be shifted to one side or the other by factors that stabilize (or destabilize) one side or the other.

If water is present, poly-vicinal carbonyl compounds usually exist as hydrates. Drawing the resonance structure of a carbonyl compound we see the the $\ce{C=O}$ bond is highly polarized with the carbon terminus being the positive end.  If we have several carbonyls adjacent to one another, then each carbonyl has this same polarization and an electrostatic destabilization results. This is particularly true for the central carbon since there is a partial positive charge on each of the carbons attached to it.

In order to avoid this electrostatic destabilization, such compounds will form a hydrate at the central (the most electrostatically destabilized) carbon. In the hydrated form there is no repulsive partial positive charge on adjacent carbons.

You've already figured out why cyclopropanone exists as a hydrate.  So the correct answer is B, since there is no driving force that would provide any special stabilization of its hydrate.


Does anyone know where I can find the reaction rate coefficient and design equation for the electrolysis of brine and sodium hydroxide solution?

You should look in some kind of reaction rate database, e.g. NIST. An example from the database:


What does it mean to equilibrate a column in size exclusion chromatography and why is a buffer used?

You want to make sure that your protein does not get damaged in the separation process. The protein travels faster than the small solutes in its solution, and will encounter the small molecules that are on the column, i.e. it will encounter different ionic strength, pH etc. So before you apply the protein solution to the column, you make sure the column is treated with an appropriate solution.

The solution is often called the buffer because it contains pH-controlling buffer substance. You can use size exclusion chromatography to separate by size, but also to switch from one buffer (or more general: one set of small solutes) to another. This is called buffer exchange, and there are short thick columns for that purpose (for separation, you would use a long thin column).


Why is tetracyanonickelate square planar?

This is the square planar splitting diagram

Many d8 metal complexes are usually square planar because a very stable electron configuration is achieved if you put the electrons on this splitting diagram -6 electrons stabilizing the compound and only two slightly destabilizing electrons-

Now you could ask, why then not d6 and d7 adopt this cinfiguration? Because for 6 or 7 electrons, an even more stable state is achieved with octahedral fields.

About hybridization... Forget about it when talking about coordination complexes. It's completely useless. It can't explain or predict anything that CFT/LFT can't


Vibration motion at the imaginary frequency for TS of the Claisen rearrangment


I asked my teacher about it and he told me that no it is not possible as these two are totally different mechanism.

Sorry for english or improper answer( if it is so) as my teacher told me this much only so I can't say much clearly about it.


Why do many pharmaceuticals contain fluorine?

A definitive reference is:
"The Many Roles for Fluorine in Medicinal Chemistry" J. Med. Chem., 2008, 51 (15), 4359–4369.
In reference to the bullet points:

Fluorine obviously affects $\ce{p}{K_{a}}$ and can make an acid stronger but also a base weaker which could be advantageous when optimising a drug candidate (lead) molecule.
I guess here they refer to $\ce{-CF_3}$ which can increase lipophilicity
That's one of the most useful roles of fluorine. A lot of times you may have a very potent and selective molecule, but when you test it in vivo let's say in rats, it has no effect because it gets metabolised i.e. chemically modified to an inactive molecule by the rats liver (and other mechanisms). This can maybe prevented by changing to an $\ce{C-F}$ bond. For example electron rich aromatic sites are prone to metabolic oxidation. You can block this by fluorination of this site, exchanging the proton for a fluorine.

Have a look at the paper on top, you will find a new world of information.


Baeyer-Villiger Oxidation restricted to Ketones

May be Baeyer Villiger oxidation is not for aldehydes in the past. But that has changed recently (Ref.2). The abstract of this reference states that:

A conceptually distinct, modern strategy for Baeyer−Villiger oxidation (BVO) was developed. Our novel method involves initial hydration of water to carbonyl compounds, followed by ligand exchange of hypervalent aryl-$\lambda^3$-bromane on bromane(III) with the resulting hydrate, yielding a new type of activated Criegee intermediate. The intermediate undergoes BV rearrangement and produces an ester via facile reductive elimination of an aryl-$\lambda^3$-bromanyl group, because of the hypernucleofugality. The novel strategy makes it possible to induce selectively the BV rearrangement of straight chain primary aliphatic as well as aromatic aldehydes, which is missing in the classical BVO: for instance, octanal and benzaldehyde afforded rearranged formate esters with high selectivity (> 95%) under our conditions, while the attempted classical BVO produced only carboxylic acids. This firmly establishes the powerful nature of new methodology for BVO.

The suggested mechanism and comparison it with classical Baeyer−Villiger oxidation is shown in following scheme $(\bf{A})$:

The part $\bf{B}$ of the scheme depicted computational calculation of energy diagram for acetaldehyde oxidation.
In addition, similar oxidation of acetal (an aldehyde derivative) to ester have been known for years. For example, conversion of cyclic acetals to hydroxy esters by mCPBA oxidation is published in 2002 (Ref.2).

Reference:

Masahito Ochiai, Akira Yoshimura, Kazunori Miyamoto, Satoko Hayashi, Waro Nakanishi, "Hypervalent $\lambda^3$-Bromane Strategy for Baeyer−Villiger Oxidation: Selective Transformation of Primary Aliphatic and Aromatic Aldehydes to Formates, Which is Missing in the Classical Baeyer−Villiger Oxidation," J. Am. Chem. Soc. 2010, 132(27), 9236–9239 (https://doi.org/10.1021/ja104330g).
Jin Yeon Kim, Hakjune Rhee, Misoo Kim, "Conversion of Cyclic Acetals to Hydroxy Esters by MCPBA Oxidation," Journal of the Korean Chemical Society 2002, 46(5), 479-483 (DOI: 10.5012/jkcs.2002.46.5.479).


In HPLC, how does column diameter affect the retention time?

The key idea is that when you change column diameter, you keep the superficial linear velocity the same. For example if I have an analyte which elutes on a 10 cm x 4.6 mm i.d column at 7 minutes at 1 mL/min. Now we change the diameter to 3 mm, we have to use a flow rate of 0.425 mL/min to keep the retention time the same. Now we wish reduce the diameter further, say 2.1 mm i.d., now you should use 0.21 mL/min so that your analyte elutes at 7 minutes.

Search the keyword: superficial linear velocity, HPLC


Two of the methods are:

Atomic Absorption Spectroscopy - similar to the flame test, but is usually quantitative. The sample is heated (in a flame or furnace) and the absorbance at a defined wavelength is measured against standardized solutions. It will work for any atom if you use a tunable laser.

ICP/MS (Inductively Coupled Plasma/Mass Spectrometry) - a solution of the compound is ionized and passed into a mass spectrometer. Some atoms (especially halogens) cannot be properly ionized and passed into the mass spectrometer. Oxygen and nitrogen (and a few others) would be difficult to quantify since it may come from impurities in the sample.


What (non toxic) light sensitive materials do not burn away at high heat?

There are a number of types of photosensitive glass which change in appearance and chemical reactivity upon exposure, and of photochromic glass, which undergoes reversible change in color with radiation.
If you are using this as a surface applique on ceramic, one key factor would be having therml expansion coefficient compatible with the ceramic, or the object will crack on cooling. Consult the literature from a manufacturer, such as Schott or AZO Materials.
There are also techniques for making a photosensitive layer on ceramics, such as Pyrofoto, that do not involve applying glass. A small Pyrofoto kit is ~US$20.
BTW, you might find that microwave oven kiln adapters do not reach high enough temperature to be useful in this work -- literature states a limit of ~900°C, cone 010.


Feasibility of another product in peroxide effect mechanism of propene

The energy cost of bromine radical abstracting hydrogen from propene is 89 kcal/mole. The gain of forming $\ce{HBr}$ is 18-28 kcal/mol (for a $\ce{H-Br}$ bond from $\ce{H2}$  and $\ce{Br2}$; variation is due to solvent effects) + 23 kcal/mol (from the $\ce{Br2}$ which was fractured into 2 Br radicals by 46 kcal/mol bond energy); so the gain is 41-51 kcal/mol vs a cost of 89, or a net cost (net energy requirement) of 38-48 kcal/mol. This would be a slow reaction with an activation energy even higher than 41-51 kcal/mol. (All data from CRC Handbook)

The alternate reaction, of adding a Br radical (with 23 kcal/mol energy above ground state) to propene, loses the pi bond and its energy (84 kcal/mol) and gains a $\ce{C-Br}$ bond (70 kcal/mol). The energy cost of forming the bromoradical is 84 but the gain is 70 kcal/mol for $\ce{C-Br}$ plus the 23 kcal/mol from the Br radical, or 93 kcal/mol. Net gain is 9 kcal/mol. This reaction will go spontaneously while the first will not (rate determined by activation energies, of course).

It turns out that allylic stabilization is only part of the story, and is, in fact, rather small: the bond strength of $\ce{H - allyl}$ is 89 kcal/mol vs the bond strength of $\ce{H - n-propyl}$ which is 98 kcal/mol.


What are the common names of this two types of plastic?

The most common types of plastics in use today are polyethylene (PE) and polystyrene (PS). Each of these comes in several variants.

It is difficult to tell just from a photograph exactly what plastics are used in the products shown.

Also, various additives are often added to plastic to enhance certain properties such as high-impact, chemical resistance, recyclability and flexibility.

The type of plastic is sometimes marked on the item itself to help with recycling, typically in the form of an extruded imprint, or you could try to contact the manufacturer.

Both PE and PS are suitable for moulding and extrusion. You can also do some simple physical and/or chemical testing to determine what the plastics contain, although this will involve destroying at least a part of the item. Low density polyethylene (LDPE) typically has a lower melting point $(105-130^{o}C)$ than polystyrene $(200-240^{o}C)$.

LDPE is also softer and slightly rubbery whereas polystyrene is typically more rigid & crackly when crushed. Without physically touching the items to get a proper feel for what they are made of, as a first guess, I would say that the colourful containers such as plastic bins and laundry baskets in your first photo look like some sort of polyethylene, whereas the clear plastic packaging in the second photo looks more like polystyrene.


Basically, balanced chemical reactions must 1) have the same number of each element on each side, and 2) have the same net charge on each side. Consider the reaction $H_2 + O_2 → H_2O$. It is unbalanced because there are two oxygen atoms on the left side but only one on the right. If we assign variable coefficients to each reactant and product, $a H_2 + b O_2 → c H_2O$, we can set up a system of equations by multiplying the variables by the number of atoms in each species. For oxygen, $2b = c$ and for hydrogen, $2a = 2c$. While the system has infinitely many solutions, we take the lowest integer solution. Thus, the balanced equation is $2 H_2 + O_2 → 2 H_2O$.

Chemical reactions involving charges are a bit more complicated. A good explanation can be found this tutorial.


You might want to have a look at free online courses, e.g. at Khan Academy. To my opinion, this can however not replace the real experience of endless, often frustrating hours in the lab ;) - chemistry is an experimental science.


Where can I find an extensive list of the preferred chemical nomenclature?

Here's a pretty extensive list on Wikipedia.

  Dictionary of chemical formulas

I doubt this nomenclature would change much depending on context, given that there is a widely used nomenclature system (IUPAC) (see here).


Use the 2nd order scheme $2A \to B$ with initial amount of $A=A_0$ so $B=A_0-A$ and B is the product you observe.  The rate of change is $\displaystyle \frac{1}{2}\frac{dA}{dt}=-k_2A^2$ and I'm not going to use [ ] for concentrations just $A,\,B,\,A_0$. Also the 2 in the equation is incorporated into the rate constant just for convenience.
Solving the equation gives $\displaystyle A=\frac{A_0}{A_0k_2t+1}$ and so the product follows $\displaystyle B=\frac{A_0^2k_2t}{A_0k_2t+1}$. This is shown as the red line in the figure below. I just guessed the rate constant a few time to fit  and as you can see the fit is good. You can also linearise this equation by isolating $A_0k_2t$.
The first order plot is also shown (grey line) and this is poor. $B=A_0(1-e^{-kt}$)


Can a free radical be created by chemical reaction of non-radical species?

As for #3, photoinduced electron transfer (PET) reactions do what you want.

Imagine that 2,4,6-triphenylpyrylium tetrafluoroborate ($\ce{P+BF4-}$) is irradiated at $\lambda = 350~\mathrm{nm}$ in the presence of an alkene and a nucleophile ($\ce{ROH}$).

In the excited singlet state, $\ce{^1P+^*}$ is a strong electron acceptor, that oxidizes the alkene to the corresponding radical cation an is itself turned to a pyranyl radical, $\ce{P^{\cdot}}$. (Here is the first radical.)

Then the alcohol adds to the alkene radical cation, the adduct deprotonates spontaneously. (Here is the second radical.)


Diluting a sample of weak base before titration. What happens to equivalence point pH of diluted vs undiluted?

At the point of equivalence, when the added molar amount of $\ce{HCl}$ is just matching the present molar amount of $\ce{CH3NH2}$, there is solution of the weak acid $\ce{CH3NH3+}(+\ce{Cl-})$, that I will denote as $\ce{HA}$.

If we neglect $\ce{H+}$ from water dissociation and consider $c_\ce{A-}\ll c_\ce{HA}$, then $c_\ce{A-} = c_\ce{H+}$
Therefore

$$K_\mathrm{a} = \frac{c_\ce{H+} c_\ce{A-}}{c_\ce{HA}} = \frac{c_\ce{H+} c_\ce{H+}}{c_\ce{HA} + c_\ce{A-}}$$

$$\begin{align}
\mathrm{pH} &= \frac12(\mathrm{p}K_\mathrm{a} - \log c) \\
  &= \frac12(14 - \mathrm{p}K_\mathrm{b} - \log c) \\
  &= 7 - \frac12(\mathrm{p}K_\mathrm{b} + \log c) \\
\end{align}$$

where $\mathrm{p}K_\mathrm{b}$ is the methylamine basicity constant;
$c$ is summary concentration of methylamine and it's a conjugate acid.

Therefore, if there is final volume 30 instead of 20 ml, the final

$$\Delta\mathrm{pH} = -0.5\log \left(\frac{\pu{20 ml}}{\pu{30 ml}}\right) = +0.088$$


The international standard ISO 105-E01:2013 Textiles – Tests for colour fastness – Part E01: Colour fastness to water includes specifications for two different solutions:

Alkaline solution, freshly prepared, using grade 3 water complying with ISO 3696, containing, per litre:

0.5 g of ʟ-histidine monohydrochloride monohydrate ($\ce{C6H9O2N3.HCl.H2O}$);
5 g of sodium chloride ($\ce{NaCl}$);
and either
5 g of disodium hydrogen orthophosphate dodecahydrate ($\ce{Na2HPO4.12H2O}$)
or
2.5 g of disodium hydrogen orthophosphate dihydrate ($\ce{Na2HPO4.2H2O}$).

The solution is brought to pH 8 (± 0.2) with 0.1 mol/l sodium hydroxide solution.

Acid solution, freshly prepared, using grade 3 water complying with ISO 3696, containing, per litre:

0.5 g of ʟ-histidine monohydrochloride monohydrate ($\ce{C6H9O2N3.HCl.H2O}$);
5 g of sodium chloride ($\ce{NaCl}$);
2.2 g of sodium dihydrogen orthophosphate dihydrate ($\ce{NaH2PO4.2H2O}$).

The solution is brought to pH 5.5 (± 0.2) with 0.1 mol/l sodium hydroxide solution.


Disrupting the chemistry of oobleck so that it behaves like a newtonian fluid

Disposal isn't a problem - add more water as you dump it down the drain.

There are two extremes to ooblek:

Almost all cornstarch and almost no water: this is a dry solid or slightly wet starch. No big deal.
Almost all water and very little starch: this is some water with a little bit of starch mixed in. Sometimes it clumps or falls out of suspension, but it's also no big deal.

The non-Newtonian mixture occurs at intermediate ranges.

Since cornstarch will degrade in the sewer system (e.g., feed bacteria), it's safe down the drain. Alternatively, I compost it.

But if you're worried about non-Newtonian behavior, you just add more water.


Before the CIP system was created, how did chemists ensure that the D/L system matched the R/S assignments?

Chemists actually still use D- and L- prefixes today, especially for carbohydrates where using it saves a lot of characters: Compare D-ribose to (2R,3R,4R)-2,3,4,5-tetrahydroxypentanal. The nomenclature is also in common use for amino acids, since all proteinogenic amino acids (except for glycine) in animals are L-configured (Bacteria, smart as they are, have many exceptions), while cysteine has the opposite absolute configuration to all other amino acids due to sulfur’s high priority. But your question revolves around history.

When Emil Fischer first devised the D- and L- nomenclature in 1891, not much was known about stereochemistry at all. It was known that carbon took four bonds and what the elemental composition of monosaccharides was. And methods were known to find out what a compound was, but much less sophisticated than nowadays. You would essentially derive it back to known products via known reactions and measure melting point (solids) or refractive index (liquids) for additional confirmation. Optical activity had been known since 1850, so that was probably performed in Fischer’s lab, too. At the same time, Pasteur performed his experiments, separating a racemic mixture of tartrate crystals into the enantiomers.

The choice Fischer made, when he assigned the D-configuration to (+)-glycerinaldehyde (the plus sign meaning that linear polarised light is rotated clockwise), coming from Latin dexter meaning right, was arbitrary. Not even essentially arbitrary, it was completely arbitrary. He probably knew he had a $50~\%$ chance of getting it wrong but in no way was he going to get proven wrong anytime soon. He decided it fitted nicely: the glycerinaldehyde that polarised to the right (clockwise) shall be the one with the asymmetric hydroxy group pointing to the right. He used that as a basis for all further sugars, knowing that all reactions he performed would take place on the aldehyde and leave the hydroxy group unchanged.

There was a fair bit of understanding how reactions worked even then. The first step to most sugar syntheses starting from glycerinaldehyde is the addition of a cyano group to give a cyanhydrine: Attack to the aldehyde, forming a new carbon-carbon bond and nothing else.

$$\ce{HOH2C-CHOH-CHO + N#C- + H+ -> HOH2C-CHOH-CHOH-C#N}$$

In any case, once a reaction like that was performed, derivatisation back to known compounds was done, and they would have realised if something had gone wrong with the hydroxy group. So we can be pretty sure they knew what wouldn’t influence it.

It wasn’t until half a century later and the X-ray crystal structure of glycerinaldehyde that one actually found out what the configuration of its carbon atom was. When it was proven that he had backed the lucky horse and the hydroxy group of D-glycerinaldehyde indeed pointed rightwards according to his convention, Fischer probably smiled a big smile from up in heaven. But if he had been wrong, he would have shrugged and thought ‘better luck next time.’

Now how does all this connect to the absolute configuration? Not at all! The absolute configuration is assigned on a per asymmetric atom basis. You need to know what the atom’s surroundings are like and can then apply a descriptor directly to it. Most notably, it was not the molecule as a whole that served as the reference but only the neighbouring groups and their priorities (read: atomic charge values). Any similarities to other stereodescriptor schemes, in use or not, was entirely accidental and unintended.

You have to remember that the retention of configuration in atomic terms does not automatically mean that the absolute configuration is retained. Take for example the Mosher acid as depicted below (image was taken from Wikipedia, where a full list of authors is available):

(R) and (S) Mosher acid.

Assigning the (R) configuration to the structure on the left is trivial. Assume you were to turn the (R) Mosher acid into the corresponding acid chloride. Note that although the asymmetric carbon was not touched the acid chloride turns out to be (S)-configured! So absolute configuration is something far more ‘wobbly’ than stereocentres are; make a remote change and influence an absolute configuration although that part of the molecule wasn’t touched.

Incidentally, that is the beauty of the Fischer nomenclature, especially when you consider the age in which it was invented. You don’t need to know which oxygen is pointing where you don’t need to figure out a long chain of priorities. You know you started at D-glycerinaldehyde so your sugar is now D-ribose. And you can take a pen and draw the lowest hydroxy group pointing to the right.

Do note make the mistake and try to correlate Fischer nomenclature with the sign of optical activity or with absolute configuration; all attempts will fail. Each has their use, their place to shine and none treats all similar molecules in the same way. Use wisely.


Difference between continuous and line spectrum of hydrogen

I will quote part of this answer:

  Quantum mechanics dictates that the energy levels of atomic hydrogen are discrete--just from the way the math works out. The reason why hydrogen's peaks are so spaced out is that it has so few degrees of freedom (just the one electron). Xenon, in contrast, has many closely-spaced peaks because it has 54 electrons (so it has many more accessible energy levels than hydrogen).

The implication is that to a certain resolution, line spectra appear to become continuous once the energy level spacing and the excitation energies become close together. If you look closely enough, they are still discrete.

However, there comes a point where the electron that's being excited is no longer bound to the atom; that would be the first ionization potential (IP) or ionization energy. In the case of the hydrogen atom, that is the Rydberg constant. See this answer for how this arises mathematically.

For more details on the hydrogen atom, see here:

  The infinity level represents the highest possible energy an electron can have as a part of a hydrogen atom. If the electron exceeds that energy, it is no longer a part of the atom. The infinity level represents the point at which ionization of the atom occurs to form a positively charged ion.

At this point it is important to distinguish between hydrogen's absorption/excitation spectrum and its emission/deexcitation spectrum (taken from here):

Since energy is inversely proportional to wavelength, the higher-energy excitations are toward the left of the spectra. This is visual confirmation that as energy level spacing decreases, lines in spectra get closer together. Past that point, what do the spectra look like? Certainly there will be peaks at the ionization energy, which is at a wavelength of about 91 nm. Applying more energy will also strip the electron from the proton, but the excitation from $n = 1$ to $n = \infty$ will still occur at the same energy, and likewise for the absorption from $n = \infty$ to $n = 1$. So there is only one peak in the hydrogen emission and exitation spectra directly resulting from the interaction between the electron and the proton (related). Any potential continuous band structure at higher energies due almost entirely to the free electron in the continuum, or you could think of any point in that spectral region being composed of two contributions, one being the fixed electron-proton interaction energy and the other being related to the kinetic energy of the free electron.

Note that this is for a single atom; for a collection of atoms approaching a bulk phase, you may have continuous spectra under other situations. Any time there are free electrons, energy levels may be continuous. There is also above threshold ionization, but without doing any calculations I am not sure how intense these effects are in terms of directly contributing to UV/vis spectra; if the $x$-axis is electron energy in eV and the $y$-axis is related to excitation probability, the effect is most likely negligible in practice.


Could a lead-acid battery electrolyte be replaced by hydrochloric or nitric acid?

In the lead battery it is very important that the electrodes transform into insoluble lead sulfate when discharged, especially at the cathode where lead(IV) oxide is turning into lead(II) sulfate. Consider the two half reactions of the lead-acid battery:

$$\ce{Pb(s) + HSO4−(aq) → \color{red}{PbSO4(s)} + H+ (aq) + 2e−}$$
$$\ce{\color{red}{PbO2(s)} + HSO4−(aq) + 3H+ (aq) + 2e− → \color{red}{PbSO4(s)} + 2H2O(l)}$$

Lead nitrate is soluble, so metal ions would go into solution. The lead(IV) oxide would not form again from these ions on charging, metallic lead from the electrode would dissolve. The battery would no longer be rechargeable.

Lead chloride is insoluble, but lead(IV) oxide on the cathode would oxidize chloride into chlorine. Thus no more battery.


Is a supersaturated solution still a solution?

You seem to be getting it wrong. A solution with crystallized solids at the bottom is not the same as supersaturated solution, and has never been called such. It is called "solution with precipitate", "heterogenous mixture of solution and something", "water with mud", or anything in the same vein.

A supersaturated solution is a solution, not just technically, but in all sincerity. It is homogenous, clear, transparent, and you'll never tell it from ordinary solution - I mean, until you touch it, at which point a precipitate would start to form.

As for your second question: yes, there are quite a few compounds that would dissolve in water incredibly well. ZnCl2 comes to mind... Wait, it's simpler than that - common table sugar (sucrose) would do! It dissolves up to 2000 g/L, how'd you like that?


The Acidity of the Hydrated Iron (III) Ion

When a water molecule coordinates to $\ce{Fe^3+}$, one non-bonded $sp^3$ hybrid orbital (with one of the 2 lone pairs) of oxygen overlaps with an empty $d$ or $p$ orbital of $\ce{Fe^3+}$. Electron density is transferred from oxygen to iron, and due to the electronegativity of oxygen, the polarization of the $\ce{O-H}$ bond in the water ligand is increased (more electron density is "pulled" from $\ce{H}$ towards $\ce{O}$). The hydrogens are more positively polarized than in a free (uncoordinated) water molecule, and this increases their electrophilic character and acidity to the point that they are readily abstracted by solvent (non-coordinated) $\ce{H2O}$ in aqueous solution:

$$\ce{Fe(H2O)6^3+ + H2O \rightleftharpoons  Fe(H2O)5(OH)^2+ + H3O+}$$

$$\ce{Fe(H2O)5(OH)^2+ + H2O \rightleftharpoons  Fe(H2O)4(OH)2^+ + H3O+}$$

Further deprotonation results in the neutral complex $\ce{Fe(H2O)3(OH)3}$, from which $\ce{Fe(OH)3}$ precipitates.

The aqueous solution of $\ce{Fe^3+}$ aquo complexes is fairly acidic, wth a pH of up to 1.5 depending on iron concentration. The first dissociation equilibrium lies mainly on the product side, which can also be seen in the color of the solution: Pure $\ce{Fe(H2O)6^3+}$ is pale violet and found in crystals of some hydrated $\ce{Fe(III)}$ salts, while aqueous solutions get their orange color from $\ce{Fe(H2O)5(OH)^2+}$.

You have mentioned ion charge (oxidation state) and charge density (which is proportional to ion size) of the metal center as potential factors of influence. One can expect that with increased charge and/or charge density, the polarization of the water ligands also increases, and thus the aquo complex should be more acidic. This is true for many aquo complexes; however, there are also some deviations from this rule, like $\ce{Cr(H2O)6^2+}$ and $\ce{V(H2O)6^2+}$ (reference).


The amount of X-rays absorbed by an element depends on the size of its atoms (its absorption cross-section, specifically, as affected by the size of core orbitals that contain electrons that can be excited by X-ray absorption, and the number of electrons in those orbitals) and how many atoms are packed into a given volume. Big atoms that are close together can absorb more X-rays than small atoms that are far apart.

Beryllium has very small atoms (it has a very small electronic core, with only two electrons in it), and very low density, so it will absorb relatively few X-rays compared to other materials.

According to this page, 37 % of X-rays with a wavelength of 0.56 nm will be absorbed by beryllium window that is 23.5 mm thick. If the window were made of aluminum, it would only have to be 1.4 mm thick to absorb the same amount of X-rays, and if it were made of lead it would only have to be 13 μm thick.


Find the maximum number of tripeptides can be synthesized by using three different amino acids

There is a misconception that runs through the questions, the answer key and the comments: For tripeptides, the peptide linkage is directional (i.e. the tripeptide has an amino terminal and a carboxy terminal), so ABC is different from CBA.

To get the number of tripeptides combined from three types of amino acids, you have three choices for the first, three independent choices for the second and another three independent choices for the third position, for a total of 3 x 3 x 3 = 27 possibilities.

  [OP] I am unsure if I've done it in a valid way, as answer given is 18.

The answer key is incorrect. I have no way of reading the question that 18 would be the answer, even considering the poor wording of the question.

  [OP] For AAZ type, we've three set, AAB, ABA, AAC, ACA. So, total 12 tripeptides.

There are three choices for Z, two remaining choices for A, and three positions for Z (AAZ, AZA, ZAA). So, total of 3 x 2 x 3 = 18 tripeptides containing two different amino acid types of the initial set of three types.

  [Orthocresol] This question is phrased very poorly. Without reading your working, I would have answered 20×19×18=6840.

This answers how many different tripeptides made from (20) standard amino acids contain no amino acid twice. The question, certainly poorly stated, says "synthesized by using three different amino acids". So your reagents are a set of three amino acids types, and you are making tripeptides from them.

  [Maurice] With two amino-acids, A and B, there is already 8 tripeptides : AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB. It makes 8 tripeptides with A and C, and also 8 tripeptides with B and C. Plus the tripeptides with ABC, which are : ABC, ACB, BAC, BCA, CAB, CBA. In total : 8 + 8 + 8 + 6 = 30 tripeptides.

That is indeed a math problem, not a biochemistry problem. You can make AAA from A and B, but also from A and C, so you counted AAA twice (and BBB and CCC). So your answer overshot by 3.

  [Rahul Verma] Are the synthesized amino acids valid? If one of the amino acid (say A) is optically active, then can BAB be two tripeptides?

The linkage does not introduce new chiral centers (different from e.g. combining sugars), so each sequence XYZ designates a single molecule (given that you start with enatiomerically pure amino acids).

  [porphyrin] The list is aaa , aab , aac , aba , abb , abc , aca , acb , acc , baa , bab , bac , bba , bbb , bbc , bca , bcb , bcc , caa , cab , cac , cba , cbb , cbc , cca , ccb , ccc which has 27 entries and removing 9 palindromic arrangements (except aaa, bbb, ccc) this leaves 18.

The list is correct, but there is no reason to remove palindromes (9 in the list) or divide non-palindromes by two (18 in the list).

Update

I missed all the comments that came in while I was writing this. They now reflect the misconceptions and how these can be resolved; what is not resolved is the poorly worded question and its incorrect answer key - maybe someone who has the original question can take on the task of sending an erratum.


adiabatic expansion vs Joule-Thomson Effect

In the Joule-Thompson experiment, there is major viscous dissipation of mechanical energy (in this case pressure) to internal energy occurring within the pores of the plug.  This is analogous to Poiseuille flow of a viscous fluid in a tube.  If the tube is adiabatic, the fluid flowing through a tube experiences a temperature rise as a result of the viscous "heating."  The work to push the fluid through the plug is totally dissipated and converted to increased internal energy of the gas.  The reason that viscous effects are so important in flow through a plug is that the pores are very small.  So flowing the gas through the plug is analogous to flowing the gas through an array of very tiny tubes in parallel.

Ordinarily when a gas expands adiabatically, it cools.  But, for flow of an ideal gas through a porous plug, the adiabatic cooling is exactly offset by the viscous heating.  So there is no temperature change.  And, as a result, there is no enthalpy change.


Why isn't nitrous acid a product of nitrite and water?

It should form nitrous acid... If it was $NO_3^{-}$, nitrate ion, then it would presumably form nitric acid $HNO_3$.

Nitrous acid reacts with water to form nitrite and hydronium:

$$\ce{HNO_2 + H_2O -> NO_2^- + H_3O^+} $$

Water is an amphoteric species. Assuming we just have water the nitrite ion, the nitrite ion accepts the hydrogen forming nitrous acid:

$$\ce{NO_2^- + H_2O -> HNO_2 + OH^-} $$

This reaction is interesting as well since nitrous acid will also dissociate again into nitrite ions and water as well:

$$\ce{HNO_2 + H_2O -> NO_2^- + H_3O^+} $$


Since OP is still in the high school, I'll try to explain it simply as possible using mathematical manipulation (hoping OP is more familiar with mathematics than chemistry).

Both gold and platinum consist of same crystal packing called body-centered cubic, which is illustrated in following diagram:

Crystal studies of both gold and platinum has revealed that the lengths of their cubic unit cells ($a$) are $\pu{407.82 pm}$ and $\pu{392.31 pm}$, respectively. If you inspect closely, you would realize that each of eight corner atoms shares with eight unit cells while each of six face-centered atoms shares with only two unit cells. Thus, total atoms per unit cell  $= 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$. Keep in mind that $a$ is depend on the atomic radius of each metal and how closely they packed against each other against some concerning forces (e.g., van der Waal's). For example, the experimental finding has been reported that the closest $\ce{Au-Au}$ separation is $\pu{288.4 pm}$ while that of $\ce{Pt-Pt}$ is $\pu{277.5 pm}$. These values imply that the metallic radius of gold is $\pu{144.2 pm}$ and that of platinum is $\pu{139 pm}$. This would cover the quote "the radii of the atoms and the bond lengths within the crystal lattice."

Based on the value of $a$ and number of atoms per unit cell for each metal, you can calculate the molar volume of each metal ($V_\mathrm{Metal}$):

$$V_\ce{Au} = a_\ce{Au}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\
= (\pu{407.82 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{10.21 cm3mol-1}$$

and similarly,

$$V_\ce{Pt} = a_\ce{Pt}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\
= (\pu{392.31 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{9.09 cm3mol-1}$$

Once we know the molar volume of each metal, we can calculate the density ($d$) by using $d=\frac{\text {molar mass}}{\text {molar volume}}$:

$$d_\ce{Au} = \frac{\pu{196.97 gmol-1}}{\pu{10.21 cm3mol-1}} = \pu{19.29 gcm-3}$$

$$d_\ce{Pt} = \frac{\pu{195.08 gmol-1}}{\pu{9.09 cm3mol-1}} = \pu{21.46 gcm-3}$$

Therefore, it is true that $d_\ce{Pt} \gt d_\ce{Au}$.

Relevant resources:

John W. Arblaster, In Selected Values of the Crystallographic Properties of Elements; ASM International: Material Park, OH, 2018 (ISBN-13: 978-1627081542).
Arnulf Maeland, Ted B. Flanagan, “Lattice Spacings of Gold–Palladium Alloys,” Canadian Journal of Physics 1964, 42(11), 2364-2366 (https://doi.org/10.1139/p64-213).
https://www.webelements.com/gold/crystal_structure.html
https://www.webelements.com/platinum/crystal_structure.html


Why does the ionic product of water remain constant after addition of non-neutral solute?

$K_w$ is the equilibrium constant of this reaction:
$$\ce{2H2 O <=> H3O+ +OH-}.$$
So, it's constant at a fixed temperature. It equals $10^{-14}$ at $25~^\circ\mathrm{C}$. At this temperature $\ce{[H3O+]=[OH^{-}]}=10^{-7}~\mathrm{mol/L}$.

If you add acid to water, the above reaction equilibrium is  displaced in order to counteract the change in the concentration of ion hydronium, i.e. the reaction is displaced to the left. This means that  ions hydroxide will react with the excess of ions hydronium to give water. As a result: the concentration of ion hydroxide will be less than $10^{-7}~\mathrm{mol/L}$, and the concentration of ion hydronium will be more than  $ 10^{-7}~\mathrm{mol/L}$. But their product is constant. $\ce{[H3O+] \cdot[OH^{-}]}=10^{-14}$.


"Volatile" usually refers to ease of evaporation, high vapor pressure, so it is a property of a pure substance while solubility is a property of the combination of two or more substances.

An intramolecular hydrogen bond makes a compound more volatile because the charge imbalances are offset internally instead of by forming an interaction with another molecule.

Interactions between different molecules of the substance would lower volatility, but intramolecular hydrogen bonds prevent the hydrogen bond donor and acceptor of one molecule from interacting with another molecule.


Why there is no need for reference standards in High Resolution Mass Spectrometry (HRMS)

The statement is not 100% correct. Ordinary mass spectrometry will not give you exact masses to 5 or 6 decimal places, say, m/z 415.14 for a singly charged molecular ion. Now there will be hundreds of compounds with this mass. On the other hand HRMS gives you a accurate and precise mass say 415.14509. This detailed number drastically reduces the number of matches in a spectral library. Not only that, the isotopic distribution pattern and their masses reduces the choice to one compound. Hence the claim that HRMS will not need a reference.

Like any claim, there are always exceptions. Structural isomers can be a problem. If we have isotopologues (isotopic isomers), even HRMS will fail.  Say you have two analytes but both are acetonitriles. Say one is

$^{13}$CH$_3$CN, and the other analyte is CH$_3$$^{13}$CN, HRMS will give the identical molecular mass to five decimal places. One might look up their fragmentation pattern but MS is not that great at low masses (high noise). Here you would need another spectroscopic technique such as microwave rotational spectroscopy. These techniques are used an in emerging field called position specific isotopic analysis.


Dependency of rate of diffusion on mean free path and velocity of molecules?

My reference is Concepts in Thermal Physics, Blundell and Blundell, Ch. 7, 2nd ed. Everything here is from there.

Consider a gas trapped in a rigid box. We begin by evaluating the molecular flux incident on one part of one wall of the box. Let there be $n$ particles per unit volume, so $n$ is the particle density of the gas in the box. Let $A$ be the area of a small patch of the wall, and let $\theta$ be the angle between the particle's direction of approach and the normal to the wall. Let $f(v)$ denote the speed distribution function of the particles, which we assume to be a Maxwell-Boltzmann distribution.

Consider all the molecules with speed between $v$ and $v+\text{d}v$ approaching the wall at an angle between $\theta$ and $\theta+\text{d}\theta$, which we write in shorthand as ($v,\theta$). There are $n$ particles per unit volume, and a fraction $f(v)\,\text{d}v$ of them are moving within the desired speed range. A further fraction $\sin\theta\,\text{d}\theta/2$ of them are moving in the desired direction, so the particle density for ($v$, $\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2.$$ (To derive the fraction $\sin\theta\,\text{d}\theta/2$, we consider a unit sphere with surface area $4\pi$. The surface area of the differential annulus is $2\pi\sin\theta$ (circumference) $\times\;\text{d}\theta$ (differential width), so the fraction of surface area occupied by the annulus is as stated.)

Now consider how many particles traveling at ($v,\theta$) hit the patch of wall in a time $\text{d}t$. These particles must be contained within a volume $$A\,\times\,v\,\text{d}t\cos\theta$$ (distance particle travels normal to the wall in a time $\text{d}t$), for all particles traveling at ($v,\theta$) must impinge on the wall within $\text{d}t$. Actually this area is slightly enlarged by a width $v\sin\theta\,\text{d}t$ uniformly, but this contribution to the area is of order $\text{d}t^2$ and vanishes in the small $\text{d}t$ limit.

All in all, then, the number of molecular collisions in $\text{d}t$ on a patch of wall of area $A$ at ($v,\theta$) is $$nf(v)\,\text{d}v\sin\theta\,\text{d}\theta/2\,\times\,Av\,\text{d}t\cos\theta.$$

The flux $\Phi$ is the rate of molecular collisions per unit area, so we divide by $A$ and $\text{d}t$, and integrate over all $v$ and $\theta$. This yields $$\Phi = \int_0^\infty\text{d}v\int_0^{\pi/2}\text{d}\theta\,nvf(v)\sin\theta/2\cos\theta = \frac{1}{4}n\langle v\rangle = \frac{p}{\sqrt{2\pi mk_BT}},$$ assuming an ideal gas and given the average speed of gases within the Maxwell-Boltzmann distribution.

By definition of the molecular flux, the rate of effusion is simply $\Phi A'$ for $A'$ the cross-sectional area of the hole out of which gas effuses, and the resulting dependencies can be read off from the equation for $\Phi$.

Interesting notes.

Due to the extra factor of $v$ in the integral for the molecular flux, the distribution of speeds of the effusing particles is $v^3e^{-\beta\epsilon}$ , rather than the standard $v^2e^{-\beta\epsilon}$. This extra factor of $v$ comes about because speedier molecules hit the walls of the box faster than do slower ones.
The hole through which particles effuse must be small, and the length scale that governs smallness is the mean free path. Make the hole too large, and the box eventually reaches equilibrium with the outside world. Consider a system of two boxes placed side by side, and a small hole between the neighboring walls of the boxes. For holes small enough that effusive behavior is observed, at equilibrium the molecular fluxes must balance and hence $$\frac{p_1}{\sqrt{T_1}} = \frac{p_2}{\sqrt{T_2}}.$$ On the other hand, if the holes are too large, then we simply have regular thermodynamic equilibrium, with $$\frac{p_1}{T_1} = \frac{p_2}{T_2}.$$


Why do gases under low pressure conduct electricity?

The cathode ray tube doesn't conduct electricity per se. It liberates free electrons that are simply moving through a vacuum from one end to the other. In practice, you can't create a real vacuum, so a low pressure gas (as low as possible) is used instead.

Air has a high enough dielectric strength that it will dampen the effect of the applied voltage. That is, it'll be harder to liberate free electrons to shoot down the tube.


Work done on a container, resulting in pressure change

Yes, the explanation

  the container does not expand, so as there is no change in volume of the container, no work is done on the container.

is right as it takes into account that the container is rigid, so its volume can not be changed. Thus, there is no work done by the gas on the container.

Also, if you say

  " pumping gas into the container require work in itself, due to the effort required to get the gas molecules into the container "

Yes, there is work done on the gas, but it is by the piston(or the pump).

You should see how the laws of Thermodynamics make clear about the boundary between system and its surroundings, so work has to be done by "something" on "something".
Also, In the question, it is clearly asked "Which statement is true of the work done on the container?" So, as the container's volume is unaltered, there is no work on it.


How can I isolate glycolic acid from sugarcane juice?

Ethylene Glycol was not used as a catalyst! Kataoka described in 2001 the production of Glycolic acid (GA) from ethylene glycol using glycol-oxidizing microorganisms isolated from soil, which is what the author of this thesis is trying to say (the writing is unclear though). It will be difficult to reproduce this in your home since you would need to cultivate those microorganisms inside your own house, chemical oxidation would also be a problem if you mean to use it for medical purposes (the oxidation of ethylene glycol releases formaldehyde, which is toxic and probably carcinogenic), besides the reaction needs potassium dichromate to happen which is also hazardous.

Fruits produce a small amount of glycolic acid and you can try to extract it from banana or sugarcane pulp by an acid-base workup, though I believe you won't get much pure GA using this method (specially if you don't have a deep knowledge of organic chemistry).

Glycolic acid is very easy to find online and it will be a lot cheaper to buy it from ebay or other source then to synthesize it if your goal is to use it for its skin effects.

Source: Glycolic Acid Production Using Ethylene Glycol-Oxidizing Microorganisms
Michihiko KATAOKA, Mié SASAKI, Aklani-Rose G.D. HIDALGO, Michiko NAKANO & Sakayu SHIMIZU
Bioscience, Biotechnology, and Biochemistry Vol. 65 , Iss. 10,2001


How to wedge d/hash notation a molecule?

If not using a Fischer projection, structural formulae with hashes and wedges allow to be more specific.
In the structure drawn in your current question, you provide information about the connectivity of atoms within the molecule, up to a point to indicate if the bond between two atoms is a single, double, or (not present here) a triple / quadruple.
Using wedges and dashes, equally provides this information; however additionally informs the reader about the sterochemical relationship of the atoms bound around an atom in a centre, addressing a specific (R)/(S) configuration.  Returning to your example, there are three sterogenic centres that I marked by an asterisk:

and for each centre you may find a pair of either configuration; for example for the hydroxyl group in $\alpha$ position to carbonyl alone

In total (for 3 stereogenic centres), there are $2^3 = 8$ distinct stereoisomers possible with different properties, which all fit into the first "line only" general representation.


If I understood what strong means, then graphene has high strength because of the 2-D layer structure of carbon atoms. The strength can also be attributed to its $sp^{2}$ hybridisation.

Any crystalline material will have a lattice structure. When we mean strength of a material its usually the ability to break apart the bond between two lattice atoms. But the observed strength is usually 2 or 3 orders lower than theoretically predicted strength due to flaws (voids, dislocations etc). But when graphene has a near flawless 2-D structure we can reach its theoretical strength. Though its is a very expensive process to actually realize "flawless". Even with flaws its strength is really high compared to steel.

So its essentially the 2-D one atom layer structure that can be responsible for its high strength.

One important thing is though graphene has high strength and elasticity (ability to get stretched)  it is relatively brittle (fracture toughness ~4 MPa$\sqrt{m}$).


How can you manage SCF convergence problems?

There can be many reasons for the SCF to not converge.

The initial guess wasn't great, and the convergence is slow. Increasing the number of SCF iterations can help here.
The SCF oscillates. This can happen if there are two orbitals close in energy. Using damping, mixing (i.e., the next iteration is 50% the old and 50% the updated iteration), Fermi broadening, etc. can help here.
The SCF strategy (e.g., DIIS) is causing problems. Usually DIIS can greatly accelerate convergence, but in some hard-to-converge systems, you need a lot of iterations with another strategy (e.g., Newton-Raphson)

Without more information from your particular file, it's impossible to know how to help. Many times, you may simply want to increase the maximum number of SCF iterations and it'll work fine.


How to determine the worst resonance structures out of a given set?

(C) is technically a valid resonance structure. You can derive it from (B) by moving the electron pair from the double bond between the carbons 2 and 3 (numbering from the left: this may not be correct from a nomenclature standpoint, but I'm using it for this answer) onto carbon 2 and moving the electron pair from the double bond between the oxygen and the carbon onto the oxygen.

(C) is least stable because it is, simply put, absolutely disgusting. Charge separation (carbons two and three) is gross, you have two cations in close proximity (carbons three and four), a terminal carbocation (carbon 1), and three atoms with empty valence shells (carbons 1, 3, and 4). So while the resonance structure is valid enough for (C) to be considered a resonance structure, that resonance structure happens to be the spawn of the devil.

That said, the compound in the answers isn't the same as the compound named in the question, like ron said.

As to the first part of your question, the filled valence shells are more strictly necessary for the second row elements because they don't have d orbitals to confound the picture.


What is the size of the rate constant dependent on?

The rate constant $k$ includes all the factors other than the dependence on concentration.  A significant factor in the rate is collision frequency and energy, so temperature has a significant influence on the value of $k$.  Other factors like the range of angles of collision necessary for a successful reaction, the strength of dipoles, solvent effects, quantum effects, and whatever factor corrects the units of the equation are not determined.

However the rate constant can be expressed by the Arrhenius equation
$$k=Ae^{-{E_a/RT}}$$
Doing this decomposes the constant into an exponential relationship including the activation energy, $E_a$, the molar gas constant, $R$, and of course temperature, $T$, which represents the fraction of total molecules that have sufficient energy to overcome the activation energy.
This equation comes with it's own new constant, the pre-exponential factor, $A$, which now includes all the factors in the rate constant that aren't the exponential relationship with $E_a$, $R$, or $T$. There may still also have some temperature dependence hidden within $A$, but usually it can be assumed constant over small temperature ranges.  Factors like collision frequency and angle, or action of a catalyst, are still obscured within this constant.

Even if these obscured effects were determined there would still be some new constant, $C$, which would include subtle effects such as interactions with surfaces in a container or non-reacting particles, temperature and concentration gradients, ionic strength, entropy changes in the solution due to changing composition, and any other factor that influences the reaction between the particles that hasn't been explicitly considered.


If alloys are homogeneous mixtures, why can't we separate their components?

If you melt the alloy and then use electrolysis, or if you use a redox reaction, you should be able to separate the components.

For example, if you used a basic acid, let's say hydrochloric acid, on a zinc-copper alloy, I'd imagine the zinc would be removed from the alloy and taking the hydrogen's place, forming zinc chloride, leaving hydrogen gas and copper metal.

This is based off of simple redox reaction memories and no personal experience. However, if you're curious, try it out!


Also we need some empty orbitals on the metal cation, since we're looking at Lewis adducts.

Ahem. Nope. $\pi$-bonding in $\ce{CO/NO+}$ complexes is from back-donation, from occupied metal d-orbitals onto unoccupied $\pi^*$ orbitals of the ligand.

Looks like homework question, so no further details provided.


According to J. Am. Chem. Soc. 1945, 67 (2), 322–324:

Planar configurations for resorcinol and hydroquinone are more acceptable on comparison of experiment and calculation than are the structures assuming either free or partially restricted rotation of the hydroxyl groups.

It states that the dipole is present because there is restricted rotation with the hydroxyl groups. The preferred conformation for hydroquinone is a flat molecule with the hydrogens in the plane.


Total number of isomers formed on monobromination of 2-methylbutane

Here are the six structures that your source is probably referencing.


Your basic strategy to involve a Friedel-Crafts acylation is correct, but your route is a bit circuitous.  You could acylate benzene directly to produce benzaldehyde in one step using carbon monoxide and hydrochloric acid to generate formyl chloride in situ.

While it is "just another" Friedel-Crafts acylation, it has it's own name, the Gatterman-Koch reaction, after its discoverers.  Once in hand, you could oxidize benzaldehyde to benzoic acid.

An alternate approach would be to 1) brominate benzene to form bromobenzene, 2) form the Grignard reagent with magnesium and 3) react it with carbon dioxide to produce benzoic acid.

In either case, with benzoic acid in hand, you can procede as you've indicated, on to benzophenone.


Wouldn’t it better to just fill them up with single electrons, as per
  Hund’s rule, and then leave them like that? Would that not be more
  energetically favorable?

You have to consider two different scenarios here.

Case 1. Orbitals of different energies.

Consider a two electron system, say, a $\ce{He}$ atom, and let us concentrate on $\mathrm{1s}$ and $\mathrm{2s}$ orbitals exclusively. Naively, one could think of the total energy of the system being simply the sum of orbital energies. Then for $\mathrm{1s^2}$ configuration we get $E_{\mathrm{1s^2}} = \varepsilon_{\mathrm{1s}} + \varepsilon_{\mathrm{1s}}$, while for $\mathrm{1s^1 2s^1}$ configuration we get $E_{\mathrm{1s^1 2s^1}} = \varepsilon_{\mathrm{1s}} + \varepsilon_{\mathrm{2s}}$. Now clearly, since $\varepsilon_{\mathrm{1s}} < \varepsilon_{\mathrm{2s}}$, we could expect that $E_{\mathrm{1s^2}} < E_{\mathrm{1s^1 2s^1}}$, i.e. that  $\mathrm{1s^2}$ configuration is indeed more energetically favourable than $\mathrm{1s^1 2s^1}$ one.

This is a bit naive way of looking at things since the total electronic energy is not equal to the sum of orbital energies. As you mentioned, pairing the electrons up takes some energy, so that $E_{\mathrm{1s^2}}$ is clearly not just the sum of $\varepsilon_{\mathrm{1s}}$ and $\varepsilon_{\mathrm{1s}}$, but this amount of energy is relatively small compared with the energy gain you get by having both electrons in the lower $\mathrm{1s}$ orbital. In other words, energy gain from placing both electrons in energetically more favourable $\mathrm{1s}$ orbital overcompensates for having to pair them up during the placement.

How do we know that? Well, we know it from the quantum theory, or, to be more precise, from its application to a chemical systems. Thus, a solid understanding of quantum mechanics & quantum chemistry is required for a quantitive discussion, but for now you can be certain that electrons fill the lowest available energy levels before filling higher levels. This is the principle known as the Aufbau principle which follows from quantum theory and generally holds.

Case 2. Orbitals of same energy.

Consider a $\ce{C}$ atom now and let us have a look at two electrons in  $\mathrm{2p}$ orbitals. You could populate the same, say, $\mathrm{2p_x}$ orbital by a pair of electrons, or you could let the electrons populate two different orbitals, say, $\mathrm{2p_x}$ and $\mathrm{2p_y}$. Naively, we could expect the total electronic energy for both configurations to be equal to each other, since $\varepsilon_{\mathrm{2p_x}} = \varepsilon_{\mathrm{2p_y}}$, so that,

$$
  E_{\mathrm{2p_x^2}}
  = \varepsilon_{\mathrm{2p_x}} + \varepsilon_{\mathrm{2p_x}}
  = \varepsilon_{\mathrm{2p_x}} + \varepsilon_{\mathrm{2p_y}}
  = E_{\mathrm{2p_x^1 2p_y^1}} \, ,
$$

but this is again "not quite right", since pairing the electrons up takes some energy. And this time there is no any gain to compensate for this effect since $\varepsilon_{\mathrm{2p_x}} = \varepsilon_{\mathrm{2p_y}}$, and as a result $\mathrm{2p_x^1 2p_y^1}$ configuration is energetically more favourable than $\mathrm{2p_x^2}$ one.


Electrolysis of copper without any copper initially in solution

It is a complex problem. Yes, NaCl solution is a conductor, so electricity will pass through the solution. Now in your situation both electrode are made of copper. Since electricity is passing through the solution, electrolysis must occur at both electrodes. You cannot have electrolysis only at one electrode.

Assuming there is sufficient salt in the solution, what will happen at the anode?

Chlorine gas will evolve. I will let you find relevant equations. Chlorine gas attacks copper very quickly and the anode starts to "degrade" quite fast, but will it pass into solution? Nope.

Let us see what is happening at the cathode?

Sodium ions cannot be reduced in water esp. on copper electrodes (on mercury they do), instead water will be reduced. What happens when water is reduced? It turns into hydrogen gas, which escapes, and leaves hydroxide ions behind. This process makes the solution basic.

Now coming back to the anode, since you have hydroxyl ions around, they precipitate copper ions. All you get is yellow orange junk in such a system. Cu(II) is blue so why an orange yellow mess? It is most likely due to Cu(I) compounds. Anode dissolves pretty fast but nothing happens to the cathode.


How does vinegar get rust off these coins

As previous comments said, this isn't a redox reaction.

The patina or corrosion is made of metal salts -- for copper, it's copper oxide, sulfide, carbonate, and perhaps other things. Vinegar readily converts all these to soluble copper acetate.

Most coins are made of metal alloys, and those alloys sometimes contain metals that are attacked by acid. Nickel and zinc are common components. Soaking in vinegar will dissolve these metals (they'll displace hydrogen, again forming soluble metal acetates). This damages the coin's surface, making it porous, and eventually reducing the coin's weight. (It also breaks collectors' hearts.)

Adding salt to the mix forms an equilibrium with hydrochloric acid, which attacks metals even more aggressively. Please don't do this unless you don't mind ruining your coins.


Omission of locants for functional groups in IUPAC names

$\hspace{6.2cm}$

There is only one possible phenylacetaldehyde, so its IUPAC name is in fact phenylacetaldehyde. 2-Phenylacetaldehyde is its systematic IUPAC name, but many chemicals' official and systematic IUPAC names differ, like propan-2-one for example. The reasoning for this is often historical and to reflect what is most commonly used in the scientific community.

$\hspace{4.9cm}$

Chloroacetaldehyde is quite obviously an aldehyde, but  switching the position of the hydrogen and the chlorine as you suggested results in a completely different class of molecule known as an acyl chloride. There is therefore only one possible chloroacetaldehyde and no need to specify the chlorine's location, though its systematic IUPAC name is indeed 2-chloroethanal.


The thickness of the attacked layer (or the corresponding loss of weight per surface area) due to alkali attack on glass mainly depends on the following parameters:

type of glass (most laboratory glass is borosilicate glass 3.3)
concentration of the alkaline solution (or the corresponding pH)
temperature
exposure time

Note that the density of typical borosilicate glass is 2.23 g/cm³ (at 25 °C). Thus, an attacked thickness of 1.00 µm corresponds to a loss of 22.3 mg/100 cm².

The following figures are taken from product information for Schott DURAN® glass:

Alkali attack on borosilicate glass as a function of pH, exposure time: 3 h

Alkali attack on borosilicate glass as a function of temperature, calculated from the weight losses, c(NaOH) = 1 mol/l, exposure time: 1 h

Alkali attack on borosilicate glass as a function of time

For example, in order to remove a layer of about 1 µm, you can expose the glass for a period of approximately 40 min to an alkaline solution with c(NaOH) = 1 mol/l at 100 °C.


Totally confused about the location of "s,p,d,f" Orbitals inside the atom

I know that electrons revolve around nucleus in their orbits.

You should "unknow" that, because that is the failed Bohr Model of the atom.

The are no fixed orbits for electrons in atoms.

The Bohr Model has been superceded by the Schrodinger model.

The atom is described in terms of probabilty density: for each point in space the probability that an electron will be at that point (when the atom is in a given state such as 1s, 2s, 2p, etc).

  my question is "how the orbitals are located in the orbits?

There are no orbits

  Do they really posses Spherical, dumb bell, double dumb bell shapes inside the orbits??

The shapes you seen in drawings that are spherical, dumb bell, etc. are meant to show regions of higher electron probabilty density.  For example, someone might draw a sphere on the basis that there is a 90% probablity of the electron in that orbital being within the sphere at a given moment.  However, in reality, for example in the 1s state, the electron could be anywhere in space. Other states, such as 2s, have nodes, with the electrons possibly being anywhere except at the nodes.

So think of the shapes as showing where the electrons in that orbital are often, but remember that they can really be anywhere excpet at the nodes.


The unit cell for graphene is a two-dimensional rhombus according to the figure shown on page 31 of this paper.$^1$ (also here.) The result is that two atoms are contained per unit cell. The upper right structure actually appearing in graphite, stacked layers of graphene.

$^1$Zhou, J; Huang, R. Internal lattice relaxation of single-layer graphene under in-plane deformation. Journal of the Mechanics and Physics of Solids, 2008, 56(4), 1609-1623. https://doi.org/10.1016/j.jmps.2007.07.013


What is/are the product when this reacts with Lucas reagent?

A shift of the methyl group is likely, because this would yield a more stable tertiary carbocation. In concentrated hydrochloric acid, the double bond can also be protonated, producing a secondary carbocation which is captured by $\ce{Cl-}$. If both happens, the product would be 2,4-dichloro-4,5-dimethylhexane.


A quick search would tell you what the 4 nucleobases for RNA are made of:

Uracil: 4 Carbon 4 Hydrogen 2 Nitrogen 2 Oxygen

Adenine: 5 Carbon 5 Hydrogen 5 Nitrogen

Guanine: 5 Carbon 5 Hydrogen 5 Nitrogen 1 Oxygen

Cytosine: 4 Carbon 5 Hydrogen 3 Nitrogen 1 Oxygen


If we consider

$\ce{F2(g) -> 2 F(g)}$ with $\Delta H_\mathrm{F, bond}$

$\ce{2 F(g) + 2 e- -> 2 F-(g)}$ with $\Delta H_\mathrm{F, aff}$

$\ce{2 F-(g)  ->[H2O] 2 F-(aq)}$ with $\Delta H_\mathrm{F, hydr}$

and respectively

$\ce{Cl2(g) -> 2 Cl(g)}$ with $\Delta H_\mathrm{Cl, bond}$

$\ce{2 Cl(g) + 2 e- -> 2 Cl-(g)}$ with $\Delta H_\mathrm{Cl, aff}$

$\ce{2 Cl-(g)  ->[H2O] Cl-(aq)}$ with $\Delta H_\mathrm{Cl, hydr}$

Then overall reaction enthalpy for $\ce{ F2(g) + 2 Cl-(aq) -> Cl2(g) + 2 F-(aq)}$ is

$$\Delta H_\mathrm{r} = \left({\Delta H_\mathrm{F, bond} - \Delta H_\mathrm{Cl, bond}}\right) + \left({\Delta H_\mathrm{F, aff} - \Delta H_\mathrm{Cl, aff}}\right) + \left({\Delta H_\mathrm{F, hydr} - \Delta H_\mathrm{Cl, hydr}}\right)$$

The differences between fluorine and chlorine enthalpies of atomization(bond breaking ), electron affinity and ion hydration lead to the overall reaction enthalpy, that is pushing reaction toward chloride oxidation.

The truth is, that the equilibrium is driven by Gibbs energy, not enthalpy,

$$\Delta G^{\circ}_\mathrm{r}=\Delta H^{\circ}_\mathrm{r} - T \cdot \Delta S^{\circ}_\mathrm{r}=-RT\ln{K}$$

but for the rough reasoning you get the picture. Futhermore, the entropy change will not be major factor, due analogical structures.


You are right assuming that μ is a metric prefix, and that the number denotes the concentration of suxamethonium (suxamethonium chloride, I presume). However, this is supposed to be small caps "M" (as $\rm \small M$), not capital "M" (as $\rm M$). Also note that use of the term "molarity", as well as its notation $\rm \small M$, is discouraged [1, p.27]:

  The term molarity and the symbol $\rm \small M$ should no longer be used because they, too, are obsolete. One should use instead amount-of-substance concentration of $\ce{B}$ and such units as $\rm mol/dm^3$, $\rm kmol/m^3$,
  or $\rm mol/L$.

And, of course, $1\,\rm {\small M} = 1\, mol/L$.

Also, check out the answer to a similar question "What does the unit M stand for?"

Reference

Thompson, A.; Taylor, B. N. Guide for the use of the international system of units (SI). NIST Special Publication 811 2008.


Relation between size of ion and hydration energy

We can look at solubility as a competition between lattice energy and hydration.

$\Delta H_\text{solution} = \Delta H_\text{hydration} + \Delta H_\text{lattice}$

If the enthalpy of solution is exothermic or slightly endothermic, the compound will be soluble (we are ignoring entropy here).

If we take a simplified view of both lattice energy and enthalpy of hydration, we can see that they are both generally proportional to both the size of the ions as well as the charges, that is:

$$E \propto\frac{|q_1q_2|}{r} $$

Where $q_1$ and $q_2$ are the charges on the species in question and $r$ is the inter-nuclear distance.  So, ions with higher charges are more attracted to each other AND form strong ion-dipole interactions with water.  At the same time, larger ions experience weaker attractions to each other AND form weaker ion-dipole interactions with water.

It is important to remember that this simplified analysis does not take lattice structure and crystal packing into account at all.

The differences in solubility between different ionic compounds follows two general trends.

Mismatch in Ion Size
In general, when there is a difference in the size of the ions in a compound, the compound is more soluble.  This probably arises from poorer packing in the solid state.  For example at room temperature, $\ce{MgSO4}$ has a solubility of 1.16 g/100mL while $\ce{BaSO4}$ has a solubility of $2.45*10^{-4}$ g/100mL$^{[1]}$.  The barium ($159$ pm) and sulfate ($242$ pm) ions are much closer in size than magnesium ($86$ pm) and sulfate$^{[2]}$.
Covalent Character
In general, the more covalent character present in an ionic compound, the lower it's solubility.  We can estimate covalent character by looking at the difference in electronegativities of the ions in the compound.  The larger the difference in electronegativities, the less covalent character in the compound.

For example, $\ce{AgF}$ has

$\Delta \mathrm{EN} = (4.0 - 1.93) = 2.07$

Solubility = $\dfrac{180}{100} \mathrm{~g~mL^{-1}}$

While $\ce{AgCl}$ has

$\Delta \mathrm{EN}= (3.5 - 1.93) = 1.57$

Solubility = $\dfrac{0.193}{100}\mathrm{~g~mL^{-1}}$

In summary, the solubility of ionic compounds is greater when there is a significant size difference between the ions and when there is a greater difference in their electronegativities.

Footnotes

[1] Tro et al, Chemistry, A Molecular Approach, 2nd Canadian Edition

[2] https://en.wikipedia.org/wiki/Ionic_radius


Acidic nature comparison; benzoic acid and phenylacetic acid

The acidity of benzoic acid $(\ce{C6H5-COOH})$ and its substituted derivatives, $\ce{R-C6H4-COOH}$ $(e.g., \ \ce{4-NO2-C6H4-COOH})$ have been studied extensively using the Hammett Equation. Similarly, acetic acid $(\ce{CH3-COOH})$ and its substituted derivatives, $\ce{R-CH2-COOH}$ $(e.g., \ \ce{Cl-CH2-COOH})$ have been studied extensively as well using similar Hammett Equation. You may read some brief discussion of Hammett Equation here.
Actually, you are comparing benzoic acid and a $\ce{R-CH2-COOH}$ where $\ce{R = Ph}$. Their is no relationship between these two acids except for both having $\ce{Ph}$ functionality. Benzoic acid has $\ce{Ph}$ directly connected to $\ce{COOH}$ group while acetic acid derivative has it connected through $\ce{-CH2\!-}$ group. Because of that, acidity of $\ce{Ph-CH2-COOH}$ is depend solely on inductive effect. It isworth noting that I cannot find any comparative studies of these two type of carboxylic aid in one single research.
The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.20$ $(K_\mathrm{a} = 6.3 \times 10^{-5})$ while the $\mathrm{p}K_\mathrm{a}$ of acetic acid is $4.75$ $(K_\mathrm{a} = 1.77 \times 10^{-5})$. Thus, benzoic acid is about 3.6 times acidic than acetic acid. That is because when you consider $\ce{R-COOH}$ acid, it is more acidic if $\ce{R}$ group is electron withdrawing group (EWG) such as $\ce{Ph}$ group, compared to $\ce{R}$ group being electron donating group such as $\ce{CH3}$ group (EDG) like in the case of acetic acid.
Now, let's increase the electron withdrawing ability of $\ce{R-CH2-COOH}$ acid compared to acetic acid (where $\ce{R = H}$): The $\mathrm{p}K_\mathrm{a}$ value of 2-chloroacetic acid (where $\ce{R = Cl}$) is $2.87$ $(K_\mathrm{a} = 1.36 \times 10^{-3})$, which is about 76.8 times acidic than acetic acid. Thus, we can conclude that $\ce{ClCH2}$ group has larger inductive effect than that of $\ce{Ph}$ group. Thus, if we change $\ce{Cl}$ in $\ce{ClCH2}$ group to $\ce{Ph}$ group, then resulting carboxylic acid should be less acidic than that of 2-chloroacetic acid.
As given in OP's question, this carboxylic acid, 2-phenylacetic acid $(\ce{Ph-CH2-COOH})$ has a $\mathrm{p}K_\mathrm{a}$ value of $4.31$. It is more acidic than acetic acid, but not quite acidic as benzoic acid because of the extra $\ce{CH2}$ group. Remember, inductive effect is weaken by additional $\ce{C-C}$ bonds (c.f., ortho-, meta- and para-inductive effects).
Source of  $\mathrm{p}K_\mathrm{a}$ values


What forces keep a cotton fiber together?

A glucose molecule is roughly 1 nanometer long (ref).  There are roughly 9,000-15,000 glucose molecules in a cellulose fiber (ref, see Table 5).  If we use 10,000 as the number of glucose units, then the length of a cellulose fiber will be
$$10^{-9}~ \mathrm{m/molecule} \cdot 10^4 ~\mathrm{molecules/fiber}=10^{-5}~\mathrm{m/fiber} $$
or ~10 microns/fiber.

The mechanical strength of the fiber comes from two sources.

Structural: As your diagram illustrates there are many walls in the cellulose fiber.  All of these walls reinforce one another and build up the mechanical stability of the fiber. Each wall is composed of highly-organized cellulose chains.  "Highly-organized" is another way of saying "crystalline".  Cotton has an average crystallinity around 75%.  This allows the cellulose fibers to pack closely together further increasing mechanical stability.  The cellulose fibers in the different walls are orientated in different directions, again serving to increase the fiber strength.
Chemical: Here is a drawing of 2 glucose molecules from a cellulose chain.

Note that each glucose molecule has 3 equatorial hydroxyl groups oriented out and away from the polymer chain.  Because of the high crystallinity mentioned above, the cellulose chains are positioned very close to one another.  This means that thousands of hydrogen bonds can form between hydroxyl groups on different chains.  As a rule of thumb, a hydrogen bond might have a strength around 5 kcal/m; if we have thousands of these bonds holding the cellulose chains together, then we have the equivalent of hundreds of normal chemical bonds holding the chains together!  These hydrogen bonds account for much of the stability of the cotton fiber.


How to calculate pH of the Na2CO3 solution given ambiguous Ka values

The  carbonate ion is the Conjugate base of the weak acid
$\ce{HCO_3^-}\ (K={4.7\times10^{-11}})$, so this solution will alkaline. Given the concentration of this solution ,the pH should be sufficiently high to preclude the formation of any significant amount of $\ce{H_2CO_3}$ , so the solution of this problem as a solution of a monoprotic weak base:
$\ce{CO_3^{-2} + H_2O <=> HCO_3^-  + OH^-}$
$$\ce{K_b}=\frac{[OH^-][HCO_3^-]}{[CO_3^{-2}]} =\frac{K_w}{Ka}=\frac{10^{-14}}{4.7\times 10^{-11}}=\ce{10^{-3.7}}$$
Neglecting the $\ce{OH^-}$ produced by the autoprotolysis of water, it is valid to make the usual assumption that $\ce{[OH^-]}={[HCO_3^-]}$,and thus
  $$\dfrac{[OH^-]^2}{0.05-{[OH^-]}}=K_b= \ce{10^{-3.7}}$$
The equilibrium expression must be solved as a quadratic and yields the root $\ce{[OH^-]}$=0.00306 Which corresponds to pOH = 2.5 or pH = 11.5