Good Introduction to chemistry textbook I want to learn basic chemistry, so which textbook do you think I should pick? It should be something both concise and clear. Thank you.

I would recommend Zumdahl Chemistry. It discusses most of the basic topics, and contains lot of exercises.


What does an electron's spin of 0.5 and minus 0.5 signify? While teaching me magnetism, my teacher told me about the spin of an electron. He told me that the spin of .5 means that if we rotate the electron twice counter-clockwise on its axis, we would have the same face(picture) of the electron. And if we rotate an electron of -0.5 spin twice in the opposite direction, we would have the same face. But if rotate it only once(360⁰), we would not get the same face/picture of an electron.
How can this be possible? Or am I just making a mistake somewhere? A simple explanation would be greatly appreciated for I am only a tenth grader.

It is very tempting (and often also very useful!) to picture electron spin as an angular momentum vector, similar to a spinning top. Using this analogy, there are two properties (or numbers) of this angular momentum vector that we need in order to describe the electron spin. The first one is the spin itself and this is often designated the symbol $s$. The second number is the projection of the spin on a particular axis (often the $z$ axis) and this is assigned the symbol $m_s$.
You probably learned already that not all values of $s$ and $m_s$ are allowed in quantum mechanics. From measurements we know that an electron has a spin of $s=1/2$ (always positive) and its projections can be $m_s=+1/2$ or $m_s=-1/2$. The quantum numbers $s$ and $m_s$ are just numbers we use to label a particular state of the electron, but when we do measurements we also measure units. The units of angular momentum are J s (Joule second) and from comparing the classical definition of angular momentum with the rules of quantum mechanics, we can derive that one unit of angular momentum in quantum mechanics has a value of $\hbar$. This means that the angular momentum of the electron has a value which is basically half of what you would expect based on classical arguments and sometimes people like to emphasize the weirdness of this by saying things like "an electron has to rotate twice around its axis to come back to its original position". Of course this not true because the angular momentum of the electron is not a consequence of a rotation at all (the electron is a point particle). In addition, the sign of the projection of $s$ has nothing to do with this hypothetical rotation either. The sign of $m_s$ does influence the interaction of the electron with an inhomogenous magnetic field. If you have a magnetic field gradient along the $z$ direction, electrons in one state move up while electrons in the other state move down and you will see two separated spots on a position sensitive detector behind the magnet. This is the basis of the famous Stern-Gerlach experiment. Note that you have to perform these kind of experiments with neutral systems that behave like a single electron because a charged electron would experience a much stronger interaction between its charge and the magnetic field (Lorentz force) than between its spin and the magnetic field (Zeeman interaction).
What I find very interesting about half-integer spin is that there is no orientation possible for which the electron spin does not interact with the magnetic field (which would correspond to $m_s=0$ or a perpendicular orientation of the electron spin with respect to the magnetic field).


What equipment to use to make liquid oxygen? I'd like to do some experiments with liquid oxygen but as a first step I'll have to make some. Unfortunately I don't have any experience designing cryogenic experiments. I'm thinking I'll cool gaseous oxygen using liquid nitrogen as the coolant and then I want to collect the liquid oxygen condensate in a small dewar of some kind.

It needs to be high purity so I'll be using 99.9% purity "scientific grade" oxygen from a cylinder.

Does it make sense to use conventional condenser glassware like a dimroth condenser for instance? Alternatively what's the best equipment to use and what's the best way to rig a cryogenic setup like this?

Also what safety precautions should I take?

Simply put an open flask in the LN2 and it will fill with liquid oxygen from the air.

Or better: DON'T DO THAT!! What the heck do you want with it?!?

LO2 is highly dangerous. If it comes in contact with anything combustible, it will blow up vigorously. Together with lab solvents, it will likely explode. Make sure the flask or dewar is dead clean.
Chemistry is science, and science is about prediction. If you cannot predict, from knowledge, not hearsay, what your experiment will likely do, don't do it.


Why does C=O have a larger dipole moment than C-O? Why does $\ce{C=O}$ have a larger dipole moment than $\ce{C-O}$?

According to me, dipole moment directly depends upon bond length and electronegativity difference. In $\ce{C=O}$ and $\ce{C-O}$, (I guess, please clear if I am wrong) the EN difference is same, and bond length of $\ce{C=O}$ is less than $\ce{C-O}$. So, $\ce{C=O}$ must have less dipole moment than $\ce{C-O}$ but it is not so. Why?

According to Wikipedia, bond dipole moment depends on:

Distance between atoms and
Overall charge difference, not just electronegativity difference.

Resonance tells us that there is some amount of charge separation in $\ce{C=O}$ bonds because of the $\ce{C+-O-}$ contributor. This difference in charge, in addition to the electronegativity difference, is more significant than the decrease in distance between atoms, hence the larger dipole moment for $\ce{C=O}$.


Do absorbances of different compounds at one wavelength sum up? Question on my practice exam. The answers to the questions were not released, otherwise I could work backwards to figure out how to solve it. As it stands, though, I can only guess the method to solve this question.

The extinction coefficients of the 4 natural nucleoside-5'monophosphates @ 260nm
(Cavaluzzi,MJ,Nucleic Acids Res., 2004, 32,e13).

*pdA = 1.506 E4 L/mol/cm
*pdG = 1.218 E4 --
*pdC = 7.10 E3 --
*pdT = 8.56 E3 --

Find the Extinction Coefficient of a single stranded, 5'-phosphorylated Dickerson Dodecamer sequence d(pCGCGAATTCGC)

If I saw this question on a test, my first instinct would be to add up the values for pdC+pdG+pdC etc...(following the sequence of letters). Or maybe average them all.

We did learn the Beer-Lambert Law used in absorption spectrum, where e was extinction coeffecient: $$A=E(extinction)*b*c$$  but I can see no way to use this equation, since i'm only given a wavelength and a whole bunch of coefficients.

Does anyone know how to solve this? Is there an equation I'm missing?

Thanks in advance.

[…] I can see no way to use this equation, since i'm only given a wavelength and a whole bunch of coefficients.

We have all the data that we need.

You have already mentioned the Lambert-Beer law.

For the solution of a single compound S in a cuvette of width d, the absorbance $E_\lambda$ at a particular wavelength $\lambda$ is proportional to the concentration $c$ of S.

\[E_\lambda = \log_{10}{\left(\frac{I_0}{I} \right)} = \epsilon_\lambda\cdot c \cdot d \]

A higher concentration of S will result in higher absorbance.

Now imagine that the light-absorbing molecules S are linked together by units that do not absorb the light at the wavelength observed.

Let's compare two equimolar solutions of
$$\ce{-linker-S-linker-S-linker}$$
and
$$\ce{-linker-S-linker-S-linker-S-linker-S-linker-S-linker}$$

The second will show a higher absorbance, since the concentration of S is higher!

Now let's have a look at equimolar solutions of different compounds S1 and S2 and their absorbance.

We have outlined above that the absorbance $E_\lambda$ is proportional to the concentration of the light-absorbing species.

But apparently, different compound to not all absorb the same at a particular wavelength!

Exactly this is expressed with the molar absorption coefficient $\epsilon_\lambda$. The higher $\epsilon_\lambda$, the stronger the attenuation of incoming light. Note that $\epsilon_\lambda$ is

always given for a particular wavelength
a compound-specific constant

Let's tie the pieces together!

When light-absorbing units are linked together by 'blind' linkers, the effective concentration of the absorbing units counts.
Different compounds can absorb at the same wavelength, but due to their specific $\epsilon_\lambda$, they will do this to different amounts.

Consequently, we can sum up the combined absorbances in the following way:

$$E_\lambda =  \left(\epsilon_A \cdot c_A + \epsilon_C \cdot c_C + \epsilon_G \cdot c_G + \epsilon_T \cdot c_T \right)\cdot d$$


How long does it take to replicate Friedrich Wohler's synthesis of urea? I’m an author, writing a story where a doctor from modern day gets dropped into the late 1800s in California. They need urea to help a patient.

Friedrich Wohler's synthesis of urea happened before the time of my story and the necessary chemicals, Silver Cyanate and Ammonium Chloride, appear to be part of a normal pharmacist chemicals of the time.

So my questions are:

With the help of a local pharmacist, could they synthesize urea?
And if so, how long would it take?

Thanks for the help.

OK, I am not answering the question but the author Kyros might be interested in how Professor Benjamin Silliman, Sr. (1779-1864) described Wohler's synthesis in his Elements of Chemistry (1831) (Vol II, pg.600) just 3 years after the landmark experiment.
Although Silliman conceded that the properties of the "artificial" urea had the properties of urea, he appears to have been skeptical. Wohler wrote to his mentor J. J. Berzelius in Sweden about his discovery on how he, Wohler, could make urea without the need of kidney, be they animal, human or dog. Berzelius was delighted because the reaction was an example of isomerism, a term Berzelius had coined.

Preparation. — By double decomposition, by muriate of ammonia and cyanate of silver; or by single decomposition, by cyanate of lead and liquid ammonia. In this second process, the oxide of lead is liberated and a compound is obtained, in colorless rectangular prisms. In their properties, they are identical with urea, and their composition is exactly the same; they give the same results in analysis as cyanate of ammonia, provided 1 equiv. of water is supposed to exist in that salt.
Still the artificial urea, although from the mode of its formation it would appear that it contains only cyanic acid and ammonia, yields neither, by chemical agents. Potassa does not evolve ammonia, nor do the stronger acids give carbonic acid and evolve cyanic acid, with the peculiar odor which appears when the cyanates are decomposed by a stronger acid. The artificial urea gives no precipitate with salts of lead or silver.*


Henry law and equilibrium Is Henry's law a consequence of the equilibrium between a substance S in aqueous phase and the same substance S in gaseous phase? What I mean is, is it correct to treat Henry law as a manipulation of the equation for the equilibrium constant, where

$$ \ce{S(g) <=> S(aq)}, \ \ \ K=\frac{[S_{aq}]}{p_S}$$

$$\rightarrow K \times p_S = [S_{aq}]$$

which is Henry law ($p_S$: partial pressure of S in the gas phase; $[S_{aq}]$: concentration of S in aqueous solution).

That's an appropriate interpretation of Henry's law, although as you've written it $K$ would be referred to strictly as the solubility constant $H$, see this link to the wikipedia page provided in the comments.

The following shows more formally how to derive the connection between Henry's law, an equilibrium constant and the standard Gibbs free energy for evaporation of the solute.

At equilibrium between the solute in the gas and solution phases, the chemical potential of the solute is equal in both phases, so that

$$\begin{align} \mu(sol)&=\mu(g)\end{align}$$

In the gas phase (assuming ideal behavior)

$$\begin{align} \mu(g)&=\mu^\circ(g)+RT\log\left(\frac{p_i}{p^\circ}\right)\end{align}$$

Now according to Henry's law, the partial pressure and concentration in the solution are related as

$$p_i=K\frac{m_i}{m^\circ}$$

(of course, as you point out, this can already be recast as an equilibrium constant expression, but read on)

so that

$$\begin{align} \mu(sol)&=\mu^\circ(g)+RT\log\left(\frac{K(m_i/m^\circ)}{p^\circ }\right)\end{align}$$

Defining

$$\begin{align} \mu^\circ(sol)&=\mu^\circ(g)+RT\log\left(\frac{K}{p^\circ }\right)\end{align}$$

we can see that

$$\Delta G^\circ_m = \mu^\circ(g)-\mu^\circ(sol)=-RT\log\left(\frac{K}{p^\circ }\right)=-RT\log\left(K_{eq}\right)$$

which is the familiar relation between the standard Gibbs free energy change and an associated equilibrium constant for the equilibrium $\ce{solute <=> gas}$ provided we define the equilibrium constant as

$$K_{eq} = \frac{K}{p^\circ } = \frac{p_i/p^\circ}{m_i/m^\circ }$$


Correlation between 1H and 13C shifts - coincidence or not? A quick glance at any table of typical chemical shifts will reveal that the chemical shifts of protons $(\delta_\mathrm H)$ tend to correlate pretty well with the chemical shifts of the carbons to which they are attached $(\delta_\mathrm C)$. This correlation is frequently taught in introductory organic chemistry and the rule of thumb

$$\delta_\mathrm C \approx 20 \cdot \delta_\mathrm H,$$

although simplistic, actually holds up fairly well. To illustrate the point further, here is a diagram taken from the textbook by Silverstein et al. on organic spectroscopy:

Image taken from: Silverstein, R. M.; Webster, F. X.; Kiemle, D. J. Spectrometric Identification of Organic Compounds, 7th ed.; Wiley: Hoboken, NJ, 2005, p 207.

To the beginner such a correlation might seem intuitive. However, delving deeper into NMR theory, one learns that the $\ce{^{13}C}$ chemical shifts are generally dictated by paramagnetic shielding, while $\ce{^{1}H}$ chemical shifts are generally dictated by diamagnetic shielding and effects due to neighbouring groups. See, for example, this question:  NMR chemical shift range of different elements.

In light of this, how much should we make out of the general correlation seen? i.e., is it best understood to be a pure coincidence, or is there more to it than that?

Maybe it would help to comprehend the coincidental changes in 1H and 13C chemical shift by understanding the origin of chemical shift. In both cases, very crudely speaking, the influence of ring currents and electronegative elements/groups cause a downfield (away from zero) shift. The factor of 20 quoted is a function of the number of electrons (usually outer ones) involved in forming the bond. So the trend will also be true for other elements e.g. 15N where aromatic N compounds i.e. pyridine have a larger downfield shift.


How does CrF₆³⁻ hybridize? My textbook says that $\ce{[CrF6]^{3-}}$ is an example of $sp^3d^2$ hybridization. But I don't really see how. Chromium has an electronic configuration of:

$$3d: \boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \ 4s: \boxed{\uparrow}$$

Give three more electrons and it might look something like this:

$$3d: \boxed{\uparrow \downarrow}\boxed{\uparrow \downarrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \ 4s: \boxed{\uparrow \downarrow}$$

This is the ground state of the electrons. I see no way for electrons to be promoted for the formation of six orbitals. And the electrons have to be promoted all the way up to the $4d$ orbital, which would require a significant amount of energy. What is the exact mechanism of bonding in this ion? Amazingly, I couldn't find anything on the Web about this.

EDIT: My textbook also says that $\ce{[Co(NH3)6]^{3+}}$ is an example of $d^2sp^3$ quite similar to $\ce{[CrF6]^{3-}}$. But when I try to understand the exact mechanism, I encounter a similar problem as a I did above.

In $\ce{[CrF6]^{3-}}$ the central atom is $\ce{Cr(III)}$ or $\ce{Cr^{3+}}$. Chromium is usually $d^6$, so now (take away 3 electrons) $d^3$.

When we start filling in the electron pairs from the ligands, we get the following:

The electrons from the metal are marked in red.

The electrons from the ligands (blue) are filled into two $d$, one $s$ and three $p$ orbitals: $d^2sp^3$ hybridization as per solution!

Now try to do the same with $\ce{[Co(NH3)6]^{3+}}$!


Why do cations have higher ionization energies than neutral atoms? I know that metals have low ionization energies and non-metals have high ionization energies, but once a metal/non-metal becomes a cation, why does the cation itself have a ionization energy higher than the original neutral atom?

What is the scientific reason behind this increase in ionization energy?

Consider a simple thought experiment. Your ionisation process is essentially removing a (negatively charged) electron from a remainderatom.

$$\ce{Atom^n+ -> Remainderatom^{(n+1)+} + e-}$$

Throughout this process, you are separating charges, thus there is an electrostatic force driving the electron back to its remainderatom.

If the remainderatom has a charge of $2\ (\Rightarrow n=1)$, the attractive force is twice as strong in simplistic theories than if remainderatom’s charge is $1\ (\Rightarrow n=0)$.

Thus, it is twice as hard to remove a second electron.

This thought experiment makes use of a number of simplifications. For example, it does not consider electron energy levels (orbitals), it does not consider orbital contraction upon ionisation and much more. But it does a good job at explaining the general trend.


Deuterium removal from water What percentage of deuterium might be removed from water after a typical electrolysis procedure? From a post on this site I read that, "In the first stage, $\ce{NaOH}$ solution (initially $0.5\rm{M}$) is subject to electrolysis, until only $\frac{1}{32}$ of the original volume remains. This increased the original concentration of deuterium by a factor of $12$."

This seems to be, for the electrolyzed hydrogen, an approximate $38\%$ decrease from the original concentration. If we assume the original concentration was $150\rm{ppm}$ then after the first pass, we might expect to have $93\rm{ppm}$ in the deuterium reduced hydrogen. If my estimate is valid, may I expect that each successive pass would reduce the deuterium content by a similar percentage?

You are quoting from my answer here.

Looking at the source reference Inexpensive Equipment for the Preparation
and Concentration of Pure D2O or of
Deuterium-Rich Water. Description of
Construction and Operation of the Former
Ohio State University Heavy Water Plant The Ohio Journal of Science. v41 n5 (September, 1941), 357-365:

  With NaOH electrolyte we obtained, with Armco iron, an electrolytic
  separation factor[footnote 8] of a little more than 8, whose magnitude was
  practically independent of current density and of the deuterium concentration
  in the electrolyte. This is the true separation factor applicable
  to the dry gas. The practical separation factor is somewhat less than
  this (about 6 or 7) due to water vapor carried out with the electrolytic
  gases.

So, like Ivan Neretin is commenting, it is not optimal to electrolyze most of the first stage solution if the goal is obtaining deuterium depleted hydrogen.  The initial hydrogen coming from natural abundance water will be deuterium depleted by a factor of "about 6 or 7".  Since natural abundance water is essentially free, it would be best to only electrolyze a small fraction of a large amount of water.  Then you will deplete deuterium down to about 25 ppm in one stage.

After the first stage, if you attempt a second stage, burning the water to make deuterium depleted water, you will then have to add electrolyte, which, if it is NaOH, would introduce more deuterium.  So if you used a hydrogen-free electrolyte it would be better.  By the second stage, the starting material is valuable, and you would want to electrolyze a significant fraction of it, but not 31/32.


What is the coordination number for BeCl2? I am confused, should it be 2 or 4?

If I consider $\ce{BeCl2}$ as a chain structure the coordination number is 4, and in the gas phase it is 2.

Coordination number is determined by compound. It is normal for an element to have several different CN. Preferable CN can change for an element depending on the atom's neighbors. For example, Fe has preferrable CN=6 for fluorine anions, but CN=4 for chloride anions. When heated or under extreme pressure preferable CN may change as well. For example, $SiO_2$ has high-pressure modification with 6-coordinated silicon and 3-coordinated oxygen, while at normal pressure they are 4- and 2-coordinated respectively.

In case of $Be$ it prefers, just like all elements of the 2nd row, to have CN=4, but in many cases is forced to live with smaller numbers. BTW, in gas phase, $BeCl_2$ exists in equilibrium between monomer with Be's CN=2 and dimer with Be's CN=3.

In your case CN for Be in cristalline $BeCl_2$ is 4, while in gas phase it is 3 and 2 for dimere and monomer respectively. There is no 'one CN for Be atom right for all Be compounds.'


Fractional index in a chemical formula of metal oxide I have just read an article in which they used $\ce{WO3}$ and $\ce{WO_{2.9}}$ as precursors, whose objective was to form $\ce{WS2}$, in an atmosphere of argon by CVD. In one boat there is the powder of $\ce{S}$ and in another boat there is $\ce{WO3}$ or $\ce{WO_{2.9}}$, in an oven with a double temperature zone. In the article they conclude that $\ce{WO_{2.9}}$ forms larger layers of $\ce{WS2}$ in the substrate giving the following justification:

  They found that $\ce{W^6+}$ cannot be directly sulfated by $\ce{S},$ unless some intermediates are formed due to the high energy of the $\ce{W-O}$ bond. The reduction from $\ce{W^6+}$ to $\ce{W^5+}$ is mandatory for the incorporation of sulfur in the $\ce{WO3}$ network. For $\ce{WO_{2.9}}$ in our case, its partial $\ce{W^6+}$ ions have been reduced to $\ce{W^5+}$ or $\ce{W^4+}$ ions. Therefore, we find that replacing $\ce{W^5+}$ or $\ce{W^4+}$ with $\ce{W^6+}$ in the initial stage facilitates the growth of the single crystal $\ce{WS2}$ film.

I didn't realize what the difference is between $\ce{WO3}$ and $\ce{WO_{2.9}}$ and where these ions come from that they talk about in this part of the article. I also don't understand what a fractional index means in a chemical formula. Could you help me to interpret that part of the article and explain these doubts that I have please?

There is nothing special about fractional indices in chemistry, it is just a pointer signifying we are dealing with a non-stoichiometric compound.
Group 6 metal are known not only for their regular oxides existing in numerous polymorphic modifications, but also for forming homologous series of nonstoichiometric oxides $\ce{\ce{M^{VI}_nO_{3n-1}}}$ and $\ce{\ce{M^{VI}_nO_{3n-2}}}$ with the general formula $\ce{M^{VI}O_{3-x}}$ $(\ce{M} = \ce{Mo}, \ce{W};$ $x \in (0; 1)).$

Now, to make things clear, notation "$\ce{WO_{2.9}}$" does not refer to a molecule; $\ce{WO_{2.9}}$ is a formula unit showing composition of a network solid.
To further illustrate the origin of the fractional coefficients here, let's refer to the crystal structures of the stoichiometric and non-stoichiometric tungsten(VI) oxides $(\ce{WO3}$ and $\ce{WO_{2.9}},$ respectively).

Depending on the temperature, $\ce{WO3}$ adopts several polymorphs with various crystal structures; at the room temperature the most common structure it adopts is the one of distorted $\ce{ReO3}$ type, being made of tilted $\ce{[WO6]}$ octahedra cross-linked via the corners (for the illustration, see e.g. Wikipedia).

On the other hand, $\ce{WO_{2.9}}$ is an oxygen-deficient compound in comparison.
To account for that and eliminate the presence of point defects, crystal structure undergoes a transformation resulting in increased number of edge-sharing groups of $[\ce{WO6}]$ polyhedra and reduced number of the corner-sharing ones, e.g. the structure gets more "packed", and, more importantly, a crystallographic shear is introduced (which is not an impurity).
Let's have a look at the asymmetric unit of $\ce{WO_{2.9}}$ of the crystal structure determined by Magnéli [1]:

Figure 1. Crystal structure of $\ce{WO_{2.9}}$ [1, ICSD-24736] showing an asymmetric unit $\ce{W10O29}.$
Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#2194D6}{\Large\bullet}~\ce{W}$.

Due to the asymmetric unit $\ce{W10O29},$ the chemical formula (formula unit) is reduced to $\ce{WO_{2.9}}.$
One can also fill the entire unit cell knowing the number of formula units $(Z)$ per unit cell; here, $Z = 2$ and this results in equal formula $\ce{W20O58},$ often seen in the literature:

Figure 2. Crystal structure of $\ce{WO_{2.9}}$ [1, ICSD-24736] showing packed unit cell.
Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#2194D6}{\Large\bullet}~\ce{W}$.

Finally, to illustrate the location of the aforementioned shear plane, let's complete coordination shells on all tungsten atoms and view the structure along the $b$ axis:

Figure 3. Crystal structure of $\ce{WO_{2.9}}$ [1, ICSD-24736] showing polyhedral representation ($[\ce{WO6}]$ octahedra) with the shear plane along $a$ axis denoted with red arrow.
Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#2194D6}{\Large\bullet}~\ce{W}$.

As for the assignment of the formal oxidation numbers for the non-stoichiometric compounds like this, this is typically considered a non-trivial task of a questionable value.
Also, see the answer of mine to the related question Oxidation state of tungsten in $\ce{W3O8}$.

References

Magnéli, A. Structure of β-Tungsten Oxide. Nature 1950, 165 (4192), 356–357. DOI: 10.1038/165356b0.


Determining the order of basicity of acetate, phenoxide, hydroperoxide, and hydroxide ions Arrange the following ions in order of basicity:
  (1) $\ce{CH3COO-}$ (acetate)
  (2) $\ce{PhO-}$ (phenoxide)
  (3) $\ce{HOO-}$ (hydroperoxide)
  (4) $\ce{HO-}$ (hydroxide)

The first thing I see is that all of them have the same donor atom, i.e. oxygen.

In both $\ce{CH3COO-}$ and $\ce{PhO-}$, the charge is stabilized by resonance. So, they are least basic, $\ce{PhO-}$ being the less basic among the two as you can see that in case of $\ce{PhO-}$, the number of resonance structures are more than $\ce{CH3COO-}$.

Next, comparing $\ce{HO-}$ and $\ce{HOO-}$ I find that the negative charge on $\ce{O-}$in $\ce{HOO-}$ maybe slightly more stable as the charge may be shared by the other oxygen, whereas $\ce{HO-}$ has no such option.

So I get the basicity order as 4 > 3 > 1 > 2. But the answer is 3 > 4 > 2 > 1. Where am I wrong?

You are partly wrong and partly right. The correct order is $4 > 3 > 2 > 1$ ($\mathrm pK_\mathrm a$ data taken from the Evans table):

$$\begin{array}{cc}
\hline
\text{Molecule} & \mathrm{p}K_\mathrm{a} \\
\hline
\ce{CH3COOH} & 4.76 \\
\ce{C6H5OH} & 9.95 \\
\ce{H2O2} & 11.6 \\
\ce{H2O} & 15.7 \\
\hline
\end{array}$$

Between acetate and phenoxide, phenoxide does have more (non-trivial) resonance forms, but all of them involve putting a negative charge on carbon which is hardly favourable. So, the stabilisation derived from resonance is not much at all.

On the other hand, in acetate, the other major resonance form has the negative charge on an electronegative oxygen. The moral of the story is that the actual number of resonance contributors is not the only deciding factor:

Between hydroperoxide and hydroxide, the $\mathrm{p}K_\mathrm{a}$ values show that you are right in saying that the second oxygen in the hydroperoxide ion stabilises the negative charge. This operates entirely via an inductive effect, though. There is no resonance involved, which partly explains why both ions are more basic than phenoxide and acetate.


Do all polymers contain either carbon or silicon? Is either carbon or silicon present in every known polymer?

If not, is there a small group of elements that, between them, are present in every known polymer, or can polymers be based on a lot of different elements?

No, neither C nor Si have to be present in a polymer.
IUPAC defines polymer simply as

  A substance composed of macromolecules.

In turn, there is no limitations on the elements the macromolecules may consist of.
In fact, there is an extensive class of inorganic polymers free of carbon and silicon.


Is precision measured by significant digits in chemistry? I was recently asked this question:

An experiment is done 3 times, with a different measuring tool used
for each time. The results of the experiment are 1.39, 0.9, and 1.0
(all centimeters). Which measurement is the most precise, and how do
you know?

On my paper, I put that the answer was 1.39 because it had the most digits past the decimal point and submitted. However, I've read online and found since then that my explanation actually referred to accuracy, not precision. My question is, would saying "1.39 is the most precise because it has the most significant digits" be a correct answer?

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 $+0.1 \pm 0.05$.


So, is baking soda actually a strong base? Based on my recent thoughts, when baking soda ($\ce{NaHCO3}$) is dissolved in water, the following hydrolysis reaction occurs:
$$\ce{NaHCO3 + H2O <=> NaOH + H2CO3}$$
However,
$$\ce{H2CO3 -> H2O + CO2 ^}$$
so the actual one is:
$$\ce{NaHCO3 -> NaOH + CO2 ^}$$
$\ce{CO2}$ escapes, so the reaction is irreversible, all the $\ce{NaHCO3}$ is converted to $\ce{NaOH}$ and the product is a strong base.

The following reaction is possible:
$$\ce{HCO3- <=> H+ + CO3^2-}$$
but it's reversible unless it results in $\ce{H2CO3}$ formation, which is the same as the first one, so it doesn't really matter.

However, I was unable to find any evidence to that but a lot of discussions on whether $\ce{NaHCO3}$ behaves like a weak or a strong base, which put me in doubt. Am I right?

There is nothing to discuss (unless you find it necessary to compare two known numbers by discussion). $\ce{NaHCO3}$ is a very weak base.

Really, do you have baking soda around? Throw some in the water and see whether $\ce{CO2}$ actually escapes. It doesn't. It will if you heat the solution to near-boiling temperatures; then you'll have $\ce{Na2CO3}$ which is a base of moderate strength. You can't convert it all to $\ce{NaOH}$ this way.

So it goes.


Equilibrium constant for the thermal dissociation of phosphorus pentachloride Thermal decomposition of $\ce{PCl5}$ is studied in a closed vessel. Partial pressure of $\ce{PCl5}$ in gaseous equilibrium mixture is $\pu{1 atm}.$ $\pu{50 \%}$ $\ce{PCl5}$ dissociates. Find equilibrium constant.

I am aware of the formula of equilibrium constant, but I am finding difficult determining concentration of reactants and products from partial pressure and $\pu{50 \%}$ dissociation.

I know the formula

$$K_c = \frac{[\ce{PCl3}][\ce{Cl2}]}{[\ce{PCl5}]},$$

but how to calculate the concentration?

Note the equation as follows: $$\ce{PCl5 <=> Cl2 + PCl3}$$

You may begin with an ICE table showing the initial and final pressures of the gases.
$$\begin{array}{|c|c|c|} \hline
 & \ce{PCl5}  & \ce{PCl3} &\ce{Cl2} \\ \hline
\text{Initial} & 2 & 0&0 \\ \hline
\text{Change} & -1 & +1 & +1 \\\hline
\text{Equilibrium} & 1 & 1 &1 \\ \hline
\end{array}$$
Then it is a simple matter to find $K_p$ with the equation $$K_p = \frac{[\ce{P_{PCl_3}}][\ce{P_{Cl_2}}]}{[\ce{P_{PCl_5}}]}$$

If you would like $K_c$ you would need the ideal gas law ($PV=nRT$) but I doubt that is what your question requires.


do strong bonds have lower energies? I was looking at a graph of the internuclear distance vs. the potential energy, and the strongest H2 bond had the lowest potential energy.

Is this always the case?

Well, yes, and this is not limited to bonds. The stablest anything has the lowest potential energy, because things just tend to find positions with lower potential energy and stay there. This is the very nature of our world.

On the other hand, in this case the energy required to get things out of this position (that is, to break the bonds, if we are talking about bonds) is the greatest.


What is the function of nitrobenzene as a solvent in Friedel Crafts alkylation reaction? I would like to know what is the function of nitrobenzene as a solvent in Friedel Crafts alkylation reaction.

Answer to this question suggests that Friedel Crafts reaction is not possible with nitrobenzene and I understand that it's because of the big loss of electron density of the benzene ring due to the highly deactivating nature of $\ce{-NO2}$

Without any more information, it's just that, a solvent. Since it is a poor substrate for Friedel Crafts alkylations, it can be considered inert in these conditions. The choice of nitrobenzene would probably be to carry out the reaction at a relatively high temperature, since it boils at 210.9 °C.


Determining the concentration after half-life period I was given the reaction

In diluted solution and with neutral pH, the reaction reacts according to a reaction velocity law of pseudo-1. order:

$$-\frac{dc_{DNPA}}{dt}=k_{obs},\quad k_{obs}=k\cdot c_{H_2O}$$

At $t=0$ we have $96\mu M$ DNPA.

How big is the concentration of DNPA and DNP after three half-time periods ($3\cdot t_{1/2}$)?

Of course, for DNPA we would just have $1/8$th of $96\mu M$ which is $12\mu M$. But what about DNP?

For me it's clear that it correlates directly with the concentration of DNPA since we have a heavy diluted solution, so the concentration of the water can be assumed as constant.

Further I know that $v_c(t)=\frac{1}{\nu_i}\frac{dd_i(t)}{dt}$ whereas $\nu_i$ are the stoichiometric coefficients. So basically, the concentration od DNPA, DNP and HOAc are the same since all have a stoichiometric coefficient of 1. Right? So they all change the same. Meaning: if DNPA changes to $1/8$-th, so does DNP and HOAc.

I am still a bit unsure, but can I now conclude, that $c_{DNP}(3t_{1/2})=12\mu M$?

I'm afraid you are mistaken. The DNPA essentially turns into DNP in the course of the reaction, so the answer is $84\ \mu M$, assuming the initial DNP concentration is zero. Formally, the stoichometric coefficient of DNPA is -1, which illustrates nicely that the sum of the amounts of DNPA and DNP is constant in this system.


Does changing the electrolyte concentration in the electrolysis of water affect the speed that hydrogen gas is produced at the cathode? I am currently completing a Year 12 Extended Practical Investigation. Due to the lock-down restrictions we are unable to complete our own practical experiment and have been given sample answers with made up lab conditions.
The experiment I chose was to investigate the difference of time taken to produce 25mL of H2 gas in an electrolytic cell when the concentration of the potassium hydroxide electrolyte was changed.
The lab conditions given were a pressure of 101kPa, a temperature of 26.0 celcius and in addition the current in the system was kept at a constant 0.600A throughout all trials of different concentrations.
When analysing the results there was a trend that showed the higher the concentration of the potassium hydroxide the faster the 25mL of H2 gas was produced. However, Through using the formulas Q=It, Q=n(e-)F and the electrochemical half equation for H2 production it seemed that due to the current and the amount of Hydrogen being formed both being constant the time taken to form the hydrogen should also remain constant.
I have found a write up for the exact experiment on https://www.qcaa.qld.edu.au/downloads/senior-qce/sciences/snr_chemistry_19_ia2_asr_high.pdf however when it comes to talk about the effect of increasing the KOH concentration it seems that there is no mention that the current was kept constant.
I would like to know if the trend shown should have occured and why.

With decreasing electrolyte concentration, either the needed voltage raises at constant current, either the passing current decreases at the constant voltage.
Both happens because at lower concentration, a higher potential gradient across the electrolyte is needed to maintain the same current, as the smaller ion count requires  ions to move faster for the same current..
This potential gradient and the resulting potential drop acts against the external voltage. It pushes forced electrode potentials closer to their equilibrium potential. The net electrode reactions are then slower.


Net ionic equation for a synthesis reaction? In the following reaction

$$\ce{S (s) + O2 (g) -> SO2 (g)}$$

there are no spectator ions because both $\ce{S}$ and $\ce{O2}$ take part in the reaction. Does that mean the net ionic equation is the same as the molecular equation?

I think there should not be a net ionic equation at all, because there are no ions in the reactants.

You are right, you can't write ionic equation for that reaction. Quick search showed me that even the mechanism doesn't really involve any kind of ions, which, I'd say, is completely logical, considering reaction conditions, which favor homolytic, not heterolytic  bond cleavage.

You can only write ionic equations for reactions where ions are in charge, and they are preferably formed in polar medium.

P. S. The mechanism of S+O2 reaction isn't researched much, so here is the study of H2S+O2 reaction .


How does pH affect the degradation of ascorbic acid (vitamin C)? I know that vitamin C degrades over time (as I understand through oxidation). However, I am having trouble understanding how changing the pH of the solution in which the vitamin C is affects the rate at which the vitamin C degrades.

Among others, Roig et al. [1] demonstrated that there is indeed a change in degradation based on pH.

References

Roig, M. G.; Rivera, Z. S.; Kennedy, J. F. A Model Study on Rate of Degradation of L-Ascorbic Acid during Processing Using Home-Produced Juice Concentrates. International Journal of Food Sciences and Nutrition 1995, 46 (2), 107–115. https://doi.org/10/fgwc5c.

Ascorbic Acid is a known antioxidant molecule and its stability has always been an issue especially for pharma industry which often makes use of AA in formulation. Now as much as pH is a culprit in AA degradation, its only one among other factors like solvent, exposure to light/uv, and the nature of solution (viscous/non-viscous). There are quite a lot of studies on stability of Ascorbic Acid like this one

I normally prepare Ascorbic Acid freshly for each experiment, but it would be an interesting experiment to check the stability of AA for an extended time period at different pH values.

Here is an interesting simulation demonstrating degradation of Ascorbic Acid.


Potassium chlorate from potassium chloride and hydrogen peroxide Why can't you make potassium chlorate using the chloride anion from the potassium chloride and having oxygen bond with it from hydrogen peroxide?

Do not attempt! Peroxide is not a strong enough oxidant to oxidize chloride ion to chlorate. Neither is oxygen. However, peroxide can oxidize chloride to poisonous chlorine gas.

Redox reactions are spontaneous if the two half reactions would produce a positive potential in a galvanic cell. Most redox half reactions have measured standard electrode potentials against a known standard so that they can be compared.

The reaction you want to do is:

$$ \ce{3H2O2 + Cl- -> 3H2O + ClO3-}$$

The two half-reactions are:

Oxidation:   $\ce{Cl- +3H2O -> ClO3- +6H+ +6e-}$
Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$

The reduction potential of hydrogen peroxide is on the table, but the reduction potential for the other reaction is not. However, the other reaction is the sum of two reactions on the table, so we can use Hess's Law.

$$\ce{ 2ClO3- + 12H+ + 10e- ->Cl2 + 6H2O} \ \ \ \ E^\circ = +1.49 \text{ V}$$
$$\ce{Cl2 +2e- -> 2Cl-} \ \ \ \ E^\circ = +1.36 \text{ V}$$
$$\ce{2ClO3- 12H+ + 12e- -> 2Cl- + 6H2O}  \ \ \ \ E^\circ = +2.85 \text{ V}$$

Since standard potentials do not scale by stoichiometric factor (it's in the Nernst equation instead), the value we just found also works for our oxidation reaction. However, since this reaction is reverse of the "reduction half-reaction", we need a negative sign:

Oxidation:   $\ce{Cl- +3H2O -> ClO3- +6H+ +6e-}\ \ \ \ E^ \circ =  -2.85 \text{ V}$
Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78 \text{ V}$
Overall: $\ce{3H2O2 + Cl- -> 3H2O + ClO3-}\ \ \ \ E^ \circ =  -1.07 \text{ V}$

This reaction is not spontaneous. Peroxide cannot oxidize chloride, at least not to the chlorate ion. Note that peroxide can and does oxidize chloride to poisonous chlorine gas in the presence of acid. This redox is spontaneous and will happen.  Do not attempt!

Reduction:   $\ce{H2O2 +2H+ +2e- -> 2H2O} \ \ \ \ E^ \circ =  +1.78
   \text{ V}$
Oxidation:   $\ce{2Cl- + 2e- -> Cl2} \ \ \ \ E^ \circ =
   -1.36 \text{ V}$
Overall: $\ce{ 2Cl- + H2O2 + 2H+ -> Cl2 + 2H2O} \ \ \ \ E^ \circ =  +0.42 \text{ V}$

Peroxide will oxidize chlorine gas to chlorate, but who wants to handle chlorine?


Are there any tasteless edible salts? I saw on Youtube a video where caffeine was extracted from Red Bull using dichloromethane. However, due to dichloromethane tendency to form emulsions with liquid, an inefficient method was used to mix the dichloromethane and the Red Bull to minimize the emulsion.
The video maker stated that an emulsion can be broken by adding a salt, but a salt would also drastically affect the taste of the final product. I was wondering, are there any tasteless edible salts that could have been used to break the emulsion without affecting the flavor?

Giving a cursory Google e-book search gave me results particularly from old textbooks of 19th century:

Phosphate of soda (Sodium phosphate, $\ce{Na3PO4}$) - was sold under the name of tasteless salt
Sulfate of lime (calcium sulfate, $\ce{CaSO4}$) - it was named agustine because this salt was considered to have no taste. It was considered to mix with acids to form tasteless salts
Phosphates of barytes (barium phosphate, $\ce{Ba3(PO4)2}$) - considered tasteless
Phosphate of strontium (strontium phoshate, $\ce{Sr3(PO4)2}$) - considered tasteless
Superphosphate of lime (calcium superphosphate, $\ce{Ca(H2PO4)2}$) - at high temperature, it melts to a semi-transparent glass which is tasteless.
Sulfate of strontium (strontium sulfate, $\ce{SrSO4}$) - considered tasteless

@Ivan said in a comment(deleted) that calcium sulfate, barium phosphate, strontium phosphate and strontium sulfate are insoluble in water, so they cannot serve the purpose of breaking the emulsion. Finally as OP clarifies, calcium superphosphate is corrosive and hence not safe to eat, it all comes down to sodium phosphate which is both tasteless and edible.
References(in order of mentioned points)

London Medical Gazette: Or, Journal of Practical Medicine, Volume 18, 1836
The Critical Review: Or, Annals of Literature, Volume 33
Tobias Smollett, 1801
Pantologia: A New Cabinet Cyclopaedia, Comprehending a Complete Series of Essays, Treatises, and Systems, Alphabetically Arranged, J. Walker, 1819
The Cyclopædia, Or, Universal Dictionary of Arts, Sciences, and Literature, Volume 39, Longman, Hurst, Rees, Orme & Brown, 1819


Formaldehyde safety levels I've obtained a simple air quality analyzer which measures several items, formaldehyde concentration among them. While I cannot vouch for its correctness, I found the values measured during the last few days worth asking about.

Wherever I go, the analyzer measures higher than ideal levels of formaldehyde. From reading a bunch of papers, I concluded that the "safe" value is around 0.1 mg/m3. However, I can only measure so little on fresh air, outside, where it is around 0.04 - 0.08 mg/m3.

Indoor locations include:

At my folks' place which has not been recently renovated or painted, has no new furniture and the levels are around 0.180-0.280 mg/m3.
In my apartment which has some newer furniture and has been painted a couple of years ago, levels are from 0.250-0.450 mg/m3. I do open windows often and try to vent but the levels easily spike over 0.1 mg/m3.
At my office, which is not renovated, has no new furniture and has a lot of air, always above 0.150 mg/m3.

There is no smoking at any of these locations.

From what I read, this values are in line with furniture factories. I'm a little bit confused and worried, because I'm not sure what to make of this. Are there ways to reduce the concentration of indoor formaldehyde? Might it make sense to get a second analyzer to corroborate the results?

Background: I'm an electrical engineer and not too much into advanced chemistry or health & safety. I've spent some time looking through papers and publications, but a lot of them are about WHS and not easily understandable by me, quoting different methodologies or abbreviations I can hardly pick up.

Two different parts to the question I think:

Are the readings you are getting on your meter meaningful?
If formaldehyde levels are high, what (if anything) can/should you do about it?

Regarding interpreting the reading from the meter, there are a few things you should consider right off:

What does the manufacturer say about accuracy? Typically the results should be what's displayed +/- some amount (say 25%). If this isn't documented, you may be able to contact the manufacturer for more information.
What other substances can cause interference with the sensor. It's very possible that there are other contaminants in the air that are causing the reading to increase. Again, check the documentation and/or contact the vendor.

My experience (especially with non-professional grade equipment) is that direct reading instruments are typically good at helping to understand concentration trends. For example if you check the readings over the course of the day, you may find higher concentrations at certain times and lower concentrations at other times - could be linked to temperature, ventilation rates, or something else. The information is useful for understanding at a high level what's going on and for figuring out why you have a problem.

I have a lot less confidence in the absolute readings that they provide. Professional grade meters for formaldehyde can cost thousands of dollars and require regular calibration. And even then, if I wanted to be confident in my results, I'd prefer to use a wet chemistry method like one of these.

Now, assuming you are still convinced that you have a formaldehyde problem, what can you do about it?

Some people will say that you can attempt a "bake-off". Get some space heaters and run the indoor temperature up to as high as you can get it and keep it there for a day or two. The idea is that heating everything up will cause whatever formaldehyde reservoirs to fully "off-gas", and when the temperature returns to baseline, there won't be any formaldehyde left. I'd be cautious on following this method, however, as some studies have found that any reductions seen are generally short lived.

Three more realistic approaches are:

if you can increase the amount of outdoor air coming into the house, this will go a long way toward diluting concentrations.
NASA has also done research and found that certain types of house-plants are good at absorbing formaldehyde from indoor air.
Now that you're more aware of formaldehyde as a potential indoor air quality problem, keep it in mind the next time you update furniture or your home interior. There's increasing awareness in the market, and you should be able to find low VOC formaldehyde-free alternatives for most relevant products.


Aspartame poisoning: Can it modify aminoacids? I just accidentally closed a webpage (and did not find it later) which described that somehow, Aspartame poisoning victims had faults in the vital proteins of their body due to the reaction of Aspartame with some aminoacids. I doubted that this is true. However, considering the high solubility of Aspartame in the liquids in the body and its digestibility it is feasible for it or one of the related compounds to be causing this symptom. The info that will be shared is welcome.

Aspartame is one of the most well studied food additives there is. It seems to be controversial in the same way that evolution is—scientific research is overwhelmingly on the side of the substance being safe to consume, even at very high levels, but a Google search on it brings up things like "Aspartame is, by Far, the Most Dangerous Substance on the Market that is Added To Foods" (from a site trying to sell natural health products). A lot of this comes perhaps from the naturalistic fallacy: aspartame is artificial and thus unsafe. A huge number of regulatory authorities have deemed it safe to consume: FDA, Health Canada, EFSA, FSA, etc.

There is also a good deal of bad biochemistry thrown about: aspartame is metabolized into the two amino acids aspartic acid and phenylalanine, as well as methanol. There have been some hypotheses that aspartic acid intake could induce some amount of toxicity, but experimental evidence is lacking. Methanol, of course, is well known to be toxic. The key is, that aspartame is very sweet and only tiny amounts are added to food. So even drinking enough diet cola to hit the FDA's acceptable daily intake of aspartame (which works out to ~7.5 L), the quantity of methanol consumed is only around 300 mg, and things like fruits are larger sources of methanol in people who drink sane amounts of diet soft drinks.

The one condition in which aspartame is known to do harm at reasonable intake is in people who have phenylketonuria (PKU). This might be what you're thinking of by "faults in the vital proteins of their body". PKU is a genetic disease caused by a defective gene for the enzyme phenylalanine hydroxylase which results in an inability to metabolize phenylalanine correctly; thus, consuming phenylalanine from any source is hazardous to sufferers of the disease.


Is the equilibrium constant calculation only applicable to homogeneous reactions? In my text book, it states the general formula for the equilibrium constant is:

Then it states that "The equilibrium constant expressions described here apply to homogeneous reactions."

Does this mean that it only applies to homogeneous reactions or that it can also be applied to homogeneous reactions among other types?

Does this mean that it only applies to homogeneous reactions or that it can also be applied to homogeneous reactions among other types?

The latter. If you consider the distribution of a solute S in a mixture of water and octanol, this is a heterogenous system (in this case, two immiscible liquids). The equilibrium reaction (or process) is:

$$\ce{S(oct) <=> S(aq)}$$

There is an equilibrium constant for this process,

$$K = \frac{[\ce{S(aq)}]}{[\ce{S(oct)]}}.$$

If I dissolve some S in the aqueous phase, some of it will go into the octanol phase until equilibrium is established. If I dilute the aqueous phase, some S will go back to the aqueous phase to re-establish equilibrium.


Citric acid chelates calcium carbonate more than ethanoic acid, why does this result in a faster rate of reaction? I performed an experiment using egg shells (calcium carbonate) to represent teeth and reacted the calcium carbonate with acids. When measuring the rate of reaction citric acid had a higher rate of reaction and from my research I saw that citric acid forms a more stable compound with calcium hence it chelates the calcium more than ethanoic acid.

In the mouth this would remove calcium ions from
$$\ce{Ca10(PO4)6(OH)2(s) + 2H+ (aq) <=> 3Ca3(PO4)2(s) + Ca^2+(aq) +2H2O(l)}$$
that equation hence the position of equilibrium shifts to the right so more tooth decay.

But with egg shells:
$$\ce{CaCO3(s) + 2H+ (aq) -> Ca^2+(aq) +CO2(g) +H2O(l)}$$

there is no equilibrium so if more calcium ions are chelated why does this increase the rate of reaction?

Acetic acid has a $\mathrm{p}K_\mathrm{a}$ value of $4.76$ according to Wikipedia.

Citric acid has a $\mathrm{p}K_\mathrm{a,1}$ value of $3.13$ and a $\mathrm{p}K_\mathrm{a,2}$ value of $4.76$, also according to Wikipedia. It is about ten times as acidic as acetic acid is.

The greater acidity translates to a lower $\mathrm{pH}$ of the resulting solution. The lower $\mathrm{pH}$ translates to bases being protonated more and quicker. Considering that the decomposition reaction of carbonate is actually:

$$\ce{HCO3- + H+ <=> H2CO3 <=>> H2O + CO2 ^ }$$

The $\mathrm{pH}$ effect alone should be enough to explain a faster reactivity.

However, you can also invoke the following equilibrium which will be right-side shifted with citric acid (but obviously not so much with acetic acid):

$$\ce{CaCO3 (s) + cit^3- <=> Ca^2+ + CO3^2- + cit^3- <=>> [Ca(cit)]- + CO3^2-}$$

Where $\ce{cit}$ is citrate or the anion of citric acid. Complexation of calcium will remove free calcium ions from solution, and the solubility product only considers free ions in solution. Thus by complexing the free ions, more calcium can be dissolved.

I would still perform control experiments to make sure it is actually the complexation effect of citrate and not its acidity. For example, try reaction $\ce{CaCO3}$ with acetic acid in the presence of $\ce{Na3cit}$ or $\ce{Na2Hcit}$. These should be equally good complexing agents.


Calculating pH of methanoic/formic acid (HCOOH) When calculating the pH of methanoic/formic acid (HCOOH) why do we not multiply the given molarity of the acid by 2 as there are 2 moles of H+ ions, as seen in chemical formula HCOOH?
However when it comes to calculating the pH of sulfuric acid (H₂SO₄), we do multiply the given molarity by 2 as there are 2 moles of H+ ions.

The reason for this is that the hydrogen atom attached to the carbon will not be lost. Only the hydrogen attached to the oxygen will be ionized. In fact, in a solution of formic acid, not all molecules will be ionized anyways. Only very strong acids will fully ionize. Even in the case of sulfuric acid, not all molecules will fully ionize and lose the second proton. It is actually a simplification.


Why is vicinal diiodopropane unstable, leading to deiodination? I came across this reaction where simple electrophilic addition mechanism will result in a vic-diiodide.
$$\ce{H2C=CH-CH2-I ->[HI, excess][CCl4]}$$
But the solution says that vicinal iodine groups are unstable and the following reaction occurs:
$$\ce{CH3-\underset{(unstable)}{\underset{\underset{\huge I}{|}}{C}H-\underset{\underset{\huge I}{|}}{C}H2} ->[][$-$I2] CH3-CH=CH2}$$
I am unable to figure out how the dihaloalkane is unstable. It might be due to steric reasons, but still the two iodine can arrange in a way that reduces the repulsions.
But even if it is unstable, how did $\ce{I2}$ leave the haloalkane to yield an alkene? This dehalogenation does happen when a metal is used though (Zn dust, Mg in acetone, I guess) but what would be the mechanism here?

That 1,2-diodopropane is unstable does not indicate whether it is as the neat halide or in solution. Breaking (BDE) two $\ce{C-I}$ bonds costs $\pu{+114 kcal/mol}$ while forming the $\pi$-bond of propene is worth approximately $\pu{-62 kcal/mol}$ and the formation of iodine, $\pu{-36 kcal/mol}$. The net reaction is endothermic by $\pu{+16 kcal/mol}$. The reaction is entropically favored and the products are volatile. Therefore, the unfavorable equilibrium can be shifted to the right.


Chlorination by SOCl2 vs SO2Cl2 My textbook mentions that chlorination of alkanes by $\ce{SO2Cl2}$ is preferred over $\ce{SOCl2}$, but I didn't understand why.

Could someone explain which of the two reagents I should use in what situations, and briefly tell me why?

There is a similar question here:
How does SO2Cl2 react with alcohols?

However the answers there did not completely clarify my doubts.

The reactivity patterns of $\ce{SO2Cl2}$ and $\ce{SOCl2}$ are quite different. $\ce{SOCl2}$ is a good electrophile, and can be thought of as a source of $\ce{Cl-}$ ions.  These ions can go on to react in their typical nucleophilic fashion. $\ce{SO2Cl2}$ however is often a $\ce{Cl2}$ source, as it readily decomposes giving off sulfur dioxide. Usually much easier/safer to use this than measuring out (and getting into solution) chlorine gas. The chlorination of simple alkanes by $\ce{Cl2}$ gas (or something that makes it in solution) happens by a radical mechanism i.e. $\ce{Cl.}$ not $\ce{Cl-}$.


Obtaining Benzene from toluene What will be the reaction for producing benzene from toluene?

My attempt:

Toluene→Benzoic acid→Sodium benzoate→Benzene(decarboxylation) ...

Is there any other ways, like a direct reaction?

There is also a disproportionation reaction that converts toluene to a mixture of benzene and xylene (Bawa et al), also, see wikipedia.  It has commercial applications (because the benzene/xylene mixture can often be sold for more than toluene) and can be licensed.  However, it is more of an industrial process than one you would use in the lab, as it is really driven by an exothermic dealkylation in vapor phase over zeolite catalyst.

Interestingly, as zeolites act as a Lewis acid, this is strangely similar to Friedel-Crafts alkylation, which uses Lewis acid catalysis in liquid phase.  The paper by Bawa et al discusses experiments in which the trans alkylation was done with Frieden-Crafts catalysts in liquid phase.


How to Protect one -OH group in 1,4-benzenedimethanol using TBDMS Actually i want to protect one -OH group using using 1.1Eq of TBDMSCl and 1.5 Eq of Imidazole. Is is possible to protect one OH group using these condition. Even if i get mixture of Products ( one protected OH and Both OH protected compounds), separation by column would be easy and what would be approximate yield of both products.

any suggestions.?

The product is symmetrical so it doesn't really matter which alcohol gets protected, you'll end up with the same product.

There are many examples of doing this kind of mono protection of simple substrates. Quite often a fairly large excess of the diol is added to try and ensure no bis protection takes place. In your case the starting material should be easily separable from product by column chromatography.

A potentially better method is to formally deprotonate the diol with 1 eq of a base such as sodium hydride or butyl lithium then trapping with a solution of the silyl chloride. This often works well in cases where bis protection may occur, as energetically its unfavourable for both oxygens on the molecule where there's molecules that are neutral


Why carbon is so special? One special branch in chemistry is allotted to compounds of only one element (carbon). Is it justified when there are more than 115 elements and their compounds but not with any special branches?

Carbon has a valency of four which helps in bonding with other atom or groups easier and adds to catenation. When one talks of Carbon's ability to bond, then Catenation property should come to your mind. Catenation is the linkage of atoms of the same element into longer chains. Catenation occurs most readily in carbon, which forms covalent bonds with other carbon atoms to form longer chains and structures. This is the reason for the presence of the vast number of organic compounds in nature. One such marvel is DNA! So, while studying chemistry, Carbon compounds form a class of peculiar natured compounds and the vast range of molecules and development of their own domain in their own scale might have made people feel that a separate category should be made for Carbon compounds because of their vastness and availability, also predominately manifestation of them in almost every thing, even life! In fact life became possible only because of Carbon, such is the importance. They have got their own nomenclature system in their domain.. Its really vast. The study of everything begins with Organic compounds first.. be it isomerism, medicinal reaction or anything. So, you may see, how important Carbon is. If you are a biology student, you must have learned about biomolecules.. DNA and gene pool, replication, transcription, translation, protein coding, defective proteins and some viral components and its lifecycle. Molecular basis of genetics is based on DNA and other organic compounds. That gives you thrilling experience to see how stuffs work and wonder about their importance and manifestation. Carbon is a wonder.


What does radiofrequency really do in NMR & MRI? In NMR, the protons’ frequencies are affected by the magnet which makes them precess around its magnetic field so now all protons will have different frequencies according to their environment including shielding and de-shielding effects. As I read, the radio frequency pulses only change the resonance of protons and make their $\omega$ angle  change  ($180^\circ$) to be induced inside the coil and make signals that can be measured.

But if this true why do we have to use different gradient coils and phase encoding gradient coils in MRI, if the protons really have different frequencies?

The radiofrequency pulses and the gradients have completely different purposes in NMR. Radiofrequency pulses are the basic building block of NMR experiments, you use them to manipulate your spin populations and coherences. While each proton has a different frequency depending on its environment, you generally can't (or don't) target individual protons with your pulses. In most experiments you hit either all protons with your pulses, or a specific group of protons.

Gradients have different purposes in NMR and MRI, what they allow you to do is to change the magnetic field depending on the position inside your sample. In NMR this allows you to remove unwanted signals and select specific coherences.

In MRI, the gradients are what makes imaging possible. They allow you to encode the spatial position of your protons into their frequency. In NMR you look at all the molecules in your sample at once, you don't care where your molecules are floating around. In MRI the position is important, as you want to know what the protons at a particular point in the body are doing. Gradients also allow you to only look at a slice of the body and not all of it at once.

I'll skip phase encoding in gradients as it is MRI-specific and I don't know much about it. Each part of your question has whole books written about it, so this is a drastically simplified answer.


Is chitin actually protein? Recently insects are featured as a protein rich source for human nutrition. That humans can really digest chitin through chitinase enzym has been only recently confirmed.

But, does the chitin shell of their body actually count to the protein family?

Wikipedia does not state a clear yes or no. It says it is derivative of glucose, and may be compared by function to keratin protein. I'd really like to exclude any ambuguity and undersrand why.

Chitin and protein are completely unrelated.
The only common thing is that they are polymers.
Chitin is a polymer of amino sugars while protein is a polymer of amino acids.
Both monomers are very different and are not converted one to the other.

In fact, chitin, like cellulose, is not fragmented by animals, so it is not absorbed.
Moreover, chitin is a homopolymer, so even if it where absorbed and transformed to amino acids, it would yield only one (or maybe a few ones) but all the others would have to be obtained through a different source.

So chitin is not a source of protein or amino acids.


Can molecular structure be determined without modern characterization methods? Molecular composition can be determined in some "primitive" ways (1, 2). What about the molecular structure? Is there a way to know what's bonded to what without spectroscopy etc.?

By "modern" I mean any method that emits and detects particles (photons, electrons, etc.).

Structure elucidation via degradative studies and decomposition chemistry has sadly all but passed from the chemists' toolbox, but this was the method once used to determine molecular structures up until about half a century ago. One of my very favourite examples that I use to teach structure determination by NMR is the story of strychnine. Strychnine (below) is by no means a trivial structure; it is hard enough to solve by 'modern' spectroscopic means. The story of how Robinson and Woodward, standing upon the shoulders of those before them, unravelled the the structure of this alkaloid over a decade and a half is an intriguing story, and one well worth investigating further. (AMBIX, Vol. 48, Part 3, November 2001)

Essentially, the structure is slowly pieced together through a series of degradation reactions. It is a methodology that relies on chemical processes, reactivities and properties only. For instance, it was worked out in the late 19th century that strychnine contained an amide as a lactam through a hydrolysis reaction to give strychnic acid. Were it not a lactam, two separate products would have been formed, rather than the one. Suggestions of the quinoline skeleton came from the reactivity of a dimethyl derivative in a series of chemical tests. Over the next 20 or so years, this was revised and superceded by more likely candidates. So on and so forth over a number of years, each little piece of the puzzle is fitted together, all through the lengthy process of chemical reactions to give smaller easier to identify products. A number of complicated framework structures were proposed, mulled over and discarded for a number of years, and final structure was tested through total synthesis reproduction. This whole process requires an incredibly talented chemist, and it simply makes my mind boggle at the thought of how these guys operated.

And though, it can be argued that UV spectroscopy did play a role in the final jigsaw piece in 1947, the correct structure was proposed by Robinson in 1942, and total synthesis confirmed this in 1954.

This is just one example of determination of molecular structure through degradative or decomposition chemistry, and many complicated structures were determined this way. Thank goodness for high field NMR.


Why are fluoroalkyl chains hydrophobic/oleophobic? I'm searching for an answer that explains the hydrophobicity / oleophobicity in terms of intermolecular forces, but can't really find one.

Below is an example fluoro-alkyl nano-particle, F-POSS. It has 8 ligands of a fluoroalkyl chain $\ce{(CH2)2(CF2)4CF3}$ with a silicon-oxygen cage in the center.

When grouped with other F-POSS molecules it forms a membrane with the fluoroalkyl chains:

Hydrogen Bonding - Wouldn't the hydrogen be attracted to the fluorine atoms? They're essentially electron sinks.

London Dispersal Forces - Why would other molecules not be attracted to these ligands? I was told that the ligands themselves aren't even attracted to each other.

Apologies for the lack of citation, but what's been given to me has mostly been word of mouth. Research articles I've read don't really go into the intermolecular forces.

Image source: http://pubs.acs.org/doi/abs/10.1021/la201545a?journalCode=langd5

You have identified a fairly common but counter-intuitive result. At least it's counter-intuitive based on the way the properties of fluorine are described in undergraduate chemistry classes. Specifically, we're told it's very electronegative and hence forms very strong hydrogen bonds, and that it is very reactive. This is, however, a poor description of what tends to dominate the behavior of fluorine when it is bonded to carbon.
Let's start with something simple but illustrative. If I told you that the boiling point of $\ce{CCl4}$ is $\pu{350 K}$ and the boiling point of $\ce{CBr4}$ is $\pu{463 K}$, what would be your guess for the boiling point of $\ce{CF4}$? Probably less based on the trend, but just how much less?

$\ce{CF4}$: b.p.=$\pu{145.5 K}$
$\ce{CH4}$: b.p.=$\pu{111.7 K}$

First time I've used one of these :)

Well hopefully you don't look this far, but you'll notice that the boiling point of $\ce{CF4}$ is much, much lower than that of $\ce{CCl4}$. Importantly, the difference between the fluorine and chlorine compounds is much larger than that of the chlorine and bromine compounds.
Additionally, the fact that $\ce{CF4}$ boils only slightly higher than $\ce{CH4}$ indicates that the molecule basically loses all interactions when going from the chlorine to the fluorine.
What does this actually illustrate? It doesn't tell us anything about hydrogen-bonding or dipole interactions (all have zero dipole), but it tells us about the dispersion forces present in the system. Namely, if this result generalizes, we have learned that a $\ce{C-F}$ bond is essentially non-polarizable. This is very important because polarization is most of what determines the strength of dispersion interactions. Okay, this basically answers the question, but let's be more systematic.
$\textbf{Dipole-Dipole Interactions:}$
The paper you linked is basically looking for a material which won't interact strongly with water and other polar substances. One important feature is to make sure the ligands on this material do not have an appreciable dipole then. It is true that the $\ce{C-F}$ is statically polarized towards the fluorine atom, so there is a dipole along the bond, but the molecular dipole is generally what we will care about. The ligands are basically cylindrical, so any dipole perpendicular to the cylinder ought to roughly cancel out. If anything there would be a dipole along the ligand backbone, but this is probably very small if it exists at all. So, dipole interactions don't particularly matter.
$\textbf{Hydrogen-Bonding:}$
Despite what you might think, it actually is quite possible for water (or other $\ce{O-H}$ containing solvents) to hydrogen-bond with the fluorine in a $\ce{C-F}$ bond. See [1] for some calculations demonstrating this.
So, if you're designing this material, you might have to worry about this a little bit, but leave it up to the experiment to tell if this matters.
$\textbf{Dispersion Forces:}$
Here is the key. In organic chemistry, you're usually sold a lie that the things determining the size of dispersion forces are the surface area and the molecular weight of a molecule. If this were strictly true, then we would expect the increase in boiling point we saw above going from $\ce{CH4}$ to $\ce{CF4}$ to be much more dramatic than it is based on the change going from $\ce{CCl4}$ to $\ce{CBr4}$. The reason this rule works is because generally organic compounds are just a bunch of $\ce{C-H}$ bonds, so most things are roughly as polarizable as the next thing.
Okay, but it's not like the $\ce{C-F}$ bonds are inherently unpolarizable. They are about as polarizable as a $\ce{C-H}$ bond, so for all consideration of physical interactions, we can basically imagine that these ligands are just lipids.
$\textbf{Why Do Lipids Form Membranes?}$
This is actually quite a tricky, but interesting question. Imagine a big fat lipid in a solvent of water. One thing we know is that water likes to hydrogen bond. This lipid disrupts the hydrogen-bonding network. Furthermore, because there has to be some water at the surface of the lipid, it is likely these waters will get stuck aligning their dipoles with whatever dipole is present in the bond it is near.
So, it seems like having the lipid there is both energetically unfavorable and entropically unfavorable. It is unclear without an experiment which of these things will dominate and if clumping all the lipids into a membrane is actually better because that's definitely a decrease in entropy for the lipid.
Well, [2] shows the experiment describing this same idea. Slightly different system but same idea. Basically the conclusion is that the energy plays almost no role ($\Delta H\approx 0$) but the entropy for forming a membrane dominates.
Shamelessly quoting from the abstract:

Possible mechanisms for the entropy increase include: (i) the attraction between discrete oppositely-charged areas, releasing counterions; (ii) the release of loosely-bound water molecules from the inter-membrane gap; (iii) the increased orientational freedom of previously-aligned water dipoles; and (iv) the lateral rearrangement of membrane components.

The lesson to learn here is that the entropy of the solvent can actually be very important when considering both the equilibrium arrangements of heterogeneous systems as well in chemical reactions.
$\textbf{Bringing it All Together:}$
So, I hope you see what I mean when I said that noticing $\ce{C-H}$ and $\ce{C-F}$ bonds have almost identical intermolecular forces behave basically answers the question.
When we consider explicitly the interactions of a substance with the ligands you have asked about (in comparison to a lipid), we see that dipole interactions might be a little stronger between the solvent and the ligand, but the ligand-ligand interactions should also be a bit stronger than in the lipid. So, again, $\Delta H\approx 0$. In contrast, the stronger dipole means that the solvent is more tied up, so the entropy change should increase (not accounting for the effect of the cube thing the ligands are mounted on).
Note that this argument for membrane formation based on entropy is not exclusive to strongly polar solvents such as water. That is, pretty much any solvent gains a lot of entropy when a membrane forms. So, as long as it does not interact strongly the ligand, the ligands should form a membrane.
Finally, based on what I've said, why not just use $\ce{C-H}$ bonds rather than have fluorine present? I only skimmed the paper, and they mention durability which makes sense because a $\ce{C-F}$ bond is much stronger than a $\ce{C-H}$ bond. Plus alkanes have the problem of being very combustible which is probably not good as these materials would likely be used at high temperatures.
Random semi-relevant note. Teflon is just an alkane but with all $\ce{C-F}$ bonds, and it too is quite slippery (it's the stuff on non-stick pans).

$\textbf{Edits:}$
I forgot to address the oleophobic nature of fluorocarbons, which is the most important part of the question, really.
$\textbf{Oleophobicity of Fluorocarbons}$
One interesting trend which I was not aware of, is that the boiling point of fluorocarbons increases more slowly than boiling point of corresponding hydrocarbons. For instance, the boiling point of perfluorooctane is $\pu{103 ^\circ C}$ while the boiling point of octane is $\pu{125 ^\circ C}$. So, this means that the dispersion forces in hydrocarbons grow more quickly than that in fluorocarbons. This makes some sense because the carbon in fluorocarbons will be more positively charged, and the fluorine atoms are comparable in polarizability to hydrogen, so the polarizability of the carbon atoms is diminished.
It seems, however, that the oleophobicity of this particular structure has a lot to do with the fact that the chains are not free to move about in solution. For instance in [3], the authors show that by attaching a fluorocarbon, which is ordinarily not oleophobic, to silica particles, the flurocarbon becomes oleophobic. This indicates to me (and this could be argued) that by restricting the translational motion of the fluorocarbons by mounting them on a large structure, as is done in the example you give, that the favorability of mixing is lost. This assumes that the energetics of having the fluorocarbons dispersed throughout the oil is roughly the same as having the fluorocarbons and oil separated into phases.
So, it seems like entropy is the dominant factor in both the hydrophobic and oleophobic characteristics of these fluorocarbons. The big difference is that basically any nonpolar organic molecule will form its own phase in the presence of a polar solvent like water. This is because the interactions between the polar substances are fairly strong, but there is also still a lot of free motion (hydrogen-bonds, etc. break and form again quite often). With the fluorocarbon, one has to manually restrict the entropy gain due to mixing by forcing the fluorocarbons onto a silica base. These things individually have the same effect of disrupting some of the degrees of freedom of the solvent, so the two separate to maximize entropy.
Obviously, some of the parts about oleophobicity are a little bit hand-wavy because really either effect could dominate, but the fact that putting it on the silica base makes the fluorocarbon oleophobic as ref. 3 points out convinces me this is due to entropy and not enthalpy.

References:
[1]: Andrew V. Razgulin, Sandro Mecozzi, "Binding properties of aromatic carbon-bound fluorine," Journal of medicinal chemistry 2006, 49(26), 7902-7906 (https://doi.org/10.1021/jm0600702).
[2]: Husen Jia, John R. Liggins, Wah Soon Chow, "Entropy and biological systems: Experimentally-investigated entropy-driven stacking of plant photosynthetic membranes," Scientific reports 2014, 4, Article number: 4142 (7 pages)(DOI: https://doi.org/10.1038/srep04142)(PDF).
[3]: H. F. Hoefnagels, D. Wu, G. de With, W. Ming, "Biomimetic superhydrophobic and highly oleophobic cotton textiles," Langmuir 2007, 23(26), 13158-13163 (https://doi.org/10.1021/la702174x).


Singlet/triplet oxygen cycloadditions Why can singlet oxygen participate in cycloaddition reactions while triplet oxygen, on the other hand, can't?

Normal dioxygen ($\ce{O2}$) exists as a ground state, triplet biradical. This is an example of a molecule that, thermodynamically should be quite reactive, yet is kinetically unreactive - once again, a case of kinetics vs. thermodynamics.

Because spin must be conserved in a reaction, if ground state, triplet $\ce{O2}$ were to react with something, it would have to produce a triplet product. For most compounds, the triplet state is considerably higher in energy than the singlet state.  Hence reaction with ground state, triplet $\ce{O2}$ would produce a higher energy, triplet product. Since ground state, triplet dioxygen doesn't have any excess energy, the activation energy required to produce a high energy triplet product just isn't there - hence, no reaction.

On the other hand, singlet $\ce{O2}$ is a spin-paired excited state with plenty of excess energy (the first singlet lies some 95 kJ/mol above the ground state triplet).  Basically, singlet $\ce{O2}$ behaves like any other electrophilic (due to the electronegativity of oxygen), reactive (due to its excess energy), compound with a double bond. It readily undergoes the Diels-Alder, ene and even 2+2 cycloaddition reactions - again, just like any other reactive olefin.


How is silver chloride more soluble in brine than water? I heard that:

  Silver chloride is more soluble in very concentrated NaCl solution (brine) than in pure water.

Is it true? I mean won't the common ion effect operate? Or may be this is due to some kind of complex formation as NaCl solution is highly concentrated ???

I mean won't the common ion effect operate?

The common ion effect always applies for pure substances. With the added sodium chloride, you have changed the effective solvent and changed the ions.

  Or may be this is due to some kind of complex formation as NaCl solution is highly concentrated?

Exactly, before your ionic equation was:
$$\ce{Ag+ (aq) + Cl- (aq) <=> AgCl (s)}$$
but with brine, the ionic equation is:
$$\ce{AgCl (s) + Cl- (aq) <=> [AgCl2]- (aq)}$$
Which has a different solubility constant, and where as before more chloride made the $\ce{AgCl}$ less soluble, if you note that you have switched regimes and $\ce{Cl-}$ has switched from being a reactant to a product, which means now with additional chloride silver chloride becomes more soluble.


Are viscoelastic materials to be considered solids or fluids? I recently stumbled upon the concept of viscoelasticity. Not being a chemist, but being interested in the topic, I started to read the wikipedia article.

The main thing I extracted from it is:
Solids are elastic, fluids are viscous, viscoelastics materials are both viscous and elastic.

I didn't get it, maybe due to my poor background. Is viscoelasticity a state of matter or just a property of materials? In the latter case, can we talk about viscoelastic solid and viscoelastic fluids?
I know the state of matter isn't an element nor compound or material property but it changes as temeprature and pressure change. The strength of the bonds between atoms/molecules and the "value" of their kinetic/vibrational energy is what determines how we see them.

Anyway, I couldn't really fit viscoelastic material in my chemistry mental scheme, so I'm here in search of some clarifications.

There are two basic models that describe the two fundamental branches of viscoelastic behaviour:

The Maxwell model is made up of a spring and a dashpot in series: For fast deformations, the dashpot does not really move, and your spring gives pure elastic behaviour. For slow deformations, the spring stays relaxed, and the damper fluid in your dashpot turns all energy into heat.
The Voigt-Kelvin/Kelvin-Voigt model is a spring and dashpot in parallel: Even at the slowest/smallest deformation, the spring never relaxes until you let go and let it return your sample to the original shape. And the faster you try to deform it (e.g. oscillatory), the more heat your damper produces.

The former, e.g. a molten polymer, is a liquid, the latter, e.g. crosslinked rubber, is a solid.


Reaction of Iron with Concentrated and dilute hydrochloric acid Does Iron react differently with dilute and concentrated hydrochloric acid?
A. I know that with dilute hydrochloric acid Iron (II) chloride is formed.
B. I read on the internet that with concentrated hydrochloric acid Iron reacts to form Iron (III) Chloride.
Could someone tell me:
1. Is FeCl3 formed as a product of reaction between HCl (conc.) and iron?
2. If yes, then what is the complete chemical reaction?

Thank you.

No, $\ce{FeCl3}$ is not formed upon reaction with Conc. $\ce{HCl}$. Both dilute and concentrated $\ce{HCl}$ give $\ce{FeCl2}$ as product.

$$\ce{Fe + 2HCl_{dil} -> Fe^2+ +  2Cl^- + H2 ^}$$
$$\ce{Fe + 2HCl_{conc} -> Fe^2+ +  2Cl^- + H2 ^}$$

Source: Vogel's Qualitative Inorganic Analysis


When Compounds are Considered Compounds and Their Types I'm a bit new in chemistry, so I need help with a couple of things. Although I don't mind complicated answers, please try to keep it around high school level, and not rocket science. Thanks!

Can someone describe exactly what makes a compound a compound? Is it just the simple definition of "multiple elements combined" or can someone get a more specific?
Sodium chloride is an ionic compound, but it's not made out of "molecules". So does that mean Sodium Chloride isn't an actual molecule (like $Na^+$ and $Cl^-$ are combined as separate ions but they can't be combined in a 1 to 1 fashion)
Can someone explain to me exactly what the law of gaseous reaction is, (it doesn't show up on internet specifically, but it's a law by Gay-Lussac).

Thank you in advance!

A molecule is formed when two or more atoms join together chemically e.g. Na and Cl.

A compound is a molecule that contains at least two different elements e.g. Na and Cl.

All compounds are molecules but not all molecules are compounds e.g. H2 and O2 are molecules but are not compounds because bond is between two of the same elements.

Im not sure what you are asking in question 3. Gay-Lussacs Law is generally associated with the combining volumes principle. e.g. 1 volume of H2(g) + 1 volume of O2(g) = 2 volumes of H2O(g). However, you may want to consider the combined gas law, not just Gay-Lussac law. This combines Gat-Lussacs law with Boyles' and Charles' Law.


Does the van der Waals force not affect the alkali metal 's trend of decreasing melting points down the table? Melting and boiling points increase further down the halogen group, but they decrease further down the alkali metal group. I know that the former's trend has to do with the van der Waals force, but I don't understand why it is not applied to the alkali metal group. Please explain why the halogens are affected by the van der Waals force and the alkali metals are not.

Because, in the condensed phases, you don't have molecules of metallic elements.  You have, as a first approximation, a "rice pudding" structure where cations are embedded (like raisins) in a "pudding" of delocalized valence electrons (metallic bonding).  The main cohesive force in the condensed metals is then electrostatic attraction between the opposing charges of the cations and electrons.  In alkali metals where each atom breaks up into the same charge cation (+1 charge) and produces the same number of delocalized electrons (one per atom), this electrostatic attraction goes down with larger atoms.


Is there any formula that can be used to find loss of mass due to binding forces in atomic and sub atomic particles? Atomic weight of Br-79 is 79.641 if you add the masses of protons and neutrons. However, in periodic table, it is less than the value given here. How is the difference arrived at for all the elements listed in the periodic table?

Yes, there is the Semi-empirical mass formula which will give you the binding energy/mass of the nucleus.

Once you subtract the binding energy/mass from that of the nucleons, you have the mass of the nucleus.  Then you should add the mass of the electrons.  Finally, in principle at least, you would subtract the binding energy/mass of the electrons (in other words the ionization energies of the electrons), but this is a very minor correction.


Why is Astatine monoatomic? I learnt that halogens always form covalent bonds to becoms diatomic molecules. So why is astatine monoatomic?

I mean they have the same properties, why shouldn't they all be diatomic?

One key problem with astatine is that it's incredibly unstable. There are no known stable isotopes, and the longest-lived has a half-life of ~8 hours. So no one has been able to (yet) prepare enough to make real measurements. Thus, we don't know for sure whether solid At is diatomic (like the other halogens) or monatomic.

On the other hand, high-level first-principals (quantum) calculations have been performed that suggest astatine has some unusual properties. The full paper by Roald Hoffmann and Neil Ashcroft at Cornell was published in 2013 in Phys. Rev. Lett.: "Condensed Astatine: Monatomic and Metallic" and also covered in multiple science news sites, including Chemistry World.

The calculations on At:

  predict that it will be a metallic solid. That’s perhaps not so surprising given that iodine, the halogen above it in the periodic table, is itself a dark silvery solid (although not truly metallic itself) and that astatine’s melting point is 302°C. But the real surprise of the new results is that the solid wouldn’t be composed of diatomic molecules, like all the other halogens, but would be monatomic.

Why would At be different than solid $\ce{I2}$?

  ... The monatomic form appeared only with spin-orbit coupling included. "These effects influence the atoms’ propensity to form certain bonds and crystal structures, usually by reducing molecular binding energies," says Hermann. Earlier calculations have indicated that in the gas phase astatine would still form a weakly bound diatomic molecule.

In other words, it's the spin orbit coupling contribution of relativistic effects that would prevent the formation of diatomic molecules in solid At.


How can I convert the number of particles per liter to gram per liter How can I convert the number of particles per liter to gram per liter? I know the size of these particle and so, I calculated the volume and then by the density I calculate the mass. After that I don't know what to do

$$x \frac{particles}{L} \times \frac {1 mol}{6.023 \times 10^{23}particles}\times \frac{MW gm}{mol}$$


Conditions for choosing quantifying and qualifying ion in LC-MS/MS In my PhD-work, we are trying to develop LC-MS/MS methods for targeted analysis. Usually, I determine my quantifying ion/fragment based on a few different criteria:

If there is a 13C internal standard of identical structure, stick with the same fragment for both.
If not, look for the lowest noise/highest area/lowest LOD
If two fragments are similar in the previous manner, I have a look if there are different levels of contamination in the method blanks

For most compounds, that works, but for a small subset I have the case where the method blanks are similar, but for the fragment with the lowest LOD, noise is highest and/or areas are lowest between the two fragments we analyze for.

Would you go for the fragment with the highest area, highest noise (assuming a large number is more robust against small changes) and the highest LOD, or the lowest one? Is there a best practice I can refer to? There is a 1-2 order of magnitude difference between the LODs, so I really don't want to misrepresent the data.

Thanks in advance!

The limit of detection (LOD) in these kinds of assays depends on both the noise and the signal.  Usually it is defined as lowest concentration analyzed such that the signal (peak height) is three standard deviations higher than the noise.  Exactly how to define "the noise" is a critical consideration -- many different sources provide different definitions and ideas.  The best measurement of "noise" is probably in a method blank at the exact RT window you are using for quantification.  Ideally the method blank would have a sample matrix identical to your real samples, but guaranteed to be missing the target compound.

It's a little tough to know from the information provided in your question how exactly you are calculating your LODs.  If for the moment we assume you are happy with how your LODs are computed, then I would pick the method with the lowest LOD.

It's also worth mentioning that a standard practice in targeted LC-MS/MS is to monitor two separate transitions per compound.  One of those is called the "quantifier" and it is this signal that is used to compute peak areas for quantification.  The other is called the "qualifier".  The ratio of the qualifier peak area to the quantifier peak area should be monitored for all analyses, and it should be approximately the same for all samples (that contain measuremable compound).  If this ratio deviates wildly, it's a sign that the quantifier ion may not be reliable for a particular sample.