Solubility experiment of different organic compounds Chemistry lab note says that:1) Aromatic acids like benzoic acid and salicylic acid are NOT soluble in water. But they are soluble in 5% NaOH,  and 5% NaHCO3.  Due to having a large hydrcarbon part (which is hydrophobic), it is understandable that aromatic acids won't dissolve in water. But why would they dissolve in base solvents.2) aromatic carbonyl compounds like benzaldehyde and acetophenone are NOT soluble in H2O, 5%NaOH, 5% NaHCO3, 5% HCl. But they are soluble in highly concentrated H2SO4. Got no idea about the reason behind it. Can somebody clarify?

Carboxylic acids easily react with bases forming salts, which are ionic compounds and are usually readily dissolve in polar solvents such as water, which obviously makes 95% of your base solution.

Carbonyl compounds are easily dissolved in strong sulfuric acid as 1) a reaction may take place, introducing -SO3H group to position 3 of aromatic ring of respective carbonyl compound; and 2) a strong acid can form hydrogen bonds with carbonyl oxygen. In both cases affinity of carbonyl compounds towards sulfuric acid is increased.


Dipole Moment of p-dibromobenzene What is the dipole moment of p-dibromobenzene?

I know that p-dichlorobenzene has zero dipole moment, but my teacher said that this is not the case for bromine. I am still confused and feel its dipole moment as zero. My teacher said that it does have some dipole moment (supported through experimental data).

PubChem gives dipole moment ($\mu$) of 1,4-dibromobenzene (p-dibromobenzene) as $\pu{1.43 D}$ in gas phase and $\pu{1.87 D}$ in liquid phase at $\pu{20 ^\circ C}$. It has given 1987 version of Handbook of Organic Chemistry (J. A. Dean, Ed.) as the reference. PubChem also gave $7.77$ as dielectric constant at $\pu{10 ^\circ C}$ and $6.7$ at $\pu{40 ^\circ C}$ with the same reference. However, I afraid to tell you that all of these values are incorrect. The 1999 version of Handbook of Organic Chemistry (same reference in different edition) listed followings for 1,4-dibromobenzene and its isomers (also included 1,4-diiodobenzene and its isomers for comparison):
$$
\begin{array}{c|ccc}
\hline
\text{Compound} & \text{Dipole moment, }\mu & \text{Dielectric constant, }\epsilon  & \text{reference page}\\
\hline
\color{blue}{\text{1,4-Dibrmobenzene}} & \color{blue}{\pu{0.0 D}} \ (-) & 2.57 \ (\pu{95 ^\circ C}) & \text{p. 5.111}\\
\text{1,3-Dibrmobenzene} & \pu{1.5 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 4.21 \ (\pu{20 ^\circ C}) & \text{p. 5.111}\\
\text{1,2-Dibrmobenzene} & \pu{2.13 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 7.86 \ (\pu{20 ^\circ C}) & \text{p. 5.111}\\
\color{red}{\text{1,4-Diiodobenzene}} & \color{red}{\pu{0.19 D}  \ (\pu{20 ^\circ C}\text{ in benzene})} & 2.88 \ (\pu{120 ^\circ C}) & \text{p. 5.113}\\
\text{1,3-Diiodobenzene} & \pu{1.22 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 4.11 \ (\pu{50 ^\circ C}) & \text{p. 5.113}\\
\text{1,2-Diiodobenzene} & \pu{1.70 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 5.41 \ (\pu{50 ^\circ C}) & \text{p. 5.113}\\
\hline
\end{array}
$$
Accordingly, 1,4-diiodobenzene actually has a dipole moment, but, every other 1,4-dihalobenzenes do not have a dipole moment. The values given in PubChem must be corrected during 15th edition of the book.
Reference:
John A. Dean, Editor, Lange's Handbook of Organic Chemistry, Fifteenth Edition; McGraw-Hill, Inc.: New York, NY, 1999, p. 5.111-5.113 (ISBN: 0-07-016384-7).


How do atoms lose their induced dipole? There exists the basic chemistry lesson with induced dipoles where the electron density of one atom changes causing a dipole to form and causing a cascading effect where other atoms adjacent to that polarized atom also become polarized.

How do those atoms lose their dipole if their wavefunction changes to "line up" with the dipole? Does a probabilistic event need to happen at the same time for all atoms in order to "lose" the dipole where the electron density goes back to its normal state?

The said lesson tends to oversimplify things to the point of getting them downright wrong. Atom dipoles do not line up with each other. They are way too weak for that. Should it be according to your description, they would indeed line up all in one direction and stay locked forever, much like magnetic particles in a magnet.

Now, I brought up the magnetic analogy because it is much better known than the similar electric phenomenon, called ferroelectricity. That's where your explanation is precisely right. Compounds that exhibit ferroelectric properties, once polarized, never lose their dipoles unless you force them to. But they constitute just a tiniest minority of all compounds.

What happens in the rest of them? Well, the molecules turn this way and that, atoms bounce and wiggle, spontaneous dipoles arise every now and then, only to vanish almost instantly. But the position where two neighboring dipoles are aligned in the same direction is energetically favorable, so they spend a little more time in this position than in the opposite. And that's what we call induced dipoles, dispersion forces, etc.

So it goes.


Units in modified Arrhenius equation? The modified Arrhenius equation is used to express the rate constant in a chemical mechanism model I'm working with. The equations is as follows:
$$k_\mathrm{f} = A\times T^b\times\exp\left(-\frac{E_\mathrm{a}}{RT}\right)$$
The paper states that "Units are Moles, cm3, Seconds, K, and Calories/Mole" so what would be the final units of the rate constant? I'm a bit confused due to the presence of the exponential function. When I simply plug the values in as given in the paper the rate becomes huge when multiplied by the molar concentration, is this because it gives molecules per second? To give you and idea for the following reaction:
$$\ce{C2H5OH + OH <=>C2H4OH + H2O}$$
The Arrhenius constants are as follows:
$$A = 1.74E+11$$
$$b = 0.27$$
$$E_a = 600.0$$
I am yet to calculate the reverse reaction using Gibbs Free Energy, is it just equally as large and thus it all cancels out or are the final units really not in $\mathrm{mol\ s^{-1}}$?

The exponential must be dimensionless.  That means if $E_a$ is in units of $\mathrm{cal\cdot mol^{-1}}$, then $R$ must be units of $\mathrm {cal\cdot mol^{-1}\cdot K^{-1}}$.  Make sure you use the right value of $R$ for these units; $R=8.314~ \mathrm{J \cdot mol^{-1}\cdot K^{-1}}$, but $R = 1.987~ \mathrm{cal \cdot mol^{-1}\cdot K^{-1}}$.  Temperature must obviously be in Kelvins.
The example reaction you gave is bimolecular, so I would think $k_f$ is meant to have units of $\mathrm{cm^3 \cdot s^{-1}\cdot mol^{-1}}$.  That way, when you multiply $k_f$ by the molar concentration of both reactants, e.g. ethanol and hydroxyl, you get a reaction rate that is in units of $\mathrm{mol \cdot s^{-1}\cdot cm^{-3}}$.
The remaining terms must therefore combine to give the right units for $k_f$. Since the units of $T^b$ will have units of $\mathrm K^b$, then the units for $A$ will be the units for $k_f$ divided by $\mathrm K^b$, i.e. the units of $A$ are $\mathrm{{cm}^3 \cdot s^{-1}\cdot mol^{-1} \cdot K^{\it-b}}$.


Which one is the stronger Lewis base: diethyl ether or tetrahydrofuran? I am trying to understand Lewis bases (bear with me, I am a physicist). I was asking myself how diethyl ether and tetrahydrofuran compare with respect to their properties as solvents in electrophilic additions. Or if there is any difference at all.

So, for example, is it better to use a (boron trifluoride + diethyl ether) complex or a (boron trifluoride + tetrahydrofuran) complex? As to my understanding this is mainly a question about how strong the solvent acts as a Lewis base. If the solvent (ether or THF) is a stronger base, it might inhibit the catalysis of the actual electrophilic addtion reaction we are targeting, right? So under this assumption it is probably better to use the weaker base.

My guess is that diethyl ether is the stronger Lewis base. It has two hydrogen atoms more, so it is also contributing two more electrons, and moreover due to the slightly lower electronegativity of hydrogen, these electrons tend to be dragged towards the oxygen. On the other hand, THF misses the two protons and electrons, and moreover it must be electroneutral in the symmetry plane where the opposite C-C bond is, so it contributes a little less electron stuff to the nucleophilic oxygen.

Is this argumentation correct?

And what about this: the THF molecule is a little more compact than the diethyl ether and moreover has a tensed 5-ring. So shouldn't this push the electrons a little more to the periphery of the molecule, especially to the oxygen, and thereby increase its strength as a Lewis base again? Is this effect relevant at all?

PS: After writing this question I found a table at Wikipedia (https://en.wikipedia.org/wiki/Lewis_acids_and_bases) which states that the enthalpy of formation of the BF3 adducts is

Et2O: 78.8 kJ/mol
THF: 90.4 kJ/mol

This is about the opposite of what I would have expected and seems to indicate that the effect of the two additional protons/electrons of Et2O is not as big as it might seem at first.

But is it really support for the "compactness of THF in the 5-ring" theory? Or are there other effects responsible for THF being the stronger Lewis base?

Which is "better"? In some specific reaction, with all other circumstances identified. Ether is a little too volatile for my preferences, but THF is not bad. I like the smell of ether, but the odor of THF is OK. Little problem with oxygen (fire and peroxides) with both of them, so you must be careful with both.
The difference in enthalpy of formation (11.6 kJ/mol) is small compared with activation energies of typical organic reactions (~100 kJ/mol), so "better" might be related to human preferences more than than chemical performance.
A chemist might already have a feel for the molecular configurations of these molecules, but a physicist could benefit from making the models or looking at them:

The ability of THF to poke the BF3 molecule better than ether can is apparent. Whether this is the controlling factor in which is "better" remains to be seen. I'll go out on a limb and say that either one will work, although the results (convenience, yield, expense, etc.) could easily be +/- 20-50%, depending on the situation. In a lab, one might work better; in industry, the other might be preferable.
Either solvent could probably be optimized for best performance under slightly different conditions, and these optimum conditions will not be the condition you choose for the first experiment. Probably a dozen experiments will suggest which solvent could be the better choice. The literature should suggest the better (or more typical) choice for your needs.
Ref. https://en.wikipedia.org/wiki/Activation_energy


Does NaOH have a higher OH- concentration than NH3? In my book it's a multiple choice question with only one correct answer.

There are two solutions of $\ce{NaOH}$ and $\ce{NH_3}$ that have similar concentrations and volumes.

These two are among the options:

$\ce{NaOH}$ has a higher $\ce{pH}$ than $\ce{NH_3}$
$\ce{NaOH}$ has a higher $\ce{OH^-}$ concentration than $\ce{NH_3}$

Wouldn't these both be correct?

Given that the solutions have equal concentrations, a solution of $\ce{NaOH}$ (a strong base) will be more basic than one of $\ce{NH3}$ (a weak base), since $\ce{NH3}$ only dissociates partially:

$$\ce{NH3 + H2O <<=> NH4+ + OH-}$$

Loosely speaking, the pH of a solution is defined to be

$$\mathrm{pH} = -\log_{10}[\ce{H+}]$$

and from the ionic product of water, we know that $[\ce{H+}][\ce{OH-}]$ is always a constant (equal to $10^{-14}\ \mathrm{M^2}$ at $25\ ^\circ\mathrm{C}$).

Therefore, a higher pH implies a lower concentration of $\ce{H+}$, and therefore a higher concentration of $\ce{OH-}$. This means that the truth of option 1 implies the truth of option 2, and vice versa, i.e. option 1 is logically equivalent to option 2.

My first suspicion would therefore be a typo in your book. In any case, your reasoning is correct.


What reaction is happening when a lithium polymer battery "is on fire" When something goes wrong with a lithium polymer battery (as are commonly used to power electric models and robots), and it "catches fire", what is actually happening?

What gas or gasses are being released?

Here's a video of such an event.

I'll take a stab at this one, but I'm not a battery expert.

What is a lithium "polymer" battery?

Wikipedia says that "lithium polymer battery" is now commonly used to refer to nearly any lithium ion battery that is encased in a flexible polymer shell instead of a hard metal casing.  There is a chance for confusion here because it apparently used to mean a battery with a polymeric electrolyte instead of a purely small-molecule liquid-phase electrolyte.  I'm assuming you mean the former "in a polymer shell" definition for the purposes of my answer.

What type of anode, cathode, and electrolyte materials?

There are many types of lithium ion batteries.  Electrolyte composition, cathode materials, and anode materials can all vary.  Perhaps the most common is a lithium cobalt oxide cathode with a intercalated graphite anode.  Organic solvents must be used as electrolyte because water is not stable to the high voltages that lithium ion cells provide.  Solvents stable over a wider potential range must be used instead to dissolve salts that can carry electric current.

What happens when it's crushed:

So for lithium cobalt oxide batteries with liquid electrolyte, I was able to find a good source describing the problem of "thermal runaway":

  The temperature of the cell rises due to the
  chemical reactions between the organic solvent and electrode materials leading to cell failure/explosion. The chemical reaction is prompted because of the coexistence of 1)
  combustible organic solvent with inorganic salt electrolyte 2) lithiated graphite anode and 3)
  partially delithiated LixCoO2 cathode in the charged state. The graphite anode intercalate
  fairly well during the initial cycles. As the cycle proceeds, the available intercalating sites in
  the graphite slowly decreases and the lithium deposition over the surface of the anode
  increases during charging process. Such a situation prompts non-even lithium deposition or
  even dendrite growth over the graphite surface. When the battery is in the charged state the
  cathode remains in a delithiated state. Flow of current over and above the
  tolerable/standard charged state causes instability to the cathode which starts to release
  oxygen into the electrolyte. Thus released oxygen reacts exothermally with the lithium
  plated over the graphite anode and increases the temperature of the cell making the cathode
  to release oxygen further. The cell which was in the dormant state in the absence of oxygen
  and heat now becomes an explosive device in the event the cell is met with any abuse.

That describes what happens when the battery is overcharged and leaks oxygen into the electrolyte.  I suspect that when the batteries are crushed, even if they aren't overcharged, that at least two things can happen leading to similar thermal runaway: First, the lithiated graphite is air- and water- reactive, heating the now-crushed battery and eventually lighting the electrolyte solvent on fire.  Second, the cell may be shorted by the mechanical disruption, so that even without air or water contact, the voltage in the battery can discharge very quickly.  This would create large heating from ohmic resistance, also eventually igniting the elctrolyte.


Conditions for Complete hydrolysis of salt My text book says that the cations (or anions) which are stronger than hydronium ion(or OH-) and their conjugate base (or acid) being very much weaker than water show complete hydrolysis.
 While the cations(or anions) which are weaker than hydronium (or OH-)and their conjugate base(or acid) being stronger than water but weaker than hydroxide(or H3O+) exhibit hydrolysis to Limited extent.

I don't quite understand why is this so?

The question is, in fact, a good one, but the concept of weaker/stronger is confused - and confusing!

A way to describe the situation, using the terms in the question is this: strong acids and strong bases undergo complete hydrolysis in water. The energy gained from hydrogen bonding with a water molecule is great enough to effect complete ionization in water. Examples would be HCl and NaOH.

Water lies in the middle here, being able to be protonated by strong acids and deprotonated by strong bases. Hydrolysis is a measure of how much the water is protonated to H3O+ or deprotonated to OH-.

There are also materials which ionize in water, but not so extensively, e.g., acetic acid and amines such as ammonia. They dissolve, but do not react so much and thus hydrolyze to a limited extent. Confusing, perhaps, that we call these materials weak acids and bases but they have strong internal bonds which prevent their total ionization. These strong internal bonds mean that if you did form the the conjugate acid or the conjugate base (in a salt), they would react strongly with a water molecule to form a considerable amount of the corresponding acid or base. Thus the salts would ionize and hydrolyze where the original acid or base would not be said to hydrolyze (much).

The "strength" of the cation may perhaps be better described as a polarity, which, if it is a "strong", favors hydrolysis, i.e., complete ionization, meaning it is strongly attracted to the oxygens of water and less so to its own anion. By this terminology, Na+ is a strong acid because it is strongly attracted to water (oxygens) at the same time OH- would be a strong base.

A "weak" cation, like NH4+, would be one which will not protonate water to a great extent. A solution of 1 M NH4Cl will give complete ionization to NH4+ and Cl-, but we do not call that hydrolysis. The ammonium ion is stable (strong internal bonds) and only weakly protonates the H2O molecules around it, so we call it a weak base, with limited hydrolysis.


Normalization of the wavefunction φ(x) = A I was trying to normalize the wave function

$$
\psi (x) = \begin{cases} 0  & x<-b \\ A & -b \leq x \leq 3b \\ 0 & x>3b
\end{cases}
$$

This is done simply by evaluating

$$
\int\limits_{-b}^{3b} | \psi (x) |^2 dx = 1
$$

I found that

$$
A = \pm \frac{\sqrt{b}}{2}
$$

This gives the normalized wave function

$$
\psi (x) = \begin{cases} 0  & x<-b \\ \pm \frac{\sqrt{b}}{2} & -b \leq x \leq 3b \\ 0 & x>3b
\end{cases}
$$

This was quite straight forward... too straight forward for my liking. My question is twofold:

1) Is my derivation above correct? And
2) How shall I deal with the ``$\pm$'' sign in front of the fraction? Should that be included in the expression for $\psi (x)$? Do I then have two different wave functions for the same particle, one negative and one positive? Maybe one of them do not carry any "physical significance", and can be dropped?

Your derivation is mostly correct, but you made a small mistake when evaluating the integral

\begin{equation}
\int\limits_{-b}^{3b} | \psi (x) |^2 \,\mathrm dx = \int\limits_{-b}^{3b} | A |^2 \,\mathrm dx = 1
\end{equation}

The antiderivative of a constant $C$ is the product of this constant with the integration variable, e.g. $x C$.
So

\begin{align}
\int\limits_{-b}^{3b} \lvert A \rvert^2 \,\mathrm dx = \Bigl[ x \lvert A \rvert^2 \Bigr]_{-b}^{3b} &= \lvert A \rvert^2 \bigl((3b - (-b)\bigr) = 4 b \lvert A \rvert^2 = 1 \\
&\Rightarrow  A = \pm \frac{1}{2 \sqrt{b}}
\end{align}

As for the $\pm$ sign ambiguity: In quantum mechanics, we know that a wave function can always be multiplied by a constant phase factor:

\begin{equation}
\psi'=\mathrm e^{-\mathrm i\alpha}\psi,
\end{equation}

where $\alpha \neq \alpha(x,t)$, because the probability distribution of a system is unaffected by the above transformation, and also the Schrodinger equation and the probability current are unaffected by the above transformation.
So - since $-1 = \mathrm e^{\mathrm{i} \pi}$ and $+1 = \mathrm e^{\mathrm{i} 2 \pi}$ are examples of such constant phase factors - it doesn't make a difference, from a physics point of view, whether you use $+\frac{1}{2 \sqrt{b}}$ or $-\frac{1}{2 \sqrt{b}}$ since both lead to the exact same physical results and are thus physically equivalent.


How do I solve oxidation numbers for unknown elements? I am doing very well with oxidation numbers but I have a few problems. I know all the rules but when I get an element where I can't use any of the constant oxidation numbers what do I do?
For example: $\ce{HCNS,Ag2S,Bi2S3,KSCN}$ etc. ?

You shouldn’t really need to use the shortcut rules (fluorine having $\mathrm{-I}$, hydrogen having $\mathrm{+I}$ and oxygen having $\mathrm{+II}$ ‘except in peroxides’) with the exception of really simple compounds where it all falls into place immediately. Instead:

If the compound is molecular or contains a molecular ion, draw the Lewis structure.
Assign each bonding electron pair to the more electronegative bond partner.
Count the electrons each atom ends up with, substract from the number of electrons it should have according to its position in the periodic table, invert that number’s sign.
If at step one you are only left with ions, just take the ion’s charge as oxidation state.
You did it right if the overall sum of oxidation states matches the molecular ion’s charge or zero for the overall structure.

(Note: I am simplifying here by assuming that compounds are either molecular or ionic. That, of course, is most certainly not the case but the very crude generalisation gets us very far in this case.)

To work through one of your examples:
$\ce{KSCN}$ is an ionic compound consisting of potassium cations and thiocyanate anions. According to step 4, we can assign potassium the oxidation state of $\mathrm{+I}$.

The structure of thiocyanate is $$\ce{S=C=\overset{-}{N} <-> \overset{-}{S}-C#N}$$
In both cases, the terminal atoms have enough lone pairs to complete their octet.
The electronegativites are $\ce{N}>\ce{S}>\ce{C}$. Thus, we assign the bonds so that:

sulphur and nitrogen each have eight valence electrons
carbon has zero

This is independent of the resonance structure we used.
Calculating:

Sulphur:
eight electrons, should have six.
$\displaystyle-(8-6) = \mathrm{-II}$
Carbon:
zero electrons, should have four.
$\displaystyle-(0-4) = \mathrm{+IV}$
Nitrogen:
eight electrons, should have five.
$\displaystyle-(8-5) = \mathrm{-III}$

Has been done before.
To check:

$\displaystyle \mathrm{+I + -II + +IV + -III = \pm0}$ (for $\ce{KSCN}$)

$\displaystyle \mathrm{-II + +IV - III = -I}$ (for $\ce{SCN-}$)


How do you perforate a teflon sheet? I have a sheet of thin unperforated teflon that I need to perforate by tomorrow for a lab I work in. None of my coworkers know how to perforate the sheet. Does anyone know how I can do this using basic tools from my house?

You should contact a specialized company, however you should also provide more infos about the diameter and the number of the holes you want. This is an excerpt from FedTech:

  laser cutting is the method of choice for small perforation, as laser
  cutting has a smaller, tighter cutting tolerance and thus has the
  capability to create tiny, intricate perforations. However, if you
  prefer, abrasive waterjet cutting can be used as well but for larger
  perforation (.030" and up, respectively). If you prefer to use
  non-abrasive waterjet cutting, you can create perforations up to .004"
  in diameter, however this form of waterjet cutting is restricted in
  cutting certain types of material. With non-abrasive waterjet cutting,
  you can cut up to about 12" thick from the following: plastic, rubber,
  foam, composites, silicone, foam, felt, cork, Teflon (PTFE), neoprene,
  and other soft materials


Optional determination techniques in Atomic Absorption Spectroscopy I have tried to find information about this case but I couldn't find anything specific.

The key point of instrumental methods of analysis, is for the student to be able to know which one of the possible methods to use for a specific determination. As for me, that is not an easy thing to do because every method has its pros and cons. For the moment I am analysing AAS (Atomic Absorption Spectroscopy), and I simply cannot tell:

What's the reason that $\ce{As}$, $\ce{Te}$, $\ce{Se}$, $\ce{Ge}$, $\ce{Sn}$, $\ce{Pb}$ is preferred to be determined only by the hydrid method? $\ce{Pb}$ for example can be determined also with the other methods in AAS

Does anyone have any tip, how to choose between methods if it's required for any element?

I replied with a similar question earlier about the mercury determination.

Remember this is "atomic" spectroscopy so it is expected that the "sample" is being totally ripped apart. So one "best technique" per element.

$\ce{As}$, $\ce{Te}$, $\ce{Se}$, $\ce{Ge}$, $\ce{Sn}$, $\ce{Pb}$ are analyzed by the graphite tube technique rather than in a flame because the technique provides more sensitivity.

$\ce{Hg}$ is an odd one since you use a cold vapor technique.

Of course depending on the particular sample matrix what you might need to change.


In regards to the addition of hydrogen to alkynes, does the amount of moles of Hydrogen matter? Is there any difference between having the addition of 1 mole of hydrogen to 2-pentyne with a platinum catalyst compared to the addition of 2 moles of hydrogen(H2). I was under the impression that both would result in an alkane yet I have 2 problems of that nature.
would one only result in a alkene, though I thought that only happens with the poisoned catalyst.

Your question is not entirely clear, but I think that you ask whether or not an excess of hydrogen gas would interfere with the reduction of an alkyne to an alkane.

This is not the case. A lot of the time hydrogen is just drawn into the reaction flask directly from a large reservoir (20L $\ce{H2}$ gas cylinder under pressure or a simple balloon). When the pressure in the reservoir does not change anymore (in the case of the balloon when it does not shrink anymore) it means that the reaction is finished, i.e. no more hydrogen can be added.

In general over-reduction cannot be prevented by limiting the amount of hydrogen present. It will just result in an awful mixture of products (alkanes/alkenes/alkynes). A poisoned catalyst is required to prevent this (e.g. Lindlar's catalyst).


Home made PCB tinning solution By reviewing a couple of articles and videos I decided to give some try to develop a PCB tinning solution at home. Based on most of the articles, the easiest way to develop this solution is by using the following chemicals:

Stannous Chloride ($\ce{SnCl2}$)
Thiourea ($\ce{CH4N2S}$)
Sulfuric Acid ($\ce{H2SO4}$)

After checking with a couple of stores, I noticed that Thiourea is not available to buy in any of the stores. Now my questions are:

What Thiourea does in this solution?
Is there any substitute available for Thiourea (for this process)?
Is it possible to prepare Thiourea from any other chemicals? (It seems using Ammonium Thiocyanate ($\ce{NH4SCN}$) can do that job, but it looks like a lengthy and complex process to do at home.)
Is there any possible way to build a tinning solution without using Thiourea?

Thiourea ($\ce{(H2N)2C=S}$) is a very important component of tinning solution because tinning is a redox process. That's because thiourea is a powerful reducing agent (Ref.1), chemically and biochemically used in redox reactions in vitro or in vivo:

$$\ce{(H2N)2C=S <=> (H2N)2C-S-S-C(NH2)2 + 2e-} \quad E^\circ = \pu{0.418 V} \tag1$$

The oxidizing agent in this process is $\ce{Sn^2+}$ ion, which reduces it self during the process:

$$\ce{Sn^2+ + 2e- <=> Sn} \quad E^\circ = \pu{-0.1375 V} \tag2$$

As a result, following redox reaction happens:

$$\ce{(H2N)2C=S + Sn^2+ <=> (H2N)2C-S-S-C(NH2)2 + Sn} \quad E_\mathrm{rxn}^\circ = \pu{0.2805 V} \tag3$$

Since $E_\mathrm{rxn}^\circ $ is positive, the reaction is spontanious in standard conditions and no electric current is necessary. When you dip your circuit in the solution (or touch surface to surface of the solution), metallic tin would deposit as a thin layer.

As an alternative to $\ce{(H2N)2C=S}$, dimethylated or diethylated thiourea can be used in those reactions, which require electrode potential within the range of $\pu{0.418 V}$ or higher (Ref.2):

$$\ce{(CH3NH)2C=S <=> (CH3NH)2C-S-S-C(NHCH3)2 + 2e-} \quad E^\circ = \pu{0.460 V} \tag4$$

$$\ce{(C2H5NH)2C=S <=> (C2H5NH)2C-S-S-C(NHC2H5)2 + 2e-} \quad E^\circ = \pu{0.450 V} \tag5$$

In tanning process, theoretically, any component with oxidative potential bigger than  $\pu{-0.1375 V}$ can be used as the reluctant. Thus, best alternative come to my mind is ascorbic acid (Vitamin C). Recently, natural source of ascorbic acid source, extract of fruits of Averrhoa bilimbi (Sinhala name: Billin), has been used to prepare silver and gold nanoparticles from corresponding salts (Ref.3):

In any event ascorbic acid didn't work, there is a way to synthesize thiourea derivatives without using thiocyanate or isocyanate. The scheme is illustrated in the bottom box of above diagram (Ref.4). The $E^\circ$ of these derivatives should be closer to that depicted in equations $(4)$ and  $(4)$. However, any attemp to synthesize these compound should be done in certified chemical laboratory under the observation of an experienced chemist. Keep in mind that thiourea is a potent carcinogen.

Note: It is well known that silver polishing reagents contain thiourea ($50\%$ by weight; Ref.5). You may able to extract thiourea from such solutions.

References:

Paul W. Preisler, Louis Berger, “Oxidation-Reduction Potentials of Thiol-Dithio Systems: Thiourea-Formamidine Disulfide,” J. Am. Chem. Soc. 1947, 69(2), 322–325  (https://doi.org/10.1021/ja01194a048).
Paul W. Preisler, “Oxidation--Reduction Potentials of Thiol--Disulfide Systems. III. Substituted Thioureas and Compounds Containing Two Thioureido Groups,” J. Am. Chem. Soc. 1949, 71(8), 2849–2852  (https://doi.org/10.1021/ja01176a076).
R. S. Rimal Isaac, G. Sakthivel, Ch. Murthy, “Green Synthesis of Gold and Silver Nanoparticles Using Averrhoa bilimbi Fruit Extract,” Journal of Nanotechnology 2013, Article ID 906592, 6 pages (https://doi.org/10.1155/2013/906592).
Wei Tan, Jianpeng Wei, Xuefeng Jiang, “Thiocarbonyl Surrogate via Combination of Sulfur and Chloroform for Thiocarbamide and Oxazolidinethione Construction,” Org. Lett. 2017, 19(8), 2166–2169 (https://doi.org/10.1021/acs.orglett.7b00819).
Mark Laing, Mark Beeston, "Silver polish formulation containing thiourea," US Patent 5,958,854, 1999 (Application No.: 08/869,685 in 1997).


Is there a method to separate iron from mercury? The mercury is contaminated with iron. Is there any way to remove iron from mercury?

As iron is one of few metals not forming amalgams,

  Almost all metals can form amalgams with mercury, the notable exceptions being iron, platinum, tungsten, and tantalum.

..it should preferably contaminate the mercury surface.

I would try, perhaps repeated,  dropping mercury, e.g via punched filtration paper to sulphuric acid bath.

It can be combined with  agitating mercury - acid phase border.

Important can be the origin of impurity and the intended purpose of purified mercury. There is possible the only reasonable way would be distillation, mentioned in the other answer, respecting the warning.


Comparison of stability of cyclic imines I am looking for the reason for the lowest stability of cyclohexane-1,3,5-triimine (D).

I was not able to find any explanation using resonance or aromaticity.
All help is appreciated.

The cyclohexane-1,3,5-triimine tautomer is benzene-1,3,5-triamine, which is aromatic. This makes this trienamine form much more stable than any of the other possible tautomeric forms.


How to separate lead condensate from aluminum A common problem in shooting sports is precipitation of lead onto aluminum suppressor baffles.  (The lead presumably comes from both lead styphnate priming compound and from the base and bearing surface of lead bullets that vaporize during propulsion.)

If the lead precipitate is allowed to build up a contiguous layer it requires mechanical removal, which tends to damage and degrade the aluminum alloys used.

A common solution for ferrous baffles is to strip the lead with H2O2-boosted acetic acid, but that goes after aluminum almost as fast as lead.

What practical electrochemical methods could be used to strip the solid lead from the aluminum?

(Bonus if you can propose a method that would also leave titanium baffles intact!)

You could try hard anodizing Al and Ti (e.g., Calphalon cookware) to give them a non-delicate surface, and go on from there.  Black is a good gun component color.  CVD deposition of titanium boride, ion plating of titanium nitride or carbonitride gives exceptionally survivable hard surfaces.

http://www.brycoat.com/coating-services/pvd-coatings/titanium-nitride-tin.html
http://www.brycoat.com/coating-services/pvd-coatings/ticn.html

Among mostly cheap and harmless diddles, I'd try 1:2 mole ratio of choline chloride  with urea or ethylene glycol ("all natural," etc.).  Mix, the deep eutectic liquefies, soak.  Both eutectics conduct electricity, so you could make lead contamination the anode and plate it out on a closely spaced but not touching copper foil cathode. Test on contaminated waste Al or Ti.  Careful with the voltage - too high and you mobilize Al and Ti.  Avoid contact with skin pro forma, do not ingest, keep away from eyes.

Try vacuum thermal deposition of a micron or so thickness of the Parylene family on the newly fabricated and cleaned metal - intrinsic, especially C, or F.  Parylene polymer surface coats survive the insides of excimer lasers, protecting the excitation electrodes.  If the lead still sticks, chemical cleaning should stop at the polyparylene film.  Or not.  Gotta test it.  You are worried about vapor deposition not abrasion, so an inert barrier film is reasonable.

Google, parylene coating services

For aluminum or titanium vs. lead in chemistry, I'd pull one out of left field to try on test coupons:  Rauchfuss' elixir: elemental sulfur dissolved in N-methylimdiazole (pyridine less so, very smelly, toxic, and don't get it on your skin or breathe the awful vapors).  It goes through polysulfide-forming metals like a buzz saw, giving soluble products.  I would expect aluminum to be inert; titanium is a toss-up.  It should hit lead hard, probably.  Test it small scale.

Inorg. Chem. 30 (11): 2514 (1991)
DOI:10.1021/ic00011a011
http://dx.doi.org/ (resolve a DOI)

Last gasp:  Weapon Shield to clean before and after firing.

Concentrated nitric acid (already a bad idea) passivates aluminum and eats lead.  Toxic and corrosive NOx fumes come off.  However...aluminum plus lead in galvanic contact make aluminum the sacrificial anode.  Homeland Severity does not like people possessing concentrated nitric acid.

Wear gloves and goggles, have ventilation.  None of this stuff is good for you.


Relationship of Dielectric constant with ionic bond How is dielectric constant related to formation of ionic bond?

You are probably referring to Coulomb's Law:

$${F\propto \frac{q_1 q_2}{d^2 \epsilon }}$$

This law, that you probably already met in physics, applies to chemistry and ionic bonds, since they are all characterized by electrostatic interactions.

You see that the strenght of a ionic bond is directly proportional to the product of the charges on the ions, inversely proportional to the square of their distance, and inversely proportional to the dielectric constant of the solvent. In vacuum, the dielectric constant is $\epsilon = 1$.

Otherwise, when the ions are in a solvent, the dielectric constant is associated to the solvent in use (you can find appropriate charts online).


When we blow on/stir a hot drink, why aren't we making it hotter? This may have already been a question on Chemistry StackExchange (in which case I will delete it if someone points me to the question), but I was wondering why providing even more kinetic energy into the system (stirring a cup of hot coffee) why are we expecting the coffee to become cooler?

If I understand correctly, heat is caused by molecules in a system moving rapidly and "rubbing each other" -- creating heat. So the more movement --> the more heat.
Doesn't this not make sense with the idea of stirring a cup of hot coffee?

Edit#1

Excuse me for the title and the body implying different things, I took blowing on a hot drink and stirring it as the same thing -- stimulating movement in the drink.

This is related to a concept called "vapour pressure". Liquid water (and any other liquid for that matter) always has some amount of water vapour that has to be in equilibrium with it. For example, the air above water of 5 °C has about 1% of water vapour, 40 °C has about 7% of water vapour and at 90 °C it has about 70% of water vapour.

If there is too much vapour in the air for the temperature it is, then water condensates. That's why you get condensation on your glasses when you walk from a hot humid room outside into the cold at winter. If there is not enough vapour in the air, water evaporates to fill in for the missing vapour. That's why taking off your shirt after you've exercised and sweated makes you dry faster. The air trapped between your skin and clothes is very humid and in equilibrium with the sweat on your skin. Once you take it off, dry air comes into contact with the skin, forcing evaporation of your sweat (even though you're below the boiling point of water).

This process of evaporation is endothermic, meaning it consumes heat. Whenever water is evaporating, the temperature of the water goes down. When you're blowing air on a hot drink, you're basically replacing the humid air on top of the drink with dry air. This allows the water to evaporate more easily, reducing the overall temperature of the drink. This is also why you're feeling cold when you got wind blowing on your when you're sweating.


Do strong acids actually dissociate completely? Do strong acids actually dissociate completely (every single molecule dissociates), or are they just assumed to do so for the sake of simplicity? That seems odd, considering weak acids, many of which are very similar to strong acids, don't fully dissociate.

Yes, for sufficiently strong acids, in sufficiently dilute conditions, in sufficiently basic solvents. However, things are hazier than you might expect, and depending on your definition, "clear-cut" examples of complete dissociation only become possible when you start swatting flies with nuclear bombs.
The problem is that chemistry works on equilibria, and at some point it becomes difficult to measure tiny concentrations - in a haystack of one billion straws, how can you be sure there isn't at least one needle? And in one trillion? And what about in a septillion?

First, let us perform a qualitative investigation by considering the generic acid dissociation equilibrium in water, as follows:
$$\ce{HA(aq)->H+(aq) + A-(aq)} \quad \quad \quad \mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}}=10^{-pK_a}}$$
To simplify the following discussion, we will make the very crude approximation where we equate the thermodynamic activity of a species with its molar concentration. In other words:
$$\mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}} \approx \frac{[H^+(aq)][A^-(aq)]}{[HA(aq)]}}$$
Our objective now is to virtually eliminate the presence of undissociated $\ce{HA}$ molecules. For clarity, we rearrange the above equation:
$$\mathrm{[HA(aq)] \approx \frac{[H^+(aq)][A^-(aq)]}{K_a}}$$
Clearly, to make $\ce{[HA]}$ as small as possible, the numerator must be very small, and the denominator must be very large. Let us set a goal of making $\mathrm{[HA]<10^{-27}\ mol\ L^{-1}}$, which implies less than $\mathrm{0.001}$ molecules of undissociated $\ce{HA}$ in a litre of solution.
At this point, I should mention that aqueous solutions of acids in ambient conditions must always have $\mathrm{[H^+(aq)]>10^{-7}\ mol\ L^{-1}}$, thanks to the self-ionization of water. We take the best-case scenario (which physically corresponds to very dilute solutions):
$$\mathrm{10^{-27} \gtrsim \frac{10^{-7}\ [A^-(aq)]}{K_a} \quad \longrightarrow \quad 10^{-20} \gtrsim \frac{[A^-(aq)]}{K_a}}$$
We now take a typical strong acid, hydrogen chloride ($\ce{HCl}$) which has an estimated pKa of -5.9 in water. In other words:
$$\mathrm{10^{-20} \gtrsim \frac{[Cl^-(aq)]}{10^{5.9}}=\frac{[Cl^-(aq)]}{790\ 000} \quad \longrightarrow \quad [Cl^-(aq)] \lesssim 8\times 10^{-15}\ mol\ L^{-1}}$$
We see therefore that meeting the criterion of full dissociation requires preparing an aqueous $\ce{HCl}$ solution with femtomolar concentration. Essentially every solution used in a chemistry laboratory will have a higher concentration, so every $\ce{HCl}$ solution anyone has ever touched will have undissociated $\ce{HCl}$ molecules. Strictly speaking, you could say the dissociation is never complete. However, if you have a mind to make a fully dissociated solution of $\ce{HCl}$, it is physically possible.

Just diluting acid solutions into oblivion until every molecule dissociates is kind of boring. The more exciting prospect is, of course, to increase the strength of the acid. The acidity scale goes far, far beyond aqueous solutions of $\ce{HCl}$.
Interestingly, for sufficiently strong acids (typically $\mathrm{pK_a<-10}$, meaning acids at least 10 000 times stronger than aqueous $\ce{HCl}$), it is possible to isolate hydronium salts - when the strong acid is mixed with water in a 1:1 molecular ratio, a solid salt is formed, where the cation is $\ce{H3O+}$ and the anion is supplied by the conjugate base of the acid. The stoichiometry can be made exact. Surely this represents full dissociation of the acid?
Well, again, it depends on your strictness. If you actually look at the crystal structures of many of these salts, it's clear that there is a strong, close  interaction between the $\ce{H3O+}$ cation and the anion, and you could argue this still does not count as complete dissociation.
That said, you can go crazy and consider some of the strongest acids we can put in a bottle, such as carborane superacids. The corresponding hydronium salt of these superacids can have extremely weak interactions between the hydronium cation and the anion. For example, when the hydronium salt $\ce{(H3O+)(CHB11Cl11^-)}$ is prepared in a benzene solution, the solid that forms actually contains three molecules of benzene surrounding the hydronium cation, which is completely detached from the anion. If this doesn't imply "complete dissociation" of the carborane superacid in water, I don't know what would.
The acidity scale goes further still than even carborane superacids. There are a few known acids which will practically always fully dissociate, no matter what. The representatives of this class are rather exotic things such as $\ce{H_3^+}$, $\ce{HN2^+}$, $\ce{HNe^+}$, $\ce{HHe^+}$, and so forth. These acids can only be detected in the gas phase, as they fully dissociate by protonating virtually anything they come into contact with (including any anion, which is why they cannot be isolated as a neutral compound).

The last way to ensure full dissociation of an acid in a solution is to change the solvent in which it is dissolved. All things considered, water is a pretty weak base. If, for example, ammonia (am) were used as the solvent, the relevant equilibrium for $\ce{HCl}$ would be:
$$\ce{HCl(am)->H+(am) + Cl-(am)} \quad \quad \quad \mathrm{K_a(am)=\frac{a_{H^+(am)}a_{Cl^-(am)}}{a_{HCl(am)}}=10^{-pK_a(am)}}$$
If I were to give a crude guess, I'd imagine that the $\mathrm{pK_a}$ for $\ce{HCl}$ in liquid ammonia would very roughly be somewhere around -15; that is, dissociation of $\ce{HCl}$ in liquid ammonia is about a billion times easier compared to water. This is because ammonia is a stronger base than water.
You know how we were talking about hydronium salts as if they were some unusual thing? Well, the equivalent concept in ammonia would be the formation of ammonium salts. And indeed, $\ce{HCl}$ does easily form a salt with ammonia in a exact 1:1 molecular ratio, namely ammonium chloride, $\ce{NH4Cl}$. Not exactly an unusual compound.
There's nothing stopping you from using an extremely strong base as a solvent. For example, in pure liquid phosphazene superbases, even very weak acids may be fully dissociated in ambient conditions, even in the most rigorous sense.


Freezing point depression to find molecular weight 0.845g of the unknown dissolved in 4.250 grams of camphor had a freezing point of 1.36C. The freezing point depression constant for camphor is 40C*kg/Mol and the melting point of pure camphor is 129C.

Calculate the freezing temperature depression, T? Tf=Kf*m

Calulate the molality? Which i know is mol/kg solvent

Calculate the number of moles present of the unknown compound?

Calculate the molar mass of the unknown compound?

Thank you in advance and please lists steps if possible. I really want to understand and to get this!! :)

Mass of solute $W_b=0.845\times10^{-3}\;kgm$
  Mass of solvent $W_a=4.250\times10^{-3}\;kgm$
  Depression in freezing point $\Delta T_f=1.36\;K$
  Depression constant for solvent $K_f=40\;K.kg.mol^{-1}$
  Freezing point of Camphor $T_f=129\;^\circ C$

Part I: $\Delta T_f$

Part II: Molality($m$): $$\Delta T_f=K_fm\implies m=\frac{\Delta T_f}{K_f}$$

Part III: Moles($n_b$): $$m=\frac{n_b}{W_a}\implies n_b=W_am=W_a\frac{\Delta T_f}{K_f}$$

Part IV: Molecular Mass of solvent($M_b$): $$n_b=\frac{W_b}{M_b}\implies M_b=\frac{W_b}{n_b}=\frac{W_bK_f}{W_a\Delta T_f}$$


Is this cotton or viscose? I have to examine the exact material of a specific fabric (non-woven), of which I know is either viscose or cotton. I know that viscose and cotton are structurally similar, but there should be a way to tell them apart.

This is the result of the microscopic examination (x 100):

My professor tells me this is viscose, but I thought of cotton because of the long fibers. But I also know that cotton has a 'twisted' look under the microscope.

My professor doesn't give any more clues, which leaves me with a bit of a problem. How can I tell the two apart? Does the picture show enough evidence for viscose (because of the linear structure)? I would like to double check this, just to be sure.

Yes, the picture should be sufficient to decide between viscose and cotton; see photo of cotton at http://whrhs-forensic-chem-spring-2010.wikispaces.com/Fiber+Analysis?responseToken=906bf6a76a5e268a00e54c8bce6927ba.

Recall that cotton was the cell wall of a plant cell, so it retains some of the shape of the cell. (Silk, on the other hand, is excreted from spinnerets, as are man-made fibers, so it looks more like viscose).


Why does a hard-boiled egg have more entropy than a normal egg? The question "Why does a hard-boiled egg have more entropy than a normal egg?" was recently asked on one of my tests.

I said that entropy will decrease as a hard boiled egg has molecules in a more organised way than an raw egg, but I was marked wrong with a remark that entropy will increase because the proteins and other molecules are decomposed into simpler substances due to heat.

I did ask why I was not correct but what I got was the same explanation along with the decomposition reactions.

I don't feel this explanation is satisfying.

For example, do not 10 objects on a table organised properly have more entropy than 5 objects thrown anywhere on the table?

The definition of entropy is the measure of disorder in the system. So clearly in my above example the 10 objects are more organised and thus less disordered than the 5 objects in the second case and same can be said for the main question.

Can somebody please explain to me why a hard-boiled egg has more entropy than an un-boiled egg (I like if the explanation has some math)?

I’ve come across several claims that a (hard) boiled egg has greater entropy than an unboiled egg. None of them cite any peer-reviewed research to back up their claim. While it is possible that entropy increases when you boil an egg, it is also possible that it does not. It is even possible that it stays the same.

I did my own literature search to see if I could find any publication that answered the question. What I found left the question very much up in the air. So I contacted Greg Weiss, a scientist who has done research on a related topic: ‘unboiling’ an egg. Dr. Weiss was kind enough to discuss the topic with me via email.

tl:dr

Whether a hard-boiled egg has greater entropy all boils down to the question of how structured are the assemblies found in aggregated ovalbumin. Furthermore, we can't just consider the assemblies found in the ovalbumin aggregate since we're talking about assemblies composed of all the types of protein in the overall albumen. (BTW, it's important to note that heated egg white doesn't just form any type of aggregate, it forms a gel, which presumably implies additional structure.)

For those who are interested in the details, here is the full email thread.


Why can toluene be chlorinated at either the benzylic position or the ring depending on the conditions? My book mentions a reaction for the preparation of benzaldehyde with toluene in which side chain chlorination of toluene gives (dichloromethyl)benzene, which upon hydrolysis gives benzaldehyde:

Why does the chlorination occur at the methyl group and not the aromatic ring in this case? Also, why chlorination with $\ce{FeCl3/Cl2}$ lead to 2-chlorotoluene (and 4-chlorotoluene), without any chlorination of the methyl substituent?

The halogenation of toluene is possible by two mechanisms. In German, they are referred to by the shorthands KKK and SSS, where KKK stands for Katalysator, Kälte, Kern (catalyst, cold and core) while SSS is Sonne, Siedehitze, Seitenkette (sun, sizzling heat and side chain).

At low temperatures and in the presence of an activating catalyst such as $\ce{FeCl3}$, halogenation of toluene will typically occur as phenyl ring halogenation following the electrophilic aromatic substitution mechanism.

Scheme 1: Catalytic chlorination of toluene with $\ce{Fe^3+}$ as a catalyst. ortho- and para-products expected; (para not shown) mesomeric structures of the intermediate not shown.

Und higher temperatures and irradiation, halogenation takes place according to a radical chain mechanism.

Scheme 2: Radical chlorination of toluene under irradiation. Multiple chlorination possible. Mesomeric structures of the benzyl radical not shown.

The electrophilic aromatic substitution is, of course, not as easily possible on the benzylic carbon; we would somehow need a free electron pair to appear there — not likely.

On the other hand, if a chlorine radical attempted to remove a hydrogen of the phenyl ring, that would result in a radical that cannot be delocalised (the $\ce{C-H}$ bond is orthogonal to the π-system) which is therefore very unstable. Another way of looking at it is to consider the $\ce{C–H}$ bond strengths: weaker bonds are more easily cleaved homolytically, as is the case in free-radical halogenation. Aryl $\ce{C–H}$ bonds are roughly $\pu{100 kJ/mol}$ stronger than benzylic $\ce{C–H}$ bonds (Wikipedia), so radical abstraction preferentially occurs at the benzylic position.


Calculate the amount of energy released when 500 cm³ of methane gas at STP reacts with excess air according to the equation: $$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)} \qquad \Delta H = -890\ \mathrm{kJ\ mol^{−1}}$$

I am not totally sure how to work this out.

What I did, though, was:
$\ce{CH4}$ $M = 16\ \mathrm{g/mol}$
So, $500\ \mathrm{cm^3} = 31.25\ \mathrm{mol}$
$\ce{CH4}$ $H_\mathrm f = -890\ \mathrm{kJ/mol}$
-> For $31.25\ \mathrm{mol}$ we have $27\,812.5\ \mathrm{kJ}$

If I am wrong, can someone explain me why?

How you find the moles from volume? Did you assume $500\ \mathrm{cm^{3}}$ is $500\ \mathrm g$? This is wrong, because the density of $\ce{CH4}$ is not $1\ \mathrm{g/cm^{3}}$. You have to use another way to calculate moles of the gas. Think about ideal gas law or how much volume for STP gas.


Sources for raw 1H-NMR data? Where can one find examples of raw data for an 1H-NMR spectrum?

This may sound as an odd request, so I'll give some context.

I have found lots of images of spectra such as this:

where you can see an image of the peaks. What I am interested is a spreadsheet
of the $(x,y)$-values used to draw the spectre. The data is needed for a student mathematics/chemistry project (high school). The mathematics section will be on calculating the area under a curve using numerical methods. To get a connection to the chemistry part, I'd like the student to use realistic data.

(I am aware that it is the ratio of the areas of the peaks that are interesting, so the actual scaling of the y-axis is unimportant.)

I might suggest that you write a short program to simulate the spectra.  The problem with using real data is the amount of points (upwards of 32k) that make publishing CSV files of raw spectra unreasonable.  Below is a quick attempt to give you what you are looking for.  I apologize for those who do not use Mathematica; however I tried an Excel version and it became unwieldy very quickly.

I have three functions:

Clear[feature];
feature[shift_, n_, coupling_, intensity_] := Module[{},
  Plus @@
   Table[Binomial[n, k]*intensity/(n + 1)*
     Cos[(x)*(shift + coupling (k - n/2))], {k, 0, n}]]
Clear[fid]
makefid[feature_, broadening_] := Module[{},
  Exp[-broadening x] Plus @@ {feature}]
Clear[transform]
transform[fid_] := Module[{resolution = 4096, $y, $x},
  $y = Re[Fourier[Table[fid, {x, 0, resolution, 0.1}]]];
  $y = Take[$y, Round[Length[$y]/2]];
  $x = Range[Length[$y]]/(resolution/(2 Pi));
  Transpose@{$x, $y}
  ]

The function feature creates a peak with a chemical shift as the first argument, the splitting as the second, the coupling constant (use 0.1 for starters) and the intensity (its integral).  makefid creates a simulated FID based on the feature that it is passed and transform makes the spectrum.  It can be used like this:

a = feature[1.2, 2, 0.1, 3];
b = feature[2, 0, 0.1, 3];
c = feature[4.1, 3, 0.1, 2];
fid = makefid[Plus @@ {a, b, c}, 0.01];
ListLinePlot[transform[fid], PlotRange -> {{0, 5}, All}]

The features {a,b,c} are simulated ethyl acetate yielding the following spectrum:

The data you want is in transform[fid] which can be exported as a CSV file:

Export["myfile.csv",transform[fid]]

Which will make it accessible to other software packages and you can perform your math lesson plan.  To make sure this is working properly in Mathematica, we can make a function out of our data and numerically integrate the three peaks:

data = Interpolation[transform[fid]];
NIntegrate[data[x], {x, 0.9, 1.5}]
NIntegrate[data[x], {x, 1.5, 2.2}]
NIntegrate[data[x], {x, 3.7, 4.5}]

This produces areas of 0.32, 0.25, and 0.33, respectively, which is what we expected.


Which is the Electrolyse Reaction? $$\ce{NaOH +H2O->H2 +O2 +NaOH}$$
We know that $\ce{H+}$ and $\ce{OH-}$ go  to the cathode and anode, but how can I write the total reaction?
Is it: $$\ce{2NaOH +2H2O->2H2 +O2 +2NaOH}$$
I've put NaOH both sides because if the mass of NaOH is given, how can I find the mass of H2 for example?

Electrolyse of water is a redox reaction, meaning that there is a transfer of electrons between atoms. Now, every redox reaction can be written as a combination of 2 half-equations. One half-equation shows one species getting oxidised (losing electrons) and the other half-equation shows the other species getting reduced (gaining electrons). When you combine these two half-equations you get the overall reaction, which is what you are asking for. Here is guide that should help you find the overall equation of any redox equation, not just this one example.

Step 1: Determine what species gets oxidised and species gets reduced

Now, what I think the reaction you are trying to describe is the electrolyse of water. Correct me if I am wrong. If that is the case, there is no need for $\ce{NaOH}$ in the equation as it appears on both side, meaning that they cancel each other out (also $\ce{NaOH}$ has nothing to do with electrolyse of water). Secondly, it is $\ce{H2O}$ that gets oxidised and reduced, not $\ce{H+}$ and $\ce{OH-}$.

Step 2: Determine  half-equations

These can be easily found by using your table of standard reduction potentials. Oxidation half-equation: $$\ce{2H2O -> O2 + 4H+ + 4e-}$$
Reduction half-equation: $$\ce{2H2O + 2e- -> H2 +2OH-}$$

Step 3: Balance each equation so that there is the same number of electrons

We can't immediately add the half-equations as the electrons won't cancel each other out. This means that we need to multiply the reduction half-equation by 2:$$\ce{4H2O + 4e- -> 2H2 +4OH-}$$

Step 4: Add the half-equations

When you add each half-equation, you get: $$\ce{6H2O   -> 2H2 + O2 + 4H+ + 4OH-}$$ The $\ce{H+}$ and $\ce{OH-}$ combine to form water which then cancel out with the water the other side. Therefore the overall reaction is: $$\ce{2H2O  -> 2H2 + O2}$$


Ripping apart plastic: is this a chemical or physical change? I was holding a piece of plastic earlier, and I ripped it in half. This caused me to start to think about what was happening at the atomic level. If the plastic is being ripped, clearly bonds are being broken and new bonds are being formed, so is this not a chemical reaction?

I realize that the plastic retains all of its chemical characteristics despite it being ripped, so I am quite confused about how to classify what is occurring.

When you rip, tear or mechanically deform a polymer (for example a piece of plastic) you are putting energy into the material.  The energy from this deformation causes the polymer chains in the vicinity of the deformation to attempt to align.  To some degree this partial alignment makes continued deformation easier.  To continue tearing the polymer apart you just need to overcome the van der Waals forces holding the chains together.

In addition to the polymer chains being pulled apart, some chemical bonds will be broken.  I don't know how many bonds are broken per chain pulled apart, but I suspect it is low.  Low or not, some free radicals (or ions) will be generated when the chemical bonds break.  If the experiment is performed under ambient conditions, the radicals or ions will quickly react with oxygen or ions in the air.  If the experiment is performed in an inert atmosphere, then the radicals will persist (dangling bonds) and can be observed in an esr spectrometer.  It has been demonstrated that cross-linked polymers, which are less prone to flow deformation and therefore more likely to undergo bond breaking, do produce larger esr signals.

So most of what is going on is molecular deformation and pulling apart of the polymer chains - but some bond breaking does occur.

It would be an interesting experiment to cut two pieces of plastic and then cut one of them into finer shreds.  Place the two samples into separate tubes and then record their esr signals.  How much more intense would the signal be in the finely cut sample compared to the bulk sample?  Polycarbonate-A, straight out of the reagent bottle, produces an esr signal.


What is a prosthetic group of protein? What are prosthetic groups? I have found this term while I studying the topic of ETS in respiration in plants.

A prosthetic group is a smaller organic molecule strongly bound to a protein. Typical examples are

haemoglobin (bound to haeme)
biotin (bound to carboxylases)
flavines (bound to flavoproteines)


Determining water of crystallisation in a hydrate from the mass percentages How would I find the amount of water in a hydrate if all I was given was the mass percentage of one element in the compound? For example, a question asks:

  $\ce{Zn(ClO3)2. $x$H2O}$ contains $21.5\%$ zinc. Find the value of $x$.

After thinking for some time I now know that I need to find the mass of $\ce{Zn(ClO3)2}$, and then find the mass and amount of water, and then divide by the amount of $\ce{Zn(ClO3)2}$. But I can't seem to find the mass of $\ce{Zn(ClO3)2}$.

In this case you only have one variable to solve for ($x$) and you have one piece of information (Zn being 21.5% by mass) so you have enough information to solve the question.

The molecular mass of the hydrate can be expressed in terms of $x$:

$$\begin{align}
M(\ce{Zn(ClO3)2.xH2O}) &= M(\ce{Zn}) + 2\cdot M(\ce{Cl}) + (6+x)\cdot M(\ce{O}) + 2x\cdot M(\ce{H}) \\
&= \pu{65.38 g mol-1} + 2(\pu{35.45 g mol-1}) + (6+x)(\pu{16.00 g mol-1}) + \,\\
&\qquad 2x(\pu{1.01 g mol-1}) \\
&= (232.28 + 18.02x)~\pu{g mol-1}
\end{align}$$

Now

$$\begin{align}
\frac{M(\ce{Zn})}{M(\ce{Zn(ClO3)2.xH2O})} = \frac{\pu{65.38 g mol-1}}{(232.28 + 18.02x)~\pu{g mol-1}} &= 0.215 \\
65.38 &= 0.215(232.28 + 18.02x) \\
x &= 3.99 \approx 4
\end{align}$$


Impossible pH For Aqueous Weak Acid Solution An aqueous solution of $\ce{HCl}$ has a $\mathrm{pH}$ of $0.00$ if $[\ce{H+}] = 1.00 \; \mathrm{M}$. I tried to see if an aqueous solution of acetic acid could have a $\mathrm{pH}$ of $0.00$, given that its $K_a$ value is $1.8 \times 10^{-5}$. I tried finding the concentration $[\ce{H+}]$ of such a solution using

$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{C2H3O2^-}]}{[\ce{HC2H3O2}]}$$

and then using molar mass and density conversions to determine whether such an aqueous solution exists.

Does it make sense to say that such a solution does not exist if the amount of substance of acetic acid in 1 liter of solution actually take up more than 1 liter of volume?

Your reasoning is fine. But a major caveat (or additional factor) is that the pH scale itself no longer applies to concentrated solutions (say, more than 1 molar in acid) where the molarity of water falls significantly below 55M (pure water).

You can extend the pH scale down below 0 by empirically measuring how effective an acid solution is at protonating a weak base, but the scale deviates from the straight line predicted by the concentration of acid present.


1-bromo-2-fluorobenzene reaction with magnesium and furan Why does bromofluorobenzene react with magnesium and furan to form a benzyne intermediate?

Does it have something to do with a Grignard being highly basic and a fluorine being electron withdrawing, thereby making elimination a likely pathway?

Then does a Diels-Alder type reaction commence, with furan acting as the diene and the benzyne acting as the dienophile?

How likely or favorable is this reaction in real life?

The overall pathway described is shown below. Treatment of the aryl bromide with magnesium gives a Grignard, which is equivalent to a carbanion. The carbanion can eliminate fluoride to give benzyne. Being highly strained, benzyne is a fantastic dienophile and reacts with furan through a Diels-Alder reaction.

Most of the techniques to generate benzyne use the same strategy: generate a carbanion next to a leaving group. The classic method taught in introductory organic chemistry is by deprotonation of an aryl halide with a very strong base (sodium amide), which eliminates the halide. The strategy shown here generates the carbanion using the milder method of forming a Grignard.

The elimination is possible because carbanions are very good at kicking out leaving groups. It's commonly taught that fluoride is a poor leaving group, but the end of that sentence is that it's a poor leaving group for SN2 reaction or compared to chloride, bromide, and iodide. Later, it's taught that enolates (a reasonably stable class of carbanions) are reactive enough to eliminate alkoxides/hydroxide. So it makes sense that an unstabilized carbanion as here should be able to kick out a relatively poor (but not that terrible) leaving group.

The Stoltz group has a nice summary of benzyne's history, generation, and reactivity.


Detecting a gas molecule moving through a small pinhole Suppose I have the four DNA base chemicals - thymine, adenine, cytosine, and guanine. Imagine I turn each of them into a gas, make a "cartridge" of the gas for each chemical, and then have a "pinhole" that is opened or closed by a semiconductor. When the pinhole is open, the gas will, of course, expand through the pinhole and exit the cartridge. The system must emit only one molecule at a time.

If I send a command to the semiconductor to open the pinhole, what kind of sensor can I use to detect a molecule passing through? If necessary, you can modify the base chemicals as long as it would not interfere with their chemical properties in any way other than to improve detection of those molecules, if that makes sense.

The cheaper and more compact the system, the better, of course. I was thinking about the way an inkjet printer works, with a piezoelectric crystal vibrating the ink into droplets, but even the smallest droplets (a picoliter or so) are still bigger than a single molecule, so that doesn't work. Other possible methods I've thought of include a MALDI like system, a system where a weak laser is set up with a light detector; when a molecule passes through, the beam is interrupted, using a scanning tunneling microscope (STM; no idea how that would work), or a system using fluorophores.

The best way to do this is to detect the molecules you want in the presence of others that carry it along. You should not expect to detect every molecule of the type you want but just one of them at a time.

The most sensitive detection technique is mass spectrometry, more sensitive than fluorescence because if a molecule fluoresces in a direction away from your detector its photon is lost. All ions can in principle be collected in a mass spectrometer.

To get the molecules into the spectrometer mix them with a solvent, (if necessary) and heat this with a pulsed IR laser. This has to be sufficient to vaporise the mixture/substrate  and it has to be positioned close to the input of the spectrometer since this is done outside the spectrometer. The vapour is at atmospheric pressure and mixed with air but because of the pressure differential efficiently enters the spectrometer (it is easy to see this). Here by differential pumping and using skimmers before the entrance to the mass spectrometer part some of the target molecules will be present in the beam produced. The mass spectrometer should probably be of the time of flight/reflectron type to increase throughput and mass resolution. By pulsing the laser it is then possible to gate the detector to observe the molecules at just the right time and so reduce noise. Of course once you have detected the molecule it is destroyed on the detector.

If you do not want to destroy the molecules then fluorescence detection is the only possibility provided the molecule does fluoresce as not all do with a reasonable yield. A second laser is thus needed after the mass selection and then a photomultiplier is used to count single photons of fluorescence as they arrive (v. common technique now for 50 years or more). I'm not sure what you could do after that but again there is time as photons travel effectively instantaneously compared to the sped the molecule will be travelling at. As I said in a comment not cheap, not trivial.


How do orbitals fill, what is the difference between them, and why is this stable? I know about the Aufbau principle and the order in which electrons fill orbitals, but how is it that atoms are more stable when they have a full valence shell? It seems that any shell can be comprised of many different energy levels(3s,3p,3d,etc) and that it would be kind of arbitrary to say that "since all the orbitals that start with three are full, the atom is stable". Are all the "3" energy levels linked in a special way that differentiates them from 2 or 4? Also, why would an electron even be inclined to join a specific orbital, and do electrons change their spin in order to fit with the pre-existing particle in the orbital.

What is stable?

In this context, stability is that it neither gains electron to form negative anions nor loses electron to form positive cations. I will only discuss noble gases because they are so exceptionally stable.

It does not lose electrons

This is not unique to noble gases. Elements in group VII such as fluorine and chlorine also do not tend to lose electrons.

This is also not entirely true. Compounds like $\ce{XeF4}$ and $\ce{Cl2O7}$ do exist in which $\ce{Xe}$ and $\ce{Cl}$ bear partial positive charges respectively, although $\ce{Xe}$ is a noble gas.

$\ce{Xe}$, being a noble gas, indeed has a high ionization energy (energy taken to remove an electron) but it is not infinite. This means, as long as we have enough energy, we can make them form compounds.

The ionization energy increases from Group III to Group 0 because there are more protons in the nucleus which exerts a higher pull to the orbitals, which lowers their energy. For example, the ionization energy of helium is twice that of hydrogen, and helium has exactly twice the number of protons than hydrogen.

It does not gain electron

This is because the next orbital has a positive energy. Orbitals with negative energy means that it takes work to free the electrons from the orbitals. Orbitals with positive energy means that no electrons can reside in the orbitals: they are already free.

For example, the electronic arrangement of argon is $1s^2 2s^2 2p^6 3s^2 3p^6$. The next orbital, if it existed, would be $4s$, which has a positive energy. Therefore, its electron affinity (energy released by gaining electrons) is 0. This is evidenced by the ionization energy of the next element, potassium, which is small, suggesting that the extra proton pulls the $4s$ orbital just enough so that it has a slightly negative energy.

Conclusion

Noble gases are stable because they have a full $p$ orbital, so the next electron would need to go to the next orbital which has a positive energy, and the extra protons pulling on the orbital lower the energy of the electrons in the orbital, making it not tend to lose electrons.


Why do some reactions require specific pressures to happen? For example, carbon monoxide reacts with hydrogen to synthesize methanol in the presence of some catalysts, but the pressure needs to be $\pu{50 atm}$ and the temperature needs to be $\pu{523 K}$.

Why the pressure? What does the pressure adds to the reaction?

ChemGuide has a good introductory article here.

The effects of increasing pressure and temperature are, to an extent, equivalent. Increased pressure leads to increased collisions and increased collision strength between molecules, allowing the (usually high) activation energy barrier to be overcome at a noticeable rate; at standard temperatures and pressures, for example, collisions between $\ce{CO}$ and $\ce{H2}$ are far too infrequent and involve far too little energy for the formation of $\ce{CH3OH}$.

Remark. How do we achieve high pressures and temperatures? We could pressurize a steel cylinder and use it as a reaction vessel, for one, but a novel approach involves sonochemistry. Ultrasound waves in solution lead to cavitation, the formation and rapid collapse of bubbles, which are highly pressurized and at high temperatures. The use of ultrasound can even open up new reaction pathways not usually available; for example, radical species are often transient, but the forcing conditions in these bubbles allow them to survive long enough to react.


Proton NMR signals and rings Does pyridine or furan have the more shielded proton NMR signals?

I know that both are aromatic, so both have rather deshielded signals due to ring current combining constructively with the applied magnetic field.

However, which would have the more shielded signal? I know that shielded protons have more electron density around them. I guess then the question boils down to whether nitrogen or oxygen is more electron-releasing, as both nitrogen and oxygen have at a lone pair participating in resonance.

Here is the proton nmr spectrum for pyridine.

Here is the proton nmr spectrum for furan

For comparison, benzene has a single proton nmr signal at 7.27 ppm.  In both pyridine and furan there are different types of protons, so multiple signals are seen in both cases.

Here are the resonance structures for pyridine

and furan.

Pyridine removes electron density from the alpha and gamma positions by resonance while furan adds electron density to all four ring positions by resonance.

In addition to resonance effects there are also inductive effects and ring currents that will modify the proton chemical shifts.  Inductively, the alpha positions in both molecules will be deshielded.  Ring currents will also deshield the protons as you've noted.

All of the signals are in the general vicinity of benzene.  The alpha-protons in both molecules are the most deshielded due to inductive effects, with furan's alpha signal upfield from pyridine's due to the electron donating resonance effect of the furan oxygen.  The beta proton in furan is the most shielded since it is not inductively deshielded and, again, the oxygen in furan is pushing electron density into the ring via resonance which will cause shielding.


Energy capture photochemistry with ionizing radiation? Chlorophyll mediates a photochemical reaction: energy from a visible light photon is used to reduce CO2 to CHO, and the energy from that photon can later be recovered when CHO is oxidized back to CO2 in an organism.

Is there any analogous photochemical reaction where energy from shorter wavelength ionizing radiation (like Xrays or gamma rays) is captured in a molecule such that it can later be recovered chemically to do work?  I do not think any biochemistry happens along these lines but maybe synthetic corroles or porphyrins have been made that can accomplish this.

The energy scales of x-ray and gamma radiation are too large to be chemically helpful in the way that the electromagnetic around visible wavelengths is.

X-rays, if absorbed, typically excite core electrons of atoms. As we know, core electrons are (to a first approximation) not affected by the details of chemical bonding in the valence, so any attempt to use a chemical bonding motif to capture x-rays (e.g. conjugation of porphyrins) is of the wrong energy scale. You could use the Auger Effect to eject a valence electron with x-ray scale energy. This does constitute radiationless conversion of the light into electronic energy, but it doesn't meet your qualifications for storing and extracting work.

Gamma rays are typically produced by the decay of one nuclear state to another, and are vastly different in energy scale than electronic states, so there's no way to do chemistry with gamma radiation, at least directly. The Mössbauer effect does allow us to use gamma rays to understand local structure in a crystal, but even then most of the energy is absorbed into exciting the nuclear state, which we cannot control chemically.


Experiment to demonstrate Dalton's Law of Multiple Proportions Is there an experiment that I can do in the lab with middle or high school students to demonstrate Dalton's Law of Multiple proportions ?
For example, the reaction of $\ce{C}$ and $\ce{O2}$ to give $\ce{CO}$ and $\ce{CO2}$.
Is there an experiment where I can demonstrate that $\ce{CO}$ and $\ce{CO2}$ have these proportions ?

Because of toxicity, I'd avoid middle or high school student experimentation with $\ce{CO}$.
You might consider $\ce{H2O}$ and $\ce{H2O2}$, if you stick to 6% (20 vol) or weaker. There is the complication of working with a dilute solution, though, in getting precise measurements on proportions.
Another possibility is demonstrating the various oxides of iron. However, there is another issue: you can produce fairly pure pure red iron(III) oxide, $\ce{Fe2O3}$, by the electrolysis of sodium bicarbonate, $\ce{NaHCO3}$ with an iron anode, but rather than pure black iron(II) oxide, $\ce{FeO}$, thermal decomposition of iron(II) oxalate in an inert atmosphere, makes a mix, wustite (a nonstoichiometric compound) that is primarily $\ce{FeO}$.
Chromium compounds, e.g. oxides, of varying proportions might make an interesting demonstration because of obvious color changes, but the toxicity of chromium compounds precludes use by students, and even safe disposal of the demo material may be problematic.
BTW, an interesting effect is observed in podzol soils, as found in pine forests. If there is iron in the soil, the reducing nature of freshly-exposed soil makes it look greenish; but after exposure to air, the iron is oxidized to a reddish color (I've observed the change in less than 15 minutes.).


Can a solid and liquid be miscible? To be specific, what I mean is: is the following situation possible? At given temperature T, there are two substances, X and Y, such that pure X is a solid, pure Y is a liquid, and X and Y mix to form a single phase in all proportions.

The closest I could find was this phase diagram for silver and gold (two very similar metals, I suppose):
http://www.crct.polymtl.ca/fact/phase_diagram.php?file=Ag-Au.jpg&dir=SGTE

But it looks like there is still a narrow region in which there are two phases.

It's not necessarily a strict definition of a single intermingling of phases, however the closest thing I can think of corresponding to this in reality would be interstitial metal hydrides.

These materials are being investigated as hydrogen storage materials, so essentially the hydrogen molecules interact with the metal to form hydrides, "$\ce{H-}$", which recombine under different conditions to reform $\ce{H2}$ as required. A review of this technique is given here.

It would be interesting, to answer your question more directly, if liquid hydrogen could be used instead of hydrogen gas (this requires temperatures of <33K). I imagine this process would give exactly the same result, albeit would happen more slowly than the gas phase.


How to get the reducible representations of the B-B and the B-Cl bond of B2Cl4 I just can't seem to get the correct answer to this symmetry question in my book. "Introduction to molecular symmetry J.S.Ogden."

I am trying to find the reducible representations of both the $\ce{B-Cl}$ bond and the $\ce{B-B}$ bonds in $\ce{B_2Cl_4}$ in the point group of $D_\mathrm{2d}$.
The correct answer for both is $\Gamma_{\ce{B-Cl}}=A_1+B_1+E$ and $\Gamma_{\ce{B-B}}=A_1$.
My reducible representations are:

$\Gamma_{\ce{B-Cl}}=4,0,0,0,2$ for the $\ce{B-Cl}$ bond in the order $E, 2S_4, C_2, 2C', 2\sigma_\mathrm{d}$.
$\Gamma_{\ce{B-B}}= 1,0,1,0,1$.

Clearly either of these are not correct but honestly I can't see why! Your help is much appreciated! (I worked them out based on the number of bonds that remain unshifted during the respective symmetry operation. I have also made a little model for myself in Gaussian but it didn't really help.

There seem to be two problems: the first is that your reducible representation for the $\ce{B-B}$ bond is wrong but your reducible representation for the $\ce{B-Cl}$ bonds seems to be correct and the second is that your book is apparently completely wrong. If I didn't make a mistake and my second assumption is correct then maybe you should think about switching to another book as it seems to raise more questions than it answers. But I'm not working regularly with group theory, so my analysis of the problem might be faulty, but maybe someone else can support or refute my answer.

Ok, here is my analysis:
First, make yourself familiar with the symmetry operations of the $\mathrm{D}_{2\mathrm{d}}$ group. Have a look at this video for example. Now, I looked at the effect each symmetry operation has on the bonds I'm looking at. For that I will use the following depiction of $\ce{B2Cl4}$:

For the $\ce{B-B}$ bond I used an arrow to indicate the direction of the bond for better emphasis as I will need that soon. In the following I will assume that you are familiar with the way characters are related to your reducible representation and how you get those characters from analyzing the changes of your basis vectors (in this case the bonds) under the symmetry operations of the group since you seem to have already used it to determine the reducible representations in your question.
Furthermore, I use the following notation: The reducible representation of the $\ce{B-Cl}$ bonds will be denoted as $\Gamma_{\ce{B-Cl}}$ and the character of the symmetry operation $C_{2}$ for this reducible representation will be denoted as $\Gamma_{\ce{B-Cl}}(C_{2})$.

Ok, let's see what the symmetry operations of the $\mathrm{D}_{2\mathrm{d}}$ group do to this structure, so that we can get the reducible representations:
The $E$ operation does nothing to the structure, i.e. all four $\ce{B-Cl}$ bonds and the one $\ce{B-B}$ bond stay where they are. Thus you get a character of $4$ for $\Gamma_{\ce{B-Cl}}(E)$ and a character of $1$ for $\Gamma_{\ce{B-B}}(E)$.

Now, we come to the improper axis $S_{4}$. It rotates along the long the $\ce{B-B}$ bond and mirrors perpendicular to it:

As you can see the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(S_{4}) = 0$. Since the mirror plane is perpedicular to the $\ce{B-B}$ bond the two $\ce{B}$ atoms' positions are exchanged and thus the $\ce{B-B}$ bond - while keeping its location - reverses its direction, and thus you get $\Gamma_{\ce{B-B}}(S_{4}) = -1$.

Next, we look at the principle $C_{2}$ rotation along the $\ce{B-B}$ bond. It converts your molecule to the following:

As you can see the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(C_{2}) = 0$. But as the rotation is conducted along the $\ce{B-B}$ bond axis the two $\ce{B}$ atoms' positions are unchanged and thus the $\ce{B-B}$ bond keeps its location and direction, and thus you get $\Gamma_{\ce{B-B}}(C_{2}) = 1$.

Now, for the two $C_{2}^{\prime}$ axes which are perpendicular to $C_{2}$. One disects the dihedral angle between the dark red $\ce{Cl}$ and the cyan $\ce{Cl}$ - its effect on the structure is shown in the left picture - and the other disects the dihedral angle between the red $\ce{Cl}$ and the blue $\ce{Cl}$ - its effect on the structure is shown in the right picture:

$\qquad \qquad \qquad$

Again the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(C_{2}^{\prime}) = 0$. But now the rotation is conducted along an axis perpedicular to the $\ce{B-B}$ bond and so the two $\ce{B}$ atoms' positions are exchanged and thus the $\ce{B-B}$ bond - while keeping its location - reverses its direction, and thus you get $\Gamma_{\ce{B-B}}(C_{2}^{\prime}) = -1$.

Finally there are the two $\sigma_{\mathrm{d}}$ planes which lie along the $C_{2}$. The one plane contains the dark red and the red $\ce{Cl}$ while disecting the angle betweeen the  cyan and the blue $\ce{Cl}$ - its effect on the structure is shown in the left picture - and the other contains the cyan and the blue $\ce{Cl}$ while disecting the angle betweeen the dark red and the red $\ce{Cl}$ - its effect on the structure is shown in the right picture:

$\qquad \qquad \qquad$

In this case either the red and dark red $\ce{Cl}$ atoms or the blue and cyan ones remain unchanged while the other $\ce{Cl}$ atoms change their places. So, two $\ce{B-Cl}$ bonds keep their location and direction which gives $\Gamma_{\ce{B-Cl}}(\sigma_{\mathrm{d}}) = 2$.
The $\ce{B-B}$ bond axis lies within the mirror plane and so the two $\ce{B}$ atoms' positions are unchanged and thus the $\ce{B-B}$ bond keeps its location and direction, and thus you get $\Gamma_{\ce{B-B}}(\sigma_{\mathrm{d}}) = 1$.

So, in conclusion you end up with the reducible representations (I kept the order of the symmetry elements you chose, i.e. $E$, $2 \, S_4$, $C_2$, $2 \, C_{2}^{'}$, $2 \, \sigma_{d}$):

\begin{align}
  \Gamma_{\ce{B-Cl}} &= 4, 0, 0, 0, 2 \\
  \Gamma_{\ce{B-B}}  &= 1, -1, 1, -1, 1 \\
\end{align}

and they reduce to

\begin{align}
  \Gamma_{\ce{B-Cl}} &= A_1 + B_2 + E \\
  \Gamma_{\ce{B-B}}  &= A_2 \\
\end{align}

As you can see both results differ from the ones your book provides.


EZ-Water - Fraud or breakthrough? Recently the quack medicine folks online have been promoting the research of a certain Dr. Gerald Pollack who claims to have discovered a "forth phase of water", and who has recently published a book on the said topic.

Although it's clear that his research is being deliberately misinterpreted, it's not clear to me whether or not his own claims are valid in the first place. Although some papers on the topic were published in peer reviewed journals, it doesn't seem like anyone else in the scientific community has acknowledged or replicated his results.

So, is "EZ-water" a real breakthrough or just the brainchild of a deluded scientist?

EZ-water is not a breakthrough. It is not new, nor is it valid. This appears to be one of the many claims about the healthful benefits of drinking "ionized water".
The original "article" linked, contains a number of misleading statements.

Water molecules make up 99% of your body. This is true, but misleading. According to various sources, the human body is between 57% and 75% water by mass. Since water has a much lower molar mass than most biomolecules, 99% of the molecules in your body may well be water.
The journal listed in which the research is published is Water, which is not a high impact journal, despite its name. Water had an impact factor of 0.973 in 2012. That means that articles published in 2011 in Water were cited less than 1 time each by anyone in 2012. Compare with Cell, which had in impact factor of 31.957 in the same period - each article in Cell is cited an average of nearly 32 times.
Water has a "fourth phase" and no single boiling point or melting point. This is true! Any student of chemistry should be able to tell you this. Take a look at the phase diagram of water below. Water apparently has at least 18 phases (15 solid phases, plus liquid, vapor, and supercritical fluid). You can also see that the solid-liquid and liquid-vapor phase boundaries are not single points. They are curves varying in temperature and pressure. Those curves represent the conditions in which the two phases have the same vapor pressure. This is not a new discovery or a startling secret. This behavior has been known for many decades (if not centuries).

Later on the "article" tells us that EZ-water is not $\ce{H2O}$ (note that they cannot be bothered to format their chemical formulas) but $\ce{H3O2}$, which is alkaline with a negative charge. Hmmm... If it has a different formula and different properties, it must be a different compound. Note that this "formula" is essentially $\ce{H3O2 = H2O + OH}$, and hints at what happens inside these water ionizers.
EZ-water has a negative charge. It is possible for objects to build up charge, but it is not possible for them to remain permanently charged. Overall, the universe is balanced in charge. Consider lightning as a consequence of attempting to maintain separation of charge. This "negative charge" comes from a misunderstanding of pH. The water definitely has a higher pH, which means more $\ce{OH-}$ ions than $\ce{H+}$ ions, but something (probably $\ce{Na+}$) needs to replace the missing $\ce{H+}$.
I can go on, but you get the point.

A water ionizer electrolyzes water, which requires electricity (not light as suggested by Mercola). I had the good fortune to be involved in the testing and dismantling of one of these ionizers recently. The device claimed to take ordinary tap water and convert it to a "super-cleaning solution that contained only negatively charged oxygen particles" or something like that. From experiments:

The ionizer increases the pH of the water.
The ionizer does not work with deionized or distilled water!
The ionizer does not produce measurable oxygen gas (a smoldering splint will not relight in the vicinity)
The device contains an electrolysis cell and an ion exchange resin

Here is what happens:

Water is electrolyzed.
Doing so requires ions in the water already to conduct the electricity, which is why the device must use tap water (and not purified water). At the cathode, water is reduced to hydrogen and hydroxide. At the anode, it is oxidized to oxygen and protons.
\begin{align}
\ce{2H2O +2e- &-> H2 + 2OH-}\\
\ce{2H2O &-> O2 +4H+ + 4e-}
\end{align}

Protons are exchanged for sodium ions
To prevent the recombination of hydroxide and protons to form water and neutralize the solution, there is an ion exchange membrane between the anode and the rest of the solution.
$$\ce{H+ + OH- -> H2O}$$
The ion exchange resin collects the protons and releases sodium ions into solution to maintain charge neutrality, which must happen or you would have a lightning gun and not a mystic device to produce super water.

Ultimately, drinking this stuff would probably have no effect on your health. But we are no doctors, so don’t listen to us either. It might help or it might hurt, depending on your condition, it also might not matter.
First, your stomach secretes acid, so any extra base would be neutralized immediately. Let's say an ionizer gets the water's pH to 8 (slightly basic, and more basic than your blood). This means that the concentration of protons is $\ce{[H+]}=10^{-\text{M}pH}=10^{-8} \text{ M}$ (remember that pH is a logarithmic scale. The concentration of hydroxide is $10^{-14} / 10^{-8} = 10^{-6} \text{ M}$. The pH in your stomach ranges from 1.5 - 3.5, which represent proton concentrations between $10^{-3.5}= 3.16\times 10^{-4}$ and $10^{-1.5}=3.16\times 10^{-2}$. If we use the higher pH stomach acid to neutralize an equal volume of EZ-water, the final pH would be:
\begin{align}
[\ce{H+}]_f &= \frac{3.16\times 10^{-4} - 1\times 10^{-6}}{2}=\frac{3.15\times 10^{-4}}{2}=1.07_5 \times 10^{-4}\\
\text{pH} &=-\log[\ce{H+}] =-\log(1.07_5 \times 10^{-4})=3.97
\end{align}
Since your stomach acid would likely have a greater volume than the ionized water, the change in pH would be even smaller. This calculation assumes your stomach acid is not a buffer, which it is.
Second, the acid-base homeostasis of your blood prevents the pH of your body from changing. Yes, your blood pH is 7.4 or so (which is slightly basic), but that does not mean your blood is negatively charged. Outside of range of 7.38 - 7.42 all of your blood proteins (like hemoglobin) start to denature, which will kill you.
Well, to be more precise, you will suffocate before that happens as hemoglobin/myoglobin oxygen binding capacity is pH sensitive. And if anything, we can tolerate acidic blood more robustly than basic blood pH (down to 6.8 or 7, but only up to about 7.6) before death. You might want to look up acidosis.


Structure of pyrophosphorous acid What is the structure of pyrophosphorous acid?

I asked this question here because the answer which I found on web and other resources are inconsistent which each other.
For example, following data is given regarding the structure of pyrophosphorus acid in my textbook, which is not consistent with the number of oxygen atoms.

On checking here,and here there are no double bonds of phosphorus with oxygen, however here there are. And here both the structures are given on the same website.

Oxyphosphorus compounds, all of which contain phosphorus-oxygen linkages, are the most dominated subset in Phosphorus Chemistry. You may find good review of oxyphosphorus compounds in Ref.1. In particular, most of these commonly known as phosphates are described in Chapter 3 of Ref. 1 (Pages 169-305) which states that:

  Oxyphosphorus compounds may be defined as compounds which contain phosphorus-oxygen linkages. They may contain up to six oxygen atoms linked to a central phosphorus atom. Pyramidal derivatives are represented by phosphite esters (la), tetrahedral compounds by orthophosphate salts (lb) and esters (lc), trigonal bipyramidal compounds by pentaoxyphosphoranes (Id), and octahedral compounds by a few hexaoxyphosphorides (hexaphosphates) of type (le). In addition, there are the comparatively rare 2-connected angular phosphenites (If) and 3-connected planar phosphenates (lg).

The term normal phosphates is used to described the compounds, in which only $\ce{P—O}$ linkages are present (e.g., structures 1a-g). Compounds such as 1b containing discrete $\ce{PO4^{3-}}$ anions are known as orthophosphates. If the neutralizing cation is $\ce{H+}$, then it is called phosphoric acid or orthophosphoric acid.

On the other hand, if some of the oxygen linkages are replaced by other atoms or groups (e.g., $\ce{H, NR2, CR3}$, etc.), the compounds can be termed substituted phosphates. For example, if one $\ce{P—O}$ linkage is replaced by a $\ce{P-H}$ linkage, the compound series are called phosphites (2a) while when the two $\ce{P—O}$ linkages are replaced by two $\ce{P-H}$ linkages, that series is known as hypophosphites (2b).  Similar to phosphates, if the neutralizing cation is $\ce{H+}$ in 2a and 2b, then they are traditionally called phosphorus acid (Wikipedia1) or hypophosphorus acid (Wikipedia2), respectively. The IUPAC names of these acids are phosphonic acid and phosphinic acid, respectively. Meanwhile, substituted phosphates containing a single $\ce{P=O}$ linkage (e.g., 2c) are usually referred to as phosphoryl compounds.

When $\ce{PO4^{3-}}$ anions are linked together with sharing oxygen atom in common, they are called condensed phosphates (e.g., pyrophosphate, 3a). If the neutralizing cation is, again, $\ce{H+}$, then it is called pyrophosphoric acid or diphosphoric acid (IUPAC name). On the other hand, one can expect when the two anions linked together with sharing oxygen atom in common are phosphites, they would be called pyrophosphites, 4a (Wikipedia3).  However, recent research showed that there are some controversies on the structures, not only on pyrophosphites, but also on phosphite itself.

Phosphorous acid, $\ce{H3PO3}$ (systematic IUPAC name is phosphonic acid), exists in two tautomeric forms: An one with tricoordinate about phosphorous $(\ce{P(OH)3})$; and another with tetracoordinate about phosphorous $(\ce{HP(O)(OH)2})$ (See Scheme below), which is the favored tautomer between two (Ref.2).

The $\ce{P(OH)3}$ form of two tautomer is rarely isolated or directly obtained, yet its presence in the equilibrium has been repeatedly postulated from kinetics studies (Ref.3,4) and thermodynamic studies (Ref.5,6).

According to most literature, the $\ce{P(OH)3}$ form of tautomers is possible only in triesters such as 1a $(\ce{P(OR)3})$. Even in such cases, there is a strong tendency to rearrange to the tetracoordinate species, $\ce{RP(O)(OR)2}$ (Ref.5). Hence, it is safe to say that pyrophosphites has high tendency to have a structure resembling of 4a (See first scheme).

References:

D. E. C. Corbridge, In Studies in Inorganic Chemistry, Volume 20: Phosphorus – An Outline of its Chemistry, Biochemistry and Uses; Elsevier B.V.: New York, NY, 1995, Pages 1-1208 (https://www.sciencedirect.com/bookseries/studies-in-inorganic-chemistry/vol/20).
F.A. Cotton, G. Wilkinson, C.A. Murillo, M. Bochmann, Advanced Inorganic Chemistry, 6th Edn.; Wiley-Interscience: New York, NY, 1999, “Chapter 10: The Group 15 Elements: $\ce{P, As, Sb, Bi}$,” pp. 380-443.
G.A. Haight Jr., M. Rose, J. Preer, “Reactions of chromium(VI) with phosphorus(III) and phosphorus(I). I. Dihydrogen phosphite, phosphorous acid, and hypophosphorous acid,” J. Am. Chem. Soc. 1968, 90(18), 4809-4814 (DOI: 10.1021/ja01020a011).
R.O. Griffith, A. Mckeown, “Kinetics of the reaction of iodine with phosphorous acid and with phosphites,” Trans. Faraday Soc. 1940, 36, 766-779 (DOI: 10.1039/TF9403600766).
J. P. Guthrie, “Tautomerization equilibria for phosphorous acid and its ethyl esters, free energies of formation of phosphorous and phosphonic acids and their ethyl esters, and $\mathrm{p}K_a$ values for ionization of the $\ce{P—H}$ bond in phosphonic acid and phosphonic esters,” Canadian Journal of Chemistry 1979, 57(2), 236-239 (https://doi.org/10.1139/v79-039).
J. P. Guthrie, “Carbonyl Addition Reactions: Factors Affecting the Hydrate–Hemiacetal and Hemiacetal–Acetal Equilibrium Constants,” Canadian Journal of Chemistry 1975, 53(68), 898-906 (https://doi.org/10.1139/v75-125).


Relation between Temperature and viscosity of Linseed Oil? I have been trying to get a relationship formula to determine the viscosity of Linseed oil. I have Absolute viscosity of the oil at 33.1 Cp (Source: http://www.engineeringtoolbox.com/absolute-viscosity-liquids-d_1259.html).

The problem is the system where I am using the fluid has a varying temperature, and I want to be able to know what the varying viscosity is. I am aware that the viscosity reduces with increase in temperature, but wasn't able to find a specific formula to relate the two.

Is there a formula to get the relationship between temperature and viscosity?

Thank you

There are many formulas!  This usually means that some of them are OK in general for estimation purposes, but none of them will be especially accurate.  Not easy to choose from the many formulas.

One of the most simple and popular seems to be the Wright equation,
$$\log_{10}\left(\log_{10}\left(\nu + \lambda \right)\right)=A-B\log_{10}(T),$$
where $\nu$ is the kinematic viscosity, $\lambda$ is a parameter, usually taken to be $0.7$, and $A$ and $B$ are compound-specific parameters.  With knowledge of linseed oil viscosity at least one other temperature than the one you provided in your question, you could fit $A$ and $B$ parameters and apply the equation.

However, the best answer to your question is to obtain some empirical data.  A 1977 paper[1] reports on linseed oil viscosity as a function of temperature, for example.  They found that a different equation (not the Wright) gave a reasonable fit to the data.  Specifically, they found
$$\ln(\eta) = -5.73 + \frac{2820}{T}$$
for viscosity $\eta$ in centipoises and temperature $T$ in K.

Another thing to keep in mind is that linseed oil is prone to oxidation.  Oxidized linseed oil may have different properties than undergraded, "pure" linseed oil.

[1]: T. Gallagher; D. E. Kline; J. Am. Oil Chem. Soc. 1977, 54 (2), 68-70. DOI: 10.1007/BF02912392