Simultaneous Equilibrium Reactions, One Goes to Completion Say I have the following reactions:

$\ce{CH4 + H2O <=> CO + 3H2}$

$\ce{C2H6 + 2H2O -> 2CO + 5H2}$

Which undergo these processes in the same container. The first reaction goes to equilibrium and the second to completion. Clearly the reactions are entangled as the products are the same. The equilibrium concentration of methane will intuitively depend on the initial amount of ethane introduced. I'm not sure how to go about calculating the equilibrium concentration of methane however. My first thought was to find the amount of carbon monoxide and hydrogen gas generated by the second reaction and use this as initial product in the first reaction and to calculate equilibrium concentration from there but I'm having trouble doing this.

Initially

Suppose that initially, before any reaction, these are the species and amounts present:

$$\ce{CH4}: m \textrm{ moles} $$
$$\ce{H2O}: w \textrm{ moles} $$
$$\ce{C2H6}: e \textrm{ moles} $$
$$\ce{H2}: h \textrm{ moles} $$
$$\ce{CO}: c \textrm{ moles} $$

After the second reaction completes

Let's assume that the second reaction is limited by the amount of ethane present, not the amount of water.  (This must be true, otherwise the first reaction could not be at "equilibrium" because one of the equilibrating species would  head to a concentration of zero.)  If the reaction is limited by ethane, then when it completes, $2e$ moles of carbon monoxide and $5e$ moles of hydrogen are formed.

$$\ce{CH4}: m \textrm{ moles} $$
$$\ce{H2O}: w - 2 e \textrm{ moles} $$
$$\ce{C2H6}: 0 \textrm{ moles} $$
$$\ce{H2}: h + 5e\textrm{ moles} $$
$$\ce{CO}: c + 2e\textrm{ moles} $$

After the first reaction equilibrates

The first reaction doesn't go to completion.  Instead, let's assume the extent of the reaction from our current state to the equilibrated state is $\xi$ moles.  That means that the species inventory becomes:

$$\ce{CH4}: m - \xi \textrm{ moles} $$
$$\ce{H2O}: w - 2 e - \xi \textrm{ moles} $$
$$\ce{C2H6}: 0 \textrm{ moles} $$
$$\ce{H2}: h + 5e + 3 \xi \textrm{ moles} $$
$$\ce{CO}: c + 2e + \xi \textrm{ moles} $$

These values can now be plugged into an equilibrium equation:

$$ K = \frac{[\ce{H2}]^3 [\ce{CO}]}{[\ce{CH4}][\ce{H2O}]}$$

$$ K = \frac{(h + 5e + 3 \xi)^3(c + 2e + \xi)}{(m-\xi)(w - 2 e - \xi)} $$

Now "all" you have to do is solve that equation for $\xi$.


Calculating partial pressure equilibrium constant Kₚ given initial pressure and equilibrium pressure A reaction
$$\ce{A(g) <=> B(g) + C(g)}$$
happens in constant volume and constant temperature. The reaction starts only with gas $\ce{A}$ (no $\ce{B}$ or $\ce{C}$) with given pressure $P_1 = \pu{6 atm}$, in equilibrium the pressure of all three gases is $P_2 = \pu{10 atm}$. Calculate $K_p.$

It seems to me like a very simple question, however it seems that I don't understand a basic concept regarding gas equilibrium. As for my understanding it is suppose to be
$$K_p = \frac{P_2 \cdot P_2}{P_2} = P_2 = \pu{10 atm}$$
But the given solution is $\pu{8 atm}$ and I really don't get what miss.

The problem is that you are using wrong pressures.
By definition for the reaction at equilibrium partial pressures can be expressed via initial partial pressure $P_1$ and conversion factor $α$

$$
\begin{array}{ccc}
\ce{&A(g) &<=> &B(g) &+ &C(g)}\\
&(1 - α)P_1& & αP_1& & αP_1
\end{array}
$$

equilibrium constant $K_p$ is to be found as

$$K_p = \frac{P(\ce{B})\cdot P(\ce{C})}{P(\ce{A})} = \frac{α^2P_1^2}{(1 - α)P_1} = \frac{α^2P_1}{1 - α}$$

Unknown $α$ can be found by equating total pressure at equilibrium $P_2$ to the sum of partial pressures of all gaseous components at equilibrium:

$$
\begin{align}
P_2 &= P(A) + P(B) + P(C) \\
  &= (1 - α)P_1 + αP_1 + αP_1 \\
  &= (1 + α)P_1
\end{align}
\quad\implies\quad
α = \frac{P_2}{P_1} - 1 = \frac{\pu{10 atm}}{\pu{6 atm}} - 1 = \frac{2}{3}
$$

Finally, all the values can be plugged into the expression for $K_p$:

$$K_p = \frac{\left(\frac{2}{3}\right)^2\cdot\pu{6 atm}}{\left(1 - \frac{2}{3}\right)} = \pu{8 atm}$$


Reaction of CaO with C $\ce C$ is a good reducing agent and hence it can remove oxygen from $\ce{CaO}$. But instead of forming $\ce{Ca}$ and $\ce{CO}$, the products are $\ce{CaC2}$ and $\ce{CO}$.
Why does this happen?

Why would it not? C is a nonmetal, and it is common for nonmetals to form compounds with metals. Oxygen gives oxides, nitrogen gives nitrides (if you succeed in making him react, which may be not quite trivial, but that's another story), chlorine gives chlorides, etc. Now, C is less eager to take extra electrons than N or O, so it would form ionic carbides only with active metals. Well, Ca is pretty active.


Nucleophilicity with respect to the solvent I am currently studying nucleophilicity of molecules and am getting a set of conflicting information, so I wanted to clarify nucleophilicity trends with respect to solvent.

Master Organic Chemistry discusses how nucleophilicity increases down the periodic table with protic solvents and decreases down the periodic table with aprotic solvents; the main argument is that a solvating shell will form with the protic solvent, making the smaller atoms less likely to form a bond. When this effect is removed, the trend is reversed.

However, during my lectures, my professor said that there is no such change in the trend: the larger atoms such as $\ce{I-}$ will always be more nucleophilic than $\ce{F-}$ due to the fact that it is more polarizable and likely to donate an electron pair. The textbook that I am using, Organic Chemistry by Wade, also does not discuss this trend reversal depending on the solvent. The only discussion talks about how $\ce{F-}$ would be a poor nucleophile as it is small and holds its electrons "tightly."

A practice problem provided during class also asked to compare the nucleophilic strength between $\ce{H2O}$ and $\ce{H2S}$ in polar aprotic solvent; similarly, the correct answer provided was that $\ce{H2S}$ would be a stronger nucleophile as it is larger and more polarizable.

I would appreciate it if someone could clarify this matter.

Actually nucleophilicity is inversely proportional to electronegativity when comparing compounds across the period, and inversely proportional to size when comparing in the same group. This condition is meant in aprotic solvent.

In protic solvents  the more electronegative ions gets easily solvated to a larger extent (i.e., like a ball with the ion in the centre and the water molecules around it) so the order across the period remains same since electronegativity is not effected but the order about the group changes since the now smaller ions become the larger ones.

For example, the trend for nucleophilicity is  $\ce{F- > Cl- > Br- > I-}$ in aprotic solvent but in protic solvent it is
$\ce{I- > Br- > Cl- > F-}$    ($\ce{F-}$ is now larger than others in size because it is solvated.)


Why is gold so popular in nanotechnology? Gold is a very popular metal in nanotechnology. It is often used as a substrate in electronic applications, as a core of functionalized nanoparticles, and more.

Why is gold so attractive? Why are other metals not as popular? What makes gold the go-to material for new applications?

First, let's look at gold's properties.

Gold is the most malleable of all metals. This malleability alone is very useful in aiding scientists to create such small gold nanoparticles.
Gold also reflects infrared radiation very well and is an excellent conductor of heat and electricity.
Generally, gold is not easily affected by oxygen and is generally nonreactive with most acids and bases. This property serves well especially in the human body where there are an abundant of acidic and basic fluids and the circulatory system is enveloped with oxygen. Because gold is generally unaffected by most acids, bases, and oxygen, there is more assurance for scientists that their golden nanoparticles for medicinal uses would not react with other things before reaching their specific target.
Going back to gold's ability to reflect infrared radiation, an essential reason why physicians, scientists, and researchers utilize nanotechnology is to track down certain cells (cancer, etc), molecules, or processes. Thus, it is important to use such a metal that can reflect radiation to tell them where their targeted molecules are or where their targets are going. In essence, gold nanoparticles are suitable for "staining," or showing contrasts in cells more visible, such as in tissue samples. This can provide higher spatial resolutions for labeling applications.
Gold, as a nice heat conductor, can also absorb heat very well, as I mentioned earlier, and in these biotechnologies, strong absorption of light energy is also another crucial aspect for allowing gold to then dissipate this absorbed light energy in its surroundings so that these heated areas can also be tracked and help with drug delivery purposes.

Here are some references:

Instant insight: A golden future
Nanotechnology Research into the Uses of Gold Nanoparticles


Is it true that allylic free radical is more stable than allylic carbocation? I have been studying reaction mechanisms involving allyl groups and it seems that the chances of forming of an allylic free radical are higher than forming a carbocation, I know that these things might be specific to reaction occurring, so I just wanted to find if there is any such relation between stabilities of radicals and cations.

This is trickier than it sounds because you have to compare things that are similar or maybe you won't get the comparison you want.

First, you can't compare the energies of two things that don't at least have the same components. In this case, the allylic radical has an extra electron over the allylic cation. Let's say you generated the radical by fragmenting propene via removing a hydrogen atom or a hydride anion, respectively, as the other product. Notice that you there is energy in the other product, and you can't really just compare the radical and cation without taking into account the other product.

Second, you're in a situation where you make one species from the other. In fact, if you ionize the allylic radical, you will make an allylic cation. Notice that since you are ionizing a neutral species, the ionization energy is positive: you're going to need to add energy into the system to remove the electron. This makes sense because that electron is stabilized by the 8 nuclei that are present.

So, while you can't really compare the two species in your question head-to-head, it's clear than you can create the cation from the radical by adding energy to remove an electron. But this doesn't mean the cation is higher in energy; it just means the cation and the electron separated at infinite distance are higher in energy than the radical.

Hope this helps.


Degree of dissociation of acetic acid at infinite dilution Afew drops of acetic acid are added to a large excess of water at $\pu{25 ^\circ C}$. In the limit of infinite dilution, what percentage of acid remains unionised? The $K_\mathrm{a}$ of acetic acid is $1.85\times10^{-5}$.

The answer is supposed to be 0.54%, but I'm not sure how to reach this value. I know that given the pH we can use the definition of $K_\mathrm{a}$ to calculate the ratio of $\ce{CH3COOH}$ to $\ce{CH3COO-}$, but how do I find the pH?

As the dilution becomes infinite for the amount of acetic acid, then the acetic acid's contribution to the solution's acidity falls, and the pH of the water is controlled by the acid dissociation of water. So the pH falls to 7.00. Thus,

$$\begin{align}
\frac{(10^{-7})[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &= 1.85 \times 10^{-5} \\
[\ce{CH3COO-}] &= 185 [\ce{CH3COOH}]
\end{align}$$

The fraction of unionised acid, $x$, is therefore given by

$$\begin{align}
x &= \frac{[\ce{CH3COOH}]}{[\ce{CH3COOH}] + [\ce{CH3COO-}]} \\
&= \frac{1}{1+185} \\
&= 0.54\%.
\end{align}$$


What is the pKa Range for weak acids and bases? Am I correct in assuming that the following is true?

$\mathrm pK_\mathrm a < 3$ is for a strong acid
$3 < \mathrm pK_\mathrm a < 7$ is for a weak acid
$7 < \mathrm pK_\mathrm a < 11$ is for a weak base
$\mathrm pK_\mathrm a > 11$ is for a strong base

$\mathrm pK_\mathrm a$ is negative log of the acid dissociation constant ($K_\mathrm a$).

$$\mathrm pK_\mathrm a = -\log K_\mathrm a$$

Acid dissociation constant is the equilibrium constant of the dissociation of ions of an acid in an aqueous solution. Consider a weak acid $\ce{HA}$. Weak acids do not dissociate completely in aqueous solution. The equilibrium for the dissociation of such acids can be expressed as

$$ \ce{HA + H2O <=> H3O+ + A-}$$

The equilibrium constant for this reaction(dissociation) will be

$$K_\mathrm a = \frac{\ce{[H3O+][A^{-}]}}{\ce{[HA]}}$$

It can be seen that the greater is the extent of dissociation, the greater will be the value of $K_\mathrm a$. Or, the stronger is the acid, the greater will be its $K_\mathrm a$.

Since, $\mathrm pK_\mathrm a$ is negative log of $K_\mathrm a$, it's values will be greater for weaker acid.

There is no sharp boundary between weak and strong acid. Wikipedia defines strong acids as acids which ionize completely in aqueous solution. So, an acid, say $\ce{HA}$ , is said to be strong if one mole of this acid dissociates in aqueous solution to give one mole of $\ce{H+}$ and one mole of $\ce{A-}$. Now $K_\mathrm a$ of such an acid will be 1.

Usually, $\mathrm{pH}$ is used to measure the acid strength, which is negative log of $\ce{H+}$ ion concentration. $$\mathrm{pH} = -\log\ce{[H+]}$$

The concentration of $\ce{H+}$ ions in water is $10^{-7}$ (It has been found out experimentally). Hence, its $\mathrm{pH}$ is $-\log(10^{-7}) = 7$.
For acids stronger than $\ce{H2O}$, the concentration of $\ce{H+}$ ions, $\ce{[H+]} > 10^{-7}$. So, $\mathrm{pH} < 7$ for acids and $\mathrm{pH} > 7$ for bases.

What you are referring to may actually be $\mathrm{pH}$ and in that case the ranges you've given are may be correct. But once again there is no clear difference between weak and strong.

There is a simple relation between $\mathrm{pH}$ and $\mathrm pK_\mathrm a$,
$$\mathrm{pH} = \frac{1}{2}[\mathrm pK_\mathrm a - \log c]$$
where $c$ is concentration of the acid. You can derive this relation using Ostwald's dilution law.

Similarly, we have $K_\mathrm b$, $\mathrm pK_\mathrm b$ and $\mathrm{pOH}$ for bases which are basic analogues of $K_\mathrm a$, $\mathrm pK_\mathrm a$ and $\mathrm{pH}$. These variables show the strength of a base.

Conclusion

The greater is the value of $K_\mathrm a$, the stronger will be the acid and the weaker will be the base.
The greater is the value of $\mathrm pK_\mathrm a$, the weaker will be the acid and the stronger will be the base.
The greater is the value of $\mathrm{pH}$, the weaker will be the acid and the stronger will be the base. For acids, $\mathrm{pH} < 7$ and for bases, $\mathrm{pH} > 7$.
The greater is the value of $K_\mathrm b$, the stronger will be the base and the weaker will be the acid.
The greater is the value of $\mathrm pK_\mathrm b$, the weaker will be the base and the stronger will be the acid.
The greater is the value of $\mathrm{pOH}$, the weaker will be the base and the stronger will be the acid. For bases, $\mathrm{pOH} < 7$ and for acids, $\mathrm{pOH} > 7$.


How many hydrogen atoms are in a sample of NH3? How many hydrogen atoms are contained in 8.66 moles of $\ce{NH3}$?
The answer is given after the log has been taken, so the final answer is 25.19.

What I have been doing is multiplying $8.66\text{ by }6.02\times10^{23}=5.21332\times10^{24}$ and when I take the log of that I get 24.71. I'm not sure what to do to get the correct answer.

Before you use log you have to multiply the answer of $8.66~\mathrm{mol}$ by $6.02 \cdot 10^{23}$ by 3 because the answer of $8.66~\mathrm{mol}$ by $6.02 \cdot 10^{23}$ gives you the number of ammonia molecules and in each molecules there are 3 hydrogen atoms present.

So it is:

$$8.66 \cdot 6.02 \cdot 10^{23} = 5.21332 \cdot 10^{24}$$

$$5.21332 \cdot 10^{24} \cdot 3 = 1.563996 \cdot 10^{25}$$

$$\log(1.563996 \cdot 10^{25}) = 25.19$$


Using real wave functions for chemical bond explanation My question related with another one here.

For explanation valence bond direction chemists are used real wave functions, for example $p_x, p_y, p_z, \ldots$. But this functions do not describe the preffred direction of angular momentum, they are not the eigenfunctions of $\hat L_z$.
The functions, that are describe the states with some preffered directions are not real, they are complex. For hydrogen, for example, they looks like:

If some other atom nearby first one, then there is some prefered direction, and good funtcions for this situation should be not real ones, but complex ones. Then why chemists used real functions for bond explanation?

Chemists use the linear combinations of hydrogen atom wavefunctions which are real because this is simply much more convenient to think about. Notice that for hydrogen, all the orbitals for a given $n$ are degenerate, so taking these linear combinations does not affect the energy, and thus thinking about the shapes of these orbitals is potentially more useful than thinking about the shape of something which is complex.

You are correct that taking the real-valued wavefunctions causes them to no longer be eigenfunctions of $L_z$. This is a bit frustrating, but in practice we never use these orbitals for anything, so it does not really matter. I used to think organic chemists took the shape of orbitals a bit too seriously (and they maybe do), but the notion of the shape of an orbital is actually recovered in a more rigorous manner.

That is, molecular orbital theory is just a qualitative form of Hartree-Fock (HF) theory. In HF, the Fock matrix is unitarily invariant, so we are free to transform our orbitals by any unitary rotation. These orbitals can be written as functions in $\mathbb{R}^3$ as desired. Thus, there exists schemes by which molecules orbitals can be "localized" and they often take the shape of orbitals which look a whole lot like the real-valued hydrogen atom orbitals you mention.

Thus, the use of these real-valued orbitals as a notion of shape is justified after the fact from HF theory and beyond.

In practice, this never really matters if one is being rigorous because any attempt at saying exactly which orbitals are involved in a bond is thwarted by unitarily transforming those orbitals. Thus, many chemists use the shape of an orbital as a heuristic because thus far, it has proven to be a very useful conceptual tool. I think pretty much all chemists know this is not completely correct, but neither is anything in physics.

Basically, the shape of orbitals provide a useful model.


Lubricants and rubber I've been reading a lot of conflicting articles/threads pertaining to which lubricant is compatible with rubber/plastics and I'm stumped.

From what I've come across, the consensus is that silicone is safe to use on rubber and that petroleum based eats at it over time. But then I heard the opposite.

What I'm wondering is, if rubber is composed of some petroleum then why would a petroleum lubricant or grease have a negative effect on it? Also what about plastics and vinyl, do they get damaged as well?

The simple version:

Petroleum-based rubbers can be dissolved by petroleum jelly eating away at the mechanical integrity.

The slightly longer version:

Petroleum jelly incorporates its way into the rubber, changing the plasticity (ie how rubbery or stiff) the rubber is.  Changes in plasticity change the mechanical strength and toughness such that the rubber is then outside the material property specifications of the product it is with/on/in, and hence fails.

Even longer version:

Most rubbers have small molecules inside them that help the molecular chain is the rubber move.  These molecules are called plasticizers (and occasionally tackifiers).  They are small molecules usually 200-500 MW which is less than the several hundred thousand of the rubber itself.  Petrleum jelly is basically hydrocarbon of roughly the size of the tackifiers.  The petroleum jelly molecules can 'wiggle' their way into the rubber, moving and adjusting how the existing plasticizers interact with the rubber.  In some cases the plasticizers may be extracted out, also changing the interaction, fundamentally changing the energies involves in intermolecular motion which is then related to toughness and strength (many steps in that derivation omitted).  In extreme cases in poor quality rubbers, some of the rubber molecules themselves can be dissolved by the petroleum jelly.  Changes in toughness, plasticity and mechanical integrity are common when petroleum jelly interacts with rubbers, and hence it should be avoided as a lubricant.

As you have pointed out silicone oil is a costly replacement, but not unbearable more expensive for most jobs.


What happens with enol and ketone What process happens between an enol and a ketone, so that the ketone will enolate and become acidic?

Please, explain what happens with them in this process?

Because the enol and ketone aren't stable, equilibrium happens; therefore the enol has a chance of losing a hydrogen ion and because of that, the enol becomes an enolate.

Also keep in mind that an acid is a molecule that can donate a proton.
The enolate is a base and it's gathering/accepting a proton from the hydrogen.

I assume this is what you meant when you asked the question.


Is water a heterogeneous molecule? My question is based upon the statement found here:

  an element may be homogeneous on a larger scale, compared to being heterogeneous on a smaller scale. This is known as an effective medium approach, or effective medium approximations.

So, is it acceptable to say that water molecule is a heterogeneous molecule?

I think you are confused. Heterogenous and homogeneous are words that are used to describe mixtures (of substances). In homogenous mixtures, the substances exist only in one phase, while heterogeneous mixtures the substances exist in two or more phases.

A molecule is homonuclear if it consists of atoms of a single chemical element ($\ce{O2}$). A molecule is heteronuclear if it is composed of more than one element ($\ce{H2O}$). Note that any pure molecule is considered a substance, and it is the combinations of these substances that form mixtures, either homogeneous, heterogeneous, or colloidal.


Cyclic carbocation rearrangement Is there a way for the 4-methylcyclohexyl cation (4-methylcyclohex-1-ylium) to undergo some rearrangement and become tertiary carbocation via 1-3 or 1-4 shift?

1,3- and 1,4- shifts don't really happen. It's just too far away. If you have a double bond at 4 or 5 carbons away, maybe you can do a ring closure or form a non-classical carbocation. The cation you've drawn could however undergo a sequence hydride shifts. Each hydride shift would move the cation closer to the methyl group, until the 3rd shift places the cation on the tertiary carbon. You'd need to out-compete any kind of elimination or cation reaction with a nucleophile...


Why there are nodes in orbital? I know that because Schrödinger equations solution for probability of that region is 0, but I was looking for an intuitive explanation.

How can you intuitively explain the existence of nodes in an orbital?

It is simply because the wave nature of the functions.

Imagine a tight rope with fixed ends (like a string in a guitar). If you shake it it will vibrate. Both ends remains fixed. If you take a high frame rate video and see it in slow motion, at a fixed horizontal length (let say $x$ positions) the rope will oscillate up and down with some amplitude. It can be seen one or more intermediate $x$ positions that remains always with null amplitude (if it is in state close to a stationary one). There you have nodes in one dimensional string.

Visual example

You would have more complex patterns going to a two dimensional system, and even more in a three dimensional case. They all have nodes, like the 1D case. The 3D case can be represented mathematically by using the spherical harmonics. The angular part of the solutions for the hydrogen atom are those spherical harmonics. The only intuitive explaining that I found is their wave nature. Of course it is not as visual as in the 1D case which is easily found in the daily life. And clearly, what is more weird is the wave nature of matter.

I hope that this answer address your question.


How does steric hindrance cause sulfur hexafluoride to have almost no reaction chemistry? Sulfur hexafluoride ($\ce{SF6}$) is, according to the Wikipedia page, a hypervalent substance.  However, according to the page in regards to $\ce{SF6}$ reactions:

  There is virtually no reaction chemistry for $\ce{SF6}$. The main contribution to the inertness of $\ce{SF6}$ is the steric hindrance of the sulfur atom, whereas its heavier group 16 counterparts, such as $\ce{SeF6}$ are more reactive than $\ce{SF6}$ as a result of less steric hindrance

The linked steric hindrance provides many examples of how that chemical process works, but specifically, what steric hindrance (or any other) mechanism results in $\ce{SF6}$ possessing virtually no reaction chemistry?

As you can see:

 Sulfur (16) is significantly smaller than Se (34). That size difference
 means that there is more room between the fluorine atoms on Se than on
Sulfur. That is what steric hindrance means. Think of steric hindrance
 as limited access. The smaller the atom the less access there is.
Oxygen is so small that less than 6 fluorine atoms can crowd around it
because of steric hindrance... probably more like four.

So because there is more room between the fluorine atoms on Se there is
more opportunity for other atoms to attack the central atom Se, but
sulfur is completely shielded by inert fluorine.


Looking for a non-toxic substance that evaporates very quickly Firstly I am not a chemist so please be kind. I am a creative technologist working on a large art installation. The challenge is to make people perceive rain without it raining. I am looking for something we can very occasionally drop from above that will feel like a rain drop but almost instantly evaporate leaving the person completely dry when they leave (they will only experience a few drops whilst in the room — the rest of the effect uses lights and sound).

My initial thought was liquid nitrogen but I assume that whilst a tiny occasional drop on the back of the hand would be fine, one in the eye for example would probably be dangerous. Plus the room would need to be heavily ventilated to prevent the risk of asphyxiation.

I'm assuming the answer is nope but if anyone has a genius idea for some way i could achieve this effect I would love to hear it. Over to you.

I'm assuming the answer is nope but if anyone has a genius idea for some way i could achieve this effect I would love to hear it.

Nope ;-)

Your concerns are justified and I wouldn't expose the audience to anything but water either.

you can

use water from all above

and let the audience get wet
and hand out rain coats

use water and only let it rain to the left and right of a marked path or
apply a more sophisticated technique, as used by the random international art studio in their rain room. Here, it rains thoughout a whole room, but the visitors are tracked by cameras and whereever they move, the water is selectively switched off.


Separating specific components of a PUREX raffinate The PUREX extraction in nuclear fuel reprocessing leaves behind a nitric acid raffinate full of (mainly) 'fission products', which are a diverse set of elements produced by fission in a nuclear reactor. The concentrations of the various components of the raffinate are shown in this image.

My goal is to separate out the group III elements from the group IV+ elements (shown highlighted in red/blue). It does not matter what happens to the rest of the components, they can go with either or neither of these groups in any combination, all that matters is that the group III elements are spatially separated from the group IV+ elements. The process can be as inexact as necessary as long as it mostly accomplishes this goal.

Any thoughts? My nuclear chemistry book says that these group III elements are "have soluble nitrates, chlorides, and sulfates while their fluorides and hydroxides are insoluble. LaF3 is frequently used as a carrier for this group. They form stable complexes with strong chelating agents such as DPTA, EDTA, and the like". Perhaps this makes the question fairly straightforward, I'm coming from a physics, not a chemistry background. I'm just confused as to what properties the group III elements have that the group IV+ elements don't which could be leveraged to achieve some kind of separation.

Sorry for the wall of info, and thanks very much in advance for any help or input!

You have chosen a difficult challenge, a word of advice unless you are working in a nuclear site I would advise you in the strongest terms possible not to attempt a separation of radioactive things in your kitchen, basement, garden shed or some other home lab.

OK warning over, now the idea of adding flouride to create a percipitate of lanthanide (and post plutonium actinides) is a nice idea but it will make any further processing of these elements harder. I also worry that any traces of plutonium or other elements such as zirconium might become incorperated into the lanthanide flouride perceipitate.

I would suggest that you consider solvent extraction as the separation method, I would suggest that you consider the TRPO process chemistry from China. If you were to dilute the PUREX raffinate to reduce the nitric acid content to about 0.5 M, then shake the PUREX raffinate with about 30 % (v/v) of cyanex 923 in an aliphatic kerosene then you will have a organic solution which is a mixture of lanthanides, zirconium, iron and residual plutonium / uranium.

I would suggest stripping (back extracting) the organic phase with strong nitric acid, the plutonium, uranium and zirconium will stay in the organic phase while the lanthanides will go back into the aqueous phase.

In some ways a better alternative would be to use TODGA or a malonamide, a solution of one of these reagents in aliphaltic kerosene will extract the lanthanides from PUREX raffiniate without adjustment of the nitric acid content of the raffinate. You can then strip the lanthanides (plus the americium / curium) from the raffinate. These types of processes have been tested at ITU in their hot cells using PUREX raffinate formed from real fuel.

In some ways I would want to use dilute nitric acid as a stripping agent for the lanthanides from a diamide based process. You can then reextract the lanthanides from the dilute nitric acid with 30 % DEHPA (bis-2-ethylhexyl hydrogen phosphate) in kerosene. In turn back extract this new organic phase with nitric acid, if you are careful it is possible to get some separation of the lanthanides into the heavy and light lanthanides by selective stripping with different concentrations of nitric acid.

Another method which would work would be to use the TRUEX process chemistry from the USA, this is based on CMPO in tributyl phoshate modified kerosene.


Coloration of substances because of F-Centers In case of metal deficient defects in the crystal lattice of a metal, the positive voids are filled by electrons.

How does this result in the coloration of a substance?
Does this condition arise only in the case of metals?

For a simple explanation, we could explain the coloration of the substance by the absorption of the energy of the electrons in the void, which leads to the color of the compliment color (as is suggested by http://rspa.royalsocietypublishing.org/content/204/1078/406)

However, a more rigorous explanation could be found here: http://ptp.oxfordjournals.org/content/4/2/181.full.pdf

For the second question, it is written in Ceramic Materials: Science and Engineering that "the term color center is now applied to any defect, including an impurity, that produces color in an insulator". It also notes that the original observation was made by the production of F centers, which suggests that the modern interpretation of the phenomenon applies to the defects which product color in an insulator.

I'm not sure if I answered you question exactly as you hoped, but if you look into the resources, there is a wealth of knowledge there.


Why is water transparent? Some substances like copper sulfate for example have vivid colors. But why is water transparent? Does it not emit any visual light from the electromagnetic spectrum?

Water is not, in fact, colourless, but is slightly blue. As you can see below, in the visible range, there is slightly more absorption toward the long wavelengths, so more blue light makes it through than redder light. However, the absorption is very low in the visible range so the blue colour is faint and not apparent unless looking through a significant thickness of water. The reason things like $\ce{Cu^2+}$ solutions appear so vividly blue is because they also absorb red light, but very strongly, so a good deal more blue light than red passes through the solution. (see here)

I'm not sure exactly what you mean by transparent here, but if you mean why isn't water opaque, because water doesn't absorb much visible light (as described above), and doesn't contain any non-uniformities (like particles or the fat globules in milk) that can cause scattering, photons can travel through water in a straight line and most make it through without being absorbed, making water colourless (when looking through a small quantity) and transparent, much as air is.

Source


Does aluminium carbonate exist? I read in a west African chemistry textbook (New School chemistry by Osei-Yaw Ababio) that aluminium trioxocarbonate (IV) (aluminium carbonate) does not exist. I need verified confirmations or oppositions.

It does not exist in nature, but you could make it in the lab by adding pressure.

From a traditional pencil and paper calculation I found the pressure needed for

Al2/3O (s) + CO2 (g) = Al2/3CO3 (s)

is about 10 GPa...

Or you could make an Al2(CO3)3 (aq) solution with concentration of $10^{-5}$ M.

Well, still not impossible. And there should be some special solvents and/or complex structures that easily allow this reaction (postulation).


How to identify an emerald green solution with possibly two anions? The solution is emerald green in colour.
When added with water, it gives a pale blue solution.
When reacted with $\ce{BaCl2}$, it gives a sky blue solution and a white precipitate.
When reacted with $\ce{AgNO3}$, it gives a sky blue solution and a white precipitate.

This does not make any sense to me. The $\ce{BaCl2}$ is used to test sulphates ions and $\ce{AgNO3}$ is used to test halides. How could you have two negative ions?

You can have two different anions if your solution is a mixture of two different metal salts, a chloride and a sulfate. The color change from green to blue upon dilution with water fits well with $\ce{Cu^2+}$ being the corresponding cation. Concentrated solutions of $\ce{CuCl2}$ are green due to the presence of chlorocuprates, like $\ce{[CuCl3]^{-}}$ and $\ce{[CuCl4]^{2-}}$. When water is added, a stepwise ligand exchange takes place, and the pale blue color of $\ce{[Cu(H2O)6]^{2+}}$ starts to appear. The image below shows copper(II) chloride solutions of different concentrations: Green when the chloride concentration is high, and blue when it is lower (Wikipedia: copper chloride).

In summary, the observations (color change, precipitates) indicate that you have an aqueous solution of $\ce{CuCl2}$ and $\ce{CuSO4}$.


How does an oscillating reaction work? I watched a video showing an orange solution that goes to clear and then back to orange (and so forth.) The reaction goes through a number of cycles before it will no longer oscillate. This tells me that the overall energy in the system dissipates over time, but how does such a reaction work?

What is happening that causes this oscillation in concentration?

I worked on a project using this reaction fairly recently. There are a significant number of intermediate species involved in the reaction with a significant number of autocatalytic steps.

In order to obey the rules of free energy, the reaction may not oscillate about its equilibrium as a pendulum does- there is always a monotonic decrease in reactant concentration and an increase in product concentration, but the multitude of intermediates oscillate in concentration as the reaction switches its reliance on certain sections of the mechanism (dependent upon the instantaneous concentration of stuff in the mixture). The indicator used dictates the colour changes- I used ferroin, and so when the concentration of Ce(4+) (the catalyst used) is high the mixture is red, and when the concentration of bromate is high the mixture is blue.

It should be noted that the reaction will always come to a stop unless you use a continuously stirred flow reactor in which products are removed and reactants added as the reaction proceeds.

There are several mechanisms such as the FKN, MBM, GTF and Oregonator. The Oregonator is by far the most simple but lacks chemical accuracy, the former 3 display real chemical conversions yet still fail to perfectly model the reaction. The mechanisms can be modelled using coupled differential kinetic equations, which I did during my project in Mathematica.

The simplest general reaction scheme for a chemical oscillator is given by the Brusselator:

A → X
2X + Y → 3X
B + X → Y + D
X → E

In this scheme, A represents the reactions and E the products. X is autocatalytic (it catalyses its own production).

Another interesting example comes into play in population dynamics, the Lotka-Volterra equations (also known as predator-prey equations). They can be used to quantify the changes in populations in a predator-prey system such as foxes and rabbits. The reaction itself has also been suggested as a very primitive form of chemical order- interesting stuff.

Disclaimer: Sorry if this seems a little off-topic; the details of the specific BZ mechanisms are far less interesting than the concept of the oscillators themselves. If you want to look into the actual mechanisms though, start with the FKN model, it's fairly easy to digest and even model.


Is it possible for one of reactants to deplete before achieving equilibrium Consider the following reaction where $\ce{A}$ & $\ce{B}$ reacts to produce $\ce{C}$ & $\ce{D}$.

$$\ce{ A(g) + B(g) <=> C(g) + D(g)}$$

Is it possible for $\ce{A}$ or/and $\ce{B}$ to run out before equilibrium at some given starting moles of $\ce{A}$,$\ce{B}$,$\ce{C}$ & $\ce{D}$.

No, that is not possible.

Consider the definition of equilibrium from the IUPAC gold book:

  chemical equilibrium
  Reversible processes [processes which may be made to proceed in the forward or reverse direction by the (infinitesimal) change of one variable], ultimately reach a point where the rates in both directions are identical, so that the system gives the appearance of having a static composition at which the Gibbs energy, $G$, is a minimum. At equilibrium the sum of the chemical potentials of the reactants equals that of the products, so that:
  \begin{align}
\Delta G_\mathrm{r} &= \Delta G_\mathrm{r}^\circ + R T \ln K &&= 0\\
\Delta G_\mathrm{r}^\circ &= − R T \ln K\\
\end{align}
  The equilibrium constant, $K$, is given by the mass-law effect.

The important thing is that the reaction proceeds in both directions. It means, that as soon as some of the product is formed, it will react back towards the reactants. The equilibrium is reached when the rates of both of the directions are identical.
From the formula in the definition, you can also see, that the location of the equilibrium, i.e. the concentration of all species or $K$, is only determined by the difference in the Gibbs energy between products and reactants.
In other words, a system that forms an equilibrium will always form an equilibrium, independent of the concentration of the starting materials.


What is the origin of the "positive-to-negative" convention concerning dipole moments in chemistry? I'm a physics professor who often teaches students who have taken a lot of chemistry (read:  pre-meds).  In physics, the dipole moment of a charge configuration is defined as pointing from concentrations of negative charge towards concentrations of positive charge.  This is embodied in the formula we use for the dipole moment of a charge configuration,
$$
\vec{p} = \sum_i q_i \vec{r}_i.
$$
It's not too hard to see that this yields a vector that points from negative charges to positive charges.  For example, if I have a charge of $-e$ at the origin ($\vec{r} = 0$) and a charge of $+e$ a distance $d$ away, this formula yields a vector that points from the negative charge to the positive charge, with magnitude $ed$.

However, the first time I taught this material in a class with a large number of chemistry students, they were very confused;  they had always learned that the dipole vector of a polar molecule points from positive charge towards negative charge, i.e., in the opposite direction of the convention from physics.  For example, by the physics convention, the dipole moment of the water molecule points away from the point of the "V" formed by the bonds, while under the chemistry convention, the dipole moment points in the direction of the point of the "V".

How did this convention arise?  Is there an advantage to defining the dipole vectors this way in chemistry calculations?  Or is it one of those scientific conventions that someone wrote down one point, and got ingrained in textbooks ever since?  (Is it all Benjamin Franklin's fault?)

One of the answers to this question notes the difference between the conventions, but doesn't really explain why the first chemist to write down a dipole vector wrote it down that way.  I would love to know how this convention arose historically!

Some sources point out the direction of the dipole moment is convention dependent. Since both conventions are mathematically correct it seems important when solving a problem to identify the convention.  Both conventions seem to be derived from the bones of dead theories.

The physics convention does seem to have Franklin's work identifying negative "resinous electricity" and positive "vitreous electricity" as one step in the progression to the modern theory.  In modern theory a positive to negative electric field will orient a dipole such that the field is in the same direction as a negative to positive dipole.

On the chemistry side, Peter Debye (who the unit for a dipole moment is named) developed the concept of the dipole moment in 1912 to explain anomalous dispersion. This dispersion was explained as being caused by a negative charge collecting on one side of a molecule leading to a dipole moment. This moment's magnitude explained anomalous dispersion, so direction did not seem to receive much attention.  Molecules during this time where described using the Thompson, or "plum pudding," model. In this model molecules were described as positively charged spheres.  (These spheres had no nucleus because the nucleus concept was in its infancy during this time period.) These spheres formed the basis of mathematical models used to describe the behavior of materials with a dipole moments.  The dipole moment in this sense went from the center of a molecule to the point of highest electron density.  With the center of the sphere as a pivot point, the rotation of molecules in an electric field could be explained. In this sense the positive to negative convention may be an artifact of the early spherical molecular models first used to describe dipole moments.

sources:

J. W. Williams   Dipole theory and the size of molecules.  Trans. Faraday Soc., 1934,30, 723-728

John W. Williams The Dielectric Constants of Binary Mixtures. VIII. The Electric Moment as a Vector Quantity, J. Am. Chem. Soc., 1928, 50 (9), pp 2350–2357  DOI: 10.1021/ja01396a005  Publication Date: September 1928

L. E. Sutton  Electric dipole moments and resonance in molecules, Trans. Faraday Soc., 1934,30, 789-801

https://www.quora.com/Why-does-a-dipole-have-a-direction-from-negative-to-positive-while-electric-field-lines-go-from-positive-to-negative

http://goldbook.iupac.org/E01929.html

http://www.ushistory.org/franklin/info/inventions.htm

http://www.wondersandmarvels.com/2011/12/accentuate-the-vitreous-eliminate-the-resinous.html


How can I synthesize 2‐amino‐3‐(chlorosulfonyl)benzoic acid from toluene or benzene? I am trying to design a proper synthesis route for 2‐amino‐3‐(chlorosulfonyl)benzoic acid, though methyl 2‐amino‐3‐(chlorosulfonyl)benzoate—using an ester group in lieu of a carboxylic acid—would also be acceptable.

I am thinking a good starting material would be toluene or benzene. Here is my proposed route.

I determined the ortho-para ratio for toluene nitration from p. 816 of "Organic Chemistry, 6e" by Loudon & Parise.

Typically, para substitution predominates over ortho substitution, but not always. For example, nitration of toluene gives twice as much o-nitrotoluene as p-nitrotoluene. This result occurs because the nitration of toluene at either the ortho or para position is so fast that it occurs on every encounter of the reagents; that is, the energy barrier for the reaction is insignificant. Hence, the product distribution corresponds simply to the relative probability of the reactions. Because the ratio of ortho and para positions is 2 : 1, the product distribution is 2 : 1.

Also, according to the same page,

The boiling points of o- and p-nitrotoluene, 220 °C and 238 °C, respectively, are sufficiently different that these isomers can be separated by careful fractional distillation.

Is there a better method? Please identify any mistakes I may have made in my synthesis and suggest improvements. Note that I install t-butyl as a blocking group in order to direct the sulfonyl chloride group ortho to the amino group.

Here is a proposed route starting from aniline.
First protect the -NH2 with t-BOC
Directed lithiation of t-BOC-aniline (sBuLi/Et2O, -75C) and quench with MeSSMe to install -SMe in the 2 position.
Directed lithiation under the same conditions quenching with Ethyl Chloroformate to install  -CO2Et in the 6 position.
Oxidise the -SMe group to sulfoxide (MCPBA, DCM)
Pummerer rearrangement of the sulfoxide (Ac2O, aq. base work up) gives the free thiol.
Oxidation of the thiol to the sulfonic acid, one example of possible conditions hydrogen peroxide with Rh catalysis
Form the sulfonyl chloride (conditions using trichlorotriazene here. Depending on your choice of conditions you can remove the -BOC group or preserve it. I think preserving it will give a flexible late stage intermediate that is reasonably easily handled. It is possible to directly form the sulfonyl chloride from the -SMe with chlorine in aq acetic acid conditions here but the conditions are harsh and risk removing the -BOC group


Capacitance and flammable liquid I'm trying to attach a sensor to some tubes to measure the level of liquids in them.

see here

One tube contains toluene and the other contains Recosol R55 (shellite).

I'm thinking of using a capacitance level sensor like this or this.

But I am concerned about the safety of doing this. Could the capacitance sensor ignite these liquids? is this even possible?

Obviously I need to check with the suppliers of the device. However, most of datasheets for level sensors omit this information. Perhaps they are not game enough to make any claims about their sensor in this regard.

I don't really understand capacitance and whether it can ignite flammable liquids.

The capacitance sensor itself will certainly not present an ignition risk to your system. The sensor will be on the outside of the container, and it does not introduce any electrical potential that would cause a spark.

The way the device works is to measure the change in the sensor's capacitance depending on the dielectric effect of the liquid, which will increase as more liquid enters the electric field of the sensor (a capacitor).


Oxidation states of iron in Roussin's salts For preservation of meat, sodium nitrite is usually added and as a result $\ce{NO}$ is then formed. Consequently, $\ce{NO}$ reacts with the sulfur and iron atoms from decomposition of proteins, forming $\ce{[Fe4S3(NO)7]-}.$ X-ray crystallography shows that the complex anion has a structure as shown below:

  a) Blacken all the circles corresponding to iron atoms and add symbols Fe(A),
  Fe(B), Fe(C) and Fe(D) beside the circles in the sequence of top → left → right.

  b) The configuration of 3d electron shell of the iron atoms has been studied with modern structural analysis. Knowing that the mean oxidation number of the
  four iron atoms is $–0.5$, give their configurations of 3d shell, respectively.
  Assume that each iron atom adopt sp hybridization.

  $\ce{[Fe4S3(NO)7]^{–}}$ anion can be reduced and a new complex $\ce{[Fe2S2(NO)4]^{2-}}$ is formed

  c) Give the oxidation state of each iron atom with Arabic numerals.

  d) $\ce{[Fe2S2(NO)4]2-}$ can be converted into $\ce{[Fe2(SCH3)2(NO)4]^{n}}$, a carcinogen. Which of the following three species is added to $\ce{[Fe2S2(NO)4]^{2-}}$ : $\ \ce{CH3+}$,$\ \ce{•CH3}$ or $\ \ce{CH3−}$?

My Attempt

a)

I don't have any real reason to why I chose those 4  atoms. I chose them since they were the 4 biggest atoms in the ring and made sense they would alternate with the sulfur atoms in the ring. Is there a more solid explanation?

b) Firstly, why do the iron atoms have a negative oxidation state, doesn't it 'give' its electrons to the sulfur and nitrogen atoms since it is more electropositive? I look up negative oxidation states of iron and I found out when bonded to carbonyl compounds, the iron atom usually has a negative oxidation state. So is the $\ce{NO}$ ligand in this case acting like $\ce{CO}$?

The answer is Fe(A) has 3d7 configuration while the others have a 3d9 configuration. Could someone please give an explanation for this.

c)

I have drawn what I think is the structure above, but I have the same problem as part b)

d) I have no idea. I am guessing the methyl cation since the iron atom has a negative oxidation state (I am guessing from part c) hence it will more likely react will a cation rather than an anion or radical.

I suppose it's asked about formal oxidation numbers of iron as there is not enough information to make assumptions about the ones that reflect electronic structure.
For instance, analysis of interatomic distances could point to a delocalized electron pair, but you are not given any linear or angular geometrical parameters to begin with.

It's only the symmetry that's evident from the crystal structure: the thioferrate cluster $\ce{[Fe4S3(NO)7]-}$ belongs to $C_\mathrm{3v}$ point group, hence the only possible arrangement of four iron atoms is a tetrahedral one.
Another conclusion that can be made is that three equatorial iron atoms are equivalent ($\ce{Fe(B)}$, $\ce{Fe(C)}$, $\ce{Fe(D)}$), and the last apical one ($\ce{Fe(A)}$) most likely deviates in terms of oxidation state.

Supposing formal oxidation number ($\text{O.N.}$) of three equivalent irons and the last one are $y$ and $x$, respectively, and the fact that the average $\text{O.N.}$ is $-0.5$, the following holds true:

$$x + 3y = -0.5\cdot 4 \implies y = -\frac{x + 2}{3}$$

As formal oxidation numbers are integers, there are several pairs giving the solution:

$$
\begin{array}{l|rrrrr}
\hline
x & \ldots & -5 & -2 & \color{green}{1} & 4 & 7 & \ldots \\
\hline
y & \ldots & 1 & 0 & \color{green}{-1} & -2 & -3 & \ldots\\
\hline
\end{array}
$$

among which the least "extreme" (e.g. the one with the absolute lowest negative $\text{O.N.}$) for a metal would be the $-1;+1$ pair; so that for $\overset{0}{\ce{Fe}}:~[\ce{Ar}]\,(\mathrm{3d})^6\,(\mathrm{4s})^2$ your answer gives two configurations, $(\mathrm{3d})^7$ for $\overset{+1}{\ce{Fe}}\ce{(A)}$ and $(\mathrm{3d})^9$ for $\overset{-1}{\ce{Fe}}\ce{(B,C,D)}$.


Negative temperature as limiting case for two-level system For N particles in a two-energy-level system, with energy levels being 0 and $\Delta$, the internal energy tends to $N \Delta /2$ as temperature tends to infinity. At this point, the occupancy of each energy level is equal, and any further increase in internal energy would lead to a decrease in the entropy of the system.

Temperature is defined as $T = (\frac{\partial U}{\partial S})_V$. Using this definition, for a state with equal occupancies described above, the temperature should be negative. Where is the contradiction?

I think you would need to describe the derivative $\frac{\partial U}{\partial S}$ at that point, which should be infinity by your logic since the entropy is at a maximum with respect to internal energy ($\frac{\partial S}{\partial U}=0$). This is obtained using $$\frac{\partial U}{\partial S}=\frac{1}{\frac{\partial S}{\partial U}}$$


Should temperature and pressure conditions for two reactions be same for them to combine for enthalpy calculations? I was solving a question in which I had to calculate Enthalpy of formation of $\ce{CS_2(l)}$ using the given Enthalpy of combustion (not in standard state) of $\ce{C(s)}$, $\ce{S(s)}$ and $\ce{CS_2(l)}$.
To calculate the $\Delta H_f$ of $\ce{CS_2(l)}$, we combine the corresponding equation of combustion of these chemicals  in appropriate order to form the equation :
$$\ce{C(s) + 2S(g) -> CS_2(l) }$$
Should the temperature and pressure conditions of the combustion equations be same for me to combine them mathematically?

yes they should be for a simple reason: when you do make C and S react together  you are going to assume that the temperature and the pressure of the box you're using for the reaction is the same. In particular you can even think of putting for example the hot solid in the box with the cold gas, when you're talking of entalphy of formation you're basically speaking of the equilibrium situation. You will reach equilibrium only once al the macroscopical quantities are perfectly equal in all the box you're working with. If this was not the case, the equations you would have to solve are much more complicated than those you are usually seeing .


Fitting titration curve To simulate the action of gastric $\ce{HCl}$, we have titrated 100 ml of a meal with 0.1 N $\ce{HCl}$. To plot the data, use (R-code)

vol = read.table("http://menne-biomed.de/uni/volph.txt",sep="\t",header=TRUE)
plot(vol$vol,vol$pH,pch=16,type="l",main="Titration of 100 ml meal with 0.1 N HCl")

I am trying to determine, for example, how much 1N $\ce{HCl}$ I need to add to get a pH of 1.5; if I understand correctly, I have to fit the Hasselbalch equation.

Being a statistics guy, the numerics is no problem, I just do not understand how to get the terms in that equation together with my data.

The best thing would be if you went back to the mass action law, considered all reactions that take place ($\ce{Ca}$ buffer!) and derived a formula for your specific problem. The Henderson-Hasselbalch equation actually does exactly that for a one-component acid/base system with a single protonation stage. Remember that the concentrations that occur in the mass-action law are those at chemical equilibrium, not the initial ones; i.e. for any component $C$, the eq. concentration is $[C] = [C]_0 - \Delta_C$. The problem at hand is then to find how the changes in concentration depend on the added volume of titer.

If you just took the Henderson-Hasselbalch equation (equating the change in concentrations with the added concentration of $\ce{HCl}$), you do not include the $\ce{Ca}$ buffer, which will probably play an role in this. (You get a $pK_a$ of around 0.7, which I guess is a bit on the strong side ...)

One more thing I just noticed: You apparently added 100 mL of titer to 100 mL of sample. That's kind of bad, because you'll also have to consider how the change of overall volume affects your concentrations (effectively the pH) during titration.


E, Z designation - what to do when substituents are identical on one side of the double bond? Here, A has a greater atomic number than B, which has a greater atomic number than C. To assign E,Z designations you would divide the double bond into left and right portions and assign priorities. On the right side, A > C. However, on the left side, B = B. A (first priority on right side) is simultaneously on the same side and on opposite sides to B (first priority on left side). How would you class this compound then?

When both the substituents are identical on one side of the double bond, the compound does not show geometrical isomerism.
There is no point in using E-Z nomenclature in that case.


Is there any evidence, *any evidence at all*, that nascent hydrogen actually exists? As far as I know, nascent hydrogen is nothing more than an especially magical unicorn entity that either makes reactions happen or at least greatly facilitates reactions happening. Many years ago, it was assumed that nascent hydrogen existed, but it was never characterized in any way because it did its work so fast as to escape detection. Now I am seeing "answers" to questions that invoke nascent hydrogen. I consider nascent hydrogen to be entirely bullcrap, pardon my language. Would someone please tell me what is going on with this?

You should read the Solution to nascent hydrogen challenge by Juris Meija & Alessandro D’Ulivo Link. The authors vehemently deny the existence of any nascent hydrogen.

You will find that the concept of nascent hydrogen and nascent oxygen is still taught in some Indian chemistry textbooks. This term has literally vanished from US & European chemistry books. I just thought to search an elite journal JACS  and this term is present in 2011 in the abstract of

Highly Selective Ammonia Synthesis from Nitrate with Photocatalytically Generated Hydrogen on CuPd/TiO2, J. Am. Chem. Soc. 2011, 133, 5, 1150-1152

  "The continuous generation of nascent hydrogen atoms on the surface of
  the CuPd nanoalloy, where Cu and Pd are homogeneously mixed, led to
  the high selectivity for ammonia."

Well, this is exactly a classical example of the usage of nascent hydrogen when Zn could reduce nitrate ions in an alkaline medium. The texts of 1940s describe those experiments very well. See Mellor's Treatise of Inorganic Chemistry.

Another 2012 RSC journal Chemical reduction of an aqueous suspension of graphene oxide by nascent hydrogen


How is this an autocatalytic reaction? I'm probably missing a simple component of this reaction, but for some reason I not seeing it.

These are two steps that compose a radical process of the Briggs-Rauscher reaction (that are autocatalytic):

$IO_3^- + HIO_2+H^+ \rightarrow 2IO_2 \cdot +H_2O$
$IO_2 \cdot + Mn^{2+} + H_2O \rightarrow HIO_2 + Mn(OH)^{2+}$

This process is autocatalytic in the sense that 2 $HIO_2$ is produced for each one consumed.  However, I'm just not seeing that for some reason.  Like I said, it's probably a simple thing I overlooked. So, I'm looking for an explanation for what's going on here.

In your first reaction, $\ce{HIO2}$ and $\ce{HIO3}$ react to form $\ce{2IO2.}$ and $\ce{H2O}$.

In your second reaction, $\ce{IO2\cdot}$ is being reduced back to $\ce{HIO2}$ by $\ce{Mn^{2+}}$. Since the first reaction produced $\ce{2IO2.}$, this second reaction needs to happen twice for each time the first reaction occurs. The pair of reactions is really:

$$\ce{ IO3- + HIO2 + H+ -> 2IO2. +H2O }$$
$$2\left(\ce{IO2. +Mn^{2+} +H2O-> HIO2 + Mn(OH)^{2+}} \right)$$

When you combine these reactions into an overall reaction, the instances of $\ce{IO2.}$ cancel and you are left with a net gain of $\ce{HIO2}$.

$$\ce{IO3- + H+ +2Mn^{2+} + H2O-> HIO2 + 2Mn(OH)^{2+}}$$

Of course, the net reaction does not hint at the autocatalytic behavior. To emphasize that the reaction is autocatalytic, you could write the overall reaction in a not quite simplest form:

$$\ce{IO3- + HIO2 +  H+ +2Mn^{2+} + H2O-> 2HIO2 + 2Mn(OH)^{2+}}$$


Classification in Stereoisomerism I was trying to understand stereoisomerism. I am having trouble creating a 'flow chart' of Stereoisomerism. While there are many sources available over the internet and books, I don't find a standard answer. Some sources mention conformers and geometrical(cis/trans) under diastereomers while others seperate them completely. According to definition of diastereomers available on Wikipedia,

Diastereomerism occurs when two or more stereoisomers of a compound have different configurations at one or more (but not all) of the equivalent (related) stereocenters and are not mirror images of each other.

While this definition includes geometrical isomers very well into it, not much clarity arises regarding conformers as it used the word configurations rather than conformation.

Also I was finding difficulty in placing Mesoisomers in the chart.

It would be really nice if you could draw the flow chart and post it here.

Thank you!

The problem with looking for information about stereochemistry on the internet is that many sources use outdated terminology, imprecise definitions or have downright erroneous information!

Here's a flow chart that I put together that I believe to be accurate, yet hopefully straightforward:

When using the flow chart, it is really important to remember that each decision on the flow chart is related to comparing two molecules, ie. it doesn't make sense to ask "what is compound A", but it does make sense to ask "how are compounds A and B related"? A molecule may be the enantiomer of another molecule, but a diastereomer of another.

Implicit in the flow chart is that stereoisomers have different 3-dimensional arrangements of their atoms, while having the same connectivity of atoms to each other.

The term "geometrical isomer" is now deprecated and its use is "strongly discouraged". It is an older term for cis-trans isomers, which are just a subset of diastereomers.

Back to your original post - the part in bold about diastereomers is problematic because the definition of diastereomers does not relate directly to stereogenic centers. It emerges that compounds with two or more chirality centers (the correct term for what are often referred to as stereogenic centers or stereocenters) can often exist as diastereomers, but chiral centers are not a pre-requisite. For example, cis- and trans-2-butene are diastereomers according to the definition.

A meso-compound is an achiral compound (the mirror image is super-posable), which is a member of a set of diastereomers, where at least one of those diastereomers is chiral. A meso-compound most commonly arises when the compound has two or more chiral centers, yet is not chiral due to an internal plane of symmetry. Meso-compounds don't fit neatly on the flow chart, which is based on comparing two molecules, but are a particular type of diastereomer.


Clarification for metallic bond Recently, I learn about bonding and my teacher only focus on ionic and covelant bond, not metallic bond. So, I go on google search but I have no idea of what is it? How electrons can suddenly appear to form the bond? How can the metal highest energy level overlap if it need to obey the octet rule

Metallic bonds depend on the Pauli exclusion principle to cause a single energy level to be separated into a sea of very closely spaced levels.

Consider metallic Ca, with its s2 filled with two electrons of opposing spin state. You cannot have another s2 electron in the same atom. Now, add a second Ca atom that can freely interchange electrons with the first (e.g. metallic conductivity). Quantum mechanically, the electrons are indistinguishable, and an electron of the first atom might be found on the second one, so the combination acts as a "super-atom" with two nuclei but a single electron cloud. To allow this without violating the Pauli exclusion principle, one s2 level "moves up" and the other "moves down", splitting the level into two bands.

As more atoms are added, the Fermi "sea" becomes populated to the extent that the energy levels merge into a continuum.

See http://fog.ccsf.cc.ca.us/~wkaufmyn/ENGN45/Course%20Handouts/15_ElectricalProps/02_ConductivityConductor.html and http://www.chemexplore.net/sodium.htm for further explanation.


Heaviest atom with which chemistry has been performed? The nucleosynthesis of superheavy atoms such as transactinides is an exciting area of nuclear physics, but not much of a playground for chemistry because

My question is: what is the heaviest atom ever used in chemistry? For this I define chemistry as “making a compound with the atom (molecule or material) and characterizing it to some extent”.

The one source I found on the topic is “Critical evaluation of the chemical properties of the transactinide elements (IUPAC Technical Report)”, dated 2003, but it seems quite old for such a rapidly developing field. Also, despite being title “chemical properties” it mostly reports on physical properties.

According to Adsorption behavior of super-heavy elements (Z
[greater than or equal to ] 112) on metal and inert surfaces :

  Nowadays the research focusses on the chemical properties of
  the element 114. So far, two gas chromatography experiments to study the interaction strength of
  element 114 with a gold surface have been performed with conflicting results. One experiment [reference 1]
  reported a weak interaction of element 114 with a gold surface, leading to adsorption only at very
  low temperatures of approximately –90°C, while in the second experiment [reference 2] adsorption on gold
  has been observed at the room temperature, indicating a much stronger bond between element 114
  and gold.

See also the earlier: Chemical characterization of element 112 :

  We therefore conclude that the stronger adsorption interaction of element 112 with gold involves formation of a metal bond, which is behaviour typical of group 12 elements


Salt solubility and solvent dissociation A textbook, describing a qualitative test for the $\ce{Al^3+}$ cation in which you use $\ce{NH4Cl}$ and $\ce{NH4OH}$ to precipitate $\ce{Al(OH)3}$, mentioned that the solubility of this precipitate decreases in the presence of the ammonium salt.
I sort of reasoned it out to the common-ion effect decreasing the $\ce{OH-}$ concentration before I wondered why that would have an effect at all. As far as I know, the solubility of a salt only varies with the solvent used, the pressure, and the temperature (and of course, on the common-ion effect if it's present).
I made up an explanation, that a more dissociated solvent wouldn't have to physically contort to solvate the salt ions; that when the bonds are broken, the ions of the solvent can more freely arrange themselves in optimal positions to stabilize the salt constituents.
Is that line of reasoning correct? If not, how can the relation between solvent dissociation and salt solubility be rationalized?

Common ion effect is the key.
A solution of $\ce{NH4OH}$ and $\ce{NH4Cl}$ is basically(no pun intended) a basic buffer solution, which has an almost  constant $\mathrm{pH}$ and thus a fixed concentration of hydroxide ions.
The solubility product of $\ce{Al(OH)3}$ as you say is also constant. So as to maintain the constant nature of both $[\ce{OH-}]$ and $K_{\textrm{sp}}$, there will be a fixed equilibrium concentration of $\ce{Al^3+}$ for a given $\mathrm{pH}$ above which it will start precipitating as $\ce{Al(OH)3}$. By increasing the hydroxide concentration we are forcing the $\ce{Al^3+}$ cation to precipitate earlier than it would have done in a normal, neutral aqueous solution.
This process of forcefully changing the $\mathrm{pH}$ of a solution to alter the solubility of different compounds is very common. In this case it was used to reduce the solubility and precipitate a substance, in other case it might be used to increase the solubility.
For example, during the same qualitative analysis schedule you mentioned, $\ce{HCl}$ is added alongside $\ce{H2S}$ during the detection of group 2 cations(like $\ce{Cu^2+, Hg^2+}$) to prevent precipitation of group 4 cations(like $\ce{Zn^2+, Mn^2+}$) in the form of sulphides. As the acidic conditions increase the solubility of sulphides, added with the fact that group 4 sulphides are not that insoluble prevents them from precipitating with the group 2 cation sulphides.


Solubility comparison of hexan-1-ol and cyclohexanol in water Is hexan-1-ol or cyclohexanol more soluble in water? The answer is supposedly cyclohexanol. However, I found a statement in a book:

  Isomeric alcohols have solubilities in the order: primary alcohol > secondary alcohol > tertiary alcohol, due to decrease in polar character.

By this statement, I guessed the answer to be 1-hexanol, as it is a primary alcohol whereas cyclohexanol is a secondary alcohol. What mistake have I done?

In general is very difficult to predict solvatation some experimental models are used such as Abraham solvation equation.
I suppose that the mistake you have done is the fact that you haven't consider one of the most important factor: the molecule structure.

In this case you are comparing a cyclic compound with a linear one, if you could have a look to the potential surface of the two molecules you will see something like this:

What makes the alchol soluble is the hydrophilic hydroxyl group while the lipophilic chain is not. The  ratio between idrophilic and lipophilic area is greater in the cyclohexanol compared to 1-hexanol.


Aromatic enol is not possible in...? How to solve this kind of problems. What determines their stability?

This is a strange question - isn't it possible to form an aromatic enol in all 4 cases? #3 is the least obvious, but will still proceed:

Maybe I'm misunderstanding the question, vote me down if so!


Expression for fugacity coefficient derived from a pressure-explicite EOS In the book, Thermodynamics for Process Simulation, the authors propose to derive an expression for the fugacity coefficient from a pressure-explicit equation of state, as an example among many others.

Assuming, $P = P(T, v)$ or $z =\frac{Pv}{RT} = z(T, v)$, the calculations lead to,

$$\left(\frac{\partial \ln \phi}{\partial v}\right)_T = \left(\frac{\partial z}{\partial v}\right)_T-\left(\frac{\partial \ln z}{\partial v}\right)_T + \left(\frac{1}{v}-\frac{P}{RT}\right) \tag1$$

where $\phi$ is the fugacity coefficient.

What I don't understand is, when they integrate, they get,

$$\ln \phi = z - 1 - \ln z + \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v}-P\right)\mathrm{d}v \tag2$$

Can someone explain me how they find $z-1-\ln z$ by integrating the first two terms of $(1)$ between $\infty$ and $v$?

When integrating the first two terms, it leads to $$\left[z - \ln z \right]_{\infty}^v$$ which I understand, but then I don't get the final result.

I suppose it's linked to the values of $z$ when the volume is very large and when the volume is "$v$" but it does't make sense for me.

I would have assume to get $$\left[z - \ln z \right]_{\infty}^v = v - \ln v - \left[z(v^{\infty}) - \ln z(v^{\infty}) \right]$$ and nothing else.

But I am missing something probably obvious.

Thank you in advance for your help!

Mindful of the guideline "don't give answers in comments", I've converted my comment to an answer, trivial though it is.

When $v\rightarrow \infty$ you get the ideal gas eos, so $z\rightarrow 1$. This gives you the result, i.e. $1$ at the lower limit (subtracted) and $z-\ln z$ at the upper one.

When you write down the integrated form, $z-\ln z$, you need to remember that it means that you are evaluating the function $z(v)$ at the two limits $v=\infty$ and $v=v$, and using those values in that expression; you should not be trying to set $z=v$ at the limits, which seems to be (partly) what you have written at the end of your question.

Edit following OP comment.

More details. At constant $T$, $z=z(v)$ is a function of molar volume $v$,
determined by the equation of state, $P(v)$ at the given value of $T$. This equation is unknown, but we can be sure that,
in the ideal gas limit $v\rightarrow\infty$, $z(v)\rightarrow 1$.

Integrating both sides of the equation from the ideal gas limit to
the desired volume $v$, and using $v'$ as the integration variable, gives
eqn (2) of the question in its full form:
$$
\left| \ln \phi(v')\right|_{v'=\infty}^{v'=v}
= \left|  z(v') - \ln z(v') \right|_{v'=\infty}^{v'=v}
+ \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v'}-P(v')\right)\mathrm{d}v'
$$
I think it's clearer to distinguish between the integration variable $v'$
and the "upper" limit of integration $v$,
but many people would be happy just to use $v$ instead of $v'$.

Anyway,
on the left,
we know that, in the ideal gas limit,
the fugacity coefficient $\phi(\infty)=1$,
so $\ln\phi(\infty)=0$ and we are just left with $\ln\phi(v)$.
On the right,
similarly,
we substitute in the upper and lower limits for $v'$.
We know $z(\infty)=1$, so
the function being evaluated is $z(\infty)-\ln z(\infty)=1$ at the lower limit,
and $z(v)-\ln z(v)$ at the upper limit.
So the final answer is
$$
\ln \phi(v)
= z(v) - \ln z(v) - 1
+ \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v'}-P(v')\right)\mathrm{d}v'
$$
An important point is that the integration variable
is the molar volume $v'$ (or $v$ if you prefer), not $z$.
In evaluating the result, the integration limits $v$ and $\infty$
are substituted for $v'$, the argument of the function being evaluated.
It is incorrect to set $z=v$, or $z=\infty$.
(This should be even more clear in this case,
since $z$ is a dimensionless quantity, whereas $v$ is not).


How is the water escaping? I have a number of toy balls filled with a liquid-presumably water or something else non-toxic. Over about a year, the balls have gone from firm to indented as shown. The water level in one is visibly low but four others have become indented as if some water were removed without allowing air to take its place. There are no indications of any leaks, balls have all been maintained at room temperature.

I posted this to chemistry because my hypothesis is that there is some sort of diffusion through the ball surface or in to the floating toy contained within. In the case of the ball on the right, the toy inside appears to have a hole so maybe it was originally full of air but the liquid has migrated inside pushing the air out and appearing to lower the water level. Explanation doesn't work for the other balls.

Thoughts?

-Engineer Dad

There is a general belief that plastics are impermeable, probably because many liquids are stored in plastic containers. This is, however, not correct. Plastics are permeable to water and also to gases. The permeability is low but is not zero and therefore, when very long periods of time are involved, a perceptible change occurs. If absolute impermeability is required, then metals should be used.

The process involved is diffusion and depends basically on the type of plastic. The ones containing fluorine (Teflon and the like) are the less permeable and the polyesters the most permeable, probably because they are more polar.

What has happened to the plastic balls is that some water has permeated out and in some cases some air has diffused in.


Why does acetyl have negative charge? "Acetylation (adding an acetyl group) and phosphorylation (adding a phosphate group) make the histones more negatively charged because acetyl and phosphoryl groups are negative. They are "glass is half empty" molecules. By making histones more negatively charged, their grip on DNA will be much looser because DNA is also negatively charged. Similar charges (negative and negative) repel one another".

I know oxygen and carbon have different electronegativities but that generates a partial negative charge and a partial positive charge. So why do we consider the negative charge prevails over the positive charge in order to histones "repel" DNA?

The acetyl group itself does not have any significant charge. But the residue that gets acetylated is lysine in this case, which has a positive charge. The acetylation results in a neutral amide, removing the positive charge.

From Wikimedia Commons

I'd add that the effect of histone acetylation is likely more subtle than just the charge effect. This kind of posttranslational modification often affects the protein structure in some way, or is recognized by other proteins interacting with that site.


Acetylation reaction In the Friedel - Crafts acetylation of bromobenzene, how is it possible to obtain a black-coloured product?

When the acetic anhydride was slowly added to the reactants - $\ce{AlCl3}$, DCM, bromobenzene - a small, hard, yellow lump formed.

Would this be $\ce{AlCl3}$?

When the acetic anhydride was added, the reflux was initiated. The lump appeared to disappear, leaving behind a dark red solution.

However, when the reflux was turned off, the solution appeared black. I'm trying to think of potential products, such as ortho and para bromobenzene or acetic acid or $\ce{Al(OH)3}$ but why so black?

Does anyone have any ideas?

I assume "DCM" represents dichloromethane, if so...

There is an old paper that describes the formation of a low molecular weight polymer from the double Friedel-Crafts alkylation of benzene by dichloromethane. That's probably what's happening in your reaction as a minor pathway.  The dissolved polymer and tar is what's producing your black solution.

I actually ran that reaction once, I wanted to prepare that polymer.  It didn't work for me, but at the end, after the aluminum chloride had been removed, I did have a dark solution.


how many electrons are in the second subshell of this element? first of all sorry if I have some mistakes in my "chemistry words", I'm Iranian and I know the Persian terms but only some of the English ones.

So suppose we have an element (x) with the atomic number of 24, so the electron configuration would be something like this :
$_{24}x : 1s^2\ \ 2s^2 \ \ 2p^6\ \ 3s^2\ \ 3p^6\ \ 3d^{5}\ \ 4s^1$

So the question is : in the element above, how many electrons are there in the subshell of l=2 (which is the subshell of d) and how many are there in the l=0 subshell (which is the subshell of s)

P.S. I just took an exam and this exact question was in the exam, my answer was 5 electrons for 'd' and 7 electrons for 's'

You're answer is correct. The s-shell has 2 electrons in the 1s subshell, 2 electrons in the 2s subshell, 2 electrons in the 3s subshell and 1 electron in the 4s subshell. The d-shell has 5 electrons in the 3d subshell.

The element with atomic number 24 is chromium, who's electron configuration can be shown diagramatically as follows:

Note that the 3d shell is 'half-filled' before the 4s subshell is completely filled.

Chromium is one of the 'exceptions to the predicted electron configurations' when applying Madelung's rule for quantum numbers and electron configuration. The 'Aufbau' principle (represented in diagram below) can be used to determine the electron structure of relatively simple atoms, by hand, as long as one remembers the anomalies (such as chromium and copper). The 'anomaly' is typically 'explained' by stating that the a 'half-filled' or 'completely filled' d-subshell is more stable and will therefore happen before 'completely filling' the 4s subshell.

In modern quantum computational chemistry, various numerical methods (such as Dirac Hyper Hartree–Fock method and density functional theory) have been devised to iteratively solve the Schrodinger equation for more accurate calculation of electronic structures of larger atoms and molecules.


What does Ae mean when used in a chemical formula? In this video on slide 38 for example, the 35 minute mark, the formula appears

HAeBe

With the explaination "some chemestry has taken place..." and an infrared spectra.

What is Ae? I tried looking it up thinking it might be a common shorthand, but the only hits are answers that say "no, did you mean Al, Ag, etc.".

HAeBe stands for Herbig Ae/Be star, which is a young star before the main sequence of its formation.

The spectra of HAeBe stars show hydrogen and calcium emission lines, which are shown in the video.

The cloud could be referring to the gas-dust envelope that surrounds the star, but I'm not an expert in this field and so I'm unsure.


Applications of magnesium sulfate monohydrate MgSO4 · H2O How is magnesium sulfate monohydrate $\ce{MgSO4 * H2O}$ used for in chemistry and how does it work as a valuable laboratory drying agent?

I could not find too much information in my book or the internet besides the basics (chemical formula, physical or chemical characteristics, etc.), but I did find what it can be used in. For example, fertilizer, feed additives for livestock, nutrient flavor enhancer, pharmaceuticals, leather and fur production, coloring fabrics, fermentation in beer, paper production, sewage treatment, cosmetics, but nothing about how or why chemistry labs would use it. What is its purpose?

Each time I tried to look it up other products that sounded similar but not the same appeared such as $\ce{MgSO4 * 7H2O}$ , manganese sulfate, manganous sulfate, etc.

Magnesium sulfate is frequently used in the laboratory, especially after aqueous work-up.
Aqueous work-up is a common technique in the lab to get rid of residual impurities after completion of a reaction.
For this the organic solution of the reaction is cooled to room temperature and mixed with water. Specific impurities will then diffuse into the aqueous phase and can be separated with a separatory funnel.
Unfortunately some residual water will stay in the organic phase and this can have negative impact for following characterization of the compound.
Therefore magnesium sulfate is added, which is able to catch the residual water in its crystal lattice. After filtration of the solid magnesium sulfate, the water is reduced to a non-significant amount and the compound ready for further investigation.


Liquid solutions and vapour pressure "The vapour pressure of water at 80° C is 355 torr. A 100 mL vessel contained water-saturated oxygen at 80° C, the total gas pressure being 760 torr. The contents of the vessel were pumped into a 50 mL vessel at the same temperature. What were the partial pressures of oxygen and of water vapour, what was the total pressure in the final equilibrated state? Neglect the volume of any water which might condense."

This question is given in my module and the solution given to the same is quite vague. Could somebody please explain it to me? Thank you very much. What I could particularly not understand is how the water vapour's pressure in the 50 mL vessel remains unchanged.

The key to understanding this type of question is that the equilibrium (saturation) vapor pressure of a pure liquid is to a first approximation (that is, assuming ideal behavior) independent of the presence of other gases, such as (in this case) oxygen. Therefore, you can think of $\pu{355 Torr}$ as being always the pressure of the vapor when in equilibrium with the pure liquid at $T=\pu{80 ^\circ C}$. For equilibrium to be established requires that there is (more than) enough substance to fill the volume with vapor at that pressure. The minimum amount of substance which will fill the volume with vapor at that pressure is, under the assumed ideal conditions, given by the ideal gas law: $$n=\frac{pV}{RT_b}$$where in the present problem $T_b=\pu{80 ^\circ C}$, $p=\pu{355 Torr}$, and V is the volume occupied by the vapor, which changes from $V<\pu{100 mL}$ to $V<\pu{50 mL}$. In the present problem you don't know how much of the water is initially a liquid, only that the inequality $V<\pu{100 mL}$ is satisfied by the vapor. During the volume reduction some of the vapor condenses into additional liquid in order to retain the equilibrium pressure. The oxygen is treated here as a spectator, although it contributes to the total pressure in the vessel. In practice it also alters the vapor pressure of the liquid, although here by a negligible amount.


Role of NaI in Formation of Silyl Enol Ethers using Chloro(trimethyl)silane I am forming silyl enol ethers using Chloro(trimethyl)silane (TMSCl), and below is an experimental protocol I am following

  Silyl Enol Ether Synthesis

  In an oven-dried 100 mL three-necked flask, Et3N (37.2 mmol) and carbonyls (30.0 mmol) were added. The mixture was stirred and heated to 30~35 ºC, and TMSCl (37.2 mmol) was added dropwise over 15 minutes. The mixture was stirred at the same temperature for 0.5 hour.
  A solution of NaI (37.2 mmol) in acetonitrile (35 mL) was added dropwise to the mixture over 0.5 hour and stirred for 1 hour. The mixture was then heated to 40-45 ºC and stirred at this temperature for 2 hours.
  The mixture was then cooled to room temperature and 30 mL of cold pentane was added. The mixture was poured into 30 mL of ice-cold water and extracted with pentane (15 mL × 2). The combined organic phases were dried over anhydrous Na2SO4, filtered and concentrated in vacuo. After distillation under reduced pressure, the products were obtained as colorless oils and stored at 0 ºC under an atmosphere of nitrogen for use.

I don't understand the mechanistic background of Step 2. What does NaI do?

Side question: when adding TMSCl in Step 1, I saw some white fume. I am guessing TMSCl was reacting with moisture in the air but what would be the product of H2O+ TMSCl? Is it HCl and TMS-OH?

The use of $\ce{NaI}$ in $\ce{MeCN}$ (a solvent in which $\ce{NaI}$ is quite soluble) is forming $\ce{Me3SiI}$ from the $\ce{Me3SiCl}$. The reason for doing this is $\ce{Me3SiI}$ is significantly more reactive with the triethylammonium enolate and will increase the yield of the TMS-enolate.

You are correct in thinking that the white fume you see when adding the $\ce{Me3SiCl}$ is $\ce{HCl}$. The product of aqueous hydrolysis of $\ce{Me3SiCl}$ is $\ce{HCl}$ plus hexamethyldisiloxane (Wikipedia):

$$\ce{2Me3SiCl + 2H2O -> 2HCl + Me3SiOSiMe3}$$


Electrolysis with two similar metals I've researching electrolysis recently, and all of the examples I've seen use two different metals as the cathode and anode.
In my attempts to make a hydrogen producing cell (table salt used as the electrolyte in water) I've used two steel nails as both my cathode and anode. When I power it up, one side bubbles and the other stays normal. The side where the bubbling occurred, corrodes and makes the water murky with what I presume to be rust. The other side does nothing, not even get electroplated with the other side's metal.
My question is this: Does electrolysis work the same way when the cathode and anode are made of the same metal? It might be a stupid question, but I can't find anything that answers this question specifically.

My question is this: Does electrolysis work the same way when the cathode and anode are made of the same metal? It might be a stupid question, but I can't find anything that answers this question specifically.

The battery or the power source does not care about the electrode material. Its job is to pump the electrons in the circuit. Now the problems originate from the chemistry rather. By definition, if a direct current passes through an electrolyte, electrolysis (=chemical decomposition) has to occur. Something has to reduce at the cathode and something has to oxidize at the anode. Here, we enter the competition, as shown below in the case of your simple aqueous NaCl set up.
At the anode (the electrode connected to the positive terminal of the DC supply), you have to think...
(a) Will the electrode oxidize itself? Possible
(b) Will the solvent i.e., water oxidize to oxygen? Possible
(c) Will the chloride ion oxidize to chlorine? Possible
(d) Na+ ions, they should be repelled
(e) Can the electrode react with the electrolytic product? Very much possible
Similarly at the cathode (the electrode connected to the negative terminal of the DC supply), think about ...
(a) Can the electrode reduce itself? If it is a metal, it cannot reduce further.
(b) Will the solvent i.e., water reduce to hydrogen? Possible
(c) Can the Na+ ions be reduced to metallic sodium? Possible, but not in plain water.
(d) Cl- should be pushed away.
(e) Can the electrode react with the electrolytic products? Very much possible
So you can see here, that depending on the electrode material, yes it can be oxidized if used an anode. It can also be oxidized indirectly, if the evolving chlorine begins to attack the metal, which happened in your case.
In any case, electrolytic cell does not care about the electrodes, the electrodes can certainly be the same but they have to be inert. People always choose inert electrodes so that the metallic cathode or anode do not interfere chemically during the electrolysis. Platinum is a good choice but it is too expensive for home experiments. Carbon/graphite electrode is another choice. As you have seen, stainless steel is not a good choice as an anode. It is readily oxidized by chlorine.


What is the electron count in this nickel complex? Our class has started learning about electron counting using the ionic method. I was having a little difficulty, especially when there are two metals in one complex, so I looked at Wikipedia for help.

The Wikipedia page for organonickel complexes says:

  In $\ce{(allyl)2Ni2Br2}$ and $\ce{(allyl)Ni(C5H5)}$, nickel is
  assigned to oxidation number +2, and the electron counts are 16 and
  18, respectively.

I understand that in $\ce{(allyl)Ni(C5H5)}$, the allyl and cyclopentadienyl ligands both have a $-1$ "charge", so the nickel has a +2 "charge". Thus, the electron count is: $4 + 8 + 6 = 18$.

However, I am struggling to calculate the electron count in $\ce{(allyl)2Ni2Br2}$. Using the same logic, $\ce{Ni}$ is in the +2 oxidation state. Since the structure (see below) does not have a metal-metal bond, we do not have to add $1$ when counting electrons. So, I think the electron count should be $\frac{4*2 + 8*2 + 2*2}{2} = 14$, not $16$. How does Wikipedia get that $\ce{Ni}$ has an electron count of $16$ in $\ce{(allyl)2Ni2Br2}$?

The bromide ions contribute an electron pair on each side of the square they form with the nickel centers (thus the bromine has a positive formal charge like a bridged bromonium ion), so in your fraction you need $4×2$ in the numerator where you have $2×2$.  Then it will come out to $16$.


Point group of cis- and trans-decalin What is the point group of of cis- and trans-decalin?

I watched a video saying that it is $C_\mathrm{2h}$ for cis-isomer, but my friend says it is $D_\mathrm{2h}.$ I'm so confused.

cis-Decalin

Here is a view of cis-decalin with the bridging C-C bond in the plane of the paper:

(It was made using the Jmol software, e.g. at https://chemapps.stolaf.edu/jmol/jsmol/simple.htm using the "Load by name" button and entering "cis-decalin".) At https://www.chemtube3d.com/stdecalinflips/, you can see that this can flip into the mirror image (much like a cylcohexane chair flip).

  What is the point group of of cis-decalin?

I see a two-fold axis (perpendicular to the paper). I don't see a center of inversion or any mirror planes; the molecule looks chiral (the flip mentioned above changes it to its mirror image). So maybe C2?

trans-Decalin

Here is a view of trans-decalin with 5 carbons closer to us and 5 further away from us:

I see a center of inversion, a mirror plane (left:right swap), and a two-fold horizontally in the plane of view. I guess that's C2h.


Finding $c$ with Dulong-Petit law on non-molecular substance The Wikipedia article states the Dulong-Petit law as follows:

$$c=\frac{3R}M$$

$c$ is specific heat capacity $[\mathrm{\frac J{kg\; K}}]$ and $R$ the gas constant. $M$ is the molar mass in [g/mole], and "grams per mole" makes us think of a number of moles of a substance, as were it a molecular material.

In crystal structures there is no such thing as a "mole" of the substance. An example could be $\ce{CoSb3}$. $\ce{CoSb3}$ is not a molecule, but the simplest formula unit of the endlessly repeating crystal. So which $M$ to plug into the Dulong-Petit formula is not obvious here.

A researcher gave me the following version of the Dulong-Petit formula for this case (which I have used successfully):

$$c=\frac {3R}{M}n$$

The $M$ still means molar mass [g/mol], treating the formula unit as were it a molecule.
The $n$ is the number of atoms in the formula unit. So here, $n=4$. Simple.

I do not see how this version is equivalent with the former version. How does "treating the formula unit as was it a molecule" balance out with "multiplying with the number of atoms"? How is

$$\underbrace{\frac{3R}M}_{molecular}\Leftrightarrow \underbrace{\frac{3R}Mn}_{crystalline}$$

equivalent?

It may be useful to start with the specific molar heat capacity $c'$ (per mole, instead of per kilogram), which is somewhat more fundamental. Here $c' = 3R$.

This is true even for your crystalline solid, though we have to decide what "moles" refers to in this case. From a physical perspective, the heat capacity is a result of vibrational motion in three dimensions, and each atom will contribute to this motion. Our "moles" should therefore be "moles of atoms"---for the purposes of this calculation, we consider the atom to be the fundamental unit.

The molar mass should then be $\text{mass of solid/moles of atoms}$, which can be interpreted as a weighted molar mass. You can see that your formula $M/n$ calculates exactly this: the mass per mole of a formula unit, divided by the number of atoms in a formula unit.

Hence $$c = \frac{3R}{M/n} = \frac{3R}{M}n.$$


Stereoisomers of octahedral co-ordination compounds of the form [Ma3b3] Does $\ce{[Co(H2O)3Br3]}$ show geometrical and optical isomerism?
According to me $\ce{Co}$ should be $\mathrm{sp^3d^2}$ hybridised giving an octahedral geometry. It is of the form $\ce{[Ma3b3]}$ type. I know it can show geometrical isomerism, namely fac and mer isomers. But can it show optical isomerism?

According to me $\ce{Co}$ should be $\mathrm{sp^3d^2}$ hybridised giving an octahedral geometry.

No, please don’t do this. The most accurate description of cobalt here would be $\mathrm{d^5}$, maybe $\mathrm{d^5s}$. But don’t extend the hybridisation concept from main group elements to the transition metals; it doesn’t belong and it doesn’t help here. Rather, there are six ligands around cobalt, so you have an extremely high chance for octahedral geometry. Which is also the case.

  I know it can show geometrical isomerism, fac and mer.

That’s a good start.

  But can it show optical isomerism?

When asking about optical isomerism, ask yourself, if the compound is superimposable on its mirror image. One method to check this is to draw the compound, draw its mirror image and rotate either until they overlap — or until you think they cannot. The second method is much more efficient though: Still draw them, but check whether you can find a centre of inversion symmetry, a plane of symmetry or an axis of improper rotation (turn by $\frac{360^\circ}{n}$, where $n$ is a natural number and then mirror the result through a plane perpendicular to the rotation axis). If you can find any of these, the compound is automatically identical with its mirror image, and if there are none of these, then it displays optical isomerism. Remember to consider 3D structures.

It should be rather simple to determine this for the mer isomer. For the fac, consider a view where three ligands of the same type are above the paper plane, the central cobalt occupies the paper plane and the three ligands of the other type are below it. For both considerations, you can ignore the two hydrogens on each oxygen.

The results are given in a spoiler (mouse over to see).

   The mer isomer has planes of symmetry that transect three meridional atoms of one kind and the middle atom of the other three.

 The fac isomer has planes of symmetry that transect the cobalt, one ligand of each kind and the middle between the other two ligands of each type.

 Therefore, neither displays optical isomerism.


Reliable automatic addition of hydrogens and charge assignment to small molecules at given pH? I have some PDB structures of small molecules with missing hydrogens. Is there a software available that places missing hydrogens and assigns a charge to the molecule that is reliable? I am looking for an automatic tool that returns a PDB with added Hydrogens and the overall charge (I can do some programming if I need to combine output from multiple programs to accomplish this or something.)

It is enough for me if you suggest a couple of FREE tools (I don't have resources to buy a commercial license, but I am academic, in case they offer free academic licenses), with associated papers that evaluate their performance that I can cite. For example, a free tool that calculates pKa values of ionizable groups would solve my problem, if you can also suggest a tool that automatically detects ionizable groups in molecules.

I have used Avogadro to add hydrogens (though I don't know how to compute the charge with it), but I've found that some hydrogens aren't well placed. Is there a better tool available for this task?

If you're looking for comprehensive protonation and evaluation of $pK_a$, I suggest my answer to a previous question. There is specialized software for evaluating protonation and $pK_a$ based on large databases.

Avogadro has some simple rules, mostly for amino acid residues, but is not intended for automated evaluation of thousands of molecules.


Is toothpaste solid or liquid? My teacher didn't answer this properly:

  Is toothpaste solid or liquid?

You can't say toothpaste is a solid because solid material have a fixed shape but toothpaste doesn't. However, you can't say it's a liquid because liquids flow easily but toothpaste needs a certain force to push it out of the tube. So is it a solid or liquid? And are there any other example just like toothpaste?

Toothpaste is what is called a non-newtonian fluid, more specifically toothpaste is a Bingham plastic.  This means that the viscosity of the fluid is linearly dependent on the shear stress, but with an offset called the yield stress (see figure below). This yield stress is what makes it hard to say whether it is liquid or solid. The fact that toothpaste is viscous alone is not sufficient to explain this, because water is also viscous, but doesn't behave like a solid (unless frozen, but that's another phenomenon).

What the yield stress does is the following. Below a certain shear threshold the fluid responds as if it were a solid, as you can see happening when you have put toothpaste on your toothbrush, it just sits there without flowing away. A highly viscous but newtonian fluid would flow away (although slowly as pointed out by @ron in his comment to the answer of @freddy).

Now if you put sufficient shear stress on the toothpaste, when you squeeze the tube of paste, it will start flowing and respond as a liquid.

Other examples, as mentioned in the Wikipedia link in my first sentence, are e.g. mayonnaise and mustard. Another example is silly putty.