Limiting Reagent Stoichiometry What mass of precipitate forms when a solution containing 6.24 g of potassium sulfide is reacted with a solution containing 19.2 g of barium nitrate?

I have already identified the limiting reagent $\left(\text{K}_2 \text{S}\right)$ as well as the mass of the precipitate.

My question, however, is: why is the Barium Sulfide formed in the product a solid and not aqueous? I thought the problem translated into the molecular equation: $$\text{K}_2 \text{S}_{\text{(aq)}} + {\text{Ba}(\text{NO}_3)_2}_{\text{(aq)}} \ce {->} 2{\text{KNO}_3}_{\text{(aq)}} + \text{BaS}_{\text{(aq)}}$$

But according to the answer key I was given, $\text{BaS}$ is a solid precipitate, not aqueous. How can that compound be a solid if, according to solubility rules, all sulfides plus an alkali earth metal are soluble? Shouldn't it be aqueous, not solid?

Please help, thanks :)

So basically this problem is bad and incomplete (they tend to be at this level) in that it doesn't give you the concentrations/volumes of the reagents/solutions, so we end up making several assumptions. First of all, precipitates cannot be aqueous, by definition they are solids. So the way you end up solving this question is by assuming one of the products precipitates out completely, and it is the least soluble one that precipitates out (barium sulfide in this case). Unfortunately, this isn't exactly accurate.

In a real world scenario, given enough water you could have no precipitate at all, but most likely you would have a mix (not all of the barium sulfide would precipitate out of solution). At higher levels of chemistry (AP/IB -> post-secondary) you'll learn that none of these reactions (including the solubility ones) ever go to completion (some just go so far that you can assume they go to completion), and what you end up doing is using solubility constants to find out how much of each product is dissolved and/or precipitated. If you're interested in more info google "solubility constants" and "common ion effect".


Can I normalize my data in order to compare it? I have three chromatograms of the same eluted protein but at different concentrations, since it got diluted during purification process) and hence peak maximums at very different absorbances.

The highest is around 100 AU and lowest 20 AU.
I want to present these and look at the difference in peak shape (because I want to look at aggregation). Can I simply multiply the data that generated the peak max at 20 AU by 5? Because then it is easier to compare the shapes of the peaks when they  overlay each other.

Is this something one would do or is it totally forbidden?

Thanks in advice.

I assume you have a UV/Vis absorbance based measurement. Then the absorbance should follow Beer´s Law which states that absorbance is proportional to concentration. Therefore you should be fine with your normalization approach. You could check if you are in the linear range by performing a calibration.


If glucose can react with phenyl hydrazine to form osazone why can't it react with 2,4-DNP to give positive test? If glucose can react with form phenyl hydrazine to form osazone why
  can't it react with 2,4-DNP?

Yes I know about cyclic form of glucose and its equilibrium with straight chain and all …

I read this, this and this Stack Exchange one too Does glucose react with Brady's reagent?

Still, didn't get an answer to my question. So requesting for clarification …

Although glucose 1a exists principally in the hemiacetal pyranose forms (α and β), nonetheless it reacts with one equivalent of phenylhydrazine to form the hydrazone 1b via the aldehydic open form. Two additional equivalents serve to form the osazone and the products of reduction of phenylhydrazine, ammonia and aniline.The glucose 2,4-dinitrophenylhydrazone 1c has been reported1 but not the corresponding osazone. In their studies on the mechanism of osazone formation, Shemyakin and coworkers2 utilized p-nitrophenylhydrazine labeled with nitrogen-15 at the terminal nitrogen (the red N in the structures). Although glucose was not examined, the p-nitrophenylhydrazones  of fructose (forms the same osazone as glucose), benzoin and 2-hydroxycylohexanone were studied. The labelled nitrogen was located in the ammonia that was liberated during osazone formation.
Structures 2-7 represent the essentials of the mechanism.  Enolization of the hydrazone 2 produces the enol 3, which, via the zwitterion 4, undergoes oxidation of the carbon framework and reduction of the phenylhydrazine moiety. The transformation 3 --> 4 is essential. The basicity of the electron pair on the nitrogen in the 2,4-dinitrophenyl case is sufficiently reduced by delocalization into the aromatic ring, thereby repressing the proton transfer. This step must be completed before osazone formation can be accomplished.

1: Lloyd, E. A. and Doherty, D. G., J. Am. Chem. Soc., 1952, 74, 4214.
  2: Shemyakin, M. M., Maimind, V. I., Ermolaev, K. M. and Bamdas, E. M., Tetrahedron, 1965, 21, 2771.


Sulfate and carbonate ions as ligands The pentahydrate salt of copper (II) sulfate exists as a blue solid while the anhydrous and monohydrate salts appear as white solids. The structures of these salts are discussed here. The blue colour of the copper (II) sulfate pentahydrate salt arises due to a similar reason why $\ce {Cu^2+}$ in aqueous solution is blue in colour. Based on Crystal Field Theory, this is due to the phenomena of d orbital splitting in the octahedral complex. As discussed in the post linked, in the pentahydrate salt, the $\ce {Cu^2+}$ ion is octahedrally coordinated, with 4 $\ce {H2O}$ molecules forming dative covalent bonds with it while the other two sites are occupied by the oxygen atoms of the sulfate ion. This leads me to wonder if the sulfate ions contribute to any significant degree of splitting of the d orbitals of the $\ce {Cu^2+}$ ion. As the colour of the anhydrous $\ce {CuSO4}$ salt is white, we can infer that the sulfate ions do not contribute to the d orbital splitting. Consequently, we can infer that the sulfate ions do not really form dative covalent bonds with the $\ce {Cu^2+}$ ion.

But why would this be the case? In the sulfate ion, there is rather poor delocalisation of the lone pair due to long $\ce {S-O}$ bond lengths and thus, ineffective p orbital overlap between the p orbitals of the sulfur and oxygen atoms. Hence, the negative charges, as well as the lone pairs of electrons, are rather localised on the oxygen atoms. This analysis all points to the conclusion that the sulfate ion should be able to act as a ligand.

Similarly, the fact that anhydrous, neutral $\ce {CuCO3}$ exists as a gray solid, suggests that the $\ce {Cu^2+}$ ion does not have dative covalent bonds with the carbonate ion , i.e. $\ce {CO3^2-}$ does not act as a ligand. Although for the carbonate ion, due to the stronger delocalisation of electrons within the ion, it may be more convincing to say that it may not be able to act as a ligand effectively.

First and hopefully obvious things first: sulphate ions do form (weak) coordinate bonds to metal centres and thus by definition cause some d orbital splitting. In a simplified crystal field model, you might imagine them as weaker negative charges but nonetheless the do interact or we would not see the structures we see.

Comparing the structures of anhydrous $\ce{CuSO4}$ and its pentahydrate is very much non-trivial. For starters, they don’t even share the same surrounding. While the copper in $\ce{CuSO4 . 5 H2O}$ is approximately in a $D_\mathrm{4h}$ environment with $\ce{Cu-OH2}$ bonds of $\pu{197pm}$ and approximately $\pu{195}$ for the two different copper atoms in the unit cell, the environment of copper in anhydrous sulphate is better described as $D_\mathrm{2h}$ because the $\ce{Cu-O}$ distances average out to $\pu{187pm}$, $\pu{215pm}$ and $\pu{236pm}$, respectively—around the same copper atom. Two symmetry-identical oxygens with a $\ce{O-Cu-O}$ angle of approximately $180^\circ$ will be equidistant, all others will not.

That already means the corresponding molecular orbital schemes will look very different, even if we treat the oxygen atoms identically. Group theory tables tell us, that while in the case of the pentahydrate at least some degenerate orbitals remain, all orbitals will be independent in the case of the anhydrous sulphate. This in turn means that we are dealing with a potentially much more complicated MO scheme and all excitation discussions will get progressively more complicated. I would not be willing to base an argument on low coordinative ability of sulphate only on the difference between these two crystal structures.

Finally note that the carbonate $\ce{CuCO3}$ itself opens another can of worms: as Seidel et al. show, the copper is in a pentacoordinated environment so we cannot even use our usual octahedral arguments there.

References:

Structure of $\ce{CuSO4 . 5 H2O}$: J. N. Varghese, E. N. Maslen, Acta Cryst. B 1985, 41, 184–190. DOI: 10.1107/S0108768185001914.
Structure of $\ce{CuSO4}$: P. A. Kokkoros, P. J. Rentzeperis, Acta Cryst. 1958, 11, 361–364. DOI: 10.1107/S0365110X58000955.
Structure of $\ce{CuCO3}$: H. Seidel, H. Ehrhardt, K. Viswanathan, W. Johannes, Z. Anorg. Allgem. Chem. 1974, 410, 138–148. DOI: 10.1002/zaac.19744100207.


What is the volume occupied by the atoms and the edge length in these cubic lattices? Consider a cubic unit cell of crystal that is composed of a single kind of atom. By placing atoms at every corner of the cube, a simple cubic lattice is formed. Assume that the atoms are perfect hard spheres with a radius $r$ and that the atoms are in close contact to minimize the volume of the cube.
In a simple cubic lattice, the volume occupied by the atoms is $(?)$ % that of the cube, and the length of the edge of the cube is $(?)r$.
To the simple cubic lattice described above, atoms are added to all face-center positions of the cube. In the resulting lattice, the volume occupied by the atoms is $(?)$ % of the cube, and the length of the edge of the cube is $(?)r$.

For the first one, I tried drawing a cube with quarters of spheres at each of its corners and finding the % by calculating the total volume of spheres over the volume of the cube, yet the answer isn't accurate. I get the length of the edge as $2r$.

This is the case of CCP (simple cubic closed packing) and FCC (face centered cubic closed packing)
For CCP
Imagine sitting in a cubical room. Now place some big spheres on each of the vertices of the room such that the centre of sphere and the vertices coincide.
You will see that some part of sphere goes outside the room itself. In chemistry, this is called sharing.
If your room is surrounded in all directions by other rooms, you share one sphere with 7 other rooms. (top 4 and bottom 4)
Thus a single particle gets shared by 8 rooms. The part in your room is 1/8
And the number of such particles are 8 (as cubic room has 8 vertices)
Net particles in your room $= \frac{1}{8} \cdot 8 = 1$
So in CCP the space occupied by all eight particles is basically equal to that of a single particle.
Now as assumed in the question, the particles are tightly packed, the
Radius of sphere equals half of the edge length. In other words edge length is twice the radius of the sphere.
$$\pu{Packing efficency = \displaystyle \frac{\text{Volume of sphere}}{\text{Volume of cube}}}$$
$$= z \cdot \frac{\frac{4}{3} \pi r^3}{(2r)^3} $$
(Here $z$ is net particle contribution; for the current example it is equal to 1)
Calculating this you'll get packing efficiency $\approx 52.4%$

For FCC
Now imagine putting one more sphere in the room, but this time, at the centers of the face. (Do this for each face). Now the particle in center gets shared only by 2 rooms. Yours and the room which is opposite to you.
Therefore, each particle has net contribution of half.
Since a cube has six faces, the net contribution of all spheres at the center the of sphere are
$$\frac{1}{2} \cdot 6 = 3$$
Add the already existent corner particles to it (whose net contribution is 1 as we know)
Thus the total number of particles in FCC are $3 + 1 = 4$

If you're good at imagining things you might have observed that now spheres won't fit in until you reduce their sizes. So we must falsify the assumption that twice the radius of sphere is the edge length. Spheres have shrunk.
I want you to imagine the face of the room now.
There is little space between two spheres on corner but not between the center sphere and corner spheres. Draw a diagonal line in the sphere passing through the diameter of central sphere. Now using Pythagorean theorem:
$$ \pu{(\text{Edge of cube})^2 + (\text{Edge of cube})^2 = (\text{Diagonal of cube})^2}$$
Let edge be 'A':
$$(A)^2 + (A)^2 = D^2$$
$$2(A)^2 = D^2$$
Since the center spheres have no space with corners ones. Therefore
4(radius of sphere) = Diagonal length
$$\sqrt{2}A = D$$
$$\sqrt{2}A = 4r$$
$$\sqrt{2}A / 4 = r$$

There you go. You've found the radius of sphere. Now put it in packing efficiency formula and take $z$ = 4
You will find it somewhere near 72%
On a sidenote: Observe that simple cubic has only 1 sphere which takes up nearly 50% of space of cube. While face centred has 6 + 8 = 14 smaller spheres which take up nearly 70 % of all space in cube. It's like comparing apples and bananas.
You might think that FCC is better than CCP as it uses space judiciously. But since in CCP, the particle size is large therefore CCP is best suited for particles of that size and FCC for smaller sizes.
Although we can discuss hexagonal packing form with CCP.


How to prove that the orientation of the atomic orbitals in the superposition $\psi= a\psi_{1} + b\psi_{2}$depends on the coefficients $a$ & $b$? Hybridisation is a quantum superposition of orthogonal atomic orbitals; for instance
$$\newcommand{\k}[2]{\langle #1 \lvert #2 \rangle}
\psi_{sp^3}= c_1\psi_{2s}+ c_2\psi_{2p_{x}} + c_3\psi_{2p_y} + c_4\psi_{2p_{z}}.$$

It is a quantum superposition which implies the coefficients $c$ to be the probability amplitudes for finding the electron in respective atomic orbital provided that it is prepared in $\lvert\psi_{sp^3}\rangle$(provided $\sum_{i=1}^4 c_1^2 =1$) that is, $$c_1= \k{\psi_{2s}}{\psi_{sp^3}}, \\ c_2= \k{\psi_{2p_x}}{\psi_{sp^3}},\\ c_3= \k{\psi_{2p_y}}{\psi_{sp^3}},\\ c_4= \k{\psi_{2p_z}}{\psi_{sp^3}}.$$

Up to this, I could easily understand this as it was just like some other superposed state often encountered in QM like $\lvert\phi\rangle = \lvert i\rangle c_i + \lvert j\rangle c_j + \lvert k\rangle c_k$ where $c_i = \k{i}{\phi}.$

But, here in the case of hybridisation, they are much more than mere comples numbers that are the amplitudes of the electron to find it in one of the base states that is the associated atomic orbitals when being prepared at the hybridised state. It is evident from Peter Atkins' words:

  The coefficients in the hybrid have been chosen to give the correct directional properties of the hybrid. The squares of the coefficients give the proportion of each atomic orbital in the hybrid.

Also, the snapshot$^1$ below shows how for different values of coefficients, the spatial orientation is changed.

That means the coefficients  direct the spatial orientation of the atomic orbitals that are the base states for the superposition or hybridisation.

Is there any mathematical deduction that shows the orientation of the atomic orbitals in the superposition depends on the coefficients $c_i$?

I'm basically not understanding how these coefficients, which are rather probability amplitudes, would involve in determining the spatial orientations of the atomic orbitals. Can anyone explain me why these probability amplitudes influence the orientation of the atomic orbitals? Also, what is the mathematical proof of this?

$^1$The snapshot, above, is taken from Principles of Physical Chemistry
 By Hans Kuhn, Horst-Dieter Försterling, David H. Waldeck.

In quantum mechanics states/orbitals are defined within the mathematical framework of the Hilbert space. A Hilbert space is a vector space that generalizes the notion of Euclidean space -  one example of the latter would be the normal two- or three-dimensional spaces you might be used to from basic geometry. So, orbitals are nothing more but abstract vectors in a Hilbert space and you can draw many analogies between them and the more familiar two- or three dimensional vectors you learn about in high school. So, let's think of quantum mechanical states as vectors for now. If you add two vectors, $\vec{a}$ and $\vec{b}$, you get another vector, $\vec{c} = \vec{a} + \vec{b}$, which if $\vec{a}$ and $\vec{b}$ are linearly independent, will have a direction different from either $\vec{a}$ or $\vec{b}$. You can also choose different coefficients for the superposition, e.g. $1$ for $\vec{a}$ and $-1$ for $\vec{b}$ yielding a vector $\vec{d} = \vec{a} - \vec{b}$ which now has a direction different from either $\vec{a}$, $\vec{b}$ or $\vec{c}$. So, the direction of a sum vector changes with the coefficients.

The same principle holds true for orbitals. If you for example think of the $\mathrm{p}_{x}$ orbital as a vector pointing in the $x$ direction, i.e. in two dimensions $\bigl(\begin{smallmatrix}1 \\ 0 \end{smallmatrix}\bigr)$, and of the $\mathrm{p}_{y}$ orbital as a vector pointing in the $y$ direction, i.e. in two dimensions $\bigl(\begin{smallmatrix}0 \\ 1 \end{smallmatrix}\bigr)$, then a superposition of both (with equal coefficients of $\frac{1}{\sqrt{2}}$ to ensure normalisation) will give you a $\mathrm{p}$ orbital pointing in the $\bigl(\begin{smallmatrix}1 \\ 1 \end{smallmatrix}\bigr)$ direction.

But as with vectors you are by no means confined to simply add them together with equal coefficients. For example, if you add $2$ parts of $\mathrm{p}_{x}$ and $1$ part of $\mathrm{p}_{y}$ you get a $\mathrm{p}$ orbital pointing in the $\bigl(\begin{smallmatrix}2 \\ 1 \end{smallmatrix}\bigr)$ direction, and that is clearly different from $\bigl(\begin{smallmatrix}1 \\ 1 \end{smallmatrix}\bigr)$.


$K_c$ vs $K$ in equilibrium reactions I don’t understand why the following question

The decomposition of nitrosyl bromide $(\ce{NOBr})$ proceeds by the following reaction:
$$\ce{2 NOBr(g) <=> 2 NO(g) + Br2(g)} \qquad K = 0.0142$$
Calculate the $[\ce{NOBr}],$ $[\ce{NO}],$ and $[\ce{Br2}]$ when $\pu{10.0 mol}$ of nitrosyl bromine is placed in a $\pu{5.00 L}$ closed vessel and allowed to decompose.

has the answers that it does:

$$
\begin{align}
[\ce{NOBr}] &= \pu{1.585 M} \\
[\ce{NO}] &= \pu{0.415 M} \\
[\ce{Br2}] &= \pu{0.207 M}
\end{align}
$$

From what I understand
$$K = \frac{\prod a_\mathrm{products}}{\prod a_\mathrm{reactants}},$$
where $a$ is the activity of the products/reactants. The activity of a gas is approximately equal to the partial pressure of the gas divided by a reference pressure (usually 1 atm). This yields unitless quantities with the magnitude of the partial pressures of each gas. Thus
$$K = \frac{P_\ce{Br2}\cdot P_\ce{NO}^2}{P_\ce{NOBr}^2}$$
Converting from a partial pressure to a concentration can be done using the ideal gas equation
$$\frac{P}{RT} = \frac{n}{V} = M,$$
where $M$ is concentration. Thus,
$$K = \frac{[\ce{Br2}][\ce{NO}]^2\cdot RT}{[\ce{NOBr}]^2}$$
However, plugging the concentrations given into the solution of this problem doesn’t yield the given value of $K.$
Am I wrong in the way I am approaching this problem, or is the problem itself wrong? If the equilibrium constant was denoted $K_c$ instead of $K,$ the given solution would be correct, but I don’t think it’s correct to assume that $K_c$ is the same as $K.$

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

According to wikipedia,

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

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

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

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

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

What is the problem with this problem?

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

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


Uv vis and fluorescence spectroscopy: sensitivity Is there any reason why fluorescence spectroscopy is said to be more sensitive than Uv-Vis?

Fluorescence is a 'zero background' or absolute type of measurement meaning that single photons can be measured against a 'dark' background so the sensitivity is huge, and limited by the fraction of light caught and efficiency of the detector.

Absorption is a relative measurement as it is the difference in two large numbers, one the light intensity with no sample $I_0$ and the other when the sample is present I. In terms of photons these intensities are huge, say $I_0=10^6$ and a 1% change would reduce the number of photons to $10^6-10^4$; still a large number and so the absorption measurement is looking for the small difference in two large numbers. This measurement is limited by noise in detectors and other electronics.   (Its a bit like trying to weigh the captain of a ship by weighing the ship with him standing on the deck and then when he is not there.)


Why can't one make C60 with a combination of hexagons and heptagons? $\ce{C60}$ has a pentagon-hexagon system. Why can't one make it from a hexagon-heptagon system, or a pentagon-heptagon system for that matter?

Well, let's see. This is not quite a chemical question, though I wouldn't recommend moving it to Math.SE either, because they might have hard time recognizing what's $\ce{C60}$.

Now, let's imagine we're considering locally planar conjugated carbon
structures. Each carbon has to reserve one $p$ orbital for $\pi$-bonding,
which leaves it with 3 bonds. Now, the bonds tend to go in cycles; let's
confine ourselves to 5-, 6-, and 7-membered cycles, because otherwise we'll
have too much of angular strain (not that other cycles are totally impossible, though). So in effect, we are considering graphs
consisting of pentagons, hexagons, and heptagons, with 3 edges meeting at
each vertex. Let's say the numbers of these n-gons are $n_5, n_6$, and $n_7$,
correspondingly.

That's when old Euler kicks in. Polyhedra must meet a certain relation
between the number of faces (F), edges (E), and vertices (V):
$$V - E + F = 2$$
OK, what is the number of faces? Why, that's quite simple: $F=n_5+n_6+n_7$.
Then what is the number of vertices? Apparently, each pentagonal face has
five of them, and so on, so it would be $5n_5+6n_6+7n_7$. But wait, we've
counted each atom three times (for it belongs to three polygons), hence we must divide by 3: $V=(5n_5+6n_6+7n_7)/3$. By similar reasoning, the number of edges would be $5n_5+6n_6+7n_7$, if not for the fact that each edge belongs to two faces and thus is counted twice, hence $E=(5n_5+6n_6+7n_7)/2$.

All in all,
$$
V - E + F = 2\\
{5n_5+6n_6+7n_7\over3} - {5n_5+6n_6+7n_7\over2} + n_5+n_6+n_7 = 2\\
-{5n_5+6n_6+7n_7\over6} + n_5+n_6+n_7 = 2\\
{n_5-n_7\over6} = 2\\
n_5-n_7 = 12
$$

That's pretty much the size of it. You can do more or less whatever you want, as long as you have twelve more pentagons than heptagons. There is no going around it.

Now chemistry kicks in, too, and brings its own limitations. An $sp^2$ carbon wants to have its bonds in one plane, at $120^\circ$ angles. You can convince it to budge, but only so much. If you strain it more, the system becomes unstable. It turns out that two pentagons meeting at one carbon is already too much (not for an $sp^3$ carbon, though; see dodecahedrane). That's why you can't have constructions made of pentagons alone (or pentagons and heptagons, for that matter). Geometry would allow that, but chemistry wouldn't.

Let's see how that translates to actual shapes. Several fused hexagons give us coronene; it's planar.

(It is tilted sideways to give it a 3D feel. There are hydrogens attached on the outer edge; I omitted them for clarity.)

Extend it further, and you'll have an infinite flat sheet; that would be graphene.

Throw in a pentagon, and here's corannulene; it's bowl-shaped.

Extend that pattern, and you'll have fullerene.

Use a heptagon, and get circulene:

See that? It's bent indeed, but it's bent the wrong way. It's a
saddle-shaped bend. An example of negative Gaussian curvature, so to say. You can't wrap it to a sphere, no matter how hard you try.

Can we extend this further? Well, we kinda can (on paper, that is).
Moreover, it can be done in a number of ways, some leading to magnificent structures (purely hypothetical, of course).

Upd. It was brought to my attention that according to some studies, so-called carbon nanofoam might be a real example of structures of the latter kind. I'm not quite convinced (to me, until you have an X-ray diffraction study, everything is just hand-waving), but let's mention that anyway.


Bond stability of C-C in ethane, ethene, and ethyne Which one of the three compounds has the highest $\ce{C-C}$ bond stability?

Ethane has $\mathrm{sp^3}$ hybridisation so less $\mathrm{s}$-character and lesser electronegativity compared to ethyne. Ethyne has high $\mathrm{s}$-character and high electronegativity. Therefore the covalent character of $\mathrm{sp^3-sp^3}$ in Ethyne should be stronger and have more stability right? the order would therefore be $\mathrm{sp-sp}$ >  $\mathrm{sp^2-sp^2}$ >  $\mathrm{sp^3-sp^3}$
Some online sites tell so, but others say since the $\ce{C}$ atoms are so close in ethyne that the nuclear repulsion would make it less stable. So ethane's $\ce{C-C}$ would be stronger and that therefore the order should be reversed
Edit: nutshell- sigma bond of which hybridisation is strongest among $\mathrm{sp}$, $\mathrm{sp^2}$ and $\mathrm{sp^3}$?
I took ethane, ethene, and ethyne just as a example

The truth is: in reality, sigma bonds, bond orders and hybridization do not exist (not even bonds or orbitals do). They are just concepts to explain the reactivity, stability and geometry of a molecule (and in the case of orbitals to build approximate many-particle wave functions). There is thus no way to compare a sigma-bond strength without choosing some model which introduces bias.
The least-biased way to get there is probably starting from the experimental dissociation energy already mentioned. These can be divided by the bond order (which, again, is not a thing, but quite well established):

ethane: 350 kJ/mol -> 350 kJ/mol per bond
ethene: 610 kJ/mol -> 305 kJ/mol per bond
ethyne: 835 kJ/mol -> 278 kJ/mol per bond

In general: for any chosen approach, the sigma bond in ethane will be the strongest even though the bond distance of ethyne is the smallest. The additional electrons of the pi bond(s) destabilize the bond.


Why does charcoal not produce smoke when heated? As I understand; the reason we have smoke in a fire is because the heat causes wood (or whatever material is burning) to dis-integrate, the disintegrated components react with oxygen to create more heat; causing a chain reaction.
Much of the smoke we see is the dis-integrated carbon in the air from incomplete combustion.
However, when we heat charcoal in a vacuum, there is no smoke. But if heat is what causes carbon to dis-integrate, then why does heating carbon (regardless of oxygen presence) not cause smoke?

You say that wood decomposes under heating and produces gases(1) which then burn. That's correct. It would be wrong, though, to assume that all combustible materials are like that. Indeed, many - arguably, most of them are. Wax, tar, and all kinds of plastic come to mind. But carbon is a different story.
See, carbon is very heat-resistant, about as much so as the most heat-resistant thing you know, only more. You heat it to $1000\;^\circ\rm C$, and nothing happens. Well, it may start burning provided there is air around, but short of that, it is nothing. You heat it a few hundred degrees more, it starts glowing red, then white, then steel melts, then you pass $2000\;^\circ\rm C$ and continue to push on, then everything else melts, by now your carbon glows so bright that you can't see a thing, but it is still intact.
Burning of carbon, unlike burning of many other materials, is a true heterophase reaction. You chip atoms off that piece one by one, hence no smoke is produced. Burning of wood or plastic is another story. Combustible gases are produced and fly away, and when some of them do not burn completely(2), they leave a solid residue (which, BTW, may or may not consist mostly of carbon). That's what we call smoke.

P.S. Thought I'd rather add two clarification notes:
(1) When heated, plastic turns into gases all but completely. Wood doesn't. Much of it remains in the form of charcoal, which is mostly carbon and burns like carbon, that is, producing no smoke.
(2) Incomplete combustion of carbon is also possible, and it results in carbon monoxide, which killed quite a number of people over the past few millenia. But it isn't smoke; it's a gas. You can't see it. You just see people dropping dead.
So it goes.


How is the selective titration testing equation simplified? When there is $\ce{N^m+}$ and $\ce{M^n+}$  metal ions in a solution with respective m and n charge,
$\ce{N^{m+}, M^{n+}}$ reacts with EDTA($\ce{Y}$) as follows:
$$\ce{N^m+ + Y^4- <=>  NY^{m-4}}$$
$$\ce{M^n+ + Y^4- <=>  MY^{n-4}}$$
Formation constant for each chemical reaction,
$$\ce{K_{NY^{m-4}}} = \frac{\ce{[NY^{m-4}]}}{\ce{[N^{m+}][Y^{4-}]}}$$
$$\ce{K_{MY^{n-4}}} = \frac{\ce{[MY^{n-4}]}}{\ce{[M^{n+}][Y^{4-}]}}$$
where as $\ce{N^m+}$ has the higher value for formation constent KNYm-4 > KMYn-4
Fraction of each ion exsists ${α}$, $\ce{[N_{total}][M_{total}][Y_{total}]}$ are the total concentraion of the ion in any form.
$\ce{α_{Y^{4-}}}$ = $\frac{\ce{[Y^4-]}}{\ce{[Y_{total}]}}$
$\ce{α_{N^{m+}}}$ = $\frac{\ce{[N^m+]}}{\ce{[N_{total}]}}$
$\ce{α_{M^{n+}}}$ = $\frac{\ce{[M^n+]}}{\ce{[M_{total}]}}$
After $\ce{N^m+}$ has reacted with EDTA,the remaining amount EDTA is used for $\ce{M^n+}$
$$\ce{[Y^{'}_{total}] = [Y_{total}] + [NY^{m-4}]}$$
Where as $\ce{α_Y^{'}}$ is,
$$\ce{α_{Y^{'}}} = \frac{\ce{[Y_{total}]}}{\ce{[Y^{'}_{total}]}}$$
These are all the definition constants but I have no clue how this equation came
$$\ce{K''_{MY^{n-4}} = K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}} - α_Y^{'}}$$
To perform a selective complexometric titration this condition need to be satisfied,
$$\ce{K''_{MY^{n-4}} = K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}} - α_Y^{'} \geq 10^7}$$
where αY' is given below,
$$\ce{α_{Y'}} = \frac{1}{\ce{(1 + K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}])}}$$
Using EDTA as for an example Y4-
How did this equation came with the assumptions (The simplified equation)
$$\ce{K''_{MY^{n-4}}} = \frac{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}}{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}]}$$
One of the assumptions would be
$$\ce{ K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] \gg 1}$$
Addition of one can be neglected.
What are the assumption do need to use to get the simplified expression
Can someone help me just using basic mathematics did not help I tried it by myself but no luck I am pretty sure I have missed some assumptions

I finally found the answer to the question I have made a huge mistake in the testing equation
In order for a titration to happen the Formation constant $\ce{K}$ should be greater than or equal to $\ce{10^7}$
What I have written in my note book is

$\ce{K''_{MY^{n-4}}} = \ce{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}-α_{Y^{'}}}$

Which is incorrect it,
Should be like this

$\ce{K''_{MY^{n-4}}} = \ce{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}α_{Y^{'}}}$

The conditional formation constant of metal $\ce{M}$ should be multiplied by $\ce{α_{Y^{'}}}$ not subtracted
Substituting the equation for $\ce{α_{Y^{'}}}$ will give the simplified equation
$$\ce{α_{Y^{′}}=  \frac{1}{(1+K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}])}}$$
$$\ce{K''_{MY^{n-4}} =  \frac{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}}{(1+K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}])}} $$
But since
$$\ce{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] >> 1 }$$
$$\ce{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] + 1 \approx K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] }$$
Which gives the following final simpified experssion
$$\ce{K''_{MY^{n-4}} =  \frac{K_{MY^{n-4}}α_{M^{n+}}}{K_{NY^{m-4}}α_{N^{m+}}[N_{total}]}} \geq 10^7 $$
Finally the testing equation is sloved for selective titrations


Why does blood get further oxidized than Iron oxide? Oxygenated blood is bright red and deoxygenated blood is dark red or brown.

If you take oxygenated blood and leave it in the air it will turn dark red, then brown, then finally a bluish green from exposure to atmospheric oxygen.

Why does it oxidize this far to get a copper oxide color to it after it has passed the phase where it is the same color as iron oxide? Why doesn't it stay at the iron oxide color?

This often happens during the last few days of my period that the bleeding is extremely light and the color of it when it first comes out is reddish brown (sort of like iron oxide) and after a few hours, it starts turning bluish green (sort of like copper oxide). However, this happens very slowly and after 24 hours of exposure to atmospheric oxygen still only a fraction of the blood is further oxidized. I don't take medications or drugs that have a lot of sulfur so I don't think it is from $\ce{H2S}$ or $\ce{SO4-}$ (which would give it a dark green color, not light bluish green) and I don't have excess copper in my blood either.

I think that it is Iron(II,III) oxide with the Iron(III) released and the autoionization of water (which happens in both the liquid and gas phase) forming Iron(II) Hydroxide which can be further oxidized to Iron(III) hydroxide which has that bluish green color along with a small amount of Copper oxide forming (Due to the fact that our bodies need some copper and like Iron there is always some copper in the bloodstream).

There are two misconception behind your question:

You assume you can directly compare the colors of different compounds and derive oxidation states e.g of a heme (porphyrin) comparing to iron-oxide. No, it is not true.
You assume this whole story has anything to do with copper. No, the blueish-green is nothing to do with copper, it is a (porphyrin) degradation product.

The color of transition metal compounds mostly can be derived from the crystal field (ligand field) around the central metal ion. In inorganic iron-oxo/hydroxo compounds have different structure and hence different field than heme compounds. In heme, the N coordination of porphyrin ring is generating a very strong planar field, and the oxygen coordinates as molecule on the top of it. Charge transfer to the ring can also be an important factor in heme derivatives. As you see it is a whole complicated mass, and not much direct connection with different inorganic salts you see in lab.

The colors you see are coming from oxo-hemoglobin (which you have in your healthy blood), and most probably several steps of enzymatic degradation of heme. The porphyrin ring is oxidized and cut open by an oxygen on the the iron. Several colorful compounds are produced, including the iron complexes themselves and biliverdin (green), bilirubin (yellow). These are the same colors you see under your skin, of you hit yourself hard and have some bleeding in your tissues. First the injured area is deep red/blue, than go to greenish, finally yellowish patch. This also happens continuously in your body, when your blood cells get old, and heme is enigmatically degraded so the whole thing can be recycled:  http://www.elmhurst.edu/~chm/vchembook/634bilirubin.html


What is the cause of decrease in volume of gaseous mixture in the following problem? I am studying Gaseous State.

The problem says: A mixture of gas contains $\ce{N2}$ and $\ce{C2H2}$. $\pu{20 \\mL}$ of this mixture is added to $\pu{70\\mL}$ of $\ce{O2}$ and combustion is allowed to take place over mercury. After cooling the volume of gas was found to be $\pu{68.5\\mL}$. When the resultant gas mixture is passed through a $\ce{KOH}$ solution, the volume of residual gas is $\pu{38.5\\mL}$. Then the percentage composition of $\ce{N2}$ is:

In the solution to the question there is this following step which I am not getting:

:Volume of $\ce{CO2} = \pu{68.5\\mL} - \pu{38.5\\mL}$.         How come?
Also please clarify why nitrogen does not undergo combustion.

Please help.

I think the fact you are missing is that after combustion, water vapors will be given away and the volume lost during combustion will be equal to the amount of water lost. $\ce{KOH}$ will absorb $\ce{CO2}$ and the volume absorbed by $\ce{KOH}$ will be equal to volume of $\ce{CO2}$ in the solution. Thus, you will have two equations and you will be able to proceed with the solution.


Fe/Ga cell has a negative potential? In the following homework problem, the answer says:

"$E ^{o}_{cell}$ is negative, which makes the $\ce{Fe}$ and $\ce{Ga}$ half-reactions non-spontaneous."

But I found $E ^{o}_{cell}$ to be positive - NOT negative:

$$E ^{o}_{cell} = E ^{o}_{cathode} -E ^{o}_{anode} =
E ^{o}_{Fe} - E ^{o}_{Ga} = -0.409\,\mathrm{V} - (-0.56\,\mathrm{V}) = +0.151\,\mathrm{V}$$

which means that these half-reactions would actually be spontaneous.

Am I solving this problem wrong or did the writer of this question make a mistake?

After all, $\ce{Fe}$ should be the cathode since it has a more positive voltage. Right?

You need two half reactions for the total redox reaction: one reduction and one oxidation. The convention is that we provide all potentials as standard reduction potentials. If you need an oxidation potential, you're going to need to reverse the reduction half reaction and negate the sign standard potential.

This means that the total potential of the cell is the raw sum of the oxidation and reduction potentials. In your specific example, you want:
$$\ce{Ga^{3+} + 3e- -> Ga}, E_{red}^{\varnothing} = -0.560\,\mathrm{V}$$
$$\ce{Fe -> Fe^{2+} + 2e-}, E_{ox}^{\varnothing} = +0.409\,\mathrm{V}$$

The EMF of the cell is the sum:

$$E_{cell}^{\varnothing} = E_{red}^{\varnothing} + E_{ox}^{\varnothing}$$

And this value is negative.

Now, as a shortcut, you can identify which of the two reduction half reactions needs to be reversed to make an oxidation, and you can add that value--negated--to the potential for the reduction half reaction, as above.

What your teachers or book meant was the following. I've changed the names of the variables so that they don't conflict with the standard notation (which is why we're talking past each other).

$$E_{cell}^{\varnothing} = E_{red,1}^{\varnothing} - E_{red,2}^{\varnothing}$$

where $E_{red,1}^{\varnothing}$ is the standard reduction potential for the reduction half reaction and $E_{red,2}^{\varnothing}$ is the standard reduction potential for the reduction that corresponds to the oxidation half-reaction. This is hellishly confusing, which why I think you are confused. The key here is that

$$E_{\mathrm{cathode}} = E_{red,1}^{\varnothing} = E_{red}^{\varnothing}$$

but

$$E_{\mathrm{anode}} = E_{red,2}^{\varnothing} = -E_{ox}^{\varnothing}$$

Note also that ordinarily, for a spontaneous reaction, iron would constitute the cathode of the cell because it would be the reduction. However, in this case, because the reaction is non-spontaneous, the iron half reaction is actually going to be an oxidation, so that means it's actually going to be the anode. This is the other factor you're confused about. You're trying to apply shortcuts for a spontaneous cell to a non-spontaneous one. Try to understand what the underlying principle is.


What is the wave function? I understand that the square of the wave function is the probability density of finding an electron. However, what is the wave function itself? I am trying to understand molecular orbital theory but this concept is getting in the way of me completely understanding it. Any help would be great. Thanks.

It is often asked "what is waving when a particle is described as a wave?" and the correct response is "existence itself is waving."
The wavefunction is most elegantly described as a description of reality for your system. You then act on the wavefuntion with a hamiltonian to yeild an eigenvalue to quantify the energy.


Equilibrium involving both gas phase and aqueous phase If I have been given an equation that involves a gas-aqueous phase equilibria, how do I write the equilibrium constant?

I have come across several questions that give equations having BOTH gas and the aqueous phase and I couldn't comprehend how to attempt them.

Should we find the concentrations of the gas too and use it for the equilibria?

But in several questions I was given partial pressure of the gas at equilibrium but I was never given the temperature to find concentration.

What is a proper way of representing the equilibrium constant?

Here is an example:

$$\ce{RNH2(g) + H2O(l) <=> RNH3+(aq) + OH-(aq)}$$

How do I write the equilibrium constant for this, if I have been given concentrations of $\ce{RNH3+(aq)}$ and $\ce{OH-(aq)}$ and the partial pressure of $\ce{RNH2}$ at equilibrium?

I dare say most equilibrium constant don't cite specific units because the appropriate units are implied implicitly. In the case of the reaction:

$$\ce{R-NH2(g) <=>[aqueous] R-NH3+ + OH-}$$

I'd write the mathematical equilibrium equation as:

$$K_{eq} = \dfrac{\ce{[R-NH3+][OH-]}}{p_\ce{R-NH2}}$$

and give the constant the appropriate units: $\pu{mol^2\cdot L^{-2} \cdot kPa^-1}$. (I just picked kPa as one of the zillion units for pressure. Use whatever units you were given the vapor pressure in: atm, torr, mm Hg, lbs/in^2, slugs/furlong^2 ... whatever)

This is obviously a "overall" equilibrium. According to Henry's law there should be an equilibrium between the vapor pressure of $\ce{R-NH2}$ and the concentration of $\ce{R-NH2}$ in solution. Since you have no way of calculating the concentration of $\ce{R-NH2}$ in solution you can't break the overall reaction into the two different equilibria.


1 mole of substance is equal to atomic/molecular mass in grams? I have been reading the answer of this question here- Why is weight of 1 mole of substance equal to atomic/molecular mass in grams

I feel that the answer writer was wrong to say that -"the ratio of g/mol to amu/atom is exactly one"

I  feel so because he proved that the ratio of g/mol C12 to amu/atom C12 is exactly one.He never proved that the ratio of g/mol to amu/atom is exactly one.If I am wrong then please correct me!

The statement is more or less correct.

But there are two caveats here:

There is no proving of these concepts since there are definitions.
In general, it's a bit strange to use the ratio of amu to atom. An atom has a well defined mass based on it's isotope, but the number of grams per mole is defined based on natural abundance and weighted average mass. Therefore, there is some discrepancy here. Of course, here, you're referring specifically to carbon-12, so there is no confusion, but just watch out in the general case.


Fractional Subscripts? What is the physical meaning when we have an empirical formula with fractional subscripts such as $\ce{LiMn_{1-x}Fe_{x}PO4}$ $(0 \le x \le 1)$?

This an example of the paper the notation above is used:

Bai, J.; Hong, J.; Chen, H.; Graetz, J.; Wang, F. J. Phys. Chem. C 2015, 119 (5), 2266–2276. DOI 10.1021/jp508600u.

These indicate "doping" or non-stoichiometric compounds.

In this case, there's some fraction of Fe, and some fraction of Mn in a lithium iron phosphate battery. If I remember correctly, you need to dope a different ion to increase the conductivity of $\ce{LiFePO4}$.

See also here: How are non-Stoichiometric compounds determined?


Which poison was used in the failed attempt to assassinate Khaled Mashal of Hamas? This question is both historical and chemistry-related, but I'm posting it here because I am more interested in the chemical part:

In 1997, Israel tried to assassinate Khaled Mashal, a high-rank official of Hamas. This was to be done using an extremely potent poison that was supposed to be sprayed on him topically and kill him within a few days without a trace (a "quiet" murder).

I searched for the name of this poison, but I was only able to find out that it was a modified version of the drug Fentanyl (here, page 16). It was probably an internal formulation of the Mossad, but I was wondering if the formula was ever publicly released or if anyone knows how Fentanyl's structure was modified in order to make it so deadly upon contact.

According to several sources online the chemical weapon used in the failed assissination of Mashal was levofentanyl.

This Wikipedia article states:

  On September 25, 1997, Mashal was injected in the ear with a toxin
  (thought to have been a derivative of the synthetic opiate Fentanyl
  called Levofentanyl).

Another source reads:

  In September 1997, an Israeli hit squad was sent to Amman to
  assassinate Khalid Mishal, the Hamas leader. The chosen instrument was
  levofentanyl, a deadly poison that leaves no traces and produces
  effects like a heart attack. It was administered by a slight physical
  touch.

And another:

  ...The agents entered the city using Canadian passports and waited for
  Mashal at his office. As he was entering, one agent injected a
  chemically modified toxin called levofentanyl into his left ear.

Concerning the structure I don't think it was ever publicised considering the case was high profile (perhaps all traces were taken offline or can be found in deep when-no one knows)

I've been trying to figure out how fentanyl could have been modified, but have no clue either.


Potassium Metavanadate Synthesis I am looking to synthesize potassium metavanadate. Possible precursors that I have are $\ce{V2O5}$ and $\ce{NH4VO3}$.

I have considered the following procedure:
Combine $\ce{V2O5}$ and $\ce{KOH}$ then adjust pH and precipitate out with $\ce{CH3OH}$. The problem is multiple species exist at various pH values, so I am looking for a method to obtain $\ce{KVO3}$. I read a while ago a source for synthesis of different alkali metal vanadates, but I cannot find it again.

Potassium Vanadate can be prepared by any of the three process:

  Potassium Metavanadate, $\ce{KVO3}$, is obtained as white or colourless crystals by dissolving vanadium pentoxide in hot, strong
  caustic potash solution. By varying the concentrations several
  hydrates have also been prepared: $\ce{2KVO3.3H2O}$;
  $\ce{KVO3.2H2O}$; $\ce{2KVO3.5H2O}$; $\ce{KVO3.3H2O}$. On being
  heated, all these hydrates lose their water and leave a white,
  nacreous mass of the anhydrous salt, which melts at 495° C. Two
  other hydrates were prepared by Rammelsberg: $\ce{KVO3.H2O}$ and
  $\ce{KVO3.7H2O}$. (source)

  [...] the reaction of vanadium(V)
  oxide from waste catalyst with potassium chloride in the presence of
  steam.
  $$\ce{2KCl + V2O5 + H2O(g) → 2KVO3 + 2HCl↑}$$
  (source)

  Ammonium metavanadate react with potassium carbonate to produce
   potassium metavanadate, ammonia, carbon dioxide and water. Potassium
   carbonate - concentrated solution. The reaction takes place in a
   boiling solution.

  $$\ce{2NH4VO3 + K2CO3 → 2KVO3 + 2NH3 + CO2 + H2O}$$ (source)


What is this red substance that has accumulated at the bottom of a plastic bottle full of petrol (see picture)? About two weeks ago I put some ordinary unleaded petrol into a plastic water bottle, as a temporary measure (long story, don't ask!). It has now formed a reddish liquid at the bottom of the bottle.

The bottle was emptied prior to adding the petrol, but probably still contained a small amount of water when the fuel was poured in. The bottle was stored in ordinary room temperature.

I presume that the fuel has corroded/reacted with the plastic and left this substance, but I have looked to Stack Exchange in the hope of a more scientific answer...

So, what is the red substance and cause of it?

Please note that the brownish 'slug-shape' is merely an artifact of the light refraction...

Maybe could be some iron oxide or hydroxide from previous storage or ( if you have a iron tap) from it ore maybe directly from the plastic.

Some fuel has red dye in it but I don't think this is the case, because I don't see how it can precipitate.

Maybe it could be a residue of petroleum resin.


What are the crystals that have formed around the top of a conc. nitric acid bottle? I was looking through my chemicals cupboard and found that this decades-old bottle of conc nitric acid had grown a beard. I can't think of any reaction between (presumably) $\ce{NO2}$ and any airborne substance that could produce this. Any ideas?

Chemicals cupboard often produce quite extraordinary reactions, much like hospitals are home to some unusual and severe infections.

Nitric acid does not normally exhibit such a bizarre behavior. Yet in your case it did. How so? My bet is on some unknown additional component which came from the same cupboard, but from a different bottle. For instance, it could be some lower aliphatic amine (Upd. or simply ammonia, as was brought to my attention by other answers). They are volatile, and they produce crystalline salts when they meet with strong acids.

A similar thing is known to happen with $\ce{NH3}$ and $\ce{HCl}$:  if they are stored nearby, especially in poorly corked bottles, they both (as well as everything around) get covered in a white residue of $\ce{NH4Cl}$. It doesn't form nice crystals, though; it looks more like a layer of hoarfrost. But that doesn't mean much. Related compounds do not necessarily have related crystal structures, let alone related crystal shapes, so it looks perfectly plausible to me that (substituted or not) ammonia nitrate would grow into fine needles.

The prevalence of amine/ammonia in the air within the cupboard (which could happen if the other bottle was corked worse than this one) would naturally lead to $\ce{HNO3}$ molecules fighting this battle at their own doorstep, that is, at the very bottleneck of this bottle and nowhere else. This would explain the accumulation of residue at one spot.


How do I know that an acid and conjugate base are present in equimolar amounts? Acetylsalicylic acid is a molecule in several popular drugs such as Magnyl and aspirin. It has a $\mathrm{p}K_\mathrm{a}$ of 3.5.

I'm about to calculate the $\mathrm{pH}$ of a $\pu{0.010 M}$ solution of acetylsalicylic acid.

I've read up on how to do this and in one video the author talks about "equimolar amounts".
He says that now that we have equimolar amounts, we can write: $$\ce{\frac{[H+][A-]}{[HA]}} = \ce{\frac{[H+]^2}{[HA]}}$$

Which makes sense as there are equally many $\ce{H}$ and $\ce{A}$ ions. However when can I assume that an acid disassociated equally? Also, I thought that when it did this then it was called a strong acid? Why would I be using this formula at all for a strong acid?

If I can assume that acetylsalicylic acid disassociates into equal acid and base ions, then I can calculate my problem, but I'm skeptical as to assuming this is true.

I'm really confused here!

No, you are not! You only (temporarily) mixed up two things, both of with you have already understood separately.

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

1. Position of the equilibrium

This is described by the equilibrium constant $K$, which is the quotient of the rate constants for the forward and the back reaction: $K = \frac{k_{forward}}{k_{back}}$. For strong acids, of which you know that they completely dissociate, the equilibrium is on the product side.

2. Equimolarity

There's nothing out of nothing! If one molecule of $\ce{HA}$ dissociates is water, there exactly one molecule $\ce{H3O+}$ and exactly one molecule of $\ce{A-}$ generated. Not more, not less.

Q: How could another molecule of $\ce{A-}$ appear out of nothing?

A: It can't and it doesn't happen - not even in the LEGO universe.

It is just a proton transfer reaction from $\ce{HA}$ to $\ce{H2O}$; the overall number of atoms does not change!

If is however possible that some $\ce{A-}$ may come from a different source, just by adding a salt $\ce{MA}$ of the conjugate base (think sodium acetylsalicylate) to the solution of your acid. jerepierre has mentioned this in his comment! But for the moment, this is another story.


What is the difference between autocatalysis and self-replication in terms of chemical terminology? I want to know does any difference between the meaning of autocatalysis and self-replication exist in modern chemical terminology?

E.g. I want to know can both terms be used interchangeably for entities requiring the presence of themselves in a reaction environment to catalyze the formation of themselves or is there any experimentally conceivable differences between molecules capable of autocatalysis and the self-replication of entities (therefore the category includes both molecules and larger complexes of molecules) in terms of the science of chemistry?

I want to know is the definition of autocatysis versus self-replication a problem of clarification of definitions-e.g. a problem of philosophy or is there any real-world experimentally identifiable difference between the 2 process in terms of chemistry?

The two terms represent very distinct concepts and are everything but interchangable.

Autocatalysis means that the product of a reaction catalyses the same reaction. But there is no focus on the catalysed reaction being exactly that or only that product catalysing it. As an example, take acidic ester hydrolysis e.g. with ethyl acetate:

$$\ce{CH3COOC2H5 + H2O ->[H+] CH3COOH + C2H5OH}\tag{1}$$

This is a reaction catalysed by acids that produces an acid in its course. This acid increases the solution’s overall acidity and thus catalyses the reaction that created it. But this feature is in on way specific to the acid we have generated. For one, we might as well add sulphuric acid to the reaction mixture to generate the same catalytic effect. For two, the acid produced in this reaction would also catalyse the hydrolysis of methyl formate. So neither is a specific structure required for catalysis nor is the catalysis restricted to itself.

Self-replication again means that a product is able to catalyse its own formation but only its own formation. The catalysis must be specific to a certain (set of) reactant to produce a certain (set of) product. Similar but different compounds bearing the same functional groups should not be able to effectively catalyse the reaction in question and the reaction product should not be able to catalyse any other reaction equally efficiently.

Typically, self-replication is achieved by having the product molecule act as a template which brings two reactive moieties closer together and allows for a more efficient reaction to itself. For example, you might consider the trimeric complex shown in the figure below:

Figure 1: Self-replicating system as described by
Terford and von Kiedrowski[1]

Here, the amine and the aldehyde shown on the right-hand half of the figure can condense to form an imine. This process can, of course, be catalysed by acids or bases. However, the condensation product shown on the left-hand half of the figure is able to form hydrogen bonds to the two compounds in such a selective way that the reactive moieties are brought close together. This eases the creation of the imine because the amino group is already in the perfect position of attack.

The imine product is therefore able to catalyse its own formation by templating but not the formation of closely related systems; and there is no other compound or functional group that would have an equally good catalytic effect on the reactants. This is not just autocatalysis but also self-replication.

Reference:

[1]: A. Terford, G. von Kiedrowski, Angew Chem. Int. Ed. 1992, 31, 654–656. DOI: 10.1002/anie.199206541.


How do I calculate the pH of this buffered solution? Calculate the $\mathrm{pH}$ of a buffered solution containing $\pu{0.5 M}$ ammonia and $\pu{0.5 M}$ ammonium chloride when $\pu{0.15 M}$ $\ce{HCl}$ is added into it. The $\mathrm{p}K_\mathrm{b}$ of ammonia is $4.75$.

This is what I think shoud be going on in the buffer solution:

\begin{align}
\ce{NH3(aq) +H2O(l) &<=> NH4+(aq) +OH-(aq)} \\
\ce{NH4Cl(aq) &-> NH4+(aq) +Cl-(aq)}
\end{align}

This is where I get stuck thinking that the question told me that I have a buffered solution containing $\pu{0.5 M}$ $\ce{NH3}$ and $\pu{0.5 M}$ $\ce{NH4Cl}$ solution, but is that even possible? How can you make a $\pu{0.5 M}$ ammonia solution? Would not that ammonia exist as ammonium $\ce{NH4+}$ and whether the equations I wrote are correct. I am just so confused!

Sorry for the delay but I'm having exams. Okay, I'll try to do this now by not using the Henderson equation.

Firstly, ammonia will refer to ammonium hydroxide, i.e. $\ce{NH3 -> NH4OH}$:

\begin{array}{cccc}
&\ce{&NH4OH &<=> &NH4+ &+ &OH-} \\
&\mathrm{I}: &\pu{0.5 M} & &\pu{0 M} & & \pu{0 M} \\
&\mathrm{E}: &(0.5 - x)\,\pu{M} & &x\,\pu{M} & &x\,\pu{M}
\end{array}

Now for the salt:

\begin{array}{cccc}
&\ce{&NH4Cl &<=> &NH4+ &+ &Cl-} \\
&\mathrm{I}: &\pu{0.5 M} & &x\,\pu{M} & &\pu{0 M} \\
&\mathrm{E}: &\pu{0 M} & &(0.5 + x)\,\pu{M} & &\pu{0.5 M}
\end{array}

$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH4OH}]} = \frac{(x + 0.5)x}{(0.5 - x)}$$

and since $K_\mathrm{b} = 1.778 \times 10^{-5}$, I can find the value of $x$ to be $1.7778 \times 10^{-5}$.

$$[\ce{H+}][\ce{OH-}] = 10^{-14} \to [\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]}$$
and $x = [\ce{OH-}]$. So

$$[\ce{H+}] = 5.624 \times 10^{-10}$$

Now I add the $\ce{HCl}$:

\begin{array}{cccc}
&\ce{&HCl &<=> &H+ &+ &Cl-} \\
&\mathrm{E}: &\pu{0 M} & &\pu{0.15 M} & &\pu{0.15 M}
\end{array}

I will ignore any common ion effect since it will be negligible (I think).
The pH will be the total $\ce{H+}$ concentration, so $[\ce{H+}] = 5.624 \times 10^{-10} + 0.15$. This is a pH of about 0.823 which is totally wrong. What am I doing wrong? Also, the volumes of the buffer solution or of the acid solution have not been given.

Okay, so I now know what I was doing wrong (thanks to someone pointing it out). I was adding in $\ce{HCl}$ without thinking that it would actually react with the ammonia in equilibrium in solution (which is just me being either dumb or ignorant).

So I have the situation that:

\begin{array}{cccc}
&\ce{&NH4OH &<=> &NH4+ &+ &OH-} \\
&\mathrm{E_\text{(no HCl)}}: &(0.5 - 1.778 \times 10^{-5})\,\pu{M} & &(1.778 \times 10^{-5})\,\pu{M} & &(1.778 \times 10^{-5})\,\pu{M} \\
&\mathrm{E_\text{(HCl)}}: &()\,\pu{M} & &()\,\pu{M} & &()\,\pu{M}
\end{array}

You can think of $\ce{NH4OH}$ as a hydrate: $\ce{NH3_{(aq)}}$; no need for complicated reactions interconverting the two.

Concerning the problem you described, at these concentrations all subtleties are negligible, and you basically only need to calculate the nominal concentrations of $\ce{NH4+}$ and $\ce{NH3}$, and plug them into the Henderson-Hasselbach equation.

You start with $\ce{0.5 M NH3}$ and $\ce{0.5 M NH4Cl}$, and add $\ce{HCl}$ to a total nominal concentration of $\ce{0.15 M}$, without altering the total volume (the question is written ambiguously, but that's what it means, most likely). As $\ce{HCl}$ is a very strong acid, it will convert all $\ce{NH3}$ to $\ce{NH4Cl}$ (actually $\ce{NH4+}$, because we are in solution). Therefore your final situation is: $\ce{(0.5-0.15)=0.35 M NH3}$ and $\ce{(0.5+0.15)=0.65 M NH4+}$. This should give you a pH of about 8.98 (lower than the one you would have without the HCl, consistently with the addition of an acid).

So your main error above was that you dissociated $\ce{HCl}$ (correctly), but then just added the resulting $\ce{H+}$ to the total, forgetting that there was a ton of $\ce{NH3}$ around that would react with it.

Another example is when you add a solution of some particular transition metal salt to a solution of a weak acid (could be iron sulfate + hydrogen sulfide for instance). Despite the fact that the salt solution isn't acidic at all, the pH drops considerably compared to the solution of the weak acid alone, simply because the metal and the anion form a very insoluble salt that precipitates, subtracting weak acid anions from the solution and forcing the previously non-dissociated acid to dissociate.

In general, my advice is stick to what you know, don't go looking for complications: if they did not teach you to do very exact/sophisticated pH calculations, they probably don't expect you to pull them out of a magic hat at the exam. And as my stoichiometry teacher used to say, always write down all species you have in solution.


Relationship between surface area and vapor pressure How is vapor pressure independent of the surface area and volume of a liquid, although rate of evaporation depends on surface area?

Please someone help me I am totally confused.

The rate of evaporation is proportional to surface area. The rate of condensation is proportional to surface area. The vapor pressure is the equilibrium pressure where the rate of evaporation is equal to the rate of condensation. Since the scaling factor is the same, the vapor pressure is independent of the surface area.


What type of bond forms between an ionic compound solute and a polar solvent I understand the idea that the solvation enthalpy (ΔH-) must exceed the lattice enthalpy (ΔH+) for a solution to be more energetically stable, but what type of bond is formed between the disassociated ions and the the polar solvent molecules? Surely it must be even stronger than the ionic bonding in the lattice if it releases more energy than the energy needed to break the ionic bonds in the ionic compound.

Firstly, There is no such bond existing inbetween the polar solvent molecule and the solute ions. It is ion-dipole interaction which exists between the polar solvent molecules and the ions of the solute.

I hope you got your answer.

I wanted to add something more which I think you got wrong.

I agree that if hydration enthalpy is greater than lattice enthalpy then solute will dissolve but it is not compulsory for hydration enthalpy to be greater than lattice enthalpy for dissolving. What i mean is: For dissolving its important to break the ionic bonds between the ions of the solute molecule but that energy required to break the  ionic bond is not necessarily from hydration only. That is sometimes hydration enthalpy couldn't meet the energy required to break the ionic bonds (lattice energy). In those cases the extra amount of energy can be from heat of the solvent. In the cases where hydration enthalpy is larger than lattice energy, ΔH is negative. In cases where hydration enthalpy is lesser than lattice energy, in those cases the more energy needed is driven from the heat of the solvent and in this case ΔH becomes positive. Negative ΔH is preferred to be better for dissolving because it helps in making ΔG more negative. But it doesn't mean solute with positive ΔH doesn't dissolve. If ΔS is highly negative, then even though ΔH is positive still it can make ΔG negative and hence dissolving possible. Dissolving of any solute depends on ΔG. For dissolving, ΔH doesn't matter as long as ΔG is negative. Say for example in case of Nacl, net solvation enthalpy is positive, but still it dissolves in water.

Nacl hydration enthalpy is not greater than enthalpy required to break the bonds in between Na+ and Cl-. As hydration enthalpy is not sufficient to release energy which is required for bond breaking incase of Nacl, hence the extra amount of energy required comes from the heat of the water ,which in addition to the hydration enthalpy is able to break the bonds in between Nacl. As energy released is less than net energy absorbed, hence net enthalpy (solvation enthalpy) is positive.The reason behind the dissolving of the salt is ΔS being highly positive which in turn ends up making ΔG negative as ΔG= ΔH −TΔS.

So for a well dissolving ionic solute, hydration enthalpy is always not greater than the energy required to break the ions apart in the solute molecule. An ionic solute with positive ΔH would still dissolve if ΔG is negative.


Is black tea a pH indicator? Today, I made a cup of lemon tea in a different way than usual. Instead of pouring hot water in to a mug containing lemon juice and black tea (my normal routine), I let the tea steep for a little while before squeezing the lemon. Before adding the lemon, I noticed that the tea was much darker than I am used to. After squeezing the lemon, the tea changed color to a lighter brown and became more translucent.

I could see only a half inch down the side of the cup before squeezing the lemon, but afterwards I could see three inches down, all the way to the bottom. I did not notice any precipitated solids on the bottom of the cup. What happened?

The colour of black tea is given by a complex mixture of theaflavins and thearubigins.

These compounds are formed by enzymatic oxidation and condensation of polyphenols of green tea, typically of gallocatechin and epigallocatechin gallate ( EGCG )

Basic dissociated forms of these compounds have strong absorption of visible light due to delocalization of phenolate electrons. This absorption  decreases a lot when phenolates are protonized in acidic environment.

So yes, in some sense, black tea infusion is a $\mathrm{pH}$ indicator, similar to litmus or juice from red beet roots. Many natural colourful compounds have $\mathrm{pH}$ dependent colour, as their molecules undergo acid-basic reactions.

But synthetic indicators are better due stability, sharper and better defined colour transition.


How to calculate the amount of water produced by the incomplete combustion of propane? I was wondering if someone could walk me through how to do the following two questions?

  The complete combustion of pentane has a enthalpy change of $-3509~\mathrm{mol^{-1}}$. Write an equation for the complete combustion of pentane. Write a possible equation for the imcomplete combustion of pentane.

For this question, would the following equation suffice:
$$\ce{ C5H12 (g) + 8O2(g) -> 5CO2 (g) + 6H2O (g)  }$$

For an incomplete reaction, the products will be water, carbon/and or carbon monoxide due to insufficient oxygen, so would this equation be correct:
$$\ce{C5H12 (g) + 11/2 O2 (g) -> 5CO (g) + 6H2O (g)}$$

  $17~\mathrm{L}$ of Propane are incompletely combusted in $34~\mathrm{L}$ of oxygen at $25~^\circ\mathrm{C}$ and $100~\mathrm{kPa}$ into only water and carbon. Write a balanced chemical question to represent this including state symbols. What is the mass of the water produced? What observable change would indicate that a chemical change occurred?

Would an equation like this work: $$\ce{C3H8 (g) + 7O2 (g) -> 3CO2 (g) + 4H2O (g)}$$

The ratio for the above equation will be 1:7:3:4.

I am given the volume of the propane to be $17~\mathrm{L}$ and the volume of the oxygen to be $34~\mathrm{L}$. Since the question is a molar calculation, I think I have to convert the $~\mathrm{L}$ into the SI units of grams. This means $17~\mathrm{L} \Rightarrow 17000~\mathrm{g}$ and $34~\mathrm{L} \Rightarrow 34000~\mathrm{g}$.

The I have to calculate the number of moles for propane and oxygen (I think).

$$
\text{number of moles} = \frac{\text{mass}}{\text{molar mass}}\\
\text{number of moles of propane} = \frac{17000}{44.1} = 385.4~\mathrm{\frac{mol}{gram}}\\
\text{number of moles of oxygen} = \frac{34000}{16} = 2125~\mathrm{\frac{mol}{gram}}$$

Finally I have to find the mass of the resulting water, which is the part I am stuck at. I know I have to find the number of moles times the molar mass of water to find the total mass of the water, but I am just somewhat stumped as to how to find the number of moles of water given what I have worked out above.

For chemical changes, I think that bubbles of gas may appear, a temperature change will occur since this is a combustion reaction, and maybe a change of mass/volume?

Yes the complete combustion equation is correct as you stated and the equation of the incomplete combustion is certainly one possibility. A more general formula is:
$$\begin{multline}\ce{C5H12 + $\left(\frac{x}{2}+y+3\right)$\,O2 -> $(5-x-y)$\,C + $x$\,CO +$y$\,CO2 + 6H2O}\\
\left\{x,y\in\mathbb{R}^{+}_0; (x+y)\leq5\right\}
\end{multline}$$
This is assuming all the hydrogen will be converted into water. In some cases you might have incomplete combustion, where ethene or ethyne (ar many others) are formed, but I think this is taking it a bit far at this point.
The above case reduces to your formula for $x=5$ and $y=0$.

For the second question, your equation has a small error, correct is:
$$\ce{C3H8 (g) + 5O2 (g) -> 3CO2 (g) + 4H2O (g)}$$
But this is not the equation you need, as the question explicitly states, that the combustion products are carbon and water. So you working equation becomes:
$$\ce{C3H8 (g) + 2O2 (g) -> C (s) + 4H2O (g)}$$
So you need twice the number of moles of oxygen than propane, to produce four times as many moles of water than you had propane.

You cannot convert a volume of gas into a mass by an easy conversion factor, you have to use the ideal gas law:
$$pV =n\mathcal{R}T$$
You can then use $m =n\cdot M$ to convert to grams, but you don't yet have to do that.

You have been given the temperature $T=25~^\circ\mathrm{C}=298.15~\mathrm{K}$ and pressure $p=100~\mathrm{kPa}$ and the volumes for the reactants, so you can calculate the amount of substance for propane and oxygen.

   $n(\ce{C3H8})\approx 0.7~\mathrm{mol}$ and $n(\ce{O2})\approx 1.4~\mathrm{mol}$

From the reaction equation you can now determine the amount of substance of water, that will be produced.

   $n(\ce{H2O})\approx 2.8~\mathrm{mol}$

You can convert that easily into a mass now.

   $m(\ce{H2O})\approx 50~\mathrm{mol}$

There are two notable changes you will observe. Since you are combusting a gas, obviously no bubbles can appear. If you do the combustion in a see-through container you will see black stains due to the solid carbon that is produced by the reaction and liquid water drops will form on colder surfaces.
You are of course right, that you will also have a rise in temperature.

Note: The here given values are crude approximations only, that might not satisfy the task at hand. You need to do the calculation yourself to find the correct answer.


What is buffer capacity and how can a buffer "run out"? I'm a little confused on something:
A buffer has a certain capacity and then runs out.
But, if the buffer is given by: $$\ce{A- + H2O <=> HA + OH-},$$ when the $\ce{A-}$ reacts with the water, the reaction shifts to the left. So how can it "run out"? If $Q$ just keeps becoming $0$, shouldn't the reaction keep shifting to maintain $K$? Does the right side, perhaps, approach $0$ as it keeps decreasing in concentration to maintain $K$, but it can never become $0$ molar, because that wouldn't maintain $K$?
I guess if $\ce{H2O}$ was "aqueous"/had a concentration and was therefore used in the $K$ expression, this the $K$ would always be maintained.
But how does the buffer "run out" of $\ce{A-}$ to react when $\ce{H2O}$ is liquid (not aqueous)? It doesn't make sense as wouldn't the reaction always adjust to make more $\ce{A-}$ to maintain $K$?

If strong acid is added to a buffer system, conjugate base removes it.  Eventually, the concentration of conjugate base falls to such a low level that its effect on pH can be neglected.  At this point, the solution behaves more like a weak acid solution.  If even more acid is added, the system behaves like a strong acid system in terms of calculating pH.

If strong base is added to a buffer system, conjugate acid removes it.  Eventually the concentration of conjugate acid falls to such a low level that its effect on pH can be neglected.  At this point. the solution behaves more like a weak base solution.  If even more base is added, the system behaves like a strong base system in terms of calculating pH.

In the equation you wrote, $$\ce{A- + H2O <=> HA + OH-},$$ think about the effect of adding acid.  Acid reacts with the $\ce{OH-}$, producing water and removing $\ce{A-}$ from solution.  The $\ce{OH-}$ and $\ce{A-}$ cannot be removed from solution but their concentrations can be driven to negligible levels by the addition of sufficient quantities of acid.


Product of primary amine and acid anhydride How does an acid anhydride react with a primary amine? I thought it'll be some sort of condensation reaction, so I made the product by removing a water molecule and got a cyclic product. However, the product given is an open chain molecule (major).

How does the reaction occur? How does the ring open? Could someone please give a detailed mechanism?

P. S. What is the general reaction for primary amines and acid anhydrides?

It should be very clear that a primary amine is a nucleophile and an acid anhydride an electrophile. In the series of carbonyl and carboxyl derivatives, an acid anhydride is classified as quite reactive: more so than aldehydes or ketones, much more than esters or amides but less than acid chlorides. Thus, we should all agree that the first step is the nucleophilic attack on the $\ce{C=O}$ π system via a tetrahedral intermediate to give N-methyl succinic monoamide.

Scheme 1: Mechanism of the monoamide formation.

With that, we can consider whether further reactions will take place. Where we originally had an anime and an acid anhydride, we now have a carboxylate (the $\ce{C=O}$ bond that wasn’t attacked) and a secondary amide. The secondary amide still has a lone pair we might consider to be nucleophilic — but that lone pair is taking part in amide resonance and is thus not available for nucleophilic attack (at least not well). And to remove the amide’s proton, we would need a strong base which we don't have. Thus, the amide side of the equation is far less reactive.

What about the carboxylate? Thus highly symmetric structure is also very unreactive towards nucleophilic attack in the same way as an amide. Due to the negatively charged oxygen donating one of its lone pairs to the π system, the carbon atom is much less electrophilic than in an acid anhydride. Thus, in this case also there is no desire to react.

Since we have two functional groups of low reactivity, the reaction ends here. However, with enough heat (supplying activation energy), it is possible for the sluggish intramolecular reaction between the the amide and the carboxylate to occur. In principle, the mechanism is the same except that you must transfer a few protons and remove water.

Scheme 2: potential onwards reaction to succinimide under harsh conditions.


Stereoselectivity of ring closure in intramolecular iodolactonisation What is the product of the reaction above?

I am mainly confused between the following two options:

How do we decide the resulting stereochemistry of the $\ce{CH_2I}$ group?
The answer given is the one in which $\ce{CH_2-I}$ is coming out of the plane, i.e. (a).

The results of the iodolactonization example presented in this question were published in 1982 by Bartlett, et al.1 While @Aniruddha Deb has captured the essence of the reaction, there is more than one product formed in the reaction. A more detailed mechanism is offered, an issue not addressed by the original authors.
Carbonate 1 is a racemate. No products derived from reaction at the remote 7,8-double bond were detected. Understandably, the rate of closure is faster for a 6-membered ring than a 7-membered ring. A transition state for ring closure for a reversibly formed iodonium species will be favored having the carbon chain R equatorially disposed in the transition state. Transition state 2 is favored over that of 5 because the latter encounters a gauche butane interaction. In the iodonium cation the C-C bond is shorter than the C-I bond thereby having the former of greater steric consequence than the latter. SN2 attack on the iodonium cation is favored by the more polarized carbonyl oxygen than the oxygen attached to the t-butyl group. In the latter instance, establishing a positive charge on this oxygen is counterproductive because it is adjacent to a partially positive carbonyl carbon. This objection is in keeping with why amides are preferentially protonated on oxygen and not nitrogen. The cations 3 and 6 are now poised to lose a t-butyl cation, which, given the solvent acetonitrile, may lead to Ritter intermediates besides t-butyl iodide. In the event, racemic cyclic carbonates  4 and 7 are produced in 69% yield with the cis-substituted isomer 4 predominating over the trans-isomer 7.

P.A. Bartlett,  J. D. Meadows, E. G. Brown, A. Morimoto and K. K. Jernstedt, J. Org. Chem., 1982, 47, 4013. https://doi.org/10.1021/jo00142a002


Ring Expamsion And Final Structure [If you don't like to read the question just look at the 3 pictures the first one has the carbocation which have to be stabilised by ring expansions the other 2 are my predicted structure so please confirm which one is correct or if both are wrong please provide the correct one]

please help me in expandin the rings in the following structure :

I tried a lot expanding the the rings twice to get this :

but the intemediate in this structure is unstable as it has a bridge carbocation.
instead we can have a intermediate with sigma pi resonace stablising the carbocation by expanding the ring once as follow  :

I cant wrap my head which one is correct?

There are mainly two factors governing this phenomenon of ring expansion and rearrangement of carbocation, mainly Stability factor and Thermodynamic factor. When you have a carbocation which can rearrange within itself, it will definitely prefer a rearranged structure which has the highest stability. Not only that, the carbocation also needs some energy for the rearrangements to occur. If it needs multiple steps of rearrangements to attain a structure which has decent stability, but the loss in the potential energy can't compensate the requirement of energy for rearrangement, it is not also possible for the molecule to undergo that many number of steps thermodynamically.

Based on these two factors, we can at least logically conclude which will be the final rearranged product. In both of your predicted compounds, first there will be a rearrangement by Hydride shift. Now it is highly difficult for a molecule to undergo two rearrangements at once unless it gets  extreme high stability by the second case. In this case, it does by forming a cyclopropyl methyl cation which is unusually highly stable due to non-classical resonance and thus ring-expansion occurs in this case. So, by these two steps you have got your first possible carbocation.

Now , if we look for the third possible rearrangement of ring-expansion, here energy requirement is extremely high as it becomes more and more difficult for a molecule to undergo rearrangement as its number increases as a single reaction intermediates. Also, the newly formed carbocation is tertiary which is obviously less stable than cyclopropyl methyl carbocation. Thus, the last ring-expansion is energetically less favourable.

So, I think the correct one should be the image you have drawn at the end in your question, as further more rearrangements on that system is both thermodynamically and stability wise less feasible.


Organic synthesis problems I am an A-level student going into university and I am looking for a source of organic synthesis problems and answers that is freely available on the internet. The difficulty level I am looking for is something like first year undergraduate level. Any links would be greatly appreciated.

Search for the latest IChO tests and preparatory problems. You can find quite a few interesting problems.

Here are some tests:
http://www.iuventa.sk/files/documents/5_icho/past%20competition%20tasks/volume3-icho41-45.pdf

http://icho2013.chem.msu.ru/materials/Preparatory_problems_IChO_2013.pdf (problems 18-21)

http://icho2014.hus.edu.vn/Preparatory-Problems/0001/01/Preparatory-Problems-224.html (problems 18-26)


What is the origin of 'tenting' or 'roofing' in NMR spectra? Given an NMR spectrum with two coupled nuclei one can often observe so-called 'roofing' – an asymmetry in the multiplet.

For example, here is a 300 MHz proton spectrum of 1-bromo-2-methoxyethane, where the inset shows the triplets corresponding to the adjacent $\ce{CH2}$ groups:

(Source: Hans Reich, University of Wisconsin)

What is the origin of that effect?

Not just coupled, but strongly coupled; these roofing effects tend to be seen when the coupling constant $J$ is on the order of $\Delta \nu$ (the difference in resonance frequencies between the two spins).

The origin is quantum mechanical in nature. I will try to give a simplified overview of what is happening without going into the explicit details. For a pair of coupled spin-1/2 nuclei, labelled 1 and 2, the Hamiltonian in a magnetic field is

$$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z} + 2\pi J(\vec{I}_{\!1}\cdot\vec{I}_{\!2})$$

If the spins are not coupled at all, then

$$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z}$$

and the eigenstates of the Hamiltonian are exactly

$$\begin{array}{ccc}
\text{Eigenstate} & \text{Value of }m_1 & \text{Value of }m_2 \\ \hline
|\alpha\alpha\rangle & +1/2 & +1/2 \\
|\alpha\beta\rangle & +1/2 & -1/2 \\
|\beta\alpha\rangle & -1/2 & +1/2 \\
|\beta\beta\rangle & -1/2 & -1/2
\end{array}$$

NMR theory dictates that only transitions with $\Delta M = \pm 1$ are observable, where $M$ is the sum of the quantum numbers $m_1$ and $m_2$. For example, $|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle$ is an observable transition ($\Delta M = \pm 1$), but $|\alpha\alpha\rangle \leftrightarrow |\beta\beta\rangle$ is unobservable ($\Delta M = \pm 2$). From the table above one can see that there are four observable transitions:

$$|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle; |\alpha\alpha\rangle \leftrightarrow |\beta\alpha\rangle; |\alpha\beta\rangle \leftrightarrow |\beta\beta\rangle; |\beta\alpha\rangle \leftrightarrow |\beta\beta\rangle$$

Each transition corresponds to a fixed energy, and hence each observable transition gives rise to one peak in the NMR spectrum. The area under the peak - or loosely speaking, the height of the peak - is related to how much "$\Delta M = \pm 1$ character" there is in each transition. Of course, there is a way to formalise this, but for the sake of simplicity that will have to be omitted.

For the transition $|\alpha\alpha\rangle \leftrightarrow |\beta\beta\rangle$, this is entirely $\Delta M = \pm 2$. So, the amount of "$\Delta M = \pm 1$ character" is exactly zero, and the transition is entirely unobservable.

On the other hand, for all four of the allowed transitions, the amount of "$\Delta M = \pm 1$ character" is exactly one, and hence all transitions have equal intensity. [Of course, because of the lack of coupling, there will be two degenerate pairs (i.e. there are two pairs of transitions with the same energy), and so you get two singlets in the spectrum.]

Now, as the coupling term $2\pi J(\vec{I}_{\!1}\cdot\vec{I}_{\!2})$
becomes larger and larger, the eigenstates are gradually shifted away from these. These can be expressed in terms of linear combinations of the four original eigenstates. Without proof I will present the results

$$\begin{array}{cc}
\text{New eigenstate} & \text{Expression in terms of old eigenstates} \\ \hline
|1\rangle & |\alpha\alpha\rangle \\
|2\rangle & \cos\theta|\alpha\beta\rangle + \sin\theta|\beta\alpha\rangle \\
|3\rangle & -\sin\theta|\alpha\beta\rangle + \cos\theta|\beta\alpha\rangle \\
|4\rangle & |\beta\beta\rangle
\end{array}$$

where $\theta$ is defined by $\tan(2\theta) \equiv J/\Delta\nu$. As the ratio of $J$ to $\Delta\nu$ goes from zero to infinity, $\theta$ goes from $0$ to $\pi/4$. It is easy to verify that the deviation from the old eigenstates increases as this ratio $J/\Delta\nu$ increases.

Let's look at the transition $|1\rangle \leftrightarrow |2\rangle$. How much "$\Delta M = \pm 1$ character" is there here? Since $|2\rangle$ is a mixture of $|\alpha\beta\rangle$ and $|\beta\alpha\rangle$, the transition $|1\rangle \leftrightarrow |2\rangle$ can be thought of to be a "mixture" of $|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle$ and $|\alpha\alpha\rangle \leftrightarrow |\beta\alpha\rangle$.

So, there is $\cos\theta$ amount of "$\Delta M = \pm 1$ character" arising from the $|\alpha\alpha\rangle \leftrightarrow |\alpha\beta\rangle$ contribution; and there is also $\sin\theta$ amount of "$\Delta M = \pm 1$ character" from the $|\alpha\alpha\rangle \leftrightarrow |\beta\alpha\rangle$ contribution, for a total of $\cos\theta + \sin\theta$.

For quantum mechanical reasons it is necessary to take the square of this quantity, i.e.

$$\begin{align}
(\cos\theta + \sin\theta)^2 &= \cos^2\theta + \sin^2\theta + 2\sin\theta\cos\theta \\
&= 1 + \sin(2\theta)
\end{align}$$

and working through the list of transitions, one finds that the amount of "$\Delta M = \pm 1$ character" for each transition is

$$\begin{array}{cc}
\text{Transition} & \text{Amount of “}\Delta M = \pm 1\text{ character"} \\ \hline
|1\rangle \leftrightarrow |2\rangle & 1 + \sin(2\theta) \\
|1\rangle \leftrightarrow |3\rangle & 1 - \sin(2\theta) \\
|2\rangle \leftrightarrow |4\rangle & 1 + \sin(2\theta) \\
|3\rangle \leftrightarrow |4\rangle & 1 - \sin(2\theta)
\end{array}$$

Aha! So, as long as $2\theta = \tan^{-1}(J/\Delta\nu) \neq 0$, the intensities of the peaks will be different. There will be two strong peaks (those with plus signs), and two weaker peaks (the minus signs).

When the ratio $J/\Delta\nu$ is small, i.e. the weak coupling limit, then $\sin(2\theta)$ is small and the difference in intensity is hardly noticeable - not something that you can tell by eye. For example, in a $\pu{400 MHz}$ spectrometer, two protons that have a chemical shift difference of $\pu{1 ppm}$ would have $\Delta \nu = \pu{400 Hz}$. If we take $J = \pu{7 Hz}$, then $2\theta = \tan^{-1}(7/400) = \pu{0.0175 rad}$ and we have $\sin(2\theta) = 0.0175$, a difference which you'd be hard pressed to spot!

But when this ratio increases, the value of $\sin(2\theta)$ will increase, and the difference in intensity will become more and more pronounced. I won't go into the maths, but it is possible to show that the middle two transitions have the higher intensity, and the outer two transitions have the lower intensity - which, of course, gives rise to the "roofing effect" which you speak of.


Equations for the oxidation reactions of tea teaflavins and EGCG with potassium permanganate For my chemistry investigation, I used Stamm's potassium permanganate method to investigate tea tannins with potassium permanganate to find the tannin content of teas. The method assumes complete oxidation of tea polyphenols to carbon dioxide.

Assuming that the tea polyphenols consist of only theaflavins, $\ce{C29H24O12},$ and Epigallocatechin gallate (EGCG), $\ce{C22H18O11},$ could someone help me with finding the equation of the reaction between the listed polyphenols and potassium permanganate respectively, i.e. theaflavin reaction with potassium permamganate and EGCG reaction with potassium permanganate (preferably 2 equations for each redox, but 1 overall equation should be fine too)?

I tried looking for these everywhere, but no source seems to give me the equations I need.

The two organic compounds are ultimately being converted to carbon dioxide, $\ce{CO2}$, and water, $\ce{H2O}$, just as they would be in complete combustion with pure oxygen gas, $\ce{O2}$, in a bomb calorimeter. Thus the two balanced combustion equations would be as follows:

$$\ce{C29H24O12 + 29 O2 -> 29 CO2 + 12 H2O} \tag{1}$$

$$\ce{C22H18O11 + 21 O2 -> 22 CO2 + 9 H2O} \tag{2}$$

So now assume that potassium permanganate, $\ce{KMnO4}$, is being used as the oxidizing agent, with manganese dioxide, $\ce{MnO2}$, as the product. Then, as per Ron's answer (https://chemistry.stackexchange.com/a/85865/79678), and assumed reasonable conditions therein, we have the following reaction:

$$\ce{4 MnO4- + 4 H+ -> 4 MnO2 + 3 O2 + 2 H2O} \tag{3}$$

Showing the spectator potassium ions, and arbitrarily assuming hydrochloric acid as the source of the hydrogen ions, yields

$$\ce{4 KMnO4 + 4 HCl -> 4 MnO2 + 3 O2 + 2 H2O + 4 KCl} \tag{4}$$

Since 29 $\ce{O2}$ and 3 $\ce{O2}$ have no common factor other than unity, multiply equation 1 by 3, equation 4 by 29, add them, cancel the equal number of $\ce{O2}$ molecules on each side and the result is

$$\ce{3 C29H24O12 + 116 KMnO4 + 116 HCl -> 87 CO2 + 116 MnO2 + 94 H2O + 116 KCl} \tag{5}$$

Since 21 $\ce{O2}$ is 7 times 3 $\ce{O2}$, use equation 2 as is, multiply equation 4 by 7, add them, cancel the equal number of $\ce{O2}$ molecules on each side and the result is

$$\ce{C22H18O11 + 28 KMnO4 + 28 HCl -> 22 CO2 + 28 MnO2 + 23 H2O + 28 KCl} \tag{6}$$


Why can't two Silicon tetraoxide tetrahedrons share an edge or a face? Our teacher told us about structural types of silicate minerals (neosilicate, sorosilicate, cyclosilicates ...) saying that those tetrahedrons can only share one corner (oxygen), but he didn't give us any explanation, could anyone help me with this?

In solid-state chemistry, we know a certain set of rules called the Pauling rules. One of them states that sharing faces or edges is unfavorable. Imagine your silicate-tetrahedra to share faces. That would equal the closest distance between two highly charged silicon cations. They both have a charge of +4, so there is really some repulsion. Same goes for the edges which is also quite close.

If you increase the overall size of the tetrahedron however by exchanging oxide by sulfide you will end up in the silicon(IV)sulfide structure which is essentially long chains of tetrahedra that share edges. Here by introducing a much larger anion the distance between the two silicon centers is increased and edges can be shared.


Concentration of expanding gases At $t = 0$ we have 10 moles of $\ce{NO2}$ and 10 moles of $\ce{N2O4}$ satisfying this reaction in a 0.75 liter tank (the barrier is closed at this time):

$$\ce{N2O4 <=> 2NO2}$$

The system is in thermal contact with the environment.

At time t=t0 the barrier is released and the gases are free to expend to the other tank ($V_2=1$ liter)

The question is what will be the concentration of $\ce{NO2}$ when they reach equilibrium.

What I tried is calculating the chemical equilibrium before based on the concentration quotient $\text{K}_c$ (13.33) and then to use it as reference after the system reaches equilibrium. I got a weird result (negative number) Where have I gone wrong?

This is a pretty straight forward equilibrium problem in disguise. If we assume the initial state represents equilibrium, we can determine the equilibrium constant by plugging in the concentrations to the law of mass action. We can get the concentrations because we have the number of moles and the volume of the container.

$$\ce{N2O4 <=> 2NO2}$$

$$K_c=Q_{eq}=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}$$

Okay, now we let the gas expand. Being in thermal contact with the surroundings is important, since it means the temperature will not change (which means the equilibrium constants does not change). Now we will find that we are not at equilibrium any more. We will probably find that $Q_c < K_c$, so equilibrium needs to shift toward $\ce{NO2}$.

Now, we can set up the ICE (initial, change, equilibrium) table:

          [N2O4]           [NO2]
I         5.71 M           5.71 M
C         -x               +2x
E         5.71 -x          5.71 + 2x

Substitute into the law of mass action at equilibrium:

$$K_c = \dfrac{(5.71+2x)^2}{5.71-x}=13.33$$

Rearrange and solve for $X$. Being a quadratic equation, you should get two roots (two solutions for $x$), and as is always the case for an equilibrium problem, one of them will be nonsense. For example, you could get one value of $x$ to be larger than the initial concentrations!


Decay scheme of sodium-24 Sodium-24 decays into an excited nucleus of magnesium-24. Magnesium reaches nuclear ground level via two consecutive gamma emissions. What are the half-lives (time scales) for these two processes? I think that those gamma photons are emitted very rapidly after Mg-24 is formed. I'm curious to know but I cannot find data. I'm a Chem undergrad with interests in nuclear chemistry.

For energy levels in nuclei (and their gamma emission) the first place to look is at the Evaluated Nuclear Structure Data Files. I usually use the Brookhaven site, but there may be a closer mirror to you. Enter 24 into the Nuclide or Mass box and hit Enter. Check the 'Adopted Levels, Gamma' box by $^{24}$Mg, and request the pdf version (back up at the top).
On the last page you will find the lower levels like this (although rotated by 90 degrees):

So, the 4122.889 keV level has a decay time of 22 fs - this is the level that the first gamma decays from. The 1368.672 keV level has a decay time of 1.33 ps - this is the second gamma in your sequence.


What an electrolyte "really" does in electrolysis? I am reading about electrolysis and I have trouble to understand why an electrolyte must be added in order for electrolysis to occur. I have read this question but still can't get it. The definition of electrolyte is a substance that when dissolved produces an electrically conducting solution.
Electrolytes in electrolysis
Why we need an electrolyte in order for electrolysis to occur? In electrolysis of water the water is consumed not the electrolyte. So why we need the electrolyte? Couldn't the reaction happen without the electrolyte?
After reading the comments I have modified the question. As I said before I can't understand the purpose of the electrolyte. For example in the electrolysis of water with sulfuric acid why we add it? Water molecules are in contact with both anode and cathode. So why we need to add the electrolyte?

Sometimes the electrolyte directly takes part in the chemistry, by being oxidized or reduced at an electrode. In those cases, it is easy to understand why the concentration would have an influence on the kinetics of the redox reactions (and with that, on the observed electrical current).
For water electrolysis, the "active" electrolytes are hydrogen ions and hydroxide ions. If you change the pH, you change their concentration, and would expect to affect the rate of reaction. You are asking about other electrolytes such as sodium sulfate, which don't participate in electron transfer.
To explain their role, take a look at this animated GIF. It shows electrolysis of water in the presence of universal indicator and sodium sulfate:

Source: CarolinaBiological
With the 9 volts applied to the graphite electrodes, you can see that the electrolysis causes local changes in pH (because one half reaction consumes hydrogen ions and the other hydroxide ions). It takes quite a bit of effort to mix the solution (with the plastic tube) to get the pH homogeneous again. This shows you that the diffusion of hydrogen and hydroxide ions takes a while. In the absence of other electrolytes, this would lead to local charge accumulation as well, stopping the reaction. With the electrolyte (at much higher concentration than the hydrogen and hydroxide ion), we get a pH gradient only.
This pH gradient also slows things down, and increases the voltage that needs to be applied (cell potential is 1.3 V at constant pH, but according to Nernst law, it would increase because of the pH differences). Because we are applying 9 volts, this still works.

What an electrolyte “really” does in electrolysis?

By moving in the appropriate directions, the ions of the "passive" electrolyte prevent large accumulations of charge, and allow the electrolysis to proceed at higher rates than without their presence. If you decrease the gap between electrodes down to sub-Debye length, they are no longer neccessary.


Why are fires smokey? Combustion, in chemical terms, is a reaction with a certain molecule and oxygen, and it produces, energy, $\ce{CO2}$, and $\ce{H2O}$ (That's how I learned it at least). So what makes a fire'smokey'? I had always thought that smoke was simply $\ce{CO2}$, but read that it was actually the result of incompletely 'combusted' material, yet I don't fully understand what this means. Is smoke made from material that starts to combust but stops before it finishes, or does it never even begin combustion? What would the chemical reaction be of a wood (Or other similar) fire that produces a lot of smoke? And finally, is there a certain element or compound that gives smoke its 'smokey' smell?

Fires need oxygen to burn properly. By that I mean if there isn't sufficient oxygen supplied to the fire, the material that is burning does not completely burn up. They come to the middle of the combustion reaction but they run out of oxygen. These bits of incomplete combustion are called soot or smoke. $\ce{CO2}$ is another by-product of combustion. There are particular woods that give off certain smells and flavours like hickory, but smokey smells are probably only caused by carbon.