Safety when handling cyclohexylamine in non-industrial amounts Today I was pipetting some cyclohexylamine into smaller bottles so that I don't have to use the liter flask every time I need a little bit of catalyst or plasticizer. I pipetted several 10-20 mL portions in the open air on a regular table.

Because I lack experience I'm extra precautious with what I do. From the MSDS's of cyclohexylamine I found that 10 mL will kill an adult male, which if you look at is isn't very much, less than a swig of water.

Of course I'm wearing a lab coat, safety goggles and chemical resistant gloves, but a few things concerned me (or at least made me wonder if it was the right/safe way to handle). According to the MSDS of ScienceLab the odor threshold is 26 ppm, while according to CAMEO, the PAC-3 for cyclohexylamine is 30 ppm, which Wikipedia defines as:

  PAC-3 : Life-threatening health effects.

When I opened the bottle the amine fish odor was very strongly noticeable, I however found no irritating effect on my eyes or lungs, although according to the documents provided above I was inhaling a lethal concentration. I'm quite confused here. More easily put as a question: as long as I don't feel irritation of my lungs or eyes, am I safe? And if I do, is there enough time to clean up (i.e. the most essential things, like shutting off electricity and closing open jars of chemicals) and leave?

Finally I accidentally touched a drop of cyclohexylamine (from the pipette) when I was cleaning up, with no gloves on. Although cyclohexylamine is listed as highly corrosive, I did not feel any irritation at all. Then I rinsed my pipette 3x with distilled water and put the waste in a jerrycan, but I still haven't figured out if I can pour the diluted mixture of water and cyclohexylamine down the drain, without poisoning myself or the neighborhood. The MSDS's keep quiet about that.

Can someone clear some confusions up for me. I will be greatly appreciated!

I once did some work with some small chain alkyl amines and they were bad news.  They would actually burn you if some got on your skin.  I always used a respirator and worked in a hood when using them.

  as long as I don't feel irritation of my lungs or eyes, am I safe?

If this is a one time occurrence, given the small quantities, I'd say the odds are way in your favor, but they call it "odds" for a reason.  My advice would be not to do it again without a respirator and hood where the exhaust is treated.

  And if I do, is there enough time to clean up (i.e. the most essential
  things, like shutting off electricity and closing open jars of
  chemicals) and leave?

If you ask an industrial hygiene professional, I'm sure the answer would be "no".  If it were me and I just began to notice some very mild sensation, I think I'd hold my breath and try to shut things down.  I know it doesn't make sense, but that's what I'd do.  If I noticed anything beyond the onset of very minor discomfort, I'd clear out fast and call the fire department.

  Finally I accidentally touched a drop of cyclohexylamine (from the
  pipette) when I was cleaning up, with no gloves on.

I'd keep washing it every so often during the day and I'd put some burn cream on it in between the washings.  I'd also take a good shower.  No harm in overreacting.

  still haven't figured out if I can pour the diluted mixture of water
  and cyclohexylamine down the drain

I'm almost positive that the answer is no.  Besides you and the neighborhood, it might react with the pipes or what's stuck in the pipes and create longer term toxicity problems when you're in that room at some later date.   Where I live, we have hazardous chemical drop-off centers (mainly for paint, oil and household chemicals, but used for everything chemical).  I'd pay the relatively small amount they charge and drop it off there.

One final note, even when using the right protective gear, some people become "sensitized" to amines.  This means that future exposure at even lower levels to a wide array of amines can be more dangerous.  Also, I'd try to open up the exposed room to as much outside air as possible, use a room fan too if you have one.


How does chlorine form more than 1 bond? How are perchlorate or chlorate or chlorite ions and their respective acids or compounds formed. $\ce{Cl}$ can't form more than one bond but still...

 $\rightarrow$'Perchlorate ion'

 $\rightarrow$'Chlorous acid containing Chlorite ion'

 $\rightarrow$'Chlorate ion'

Chlorine is making more than 1 bond... How is it possible?

The order of bonding, and so the valence state of Cl in $\ce{ClO_{x}-}, x>1$ compounds is very debatable. Generally, two models exist.

$\ce{Cl}$ atom, just like $\ce{S, P}$ and some others has unoccupied $d$-orbitals in the valence shell. It is possible to move some electrons from $p$-orbitals to $d$-orbitals, producing half-occupied orbitals that can participate in covalent bond formation
$\ce{O}$ atom has 6 electrons at the valence shell with four orbitals. It is possible to move one electron from chlorine atom to oxygen atom to form $\ce{O^{\dot{-} }}$ ion, that can than form a bond with newly formed half-filled orbital of chlorine. However, being electronegative and having high positive charge, chlorine partially pulls back the electron pair of newly formed bond.

Currently the second option is in favor of theoreticians, both by electron population analysis and relatively high charges in the oxochloric compounds, however, in school-level chemistry people traditionally resort to the first option. On the other hand, the second approach is the only option to adequately describe nitric acid and nitrates as nitrogen does not have $d$-orbitals on the valence shell.

Now, the stability of oxochloric compounds is a completely different matter. All of them are not very stable compounds, easily freeing their oxygen. Still, the process has significant activation energy, the state in between is even less favorable then oxochloric molecule. both oxygen and chlorine have electron-rich shells that repel potential reactants. Perchlorate ion is especially stable, at its surface is completely covered by outer negative shells of oxygen, so electrons cannot reach positively charged chlorine.


Significance of Temperature in Gibbs Free Energy Why is it important to know the temperature of both reactants and products of a reaction in order to determine change in the Gibbs free energy of the reaction?

Is it because temperature has an effect on both enthalpy and entropy? Or is it because of some other reason I am missing?

Any help on this would be great. Thanks!

Change in the Gibbs free energy ($\Delta$$\ce{G}$) of a reaction at any temperature can be determined by the equation <$$\ce{$\Delta$G = $\Delta$G^0 + RTlnQ_c (Q_p if the reactants and products are gaseous)}$$ Here, $\Delta$$\ce{G^0}$ is the standard Gibbs-free energy change of the reaction which is determined by,$$\ce{$\Delta$G^0 = $\Sigma$a_iG^0_i_{products} - $\Sigma$b_iG^0_i_{reactants}}$$ where $\ce{a_i}$s and $\ce{b_i}$s are proper stoichiometric coefficients. Now, $\ce{G^0}$ of products or reactants are determined by producing 1 mole of ech of them from their corresponding constituting elements in their standard states and at a FIXED TEMPERATURE(generally at $\ce{298.15K}$). So,  $\ce{$\Delta$G^0}$ is determined also at that particular temperature,So, it is independent of Temperature.  But,when we have to calculate $\ce{$\Delta$G}$ for any reaction at any temperature($T$) by the above given relationship , we have to know the temperature of the reaction condition as you can see clearly, $\ce{$\Delta$G}$ is directly  dependent on $\ce T$.That's why, it's important to know the temperature.


Can salt and water be obtained from a reaction between a gas and an acid? Is it possible to get salt and water (neutralization reaction) by reacting a gas and an acid?

I don't know if it's possible or not. I'm unable to find any examples in this regard.

Would anyone be able to provide an example in support of the question?

If you want to get salt, then you should look for a volatile base.
$\ce{NH3}$ does our job here.
$$\ce{2NH3 + H2SO4 -> (NH4)2SO4}$$

However, I am not sure if water could be formed. If there exists something like $\ce{NH4OH}$ vapours, then sure, water will be formed.


Product of this Michael addition I'm having problems when trying to know the outcome of this reaction.

The first step must be a Michael addition of the 1,3-dicarbonyl compound to the  α,β-unsaturated carbonyl compound, followed by a acid hydrolisis and a decarboxylation of the ester.

LDA must deprotone the least hindered hydrogen, that must be the $\alpha$-$\ce{CH3}$ to the non cyclic ketone. But at this point, I don't know how this reaction can continue to give one of the above products.

Although the work of Wilds and Werth1 does not align perfectly with the conditions of the OP's question, it does illustrate that Robinson annulation and decarbomethoxylation can occur under alkaline conditions. Catalytic base was found to be optimal to effect the Michael addition with methyl vinyl ketone (MVK). The aldol condensation was conducted in aqueous KOH at reflux.
Whether decarbomethoxylation occurs prior to aldolization (2-->3-->5) or subsequently (2-->4-->5) was not addressed. α'α-Dialkylated β-ketoesters generally undergo attack at the keto group rather the carboxylate by base. This generality may be tempered by the aromatic nature of the keto group in 2. Initial aldolization of 2 to 4 affords a structure that is a vinylogous β-ketoester; it is susceptible to base initiated decarboxylation.

1) A. L. Wilds and R. G. Werth, J. Org. Chem. 1952, 17, 1154.


Tosyl Chloride? why not anything else Why do we encounter only Tosyl groups and not anything at the end of the benzene group (ie not to replace the SO2Cl part..rather the Me part at the other end)
I mean something like a CF3 or NO2 would help make it a better leaving group right?

Part of the reason why p-tosyl chloride is seen so often is not only the ease of its preparation, yet that is a side product in the preparation of o-tosyl chloride towards saccharin.  Ease in storage and addition / use are beneficial.  While a p-nosyl chloride eventually may yield a better nucleofuge than the p-tosyl chloride, the very same nitrogroup may ease the hydrolysis of the chloride as well.


Why do we call a benzene ring attached to an hydroxyl functionality an alcohol and not a benzene? But if an alkyl group is attached to a benzene ring, we call the parent chain the benzene ring, no matter how many carbons the alkyl group has. Why? For example: heptylbenzene.

For another example: when 1-propenyl-benzene reacts with sulfuric acid we get 1-phenyl-1-propanol. We don't call this product 1-propanol-1-benzene.

In your first example, heptylbenzene, the benzene ring is the parent structure for the preferred IUPAC name because the ring is always the parent structure if the ring and the chain have the same number of substituents that are cited as the principal characteristic group (by means of a suffix or a class name, or implied by a trivial name). The corresponding rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

In heptylbenzene, neither the benzene ring nor the heptane chain has any substituents that could be cited as the principal characteristic group. Therefore, the ring is selected as the parent structure.
In 1-phenylpropan-1-ol, however, the propane chain has an $\ce{-OH}$ group, which is the principal characteristic group of this compound and expressed as suffix (‘-ol’). Thus, Rule P-52.2.8 cannot be used. If the ring and the chain have different numbers of substituents corresponding to the principal characteristic group, another rule comes first, which stipulates that the senior parent structure is the one that has the maximum number of substituents cited as the principal characteristic group.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

Therefore, the propane chain is the parent structure in 1-phenylpropan-1-ol.


What would be the unit for the calculation of the rate of reaction? In a reaction between magnesium and hydrochloric acid, I measured the change in mass of the magnesium strip (Initial weight: 0.03g). I calculated the the time for the reaction to reach completion. A sample set of data we collected for during our investigation was:

With 8 molar of HCl the reaction took an average of 15.49 seconds to completely evolve into hydrogen gas.

To calculate the rate of reaction, I used the formula:

I suppose, change in mass would be 0.03 g as we went to the completion of the reaction, and change in time would be noted as the average time 15.49. My question is, what is the unit supposed to be in this formula?

The formula has been stated and you simply need to look out for the units of the quantities
 from the formula. Mass(g)/Time(s). Which is g/s.


Assumptions to be taken when rearranging terms in first law of thermodynamics Assuming a closed system, show how the first law of thermodynamics.
$$\mathrm dU = \mathrm dQ + \mathrm dW$$
becomes,
$$\mathrm dQ = \frac{C_V}{R}V\,\mathrm dP + \frac{C_p}{R}P\,\mathrm dV$$

I have an answer, but am hesitant to say it is a final answer or even the correct one at that.
I am concerned with the partial derivatives when rearranging this equation. How do certain parts cancel out or how does one approach the problem without them? Is it safe to assume that since you know how the state variables will act in a certain process, that you can use that knowledge to provide a better or more exact answer to the problem?
Attached you will find a picture of the problem that I have done to the best of my ability (which is limited I must say).
How do I make my assumptions?
Do I assume holding one part constant to solve the equation in one way - that is via an adiabatic process, isothermal process - do the same for another path and compare?

This equation can not be applied in general because it can't describe the amount of heat transferred to an incompressible solid or liquid at constant pressure.
For these cases, it would predict that $\mathrm dQ=0$ (which is not necessarily correct).  If it is not generic, it also cannot describe real gases beyond the ideal gas region.  Therefore, the equation must apply exclusively to an ideal gas. The presence of the ideal gas constant $R$ in the equation is a dead giveaway.
For a reversible path,
$$\mathrm dQ=T\mathrm dS=C_v\mathrm dT+P\mathrm dV$$
But, for an ideal gas,
\begin{align}
&\mathrm dT =\frac{P\mathrm dV+V\mathrm dP}{R}\\
&C_v+R = C_p
\end{align}
Therefore,
\begin{align}\mathrm dQ&=C_v\frac{P\mathrm dV+V\mathrm dP}{R}+P\mathrm dV \\ &=\frac{C_v}{R}V\mathrm dP+\frac{(C_v+R)}{R}P\mathrm dV \\
\implies\mathrm dQ&=\frac{C_v}{R}V\mathrm dP+\frac{C_p}{R}P\mathrm dV
\end{align}
This equation applies only to an ideal gas and a reversible path.


Oxidizing agent, reducing agent, acid or base In concrete reactions, how can you define whether Ammonia reacts as an oxidizing agent, reducing agent, an acid or a base? I'm confused because in the examples given it's not an acid/ base when an $\ce {H+}$ is donated/ accepted.

$$\ce {2NH3 + 3Cl2 -> N2 + 6HCl}$$ (oxidizing agent)

$$\ce {Na + NH3 -> NaNH2 + \frac{1}{2}H2}$$ (reducing agent)

$$\ce{NH3 + H2O + CO2 -> NH4HCO3}$$ (base)

$$\ce{NH3 + LiCH3 -> LiNH2 + CH4}$$ (acid)

The use of the terms "oxidizing agent" and "reducing agent" in this example apparently refer to the overall reaction. Ammonia certainly can act as an acid in the process of reacting with sodium metal, and as a (Lewis) base during the reaction with chlorine. However, the net effect of the reaction is a reduction or oxidation, rather than a simple proton transfer.


Conservation of angular momentum in rotational spectroscopy In spectroscopy, transitions between rotational energy levels obey the selection rule $\Delta J = \pm1$. The rule follows from the fact that the photon absorbed has an angular momentum of $\pm \hbar$, so the molecule must change its angular momentum for the total angular momentum to be conserved.

However, the angular momentum of a molecule is given by $\hbar \sqrt{J(J+1)}$, so the change in molecular ang. mom. is $\hbar\sqrt{(J+1)(J+2)} - \hbar\sqrt{J(J+1)}$. As $\hbar\sqrt{(J+1)(J+2)} - \hbar\sqrt{J(J+1)} \neq \hbar$, it seems that the total angular momentum is not actually conserved.

Questions:

Does the conservation of angular momentum apply to quantum systems?
If yes, what is the source of the contradiction above?

Sorry for the late answer, but I could not find the time to write something earlier. A concept that you encounter often in quantum mechanics is the so-called coupled and uncoupled basis. Your question is a variation on that theme: before the absorption the uncoupled representation is the more natural where the angular momentum of the photon and the angular momentum of the molecule (and their projections) are good quantum numbers, whereas after the absorption the coupled representation is more suitable and the total angular momentum of the system (and it's projection) is a good quantum number. In Dirac's braket notation we have

$$
|jm_j\rangle |j_\text{ph}m_{j_{\text{ph}}}\rangle
$$
for the uncoupled representation and
$$
|JM_J\rangle
$$
for the coupled representation.

Your question concerns the conservation of the total angular momentum of the system $\hat{J}^2$ and in the coupled representation the expectation values are easily obtained because $J$ is a good quantum number
$$
\hat{J}^2|JM_J\rangle=\hbar^2 J(J+1)|JM_J\rangle
$$

In the uncoupled representation the total angular momentum is given by $\hat{J}^2=\left (  \hat{j}+\hat{j}_\text{ph} \right)^2=\hat{j}^2+\hat{j}_\text{ph}^2+2\hat{j}\cdot \hat{j}_\text{ph}$

The first two terms of this expression are again easy, but the thrid term is a bit tricky. Using the ladder operators you can write it as
$$
2\hat{j}\cdot \hat{j}_\text{ph}=2(\hat{j}_x\hat{j}_{\text{ph},x}+\hat{j}_y\hat{j}_{\text{ph},y}+\hat{j}_z\hat{j}_{\text{ph},z})=\hat{j}_+\hat{j}_{\text{ph},-}+\hat{j}_-\hat{j}_{\text{ph},+}+2\hat{j}_z\hat{j}_{\text{ph},z}
$$
where
$$
\hat{j}_\pm|jm_j\rangle=\hbar\sqrt{j(j+1)-m(m\pm1)}|jm_j\pm1\rangle
$$
and it thus couples states of different $m_j$. Let us, for example, couple $|jm_j\rangle=|1m_j\rangle$ and $|j_\text{ph}m_{j_\text{ph}}\rangle=|1m_{j_\text{ph}}\rangle$ so that we have $(2j+1)(2j_\text{ph}+1)=9$ basis states:
$$
|1,-1\rangle|1,-1\rangle,\\|1,-1\rangle|1,0\rangle,\\|1,-1\rangle|1,1\rangle,\\
|1,0\rangle|1,-1\rangle,\\|1,0\rangle|1,0\rangle,\\|1,0\rangle|1,1\rangle,\\
|1,1\rangle|1,-1\rangle,\\|1,1\rangle|1,0\rangle,\\|1,1\rangle|1,1\rangle
$$

We determine all expectation values of the total angular momentum operator using the 9 basis states which gives us a total of 81 values that we can order in a 9x9 matrix (in units of $\hbar^2$)

6   0   0   0   0   0   0   0   0
0   4   0   2   0   0   0   0   0
0   0   2   0   2   0   0   0   0
0   2   0   4   0   0   0   0   0
0   0   2   0   4   0   2   0   0
0   0   0   0   0   4   0   2   0
0   0   0   0   2   0   2   0   0
0   0   0   0   0   2   0   4   0
0   0   0   0   0   0   0   0   6

We can diagonalize this matrix and find the eigenvalues 0, 2, 2, 2, 6, 6, 6, 6, 6. These are the same eigenvalues (including the $2J+1$ degeneracy) of $\hat{J}^2$ in the coupled representation!

So the expectation value of the total angular momentum is the same in both cases, the difference is that in the uncoupled case the matrix has off-diagonal elements and it is a bit more work to find its eigenvalues. In the coupled case, the matrix for $\hat{J}^2$ is diagonal and you get the expectation values directly. You might ask if there is an easier way to convert from the coupled to the uncoupled case and back. As you might have guessed there is such a way and it involves the so-called Clebsch-Gordan coefficients.


Fuel cell potential for the electrolysis of ammonia I am trying to calculate the fuel cell potential for the electrolysis of ammonia with the following equation:

E = Gibbs free energy for the reaction/(# electrons X Faraday's Constant)

The half reactions are as follows:

$$2NH_3 + 6OH^- -> N_2 + 6H_2O + 6e (Anode) E^0 = -0.77 V $$
$$ 6H_2O + 6e -> 3H_2 + 6OH^- (Cathode) E^0 = -0.83 V $$

I obtained the potencial values from the following paper and (double checked online as well): https://www.researchgate.net/publication/327875102_Review-Ammonia_Oxidation_Electrocatalysis_for_Hydrogen_Generation_and_Fuel_Cells
Equations (4a,4b and 4c)

The overall reaction is then:

$$2NH_3 -> N2 + 3H_2 $$ $$E^0 = +0.06 V $$ ** in the paper they seem to imply that the cell voltage is $$ + 0.06 V$$ which I don't understand why, since the oxidation takes place at the anode, therefore for the anode the potential should be $$ E^0 = +0.77 V$$ from my understanding.

I have doubled check my standard enthalpies of formation and entropies of formation. However, when I calculate the Gibbs free energy of the reaction I keep getting $$ E^0 = -0.06 V$$ for the overall reaction.
My Gibbs free energy is positive and the temperature I am using is 298 K, also NH3 is in a gaseous state. I have tried calculating the $$ E^0 $$ for $$ H_2 + 0.5 O_2 -> H_2O $$  and I obtain a cell voltage of $$ E^0 = 1.23 V $$ which corresponds to the reaction Gibbs free energy of $$ - 237 kJ/mol $$ , I just don't know where I am getting confused in the ammonia reaction.

In the paper they mention that the reaction is thermodynamically favorable, but I don't see how when the Gibbs free energy for the reaction is positive.

You can see my detailed response on the sign of electrode potentials here:
Still taught to reverse oxidation half cells in electrochemistry?

To summarize, electrochemists all over the world have decided to quote all electrode potentials as reduction potentials. This tug of war of signs between the so-called European convention and American convention of electrochemistry for more than a century, resulted in this decision. Basically, the current convention is the same as that of Ostwald.


How can I determine the pH of solution? Could anyone tell me how I would go about calculating the pH of 100.0 mL of 0.100 M $\ce{H2NNH2}$ ($K_\text{b} = 3.0 \times 10^{-6}$).

Furthermore, how would this pH change when 20.0 mL of 0.200 M $\ce{HNO3}$ is added to the solution?

I’m not looking for only an answer, but rather the general method to solve these types of problems

Just to include what I’ve tried:

I first thought that (for the initial solution) $\mathrm{pH} = 14 - \log(\sqrt{K_\text{b} [\ce{H2NNH2}]})$, but that doesn’t seem to be the case.

Concerning the first part of your problem: In the general case, which you requested, we start with the mass action law:

$$
K_\text{b} = \frac{[\ce{HB+}][\ce{OH-}]}{[\ce{B}]} = \frac{[\ce{OH-}]^2}{[\ce{B}]}
$$

That’s essentially what you just reformulated to obtain your expression for $\mathrm{pH}$ above. At this point, you made an error in the sign of the second term on the right-hand side – it should be added, not subtracted, i.e. $\mathrm{pH} = 14 + \frac{1}{2} \lg \left( [\ce{B}]K_\text{b} \right)$.

Now comes the important thing: The mass action law uses the concentrations of our components in the equilibrium! What is given as 0.1 M is, however, the starting concentration $[\ce{B}]_0$ before equilibrium, which will of course decrease; $[\ce{HB+}]$ and $[\ce{OH-}]$ will increase by the same amount, which we will denote $x$. Thus, we rewrite:

$$
K_\text{b} = \frac{[\ce{OH-}]^2}{[\ce{B}]} = \frac{x^2}{[\ce{B}]_0 - x}
$$

Now, you just solve for $x$, which is the concentration $[\ce{OH-}]$.

One more thing. You will see that $x$ is actually pretty small in this case, so $[\ce{B}]_0 - x \approx [\ce{B}]_0$. By using $K_\text{b} = \frac{x^2}{[\ce{B}]_0}$, we essentially obtain the same result. You can’t always rely on this, however, and if you have stronger acids or bases, this will get important.


Does hydrogen react with air at room temperature and pressure? I have a container with 3% hydrogen and the rest is air. How fast will the hydrogen get oxidized by air?

How much will the reaction speed up with increased temperature?

According to the Encyclopedia Britannica, to somewhat paraphrase this non-technical source, the reaction rate of $\ce{O2(g)}$ and $\ce{H2(g)}$ at atmospheric pressure is not measurable at any mixing ratio at temperatures below $\ce{300^oC}$.  Unfortunately that source does not report what kind of experimental detection limits define what "not measurable" means.
These class notes from the University of Toronto Institute for Aerospace Studies, AER1304 course state this in more dramatic fashion:

The reaction between hydrogen and oxygen is a good example of a
multicomponent kinetic system. To describe the system properly, we
should consider eight major species and at least 16 reactions.
[The overall] reaction is exothermic, but mixtures of gaseous hydrogen and oxygen are quite stable at
atmospheric conditions. Any conceivable direct reaction between the
two gases is zero.

The reaction half-time at atmospheric [pressures] has been estimated to be much larger than the age of the universe.
If the reaction is initiated by some free-radical species, then the reaction proceeds very rapidly and violently.

So the manner in which the reaction rate of $\ce{O2(g)}$ with $\ce{H2(g)}$ increases with temperature is very complex, though basically negligible in the absence of a catalyst until you reach the auto-ignition temperature.
According to this Wikipedia article:

The hydrogen autoignition temperature, the temperature of spontaneous
ignition in air, is 500 °C.

Notice that this says air not oxygen. The $\ce{H2}$ concentrations for the basis of this statement range from 4% to 74%.  This seems like quite a large range to give a single result for autoignition temperature, but given the great complexities of the $\ce{H2}$ + $\ce{O2}$ reaction, some estimations are clearly necessary in order to discuss this question in a reasonably concise fashion.
One note about the question regarding the composition and contents of the container; I'd be sure that it never contained any high surface-area platinum powder or be coated with any sort of perchlorate residues or any such ridicules thing!  The whole issue with the difficulty in getting this highly exothermic reaction to proceed is kinetics.  As eluded to in some of the comments, there are chemical catalysts that can help the reaction get over this large energy barrier, it doesn't have to be a spark!


How do polymer properties change when moving from bulk to nanosized fibers? Polymer are used in many applications, but how do the properties change when they are fabricated on the nanolevel? Im especially interested in polymeric and polymer derived nanofibers, and why they excel at their respective applications. Thanks!

Nanofiber polymers in contrast with bulk polymers have high ratio of surface area to volume.
So, nanofiber polymers could be useful for any application that needs to maximize the ratio of surface area to volume.

Let's consider their application as scaffolds for tissue engineering: increasing the surface area of artificial-organ scaffolds gives cells more points at which to adhere.

Another application of nanofiber polymers is their use as electrodes  in fuel cells:  the greater the electrodes surface area, the more efficiently they catalyze the reactions that drive the cell.

Another interersting application of polymer nanofibers is in meshes that allow only nanoscale particles to pass through: ultrafine air, water filters, membranes for fuel cells.

Finally, as the capacitance is proportional to the surface area,  capacitors that store electricity, are one example of the possible applications of  polymer nanofibers.

For more details, please see this site and this article.


Which is the most stable out of the following linked cyclohexadienes? I am unable to find the answer to this question. I was guessing it should be (2) due to symmetry but my book says it is (3) without any explanation.

I know that stability is determined by how low the energy of formation is. Does this have something to do with the different sorts of dienes that are present? Like, one type of diene would have less energy than another?

If nothing else, it would be helpful if someone could indicate the names of these compounds, because I don't even have the faintest idea where to start searching for more information.

You must be aware about the concept of resonance. Similar ly over here as you can see in option number 3 there is longer conjugation between double bonds then compared with any other option because Three double bonds are in conjugation and this conjugation supplies stability to the structure


Avogadro's number and percent error help? Calculate the percent error in Avogadro’s number calculated by using the following experimental data.
  Pipette calibrated as 30 drops/mL
  Diameter of the watch glass: $15.5\ \mathrm{cm}$
  Number of drops to complete monolayer: 23
  Concentration of stearic acid solution: $1.2\times10^{-4}\ \mathrm{g/mL}$

Updated: My thought on this: I would take $23/30=0.77\ \mathrm{ml}$. Then take $0.77\times1.2\times10^{-4}=9.24\times10^{-5}\ \mathrm g$ of stearic acid. Then $(9.24\times10^{-5})/284.48\ \mathrm g=3.25\times10^{-7}\ \mathrm{mol}$ of stearic acid. Then how do I find the Avogadro's number (molecules per mole)?

When a problem requires calculations using values, always write the values with the correct units and carry the units through the calculation. Do not omit the units while performing intermediate steps and do not just reintroduce units at the end of the calculation.

Nevertheless, your approach is correct. You calculated the volume of the solution as
$$\begin{align}
V_\text{solution}&=\frac{23}{30\ \mathrm{ml^{-1}}}\tag{1}\\[6pt]
&=0.77\ \mathrm{ml}\tag{2}
\end{align}$$
the concentration as
$$\begin{align}
c&=\frac n{V_\text{solution}}=\frac m{M\cdot V_\text{solution}}\tag{3}\\[6pt]
&=\frac \gamma M\tag{4}\\[6pt]
&=\frac{1.2\times10^{-4}\ \mathrm{g\ ml^{-1}}}{284.48\ \mathrm{g\ mol^{-1}}}\tag{5}\\[6pt]
&=4.2\times10^{-7}\ \mathrm{mol\ ml^{-1}}\tag{6}
\end{align}$$
and the amount of stearic acid as
$$\begin{align}
n&=c\cdot V_\text{solution}\tag{7}\\[6pt]
&=\frac{1.2\times10^{-4}\ \mathrm{g\ ml^{-1}}}{284.48\ \mathrm{g\ mol^{-1}}}\cdot\frac{23}{30\ \mathrm{ml^{-1}}}\tag{8}\\[6pt]
&=3.2\times10^{-7}\ \mathrm{mol}\tag{9}
\end{align}$$

Furthermore, you can calculate the area of the surface as
$$\begin{align}
A&=\pi r^2\tag{10}\\[6pt]
&=\frac{\pi d^2}4\tag{11}\\[6pt]
&=\frac{\pi \left(15.5\ \mathrm{cm}\right)^2}4\tag{12}\\[6pt]
&=189\ \mathrm{cm^2}\tag{13}
\end{align}$$

For simplicity’s sake, you may assume that a single stearic acid molecule is a small block with a length $l$ and a width $w$.

Depending on the used model, you may find that the ratio is roughly

$$l=6w\tag{14}$$

Assuming that the surface is densely packed with stearic acid molecules, the number $N$ of molecules can be estimated as
$$N=\frac A{w^2}\tag{15}$$

The corresponding volume of stearic acid is

$$\begin{align}
V&=A\cdot l\tag{16}\\[6pt]
&=A\cdot 6w\tag{17}
\end{align}$$

You also know that
$$\begin{align}
V&=\frac m\rho\tag{18}\\[6pt]
&=\frac {n\cdot M}\rho\tag{19}
\end{align}$$

where $\rho$ is the density of stearic acid. Instead of using the density of solid stearic acid at room temperature, you may want to assume the density of liquid stearic acid close to its melting point, which is

$$\rho=0.847\ \mathrm{g\ cm^{-3}}\tag{20}$$

Solving $\text{(17)}$ for $w$ and inserting the result into $\text{(15)}$ yields

$$N=A\cdot\left(\frac{6A}{V}\right)^2\tag{21}$$

And inserting $\text{(19)}$ for $V$ yields

$$N=A\cdot\left(\frac{6A\cdot\rho}{n\cdot M}\right)^2\tag{22}$$

The Avogadro constant is defined as
$$N_\mathrm A=\frac Nn\tag{23}$$

Inserting $\text(22)$ for $N$ yields

$$\begin{align}
N_\mathrm A&=\frac{A\cdot\left(\frac{6A\cdot\rho}{n\cdot M}\right)^2}{n}\tag{24}\\[6pt]
&=\left(\frac{6\rho}{M}\right)^2\left(\frac An\right)^3\tag{25}\\[6pt]
&=\left(\frac{6\times0.847\ \mathrm{g\ cm^{-3}}}{284.48\ \mathrm{g\ mol^{-1}}}\right)^2\left(\frac{189\ \mathrm{cm^2}}{3.2\times10^{-7}\ \mathrm{mol}}\right)^3\tag{26}\\[6pt]
&=6.6\times10^{22}\ \mathrm{mol^{-1}}\tag{27}
\end{align}$$

Even when taking the uncertainty of the used values into account, this result is significantly smaller than the literature value of the Avogadro constant. Therefore, you may conclude that a large part of the difference may be attributed to experimental error. I guess, you did not use the floating device method to push the stearic acid molecules together, which makes sure that the molecules are densely packed and lined up on the surface of the water with polar heads on the surface and nonpolar tails sticking up away from the surface.


Moles of Reduction Product in the Oxidation of Aluminium by Nitric Acid If, in 2 M $\ce{HONO2}$, 2 mols $\ce{Al (s)}$ are oxidized to $\ce{Al(OH2)6^3+ (aq)}$, mols of reduction product hydrogen is
  (a) two;
  (b) three;
  (c) four;
  (d) five;
  (e) six.

Regarding the above question, how would one solve it?

Would it be valid to simply say that I know that 6 electrons are being transferred because we have this skeleton equation (balanced for charge):

$$\ce{Al -> Al(OH2)6^3+ + 6 e-}$$

And given that 6 electrons are being transferred and that hydrogen is a reduction product, we must obtain hydrogen from some source in this system. The source would probably be hydronium ion, since hydronium ion is less stable than water. Plus abstracting protons from water to form hydrogen gas wouldn't make much sense in strongly acidic solution (we'd be forming hydroxide ions as well)! And knowing the following reaction:

$$\ce{2 H+ + 2 e- -> H2}$$

Can we just say since we know that 6 electrons are being moved, and that hydrogen gas is a product, we know that the immediately above reaction is occurring, and according to its stoichiometry, 2 moles of electrons yields 2 moles of hydrogen gas?

You simply build the half reactions and add them up so you can see how many moles of hydrogen gas are produced by a mole of Aluminium being oxidised. (Note: Your first equation is unbalanced.)
\begin{aligned}\ce{
(1)&& Al + 6H2O &-> Al(OH2)_{6}^{3+} + 3e^{-}&|\cdot2\\
+ (2)&& 2H3O+ + 2e^{-} &-> H2 ^ + 2H2O &|\cdot3\\\hline
(3)&& 2Al + 12H2O + 6H3O+ &-> 2Al(OH2)_{6}^{3+} + 3H2 ^ + 6H2O\\
(4)\equiv(3)&& 2Al + 6H2O + 6H3O+ &-> 2Al(OH2)_{6}^{3+} + 3H2 ^ \\
(5)\equiv(3)&& Al + 3H2O + 3H3O+ &-> Al(OH2)_{6}^{3+} + \color{\green}{3/2}H2 ^ \\
}\end{aligned}

There are $\color{\green}{\text{one and a half}}$ mole Hydrogen gas formed for each mole of Aluminium.
Hence, three mole of Hydrogen gas are formed by two mole of Aluminium.

Edit upon Dissenters follow up question "Why exactly would hydronium ion be the source?"
It says in the question that the solution is $2\mathrm{M}~\ce{HONO2}$ (nitric acid, although this is a strange way of writing it). Therefore there will be an excess of $\ce{H3O+}$. And cations travel to the cathode. Water could be a proton source, but not in that scenario.

It is necessary to acidify the solution, because otherwise you would passify the aluminium surface with $\ce{Al(OH)3}$


Temperature of an atom I read somewhere that the temperature of an atom is not defined. The definition of temperature is only for larger systems. Why is this so?

Thermodynamic functions are strictly defined only for macroscopic systems (systems that have an essentially infinite number of atoms). You can't apply them to individual atoms because that would be confusing large-scale averages with individual microscopic values.

Here's an analogy: the average speed of cars on a stretch of highway might be 55 mph, but it's possible that no individual car is traveling at that exact speed.

Another analogy: the average global temperature is rising over long periods of time, but if I look at the thermometer outside my window, the local temperature is falling over a short period of time... I can't say that my local temperature drop disproves global warming, because weather is a different thing from climate.

Suppose that you knew the speed of a single atom. Now, for a macroscopic ideal gas, the average speed of molecules $\overline{v}$ is $$\overline{v} = \sqrt{\frac{8 R T}{\pi M}}$$ It would be wrong for you to plug in the speed of your single atom and solve for its temperature, because that equation was derived by assuming that you had a macroscopic system (with a certain speed distribution over many molecules); it only applies to average velocities, not individual velocities.

This applies to other thermodynamic functions, too. Let's look at a simple chemical example. For a macroscopic crystal, the Gibbs free energy might be written as $$G = Ng(P,T)$$ where $N$ is the number of atoms and $g$ is the Gibbs free energy per atom. Little $g$ is a function of pressure and temperature.

So can we say that little $g$ is the Gibbs free energy of a single atom? No. It's the average Gibbs free energy per atom for a huge number of atoms. To see the difference,  consider a small cluster of atoms from that same crystal, the previous equation will need some correction terms:

$$G = Ng(P,T) + a(P,T)N^{2/3} + b(T)\ln{N} + c(P,T)$$
where the "a" term is a surface free energy and the last couple of terms might come from free energy contributed by things like rotation of the cluster (Source: T. L. Hill, Thermodynamics of Small Systems, Dover, 1962.) You can't say that $g(P,T)$ is the Gibbs free energy of a single atom in the cluster because of those correction terms.

TL;DR: Thermodynamic functions are averages over very large numbers of particles; random fluctuations are tiny compared to the mean values. But with small systems, those fluctuations become quite important, and there are additional effects that cause macroscopic averages to poorly describe microscopic systems.


Mechanism for simultaneous reduction of nitro group (into aniline) and aromatic substitution (into benzonitrile) by Ethyl cyanoacetate? I stumbled upon this paper when I'm trying to look at established protocol to synthesize one of the compounds (compound IIk) mentioned in the paper.
Studies on Aromatic Nitro Compounds. V. A Simple One-Pot Preparation of o-Aminoaroylnitriles from Some Aromatic Nitro Compounds

The paper described transformation of aromatic nitro compounds into o-aminoaroylnitriles by ethyl cyanoacetate in the presence of strong base. While the general scheme showed fused aromatics, this reaction also works for any electron-withdrawing (in addition to the nitro) containing single aromatics.
I have replicate the reaction for compound IIk and I was abled to purified the compound with yield consistent with what was described in the paper.
I have never seen this kind of transformation before, especially with ethyl cyanoacetate. The paper itself mentioned that the mechanism is unknown and I haven't found any subsequent papers that try to elucidate the mechanism.
Anyone has any idea how this reaction works?

This reaction was conducted using a 3- to 5-fold excess of ethyl cyanoacetate (ECA) and KOH over the arylnitro compounds in dimethylformamide (DMF). It is also important to realize that most KOH is 15% water, which allows for a source of protons in this basic medium. Steps 1 --> 3 (Scheme 1) illustrate a possible route to nitrone 4. In so doing the ECA moiety is being oxidized as the arylnitro compound is being reduced. Addition of hydroxide to nitrone 4 forms 5, which can collapse in two ways. One pathway (another will be mentioned later) leads to reactive ethyl oxocyanoacetate 6 and an aryl hydroxylamine. Base deprotonates the hydroxylamine which adds to 6 to afford species 7 and liberates cyanide anion. Addition of cyanide to the aromatic ring of 7 affords arylamine 8 with loss of the half ester of oxalic acid. A second equivalent of 6 reacts with amine 8 to produce target compound 9.

Condensation (Scheme 2) of the anion 2 of ECA with ethyl oxocyanoacetate 6 provides ketodiester 10 which condenses with arylamine 8 to form the second targeted product 11. Scheme 3 provides another mode of collapse of intermediate 5 which leads to 9. The hydrolysis of 9 and 11 is routine. There are undoubtedly other variants of these Schemes that may apply.

Update: The reference provided by the OP was not the original paper in the series and the picture provided referred to reference 5. The original[1] reactions with 6-nitroquinoline used KCN as the base providing yields of 9 (IIa) in the 50% range and trace amounts of 11 (IIIa) in reaction times up to 5 hours.  After 24 hours the yield of 9 is greatly reduced with the formation of the oxamic acid derived from ester 9 by an unspecified mechanism of formation. The appearance of 16.5% of 11 should got be construed that 11 is derived from 9. Significantly, the use of KOH as a base produces 9 in 66.5% yield without the formation of 11.

Y. Tomioka, A. Mochiike, J. Himeno and M. Yamazaki, Chem. Pharm. Bull., 1981, 29(5), 1286.


What are the physico-chemical processes at play in ironing of clothes? The process of ironing of clothes removes (to a large extend) the wrinkles that where introduced in the clothes while drying. Obviously this has to be some physico-chemical process that is initiated/sustained by the heat of the iron and (perhaps) the pressure applied with the iron.

My question is: what is the actual process that is going on when ironing? Let's assume we have a cotton shirt, do the polymers only physically rearrange or do they actual undergo a chemical change? If it is the latter, does ironing your clothes often cause more rapid degradation?

Ironing of clothes is a physical change (if you believe that physical changes and chemical changes are not part of a single continuum of change).

Most polymers (including the cellulose fibers that comprise cotton) have a glass transition. The glass transition is a reversible phase change from a rigid fixed solid amorphous "glass" to a solid phase where there is some mobility of the polymer chains, but not enough to be a liquid (this phase is sometimes called a "rubber," "plastic," or "elastomer" phase). The glass transition is sometimes used to refer to the phase change between an order crystalline (or semicrystalline) solid to a disordered amorphous solid.

Above the glass transition temperature, the solid material has some mobility - it can be shaped, molded, flattened, whatever. When it is cooled down again, it becomes rigid and holds its shape.

Ironing raises the temperature of the fabric to above the glass transition. The action of ironing presses out the wrinkles in the now more flexible fabric. When the fabric cools, it retains its new wrinkle-free shape.

Water is used during ironing for one important purpose. I tried to find the glass transition temperature of cellulose, but am unable to do so. Presumably it is close to the temperature at which cellulose begins to burn. Water prevents the burning, but not just in the way you would expect. Water does absorb heat from the iron in order to keep the fabric from getting too hot. However, if this were the only thing water was doing, the fabric would not reach the glass transition and ironing would not work. Water also acts as a plasticizer, lowering the glass transition temperature. Most fabrics are hydrophilic (and cellulose as a carbohydrate is especially so). The water binds to the chains, swells the fibers, and changes the physical and chemical properties of the material.


ΔH of reversible expansion of ideal gas Since its isothermal, $\delta U$ is zero. But enthalpy change $\delta H$ is $\delta U + \delta n(\text{gaseous})RT$ or $\delta H = \delta U + \delta (pV)$.

For reversible isothermal expansion of ideal gas, what is the $\delta H$, is it positive, negative or zero.

Ref: I've checked my study material and it says positive, which makes sense but my exam worksheets state it to be zero.

Recall that, for an ideal monatomic gas, $$U = \frac{3}{2}nRT.$$ (Diatomic and polyatomic gases have a larger coefficient.) For an isothermal expansion, reversible or otherwise, we therefore have $$\Delta H = \Delta(U+PV) = \Delta(U+nRT) = \frac{5}{2}nR\Delta T = 0,$$ since $\Delta T = 0$.


Beryllium carbide formation Why does beryllium form carbide of formula $\ce{Be2C}$ and not $\ce{BeC2}$ as the other elements of its group?
I am not able to understand the change in formula of beryllium carbide.

Briefly, it's a matter of cationic radii values and crystal packing at given $p, T$. Without referring to the detailed review of the corresponding crystal structures, you can imagine $\ce{M^{II}2C}$ as the derivative of methane $\ce{CH4}$, whereas $\ce{M^{II}C2}$ can be seen as the derivative of acetylene $\ce{C2H2}$. Smaller cation is the best fit for the "richer" tetrahedral coordination environment of carbon, whereas larger cation of the same charge won't be able to do so (sterical hindrance). That's why $\ce{Be2C}$ is the primary carbide of beryllium, $\ce{Mg2C}$ is also known to exist at higher pressures, (magnesium, along with lithium, otherwise forms an allene-based $\ce{Mg2C3}$ or $\ce{Li4C3}$ 2), but bulkier cations $\ce{Ca^2+, Sr^2+, Ba^2+}$ only form 1/2-carbides. Aluminium, on the other hand, forms $\ce{Al4C3}$ that hydrolyses to methane. This is a good example of the diagonal relationship between $\ce{Be}$ and $\ce{Al}$, and also between $\ce{Li}$ and $\ce{Mg}$.


Is this an aluminum screw dislocation? Is this an aluminum screw dislocation?
https://youtu.be/zbUPjHHml1E?t=55s

I have heard of zinc pest which looks similar.

No, what is occurring is the mercury is forming a eutectic liquid (also called an amalgam) with the aluminum. Aluminum is quite reactive with air/water, but typically aluminum can form an oxide coating that protects it from a complete reaction with air/water (passivation). When the mercury is added and the surface of the aluminum is damaged (by the drill) the two are free to mix per the phase diagram below now the aluminum in the liquid may react with the air/water, but now on the liquid the oxide formed is easily disrupted, and thus ineffective in protecting the aluminum from oxidation. This allows more aluminum to oxidize and the mercury to dissolve more aluminum as aluminum is depleted from the liquid. This creates a steady state condition where aluminum is continually dissolved and oxidized until the aluminum is completely consumed.

A screw dislocation is a microscopic feature only visible with atomic scale imaging such as with a tunneling electron microscope (TEM).


Why sp3 hybridisation in NH3 but bonding with pure p orbitals in case of other hydrides of group 15 According to my book, the following line was given relating to the bond angle in Group 15 elements

  The bond angle in PH3, AsH3 and SbH3 is close to 92° which suggests that the orbitals used for bonding are close to pure p-orbitals.

I wanted to ask if there is any reason for having sp3 hybridised orbitals for N, but not so for other elements of that group in their hydrides. Or is it completely based on experimental observations?

The answer to this question is actually much more complicated than most textbooks make it seem. There's a great explanation in Abright, Burdett and Whango "Orbital Interactions in Chemistry" Ch. 9, which I'll try to summarize here.

As explained in this answer: Why is the inversion barrier larger in PH3 than it is in NH3?, the lone pair orbital in a planar XH3 molecule would be pure $p$. Thus the bonding orbitals average to $sp^2$. As the bond angle decreases from 180, the loss of symmetry makes possible mixing between that lone pair orbital (originally a pure central atom $p$ orbital) and an anti-bonding orbital involving the central atom $s$ orbital. This "s-p mixing" stabilizes the lone pair orbital at the expense of the (empty) antibonding orbital. This mixing necessarily also involves the bonding orbital that involves the central atom $s$ atomic orbital.

In nitrogen, the low energy of the atomic $s$ orbital and the strong overlap interaction of that $s$ orbital with the hydrogen $s$ orbitals means that the bonding orbital made from the atomic $s$ orbital is quite low energy and the antibonding orbital is quite high energy. As a result, mixing with the $p$ lone pair orbital is less favorable, so less mixing occurs and the angle does not deviate as far from 180.

With the larger central atoms, the interaction of the $s$ orbital with the hydrogen $s$ orbitals is less, so the antibonding orbital is not as high energy. Likewise, the $s$ bonding orbital is not as low energy. So mixing with the lone pair $p$ orbital is more favorable, and the bond angle deviates farther from 180. The figure below (from Albright, figure 9.9) plots the energy of the occupied orbitals. The lowest line represents the $s$ bonding orbital and the top line the lone pair.

The key point here is that more $s$ character is mixed into the lone pair orbital as you move down from NH3.


Could you suggest me a good coupling method for amide synthesis between tetrahydroisoquinoline and a carboxylic acid? I want to synthesize an amide using tetrahydroisoquinoline and an unsaturated carboxylic acid which is light sensitive. I tried using DCC as a coupling agent and DCM as the solvent at room temperature, but it didn't work for me. The tetrahydroisoquinoline is in hydrochlorated form.

Do I have to do acid extraction prior to synthesis, or I can use the hydrochlorated form itself? Please suggest a good method along with the conditions for the reaction to be carried out.

A SciFinder search gave 200 or so examples of coupling tetrahydroisoquinoline with an unsaturated carboxylic acid (including benzoic acids and the like). Surprisingly dicyclohexylcarbodiimide (DCC) was not one of the reagents used, although N-(3-dimethylaminopropyl)-N′-ethylcarbodiimide (EDC) was used in more than 50 cases. Other common amide coupling agents including PyBOP, BOP, HBTU, and HATU were reported. Another related strategy through a mixed carbonate/anhydride by activating the acid with isobutylchloroformate.

I'm not sure if you have to isolate the free base of your amine first, but you should certainly include a base to liberate the active nucleophile. If you haven't tried it yet, add an equivalent of triethylamine or diisopropylethylamine.

Aapptec, a supplier of reagents and building blocks for peptide synthesis, has a nice compilation of the different amide coupling reagents that are used:
Aapptec coupling reagents.


Can you literally separate air from soil? My teacher poured water over some soil and said that the air bubbles prove that soil contains air. However I was wondering if there is a more direct approach to proving that soil contains air, like explicitly separating one from the other. Is there an experiment like this? Also, how would the air look like? I've never exactly "seen" air, so how does one know when the air has been separated? Maybe by making it a liquid (liquid nitrogen)?

Your teacher's method of pouring water over soil is an excellent way to demonstrate that there is air in the soil. If you're still unconvinced, it's possible to directly separate the air out. Here's how:

You fill a vacuum chamber with your soil. All of it. So it's all soil and no free space is left. Now you just pump the air to a different chamber. There are several ways to do it so I will not elaborate on the mechanics. This method extracts the air from the intergranular porosity (i.e. the space between the soil grains) to your other container.

  Also, how would the air look like?

Well, it doesn't look like anything, because it is after all air. It is possible to liquefy it by cooling or compressing. It's also possible to measure the composition of the gas you extract out of it and see that it matches the composition of air. It will not be exact, depending on the soil. Some soils have decomposing organic matter (dead bacteria, dead plants, dead other things) so they might have slightly less oxygen and slightly more carbon dioxide, methane, sulfur dioxide, hydrogen sulfide, and a bunch of other things that usually also smell.


pH of a buffer made with boric acid and potassium dihydrogen borate The Problem

  Calculate the pH of a buffer that is $\pu{0.200M}$ $\ce{H3BO3}$ and $\pu{0.122M}$ $\ce{KH2BO3}$. The $K_\mathrm{a}$ for $\ce{H3BO3}$ is $7.3\times10^{-10}$

What I Tried

For each of these methods, I used $\pu{0.200M}$ as the concentration for the acid, $\ce{H3BO3}$, and $\pu{0.122M}$ as the concentration of its conjugate base, $\ce{H2BO3-}$.

Method one – Henderson–Hasselbalch equation

$\mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a}) = 9.13668$

${\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{0.122}{0.200}\right) = 9.13668 + (-0.21467) = 8.92201}$
Method two – Rice Table
$$
\begin{array}{c|lcr}
 & \text{$\ce{H3BO3}$} & \text{$\ce{H3O+}$} & \text{$\ce{H2BO3-}$} \\
\hline
\text{I} & 0.200 & 0 & 0.122 \\
\text{C} & -x & +x & +x \\
\text{E} & 0.200-x & x & 0.122+x
\end{array}
$$
use x is small approximation
$K_\mathrm{a} = \ce{\frac{[\ce{H3O}][\ce{A-}]}{[\ce{HA}]}}$
$\ce{K_\mathrm{a} = \frac{0.122x}{0.200}}$
$x = 0.200 \times \frac{K_\mathrm{a}}{0.122}$
$x = 1.19672*10^{-9}$ – This equals the number of moles $\ce{H3O+}$
$\mathrm{pH} = -\log([\ce{H3O+}]) = 8.92201$

Check $x$ is small approximation: $\frac{x}{0.122} = 9.081\times10^{-9}$ which is less than $5\ \%$, so the approximation is valid.

My question

The answer I got using both of these methods, $\mathrm{pH} = 8.92$, is not correct. The only conclusion I can come to is that I have missed some nuance of this question or I am making a careless mistake. Can somebody please tell me what I'm overlooking here?

Update: Per the advice of @DavePhD , I decided to trust my work and I approached my teacher with the problem. He just emailed me to say there must have been some error in the online system; 8.92 is the correct answer.

Thanks so much to everyone who took the time to help me with this!


Electrosynthesis of HCl? I was attempting to synthesize hydrochloric acid and it seems to have formed another form of chlorinated acid. The smell is more of chlorine and is extremely powerful. I got some on my jeans and it bleached them white but caused me no burn. In fact as an acid it seems pretty weak by comparison to the smell which is overpowering and appears to be extremely toxic. The reaction was performed in a membrane separated electrolytic cell with $\ce{NaCl}$ as the electrolyte, $\ce{PbO2}$ anode, $\ce{Cu}$ cathode, $\pu{19.5 V}$.

I'm pretty sure it's not hydrochloric acid, I've worked with $\ce{HCl}$ a lot and I'm sure thats not $\ce{HCl + H2O}$, so what kind of acid did I form?

Is it just chlorine and water. If it is just $\ce{Cl}$ shouldn't it react with water to form a compound?

MODERATOR : I do not see how a chemical synthesis question is off topic. However, I rarely receive accurate answers to my questions, do feel free to delete this question and all of my previous questions and even my profile for that matter as I find this entire network rater useless. I get much better assistance from other venues.
I do thank anyone who has provided any assistance in the past, it's appreciated.
Back to meandering around the internet for answers.
Good Bye!

Above is what I distilled from the product.

You've described the classic preparation of sodium hypochlorite, better known as "bleach".

If you electrolyze sodium chloride solution, you form hydrogen at one electrode and chlorine at the other, and sodium hydroxide in solution. But chlorine tends to dissolve in alkaline solutions to form chloride and hypochlorite ions. So you get "sodium hypochlorite" solution, which is actually a complex equilibrium between sodium chloride, sodium hydroxide, sodium hypochlorite, free chlorine, and (of course) water.


Are biological macromolecules organic or inorganic compounds? We know that organic compounds are compounds which contain carbon atoms. I want to know whether all biological macromolecules (such as proteins, lipids, carbohydrates, and nucleic acids) are organic compounds, or are only a subset of them considered organic?

The definition of "organic" is somewhat vague: it can mean containing carbon, or it can mean coming from living things, particularly as defined by the United States Department of Agriculture.

To further confuse the issue, some carbon compounds, such as sodium carbonate or tungsten carbide, are sometimes defined as inorganic.

So state a particular definition, and examine those compounds according to that definition.


If for reversible process, ΔS= 0 for universe (system+surroundings), why would there be non-zero change in gibbs energy? As Second Law says, the entropy must increase for universe

$$\Delta S_\textrm{universe}\ge 0$$

Now, we know, $$\Delta G= -T\Delta S_\textrm{universe}\le 0^\dagger$$

For reversible process, $$\Delta S_\textrm{reversible, universe} = 0$$

That would mean $$\Delta G_\textrm{reversible}= 0$$

But does it happen so?

Peter Atkins, in his book writes:

  At constant temperature and pressure, for a reversible process: $\mathrm dG = \mathrm dw'_\rm{rev}\;.$

Now, isn't it contradicting that $\mathrm dG\ne 0$ for reversible process? If $\Delta G= -T\Delta S_\textrm{universe}$, then wouldn't $\mathrm dG= 0$ for $\mathrm dS_\textrm{universe}= 0$ during reversible process?

\begin{align}^\dagger \Delta S_\textrm{universe}&=\Delta S_\textrm{system}+ \Delta S_\textrm{surroundings} \\ &= \Delta S_\text{system}+ \frac{-\Delta H_\text{system}}{T} \\ \implies   -T\Delta S_\text{universe} &= \Delta H_\text{system} - T\Delta S_\text{system} \\&=\Delta G  \end{align}

The equation in Atkins' book refers to the system only, and not to the universe, and, moreover, in this equation, w' specifically refers to non-PV work, like electrical.


are half of "H2SO4 pH sample problem" webpages just wrong, or am I wrong? I did a Google search for "h2so4 pH sample problem" and went through the results, and it looks like about 50% of the webpages "explaining" how to do this problem give the wrong answer when calculating the pH of a solution of $\ce{H2SO4}$.  The incorrect sources either (a) assume that each molecule of $\ce{H2SO4}$ donates only one proton (approximately true only at high concentrations); or (b) assume that each molecule of $\ce{H2SO4}$ donates both protons (approximately true only at low concentrations).  Only about half of the sources do the problem correctly by assuming that all $\ce{H2SO4}$ molecules donate the first proton and then using the second dissociation constant to compute the degree of dissociation of the second proton.

I'm no expert, so am I missing something, or are half of the sources out there just wrong?

Googling "h2so4 pH sample problem", here are the first eight hits after removing pages that did not contain an $\ce{H2SO4}$ sample problem, and how I scored them:

http://che1.lf1.cuni.cz/html/Calculations1112_II_handout.pdf -- says "For $\ce{H2SO4}$: $\ce{[H+]}$ = 2 $\times$ $\ce{[H2SO4]}$" -- incorrect.
https://www.quora.com/How-do-I-calculate-pH-of-10-3-M-H2SO4 -- the top-voted answer also says the concentration of $\ce{[H+]}$ is twice the molarity of $\ce{[H2SO4]}$ which 0.001, which is incorrect.  (This gives the incorrect answer of pH = 2.69; the correct answer is 2.73.)
https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Acids_and_Bases/Monoprotic_Versus_Polyprotic_Acids_And_Bases/Calculating_the_pH_of_the_Solution_of_a_Polyprotic_Base%2F%2FAcid -- uses the second dissociation constant correctly and an ICE chart -- correct
https://www.youtube.com/watch?v=gd8mYfiURx0 -- "Calculating the pH value of 1 M H2SO4" -- also assumes both protons are donated -- incorrect
https://www.youtube.com/watch?v=bfdbmT27L1c -- "Calculate pH of a Strong Acid" -- also assumes both protons are donated -- incorrect
https://web2.0calc.com/questions/what-is-the-ph-of-a-0-00050m-solution-of-h2so4_1 -- uses the second dissociation constant -- correct
If the pH is zero, what would be the normality of a sulphuric acid solution? -- I'm counting this as HALF-correct, because it is assumed that each $\ce{H2SO4}$ molecule donates only 1 proton.  This is approximately true at 1 M concentration, but I think it should be stated explicitly that the second proton doesn't get donated because the concentration is so high
http://www.sparknotes.com/chemistry/acidsbases/phcalc/problems/ -- correctly explains the partial donation of the second proton, so even though they don't show the working out, it's still correct

So it looks like even for just a slightly-non-trivial problem, almost a majority of free online help resources get the answer wrong.  Is there some subtlety I'm overlooking?

...and you didn't even get into the fact that almost all explanations will ignore the autodissociation of water!

In some cases your online resources will be wrong, of course, and it would be odd to expect uniform correctness in something people donate for free on the Internet (including here BTW).  So that's worth keeping in mind.

However, it's also the case that it is not always important to treat a problem in all its subtlety.  For example, the difference you get in the calculated pH of 1M H2SO4 by treating or ignoring the second dissociation equilibrium is likely to be smaller than your measurement error in a typical undergraduate lab.  So why bother?  On the other hand, if you were running a very careful experiment with 0.01 M H2SO4 you might well want to take those other equilibria into account, and if you were doing something super sensitive with 10^-7 M H2SO4 then you'll need to worry about the autoionization of water, too.

Generally we use the least complex model that gets us the answer consistent with our measurement limitations.  That maximizes insight and reduces the obfuscation of complex calculation, which can easily make it so we fail to recognize an erroneous answer.   This runs all through science and engineering, and it's why simple models have such great staying power even when they are superseded by more sophisticated understanding.  Naturally a concomitant is that the good scientist must become very adept at identifying the least complex model he can use.  That takes deep insight into the limitations of each model, the probable sources of error, the limits of measurement, and the quality of data on hand.  Being good at this is a much more important skill in the senior scientist than mastering the details of the most complex and realistic model there is.


How to detect H₂SO₄ (aq) and NH₃ We have two bottle named $\ce{H2SO4~(aq)}$ and solution of $\ce{NH3}$ but we don't know their name actually. What I have to do to detect these solution  by only using :

$\ce{CuCl2}$

or

$\ce{NaCO3_{(s)}}$

On mixing 2 parts of $\ce{NH3}$ with 1 part of $\ce{CuCl2}$, following reaction will occur.

$\ce{CuCl2 + 2NH3 -> Cu(NH3)2^{2+} + 2Cl-}$

This will give you soluble deep-blue copper-ammonia complex: $\ce{Cu(NH3)2^{2+}}$

and also, $\ce{NH3}$ does not react with $\ce{Na2SO4}$.

And $\ce{H2SO4}$ with $\ce{Na2CO3}$ will follow the reaction below :

$\ce{H2SO4_{(aq)} + Na2CO3_{(s)} -> H2O + CO2_{(g)} + Na2SO4_{(aq)}}$

As $\ce{Na2SO4}$ solution is colorless, so you can perform any appropriate test to identify the evolved gas i.e. $\ce{CO2}$.

A simple method used to test $\ce{CO2}$ is :

  Bubble the gas through limewater (calcium hydroxide). If carbon dioxide is present, the limewater will turn cloudy and a milky white precipitate will form.


basics of NMR (spin-spin splitting) here's what I wrote from lecture,

what i'm confused about is, if two magnetic moment(B0 and the one of red H) have the same direction. wouldn't the magnetic field by red H cancel out the external magnetic field? I tried to visualized it in picture below.

2.1  if the directions of the two magnetic moment are different, is it because aromatic compound can flip upside down?

2.2 If the spin state is always same whether or not compound flip upside down, is it because of E from NMR machine?

Why do protons interact only with themselves? there are also O,C and Cl

hope someone would answer me~~

Regarding Question 1:

NMR measures the energy difference between the field-aligned energy of a nuclei and the field opposed energy of the nuclei.

The biggest field in the situation is the external magnet.

The next biggest field is the magnetic moments of the electrons.  Electron density leads to what we refer to as chemical shift.

Lastly, there are the magnetic moments of other nuclei.  But, the other nuclei are also undergoing their own spin flipping and such like.  So if a hydrogen has its molecular neighbor with a spin aligned with the external magnetic field while relaxing, the total field it experiences is stronger.  Otherwise it is weaker.  The difference between stronger and weaker is essentially the distance between the two peaks on the NMR spectrum.  It is important to remember the external magnetic field is the strongest here, so when the nuclear spins oppose and weaken (or align and strengthen) the magnetic field, it is only a tiny amount.

Regarding question 3:

Carbon and Oxygen have complete Nuclear quantum shells, and as a result have no net magnetic moment. The concept of 'spin up' or 'spin down' means nothing to these nuclei.

Since splitting is the result of a magnetic interaction between spins, there will be no splitting.

Regarding chlorine, The coupling constants are small, and generally not observed due to limitations in the machines making the measurements.


Piperic acid oxidation There are 2 double bonds $\ce{C=C}$ present within each piperic acid molecule,  thus when piperic acid reacts with hot excess $\ce{KMnO4}$, it oxidised into a carboxyllic acid. Why is $\ce{CO2}$ produced?

The carbon atom bonded directly to the benzene ring will get oxidised into a carboxylic acid and also that the carbon bonded to the $\ce{COOH}$ group will get oxidized also to a carboxylic acid to form a dicarboxylic acid. What changes will take place to the other 2 carbon atoms.

this paper stated that $\ce{CO2}$   could be formed as well, how?

Usually when we consider permanganate oxidations we think of

oxidation of alkyl benzenes to benzoic acid
cleavage and oxidation of a double bond to an acid, through the aldehyde, if there is one hydrogen on the double bond
cleavage and oxidation of a double bond to a ketone if there are no hydrogens on the double bond

However, there are a few special cases such as formic acid and oxalic acid, or their corresponding aldehydes.  These compounds are oxidized by permanganate to $\ce{CO2}$ (for an idea as to why these compounds react differently than, say, acetic acid, see Scheme 1 here).

With this information in hand we can re-examine the permanganate oxidation of piperic acid.

We see that oxalic acid is one of the primary products produced in this reaction.  As discussed above, oxalic acid can then undergo further reaction to produce carbon dioxide.


How do we find the angular, radial nodes and the quantum numbers when the radial probability distribution curve's equation is given? I am having serious conceptual issues with the Schrödinger equations. Please note that I only have basic school level (11th grade) knowledge of these equations and we are expected to know very little about it. But we are given these type of questions so I guess I will be able to understand things within the boundary of this question.

$$\psi_r = k_1 * e^{-r/k_2} * (r^2 - 5k_3r + 6k_3^2)$$

Solving the equation we get $r = 3 k_3 , 2 k_3$

It is also given that $k_3 =1$ . So $r = 3, 2$

Now, since there are 2 nodes therefore, $n = 3$ . Does the equation $\psi_r$ represents only radial nodes or all nodes?
What is the azimuthal quantum number?
How many angular nodes are present?
What is the angular momentum of the given orbital?
Is this equation of an atom or an orbital?

This equation was given as an MCQ question and we were asked to find the quantum states of the given orbital on the basis of the equation. But I guess we must have conceptual insight too to answer the question so asked it here.

For a given orbital, there are two types of modes: radial nodes and angular nodes.

The total number of nodes is $n-1$
Number of angular nodes is $l$
Number of radial nodes is $n-l-1$

In this question: If the given function represents the radial part of the wave function of an atomic orbital, then we get the number of radial nodes: $n-l-1=2$. i.e.  $n-l=3$:

We have several possibilities:

$n=3$, $l=0$: then we have the orbital $\ce{3s}$
$n=4$, $l=1$: then we have the orbital $\ce{4p}$
$n=5$, $l=2$: then we have the orbital $\ce{5d}$

and so on.


Calculating the Presence of Anthocyanin in a Solution I've produced a red cabbage solution (boiling red cabbage in water basically), and would like to calculate the concentration of the red cabbage/anthocyanin in the solution. How is this accomplished?

Thanks

It is not a trivial exercise because it is not a single compound in red cabbage which causes this color (anthocyanin is a class not a single compound). There is a whole master's thesis dedicated for isolating red cabbage pigments from the University of Ohio. https://etd.ohiolink.edu/!etd.send_file?accession=osu1437655932&disposition=inline

One can determine monomeric anthocyanin using UV-Vis spectrophotometry by measuring absorbances at two wavelengths. It is always good to mention what apparatus your school has.

Giusti MM, Wrolstad RE. 1996. Characterization of red radish anthocyanins. J Food Sci 61(2):322-6.

Giusti MM and Wrolstad RE. 2005. Characterization and measurement of anthocyanins by UVvisible spectroscopy. In: Handbook of Food Analytical Chemistry. RE Wrolstad, SJ Schwartz (eds). John Wiley & Sons Inc. New York .p: F1.2.1-F1.2.13


What is the relationship between something magnetic and unpaired electrons? I don't get how some elements are magnetic. Can someone please explain the relationship between something magnetic and unpaired electrons?

Why is cobalt magnetic?

There are five forms of magnetic behavior. Ferromagnetism is when a material is attracted to a magnetic field and can spontaneously hold a magnetic field alone (the typical form of magnetism).  Paramagnetism is when a material is attracted to a magnetic field but cannot spontaneously hold a magnetic field alone (the attraction may be too weak to observe practically).  Diamagnetism is a material that expels magnetic fields (essentially non-magnetic). there is also ferrimagnetic and anti-ferromagnetic but though these are interesting they not incredibly relevant to the answer.

Electrons form the magnetic fields of materials.  If a metal's atom have unpaired electrons the metal can attract a magnetic field.  If all of the electrons are paired in opposite spin orbitals then the spins cancel each other out and no response to a magnetic field is observed.

Cobalt has unpaired electrons making it at minimum demonstrate paramagnetic behavior. It is however able to hold a magnetic field making it ferromagnetic.  Most metals have unpaired electron and are paramagnetic (for example: aluminum, zirconium, titanium, magnesium and potassium)  Ferromagnetic metals become paramagnetic at elevated temperatures and inversely many paramagnetic metals become ferromagnetic at low temperatures.


Is sedimentation coefficient constant and why? Biochemistry by Voet writes the sedimentation rate of a macromolecule in a  centrifuge this way:
$$s=\frac{v}{\omega^2r}=\frac{M(1-\bar{V}_\rho)}{Nf}$$
where $\omega$ is the angular velocity of the rotor, $r$ is the position of the macromolecule with respect to the rotational axis, $M$ is its molar mass, $\bar{V}_\rho$ its partial specific volume, $N$ the number of moles, and $f$ the frictional coefficient.

The book further discusses the uses of $s=M(1-\bar{V}_\rho)/(Nf)$ and the information it reveals. Since they haven’t discussed any changes to $s$, it seems to me that they are holding it constant, but why should that be? If $s$ is a function of $r$, and $r$ changes, then $s$ should change too. Is my reasoning wrong? Is $s$ considered to be constant and, if so, why?

Sedimentation due to Gravity

The sedimentation coefficient of a particle falling in a fluid due to gravity is easier to understand. It is the ratio of the settling velocity ($v_{t}$) to the acceleration causing the settling($a_{g}$).

$$s=\frac{v_{t}}{a_{g}}$$

In fluids on earth $a_{g}$ is acceleration due to gravity which is known to be constant.

$v_{t}$ , which is also referred to as terminal velocity, is constant.  It is constant because the force driving the acceleration of the particle through a fluid ($F_{g}$) is equal in magnitude and in an opposite direction to the drag force ($F_{d}$).  Since these two forces cancel out the net force ($F_{n}$) is 0.

$$F_{g} - F_{d} = F_{n} = 0$$

There is no net acceleration $a_{n}$ since $F_{n}=m_{p}a_{n}=0$.  The particle mass ($m_{p}$) cannot be zero so there is no net acceleration of the particle, thus the terminal velocity is constant.

A ratio of two constants are constant so the sedimentation coefficient is constant.

Sedimentation due to Centrifugal Acceleration

Radial acceleration, $a_{r}$, equals the angular velocity $\omega$ squared times the radius, $r$.

$$a_{r} =\omega^{2}r$$

If the angular velocity is constant, which it is for most of centrifugation, and the radius is constant, then the radial acceleration, like the acceleration due to gravity, is constant.

The sedimentation velocity in centrifugation, just as in gravity settling, is constant.  It is constant because the drag force of the particle in the fluid is balanced by the centrifugal forces generated by the centrifuge. A centrifuge can generate forces several times higher than gravity so settling occurs much faster during centrifugation.

Again, the ratio of two constants are constant so the sedimentation coefficient is constant.

The other equation looks like it is a derivative of Stoke's law, which is used to determine the drag force of a particle in a fluid.


Why doesn't fast-scan cyclic voltammetry work with metal electrodes? Every paper I've seen discussing it uses it with carbon nanofiber or some other carbon material like a diamond. If the physical principles by which it works are based on sweeping through voltages, I don't understand why it can't be used with metal electrodes.

I've tried to search through the literature but I haven't come across any clear answers.

It actually does work. Carbon fiber electrodes are just more useful for the neuro work that fast scans are popular for these days.

Here is an example from the literature of a metal electrode used up to $3*10^6$ V/s , Amatore et. al. JEAC


Will the water at the bottom of the tube boil at the same temperature as the water at the top of the tube? Consider a long, vertical tube of water, as shown below:

Where $P_\mathrm{atm}$ is the atmospheric pressure and $P_\mathrm{bottom}$ is the pressure at the bottom of the tube.

The tube is long enough such that $P_\mathrm{atm}$ is much less than $P_\mathrm{bottom}$.

It is known that water at the surface (top of the tube) will boil at a temperature such that the water's vapor pressure equals $P_\mathrm{atm}$. Let's call this temperature $T$. For example, $T=100~^\circ\mathrm{C}$ when $P_\mathrm{atm}$ is about $1\mathrm~{atm}.

Question: At what temperature will the water at the bottom boil? Will it be higher than $T$ or equal to $T$? Why?

References, if available, would also be greatly appreciated.

Here's a cool example that should help illustrate the correct answer to your question.  As for why it works that way, you might think about it in terms of bubble formation .. in order for a liquid to boil, it needs to form bubbles.  Is it harder or easier to do that when the ambient pressure is high?

As for the calculation, you can use the Clausius-Claperyon equation to calculate the correction as described here.  You will need some values for $P_0$, $T_0$ and $\Delta H_{vap}$; you can use 1 atm, 373 K and 40.68 kJ/mol.


Yellow-amber debris above water filter: What is it? I have some weird, yellow-amberish pebbles that have been filtered by my Zero waterfilter. Any idea what they are? Copper maybe?

The black stuff is charcoal from the filter.

The quality of the water in my area is fairly good, with a TDS of 40-60. With the filter, I have a TDS of 0, which is excellent. But I'm still curious as to what those things are.

It looks like ion exchange resin, it is used to specifically capture ions from water and replace them by others (either $\ce{H^+}$ or alkali metal ions.

As such, it is probably also part of the filter.


Ortho-para directing and meta directing groups: Activating or deactivating the aromatic ring? Determine the reactivity of the above compounds (activating or deactivating; ortho-para or meta directing). Explain your answers. You may use resonance structures in your explanation. Which molecule will have the highest rate of reaction?
What are the products formed when the above compounds are reacted with $\ce{Br_2}$ and $\ce{FeBr_3}$. Predict the major and minor compound, if there is any.

My answers:
For Question (1):
E is an activating group because it has a nitrogen with a lone pair. F is a deactivating group because $\ce{C=O}$ is directly attached to the benzene ring. In G, $\ce{N}$ is an activating group but $\ce{C=O}$ is deactivating, so I'm not sure if G is an activating or deactivating compound. Also, How can resonance structures help to explain the reactivity of the compound? I'm not sure how to draw the resonance structures. How do we draw resonance structures that has two substituents such as G. For F, does the lone pair on $\ce{N}$ affect the resonance?
For Question (2):
E: $\ce{Br}$ will attach at ortho and para, so it has two products.
F: $\ce{Br}$ will attach at meta with 93 % (major) and ortho 7 % (minor).
G: I think $\ce{Br}$ will attach para to $\ce{N}$ because $\ce{N}$ is a strong activating group and it will not go to between the two substituents because it is crowded, is this true?

I'm not sure if G is an activating or deactivating compound.

You may want to look at the relative order of activation/deactivation of some substituents,
(source: www.chem.ucalgary.ca)

As per the image, $\ce{-NR2}$ is a strongly activating group, while $\ce{-CONH2}$ is a moderately deactivating group (bcoz, it lies somewhere in between $\ce{-COOH}$ and $\ce{-COCl}$). Hence, the net effect would be activating.

How can resonance structures help to explain the reactivity of the compound?

Resonance structures can tell whether the substituent is o,p -directing or m -directing. It can't say anything about rate of reaction.

How do we draw resonance structures that has two substituents such as G?

Just like you've drawn for a single substituent. The only difference is that the other substituent has replaced one of the $\ce{H}$ from the resonance structure drawn with single one. So, the other may participate in delocalization.

For F, does the lone pair on N affect the resonance?

Obviously, but not significantly, as it will participate in resonance too.

I think Br will attach para to N because N is a strong activating group and it will not go to between the two substituents because it is crowded, is this true?

Yes, but it is partially true.
para product (wrt. $\ce{-NR2}$) will be major, however, ortho product (wrt. $\ce{-NR2}$) will be also present as the minor one.


Predict the number of peaks in H-1 NMR of sesamolin Would all of the Hs in an NMR of sesamolin (5-{(1S,3aR,4R,6aR)-4-[(2H-1,3-benzodioxol-5-yl)oxy]tetrahydro-1H,3H-furo[3,4-c]furan-1-yl}-2H-1,3-benzodioxole; shown above) be non-equivalent to each other? So even for the two Hs at A, they would be non-equivalent to each other.

Follow up question (about a different molecule that is similar): for sesamin (5,5'-[(1R,3aS,4R,6aS)-tetrahydro-1H,3H-furo[3,4-c]furan-1,4-diyl]bis(2H-1,3-benzodioxole); shown below), would there be 8 peaks in an H-1 NMR?

For sesamolin, I think the pairs of protons at A and N would have the same chemical shift. In the spectrum this would appear as a singlet with an area corresponding to 4 protons due to the accidental equivalence of the A and N site. It looks like you did this in assignment of sesamin, which looks good to me.

Assigning shifts to geminal protons can be tricky. For example, the differing chemical shifts of the axial and equatorial protons in cyclohexane can only be resolved at low temperature:

Some more examples below:

Both of the images and more information on the topic are available here:
https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-02-delta.htm


Why is the spectroscopic entropy of carbon monooxide higher than that of nitrogen? While solving a question set, I found a question with this data table (at $\pu{298K}$):
$$
\begin{array}{lrrr}
\hline
\text{Substance} & T_\mathrm b/\pu{K} & S^⦵_\mathrm{cal}/R & S^⦵_\mathrm{spec}/R \\
\hline
\text{Nitrogen} & 77.35 & 18.3 & 18.30 \\
\text{Carbon monoxide} & 81.61 & 18.6 & 19.22 \\
\hline
\end{array}
$$
The first part of the question asked why $\ce{CO}$ has higher $S_\mathrm{spec}$ (spectroscopic entropy) than $S_\mathrm{cal}$ (calorimetric entropy). I know that this is due to the residual entropy of $\ce{CO}$.$^*$
However, the next part of the question asked why the value of spectroscopic entropy of $\ce{CO}$ is larger than that of $\ce{N2}$. The question supplied the information that the rotational reduced temperature ($\theta_T$) was same for both molecules (I think that means the rotational partition function $q_\mathrm{rot}$ will be same for both); and that the contribution to entropy from the vibrational modes to entropy is negligible in both cases.
Now the spectroscopic entropy can be written in terms of the contribution of each degree of freedom:
$$S_\mathrm{spec}=S_\mathrm{trs}+S_\mathrm{rot}+S_\mathrm{vib}+S_\mathrm{elec}$$
The translational entropy contribution depends on the translational partition function $q_\mathrm{trs}$ , which depends on the mass of the molecule. $\ce{CO}$ and $\ce{N2}$ has almost the same mass (28.01 u), so I don't think there would be any difference in $S_\mathrm{trs}$.  The rotational and vibrational terms are assumed to be same in the question. The electronic term should also be zero in both cases, as the electronic partition function would be 1 for both. (I think the excited states are all inaccessible at room temperature).
So where does the difference in the calculated entropy come from?
I feel like the answer is something simple that I have missed.
$^*$ For the concept to discern calorimetric entropy from spectroscopic entropy, see for example this Chemistry LibreText; an literature example e.g., in  1958JAmChemSoc1054.

One contribution is the number of identical conformations of a molecule that can be generated by symmetry operations, which is described by a symmetry number $\sigma$. In the high temperature limit
$$S_\mathrm{rot} /R = 1 + \ln\left( \frac{8\pi^2IkT}{h^2\sigma} \right) $$
A larger symmetry number decreases the entropy. This number is 2 for nitrogen and 1 for carbon monoxide. That alone can account for $\log(2)=  0.69 \times R$. The larger symmetry number for nitrogen decreases the entropy by this amount.
In addition the moments of inertia (inversely proportional to the rotational temperature) of the two molecules are not really identical (it is ~5% larger for $\ce{CO}$), but the difference accounts for only ~$0.05 \times R$.


What is Nesbet's theorem? In a talk by Frank Neese there was a slide with title "Nesbet's theorem" and an equation indicating something like "if your wavefunction has the double excitation amplitudes correct, then you can get the correct FCI energy for free". The caveat being that to get the doubles exactly correct you need the triples and quadruples, and to get the quadruples you need the pentuples and hexuples, etc.

Unfortunately I am not able to find this theorem stated precisely anywhere with a Google search. I have been told that the theorem can be found in the book Molecular Electronic Structure Theory by Helgaker, Jorgensen, and Olsen, however I do not have the book and need the precise statement of this theorem.

Does anyone know the precise statement of Nesbet's theorem and the original source?

Section 4.1.1 (p 237) of Szabo and Ostlund's Modern Quantum Chemistry also describes this issue.

Using intermediate normalisation the correlation energy can be written as

$$\begin{align}
\langle \Psi_0|\hat{H}-E_0| \Phi_0\rangle &= \langle \Psi_0|\hat{H}-E_0 \left(| \Psi_0\rangle + \sum_{ar}c_a^r |\Psi_a^r \rangle + \sum_{abrs}c_{ab}^{rs}|\Psi_{ab}^{rs} \rangle +.... \right) \\
&= \sum_{abrs}c_{ab}^{rs}\langle \Psi_0|\hat{H}| \Psi_{ab}^{rs}\rangle
\end{align}$$

$\langle \Psi_0|\hat{H}| \Psi_{a}^{r}\rangle = 0$  according to Brillouin's theorem.
$$\langle \Psi_0|\hat{H}| \Psi_{abc}^{rst}\rangle = \langle \Psi_0|\hat{H}| \Psi_{ab}^{rs}\rangle \langle \phi_c | \phi_t \rangle = 0$$

and similarly for higher excitations, because $\hat{H}$ is a 2-electron operator.


Are there any observable effects of casein binding to capsaicin? I would like to complete an experiment on balancing capsaicin. Since both capsaicin and milk are bases on the pH scale, I prefer not to use the common term "neutralizing".

I will be extracting the capsaicin and I do not wish to taste-test it. Is the bound casein protein with the hydrophobic capsaicin oil visible? Will it curdle in order for me to obtain quantifiable amounts?

Capsaicin doesn't dissociate in water and can't act as a base (like well-known bases such as Sodium Hydroxide or Ammonia) and won't react with the casein in milk. So there won't be any curdling/precipitation of casein protein. Casein will precipitate through pH adjustment if you lower its pH to the isoelectric point (from 6.6 to 4.6).

Notes:

Casein isn't a base, neither an acid, it has an amphoteric nature.
Bases generally don't react with each other.
Pure capsaicin is a powder, not an oil.


Are some alkali metals better conductors than others? I want to know if there is any difference at all between the alkali metals as far as conducting electricity goes.

Li
Conductivity 1.08×10^7 S/m
Resistivity  9.26×10^(-8) mΩ
Na
Conductivity 2.18×10^7 S/m
Resistivity  4.59×10^(-8) mΩ
K
Conductivity 1.43×10^7 S/m
Resistivity  7.0×10^(-8) mΩ
Rb
Conductivity 8.0×10^6 S/m
Resistivity  1.25×10^(-7) mΩ
Cs
Conductivity 5.3×10^6 S/m
Resistivity  1.9×10^(-7) mΩ
Fr
Conductivity 1.4×10^6 S/m
Resistivity  7.1×10^(-7) mΩ


What is the hybridisation in BrF5 What  is the hybridisation of $\ce{BrF5}$ ? I find different sources giving different answers.

When I approach this problem , I don't find any exceptional case like $\ce{SH6}$ (in which hybridisation doesn't take place.) So it follows normal rules, but I find the opposite in many different books ?

I think it square pyramidal $\ce{sp^3d}$. I also face the same problem for other compounds like $\ce{H2S}$ and $\ce{PH3}$.

Hybridisation of $\ce{BrF5}$ is $\ce{sp^3d^2}$ (involving one 4s, three 4p and two 4d orbitals) giving rise to octahedral geometry. But one hybrid orbital is occupied by lone pairs. So the effective shape of molecule is square pyramidal.


Is it possible to have electric field in water without having electrolysis? Is it possible to have electric field in water (using electrodes with voltage difference) without having electrolysis in the water (or any other reaction)?

Well, if the voltage is lower than needed for any possible electrolysis reaction, there would be no reaction, just the field. So the answer may seem to be yes.

Now, in fact there is a catch. Once you turn on the voltage, even if the reaction is not possible, the ions will rush to the respective electrodes and form the so-called electric double layer. Its thickness, known as Debye length, is usually quite small. As for the rest of the solution, it will be effectively shielded from the electric field.

Whether or not this counts as a positive answer is up to you.


Voltaic and electrolytic cells Why more attractive metals will be oxidized? Aren't they being reduced, because they attract and receive electrons from negative polyatomic ions? So they will be anodes, where oxidization occurs. Why anodes are negative in voltaic cells and positive in electrolytic cells? Same for cathodes.

Metals want to lose electrons and become positively charged. When the metal loses the electron, it is called oxidation. The substance that gets the electron is reduced.

Oxidation could occur at the anode if it takes electrons from a metal (i.e. $Fe^{2+}$ to $Fe^{3+}$).

Anodes are negative in voltaic cells since there are excessive electrons available to power the circuit. Cathodes give electrons away to the solution.


Why can Bi not achieve its 4+ oxidation state? Bi can reach its 3+ and 5+ oxidation state. Why not 4+? Can you use the inert pair effect here to explain this?

Actually bismuth(IV) is known, and in not so exotic conditions at that. In fact both even oxidation states +2 and +4 appear when acidic bismuth(III) solutions are irradiated:

Bi(IV) has been observed; see A. I. Aleksandrov, I. E. Makarov (1987). "Formation of Bi(II) and Bi(IV) in aqueous hydrochloric solutions of Bi(III)". Bulletin of the Academy of Sciences of the USSR, Division of Chemical Science. 36 (2): 217–220. doi:10.1007/BF00959349.

From the reference cited above:

The formation of bismuth(II) and bismuth(IV) under the action of ionizing radiation on hydrochloric acid solutions of bismuth(III) has been shown by low-temperature and pulse radiolysis. The kinetics of bismuth(IV) disappearance at 300°K have a complex nature and would seem to be caused both by the disproportionation of bismuth (IV) and the reaction of bismuth(IV) with bismuth(II), which leads to bismuth(III).

The properties of bismuth (IV), which is formed from the reaction of Cl2 with bismuth (III) agree with previously established relationships on the variation in optical transition energies and the unpaired electron density at the s atomic orbital of the ion for the series of isoelectronic ions mercury (I), thallium (II), lead (III), and bismuth (IV) in the $^2S_{1/2}$ state.


How do you explain the formula Fe3O4 using the ionic compound theory? I don't understand at all how the ionic bonding in $\text{Fe}$$3$$\text{O}$$4$ works. The oxygens all together have $4(–2e)=–8e$ net charge. But we cannot give the three irons equal positive charges to balance this: $8/3$ isn't an integer!
And anyways, what kind of iron is in $\text{Fe}$$3$$\text{O}$$4$? It can only be either Fe(II) or Fe(III), and none of them fits in here.

A diagram demonstrating the chemical structure of $\text{Fe}$$3$$\text{O}$$4$ with supporting explanation would be preferable. I'm only an 8th grader, so advanced chemical concepts won't be comprehensible to me. The lattice structure would be too huge for me to handle, so pointing out a single molecule/piece of that compound will be more helpful.

Iron forms both $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions. The first combine with oxygen ions to $\ce{FeO}$; the second $\ce{Fe2O3}$.

$\ce{Fe3O4}$ can be thought of as a mixture of both compounds in equal amounts. So $\ce{Fe3O4}$ contains both Fe(II) and Fe(III) ions; twice as much Fe(III) as Fe(II).

In the picture below, showing the crystal structure, green atoms are Fe(II) and brown atoms Fe(III). All Fe(III) atoms are linked with six O atoms, sharing 'half an electron' with each oxygen; all Fe(II) atoms are linked with four O atoms, and all O atoms are linked with four Fe atoms.

(source: Manoj B. Gawande, Paula S. Branco, and Rajender S. Varma, Chem. Soc. Rev. 2013, 42, 3371-3393. DOI: 10.1039/C3CS35480F See also: ResearchGate
)


Conductivity of Wüstite "Wüstite is often represented as $\ce{Fe_{1-x}O}$, where $x$ is some small fraction less than unity. In this nonstoichiometric $\ce{Fe_{1-x}O}$ material
conduction is electronic and the material behaves as a p-type semiconductor, i.e. the $\ce{Fe^3+}$ ions act as electron acceptors, generating holes.

Determine the electrical conductivity of a specimen of wüsite that has a hole mobility of $\pu{3.1E-5 m2 V-1 s-1}$ and for which the value of $x$ is 0.030. Assume that the acceptor states are

saturated, i.e. one hole exists for every $\ce{Fe^3+}$ ion. Wüsite has a sodium chloride crystal structure with a unit cell edge length of 0.437 nm."

My Solution:

4 units of $\ce{FeO}$ for rocksalt structure but for every $\ce{Fe^2+}$ vacancy $2$ $\ce{Fe^3+}$ are needed for electroneutrality, this implies $0.003:1\rightarrow0.24:4$

$$\mathrm{mass}_{\ce{Fe(III)}}=((0.24)(55.85))/((N_a)(10^3))$$
$$=\pu{2.225838592E-26 kg}$$
$$\rho_{\ce{Fe(III)}}=2.225838592\times10^{-26}/(4.37\times10^{-10})^3$$
$$=\pu{266.7 kg m-3}$$
Conductance for a p-type semiconductor is given as: $$\sigma=p(q)(\mu_{h})$$
$$\implies\sigma=(266.7)(1.6\times10^{-19})(3.1\times10^{-5})$$
$$\approx1.3\times10^{-21}\:(\Omega.m)^{-1}$$

I have the feeling this is wrong so any help would be appreciated.

after revisiting the problem I realised where I went wrong. The main mistake with not being careful with units and not reading the qurstion and having incorrect values for hole mobility.
$$\text{mass}_{\ce{Fe(III)}}=((0.24)(55.85))/((N_\mathrm{a})(10^3))$$
$$=2.225838592\times10^{-26}\: \mathrm{kg}$$
$$\rho_{\ce{Fe(III)}}=2.225838592\times10^{-26}/(4.37\times10^{-10})^3$$
$$=\mathrm{266.7\: kg\: m^{-3}}$$
$$\text{No. of holes}=\rho_{\ce{Fe(III)}}/1\:\text{AMU}_\ce{{Fe(III)}}$$
$$\Rightarrow \text{No. of holes}=266.7/(9.296\times10^{-26})$$
$$=2.868975904\times\pu{10^{27}\:holes/m^3}$$
Conductance for a p-type semiconductor is given as: $$\sigma=p(q)(\mu_{h})$$
$$\implies\sigma=(2.868975904\times10^{27})(1.6\times10^{-19})(3.1\times10^{-5})$$
$$\approx1.4\times10^{4}\:(\Omega.m)^{-1}$$


Which gas is easier to compress, the ideal gas or a real gas? Since the pressure of a real gas is less than that of the ideal gas  and its volume is more than that of ideal gas, I am assuming that the real gas is difficult to compress in comparison to an ideal gas.

Is my logic right? Also I'm looking for a clearer explanation!

Let's use the van Der Waals equation for this. (Real gasses also described elegantly by the Viral equation)

You were false that "pressure of a real gas is less than that of the ideal gas". It is greater. Also the volume is smaller.

Thus if it has a higher pressure and smaller volume, it typically is harder to compress (i.e. the compression factor increases) but it also depends on temperature. Here is a nice plot of compression factors vs temp for some common gasses from the cambridge chem guide.
http://www.chemguide.co.uk/physical/kt/realgases.html


Comparison Of Lewis Acidity Which is a stronger Lewis Acid $\ce{AlCl_3}$ or $\ce{SiCl_4}$?

I am confused as in $\ce{AlCl_3}$ , Al has an incomplete octet so it must be a stronger Lewis acid as compared to $\ce{SiCl_4}$, but the answer is just opposite.

AlCl3 + RCl → AlCl4(-) indicating that AlCl3 is a Lewis Acid
but
SiCl4 + RCl → SiCl5(-) ??????  So, this tells me that AlCl3 > SiCl4 in terms of electron accepting behavior.
Now, don't get me wrong: pentacoordinated Si is well known. One of the failures of the Lewis Acid concept is that, as far as I know, there is no unambiguous way to measure it. It is much more qualitative than quantitative, despite years of attempts to do so. You may know that Phosphorous is typically pointed to as the first example of the breakdown of the octet rule. (see Hypervalence) After Period 3, the concept has little use, imho. Since Lewis Acidity is tangled up with Lewis structures, perhaps it is not too surprising that the concept has some serious problems. Usually, we condsider both electronic effects as well as steric effects to 'explain' Lewis Acid/Base strength. One popular alternative is Hard.Soft.Acid.Base (HSAB) "theory". Anyway, comparing a tri-coordinated species with a tetracoordinated species and predicting which will form a bond with an electron donor is really difficult. Did you know that AlCl3 doesn't really 'exist' in the solid or liquid state? And even in the gaseous state, the dimer is common. The structure of the (solid) dimer (Al2Cl6) has two Al-Cl-Al bonds. Anyway, as far as I know AlCl3 is the stronger Lewis Acid in most apple-to-apple comparisons. To be honest, I don't think that if the Lewis Acid concept weren't a historically established meme, that it wouldn't be invented today. Personally, scales should actually be founded on measurement. The Lewis Acid strength scale is just sorta kinda correlated to what we can actually measure.


What does the term "simple" in the phrase "simple whole number ratio" mean in Gay Lussac's law? What does the term "simple" mean in the phrase "simple whole number ratio" in Gay Lussac's law?
I have searched this site as well as different sites on the internet. The answer to one question on this site states that it indicates that the numbers are small whole numbers. Another site stated that the terms of the ratio are not decimal numbers while another site states that the terms of the ratio are in simplest terms that is reduced form.

3/2 is simple. 45498334/9232493 is not.

The reason it works is because reactions of gases producing gases involve relatively simple compounds. If they were complicated, the molecular weight would likely be high enough that the substance would not be a gas.

This is akin to Loong's comment about the Law of Multiple Proportions, which also mentions "small whole number ratios." The difference is that for the Law of Multiple Proportions, "small" ratios doesn't apply in the case of something like palytoxin ($\ce{C129H223N3O54}$).

But palytoxin isn't a gas. For the reason mentioned above, you're almost certain to get simple coefficients for gas to gas reactions.