Question about relation between two time constants I am not a chemist, but a neuroscientist, so bear with me. I have struggled with this problem for over a week now and have realized it's a chemical question. I'll try to explain it so no neuroscience background is needed.

I am trying to fit a model to experimental data. The data describes how a certain channel opens and closes in response to a change in membrane potential of the cell the channel sits in. This is captured in two datasets:

The fraction of open channels at equilibrium for different values of the membrane potential (an example of a plot of this can be found here. This fraction I call G/G$_{max}$
The time constant describing the time it takes for the fraction to come to its equilibrium. This time constant I call $\tau_{G/G_{max}}$

It is thought that for each open channel, 4 gates have to be in the 'open' state. That is, it is thought that:

G/G$_{max}$ = n$_{\infty}^4$

The gate n, too, can be described by its fraction of open-state/total.
The rates of n moving between these two states are described by $\alpha_n$ and $\beta_n$. Once there are four gates in the open-state, the channel is instantly open.
 If I try to describe this in a chemical reaction, I would guess something like:

$$\ce{C_n <=>[\alpha_n][\beta_n] O_n}$$

where $C_n$ and $O_n$ depict 'closed' and 'open' n-gates respectively. And

$$\ce{4\cdot O_n <=>[instant][instant] O_{ch} }$$,
where $O_{ch}$ describes the whole channel being open.

Now is the data I've got for G/G$_{max}$(V) and $\tau_{G/G_{max}}$ (this is the only thing the experimentalists can measure). However, to model this behavior, I need $\alpha$(V) and $\beta$(V). I already know that

$n_{\infty} = (G/G_{max})^{1/4}$

but I cannot figure out how $\tau_n$ and $\tau_{G/G_{max}}$ are related to each other. Even though the reaction 4O --> open channel is instant, the shape of approaching the equilibrium differs and thus the point that they cross 1-1/e would as well. n will reach its equilibrium with a single exponential, while G/G$_{max}$ reaches it with an exponential to the fourth power.

I think the answer should be quite simple, and should only depend on the 4'th power.

I hope my question is clear, but if there are more details that will help solve this I am more than happy to clarify certain points.

Edit: Here is a figure that shows how the 'measured' time constant and the one for the differential variable n differ. Also, it shows that there is not a linear relationship.

Just to be clear about what I did:

The differential system describing this model looks like:
G/G$_{max}$ = n$^4$
with
$\frac{dn}{dt} = \alpha_n(V)(1 - n_{\infty}(V)) - \beta(V) n_{\infty}(V)$

As one of the answers pointed out (sort of), a solution to this system with initial condition V1 is:

$n(t) = n_{\infty}(V) \cdot e^{-(\alpha_n(V) + \beta_n(V))} + n_{\infty}(V) \cdot \Big(1 - e^{-(\alpha_n(V) + \beta_n(V))}\Big)$

and thus

$G/G_{max}(t) = n(t)^4$

To find the $\tau$ of either of these traces (the time constant being the time it takes to reach $1-1/e \cdot$(final value) (see link), I plotted:

$\frac{n - min(n)}{max(n) - min(n)}$

and

$\frac{n^4 - min(n^4)}{max(n^4) - min(n^4)}$

I think I understood your question now. I will reformulate it in more 'chemical' terms and give you my solution. Not nicely formatted for now, I will do that later if I find the time.

You have a cell with $G_{max}$ gates on it. They can open and close reversibly, depending on a voltage V. Closed gates open with rate constant $\alpha(V)$, and open gates close with rate constant $\beta(V)$.

This implies one differential equation. If $G_o$ is the number of open gates at time t, and $G_c$ is the number of closed gates at time t, then:

$\frac {dG_o}{dt} = \alpha G_c - \beta G_o$

And as the total number of gates is $G_{max} = G_o + G_c$:

$\frac {dG_o}{dt} = \alpha (G_{max} - G_o) - \beta G_o$

At steady-state, i.e. when $G_o$ stops changing, you have:

$\frac {dG_o}{dt} = \alpha (G_{max} - G_o) - \beta G_o = 0$

which, solved for $G_o/G_{max}$, yields:

$\frac {G_o}{G_{max}} = \frac {\alpha}{\alpha + \beta}$

The ratio $G_o/G_{max}$ is the probability that a single gate is open. As you say, a 'channel' is open when 4 of its gates are open. If we consider each gate being open or closed independently, the probability that 4 gates in a channel are open is indeed $(G_o/G_{max})^4$

What you measure is the time it takes, when the voltage is abruptly changed from an initial value to a final value, for the number of open channels to go from an initial value to 'equilibrium'.
This time is in theory infinite, but in practice it is probably the time it takes for the number of open gates to go from its steady-state value at voltage $V_1$, to a value sufficiently close to the new steady-state value at voltage $V_2$.

If so, it is sufficient to solve the above differential equation for $G_o$ with initial conditions $(t=0,G_o=G_{o,V_1})$. The solution is:

$\frac {G_o(t)}{G_{max}} = \frac {G_{o,V_1}}{G_{max}} \cdot e^{-{(\alpha+\beta)t}} + \frac {G_{o,V_2}}{G_{max}} \cdot [1-e^{-{(\alpha+\beta)t}}]$

where:

$\frac {G_{o,V_i}}{G_{max}} = \frac {\alpha(V_i)}{\alpha(V_i) + \beta(V_i)}$

The difficulty I see is the arbitrary nature of the 'time' you consider sufficient to reach steady-state. I would rather fit time points, i.e. number of open gates vs time. And my earlier suggestion to use numerical fitting may still be valid, actually.


How can I synthesize an (E)-configured enone using the Wittig reaction? I have been asked to synthesize the compound (E)-5-methylhex-3-en-2-one:

How can this be achieved using the Wittig reaction?

The Wittig reaction allows for the construction of alkenes from a phosphorus ylide and an aldehyde (or ketone). Shown in the retrosynthetic sense, keeping in mind that there are two possible Wittig disconnections:

Applying this idea to (E)-5-methylhex-3-en-2-one gives the following possibilities:

Which option is chosen often comes down to which gives the most readily accessible starting materials. However, in this case, the stereochemical outcome is critical, and that is controlled by the nature of the ylide used. Looking now in the forward direction, the top option uses an unstabilized ylide, which tend to give the cis-alkene as the major product. The bottom option uses a stabilized ylide, which changes the stereochemical outcome to favor the trans-alkene.

In practice, many chemists would opt for the Horner–Wadsworth–Emmons variant, which gives the trans-alkene as the major product and avoids the production of triphenylphosphine oxide, a byproduct that can be difficult to remove.


Joule–Thomson effect: why does temperature decrease during expansion When a gas under a high pressure in a piston (isolated from surrounding is allowed to pass through a valve), why does its temperature decrease once during expansion?

I haven't still studied thermodynamics but would like to know why exactly the temperature decreases.

Is it because under high pressure the heat was concentrated under small volume once expanded the heat is distributed?

Once the gas molecules are allowed to expand through the valve its kinetic energy increases. Where is the source of energy coming from, is it from its internal energy?

In order to expand, the molecules may do work, such as pushing against a piston, pushing against the container (as in water rockets), spinning a turbine, etc. Energy is transferred from molecular motion to another form, so the temperature drops.

In free expansion, however,

  the gas does no work and absorbs no heat, so the internal energy is conserved. Expanded in this manner, the temperature of an ideal gas would remain constant, but the temperature of a real gas may either increase or decrease, depending on the initial temperature and pressure.

[from Wikipedia].


What is the specific heat of copper through its phases? Specific heat is the amount of heat required to change the temperature of 1 gram of a substance by 1 Celsius.

For instance, the specific heat of water in the solid phase is 2.059 J/gC and 4.184 J/gC in the liquid phase.

That being said, what is the specific heat of copper in all three of its phases (liquid, solid and gas) in J/gC st standard pressure? How would one figure this out?

The NIST Webbook is generally a good place to look for thermochemistry data. You can see here that there are data for all three phases. It looks a bit daunting, but because heat capacity is not constant with temperature, it's expressed as a Shomate equation. Basically, you find the equation that starts with $C^°_p$ and substitute the parameters from the table underneath along with the desired temperature and you can calculate the heat capacity for any temperature within the range shown at the top of the table. (note that these are molar heat capacities and you'll have to convert if you want it in terms of mass)

If you prefer, you can just click View table to get computed values at certain temperatures, which will save you the calculation.


Reaction of a non-metal with alkali to produce hydrogen gas I want to ask that is there any non-metal which reacts with an alkali (NOT alkali metal) to produce hydrogen gas?

This is the extension of Alaska Ron's answer. All reactions taken from this site.

Group 13 elements

All group 13 elements except indium and thallium reacts with aqueous alkali to produce a alkali salt and hydrogen. I'm taking the example of aluminium.

$$\ce{2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2}$$

$$\ce{2Al + 2(NaOH•H2O) ->[400-500°C] 2NaAlO2 + 3H2}$$

$$\ce{2Al + 6NaOH ->[450°C] 2NaAlO2 + 3H2 + 2Na2O}$$

$$\ce{2Al + 6H2O + 6NaOH → 2Na3[Al(OH)6] + 3H2}$$

Group 14 elements

Silicon react with sodium hydroxide to produce sodium orthosilicate and hydrogen.

$$\ce{Si + 4NaOH → Na4SiO4 + 2H2}$$

Sodium hydroxide react with silicon and water to produce sodium silicate and hydrogen.

$$\ce{2NaOH + Si + H2O → Na2SiO3 + 2H2}$$

Tin and lead however forms a different product but eventually gives hydrogen gas. Lead only gives the first reaction.

$$\ce{Sn + NaOH + 2H2O → Na[Sn(OH)3] + H2}$$

$$\ce{Sn + 2NaOH + 2H2O → Na2[Sn(OH)4] + H2}$$

$$\ce{Sn + 2NaOH + 4H2O ->[\Delta] Na2[Sn(OH)6] + 2H2}$$

Group 15 elements

Arsenic react with sodium hydroxide to produce sodium orthoarsenate and hydrogen.

$$\ce{As + 3NaOH ->[\Delta] Na3AsO3 + H2}$$


Is it possible to generalize whether an organic molecule is chiral or not? Can we generalize on the basis of number of chirality centers  if an organic molecule is chiral or not?

There is a general statement. If the molecule has any improper axis of rotation, it is achiral.


Electron orbital filling order In the p sub shell there are 3 orbitals,
m=-1, m=0, m=1
which are listed by axis x,y, and z.
Some show as -1=x,0=y,1=z
Another shows -1=x,0=z,1=y
Some pictures and text also show the filling order as ‘xyz’ and others state that the z axis is filled first m=0 and then the x axis as m=-1 making the filling order z,x,y

What is the correct filling order?
If I know this then I can figure out the d and f sub shells.
d~m=-2,-1,0,1,2
F-m=-3,-2,-1,0,1,2,3
Thanks, HJS

There is no order for filling atomic orbitals of the same azimuthal quantum number (l) such as px, py, and pz since they are degenerate ie. have the same energy (however they lose their degeneracy in a magnetic field, see Zeeman effect). Therefore, one electron in a p sub-cell can be in either 3 of them and the order of filling doenst matter. For when more electrons are added in degenerate orbitals look up Hund's rule and the Pauli principle.


How much NaOH required to change the pH of a weak acid solution by 2 If I have a $\pu{20mL}$ $\pu{0.1M}$ solution of sodium dihydrogenphosphate, how much $\pu{0.1M}$ $\ce{NaOH}$ do I need to add to raise the $pH$ by 2.

Here's what I've done so far. I'm doubt it's right and I get stuck with respect to the change in volume.

Edit: I realize that this wrong as sodium dihydrogenphosphate is polyprotic.

I found using the $K_a$ that a $\pu{0.1M}$ solution will lead to a $[\ce{H+}]= 3.47\times 10^{-5}$

Since $-log(3.47\times 10^{-5}) = pH = 4.46$, the pH of the final solution is $4.46 + 2 = 6.46$.

The change in $[\ce{H+}]$ is $3.47\times10^{-5} - 10^{-6.46} = 3.44\times10^{-6}$, so there must be a concentration of $3.44\times10^{-6}M$ $\ce{OH-}$ added to the solution.

Since my solution is $0.02L$, $0.02 \times 3.44\times10^{-6} = 7\times10^{-8}$ moles of $\ce{OH-}$.

The ratio of $\ce{OH-}$ to $\ce{NaOH}$ is $1:1$, so:

$\pu{0.1M}$ $\ce{NaOH} = 7\times\dfrac{10^-8}{x}$, $x = 7\times10^{-7} mL$ of $\pu{0.1M}$ $\ce{NaOH}$ added

I know this is a VERY small amount of $\ce{NaOH}$ but is there a way to account for the fact that this change in volume will affect the $pH$ of the solution?

RANT - I hate problems that don't use significant figures properly. So is the answer supposed to be +/- 1 ml, +/- 0.1 ml, or +/- 0.01 ml?
First your overall approach is right. There are three parts to solving the problem:

(1) Calculate pHs
(2) Calculate how many moles NaOH
(3) Calculate how many ml of 0.1 molar solution.

So the problem is that there are four phosphate species, $\ce{H3PO4, H2PO4^-, HPO4^{2-},}$  and $\ce{PO4^{3-}}$. The precision needed for the answer is somewhat of a mystery. So what simplifying assumptions are valid?
I'm just going to guess that the volume of NaOH should be accurate to 0.01 ml.
So let's look at the problem another way. I start off with 20.00 ml of 0.1000 molar solution of phosphoric acid and titrate with 0.01 ml at a time with a solution of 0.1000 molar sodium hydroxide. As every aliquot of NaOH is added the pH is measured. So:

(1) when I have added exactly 10.00 ml I have effectively made sodium dihydrogenphosphate. I add 2 to this pH to get the target pH.
(2) I look to see volume of NaOH that gives the target pH and subtract 10.00 ml.

SIDEBAR

Just looking at the problem I'm guessing a few ml. Let's say 5.00 ml. Measuring the NaOH to 0.01 would then mean 1 part in 500 precision (2 parts per thousand). Let's say that the initial $[\ce{H+}]$ is $3.47\times 10^{-5}$. 1 part in 347 is about right. But a pH of 4.46 only has two significant figures, that is 1 part in 46 which is too little precision for my guess of about 2 parts per thousand being needed. So the pH should be noted as 4.460.

Looking at Wikipedia we can see that for phosphoric acid that pKa1 = 2.148,
pKa2 = 7.198, and pKa3 = 12.319.
(1) Given these pKa's we can't do better than 2 ppt.
(2) Only a small part of the $\ce{H2PO4^-}$ will ionize to $\ce{HPO4^{2-}}$ and $\ce{H^+}$. Thus the initial pH will be lower than 7.198 but quite above 2.148.
(3) Given (2) enough $\ce{H3PO4}$ might form to lower the pH when calculating to 2 ppt. This is right on the edge for 2 ppt precision...
(4) The species $\ce{PO4^{3-}}$ can safely be ignored in all the calculations. i.e.  $\ce{[PO4^{3-}] << 0.1000 molar}$
1.0 Quick and dirty method...
To avoid solving a 3rd or 4th order polynomial...
1.1 Calculate pHs
The reaction of interest is:
$\ce{H2PO4- <=> H+ + HPO4^{2-}}\quad\quad 6.339 \times 10^{-8}$
We'll let $\ce{[H+] = [HPO4^{2-}]}$ and assume that $\ce{[H2PO4-]} = 0.1000$ so
$ 6.339 \times 10^{-8} = \dfrac{\ce{[H+]^2}}{0.1000}$
$ \ce{[H+]} = 7.96 \times 10^{-5}$
pH = 4.099
Therefore end pH = 4.099 + 2.000 = 6.099

Checks
$\ce{[HPO4^{2−}]}$ of $7.96\times10^{−5}$ is just less than $1\times10^{-4}$ so the assumption that $\ce{[HPO4^{2−}]}$ is not appreciable to  $\ce{[H2PO4−]} \approx 0.1000$ is just ok.
We've also assumed that an insignificant amount of $\ce{H3PO4}$ will form which isn't true...  $\dfrac{\ce{[H3PO4]}}{\ce{[H2PO4^−]}} = \dfrac{\ce{[H^+]}}{7.11\times 10^{-3}} = 0.0112$

Da fix
We have 3 species of phosphate:
0.0008 $\times \ce{[H2PO4^{−}]}$ = $\ce{[HPO4^{2−}]}$
1.0000 $\times \ce{[H2PO4^{−}]}$ = $\ce{[H2PO4^{−}]}$
0.0112 $\times \ce{[H2PO4^{−}]}$ = $\ce{[H3PO4]}$
$\text{--------}$
1.0120 total $\ce{H2PO4^{−}}$
Normalizing $\ce{H2PO4^{−}}$ to 1.0000 total we have for the concentrations
0.0001 = $[\ce{HPO4^{2−}}]$
0.0988 = $[\ce{H2PO4^{−}}]$
0.001107 = $[\ce{H3PO4}]$
We rearrange the equilibrium for the reaction
$\ce{H3PO4 <=> H^+ + HPO4^{2−}}$
to solve for $\ce{[H^+]}$
$\ce{[H^+]} = \dfrac{\text{K}_{a1}\ce{[H3PO4]}}{\ce{[H2PO4^-]}} = 7.97 \times 10^{-5}$
so everything now checks...
1.2 Calculate how many moles NaOH
Given:
(1) $\ce{[H^+]} = 7.96\times 10^{-7}$
(2) 0.1 molar = $\ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$
Assume:
(1) $\ce{[H3PO4]} \approx 0$
(2) $\ce{[PO4^{3-}]} \approx 0$
(3) moles change in $\ce{H^+}(\text{aq})$ is insignificant
$6.339\times10^{−8} = \dfrac{\ce{[H+][HPO4^{2-}]}}{\ce{[H2PO4^-]}}$
or
$\ce{[HPO4^{2-}] =} \dfrac{6.339\times10^{−8}\ce{[H2PO4^-]}}{\ce{[H^+]}} = 0.07962\ce{[H2PO4^-]}$
We also know that
$0.1000 = \ce{[H2PO4^-] + [HPO4^{2-}] = 1.07962[H2PO4^-]}$
$\ce{[H2PO4^-] =0.09263}$
$\ce{[HPO4^{2-}] =0.00737}$
$\ce{H2PO4^- + OH^- ->  HPO4^{2-}}$ so
Moles NaOH = $(0.020 \text{ liters})[0.00737] = 1.47\times10^{-4}$
1.3 Calculate how many ml of 0.1 molar solution.
liters 0.1 molar NaOH = $\dfrac{1.47\times10^{-4}}{0.1000} = 1.47\times10^{-3}$
ml 0.1 molar NaOH = 1.47


Strong or weak force between protons and neutrons? I had learned that what keeps the protons and neutrons bound is the strong force and the weak force has to do with radioactive decay. but today I read an article from Science Daily that contradicts this. Here is the quote:
“Protons and neutrons are made of smaller particles called quarks that are bound together by the strong interaction, which is one of the four known forces of nature: strong force, electromagnetism, weak force and gravity. The weak force exists in the tiny distance within and between protons and neutrons; the strong interaction confines quarks in neutrons and protons.”
Can this be clarified by the community? Thanks!

I think it is a poor phrasing. The strong force is responsible for keeping quarks together inside a proton or a neutron, but also between quarks belonging to different proton/neutron (but I am not sure for proton-proton interaction because of the repulsive electromagnetic force).

By contrast, the weak force is responsible for turning one type of quarks into another type or quarks, thus transforming a proton into a neutron in the case of beta+ radioactive decay, with emission of a positron (the overall charge is always conserved).


The temperature rise after neutralisation The answer to this question is B, however, I don't quite understand.
If considering Q= mc delta T, the heat released is to some extent being influenced by the no. of moles. If using that line of thought, wouldn't we want C where it has also the same no. of moles?
If we really have to arrive at B, my other proposition would be that the increase in heat is dissipated in a larger volume, hence giving a temperature rise of 6 degrees C.
I am a bit confused by this question. Any comments would be very helpful. Thank you in advance

One calorie is the amount of energy (heat) needed to raise 1 gram of water 1 degree Celsius at a pressure of one atmosphere. (It is almost constant to 3 significant figures, but there is some temperature dependence...) Also assume that 1 cm^3 of water is 1 gram which is more than good enough for the calculations needed for the problem.

So for the reaction in the problem given, 50 mL of water by 6 degrees is 300 calories.

For the other reactions I'll give calories and temperature rise for product solution.

(a) 600 calories, 12 degrees C

(b) 600 calories, 6 degrees C

(c) 300 calories, 3 degrees C

(d) 300 calories, 1.5 degrees C


Are there any solvents that dissolve talcum powder? I understand that talcum powder is insoluble to water, would alcohol or something be able to dissolve it?

Talcum powder is made from talc, a silicate mineral with formula unit: $\ce{Mg3Si4O10(OH)2}$. Silicates tend to be pretty much insoluble in water, but it is possible to dissolve them in strong acids. I don't know about talc specifically, but many silicates can be dissolved in concentrated nitric acid. Hydrofluoric acid is also commonly used for dissolving minerals for chemical analysis. Both are very very dangerous substances that really shouldn't be used without proper training and safety equipment. I don't think there's really anything accessible for a home experimenter, unfortunately.


Is is possible to add NO2X, AlX3 to achieve nitration of benzene? According to Organic Chemistry as Secondary Language (2nd semester), for nitration of benzene, the general way is to add $\ce{HNO3}$ and $\ce{H2SO4}$ and it said to make $\ce{NO2+}$. $\ce{(AlBr3 + NO2Br)}$ is impossible since it is a 'nasty' stuff? Why is it 'nasty'?
I searched online, there is no info about $\ce{NO2Br}$. Is it too unstable? If yes, may I ask why? However, there is $\ce{NO2Cl}$ present, so is it possible for it to react with benzene ($\ce{+AlCl3}$) to achieve nitration?

Nitration using nitryl chloride($\ce{NO2Cl}$) was studied extensively by Price and Sears where they found $\ce{AlCl3}$ to be a suitable reagent for the reaction. They also noted that nitration was difficult for deactivating aromatics such as halobenzene and benzotrifluoride and hence this method was considered of limited value.
On further experimentation, it was found that $\ce{TiCl4}$ was the best suitable catalyst. Other reagents used were $\ce{FeCl3,ZrCl4,AlCl3}$ and $\ce{AlBr3}$ but the reaction conditions were difficult to control. $\ce{BCl3}$ gave smaller yield with possible ring chlorination and $\ce{SbCl5}$ were effective for only aromatic compounds. The following compounds were nitrated using nitryl chloride and $\ce{TiCl4}$ as catalyst and data was noted:
\begin{array}{c|c}
\mathbf{Compound} & \mathbf{Yield (percent)} \\\hline
\text{benzene} & \text{88%}\\
\text{toluene} & \text{81.5%} \\
\text{ethylbenzene} & \text{79%}\\
\text{fluorobenzene} & \text{91%}\\
\text{chlorobenzene} & \text{89%}\\
\text{bromobenzene} & \text{86.5%}\\
\text{o-dichlorobenzene} & \text{41.5%}\\
\text{benzotrifluoride} & \text{32%}\end{array}
An excess of solvent was used for the reaction. Tetramethyl sulfone was suitable for the reaction. Other solvent experimented with was carbon tetrachloride but many chlorinated byproducts were observed. The possible mechanism for the reaction (with $\ce{AlCl3}$ as catalyst) was noted:

or a possible salt formation mechanism was indicated:
$$\ce{NOCl2 + AlCl3 <=> NO2+AlCl4-}$$
Nitration using Nitryl chloride was once important in making chloronitrosoalkane where carbon-carbon double/triple bond were broken by nitryl chloride and an free radical addition reaction occured but the yield was not that high. Dinitromethane was used as solvent as nitryl chloride breaked into chlorine and nitrous oxide:
$$\ce{NO2Cl -> NO2 + Cl.}$$
$$\ce{CH2=CH2 ->[NO2Cl] Cl-CH2-CH2-NO2(45 {%}) + Cl-CH2-CH2-ONO(35 {%}) + CH2Cl2 + N2O4}$$
$$\ce{C2H5-C#C-C2H5 + NO2Cl -> C2H5-ClC=CNO2-C2H5}$$
As for nitryl bromide, it was found to be unstable. Zuskin et. al. claimed to have synthesized the product$\ce{^{[4]}}$ but were invalidated due to their instability. Many attempts$\ce{^{[5]}}$ of making nitryl bromide were made but failed as they decomposed to nitric oxide and bromine. So, nitryl bromide was observed as a combined entity of (nitryl chloride + $\ce{KBr}$ in sulfur dioxide) at $\pu{-20 ^{\circ}C}$ and using $\ce{TiBr4}$ as catalyst, nitration was done but yields were low(lower than that observed with nitryl chloride) with possible ring-brominated side products. This is due to interference of side products from partly decomposed nitryl bromide.
Notes and References

Across Conventional Lines: Selected Papers by George A Olah, G K Surya Prakash

Chemistry of Energetic Materials by George A. Olah, David R. Squire

Nitrate Esters Chemistry and Technology by Jiping Liu

Bromine vapor saturated with nitrogen peroxide passed through a tube filled with bone-ash at $\pu{200-250^{\circ}C}$.

Halogen exchange of nitryl chloride with $\ce{KBr}$ in liquid $\ce{SO2}$, reaction of $\ce{BBr3}$ with anhydrous niyric acid or dinitrgen pentoxide, reaction of bromosulfonic acid with nitric acid and ozonolysis of nitrosyl bromide.


Do individual crystals resulting from evaporating a mixture of two ionic compounds contain both cations/anions within the crystal lattice? I have wondered this for awhile. Let's say I dissolve, without loss of generality, potassium chloride and sodium chloride in water, then evaporate the water. Would any of the resulting crystals consist of randomly dispersed potassium and sodium cations within the lattice or would each resulting crystal be purely potassium chloride/sodium chloride?

Crystals with randomly intermixed ions only occur when the ions have very similar diameter (really the same spatial charge distribution in its exterior, but alkali ions are in very good approximation spherical) and same charge. That would be called a single phase solid solution.

However, $\ce{Na+}$ and $\ce{K+}$ have quite different diameters ($r=102$ resp. $\pu{138 pm}$). Your mixed solution will crystallise as NaCl with a low amount of potassium ions and KCl with some potassium replaced by sodium ions. In other words, your isotropic solution decomposes into a solid two-phase mixture.

The two substances will surely have a different tendency (also temperature dependent) to include such defect ions, and the two compounds have different solubility, so one will precipitate first, so the concentrations in the residual solution changes greatly during the process, and consequently also the defect concentration in the single crystals.


Fourth principle of Molecular Orbitals The fourth principle of Molecular Orbitals state that:

Molecular orbitals are best formed when composed of Atomic orbitals of like energies.

I'm not sure about how I should interpret this; which is the correct?

It means that the molecular orbitals that I need to consider are the ones formed by the same orbitals on each atom (like a pair of $2p$, two $3s$, etc).
It means that I need to evaluate the actual energy on each orbital using: $$E = \frac{-13.6 z^2}{n^2}$$ And then compare the energies of every orbital, to match each orbital on Atom 1 to the closest one on Atom 2? Meaning that I could possibly have a $2s$ pairing with a $3p$, depending on both atoms? (Note that the formula gives different results for each Atomic Orbital depending on the atom's $z$).

I known that every possible orbital combination exists, but I'm asking about the ones that are more relevant and thus need to be considered.

Both of the statements have merit. When the orbitals are of the same type, the overlap is maximized. When the orbitals are nearly the same energy, the atoms are closer together (and give better overlap) than when the energy difference is larger. You don't really need to calculate the energies of the orbitals, just look at the size of the ions. The larger the ion, the higher the energy level of the bonding electrons.


Which one is more acidic – 3,5-dinitrophenol or 4-nitrophenol I’m confused because in 3,5-dinitrophenol, the inductive effect is dominating whereas in 4-nitrophenol the mesomeric effect dominates. So which one should we consider for acidity?

As Zhe's comment points out, intuition does not count for much here, unless you have enough actual experience working with these systems to know what is more acidic. There are some things which cannot be predicted a priori using simplified models.

The only way to find out is to look up the data. The acidity constants of some of these phenols are tabulated in J. Org. Chem. 1998, 63 (19), 6680–6688. 4-Nitrophenol has a $\mathrm pK_\mathrm a$ of $7.14$, and 3,5-dinitrophenol a $\mathrm pK_\mathrm a$ of $6.73$, in aqueous solution.


What is the mass fraction of C2HCl3 in the liquid phase? Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene (C2HCl3) and trichloromethane (CHCl3) is established at 25 °C. The mole fraction of CHCl3 in the vapour phase is 1. What is the mass fraction of C2HCl3 in the liquid phase?
The vapor pressure of trichloroethene and trichloromethane at 25 °C are:

Pvap, C2HCl3 = 73.0 mmHg

Pvap, CHCl3 = 199.1 mm Hg

I'm confused about this problem, if the mole fraction of one element equals to 1, would not be the mole fraction of another element be 0? Can someone advise me on how to approach this problem?

I'm sorry. I mislead you. Temperature changes, and the boiling points of the liquid have nothing to do with this problem as I erroneously commented.

The correct approach is to simply apply Raoult's law. For some temperature T, let

    $p^*_i =$  partial pressure of some pure compound $i$
    $x_i =$ mole fraction of the compound $i$ in solution.

then let $p_i$ be the partial pressure of compound i above a liquid mixture. For an ideal system then:

    $p_i = p^*_ix_i$

The problem states "The mole fraction of $\ce{CHCl3}$ in the vapor phase is 1."

As you deduced, since there is no $\ce{C2HCl3}$ in the vapor phase there can't be any $\ce{C2HCl3}$ in the liquid phase either.


How to measure the speed in a catalized reaction? What is cheaper?

Set up a constant time of reaction and measure the concentration of the products or;
Set up a constant quantity of reactants and, with an indicator, measure the time until the reaction stops.

If you know another protocol, Can you explain it, please?

It depends on scenario.

If time is cheap, while measurement is expensive ( e.g.high fixed price for each value ), then constant time method is probably cheaper.

If time is expensive, while measurement cheap ( e.g. continuos measurement ), the constant products method may be cheaper.


What is the electrochemical reaction of this battery? I've watched a YouTube video "Soda Can Battery - Power From Trash - DIY Project" by Source Port Services. I understand that the anode can be oxidised, but what is the reduction reaction at the cathode? They are using a salt water solution $\ce{Na+}$ $\ce{Cl-}$ and a copper cathode.

$$
\begin{array}{cc}
\hline
\text{Half-reaction} & E^\circ/\pu{V} \\
\hline
\ce{Cu^2+ + 2 e− <=> Cu(s)} & 0.337 \\
\ce{Na+ + e- <=> Na(s)} & -2.71 \\
\ce{Al^3+ + 3 e− <=> Al(s)} & -1.662 \\
\hline
\end{array}
$$

Thus we can see that the aluminium is oxidised, but there are no $\ce{Cu^2+}$ ions to be reduced also $\ce{Na+}$ reduction would lead to a negative $E_\mathrm{cell}$ standard and a positive Gibbs energy. Is it an air cathode where $\ce{O2}$ is being reduced?

Thus we can see that the Al is oxidised but there are no Cu2+ ions to
  be reduced also Na+ reduction would lead to a negative Ecell standard
  and a positive Gibbs. Is it an air cathode where O2 is being reduced?

Well, there is no free lunch in this world. This is not a conventional cell in which each electrode is sitting in its own ions and one can easily calculate their Nernst potential. What we are probably seeing here is the very concept of contact potential. When two dissimilar metals are brought into contact via some electrolyte, a potential difference appears between the two. Recall the first experiment of Volta. He made a long series of
-silver[wet cardboard]zinc-silver[wet cardboard]zinc-silver[wet cardboard]zinc- cells, where the cardboard was wetted with salt-water. He was able to generate large potential differences this way by making very long series of Ag-Zn couple. Is this set-up good for drawing current-probably not.

These experiments were done in the 17th century by Volta. Atoms and ions did not exist then. These are contact potentials and you do not need a redox system to begin the experiment (metal dipping in its salt). Let us say you start with Al, Al will start to dissolve in a corrosive medium (salt is very corrosive) to form Al ions. You have a valid point, there are no copper ions to reduce? What else do we have in the system? Water, it has some ionized $\ce{H+}$, they start to reduce, one should see hydrogen bubbles at the copper electrode. Certainly you can draw small currents to power a LED from such batteries but not for long.

I do have some doubt about the design of the experiment. All soda cans are lined with a very very thin polymer layer. So the contact of Al is not that good with the copper coil. If one scrubs the can very well, only then that polymer layer would come off. The claim of seven LEDs is a stretch.


Does Wurtz-Fittig-reaction involves Sn1 or Sn2? Following mechanism is commonly proposed for the Wurtz-Fittig-Reaction:

Picture from Wikipedia

The red marked reaction steps is a nucleophilic substitution (SN). I suppose that this is a SN1 reaction, with an intermediate carbocation, rather than a SN2 reaction. In my opinion the resulting carbocation is good enough stabilized throughout the aromatic ring and therefore SN1 should be preferred, moreover because of the steric hindrance of the aromatic ring.

Is this assumption right? Thanks a lot for your help.

Aryl carbons cannot go through SN1 nor SN2 substitution reactions. SNAr involves the creation of benzyne intermediate. The mechanism is this:

The benzyne has sp hybridized carbons, which should be linear, but forced to be bent. Therefore it can be attacked by a nucleophile, and become sp2 which isn't linear.

Notice that the carbanion can also attack the adjacent carbon. It's a special phenomena for this type of reaction. The mechanism was confirmed by carbon-14 labeling (Organic Chemistry/Paula Bruice, sixth edition, page 692).


Limiting radius ratio I recently came across this question while studying Solid State.
The coordination no. of [Ni(CN)4]2- is 4  as Ni is bonded to 4 CNs and  is planar in structure.Hence its Limiting Radius Ratio must be between 0.225 and 0.414 . But the answer given in the book is 0.414 to 0.732 (which is the limiting radius ratio for coordination no. 6) .
How is this true?Some sites agree with my answer .
Is the answer given in the book wrong or have I made any mistake in the reasoning?

Here, limiting radius ratio is for geometry of this complex is square planar because $\ce{CN-}$ is a strong field ligand and causes pairing of unpaired electrons. So we can relate that $\ce{Ni^2+}$  occupies a square planar void of the $\ce{CN-}$ lattice. The square planar void is a cross section of the octahedral void.


What is the reason behind the phenomenon of Joule-Thomson effect? For an ideal gas there is no heating or cooling during an adiabatic expansion or contraction, but for real gases, an adiabatic expansion or contraction is generally accompanied by a heating or cooling effect. What is the reason behind such a phenomenon? Is it related to the property of real gases or is it something else?

In a reversible adiabatic expansion or compression, the temperature of an ideal gas does change.

In a Joule-Thompson type of irreversible adiabatic expansion (e.g., in a closed container), the internal energy of the gas does not change.  For an ideal gas, its internal energy depends only on its temperature.  So, for an irreversible adiabatic expansion of an ideal gas in a closed container, its temperature does not change.  But, the internal energy of a real gas depends not only on its temperature but also on its specific volume (which increases in an expansion).  So, for a real gas, its temperature changes.  The Joule-Thompson effect is one measure of the deviation of a gas from ideal gas behavior.

ADDENDUM

This addresses a comment from the OP regarding the effect of specific volume on the internal energy of a real gas.

Irrespective of the Joule-Thompson effect, one can show (using a combination of the first and second laws of thermodynamics) that, for a pure real gas, liquid, or solid (or one of constant chemical composition), the variation of specific internal energy with respect to temperature and specific volume is given by:
$$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$The first term describes the variation with respect to temperature and the second term describes the variation with respect to specific volume.  For an ideal gas, the second term is equal to zero.  However, for a real gas, the second term is not equal to zero, and that means that, at constant internal energy (as in the Joule-Thompson effect), the temperature will change when the specific volume changes.  This is a direct result of the deviation from ideal gas behavior.


Products of o-aminobenzaldehyde thermal treatment with acetone and further with sulfuric acid o-Aminobenzaldehyde is heated along with acetone in dilute $\ce{NaOH}$ solution to produce the major organic product (A), which on treatment with a catalytic amount of $\ce{H2SO4},$ produces another major organic product (B) formed via intramolecular reaction.
What will be the products A and B formed in this reaction?

I have tried solving it and I think A will be o-aminobenzoic acid. But I want to ask that what happens when o-aminobenzoic acid is reacted with catalytic amount of $\ce{H2SO4}?$

This is an example of the Friedländer synthesis of quinolines Wikipedia article  here.

Under catalysis by hydroxide acetone undergoes an aldol condensation with the aldehyde group to give 3, which is your compound A ($\ce{R2=Me, R3=H}$). Under the catalysis of sulfuric acid A loses water to give the quinoline 7 (for your B $\ce{R2 = Me, R3=H}$).

Several reviews of the Friedländer exist, this is a good one Chem. rev. here.

Image is from ref 1.


How do I calculate the ensemble-average net charge of an amino acid at given pH? I am given an amino acid with an ionizable side chain at a certain pH. How do I determine the net charge of that amino acid when there are mixed protonation states of one or more of the groups at that pH (pKa of side chain, for example, is really close to the pH)?

Amino acids have terminal carboxyl and amino groups; some amino acids have ionizable side chains. When determining the charge of an amino acid, you have to take into consideration the pH and the pKa's of each of these groups. When the pKa of one group (or more) is close enough to the pH, a fraction of the amino acids will be deprotonated at that group and the other fraction of amino acids will be protonated at that group in solution. Thus, when determining the average net charge across the ensemble (or a time-averaged charge of a single particle), you have to take this into account.

I am asking for the expected value of the net charge (which would not be an integer); this number is relevant, for example, for the migration speed of the amino acid (or a protein) in gel electrophoresis or the strength of interaction with ion exchange chromatography media.

For example, a carboxylic acid/carboxylate group at a pH equal to its pKa would have an average charge of minus one half because half of the functional groups would be protonated (charge of zero) and half would be deprotonated (charge of minus one).

The Henderson-Hasselbalch relationship describing each ionizable group is:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[A-]}}{\ce{[AH]}}$$

We can solve for the ratio:

$$10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})}=  \frac{\ce{[A-]}}{\ce{[AH]}}$$

However, we really want the fraction of protonated among the total (not the ratio of deprotonated to protonated).

$$10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})} =  \frac{[\mathrm{total}] - \ce{[AH]}}{\ce{[AH]}} = \frac{[\mathrm{total}]}{\ce{[AH]}} - 1$$

Add one to both sides:
$$10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})} + 1 = \frac{[\mathrm{total}]}{\ce{[AH]}}$$

Take the reciprocal:
$$\frac{\ce{[AH]}}{[\mathrm{total}]} = \frac{1}{10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})} + 1}\tag{1}$$

This is still general for any acid/base group. For example, we could use it to calculate the charge of ammonia/ammonium ($\ce{NH3(aq) + H+(aq) <=> NH4+(aq)}$). At very basic pH, the charge would be zero, at very acidic pH, +1. To get the average charge at any pH, we take the charge at very basic pH and add the result of equation [1] using the $\mathrm{p}K_\mathrm{a}$ value of ammonium.

For any amino acid (or any other molecule with ionizable groups with $i$ different $\mathrm{p}K_\mathrm{a}$ values), you take the charge of the species at very basic pH (all groups deprotonated), plus the following:

$$\sum_i \frac{1}{10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a,i})} + 1}\tag{2}$$

This is only an approximation because there might be some cross talk between ionizable groups (i.e. if one group becomes negatively charged, it becomes more "difficult" for the neighboring group to become negatively charged). It also becomes more complicated for polyprotic groups, but all the groups in amino acids are monoprotic with water as a solvent.


Compound having highest standard reduction potential Which compound has highest SRP (standard reduction potential) value, e.g. the strongest oxidising agent? In most tables $\ce{F2}$ is at top with $E^\circ = \pu{2.87 V}$, but I've found a table listing $\ce{H4XeO6}$ at top with $E^\circ = \pu{3.0 V}$.

Is there a more strongest oxidising agent on basis of $E^\circ$ value and is $E^\circ$ value defined for every compound? If yes, what is $E^\circ$ value for $\ce{ClF3}$?

Top 5 oxidants from the CRC Handbook of Chemistry and Physics [1, p. 5-83]:

TABLE 2. Reduction Reactions Having $E^\circ$ Values More Positive Than That of the Standard Hydrogen Electrode
$$
\begin{array}{ll}
\hline
\text{Reaction} & E^\circ/\pu{V}\\
\hline
\ce{F2 + 2 H+ + 2 e- ⇌ 2 HF} & 3.053\\
\ce{Tb^4+ + e- ⇌ Tb^3+} & 3.1\\
\ce{Pr^4+ + e- ⇌ Pr^3+} & 3.2\\
\ce{Cf^4+ + e- ⇌ Cf^3+} & 3.3\\
\ce{XeF + e- ⇌ Xe + F-} & 3.4\\
\end{array}
$$

Note that standard reduction potential is not characteristic for a single compound. Instead, it is only makes sense for a given redox pair – of course, assuming standard conditions ($\pu{298.15 K}$, $\ce{101.325 kPa}$), activity of all soluble species $\pu{1 .000 mol L-1}$ and measured against the standard hydrogen electrode.
For example, iodate $\ce{IO3-}$ has very rich redox chemistry and many SRP values [1, p. 5-79]:

$$
\begin{array}{ll}
\hline
\text{Reaction} & E^\circ/\pu{V}\\
\hline
\ce{2 IO3– + 12 H+ + 10 e- ⇌ I2 + 6 H2O} &1.195\\
\ce{IO3– + 6 H+ + 6 e- ⇌ I– + 3 H2O} &1.085\\
\ce{IO3– + 2 H2O + 4 e- ⇌ IO– + 4 OH–} &0.15\\
\ce{IO3– + 3 H2O + 6 e- ⇌ IO– + 6 OH–}& 0.26\\
\end{array}
$$

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.


How do I achieve same crystallinity when cooling polymers? I am trying to make "films" of polyethylene glycol by melting them, pouring them into a mold and then cooling the mold down. The plan is to do this with different molecular weights. The problem is that depending on the molecular weight, the crystallinity will be different. Is there a way of ensuring that the crystallinity of the polymer is the same for all of them?
Thanks

Is there a way of ensuring that the crystallinity of the polymer is
  the same for all of them?

First you need a method to assess the degree of crystallinity.  Samples of the same material that differ in degree of crystallinity should show different light transmission intensities due mainly to increased light scattering in the more crystalline samples.  If you have access to a spectrophotometer you could insert the films and measure the light transmission.  If you don't have access to a spectrophotometer, a "poor-man's" approach would be to use visual comparison of the transmitted light (use a black piece of paper and a pencil beam flashlight in a dark room).  It is important that the film thicknesses all be the same.  If there is variability in your film thickness, then cast thicker films so that the percentage variation will be smaller.

By cooling your sample more quickly, you will tend to preserve more of the amorphous nature found in the solution.  Conversely, cooling more slowly or annealing your sample will tend to increase the degree of crystallinity.

Adjust your rate of cooling on each of the films until they all produce the same transmission intensity, then the degree of crystallinity should be similar in your films.


Spin orbital coupling and total angular momentum The Sodium emission spectrum has 2 bright yellow lines called the Na-D lines. Superficially I understand the cause; there's spin-orbit coupling and the 3p orbital is split into 3p(3/2) and 3p(1/2).
But I don't understand why those numbers (3/2 and 1/2)?
Are these the total angular momentum values or the orbital angular momentum values?
And how does one work them out?

J is the total angular momentum. L is the orbital angular momentum and S is the intrinsic total electron spin angular momentum. J = L + S
L = $\sqrt{L\left(L+1\right)}\frac{h}{2\pi }$
--> L can be n-1 where n is the Principal quantum number
S = $\sqrt{S\left(S+1\right)}\frac{h}{2\pi }$
--> S is an integer or half an odd integer, depending on whether the number of electrons is even or odd
J = $\sqrt{J\left(J+1\right)}\frac{h}{2\pi }$
J can be L+S ... L-S
For the  Na-D lines the J = 3/2 is caused by L=1 and S=1/2 for the 3/2 value and L=1 and S=-1/2
The values for $J,L,S$ have a range of values e.g. $-X, -X+1,\cdots X-1, X $, e.g. if $L=5/2$ the range is $-5/2,-3/2,3/2,5/2$


Lewis acids as ligands Usually Lewis bases as my textbook says:

  2.3 Ligands
  The ligands are the ions or molecules bound to the central atom/ion in the coordination entity. This is better visualized as the combination of a Lewis acid (the central atom/ion) with a number of Lewis bases (ligands). The atom of Lewis base that forms the bond to the Lewis acid (central atom/ion) is called donor atom ...

According to me its not compulsory for a ligand to be a Lewis base, as $\ce{BF3}$ is a Lewis acid and can act as a ligand by accepting electrons from a metal in vacant orbital of boron. Is it true?

Yes it can happen.  This paper offers an example, in which $\ce{BF3}$ or $\ce{B(C6F5)3}$ adds to a low oxidation state iridium complex.  There is really not a lot of mystery behind it.  The iridium in its low oxidation state has electrons to offer and the boron with its vacant orbital can take them up.


Concerted hydride and alkyl shifts I noticed while looking through a couple of books (specifically, Wade, Carey, and Vollhardt/Schore) that the mechanism of concerted hydride and alkyl shifts is always shown with alcohol leaving groups.  I understand that these are difficult, equilibrium-driven processes.  But, I would assume they can happen with other leaving groups, such as a halogen, right?  Is it just the matter of fact that dehydrations are usually done at elevated temperatures while SN2 is not?  If that's all it is, though, we would expect to see rearranged primary E2 products, which Carey mentions does happen, but the other books remain silent on.

The SN2 reaction involves an attacking group and a leaving group moving in concert. A pentacoordinate transition state is involved.  Typically, there is no rearrangement in an SN2 reaction.  Similarly for an E2, as a base removes a proton. a leaving group is ejected in concert to create an olefin, no rearrangement products are observed.

It is the SN1 and E1 reactions that proceed stepwise through distinct carbocation intermediates.  As the carbocation is being formed in these reactions, a hydride or alkyl group can migrate to produce a more stable cation.

The leaving group can be things other than a hydroxyl group, for example chloro, bromo, or iodo groups make good leaving groups.  Back in the early days of physical-organic chemistry many alcohols were used in reactions simply because they were readily available as natural products.  Alcohols could also be readily converted into acetates,  tosylates, brosylates, or other sulfonate esters, which were even better leaving groups and allowed for reactions to proceed under even milder conditions.


How can I draw 2,3-diethylheptane? This question was given by my teacher in class:

  Draw the structure of 2,3-diethylheptane.

So I drew the following:

But now if I name the compound according to the longest chain, it comes out to be 3-methyl-4-ethyloctane! How can I properly draw these structures?

What I usually do when I start numbering is look for branching and start numbering at the end with the closest branch. Also, keep in mind that with branching you'll want to consider numbering from one of the branches (if it'll yield a long chain.)

I'm using the structure you made as an example:

Here I numbered from the side with the most branching. But then I numbered it in two different ways. One way was starting with the straight chain which gave me a 7 carbon chain. Then I renumbered it (the image on the right) starting from one of the branches. This gave me an 8 carbon chain. (You could also number starting from the other branch, but that gives you 7 as well.)

Choose the longest chain. Since stuff that looks like a side chain can actually be part of the main chain, seriously keep this in mind during exams. It's totally okay to number the same alkane in different ways and play around with it to find which is the longest chain.

4-ethyl-3-methyloctane

Now, if you're having trouble drawing them:

I would draw just a zig-zag pattern (like I drew above) containing the number of carbons you need. You need a heptane, so start out with a 7 carbon chain.

This will just be your bare, skeleton structure. Then number the carbons, then add the ethyls to carbons 2 and 3.

HOWEVER, the question is flawed. You actually can't have an ethyl on carbon number 2 here, because it just makes the chain longer.


Can you call H_2 an Atom? I am currently studying to complete the Chemistry CLEP exam, I ran into this question and was thoroughly confused by the phrase

  12.04x10^23 atoms of H_2

According to the question, this corresponds to 12.04X10^23 molecules (2 moles) of H_2, or 24.08x10^23 atoms (4 moles) of Hydrogen. The use of the word "atom" over "molecule" seems useless, confusing, and possibly wrong.

I'm not sure if this question is written poorly or if I have made a mistake in my grammar either way, any help is appreciated. Thank you!

The question is indeed poorly worded. What the question seems to be saying is that there are $12.04\times10^{23}$ molecules of hydrogen gas/dihydrogen ($\ce{H2}$).

I'm also noting that the correct answer ends with "... $\pu{44.8L}$ of $\ce{H2O}$ present", while there was absolutely no mention of $\ce{H2O}$. The question is quite misleading.


Fermentation in beer (reaction schemes)? I've been struggling with this for a while, and I can't seem to solve the problem. Alcohol is made in the fermentation process in beer:

  Sugars formed during mashing (maltose and maltotriose) will undergo hydrolysis and form multiple molecules of the simple sugar glucose. This glucose will then be fermented by the yeast under anaerobic conditions into ethyl alcohol and carbon dioxide

I understand that perfectly. My problem is to figure out the reaction schemes for the reactions, for each step. Can anyone help me with that?

We can break the entire process into three parts:

breakdown of maltose and maltotriose into glucose
glycolysis to give pyruvate
reductive decarboxylation to give the alcohol.

Writing out full equations for each actual step should hopefully not be required; especially glycolysis is explained in detail in many places across the internet and practically every metabolism textbook. Hence I will restrict myself to generic equations for each part of the process. Furthermore, each set of reactions is catalysed by a specific set of enzymes; I will be ignoring these.

Maltose breakdown

$$\ce{\underset{maltose}{C12H22O11} + H2O -> 2 \underset{glucose}{C6H12O6}}\tag{1}$$
Glycolysis

$$\begin{multline}\ce{\underset{glucose}{C6H12O6} + 2 ADP^3- + 2 HPO4^2- + 2NAD+ -> } \\\ce{2 \underset{pyruvate}{C3H3O3-} + 2 ATP^4- + 2 H2O + 2 NADH + 2 H+}\end{multline}\tag{2}$$
Alcoholic fermentation

$$\begin{align}\ce{\underset{pyruvate}{C3H3O3-} + 2 H+ &-> CO2 + \underset{acetaldehyde}{C2H4O}}\tag{3}\\
\ce{\underset{acetaldehyde}{C2H4O} + NADH + H+ &-> \underset{ethanol}{C2H6O} + 2 NAD+}\tag{4}\end{align}$$


Why does iodine have different colors in different media? It is violet in trichloromethane, but brown in water. why? In the Shriver Atkins book, they tried to explain this using Molecular Orbital Theory which was not clear to me

color is result of electron transfer between orbitals and anything that affect energy of these orbitals can change color of that specie, so iodine in different mediums have different interactions and because of that iodine orbitals have different energies in different medium,so iodine have different color in different mediums.


Why does Tyndall scaterring produce conical shape? When light is passed through a collidal solution, due to Tyndall scattering, the path of light appears to be a bright cone. Why does it have to form only a conical shape?
The conditions for Tyndall effect are that the diameter of dispersed particles should not be much smaller than wavelength of light used and refractive index of dispersed phase and dispersed medium should differ greatly.
This is perhaps because colloidal particles scatter light in all directions. Normal refraction shouldn't cause such a shape.
Is there any logical reson for the conical shape?

Source: NCERT class 12

It is simple.
Light always travels in a straight line and if a lens is present it would diverge or converge ( in a straight line ) accordingly.

The question is why no other shape like maybe a hemisphere or a paraboloid ?

Now let me ask the question, have you seen such thing?
Check out this diagram to make everything clear.

If the beam of light increases in width ( diverging beam) linearly with distance. It is because all the particles scatter equally in all directions and some of them reach your eyes and this reveals the shape of the beam. The adjacent particles does the same and so on.
Therefore you can also see the light getting more and more diffused linearly as more and more light is scattered.

It also depends on the refractive index of the medium where the incident light ray will either move closer to the normal or move further. Again this would lead to a conically diverging beam.

EDIT

  This is to be noted that the colloidal solution does not affect the shape of the beam of light ( converging or diverging ).

This is also evident from your diagram.

From your question I feel that you think that the colloidal solution is responsible for the cone shape, it is not so. It is simply the geometry of the light rays

What the solution does is scatter the light in all directions and makes the "beam" visible.  A straight beam of light in perfect vacuum would not be visible as a stalk of light ( that we see when light enters a dark room thought a window )


Why does Devarda's alloy include copper? Devarda's alloy is used in the detection of nitrates. However, judging by the reaction equation, only aluminium plays the role of reducer:

$$\ce{3NaNO3 + 8Al + 5NaOH + 18H2O -> 8Na[Al(OH)4] + 3NH3(g)}$$

Why does it contain copper? Is copper also capable of acting as a reducer here?

I know that zinc, the third component, is capable of that.

Virtually every analytical chemistry textbook's chapter on anion determination mentions that Devarda's alloy is very brittle and can be easily pulverized in a mortar, unlike $\ce{Al}$ and $\ce{Zn}$ metals.
This is quite handy property as for qualitative analysis of nitrates in a heterogeneous media the reducing phase should be ground to fine powder in order to:

assure micro-scale precision of the added reactant, which is tricky to achieve with aluminium or zinc, as both metals are usually shipped in granules, and the fine metal powders are prone to oxidation, increasing the amount of unwanted oxides (plus, their powders are sticky);
increase the available reaction surface to speed up the analysis and increase its completeness (especially important for isotopic nitrate analysis).

On top of that, German Wikipedia
also supports both comments by Ivan Neretin

  Galvanic couple, I guess.

and aventurin

  Catalyst, I guess. ;-)

and suggests that copper facilitates catalytical formation of a local electrochemical cell.
A reduction with Devarda's alloy can be carried out in weakly alkaline solution or even in neutral solution, whereas reduction with aluminum often requires a strongly alkaline solution.
At the same pH, the reaction with Devarda's alloy is overall significantly faster than when using zinc or aluminum.

This was experimentally demonstrated by Mertens et al. [1]:

  The superiority of Devarda alloy compared to aluminum dust as a reductant in alkaline medium is illustrated best with the following experiment (see Figure 3).
  Twenty ppm $\ce{NO3-}$ is subjected to reduction, first with aluminum dust and the potential of the ammonia probe is recorded continuously.
  One observes a slowly rising potential.
  On addition of Devarda alloy to this reaction mixture, a potential jump is obtained and the potential stabilizes rapidly (after 5 minutes).
  This shows clearly that the kinetics of the reduction with Devarda alloy are much more rapid than with aluminum dust.

  Figure 3. Potential observed with the ammonia probe during the reduction of 20 ppm $\ce{NO3-}$ as a function of time
  From $t = 0$ to A: aluminum dust. At point A, Devarda alloy was added.

References

Mertens, John.; Van den Winkel, Pierre.; Massart, D. L. Determination of Nitrate in Water with an Ammonia Probe. Anal. Chem. 1975, 47 (3), 522–526. DOI: 10.1021/ac60353a009.


Is HMgX also a Grignard Reagent like RMgX? We do know that the Grignard Reagent is $\ce{RMgX}$ where R is some alkyl group & X is any halogen group.
So, my question is that, whether $\ce{HMgX}$ is also a Grignard Reagent like $\ce{RMgBr}$?

According to this reference $\ce{HMgX}$ compounds are a thing.  You can't make them with magnesium +$\ce{HX}$, of course (what actually happens?), but you can get them from magnesium halide salts plus an "active" form of magnesium hydride in THF.  They act as hydride ion sources for reaction with $\ce{AlH3}$ and $\ce{BH3}$ (forming mixed tetrahydridoaluminate and tetrahydridoborate salts, respectively), so we may reasonably expect the same sort of hydride transfer with organic substrates.  Such a hydride transfer would be analogous to conventional Grignard reagents transferring their alkyl or aryl anion moieties.


What value of temperature does T represent in Gibbs Free Energy Equation? Temperature before, or after reaction? Temperature change? Or what is it?

We can not calculate the exact amount of G (Gibbs free energy), only we can calculate change of that,so we must put temperature change on this equation:
$$\Delta G = \Delta H - \Delta TS $$


Why is some ionic substance insoluble, such as BaSO4? If Ba is less electronegative than Mg, then why is BaSO4 insoluble in water and MgSO4 is soluble? I thought the greater the EN difference the more soluble the compound is because of how unequally shared the electron is.

The electronegativity is not the main factor. The entropy factor is much more important. When a salt like $\ce{BaSO_4}$  tries to get dissolved, its ions are attracting water molecules that get fixed or attached around each ion. This phenomena creates order, and decreases the entropy of the solution. The same thing happens around sulfate ions $\ce{SO_4^{2-}}$. If there is too much water attached around each ion, the substance does not dissolve. This is the case for $\ce{BaSO_4}$, because the ion $\ce{Ba^{2+}}$ is bigger than $\ce{Mg^{2+}}$, and so creates more order in attracting more water molecules around it. The entropy decreases too much if $\ce{BaSO_4}$ gets dissolved. This salt is practically not soluble. This is not the case for $\ce{MgSO_4}$.


Debye-Huckel limiting law I encountered this equation when studying the Electrolyte solutions chapter in my analytical Chemistry textbook, Actually the terms in summation are giving a physical clue of what $I$ is, But what's the point of that $\frac{1}{2}$ ?
I've searched the Internet and a molecular Thermodynamics book that I have but I found nothing but the equation itself.

$I=\dfrac{1}{2}\sum_{i}m_{i}z_{i}^{2}$

$I$ is the ionic strength of a solution, and is a measurement of the concentration of ions in a solution. $m_i$ is the concentration of the ion i, and $z_i$ is the charge of that ion. The $\frac{1}{2}$ in front of the summation is a result of the fact that both positive and negative ions are considered when calculating the ionic strength. $m_i$ could either be a measurement of Molarity or molality, and your book should state which definition you are using (I assume molality due to the lowercase m).


How is bromothymol blue synthetised? In a classical reaction, phenolphthalein is synthetised from phthalic anhydride and phenol with acid catalyst. When one uses substituted phenols, one obtains similar compounds (e. g. cresolphthalein, thymohlphthalein).

However, some acid-base indicators contain a sulfo- group instead of a carboxy- group of phthaleins. Does that mean that they are synthetised from phenols and the anhydride of o-sulfobenzoic acid? Phthalic anhydride is fairly cheap, but how is o-sulfobenzoic anhydride prepared?

Now a days, sulfonephthaleins can be prepared by reaction of readily available saccharin and the desired plenol compound (Ref.1 & Ref.2). In this method, active reagent, sulfobenzoic anhydride will be prepared in situ as depicted in following diagram:

In this method, it's described the preparation of phenolsulfonephthalein (Phenol Red). If you use thymol instead of phenol, you will get thymolsulfonephthalein, which can be subsequently brominated by bromine/acetic acid to obtain dibromothymolsulfonephthalein (Bromothymol Blue). This procedure is described in Ref.2.

References:

V. H. Tillu, D. K. Dumbre, H. B. Borate, R. D. Wakharkar, V. R. Choudhary, "Solvent-Free One-Pot Synthesis of Sulfonephthaleins from Saccharin and Phenols," Synthetic Communications 2012, 42(8), 1101-1107 (https://doi.org/10.1080/00397911.2010.535946).
B. S. Rao, J. B. Puschett, B. M. Karandikar, K. Matyjaszewski, "Synthesis of Functional Bromothymol Blue Dyes for Surface Attachment to Optical Fibers," Dyes and Pigments 1991, 16(1), 27-34 (https://doi.org/10.1016/0143-7208(91)87018-I).


Thermodynamics of Solid-Liquid If you have an ice cube in water and the ice cube does not melt after reaching equilibrium with the water, does the water have temperature 0 degrees Celsius?

Yes, when ice is present at equilibrium with liquid water (at 1 atm), the system is at 0 °C.

If you continue to remove heat from the water (such as in your freezer), the temperature of the system will remain at 0 °C until all the liquid (water) becomes solid (ice) after which, continuing to remove heat will result in a drop of temperature below 0 °C. Conversely, if heat is added to the system, the temperature of the system will remain at 0 °C until all of the solid (ice) has melted into liquid, after which, the temperature will increase above 0 °C. The amount of (latent) heat required to melt ice is called its enthalpy of fusion (334 J/g). This energy is converted into intermolecular bonds (a change in internal or potential energy of the system) rather than a change in kinetic energy (that is, a temperature change).

Note that at a much lower pressure (0.006 atm) and slightly higher temperature (0.01 °C), pure water will be in three states or phases (gas, liquid and ice). This is known as the triple point of water, as illustrated in the phase diagram of water, shown below:


Why do helium balloons expand in volume as they go higher? I realize as balloons go higher, the atmospheric pressure decreases, doing less to counteract the force of the gas particles pushing against the inner walls of the balloon. But at the same time, doesn't the outside temperature decrease, causing the gas particles in the balloon to lose energy and have less impactful collisions and a lower velocity? Is the reason why the volume nonetheless increases because the pressure outside decreases faster than the temperature? Or am I overthinking this?

I didn't know that balloons expanded during the fly because of thermodynamics, and I didn't know how high they can fly, but a rapid search tells that a partially unfilled regular balloon can fly until an altitude of around $\pu{25 km}$.

Now, $\pu{25 km}$ means that it reaches the first part of the stratosphere, with temperatures of $\pu{-60 ^\circ C}$, that gradually increase until $\pu{0 ^\circ C}$ at $\pu{50 km}$.
As for the pressure, it goes from around $\pu{40 mmHg}$ to $<\pu{1 mmHg}$ in the range $25\text{–}\pu{50 km}$.

If you try a $pV=nRT$ calculation on these data, you see that the gas is already at around 10 times its initial volume with $\pu{40 mmHg}$ pressure and a $\pu{213 K}$ temperature, and that at the $\pu{50 km}$ point the volume is increased 700 times!

Also: while the trend of the pressure is quite logical, that of temperature is caused by complex interations (eg: sun rays that heat particles). You can find this image quite interesting:


A simple equilibrium problem Take the following simple equilibrium:
$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$

Ethanoic acid will act as a typical Brønsted-Lowry acid and donate a proton to water, forming a conjugate acid-base pair. In this case, H3O+ is the conjugate acid, and CH3COO- is the conjugate base.

Logic tells us that if we have a solution of ethanoic acid and water, increasing the amount of acid will decrease the solution's pH. This makes sense, since there'd literally be a higher acid concentration.

However, the interpretation I get due to the equilibrium is the opposite. I see it as increasing the concentration of ethanoic acid will mean there's a larger concentration of conjugate base, hence the solution will become more basic.

Is this flawed logic or simply a misunderstanding of Le Chatelier's principle?

You are neglecting the hydronium ions produced during dissociation of acetic acid.

We can consider the following sequence of events when we add more acetic acid to solution. In reality they occur simultaneously, but it's easier to think about it sequentially.

We have more than the equilibrium concentration of acetic acid, so some of it dissociates to produce $\ce{H+}$ and acetate ion.
Now we have more than the equilibrium concentration of acetate ion, so some of it reacts with $\ce{H+}$ to form acetic acid again.

Your argument is that step 2 makes the solution more basic, but you are neglecting the effect of step 1. Every extra acetate ion produced in step 1 also involves the production of a $\ce{H+}$. Furthermore, step 2 has to occur to a lesser degree than step 1, because step 2 consumes the acteate ions produced in step 1. The acidity of the solution must therefore go up as a whole.*

*Of course it's also possible that steps 1 and 2 both occur at nearly the same rates, and then the pH of the solution doesn't really change much as a result. This is what happens in a buffer.


Does Le Chatelier's principle apply to the gas inside an air mattress? While sleeping on an air mattress yesterday a thought occurred to me. I had to re-inflate the air mattress since it was getting rather low and after it had a good stiffness to it, I went to sleep. The next morning I noticed that a lot of the pressure that I put into the air mattress had left and I remembered Le Chatelier's principle which says that gases in a closed system will convert to the least number of molecules possible if the pressure is increased.

So this is my thought:

Breathing into the air mattress had added moles of gas to a closed system (the air mattress).
Lying on the air mattress increased the pressure of the system.
Given Le Chatelier's Principle, the gases should have transitioned to the lowest number of moles to relieve the pressure introduced on the system.

The thing I'm not sure about is if there are any equilibrium reactions for the gases in our atmosphere that will actually do this. Considering the makeup of our atmosphere, does Le Chatelier's Principle apply to the air inside an air mattress?

No, the air we breathe is about 78% nitrogen, 21% oxygen, 0.5% carbon dioxide and other gases. Nitrogen doesn't react with anything at room temperature and there is nothing to burn inside the air mattress.

What probably happened was that there is a very small hole that the air leaked out. About the only way to find the hole would be to submerge the air mattress in a pool.


Distinguishing pentanol isomers in 13C DEPT NMR spectroscopy Pentanol has eight isomers as shown below. The circles represent carbon atoms that I think are equivalent.

Below there are eight DEPT-spectra that should represent these eight substances.

As you can see some won’t line up. What’s the problem?

In an ideal non-moving world at $0\,\mathrm{K}$, every atom would give exactly one peak in NMR spectra. So the actual question is why some of these spectra have less signals. Partly this is due to local symmetry. If, e.g., two methyl groups are attached to the same carbon atom, these two are said to be chemically equivalent, i.e. one can rotate bonds to overlap one over the other. A different example of chemical equivalence is pentan-3-ol which you can rotate around the $\ce{C-OH}$ axis to give an almost identical molecule, putting the $\ce{CH2}$ groups on top of each other and the methyl groups, too.

With that, you did the circling right, and we should expect:

3 spectra with five signals;
3 spectra with four signals; and
2 spectra with three signals.

However we see one spectrum with two signals and are lacking one with four signals. (And technically, if you look at spectrum 2 closely, that could be interpreted as having five signals, not four.)

In a standard $\ce{^{13}C}$-DEPT spectrum (i.e. the ones you showed), carbons from $\ce{CH}$ and $\ce{CH3}$ groups point upward and those from $\ce{CH2}$ groups downward — and there is your answer. There are carbons in the structures that do not have hydrogens bound to them and which thus do not show up in the DEPT spectra.

In case you’re interested, I will give you the appropriate IUPAC names of the pentanol isomers that were measured in the following spoiler tag.

   2-methylbutan-2-ol. Carbon 2 cannot be seen.

   3-methylbutan-2-ol. The two methyl groups at the end are not completely identical and thus give two signals.

   pentan-3-ol

   pentan-1-ol

   2,2-dimethylpropan-1-ol. Carbon 2 cannot be seen.

   2-methylbutan-1-ol

   pentan-2-ol

   3-methylbutan-1-ol


Carbocation rearrangement in alkene addition reaction This is a question from my textbook:

  Determine a step-wise $S_N1$ mechanism for the overall reaction:

The solution shows the following carbocation rearrangement after Bromine leaves in the first step:

Why does the carbocation rearrange like this, going from a tertiary allylic carbocation to a secondary allylic carbocation? I thought tertiary carbocations are more stable and thus favored to be the major product? I considered Zaitsev's Rule in which the more substituted alkene is the major product, but I thought that only applies in elimination reactions.

Any clarification is appreciated!

The mechanism shown is not a carbocation rearrangement. Those two structures are resonance structures, which means that the chloride anion can attack at either of the carbons that bear a positive charge. Because there are two different alkene products that are possible, the more substituted alkene will be favored. Strictly speaking, you're right, Zaitsev's Rule doesn't apply here, because this isn't an elimination reaction. However, the reasoning is the same: the most substituted, most stable alkene is the favored product.


Is this the correct relative Brønsted acidity in these four acids? We are asked to sort these four acids by increasing Brønsted acidity:

The problem is, I am confused why $\ce{H2F+}$ is the strongest acid of them all. I understand that stronger acids will have weaker bonds, and the $\ce{F-H}$ bonds should be the most electronegative of them all, leading me to believe that they are the strongest bonds. So then why are these bonds the most easily broken? What have I forgotten here? Is the correct order really from weakest to strongest acid as follows? $\ce{CH3OH}$, $\ce{(CH3)2OH+}$, $\ce{CH3SH2+}$, $\ce{H2F+}$

Fluorine being the most electronegative element in the periodic table, it doesn’t like bearing a positive charge. Thus $\ce{H2F+}$, which has the positive charge on the fluorine nucleus, is not very stable and willing to expel the extra proton. So, it is indeed a strong acid.

$\ce{H2F+}$ is formed, e.g., by autoionization of liquid $\ce{HF}$, but only because that reaction also involves creating a $\ce{F-}$ ion (associated with another molecule of $\ce{HF}$ to form $\ce{HF2-}$), which is very favorable:

$$\ce{3 HF <=> H2F+ + HF2-}$$


Why is hypochlorite a stronger oxidizing agent than other oxychlorides? The reduction potentials of perchlorate, chlorate, chlorous acid and chlorine dioxide in acidic and basic solutions are listed in the table below:
$$
\begin{array}{lcc}
\hline
\text{Oxychloride} & E^\circ_\text{acidic}/\pu{V} & E^\circ_\text{basic}/\pu{V} \\
\hline
\ce{ClO4-} & 1.19 & 0.56 \\
\ce{ClO3-} & 1.21 & 0.63 \\
\ce{HClO2} & 1.65 & 0.78 \\
\ce{ClO2}  & 1.63 & 0.89 \\
\hline
\ce{}
\end{array}
$$

Why is $\ce{ClO-}$ such a strong oxidizing agent compared to other chloride oxyanions, despite its low oxidation number? Is oxidation number representative of oxidizing strength?

First, hypochlorite ($\ce{ClO−}$) is not a strong oxidizing agent compared to other chloride oxyanions, at least in acidic medium. Under standard conditions in acidic medium, chlorous acid ($\ce{HClO2}$) is the best oxidizing agent among them due to the largest positive number for standard electrode potential:

$$\begin{align}
\ce{HClO2 + 2H+ + 2 e- &<=> HClO + H2O} &\quad E^\circ &= \pu{1.645 V} \label{rxn:1}\tag{1}\\
\ce{HClO2 + 3 H+ + 3 e- &<=> 1/2 Cl2 + 2 H2O} &\quad E^\circ &= \pu{1.628 V} \label{rxn:2}\tag{2}\\
\ce{HClO2 + 3 H+ + 4 e- &<=>  Cl– + 2 H2O} &\quad E^\circ &= \pu{1.570 V} \label{rxn:3}\tag{3}\\
\end{align}$$

Keep in mind that in the reduction half-reaction \eqref{rxn:1}, $\ce{HClO2}$ reduced to $\ce{HClO}$ with its largest positive standard electrode potential $(E^\circ = \pu{1.645 V})$ among chlorine oxoacids. For comparison, we should look at half-reactions of which each of these two reagents reducing to the same product. Thus, look at half-reactions \eqref{rxn:2} versus \eqref{rxn:4}, and \eqref{rxn:3} versus \eqref{rxn:5} where $\ce{HClO2}$ shows its superiority over $\ce{HClO}$:

$$\begin{align}
\ce{HClO + H+ + e- &<=> 1/2 Cl2 + H2O} &\quad E^\circ &= \pu{1.611 V} \label{rxn:4}\tag{4}\\
\ce{HClO + H+ + 2 e- &<=>  Cl– + H2O} &\quad E^\circ &= \pu{1.482 V} \label{rxn:5}\tag{5}\\
\end{align}$$

Perchlorate ($\ce{ClO4-}$) and chlorate ($\ce{ClO3-}$) ions, on the other hand, are also good oxidizing agents, but not as strong as $\ce{HClO2}$ or $\ce{HClO}$, yet closed. This fact is clear when compared to the half-reactions where they directly reduce to $\ce{Cl2}$ or $\ce{Cl-}$ similar to $\ce{HClO2}$ and $\ce{HClO}$:

$$\begin{align}
\ce{ClO4- + 8H+ + 7e- &<=> 1/2 Cl2 + 4H2O} &\quad E^\circ &= \pu{1.39 V} \label{rxn:6}\tag{6}\\
\ce{ClO4- + 8H+ + 8e- &<=> Cl– + 4H2O} &\quad E^\circ &= \pu{1.389 V} \label{rxn:7}\tag{7}\\
\ce{ClO3- + 6H+ + 5e- &<=> 1/2 Cl2 + 3H2O} &\quad E^\circ &= \pu{1.47 V} \label{rxn:8}\tag{8}\\
\ce{ClO3- + 6H+ + 6e- &<=> Cl– + 3H2O} &\quad E^\circ &= \pu{1.451 V} \label{rxn:9}\tag{9}\\
\end{align}$$

However, the standard reduction potentials OP is mentioning is for different reduction half-reactions $\ce{ClO3-}$ and $\ce{ClO4-}$ involved (\eqref{rxn:12} and \eqref{rxn:10}, respectively):

$$\begin{align}
\ce{ClO4- + 2H+ + 2e- &<=> HClO3 + H2O} &\quad E^\circ &= \pu{1.189 V} \label{rxn:10}\tag{10}\\
\ce{ClO3- + 2H+ + e- &<=> ClO2 + H2O} &\quad E^\circ &= \pu{1.152 V} \label{rxn:11}\tag{11}\\
\ce{ClO3- + 3H+ + 2e- &<=>  HClO2 + H2O} &\quad E^\circ &= \pu{1.214 V} \label{rxn:12}\tag{12}\\
\end{align}$$

Accordingly, when consider Latimer diagrams $(\ce{ClO4- -> ClO3- -> HClO2 -> HClO -> Cl2 -> Cl-}),$ you'd see, $\ce{HClO2}$ cannot oxidizes $\ce{ClO3-},$ although it is the strongest oxidizing agent. However, $\ce{HClO2}$ can oxidize another species, which $\ce{ClO3-}$ cannot oxidize.

Note that all reduction potential values are from: http://depa.fquim.unam.mx/amyd/archivero/TablasdepotencialesREDOX_26700.pdf (Petr Vanẏsek), which cited the following references:

G. Milazzo, S. Caroli, V. K. Sharma, Tables of Standard Electrode Potentials; Wiley, Chichester, 1978.
A. J. Bard, R. Parsons, and J. Jordan, Standard Potentials in Aqueous Solutions; Marcel Dekker, New York, 1985 (Series: Monographs in electroanalytical chemistry and electrochemistry).
S. G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at $\pu{298.15 K}$,” J. Phys. Chem. Ref. Data 1989, 18(1), 1-21 (https://doi.org/10.1063/1.555839).


Is it possible to synthesis adenine, thymine, cytosine, or guanine at home? I'm wondering if there is any way to create adenine, thymine, cytosine, or guanine from chemicals available to anyone, using basic lab equipment.

Based on jonsca's comment, extract DNA from something. The protocol will involve detergent and isopropol alcohol. The details are available in several places and is fairly simple, but you won't get much DNA per onion, so be ready to prepare a lot of them. To do the degradation, there are several things you could try. If your detergent contains nucleases ( many contain enzymes but do a poor job of labeling which enzymes ) you could mix that in and react it at 37C for a while. Can also try high pH solutions, if you happen to have NaOH around. If you do the DNA extraction differently, you can get the RNA instead of DNA. RNA is very unstable at high pH, and RNAse enzymes are EVERYWHERE. So the RNA should degrade into nucleotides pretty readily if you just use sloppy lab technique. When I extract RNA in the lab, I use phenol/chloroform to pull the proteins and genomic DNA away from the RNA, which stays in the aqueous layer. You probably don't have access to those at home, but cooking oils may be hydrophobic enough to pull it off. Also, it helps to have a centrifuge.

Don't ask me how to purify the nucleotides from your solution and ask me even less about how to verify what you have at home. Verification would require something like a UV-Vis or HPLC or NMR. Good luck getting those at home.


Why is oxygen more stable than oxygen dication? According to molecular orbital theory (MOT), $\ce{O2^2+}$ has a greater bond order than $\ce{O2}$ and two less antibonding electrons. So why is molecular oxygen $\ce{O2}$ more stable than the molecular ion $\ce{O2^2+}?$
One possible reason that comes to mind is that the antibonding (AB) orbitals, although higher in energy than the constituent atomic orbitals, are still negative in energy, so adding electrons to AB orbitals still lowers the energy. But then, by that same argument, $\ce{O2-}$ would increase stability further, which it doesn't. Am I overlooking anything?
Note: I know that from an electrostatics point of view, $\ce{O2}$ is obviously more stable as bring two negative electrons close to positive $\ce{O2^2+}$ decreases the electric potential energy of the system. I am interested in how MOT explains relative stability (or if it actually even does).

You have to consider the system as a whole - you can't directly compare $\ce{O_2}$ and $\ce{O_2^2+}$ because they have different numbers of particles. To put it another way when you consider the relative stability of two interconvertible specifies you really have to write down a chemical reaction that connects them, and then consider which direction is thermodynamically favoured, and then consider whether it is kinetically feasible. Stability is a really tricky and more often than not misused word!
So here we have to look at $$
\ce{O_2 -> O_2^2+ +2e^-}
$$
As you note $\ce{O_2^2+}$ has a stronger bond than $\ce{O_2}$. But that is not all that is going on. We are also binding two electrons to the Oxygens - how does that affect the energetics? Well the reaction above is simply the sum of the following processes
$$
\ce{O_2 -> O_2^+ +e^-}
$$
$$
\ce{O_2^+ -> O_2^2+ +e^-}
$$
In other words the energy required is the sum of the first two ionisation energies of molecular oxygen. Do we know these? Well after a good 10 seconds looking I can't find the second energy, but the first is very well known, it is after all related to the discovery of compounds of Xenon and the start of noble gas chemistry. It is 12.0697 eV (from https://webbook.nist.gov/cgi/cbook.cgi?ID=C7782447&Mask=20) which in more familiar units is 1165 kJ/mol - a huge energy, bigger than any bond energy I am aware of, and note there is at least this much to come as the second ionisation energy will be even larger, as it is pulling a negative electron away from an already positively charge centre. Thus the energy gained by attaching the two electrons will massively out weigh any increase in bonding in the molecule.
Can we relate this to MO theory? Well as Koopman's theorem tells us the ionisation energy is an approximation of the orbital energy, this is telling us that relative to having a free electron it is more energetically favourable to put the electron in the lowest energy unoccupied orbital in the molecule. That this orbital is antibonding is irrelevant. Antibonding simply means it weakens any bonding in the molecule; it doesn't say anything about the energy of that orbital relative to the free electron. And you know this must be true - many molecules have electrons in antibonding orbitals and those electrons don't just rush away to infinity, hence occupying the orbital must be a lower energy proposition for the electron compared to being unbound. Thus $\ce{O_2}$ is thermodynamically more stable than $\ce{O_2^2+}$ and two free electrons because the decrease in energy caused by the electrons occupying the highest unoccupied molecular orbitals is (much) greater than any possible decrease in the bond energy.
That's the thermodynamics. Kinetics is difficult in general, but very unlikely to be an issue here. At this point I chicken out.


In which nucleophilic aromatic substitution rate is faster m-fluoronitrobenzene or p-fluoronitrobenzene? In which nucleophilic aromatic substitution rate is faster: m-fluoronitrobenzene or p-fluoronitrobenzene?

The rate of reaction depends on stability of intermediate. The given reaction follows $\ce{S_N2Ar}$ mechanism through "Meisenheimer intermediate". So, the stability of "Meisenheimer" primarily drives the reaction.

ortho and para isomer are almost same except their position, which is the main reason why is the former reacts much slowly than the latter.

In the ortho isomer, the negative charge of the ring isn't delocalised with $\ce{NO2}$, whereas in the para isomer, it is delocalised, as seen in the following image,

(Source: Master Organic Chemistry)


Is aluminium a metal or metalloid? Aluminium is along the dark line of the Periodic Table and it is $p$-block metal.

Is it metal or a metalloid?
Why?

Aluminium is usually considered to be a metal, as described in the Wikipedia article Metalloids: Aluminium:

  Aluminium is ordinarily classified as a metal. It is lustrous, malleable and ductile, and has high electrical and thermal conductivity. Like most metals it has a close-packed crystalline structure, and forms a cation in aqueous solution.

The reason why, in a few cases, aluminium is referred to as a metalloid, from the article:

  It has some properties that are unusual for a metal; taken together, these are sometimes used as a basis to classify aluminium as a metalloid. Its crystalline structure shows some evidence of directional bonding. Aluminium bonds covalently in most compounds. The oxide $\ce{Al2O3}$ is amphoteric, and a conditional glass-former. Aluminium can form anionic aluminates, such behaviour being considered nonmetallic in character

The line between metals and non-metals is somewhat arbitrary with

  Elements to the lower left of the line generally display increasing metallic behaviour; elements to the upper right display increasing nonmetallic behaviour.

Boron, carbon and silicon (all on the other side of the boundary line) are hard and brittle and are semiconductors, which is a major reason why the majority of resources list Aluminium as a metal, largely due to the metallic properties listed in the first quote.

A related question is Metal Compounds that bond covalently


What chemical could be used as carbon monoxide scrubber? I am trying to produce nitrogen by passing air over a heated tube of coals. This leaves a mixture of $\ce{CO},$ $\ce{H2O}$ and $\ce{N2}.$ Only $\ce{CO}$ is unwanted in this case.
Could I use calcium hydroxide to scrub carbon monoxide out of the gas? What other substances could be effective?
edit: this is part of challenge to make other chemicals from natural resources, therefore i have to obtain nitrogen myself instead of buying it

Air reacting with charcoal will produce $\ce{N2},$ $\ce{CO}$ and $\ce{CO2}.$ $\ce{CO2}$ will be absorbed by calcium hydroxide or sodium hydroxide.
However, $\ce{CO}$ cannot be absorbed at ordinary temperatures and pressures. Only at $\pu{200 °C}$ and $\pu{10 atm},$ it reacts with $\ce{NaOH}$ to produce sodium formate $\ce{HCOONa}.$
Finally, $\ce{CO}$ can be oxidized into $\ce{CO2}$ by passing over hot $\ce{MnO2}$ or $\ce{Fe2O3},$ and then be absorbed by a hydroxide.


Discrepancy when calulating mol weights with ChemSketch and Python RDKit I am an IT guy with little knowledge of Chemistry, so please bear with me...
For context: I am working with a system that has a lot of structures, their .mol files and some metadata (like mol weights) in a database. Due to an error, some of the metadata was lost and needs to be recalculated. Since there are a lot of structures, I need to automate this.
Originally, the mol weights were calculated using the Freeware version of ACD ChemSketch. Since I did not find a way to call ChemSketch and have it return the mol weight or otherwise automate ChemSketch itself, I looked at RDKit for Python as an alternative.
In my tests, I noticed that ChemSketch and RDKit produced slightly different results when calculating the mol weights. I used m = Chem.MolFromMolFile('amisalin.mol') to load the file and then MolWt(m) respectively ExactMolWt(m).
Example mol file amisalin.mol:

  -ACD/LAB-

 18 17  0  0  0  0  0  0  0  0  0 V2000
   18.3905   -6.6892    0.0000 C   0  0  0  0  0
   19.4429   -6.0748    0.0000 C   0  0  0  0  0
   18.3905   -8.0661    0.0000 C   0  0  0  0  0
   17.2059   -5.9960    0.0000 C   0  0  0  0  0
   20.5930   -6.7427    0.0000 N   0  0  0  0  0
   19.4429   -4.2757    0.0000 O   0  0  0  0  0
   17.2059   -8.7560    0.0000 C   0  0  0  0  0
   16.0211   -6.6892    0.0000 C   0  0  0  0  0
   21.7398   -6.0748    0.0000 C   0  0  0  0  0
   16.0211   -8.0661    0.0000 C   0  0  0  0  0
   22.5401   -6.5380    0.0000 C   0  0  0  0  0
   14.8742   -8.7309    0.0000 N   0  0  0  0  0
   23.6965   -5.8762    0.0000 N   0  0  0  0  0
   24.8434   -6.5380    0.0000 C   0  0  0  0  0
   23.6965   -4.0709    0.0000 C   0  0  0  0  0
   25.9934   -5.8762    0.0000 C   0  0  0  0  0
   24.8434   -3.4124    0.0000 C   0  0  0  0  0
   21.0845  -12.6663    0.0000 Cl  0  0  0  0  0
  1  2  1  0  2  0  0
  1  3  2  0  1  0  0
  1  4  1  0  1  0  0
  2  5  1  0  2  0  0
  2  6  2  0  2  0  0
  3  7  1  0  1  0  0
  4  8  2  0  1  0  0
  5  9  1  0  2  0  0
  7 10  2  0  1  0  0
  9 11  1  0  2  0  0
 10 12  1  0  2  0  0
 11 13  1  0  2  0  0
 13 14  1  0  2  0  0
 13 15  1  0  2  0  0
 14 16  1  0  2  0  0
 15 17  1  0  2  0  0
  8 10  1  0  1  0  0
M  END

Results:

Average molar weight
Exact molar weight

Chemsketch
271.78628
271.14514

RDKit
271.7919999999999
271.145140004

As you can see, the results are slightly different. Since the input data is identical, I assume this is due to some internal differences when calculating the weights, possibly due to rounding (?).
My question now is: Is the difference negligible? Does anyone know where it comes from? Is one of the results "correct" / more accurate than the other?

In first place, you must clarify with the users of the database the purpose of the entries about molecular weight because average molecular weight and exact molecular weight differ in their meaning and subsequent use.  It may be that the documentation of the database provides this information.  Later, once this target is set, use either one of the program to compute the entries for all compounds of this database to ensure consistency in the data.  RDKit's current release 2021_03_1 is just some weeks old and probably the easier route to cover eventually all entries in your database.

Molecules consist of atoms (like oxygen, nitrogen, etc).  For some of them, say nitrogen, despite they are all nitrogen, there might be more than one form and these differ slightly in their mass per individual atom (isotopes).

The mass of these isotopes is published.  Equally, biannually, there are updated public records about the probability that an atom of e.g., nitrogen either is $\ce{^{14}N}$ (about 99.6%) or $\ce{^{15}N}$ (about 0.4%).

These probabilities are equally valid for molecules, if you have 1000 molecules each containing one atom of nitrogen, than 4 of them should contain the heavier isotope $\ce{^{15}N}$.  Similar there are such probabilities about other atoms, like carbon, oxygen, etc., which are all independent of each other.  Among carbon atoms, for example, 98.9% are $\ce{^{12}C}$, and 1.1% are $\ce{^{13}C}$, respectively.  (Except in radiochemistry / nuclear chemistry.)

Exact molecular weight is used, e.g. in databases eventually used for e.g., mass spectroscopy.  An analysis of your sample will recognize these different probabilities and may yield different, separate signals about the molecules with $\ce{^{14}N}$ from the ones about $\ce{^{15}N}$.  Thus exact molecular weight is about molecules of one isotopic composition.
Say our sample were $\ce{(CN)2}$ of two carbon atoms and two nitrogen atoms per molecule, than there would be four mass signals for all combinations of the two isotopes of carbon and nitrogen possible.  Their intensity  correlate with these probabilities.  Exact molecular weight is a monoisotopic mass, typically to consider only the mass of the molecules consisting with the most frequently observed isotopes per element, i.e., only about $\ce{^{12}C}$ and $\ce{^{14}N}$.  Depending on the intended analysis, it may be worth to retain six (or even more) decimals.

Average molecular weight now accounts not only for the molecules consisting of the most frequently seen isotopes per element, but for any combination of the elements with each other.  These weights are are averaged by the probability that these combinations occur, e.g that there is a molecule $\ce{(^{13}C^{14}N)2}$ consisting only of the heavier carbon isotope and the lighter nitrogen isotope among all $\ce{(CN)2}$ molecules of your sample.  Average molecular weights are typically used in to set up a reaction in the lab, to know how many grams of compound A shall react with how many grams of compound B.  Depending on the scale of the reaction and the needs of your clients, to retain two decimals of this value may suffice.

The pace these IUPAC publishes the data about these isotopes is independent from the one the programs are updated by the developers, and later by the users. This contributes why an old program still may include outdated values.


How to remove stain from pavement after having dropped sulfuric acid on it? I had this car battery that flipped in my car trunk.
When I got it out of there it spilled some acid on the pavement. It's a sulfuric acid battery. I added some baking soda on it and poured some more water+baking soda to make sure the acid was neutralized.

However there is still a brown-rusty stain on the pavement. I tried water and a big broom brush but it still there.

Any idea what I can use to clean up the pavement to its original color?

You did the right thing by neutralizing with a base.

The best solution is to let time heal the stain. The surface is obviously damaged, but there is no way to reverse that damage. The stain will only go away if someone keeps on scrubbing the area until a new surface is exposed. The good news is that dirt, soil, and water will slowly remove and even scrub away the (temporary) surface stain. Maybe keep a flower pot covering the area for the time being.


How do you determine steric hindrance? Steric hindrance is important for understanding regio-selectivity and stable conformation of molecules. I know the general definition, but can somebody explain to me in detail how does phenomenon of steric hindrance come about?

Like, to give a simple, hypothetical example, $\ce{CH4}$ is said to have no steric hindrance. But why does attacking molecule have no trouble squeezing past $\ce{H}$ atoms and attack center carbon atom. I know hydrogen atom is said to be small, but seriously how small is it?

I am asking this because when examples go beyond $\ce{H}$ atom and alkyl atoms I can't judge whether steric hindrance is present or not and thus what products are formed. Like in $\ce{(CH3)3CHOH^+}$, is there steric hindrance. It is a primary carbon cation but instead of two hydrogen then is $\ce{OH^-}$ group?

Normally we don't discuss quantum mechanics in this context, but an answer to your question requires it.

Taken from:

http://chemistry.umeche.maine.edu/CHY252/HOMO-LUMO.html

What we see here is the LUMO for CH3Cl.  This is where an incoming nucleophile will 'stick' its electrons to form the new bond.  Since the orbital is antibonding, electrons in the orbital will destabilize a bond, in this case the C-Cl bond.  The actual meaning of the size of this orbital, rigorously, is a little unclear.  However you can roughly infer that if a nuclephile gets its electrons here, a reaction is 'likely'.  The reddish region to the left encompasses the hydrogens.  That is also where the nucleophile has to go.

Compare the next from:

http://chemistry.umeche.maine.edu/CHY252/EtCl.jpg

This is for ethyl chloride.  There is an extra minor node on one of the hydrogens.  Because it is small, it is less important than the other regions.  Notice the methyl group is partly obscuring the backside of the orbital lobe near the electrophilic carbon.  There are less ways a nucleophile can approach this to cause a reaction.

I'm no longer practicing so I do not have software to produce analagous images, so you will have to imagine the extrapolation.  1-chloroopropane would not add much more to the steric hindrance, because the 'extra' methyl group would be dangling on the same of the molecule as the methylene group.  However, if we had two methyl groups, such as 2-chloropropane, the hinderance would be much more.  Three groups, as in t-butyl chloride?  Give it up, to much hinderance for a backside attack reaction.


Byproducts in Indigo Synthesis? I have been messing around with making Indigo dye while waiting for bacteria to grow. I have been following the Baeyer-Drewson method. Basically, you mix nitrobenzaldehyde with an acetone water mix and slowly add NaOH. The solution turns black and you get a dark sludge. I have been separating the sludge from the liquid by centrifugation. I have noticed that the supernatent has a dark reddish brown color.

I was wondering if anyone knew what the red byproduct was?
My indigo is black when dried, wikipedia shows nice bright blue chunks. What could be causing this color difference? Maybe the red byproduct?
Does indigo dissolve in anything without needing to be reduced first? I have had no luck with the organic solvents available in our lab.

After looking further into indigo production, I came across a related structure called indirubin.

Indirubin:

Indigo:

Indirubin is a byproduct of indigo synthesis and has a reddish color. It may also be more soluble than indigo, and indirubin derivatives have some anticancer activity. If you trust this supplier, you can buy indirubin for pretty cheap to compare with your lab's indigo reactions.

I also found a study done on old native american dyes that studied indigo and indirubin in fabric fibers, among other dyes. They provide a UV-VIS spectrum of both compounds, with Ind = indigo and Inr = indirubin. We can see that indigo has an absorption peak in the red region, so blue is reflected. Indirubin has a peak at around 540 nm, which is yellow-green, allowing some blue and red to reflect, but probably more red than blue. This paper also references a 1910 paper** studying indirubin levels in commercially available indigo samples, but I can't find that article. They suggest that higher levels of indirubin are produced when more indoxyl is converted to isatin. If you can find this 1910 article, it may provide some insight into reducing isatin and indirubin levels.

** William Popplewell Bloxam and A. G. Perkin, J. Chem. Soc., Trans., 1910, 97, 1460-1475.


Why does Ga improve conductivity of Ge? This is for my AP Chemistry course and the question is:

Which of the following elements is important in the semiconductor industry to improve the conductivity of Ge?

K
Ga
Al
Mg

It seems like any of the above could be a correct answer. The answer given was Ga. What about gallium would make it the most appropriate answer?

This questions has two parts if you look closely:

  Which of the following elements is important in the semiconductor industry to improve the conductivity of Ge?

All of the above will improve the conductivity of $\ce{Ge}$.  This is because germanium is a semiconductor which conducts electricity via the movement of holes and electrons.  Since germanium is tetravalent and gallium is trivalent, doping germanium with gallium creates a p-type semiconductor which is exponentially rich in holes, thus increasing the conductivity. This mechanism holds true for aluminum, potassium and magnesium meaning all of them can increase the conductivity of germanium

  Which of the following elements is important in the semiconductor industry to improve the conductivity of Ge?

This part is what makes gallium special.  Gallium and germanium are sequential elements in atomic number.  They have very similar atomic radii.  Gallium is also very much less reactive than the other elements mentioned. These characteristics allow gallium to dope the germanium to a higher concentration without forming a new  undesired solid phase.  Sodium and magnesium form germanides with germanium and will not dope as shown by the phase diagram.  Aluminum will dope, but its smaller atomic radius can alter and strain the crystal structure of germanium much more than gallium which can affect band gap.