text:
How can I neutralise H2 gas without burning it?
I was thinking of maybe making it bubble into a solution containing something that would react with H2 yielding either a dissolved substance or a precipitate from the reaction.

Maybe a solution with a small amount of H2O2 in it?

What do you mean by "neutralise"? In chemistry this usually suggests an acid-base reaction, or decomposition of dangerous/hazardous substances.

@andselisk It is to avoid the dangerous accumulation of H2 gas (produced as a byproduct of a reaction) into the room where I carry out the reaction. I'd like to avoid explosion/fire hazards.

If it would react, it could as well react dangerously  If you are not going to burn it, you may to mix it with excess of air or nitrogen, keeping hydrogen safely below explosive concentration.

H2O2 won't help at all.

Thanks for your answers. If it's going to necessitate complex catalytic devices I might just as well extract it from the building in a safe way (for instance, as Poutnik says, diluting it before extraction). I was just thinking maybe there's a simple reaction which makes I can transform it simply bubbling it through a liquid and therewith incorporating hydrogen atoms back into a safe molecule (like water, except without burning it not to risk detonation).

Just out of curiosity (and to try to learn more about chemistry), why is it that H2O2 + H2 doesn't yield 2 H20? Would it work if my H2O2 solution was hot enough (all the while keeping it sufficiently diluted to avoid problems with it)? Does H2O2 non-catalytically decompose into H2O and O2? and if so would heating the solution to the required temperature to make it react with H2 significantly speed up that decomposition? (I can see the risks involved with releasing concentrated O2 around H2)

Could H2O2 + H2 ⤑ 2 H2O at a temperature at which H2O2 would not release O2 if I used a catalyst in the solution, as say ionised iron or copper? (wild guesses knowing they catalyse some oxidation reactions)

Or maybe it would then also catalyse the release of Oxygen? If so, would there then maybe a catalyst that would strongly favour the reaction between H2 and H2O2 vs the decomposition of H2O2?

You really need to contact your local ES&H/Safety person and follow their instructions. If you have no such person, stop and do not proceed.

@Veritas Jon was merely trying to help.  No one is doubting your education level or accomplishments, but performing something potentially dangerous without knowing exactly what you are doing could turn out to be a recipe for disaster.

@jonsca Knowledge and knowledge only can help people ignorant on a topic, not suggesting they reach out to luxury resources (given the proportion of people who have access to an ES&H/Safety person the comment was outright delusional if not advocating for the exclusive practice of chemistry to the tiny group of people who do, which would severely impair technological development & innovation/evolution). You can give people knowledge and warn them at the same time, that's perfectly reasonable, but telling them to "stop and not proceed" without giving them any knowledge is all but helping others.

Many nuclear power plants use so-called "passive autocatalytic recombiners" for the catalytic oxidization of hydrogen that could be released into the containment in the event of severe accidents. (Note that the affected units of the Fukushima-Daiichi plant did not have such equipment.) The catalytically active materials, typically platinum and/or palladium, are doted on carriers, preferably designed as catalyst plates orientated vertically to support the flow.


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How to get the efficiency of a heat engine which undergoes an elliptical cycle?

An ideal diatomic gas undergoes an elliptic cyclic process characterized by the following points in a $PV$ diagram:
$$(3/2P_1, V1)$$
$$(2P_1, (V1+V2)/2)$$
$$(3/2P_1, V2)$$
$$(P_1, (V1+V2)/2)$$

A rough sketch:

This system is used as a heat engine (converting the added heat into mechanical work).
Evaluate the efficiency of this engine

We know that the efficiency is defined as the benefit/cost ratio:
$$e = \frac{W}{Q_h}$$
Let's focus first on the work done by the engine; taking into account the quasistatic approximation, $W=PV$. Then:
$$W = (P_2 - P_1)(V_2 - V_1)$$
Note that from the given points we can guess that $P_2 = 2P_1$. Then:
$$W = P_1(V_2 - V_1)$$
Now let's focus on $Q_h$. I have the following issue here: none of the 4 steps of the cycle has either P or V constants. This means that the strategy of using:
$$Q = nc\Delta T$$
Won't work because you cannot use neither $c_p$ nor $c_p$.
However, when we deal with a rectangular cycle:

It would be really easy to derive expressions for both $Q_a$ and $Q_b$ and then the efficiency for the system would be obtained. That is because in each step $Q_h$ is added, either $P$ or $V$ are constant (and thus $Q = nc\Delta T$ works).
What to do with the elliptical cycle to get $Q_h$?
EDIT
My bad, the work done by the working substance is the area under the PV graph. So as Chet Miller pointed out, the work is:
$$W = \pi (P_2 - P_1)(V_2 - V_1)$$
I have been trying to solve the heat equation so that we get the two angles.
So what I did was deriving $P$ and $V$ wrt the angle:
$$dP = (P_{max} -P_0)\cos \theta d\theta$$
$$dV = (V_{max} -V_0)\sin \theta d\theta$$
And plugging it into 3:
$$0=[-(C_v+R)P(V_{max} -V_0)\sin \theta+C_VV(P_{max} -P_0)\cos \theta]$$
This above equation is satisfied by two $\theta$ angles. But how to solve it?

Your sign on dV is incorrect.  And, please include the d theta's.  You also need to substitute for P and V.  See my comment after my answer.

Your equation above for $W$ is still wrong because you are using the full major and minor axes, not the semi-major and semi-minor axes. It's like saying the area of a circle is $A=\pi D^2$ instead of $A=\pi r^2$

@user5713492 I don't see what you mean. The area of an ellipse is $A = \pi a b$

Look at your expression for the area of a rectangle. Then look at your expression for the area of an ellipse contained completely within that rectangle. Which is bigger? Which should be bigger?

@user5713492 I see what you mean now. Btw thank you very much for your answer at MSE; for anyone interested: https://math.stackexchange.com/questions/3262034/solving-an-equation-for-angle-theta/3262825#3262825

Let $(V_0,P_0)$ represent the coordinates of the center of the ellipse, and let $P_\mathrm{max}$ and $V_\mathrm{max}$ represent the maximum pressure and maximum volume, respectively over the cycle.  Then, the work is the area of the ellipse, which is given by:  $$W=\pi(P_\mathrm{max}-P_0)(V_\mathrm{max}-V_0)$$This is, $\pi$ times the product of the semi-major and semi-minor axes.  This differs from the result which you gave.
The shape of the ellipse can be represented parametrically in terms of the angle $\theta$ around the cycle, assuming $\theta$ is measured clockwise from the point $(V_0-(V_0-V_\mathrm{max}),P_0)$:
$$P=P_0+(P_\mathrm{max}-P_0)\sin{\theta}\tag{1a}$$
$$V=V_0-(V_\mathrm{max}-V_0)\cos{\theta}\tag{1b}$$
Application of the first law of thermodynamics to the working fluid over a differential portion of the cycle gives us:
$$dU=nC_vdT=dQ-PdV\tag{2}$$But from the ideal gas law, $$nRdT=d(PV)$$Substitution of this into Eqn. 2 yields:
$$\frac{C_v}{R}d(PV)=dQ-PdV$$
So the differential heat added during an arbitrary portion of the cycle is given by:$$dQ=\frac{1}{R}[(C_v+R)PdV+C_VVdP]\tag{3}$$So the differential heat is zero when dQ = 0, or, equivalently, when $d\ln{P}+ d\ln{V^{\gamma}}=0$, or, equivalently when $PV^{\gamma}=\mathrm{Const}$.  This is the equation for an isentropic line tangent to the ellipse.
There are two angles $\theta$ at which adiabats are tangent to the ellipse.  These two angles can be obtained by substituting Eqns. 1 into Eqn. 3 with dQ = 0.  Once these angles have been determined, the total positive heat added Q during the cycle can then be obtained by integrating Eqn. 3 with respect to $\theta$ between the two angles.

Nice answer,To my prior comment, I forgot it is not so simple for general p-V diagram. :-)

Thanks for the answer. Why are you using $c_v$ in $dU =n C_v dT=dQ-PdV$? Volume is not held constant in any step of the cycle.

You are aware that the internal energy U of an ideal gas is a function only of temperature, and is equal to nCvdT irrespective of whether the volume is constant, correct?

@ChetMiller True, thanks. I will ask more questions when I study the answer better.

@ChetMiller I have been trying to solve the equation to get the angles $\theta$. Please see my edit. I know this is more Math; I have also asked it in MSE but got no answer yet: https://math.stackexchange.com/questions/3262034/solving-an-equation-for-angle-theta

You also need to substitute for P and V.  It then reduces to $$\gamma \frac{P_0}{(P_{max}-P_0)}\sin{\theta}+\frac{V_0}{(V_{max}-V_0)}\cos{\theta}+\gamma \sin^2{\theta}-\cos^2{\theta}=0$$I don't think that this has an analytic solution.

Definitely has an analytic solution: let $z=\tan\left(\theta/2\right)$ and the above reduces to a quartic equation in $z$. The general quartic equation has a solution, not necessarily a pretty solution, but a solution.

@ChetMiller what if were were to set $P_1 = 1$ and $V_2 = 3 V_1$? Like that the final answer would only depend on $P_1$ and $V_1$... Isn't there a way to approach this problem with a simpler method? I have been thinking about how I could make an analogy with the same problem but with a rectangular shape; that would be great.

Well, @user5713492 indicates a method for solving the equation analytically.  I haven't tried this yet, but, if so, you can get the cosines and sines of the two angles, and then integrate to get the heat analytically.  Don't bother with the rectangular shape.  In my judgment, that will lead to nothing.

We already know that $P_0=\frac{3}{2}P_1$l and $P_{max}=2P_1$, so $$\frac{P_0}{(P_{max}-P_0)}=3$$If you are now saying that $V_2=3V_1$, then $V_0=2V_1$ and $$\frac{V_0}{(V_{max}-V_0)}=2$$But I don't think that's going to lead to much simplification.

@ChetMiller: I have a program that does all this with numerical rather than analytical solution in the companion thread.

Rather than continuing to speculate about how to solve the equation analytically, within 15 minutes, I solved the equation numerically (by hand) for the specific case of the parameter values specified in my previous post, and obtained for the two angles (nearly exactly) -15 degrees and +150 degrees (or $\pi/12$ radians and $5\pi /6$ radians).  This gives you enough information now to get the engine efficiency for this case.

@user5713492 You go boy!!!  Nice!!!!  Dazzling job!!!  Do you confirm my results for the two angles?

I got $\theta_1=195.0948°$, $\theta_2=30.2732°$ from my Matlab program, so that agrees with your results because $\theta_{\text{mine}}+\theta_{\text{yours}}=180°$ since you are measuring clockwise from the left end and I am measuring counterclockwise from the right end.

@ChetMiller Maybe I am asking too much but could you please add how you solved the equation numerically (by hand) for the specific case of the parameter values specified in my previous post, getting -15 degrees and +150 degrees?

Just looking at the equation, you can tell that there is a root between -90 and 0, with the function being negative at -90 and positive at 0.  So I evaluated the function at -45, and it came out negative.  So I knew the root was between -45 and 0.  So I tried -15, and it satisfied it pretty  closely.  I did basically the same thing for the interval between +90 and +180.

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Is there any formula for finding out probable distance of orbitals or do I just remember the values?
Like, do I need to just remember that it is less in 3d than 3s and so on?

Welcome to ChemSE.  Please add some further details to your question which, at present, is very broad (Cf. https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#Example:_3D_isotropic_harmonic_oscillator).

I'm voting to close this question as off-topic because your question lacks context. This question needs revision before it is ready for a great answer. Please edit to include how this question came up and how you tried to answer it. This will help writing an answer that is useful for you and for others.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Epoxidation with peroxyacids
In this particular epoxidation/hydroxylation reaction, why does the epoxide only form at one of the double bonds and not both?
The compound initially reacts with the peroxyacid to form an epoxide, follow by an acid catalyzed epoxide opening resulting in syn-stereochemistry. My initial guess was that the epoxide would only form by the double bond that is not sterically hindered. I don't understand why the epoxide only forms at the double bond flanked by two methyl groups and not both.

It might help to consider which reactant is the electrophile and which is the nucleophile...

This source states that MCPBA gives preferential epoxidation of the more substituted alkene. This source explains that due to intra-molecular hydrogen bonding in solution, the high degree of polarisation results in an electrophilic oxygen atom that adds to alkenes. The most substituted alkene will be the most electron-rich alkene and hence the most nucleophilic. This is supported by the data shown here and here 4 on the relative rate of reaction of unsubstituted/mono/di/tri/tetra subsituted alkenes.

No working for me either. I have replaced it with a different reference


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Aldol Reaction on an intermediate
I was given the following reaction to prove a mechanism for:

and I managed to prove the first step with ease.

The second step, forming the "base-catalysed aldol" is the part that's rather confusing me.
Consider the mechanism of a base-catalysed aldol reaction below.

For this reaction to take place, I need a second carbonyl molecule, but the $\ce{CO2Me}$ on the right hand side ring appears untouched, and likewise so does the carbonyl group on the left hand side too.
I tried the reaction a second way (giving the mechanism below) but am unsure if this is correct.
Is this the correct mechanism?

Looks mostly reasonable to me. There’s an alkene missing from your final product though. Also, in the final elimination you’ve drawn an E2 reaction, but OH typically isn’t a good enough leaving group to do this. Can you think of an alternative elimination pathway?


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Significant figures of the product
How is
$$\frac{8.925 − 8.904}{8.925} × 100 = 0.24?$$
When I solved I got
$$\frac{0.021}{8.925} × 100$$
Now it will have to be $2$ significant numbers $(0.021)$ time $1$ significant number $(100)$ and the multiplication rule tells us that the answer will be of $1$ significant number, right?
But how come we ended up with $0.24?$

I'm voting to close this question as off-topic because your question lacks context: This question needs revision before it is ready for a great answer. Please edit to include how this question came up and how you tried to answer it. This will help writing an answer that is useful for you and for others.

In fact, the constant 100 in percentage calculation context has infinite number of significant digits, as 0 is significant digit as well, so 100=100.0000000000000....

According to Columbia University Website, Rules for Significant Figures are as follows:

All non-zero numbers ARE significant: The number 33.2 has three significant figures because all of the digits present are non-zero.

Zeros between two non-zero digits ARE significant: The number 2051 has four significant figures. The zero is between two non-zero digits, 2 and 5.

Leading zeros are NOT significant: They're nothing more than "place holders." The number 0.54 has only two significant figures while 0.0032 also has two significant figures. All of the zeros are leading.

Trailing zeros to the right of the decimal ARE significant: For example, there are four significant figures in 92.00. Keep in mind that 92.00 is different from 92: A scientist who measures $\pu{92.00 mL}$ knows his value to the nearest 1/100th milliliter; meanwhile his colleague who measured $\pu{92 mL}$ only knows his value to the nearest $\pu{1 mL}$. It's important to understand that "zero" does not mean "nothing." Zero denotes actual information, just like any other number. You cannot tag on zeros that aren't certain to belong there.

Trailing zeros in a whole number with the decimal shown are significant: Placing a decimal at the end of a number is usually not done. By convention, however, this decimal indicates a significant zero. For example, "540." indicates that the trailing zero is significant; there are three significant figures in this value.

Trailing zeros in a whole number with no decimal shown are not significant: Writing just "540" indicates that the trailing zero is not significant, and therefore, there are only two significant figures in this value.

Exact numbers have an infinite number of significant figures: This rule applies to numbers that are definitions. For example, $\pu{1 meter} = \pu{1.00 meters} = \pu{1.0000 meters} = \pu{1.0000000000000000000 meters}$, etc.

For a number in scientific notation: Consider a number $N \times 10^x$. All digits comprising $N$ are significant by the first 6 rules; "$10$" and "$x$" are not significant. For example, $5.02 \times 10^4$ has three significant figures: "$5.02$." "$10$" and "$4$" are not significant.

Rule 8 provides the opportunity to change the number of significant figures in a value by manipulating its form. For example, let's try writing 1100 with three significant figures. By rule 6, 1100 has two significant figures; its two trailing zeros are not significant. If we add a decimal to the end, we have 1100., with four significant figures (by rule 5). But by writing it in scientific notation: $1.10 \times 10^3$, we create a three-significant-figure value.

By rule 7, $100$ in your answer is exact number, hence carries infinite significant figures. Thus, your answer should have been $0.24$, with two significant figures (I assume you are familiar with rounding rules).
Now, let's apply Rule 8 to your answer:
$$\frac{0.021}{8.925} \times 100 = \frac{0.021}{8.925} \times 10^2 = 0.0024 \times 10^2 = 2.4 \times 10^{-1}$$
This final answer is a number in scientific notation, and hence has two significant figures.

This is the chemistry part of stackexchange but I can attempt to help you with your question regardless.
Your question was
$$\frac{8.925-8.904}{8.925} \times 100$$
which can be simplified to
$$\frac{0.025}{8.925} \times 100$$
$$ = 0.00235294117... \times 100 $$
$$ = 0.235...$$
which rounds up to $0.24$.
Multiplying by $100$ moves the decimal point two places to the right, one for each $0$ in $100$.
This is a 2 sig.fig answer as the first zero does not count towards significant figures. They only count from your first non-zero digit from left to right.
I hope that helps!
(This is my first answer by the way too!)

why is it a 2 sig fig answer and not 1? why did we ignore that 100 is a 1 sig fig number

is it because its an exact number so we ignore it?

You can't answer that question without more context.

The equation looks like a formula to calculate a percentage. If so, then yes, the 100 is an "exact number."


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How long would it take for sucrose to undergo hydrolysis in boiling water?
I was reading the book Cocktail Codex and there was a snippet on avoiding boiling sugar and water to make simple syrup that made me a bit skeptical. Here it is:

Heat also affects the molecular structure of sugar. For example, if sucrose, the disaccharide commonly known as white table sugar, undergoes an extended period of boiling, it will eventually convert to the monosacchariedes glucose and fructose, which taste sweeter (and cause nasty hangovers).

Does 100 °C water hasten hydrolysis, and if so, how long would you need to boil for the breakdown to occur?
(the hangover bit is a bit weird too considering your body breaks down sucrose into glucose+fructose?)

Boiling at a hundred+ degrees (syrup has a high sugar content), sugar will slowly caramelise, i.e. degrade towards carbon. I wouldn't bet that it hydrolyses first. The hangover bit is indeed weird. I think the warning in that book is to avoid that caramel note in the syrup.

How extended of an extended period of boiling did Cocktail Codex mean?

The hangover bits may come from the fact that sweet drinks make you drink them more.

This is a nice well-defined question, and luckily there is excellent data for which we can provide a quantitative answer.
Richard Wolfenden's research group has sought for many years to characterize the spontaneous (i.e. not enzyme catalyzed) rate of many enzymatic reactions.  In general this is so that the spontaneous rate can be compared to the enzyme-catalyzed rate, so that the catalytic proficiency of enzymes can be calculated.
All that's just prefatory remarks to introduce this paper, "Rates of Spontaneous Cleavage of Glucose, Fructose, Sucrose, and Trehalose in Water, and the Catalytic Proficiencies of Invertase and Trehalase".
The abstract's first sentence is:
The half-lives for spontaneous hydrolysis of trehalose and sucrose at 25 °C are 6.6 × 106 years and 440 years.
Thus, a solution of sucrose in water should be 50% decomposed into monomers after 440 years -- at 25 °C.  That value comes from thermodynamic extrapolation of experiments done at higher temperatures.  (I don't think they wanted to wait hundreds of years to do an experiment at 25 °C.)
But higher temperatures is exactly what you're after.  Figure 1 from the paper has the data.

The lower-right datapoint is very close to 100 °C, and shows the first-order rate constant for sucrose hydrolysis is about $10^{-6}$ per second.  That means that 63% degradation would take $10^{6}$ seconds, or about 11.5 days.  Getting 99% degradation would take about 53 days.
edits
The above has been edited to fix sloppy math errors, as noted by @gremin and @John Bentin in the comments.  Thank you both.

One complicating factor perhaps worth mentioning is that the hydrolysis can be catalyze by acid, so the half-life can be reduced substantially if your water has any acid in it.

Thanks for your answer! I'm curious, how are you calculating the seconds required for a given percentage of degradation?

It's just exponential kinetics.  $\frac{dS}{dt} = -kS$, so $S = S_0 e^{-kt}$.  If we wait $\frac{1}{k}$ time units, then we plug in $t = \frac{1}{k}$ to the formula, and get $S = S_0 e^{-1}$.  $e^{-1} = \frac{1}{e} = \frac{1}{2.71828...}=0.368$.  That's how much is left, so $1-0.368=0.63=63%$ is left.  Actually $1 - \frac{1}{e}$ if you want to be precise.

Half-life isn't the same as lifetime ($1/k$ in the kinetic equation). $T_{1/2} = k \times ln 2$. Shouldn't it be 50% in this sentence "Thus, a solution of sucrose in water should be 63% decomposed into monomers after 440 years -- at 25 °C"? The numbers from the graph are ok.

I don't see you final statement (in bold). If it takes 11.5 days to reach a sucrose-residue ratio of  e^{-1}, then (neglecting back-reaction) 34.5 days would take it to e^{-3}, or about 5%. To reach 1% would require 11.5 ln 100 or about 53 days.

Thanks to both Gremlin and John Bentin for the corrections.  I'm editing to fix these sloppy math errors.


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Kolmogorov backward equations in active biological systems
I am trying to understand the following equations from a paper by Wang et al. [1,  p. 5 at the SI]:

Michaelis–Menten Representation of the Kinesin Cycle
We assumed that there is a strong coupling between ATP turnover rate of kinesin and kinesin stepping on an MT. Each chemical step in the chemomechanical cycle of kinesin was assumed to be irreversible, except ATP binding and releasing (Fig. 1, step i). Consider the following scheme:
$$\large\ce{0 <->[$k_\mathrm{ATP}\lbrack\ce{S}\rbrack$][$k_\mathrm{-ATP}$] 1 ->[$k_\mathrm{s}$] 2 ->[$k_\mathrm{attach}$] 3 ->[$k_\mathrm{ADP}$] 4 ->[$k_\mathrm{h}$] 5 ->[$k_\mathrm{-P}$] 6}$$
Assume that $F_i(t)$ is the probability for the system to reach state $6$ if, at time $t = 0$, the system is in state $i.$ This function is governed by the backward master equations:
$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)\label{eqn:1}\tag{S1}$$

Why does equation \eqref{eqn:1} read as
$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)$$
and not
$$\frac{\mathrm{d}F_0(t)}{\mathrm{d}t} = k_\mathrm{-ATP}[\ce{S}]F_1(t) - k_\mathrm{ATP}[\ce{S}]F_0(t)?$$
I would like to change/modulate $S(t)=\cos(t)$ in time and see how it changes the states $F_i(t)$
References
1 Wang, Q.; Diehl, M. R.; Jana, B.; Cheung, M. S.; Kolomeisky, A. B.; Onuchic, J. N. Molecular Origin of the Weak Susceptibility of Kinesin Velocity to Loads and Its Relation to the Collective Behavior of Kinesins. PNAS 2017, 114 (41), E8611–E8617. https://doi.org/10/gdj4h5.

I'm still not satisfied that my answer gives a clear explanation, so I changed the title of the question in the hopes of attracting attention from someone more expert. This really isn't about Michaelis-Menten kinetics in the traditional sense.

@Andrew I started a bounty. I hope it will help us to get a canonical answer.

@Andrew Here is a paper that explains forward Master equations (very similar to chemical kinetics) and backward Master equations (used for first passage problems, and looking unfamiliar: https://arxiv.org/pdf/1503.00291.pdf

UPDATED: I wrote a first answer assuming that $t>0$ which got close to what's in the paper, but not quite the same. Thanks go to Karsten Theis for pointing out that $t<0$ in these "backward equations". Here's a corrected explanation:
The scenario is that we have a system that can be in any of seven consecutive states (numbered 0 through 6). At some time $t<0$, we want to know something about the probability of a system in each state reaching state 6 by time $t=0$, which is expressed as $F_i(t)$ where $i$ indicates the state number.
We start with a system in state 6 at time $t$. Since reversal back to state 5 is not permitted and there is no state beyond state 6, any system that is in state 6 will remain in state 6 up to (and past) time $t=0$. Thus the probability $F_6(t)=1$. In the paper, they use a Dirac $\delta$ function to indicate this, because that is necessary when $F_6(t)$ is used in the other equations.
For the other five states, the equations do not give explicit expressions for the probabilities, but instead express the change in probability over time as a function of the probability in a differential equation.
Let's consider a system that at time $t$ is in state $5$. During a time interval $\Delta t$, two things can happen: Either (1) the system remains in state $5$ or (2) the system switches to state $6$.
One assumption they seem to make is that the only way the probability can change is if the system changes state. So if the system remains in state 5, its probability $F_5(t+\Delta t)$ remains equal to $F_5(t)$. If it changes to state 6, then its probability increases to $F_6(t)$. The change is thus a weighted combination of these two.
In more detail, if we consider the probability of switching from state 5 to state 6 during the interval $\Delta t$ to be $P_{56}$, we can say that our new probability is the weighted sum
$F_5(t+\Delta t)=P_{56}F_6(t)+(1-P_{56})F_5(t)$,
so our change in probability is
$\Delta F_5=P_{56}F_6(t)+(1-P_{56})F_5(t)-F_5(t)=P_{56}[F_6(t)-F_5(t)]$.
Our last step is to figure out what $P_{56}$ is, and it is given by the rate constant for the change from state $5$ to state $6$. The rate constant is the average frequency of the reaction, so the probability of the reaction occurring in a given time interval is the rate constant times the length of time. So we replace $P_{56}$ with $k_{-P}\Delta t$ and get
$\Delta F_5(t)=k_{-P}\Delta t[F_6(t)-F_5(t)]$.
Dividing by $\Delta t$, we have
$\dfrac{\Delta F_5(t)}{\Delta t}=k_{-P}[F_6(t)-F_5(t)]$.
As $\Delta t$ goes to $0$, this becomes the derivative:
$\dfrac{dF_5}{dt}=k_{-P}[F_6(t)-F_5(t)]$.
You can repeat this for the other five states to generate the other equations in the paper. The key point is that even when back reactions are permitted (such as from state 1 to state 0), the probability function is only for the systems that began in the indicated state, so the rate constant for a system coming back to that state does not enter into the equation. An implicit assumption is that the system does not make two moves during the time interval $dt$. That is, it cannot move from state 0 to state 1 and back to state 0.

They used δ in two contexts: δ(t) is the delta function, so it is a probability of one at time zero, just like you thought it should be. δ is also the step size, the multiplier in the $V$ equation.

How would it not reach state 6? The first step never reaches equilibrium because product is continuously removed. Therefor, I would think the reaction would go to completion.

@KarstenTheis - I agree that the scheme as written suggests that everything eventually gets to state 6, but in that case all of the probabilities would be 1, so there must be some leakage that isn't shown. Perhaps the kinesin irreversibly detaches before reaching state 6? I didn't read through the paper.  There could be an explanation in there somewhere.

@KarstenTheis - also thanks for the clarification on the delta function. But wouldn't that be equal to 0 at $t \neq 0$? I would expect $F_6(t)$ to be 1 for all t, not just t=0.

These are not cumulative probabilities to be in a state, but probabilities of state transitions. t = 0 is the reference point where all species transition from state 5 to state 6. The other equations describe transition probabilities for t < 0, hence reverse master equation, see https://en.wikipedia.org/wiki/Kolmogorov_backward_equations_(diffusion)

Now I understood that part, I think you are mistaken about Fn(t) vs Fn(0), and the paper does not have a typo.

Nice job figuring that out! Do you want to write up a corrected answer, and I’ll delete mine? Feel free to borrow any text from mine that’s useful.

I don't really understand it fully yet. I was hoping someone else is familiar with master equations (most likely on the physics side of things).

Isn't it all basically just a Markov Chain?

First passage problems
This is a first passage problem, asking when a system reaches a certain final state for the first time . If you were to use a simulation to explore this, you would erase parts of trajectories right after reaching the final state before summing up results. The Backwards Master Equation is especially suited for first passage problems (one way to think about it is that there is complete knowledge at the end of the trajectory, and less and less knowledge as you go backward in time). For folks familiar with chemical kinetics, this is very confusing because the typical case is maximal knowledge at the initial state, with stuff diverging as time passes.
Sketch of Derivation
This follows https://arxiv.org/pdf/1503.00291.pdf, equations (15) - (17). I changed the name of the conditional probability to stay closer to the paper the OP asked about (the conditional probability is called $S$ and the transition probabilities are called $\alpha$ in the derivation I am paraphrasing, but here I am using $F$ and $k$ instead.)
$F_i(0,t)$ is the conditional probability that given a system that is a state $i$ at time 0 will transition into the final state 6 at time t. To derive the Backward Master Equation, we consider two exclusive events shortly after $t = 0$.

The system transitioned into a different state $j$. The probability that this happened is governed by the reaction rates $k_{ij}$ with state $i$ as reactant and state $j$ as product. This is where the asymmetry in the final equations comes from. The conditional probability to reach the final state at time = $t$ for a system that transitioned into state $j$ is $F_j(\Delta t,t)$.
The system is still in state $i$. The probability that this happened is one minus that the system transitioned into a different state $j$ during that short time. The conditional probability to reach the final state is now $F_i(\Delta t,t)$.

From this, we get:
$$F_i(0,t) = \big(\sum_j{k_{ij} \cdot F_j(0 + \Delta t,t)}\big) + (1 - \sum_j{k_{ij})} \cdot F_i(0 + \Delta t,t)$$
Again, we only consider reaction out of state $i$, not into state $i$. The conditional probabilities are stationary (not dependent on time, the system has no memory), so
$$F_i(t_1, t_2) = F_i(0, t_2 - t_1)$$
Using that, we can rewrite as:
$$F_i(0,t) = \big(\sum_j{k_{ij} \cdot F_j(0,t - \Delta t)}\big) + (1 - \sum_j{k_{ij})} \cdot F_i(0,t - \Delta t)$$
Now we can make $\Delta t$ infinitesimally small and write the differential:
$$\frac{dF_i(0,t)}{dt} = \big(\sum_j{k_{ij} \cdot F_j(0,t)}\big) - \sum_j{k_{ij}} \cdot F_i(0,t)$$
In the paper, they used $F_i(t)$ instead of $F_i(t_1, t_2)$, but we needed to allow the first time point to be different from zero in the derivation, so I included it as a argument in the function $F$.
Comparison to (Forward) Master Equation
The Forward Master Equation matches what we usually do in chemical kinetics. The variable $t$ is conventional time, and we figure out the concentration of a species at time $t + dt$ by asking about the rates to make that species and to use up that species. So both $k_{ATP}$ and $k_{-ATP}$ would appear in the same differential equation. The way the Backward Master Equation is derived (we are asking about state $i$ at a certain time $t$, and getting information information about it by going backwards from time $t + \Delta t$), we only have to consider paths from that state to other states.
Modulating [ATP]
This is just a guess. If the ATP concentration changes rapidly compared to the timescale of the reaction steps, you can just use an average rate. If it is changed very slowly, the overall speed of the motor would change proportionally (because the system has sufficient time to "settle in"). If the modulation matches the timescale of the individual steps, it will be complicated. An analogy is NMR and the timescale of conformational change (if the conformational change is fast, you get one averaged signal; if it is slow, you get two separate signals; if it is intermediate, it gets complicated and you lose it).


text:
What Are Examples of Reducing Sugar in Mailliard Reaction?
I'm interested in food science, and I've been reading about Maillard reaction.
All of the articles / videos about Maillard reactions that I found online mention that reducing sugar and amino acids are required for the reaction to occur, but never talks about what kind of reducing sugars are present while searing a steak, roasting a chicken, etc.
When we are cooking a steak with just olive oil, for example, what kind of reducing sugars are present? I initially thought glycogen and ribose in the meat might be acting as reducing sugars, but I learned that most meat we buy don't have glycogen, and there is only a very small amount of ribose present in the meat. Similarly, what kind of reducing sugar / amino acid is present when baking bread?
Lastly, my understanding is that table sugar we buy from stores is sucrose, which is a non-reducing sugar. Does this mean that adding sugar while cooking meat wouldn't help if your goal is to start a Maillard reaction? I know that sucrose does break down to glucose and fructose, which are reducing sugars, when heat is added, but does that break-down occur fast enough for it to be meaningful?
Any input would be greatly appreciated!

Quoting from medium.com

If it wasn’t for Maillard, we would not have juicy seared steak, deep fried chicken wings, nor grilled sausages. All these browning, aromas and flavor bursts are a result of simple sugars reacting together with proteins. However, Maillard is not the only reaction occurring between these different molecules in food.

This is one of the chemical reaction that whether you have an interest in chemistry or not, we all carry out on a regular, maybe even in daily basis.The Maillard Reaction${^1}$ causes two types of browning of foods: enzymatic & non-enzymatic browning. Non-enzymatic browning can again be split in two types: the Maillard reaction and caramelization${^2}$.
The Proteins contained in meat and other comestible goods are usually rich in the amino acids arginine and lysine. The side chain func- tional groups of Arg and Lys react with reducing carbohydrates such as glucose or lactose to form Amadori reaction products${^3}$.

Maillard reaction chemistry
The Maillard reaction is a chemical reaction between an amino acid and a reducing sugar, usually requiring the addition of heat.The Maillard reaction is a ‘non-enzymatic browning' reaction involving reduced sugars with compounds possessing free amino groups. A reactive sugar, such as glucose(a reducing sugar) can react with amino groups in amino acids, peptides, and proteins as well as with other molecules that contain free amino groups${^4}$.
Why a Reducing sugar?
Freshly prepared solutions of glucose in water gradually change by the opening of the pyranose ring to the free aldehyde form. This reaction is a reversal of a hemiacetal formation reaction. Reducing sugars behave like glucose${^5}$.

Following are the steps involved.

Initial reaction between a reducing sugar and an amino group forms an unstable Schiff base${^6}$.

The Schiff base slowly rearranges to form Amadori products${^7}$.

Next, oxidative decomposition of Amadori products can lead to the formation of a wide range of reactive carbonyl and dicarbonyl compounds.The complex reactions are summarised in figure below${^7}$.

Different amino acids produce different amounts of browning.
The Maillard reaction is responsible for many colors and flavors in foodstuffs${^8}$.

The Maillard reaction does not just make food taste better. It has a science involved in it. Understanding the key facts allows you to manipulate the process of cooking according to what your taste buds crave for. It is the difference between blindly following a recipe and being able to make a recipe work for your taste${^9}$.

UPDATE

The Maillard reaction occurs most readily between temperatures 300°F-500°F. When the meat is cooked, the outside temperature becomes hotter than the inside, which triggers the reaction. Protein molecules in the meat are bonded in coils and when heat is applied, those bonds begin to break and the coils begin to unwind. Most of the water content in the muscle fibers then leak out. Reactive carbonyl groups of the sugars react with the nucleophilic amino acids, which produces glycosylamine and water. Pentose sugars react more than hexoses

The pentose sugars might be part of Nucleotide of DNA and RNA(In case no external reducing source is added).
That said, Reducing sugar free xylose, has been reported in onion, strawberries, prunes, apples, pears, grapes, juniper berries, barley malts, brewhouse worts, maple syrup, asparagus, the white and the yolk of eggs, corn, tomatoes, apricots, bamboo shoots, potatoes, beans, alfalfa, beer, and mangoes${ ^{10}}$.
Most of the carbohydrate of the carrot was found to be in the form of soluble reducing substance(You could try as additives for steak prepration)${^{11}}$.
References
${^1}$:https://www.compoundchem.com/2015/01/27/maillardreaction/
${^2}$:https://foodcrumbles.com/maillard-reaction-mechanism-hard-core-chemistry/
${^3}$:https://www.iris-biotech.de/global/knowledge-base/custom-synthesis/maillard-reaction-products
${^4}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx
${^5}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx
${^6}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx
${^7}$:https://www.biosyn.com/tew/The-Maillard-reaction-and-Amadori-rearrangement.aspx
${^8}$:https://www.scienceofcooking.com/maillard_reaction.htm
${^9}$:https://medium.com/@vijinimallawaarachchi/maillard-reaction-the-science-of-browning-flavoring-and-aroma-1605133960ae
${^{10}}$:https://www.researchgate.net/publication/268009365_Free_Sugars_in_Fruits_and_Vegetables
${^{11}}:$http://www.jbc.org/content/46/3/537.full.pdf

Thanks for the resources! But the information above doesn't describe where the reducing sugars such as glucose is coming from. When cooking a steak with oil, where is glucose coming from?

@Sonul I have update regarding reducing sugars in the post.


text:
Why is the titratable acidity of my wine going down as it ages and oxidises?
I am currently studying chemistry in grade 12 studies and I have come across a halt in my investigation.
The experiment was to see how the increase in ethanol would lead to a more acidic wine, because ethanol oxidises to ethanal and then to acetic acid. However, the inverse has appeared to happen throughout the 15 days of the test. Inbetween the tests, the wines were stored in an airtight bottle however they had a chance to oxidise while we opened them to test them.
Why is the acidity of the wine decreasing when the increase in reactants should also result in the bias towards the product side? My tests have yielded the opposite to what I expected (titratable acidity going down by almost half of what it was initially by the end of the 15 days) and I cannot explain it in my discussion.

Have you considered esterification ?

@Poutnik I haven't learned about that, however i will google it and try to learn for myself.

$$\ce{EtOH + R-COOH <=> R-COO-Et + H2O}$$ Typical esterification constant in water is 4, AFAIK. If you add ethanol to mixture previously in equilibrium, the concentration of organic acids would decrease ( in absence of oxygen demanding ethanol to acetic acid fermentation )

@Poutnik Cheers, pretty sure that's what happened, I don't fully wrap my head around it but as long as the theory is there it should make sense in general.

If you let wine and alcohol enriched wine open to air, then the later should be more acidic. But there may be the catch the alcohol concentration must not be too high, otherwise fermentation to vinegar may not run well. Note that it is rather metabolic then chemical process.


text:
Why is the structure of HNO3 have a central atom rather than being linear?
Why is the structure of $\ce{HNO3}$ have a central atom rather than being linear?
I know formal charges need to be reduced as much as possible, and if there was a structure with two oxygens and a hydrogen dangling off the nitrogen on one side and with a double bonded oxygen on the other, wouldn't that make all atoms in it have a formal charge of $0$ while obeying all the rules?
To clarify, the order is hydrogen, single bond, oxygen, single bond, oxygen, single bond, nitrogen, double bond, and the last oxygen.

https://chemistry.stackexchange.com/questions/48641/is-peroxynitrous-acid-an-isomer-of-nitric-acid

What you describe is not nitric acid $\ce{HONO2}$, but it's much less stable existing structural isomer $\ce{HOONO}$. From Wikipedia:

Peroxynitrous acid $(\ce{HNO3})$ is a reactive nitrogen species (RNS). It is the conjugate acid of peroxynitrite $(\ce{ONOO−}).$ It has a $\mathrm{p}K_\mathrm{a}$ of $\sim 6.8.$ It is formed in vivo from the diffusion-controlled reaction of nitrogen monoxide $(\ce{ON^.})$ and superoxide $(\ce{O2^{.−}}).$ It isomerises with a rate constant of $k = \pu{1.2 s−1},$ a process whereby up to $5\%$ of hydroxyl and nitrogen dioxide radicals may be formed. It oxidises and nitrates aromatic compounds in low yield. The mechanism may involve a complex between the aromatic compound and $\ce{ONOOH},$ and a transition from the cis- to the trans-configuration of $\ce{ONOOH}.$ Peroxynitrous acid is also important in atmospheric chemistry.

"Formed in vivo"... Surely they meant in situ

A potent oxidant synthesized by the cell during its normal metabolism. Peroxynitrite is formed from the reaction of two free radicals, NITRIC OXIDE and the superoxide anion (SUPEROXIDES).

pubchem.ncbi.nlm.nih

@electronpusher No, in vivo, see other comment and this reference: https://www.ncbi.nlm.nih.gov/pubmed/15647782

Tech detail: please note that copy-pasted text (also from Wikipedia) lacks most of the intended formatting and markup features. Whenever possible, those need to be recreated manually with MathJax and Markdown (see my edit).

@andselisk Well, I have done so intentionally. I admit I have troubles in accepting this quoting approach I am aware of, seeing it being discussed on meta. I still consider it rather Wikipedia content forwarded to SE than imported to SE. Like an ambassy territory the hosting land laws do not apply to.

@andselisk But OTOH, I usually copy the source code, mostly directly usable , replacing  by dollars, what I did not do this time.

@Poutnik If you open Wikipedia editing panel, you'll see numerous markup and formatting marks that actually make the text literate and readable. Unfortunately, one cannot just copy-paste chunks of source text between SE and Wikipedia as both are using different syntax, so that you have to adapt the formatting. While your intention to preserve the original is commendable, it's readability and literacy that matters as long as the context hasn't been changed.

@andselisk When I copy from the Wikipedia source code as the quote, what I usually do, I do necessery adjustment for the formatting functionality.  Formulas are usually 1:1 transferable. But I do so rather to reproduce Wikipedia look than the SE expected standard, even if it is usually the same. Thank you, I have read the discussion before. Unless it is required by moderators and site policy, I will change in quotes only what must be changed, but not more, As there is always a chance the correction may lead to alternate interpretation, even if it did not looked so in time of the correction.

@Poutnik Well, this is precisely the case when formatting must've been applied, if you were to read that Meta topic carefully. No meaning is (or hardly could be, in this particular example) changed, but the commonly accepted and standardized notations are used in place of plain text. If you want to preserve the original's meaning, this is exactly what you want to do and not just through out in the open what your browser lets you to copy from a web page.

I do not dispute this particular case now, as I should have copied the source code instead of displayed code, to reproduce the Wikipedia look, or to reformat the displayed text copy. I do agree in this with you and I do thank you for the correction.

As discussed, there is nothing in principle to stop HNO3 having the structure you have proposed. However, it is also shown that the common anion derived from this has all oxygens equivalent (see e.g. crystal structures, or NMR studies) so that the structure like a fidget-bit is more common.

How exactly linear.First of all atoms inside the atom must be share as many electrons as their electron configuration "tells "them to share(Oxygen must share 2 electrons).Another issue is that nitrogen and oxygen have lone pairs of electrons at their valence orbital and those lone pairs repel the bonding pairs which gives the molecule its shape.

If you're disagreeing with the possibility of existence of a linear isomer, what about the literature quoted by Poutnik? Please further elaborate your stance.

Nitrogen makes 5 bonding pairs in HNO3 . Those 5 bonding pairs repel one another forcing the creatipn of non-linear molecule.Also oxygen has two lone pairs , which repel the bonding pairs of the oxygen , creating a non-linear molecule.

I'm sorry, I don't quite understand your argument. Which structure are you referring to? The typical structure or the peroxo one?

HNO3 molecule   OONOH

5 bonding pairs 2 double bonding pairs 1 single bonding pair.

And I will another argument O-O single  bond is weaker than the N-O  double bond and weaker than N-O double bond so the molecule has lower energy.

That way formed.

I don't find that satisfactory enough as a relevant argument. And anyway, nitrogen has 3 bonds in the peroxo structure, please verify from the link in Poutnik's answer.

In HNO3 Nitrogen has 2 double covalent bonds and 1 single covalent bonds = 5 bonds.

@4HonorNDFame No. Please finish General Chemistry.

What do you mean general chemistry.There is no such thing as general chemistry.


text:
What does the number of orbitals in a subshell represent?
s subshell has 1 orbital and p subshell has 3 orbitals. Does that mean that there is a high probability that the electron is in the p subshell? As orbital are the space where there is a high probability of finding electrons.

Short answer
Number of orbitals does not have direct relation to either probability of  electrons occupying them, either the space they are describing.
The related probabibility is to be understood as:
IF the electron is in the quantum state, belonging to the particular orbital, THEN the wave function of the orbital describes spatial distribution of the electron  occurence probability, what is geometrically illustrated by the orbital 3D shape.
Longer answer
There are 3 meanings of the term orbital:

A complex wave function $\Psi$ being a particular solution of the quantum wave equation
A quantum state of an electron, following 1., described by particular quantum numbers $n$, $l$ and $m$, representing discrete values of electron energy, orbital angular momentum and one of component of angular momentum, respectively.
3D shape, following 1. and 2., representing statistical  probability of the electron presence at given point by$\Psi^2$, determined by the conventional value by residual external probability and by the isoprobability surface.

If there is 1 orbital 2s and 3 orbitals 2p, it means,
for  $n=2$  and $l=0$, there is 1 value $m=0$.
But for $n=2$  and $l=1$,
there are 3 values $m=-1,0,+1$
Electrons are fermions with half integer 4th spin quantum number $m_s$, and every electron in the atom must have unique set of quantum numbers.
Therefore there can be 2 electrons in the orbital 2s, having $n=2$, $l=0$.
But there can be up to 6 electrons in 3 2p orbitals, having $n=2$, $l=1$.

Good answer, but I have a feeling this explanation is not suitable for OP's level.

You are right, I sometime let the topic to lead myself too far.


text:
Mixing cobalt(III) hydroxide and thallium(III) hydroxide
A simple question on solutions is driving me crazy. At this point, I think the textbook's wrong. Here's the question:

A saturated solution of cobalt(III) hydroxide $(K_\mathrm{sp} = \pu{1.6e-44})$ is added to a saturated solution of thallium(III) hydroxide $(K_\mathrm{sp} = \pu{6.3e-46}).$ What is likely to occur?

The answer given is that $[\ce{OH-}]$ is above saturation levels for both the cobalt and thallium in the solution, so that both will precipitate (thallium(III) hydroxide first, since it has a smaller $K_\mathrm{sp}).$
What I don't understand is how they can make such a statement. We don't know the amounts of each solution originally. Furthermore, wouldn't $[\ce{OH-}]$ be somewhere between the two original $[\ce{OH-}]$ depending on amount of each solution? And if so, wouldn't it not trigger precipitation of the cobalt (III) hydroxide (larger $K_\mathrm{sp},$ larger $[\ce{OH-}]$ at full saturation than $[\ce{OH-}]$ of the mixed solutions)?
Because of this, my answer would be that thallium(III) hydroxide precipitates, but cobalt(III) hydroxide remains dissolved.

Your reasoning is fine, and the book says nonsense.

The reasoning is a little suspect to me. My calculations, paralleling those of @KentdelosReyes lead me to conclude that there is no concentration of $\text{Tl}(\text{OH})_3$ that leads to precipitation because when $[\text{OH}^-]$ is high, $[\text{Tl}^{3+}]$ is low. Quantitatively you get a fourth-degree equation with $3$ negative roots and one positive root corresponding to pure $\text{Tl}(\text{OH})_3$ saturated solution.

Oops, I take that back. With a corrected fourth-degree equation I get a precipitate when the fraction of $\text{Tl}(\text{OH})_3$ is between $10.6$% and $100$%.

The gist here is that you must assume that there is an excess of both cobalt(III) hydroxide and thallium(III) hydroxide after the solutions are mixed. Thus since both ppt's are in the final solution, it doesn't matter what the initial relative proportions are.

This should be a comment, but I decided to post it as an answer, since from my point of view, the present answers fail to shed light on the absolute simplicity of the truth.
We can state from the present answers that there will be no precipitates, but not because the concentration is smaller than an atom per liter, we don't know the volume used since it was not stated.
The simple fact is that the $[\ce{OH^-}]$ concentration stays the same, it will be $10^{-7}$ before and after the mixing of the solutions, no matter how much from each solution we will mix together, since the solubility product is so small and the dominating factor for the $[\ce{OH^-}]$ concentration comes from the autoprotolysis of water. Accepting this little knowledge, no matter what we will never reach the saturation concentration after mixing. The saturation concentration will stay the same always (in this problem), so we will just dilute the initial concentration.
Imagine we mix an equal amount of both solutions, then the before saturated concentration of the substances will be halved. So in the end in this specific case for an example the solution will be half saturated with both hydroxides. That is why there is no precipitation.
Edit: Also there is no sense talking about common ion effect, since this phenomenon arises from the charge balance equation.
$$3[\ce{Tl^{3+}}]+3[\ce{Co^{3+}}]+[\ce{H^+}]=[\ce{OH^-}]$$
As we postulated before $[\ce{H^+}]=[\ce{OH^-}]$, since the autoprotolysis is the domination factor, both of the concentration of the ions will be zero (simplified). In truth the $[\ce{OH^-}]$ concentration will be just slightly higher than the $[\ce{H^+}]$ concentration, but we can neglect this equation since we assumed the only $[\ce{OH^-}]$ concentration comes from autoprotolysis, meaning we don't have to solve a system of equations, just plugging in the value in the solubility product.

You don't need to know the original amount of the substance because the $K_\mathrm{sp}$ will give you the saturated concentrations of the ions present.
For $\ce{Co(OH)3},$ the $K_\mathrm{sp}$ is $\pu{1.6e-44}$ which means the saturated concentration of the solution is $\pu{4.93e-12 M}$ for $\ce{Co^3+}$ and $\pu{9.86e-12 M}$ for $\ce{OH-}.$ You can calculate this using the equation
$$K_\mathrm{sp} = x\cdot (3x)^3$$
The same applies for $\ce{Tl(OH)3},$ the $K_\mathrm{sp}$ is $\pu{6.3e-46}.$ The saturated solution contains $\pu{2.20e-12 M}$ of $\ce{Tl^3+}$ and $\pu{6.6e-12 M}$ of $\ce{OH-}.$
Now that we clarified the initial concentration dilemma, now we can address the real issue at hand which is the common ion effect. The two slightly soluble solids has a common ion of $\ce{OH-}$ that decreases the solubility of both cobalt(III) hydroxide and thallium(III) hydroxide. This is a complex topic so I recommend you to dig deeper on it.

True, if we saturate the solution with both solids together, they both will reach the equilibrium concentrations slightly less than their corresponding solubilities when measured one by one. But that's not what the problem asks for.

@Kent de los Reyes: Please check your calculations. The $[\ce{OH-}]$ should be much higher than what you have calculated.

The calculations made by Kent de los Reyes are based on the hypothesis that the solvent (water) does not produce any $\ce{OH-}$ ions. He calculates that the concentrations of $\ce{OH-}$ is about $10^{-12}$ in saturated solutions of $\ce{Tl(OH)3}$ or $\ce{Co(OH)3}$. But the real concentration of $\ce{OH-}$ is much higher, because of the water dissociation. If the saturated solution is prepared in pure water, $[\ce{OH-}]$ should be equal to $10^{-7}$ M. And the concentrations of the ions $\ce{Co^{3+}}$ and $\ce{Tl^{3+}}$ are of the order of $\dfrac{10^{-45}}{({10^{-7}})^3} = 10^{-24}$.
Well! Such a concentration is smaller than one atom per litre. So it is of no use discussing any precipitate whatsoever.


text:
natural staining on wood from human skin
I have a question about the dark staining that occurs on my wooden spectacle frames where they rub constantly with my skin around the ears and nose. Can anyone tell me what is causing this? And could it be reproduced artificially?
I think you can see what I'm talking about in the photo. (Excuse poor image quality).
I make the glasses, and one of the difficulties is finding a good way to stain them - dark frames suite many people best but the woods from around here are mostly light. If I could only make them go dark like they do from rubbing with skin. But all over. That would be perfect. Anyone think it could be done?

This darkening is most likely caused by skin oil. A polish of oil (just for trial even olive oil) may make the wood appear darker.

I wanted to comment as above. Also wetting should do the same.

Wetting would darken the wood, but not nearly as much as this.

Application of oil (olive oil, linseed oil) does darken the wood, but not nearly this much. I have applied something like 17 coats of linseed oil to some oak wood, with nothing like this much darkening.


text:
Why is cyclopropane-1,1-diol stable
The reason given me in my book is that it's stable due to steric relief.
The answer given on quora said "there is a lot of angle strain."
What exactly is steric relief ?
Can anyone help me elaborate how these points are contributing to the answer.Also what is the minimum angle to provide strain so that geminal diols are stable

Think about how strained the bond angles are in cyclopropanone v the 1,1 diol.

How do strained the bond angles help in stabilizing the diols?

The bonds in cyclopropane 1,1-diol are less strained than the cyclopropanone.
The preferred/lowest energy for tetrahedral bond angles is 109.5 degrees so the closer they can get to that the lower the energy. In cyclopropanes the angles are 60.
The lowest energy angle for the R2C=O system (SP2) is 120 degrees, in cyclopropanone it is 60. This is a very strained and unstable system, anything that reduces that strain is a positive change energywise.
This problem is discussed here see problem 2

Are you suggesting that -Oh group reduces strain in cyclopropanone thus increases it's energy?

Decreases its energy

Did any one find data (I mean thermodynamic data, equilibrium data or something like this) on cyclopropanone / cyclopropane 1,1-diol system? I had a quick look when I saw the question and could not find much informative information. I thought it would be worth it to go beyond just using this example as an example of bond straining.

Good question, which I can't answer. I don't have database access any longer so I rely on what Google turns up.


text:
Does adding solutes to water alter its triple point?
Does adding solutes to water alter its triple point? If yes, how do we find the new temperature and pressure of this point?

Adding things to water changes much more than you expected. It is not that the triple point will just change. No, it will cease to be a point.

@IvanNeretin Are you implying that when I dissolve solvents in water, it will not have a triple point at all?

It neither "yes" nor "no"; it is complicated.

@IvanNeretin  Can you provide a few sources/websites/papers which are related to this question?

Triple points are typically defined as points on the phase diagram of a pure substance where three phases coexist. Their existence requires that the phase rule be observed, which is $$f=C+2-P$$ where $$\begin{align}f &: \text{number of degrees of freedom} \\ C &: \text{number of components} \\ P &: \text{number of phases}\end{align}$$ The number of degrees of freedom is the number of intensive variables that can be altered independently while remaining in a given phase.
In general, when three phases coexist, $$f=C-1$$ If the system consists of a pure substance, $f=0$, as expected (the coexistence region corresponds to a point). If you have a mixture of $C$ components (labeled $i$), you will observe a different triple point for each choice of $C-1$ independent mole fractions $\chi_i$. But for each combination of mole fractions, coexistence of three phases will still occur at regions corresponding to single points.

Interesting...But, I found this(https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch15/colligative.php) website which says that there will simply be a depression in the boiling and melting points. So, the triple point will just decrease. Im confused.

Both (my answer and that website's explanation of freezing point depression) are correct and consistent. The explanation in the link answers your second question (I only answered the first - sorry!!).

I would read that page very carefully, there is a lot of information crammed into the phase diagram, but if you parse it apart you should come to understand what's going on.

How can they both be the same? According to the gibbs equation, there must be a line representing triple points. But, in the website there is only one point..

Does the equation only apply to large concentrations of solute?

What the Gibbs equation says is that you can expect a new TP for each solution with a different mole fraction of the solute. Provided that solution composition is constant, you will get a phase diagram that looks qualitatively just like that of a pure substance, as shown in the link. Guaranteeing constant solution composition might not be simple though, but that's not necessarily a problem with thermodynamics.

Yes, the triple point of the system will change. Instead of a single point specified by temperature and pressure at which multiple phases of water coexist, the addition of solutes (not solvents) will cause the conditions for 3 phases to exist to become a range of temperatures and pressures.

But how do we find this range of temperature and pressures?


text:
Ab Initio and Molecular Orbitals
What is the difference between ab initio calculations and molecular orbital theory?

What is the difference between sunrise an chair?

"ab initio" literally means "from first principles" or "from the beginning" in latin. It therefore (literally) depends what you consider the most fundamental theory (what mirhoron calls high level theory). https://en.wikipedia.org/wiki/Ab_initio_quantum_chemistry_methods

Ab initio means calculations that are strictly from the principles of quantum mechanics. This contrasts with a non ab initio method which can involve parameters that do not arise directly from principles (i.e. results from experiment and/or 'better' calculations). One typical example of something that is not ab initio is a functional for Density Functional Theory. Because the exact functional is unknown, different functionals are generated using large datasets of high level calculations. These are often good at reproducing experimental data in many situations however you cannot derive them directly from the postulates quantum mechanics.
Molecular Orbital Theory (which can be implemented with the Hartree-Fock method) is an example of an ab initio method. One can (very roughly) come up with MO theory by approximating each electron as only feeling the effect of the average positions of the other electrons. So, if an orbital has an electron in it all the other electrons will feel a repulsion from the space occupied by that orbital. This results in a set of coupled one particle problems the solutions (eigenfunctions) which can be solved to find the Molecular Orbitals for the system.
More generally, ab initio calculations also refer to the more accurate post Hartree-Fock methods such as Configuration Interaction (CI), and Coupled Cluster (CC).

Ab initio can mean pretty much whatever people consider relatively high level of theory.

This is misleading at the very least. Regarding DFT: https://chemistry.stackexchange.com/q/33764/4945 It is paramount to understand, that DFT is exact in principle. HF is not the same as MO theory; it is one attempt at mathematically formalising MO theory.

DFT is of course exact in principle, but also the perfect example of a method that needs parameters from another source for its practical usage. This is why I referred to specific functionals. For this question, I think directing towards HF is the most helpful direction for future inquiry.

The way you wrote this is a bit misleading; at least I obviously misunderstood what you meant. Sure, there are density functional approximations which contain empirically fitted parameters, but there are also some that do not. The usage of this terminology may lead to confusion, and judging from the question, there is not a solid foundation to clear these up. I am not sure how helpful it is to direct someone to HF, when MO theory is not a solid foundation.

Thank you for those comments Martin, I have incorporated them into the answer to (hopefully) improve clarity.


text:
What are the chemical principles behind paper/ink deterioration over time and how do I prevent them?
I have been working on a time capsule for the past few months, in which I decided to not include any electronics (ie: no USB). I am done writing the content of the capsule, but I have some concerns regarding the surface on which it'll be printed and the casing it'll be trapped in.
I know for a fact that all kinds of paper do not age well, and those that do only do so in specific conditions. An example of that would be the old report cards of my parents that have turned yellow-ish and whose paper is surprisingly light and fragile. The ink started disappearing in some areas of the report cards, and left the paper directly under it more damaged than in other, uninked, areas.
These report cards being nearly 40 years old and already slightly damaged, I am quite concerned about leaving printed paper inside a time capsule for decades (if not longer). Since it will most probaby be an encased stack of paper, I am also worried that the pages at the bottom of the stack will get damaged even faster because of the pressure.

In this case, a damaged piece of paper means either:
A) The paper has become too fragile to be held with bear hands
B) The ink has disappeared and the paper has become unreadable to human eyes
I have thought of some of the factors that I should probably account for like temperature, material of the casing, oxydation, and humidity, but besides putting the paper in a vacuum (which might not even be that good of an idea), I don't know what to do.
So, my question is: What are the chemical principles behind the deterioration over time of both ink and paper when put together, and how do I prevent/slow them over long periods of time ?
I have control over these variables:

Type of paper*
Type of ink*
The ink can either be printed using a regular/office printer or placed using writing tools
Type of paper coating*
Material of the casing*
Environment surrounding the paper: it could be a vacuumed space or any gas or liquid*
Location of the time capsule: I can place it/burry it anywhere around the world, as long as it is not unrealistically deep or in a militarized zone

*: I will accept any answers for these materials, but if cheaper alternatives do exist it's better.

Modern acid-free paper lasts forever if kept out of the sun. Write on it with an ordinary old-school pencil. Pencil lasts literallly forever, the paper I guarantee five thousand years in reasonably dry air.


text:
Why isn't aluminum ferromagnetic?
I know that the ferromagnetism of iron is because of its unpaired d electrons, but I also know that there are other metals with unpaired electrons like aluminum that are not ferromagnetic.  Is the difference due to the molecular orbitals or whatever the name is for the orbitals in a solid metal?  I have a suspicion but I'm not sure that in bulk aluminum, the lone unpaired electron of one aluminum somehow shares a molecular(?) orbital with another aluminum atom with a lone electron of the opposite spin so that the net result is that there are no unpaired electrons.
I think this based on a calculation I read in a very well-regarded physics book by Purcell.  In the book he calculates the magnetic dipole moment of a magnet that he doesn't identify except to say that it is a new type of magnet that can be formed into a disc, and since Purcell came out in the 1960's, I suspect he is referring to the ferrite ceramic magnets.  In his calculation, he seems to use one unpaired electron per "molecule".
If the ferrite magnets do have one unpaired electron per "molecule", in contrast to the iron atom in textbooks having 4 unpaired electrons, I'm suspecting that some of the spins are being cancelled out in the actual magnet, and thus I suspect that something similar is happening in aluminum.  Is this correct?
Thanks

Unpaired electrons can give paramagnetism. Ferromagnetism is like a dire version of paramagnetism, which is quite rare in comparison. Read up about magnetic domains.

I will comment that aluminum contributes 3 electrons per unit cell to the conduction band of the metal. While not leading to ferromagnetism, it does cause some weird effects in the end, including a positive Hall coefficient at high electric fields...


text:
Enol content tautomers

I have figured out that 4 one will have the highest enol content of all, and the 1 structure would come after 4 in enol content, i.e 4 > 1. Then I got confused between the 2nd and the 3rd structures.
Please give a brief explanation for the answer as well.

What are the factors that determine enol formation? Is it easier to deprotonate an Ester or a ketone? Those are the sort of questions you need to consider

Enol content in 3 is greater then enol content of 2

Greater the acidic nature of alpha Hydrogen,greater is the tendency for Tautomeric behavior. H at alpha position in ketone is more acidic then H at alpha position in an ester.Pka values are shown below.The withdrawing power of ester is less compared to carbonyl group in ketone. Since lone pair on oxygen is in conjugation with C=O in ester,its withdrawing power decreases(as shown below).A more acidic α-hydrogen implies a weaker C−H bond. Summarizing,the enol content increases with the acidity of the enolic hydrogen.
$$\ce{carbonyl <=> enol}$$
$$\mathrm{K_{eq}=\frac{[enol]}{[carbonyl]}}$$
The equilibrium between keto and enol form is tipped to enol if H at alpha carbon is more acidic.
Hence enol content in 3 is greater then enol content of 2.
[]

Pka of cyclohexanone is as per this link: https://www.drugbank.ca/drugs/DB02060

Enol of 4th compound is aromatic and should be most stable.
Therefore extent of enol content must be 4>1>3>2.
As a general guide line the following can be used for predicting enol content.

The enol form of cyclohexanone  has bond angles around double bond other then 120${^o}$ as shown below using avagadro.

Pretty sure cyclohexanone doesn’t have a pKa of 11.3! According to Bordwell (https://www.chem.wisc.edu/areas/reich/pkatable/) it’s 26.4, and acetone is 26.5...

@ PCK i will give source shortly

@ PCK , https://www.drugbank.ca/drugs/DB02060 ,iwill try to include snapshot of the pka ,which you will find here.

Can you give a plausible explanation why (according to your numbers) cyclohexanone is a billion times more acidic than acetone? Is the solvent the same for your acetone and cyclohexanone values?

@PCK not according to my numbes,but from references as ptesented.

Shouldn’t we critically analyse data rather than just accepting it blindly? 11.3 is implausible

But maybe I’m being an idiot here - anyone else care to comment?

@ PCK https://owl.oit.umass.edu/departments/OrganicChemistry/appendix/pKaTable.html gives Pka of cyclohexanone as 16.7

So what’s the actual value? ¯_(ツ)_/¯

@ PCK https://pubchem.ncbi.nlm.nih.gov/compound/Cyclohexanone#section=Computed-Properties pka 11.3


text:
Synthesis for turning synephrenine into N-methylphenethylamine
I am trying to devise a synthesis to change synephrine (4-[1-hydroxy-2-(methylamino)ethyl]phenol) into N-methylphenethylamine.
I was figuring that I could remove the hydroxyl group attached to the ethyl part using thionyl chloride to convert the hydroxyl into a chloride and then remove the superior leaving group using $\ce{LiAlH4}$ as a reagent and $\ce{AlCl3}$ as a catalyst, but thionyl chloride affects amine groups and I couldn’t find any information about how to protect the amine group when performing this reaction to reduce the hydroxyl.
I also researched how to remove the hydroxyl from the benzene ring but that involved using zinc and distilling the benzene which wouldn’t work for this because the boiling point of the substance is around 341 °C for synephrine.
Any insight would be really helpful.

Can you make an aryl triflate then do Pd-insertion in the presence of a reducing agent like a silane? I dont know if I just made that up or if it’s a thing...

Yes, you can. Relatively straight forward catalytic transfer hydrogenation Example here  https://www.sciencedirect.com/science/article/pii/S0040403900852624

Suggested route - step 1 acetylate the amine with 1 eq of acetyl chloride in pyridine.
step 2 hydrogenation of the benzylic alcohol over Pd/C in the presence of acid or use TFA/triethylsilane
step 3 form the triflate of the phenol
step 4 catalytic transfer hydrogenation of the triflate, conditions here
Step 5 acid hydrolysis of the N-acetyl group with cHCl to give product
Thanks to @PCK for the suggestion of forming and reducing the triflate


text:
Hydrolysis of an amidine derivative under acidic conditions
I know that an imine can undergo acid hydrolysis to yield a corresponding amide:

https://chem.libretexts.org/@api/deki/files/13283/10.jpg?revision=1
But I am looking at a mechanism for acid hydrolysis of a drug where the doubly-bonded N remains, and instead a different R-group (NH3) is cast off.
 https://i.stack.imgur.com/ZAxZN.jpg
Empirically, this the primary degradant produced after exposure to acid, so I know it's 'correct', but I don't understand why the general imine-to-amide mechanism is not occuring. Does it have something to do with the relative properties of the NH3 and NHSO2NH2 R groups?

The first example you drew shows the hydrolysis of an imine to a ketone (not an amide). The second example isn’t an imine to start with, it’s an amidine. Basically, in both cases the C=N bond is hydrolysed to C=O, but since the starting materials are different functional groups, the products are different too.

Thanks.  I found this regarding amidines: https://i.stack.imgur.com/jYLC2.jpg.  I can't find any general information about relative rates but is it plausible that when you limit the amount of acid exposure, there is a resonance that happens that flips the double bond position and then that become the site of the hydrolysis?


text:
Applying vacuum at the end of a flash chromatography column
I know that flash chromatography uses a positive or pushing pressure to achieve the   goal which is separation. Would the same goal be achieved if you had a vacuum on the other end instead of a pressure at the top, a vacuum on the bottom?
I have used this technique for ion exchange column with efficient and quick results and wonder if this can be achieved with flash chromatography.

It should just work the same, but I guess it might limit the solvents you  could use (e.g. Et2O might be too volatile)

All chromatography needs to work is some pressure differential, no matter how it's established. Gravitational chromatography uses hydrostatic pressure from the weight of the solvent, and a typical gravitational column can have a roughly ~20-50 mbar pressure difference between the top and bottom of the column; flash chromatography also has the hydrostatic pressure, but has an additional positive gas pressure at the top from e.g. pumped air, and can typically reach ~100 mbar pressure differences.
The principle of chromatography is to force compounds in a mixture to stay at the interface of a stationary phase (e.g. silica) and a mobile phase. The slight differences in how molecules behave in this boundary are added up over time, creating the observed separation.
Since this is a surface process, having a stationary phase with higher specific surface area and better packing effectively means molecules spend more time undergoing separation at the interface than being dissolved in the bulk of the eluent, where no separation happens. However, a more tightly packed stationary phase also restricts eluent flow. Under a simplified set of assumptions, decreasing the stationary phase particle size by half (and therefore increasing specific surface area by a factor of 4) requires 4 times the pressure differential to produce the same flow rate. So getting better separation in reasonable timeframes requires larger pressure differentials.
This is where vacuum chromatography techniques come in. Vacuum pumps are a staple laboratory instrument, and can easily achieve negative pressures of ~900 mbar with common glassware. Therefore, without too much hassle, it is possible to obtain good flow rates using finer stationary phase particle sizes and get improved separation.
Of course, larger pressure differentials can be achieved with positive pressure at the top of the column. Medium, high and ultra-high pressure chromatography (MPLC, HPLC, UHPLC) make use of stationary phases with very fine particles (from ~10 μm to ~1 μm diameter) which start to look more like silt or clay. The only way to force liquids through these stationary phases is with very large pressure differentials, from ~5 bar to as much as ~1000 bar, but they can achieve incredible resolution of compounds during separation. The reason UHPLC isn't used for every single separation is cost and scalability; it's an engineering challenge to control such enormous pressures safely. Glass is very resilient under compression (which is why glassware can commonly hold a vacuum), but weak under expansion. This means such positive pressures need much more rigid containers, such as thick steel columns. This also limits their feasible sizes and therefore the amount of stationary phase/eluent/compound they can handle.
As an example of a vacuum chromatography technique, I point towards the excellent video by Daniel Pedersen detailing dry column vacuum chromatography (DCVC). There is also a preceding publication which goes into further details. For what it's worth, I've found this vacuum chromatography technique to be far superior to flash chromatography.

In order to make the fluid flow, you just need the pressure differential. At a given flow rate, the head pressure (at the top) and bottom pressure can be any number as long as the difference in pressure is the same.
$$
\begin{align}
P_\mathrm{head} &= \pu{1.0 atm} \\
P_\mathrm{end} &= \pu{0.1 atm}
\end{align}
$$
will have the same flow rate as
$$
\begin{align}
P_\mathrm{head} &= \pu{1.9 atm} \\
P_\mathrm{end} &= \pu{1.0 atm}
\end{align}
$$
Mechanically it is more convenient to generate pressure rather than generate vacuum.

At low absolute pressure, you have a much larger danger of rupture of the static phase of your column. Especially during startup, the form of the pressure gradient in the column is likely fatal.

Very detailed answer by Nicolau Neto. I wish to add a clarifiction "The reason UHPLC isn't used for every single separation is cost and scalability; it's an engineering challenge to control such enormous pressures safely. " is not entirely correct. I have had good discussions with the person who invented the ultrahigh pressure liquid chromatograph (UHPLC). Although 1000 bar may sound dangerous, the actual force to generate this high pressure very small.* In a UHPLC the typical flow rate is on the order of 0.2 to 2 mL/min (max). Higher end or the latest ones have upto 8 mL/min. All they do is make the piston diameter smaller. Recall that pressure =force/unit area. If the cross-sectional area of the pump head bore and the piston bore decreases but the force pushing the pistons remains the same, then you have a UHPLC pump. The conceptual issues with a UHPLC is not the safety concern at all, but frictional heating with very small particles, heating of the mobile phase during compression/expansion, and leakages at pressure extremes, at every possible tubing connection. Eventually, many companies started to use gold seals, as well as gold coated connections. Secondly, because of viscous frictional heating, narrow columns such as 1 to 2.1 mm i.d. must be used on a UHPLC. Radial temperature gradients inside a packed bed ruins the separation.

A large pump such as a pneumatic pump at 1000 bar is very very dangerous. You need a lot of safety measures. These pumps are used for packing very small particles in stainless steel columns.

This is an interesting addition, but to me it looks like this might be better organized as an edit to your another already existing answer with a brief comment below Nicolau's answer.


text:
Band MO of Sodium Filling of electrons confusion
I wanted to ask a question about metallic bond - band electron theory.
Consider the diagram of Sodium below:

Circled are orbitals both containing $2$ electrons each which combine using LCAO to give a set of bonding and antibonding electrons.
However, the $4$ electrons involved in this bonding  appear to only give $3$ electrons in the bonding and antibonding MO's produced.
This confuses me. Surely using $4$ electrons, I expect all $4$ of these electrons to also be present in the rectangle drawn.
Why are there only 3 electrons shown? Is one electron being used for the next addition of AOs?
Any answers would be greatful!

From left to right, it shows 1, 2, 3, and 4 atoms of sodium, with a total of 1, 2, 3, and 4 valence electrons. So this diagram is different from others which have the uncombined AO's on the left and right, and the combined MO's in the middle. Here, we are adding more and more atoms to go from discreet energy levels (left 4 columns, up to four atoms) to a band structure (rightmost column, many atoms in bulk metal).
The only column that shows the AO is the left-most because for that one, there is only one atom, which means that there can't be any MOs.
Maybe this diagram makes more sense to you (the full and empty circles on the right show how the AO's are combined to make the bonding, non-bonding and antibonding MO's):

Source: http://xbeams.chem.yale.edu/~batista/vaa/nabands.gif

Oh right! The above diagram makes sense. So, one question - why are there then the dotted lines between each MO following on from the leftmost AO towards the right?

@vik1245 You could imagine adding another and another AO one by one, or you could take the image I posted and rearrange it to get the other one. In any case, all the MO's are base on the one AO that is shown on the left.

Oh right so as I'm moving one to the right, I'm adding simply one AO to the right to get a new MO set, similar to what I can get if I simply splice your diagram and rearrange it to the image I posted? I think I get it then if I'm correct! Just confirm that please!

@vik1245 yup confirmed

Ticked! By the way in your diagram for $\ce{Na3}$ the middle energy set of orbitals, is that simply missing a black dot by mistake?

@vik1245 No the coefficient for the middle AO happens to be zero for that one (and it is none bonding, energy the same as AO).

@vik1245 See https://chemistry.stackexchange.com/questions/116840/why-do-solutions-of-electron-in-a-box-and-in-a-corral-predict-coefficients-for

Make sense fabulous thanks very much.


text:
How are neutral amines effective bases in organic chemistry?
I have found it to be rather strange that neutral amines, such as triethylamine  or DIPEA, find applications in organic chemistry in deprotonating alcohols or even α-carbonyl compounds. Amines are basic but they aren't really that basic, compared to negatively-charged hydroxides or alkoxides. According to Wikipedia, the $\mathrm{p}K_\mathrm{a}$ (in water) of the conjugate acid of triethylamine is $10.75.$ The same values for alcohols are in the range of $16 - 18.$ For water, it is $14.$ Thus, hydroxides and alkoxides are superior alternatives to these amines.
My motivation for asking about this is that I recently encountered a question in my school chemistry course on why the protection of alcohols with $\ce{MOMCl}$ uses large bulky amines to deprotonate the alcohols, rather than using hydroxides. The answer given was that hydroxides would not be able to deprotonate the alcohols to a significant extent as alcohols are not acidic enough. I find this answer very strange and inconsistent with what I know since as shown by the data above, amines are less basic than alkoxides and hydroxides.

After reacting with lithium, e. g. LDA? https://images.app.goo.gl/Rx379FHyus1i5kMc9

@KarstenTheis No. I specifically refer to neutral amines, such as the two I have written above.

Alkoxides and hydroxide are too nucleophilic. The textbook explanation does not make sense.

In spite of the downvote on my answer to your concern about the reaction of MOMCl with alcohols, in my experience it is an SN1 process. The hindered base's role is to scavenge HCl. MOMCl is unstable in water.

@user55119 I see... It is unstable in water. I suppose there is neighbouring group participation in the reaction of the chloride with water, hastening the reaction. Thank you for your insight.

CH3OCH2Cl --> [CH3O=CH2]+ and chloride. The oxygen facilitates the loss of chloride. The alcohol adds to the cation and the amine picks up HCl.

MOMCl reacts via an SN1 reaction and alcohol is the nucleophile, not an alkoxide. The amine serves to scavenge HCl. ADDENDUM: Typically, a generic alcohol in an inert solvent is treated with MOMCl 1 in the presence of a hindered tertiary amine such as Hunig's base 2. MOMCl, which is prone to ionization by virtue of the participation of the oxygen atom, forms intermediate 3. This species reacts with the substrate alcohol to form the mixed acetal 4 and the amine hydrochloride 5. MOMCl is a very reactive "alkyl" chloride. It reacts with water to form hemiacetal 6 and, in the presence of liberated HCl, forms formaldehyde and methanol. Under these conditions the formaldehyde will be present as its hydrate 7.

Oh I see what you ate trying to say finally! Can you also include your comment on MOMCl being unstable in water for a complete answer?

Did you mean what I said and you were trying to understand?  ;) I expanded the answer per your request.

Although your answer doesn''t directly answer my question, I think it tackles the root cause of my confusion.

To my mind you had two questions, one in each paragraph. I was addressing your specific question about MOMCl.

There are several interesting aspects to your question.

How are neutral amines effective bases in organic chemistry?

First of all, recall the definition of a base. Anything which accepts a proton (or donates a pair of electrons) is a base. Organic bases are very strong and corrosive! These bases should not be undermined, even though they may sound weak (and their stench is worse than inorganic bases). Your lye-free oven cleaner has an organic base, an neutral organic amine such as ethanolamine, which dissolves dirty grease stuck in the oven. This molecule has no charge yet it is a strong base, acting just like sodium hydroxide based oven cleaner in terms of performance.

Thus, hydroxides and alkoxides are superior alternatives to these
  amines.

The second reason to avoid hydroxides and alkoxides is their solubility in organic solvents. Neutal organic amines easily dissolve in toluene and a variety of organic solvents. NaOH or NaH remain as a suspension.

MOMCl  uses large bulky amines to deprotonate the alcohols, rather
  than using hydroxides

Again you have to think about the reaction solvent. Is an inorganic hydroxide soluble in your solvent? Then comes an important point. Why use bulky amines? Their sole purpose is to act as a proton sponge in the reaction mixture and prevent the formation of free acid. Bulky amines are preferred so that they don't react easily with the alkyl chlorides or acyl chlorides, thus prevent the formation of their own undesired side products.


text:
Testing thermite for chemical properties
I'm writing a movie in which a bunch of military explosive ordnance experts destroy a turbine with thermite and the local fire chief has to work out whether it is plain $\ce{Fe2O3}$ and aluminium powder thermite or military-grade thermite with other stuff in it.

How would he test for these ingredients and better be able to discern between the two?
And what sort of language would he use to translate his finding?

The question is:  Is there any military grade thermite ? I could imagine an addition of magnesium, but not sure, if it would make any additional effect.

@Poutnik, if they added magnesium, it would be for the purpose of making it easier to ignite.  For all its reputation as a ferocious incendiary, thermite is surprisingly non-flammable.

Typical military-grade thermite mixtures (e.g. used in incendiary hand grenades) contain a large amount of barium nitrate (e.g. 29.0 % in Thermate-TH3, see Patent No. US 6766744). It should be possible to detect the barium oxide that is left over after the fire, maybe first with a simple flame test on site followed by AAS (atomic absorption spectroscopy) in a laboratory.

According to this Wikipedia article, there has been military research into nano-thermite, which can contain molybdenum, bismuth or tungsten oxides.  It would be possible to test for residues of those after the fire.
Also, "nano-thermite" sounds awesome!