text:
How to find volume needed from a stock solution
I am making a media that is amended with substrates after being autoclaved in order to keep the media anaerobic. In order to do this, we make stocks of the solutions from a known concentration. I am having trouble figuring out what amount of liquid to add to one liter. For example, I make a 100 ml stock of sodium fumarate (160.04 g/mol) using 0.6402 grams in 100 ml of MQ H20. The final desired concentration in the liter media is 40 mM. How do I figure out how much of the liquid stock of sodium fumarate I am supposed to add? Thanks in advance for the help!

This is calculated by equating the moles of the two solutions.
You are using $\pu{0.6402g}$ of sodium fumarate in $\pu{100mL}$ of solution.
$$ \text{No. of moles} = \frac{\text{Given weight}}{\text{Molecular weight}} = \frac{0.6402}{160.04} = \pu{0.0040 moles} $$
$$ \text{The concentration of the solution can be found easily} = \frac{0.0040}{0.1} = \pu{0.04 M}$$
If you require a solution of concentration $40$mM, the stock solution that you have prepared is exactly that.
If you need a more diluted solution, you may use the formula:
$$ M_1 V_1 = M_2V_2$$
where, $M_1$ and $M_2$ are the initial and final concentrations, $V_1$ is the volume of stock you would require for the dilution, and $V_2$ is the volume of final desired concentration.
Let us assume you require $\pu{50 mL}$ of $\pu{10 mM}$ solution, and you have a stock of $\pu{100 mL}$ of $\pu{40 mM}$.
Then
$$
M_1 = \pu{40 mM}\\
V_1 = ? \\
M_2 = \pu{10 mM} \\
V_2 = \pu{50 mL}
$$
Plugging in the values, you will find out that you require $\pu{12.5 mL}$ of stock, and dilute it to $\pu{50mL}$ using the solvent, i.e., add $\pu{37.5 mL}$ of water to $\pu{12.5 mL}$ of stock solution.


text:
Calculating ΔG at the extremes of reaction extent
This question follows on from a great post on comparing $\Delta_\mathrm{r} G^{\circ}$ to $\Delta_\mathrm{r} G$ answered here
The G vs ξ graph that is used in the answer is an excellent tool to understand how $\Delta_\mathrm{r} G$ changes with the extent of reaction:

If $\Delta_\mathrm{r} G$ is found on this graph by the slope of the line, what happens at the two extremes, where we have pure reactants on the left ($Q = 0$) and pure products on the right ($Q ⟶ ∞$)?
The equation to calculate $\Delta_\mathrm{r} G$ is: $$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^{\circ} + RT\ln Q$$
This would give  $\Delta_\mathrm{r} G ⟶ -∞$ (vertical slope asymptotic to the y axis) when $Q=0=ξ$ (i.e., when we have pure reactants). Logically, that would mean that absolutely pure reactants have infinite G (which doesn't seem right).
A similar problem occurs as we approach pure products: $\Delta_\mathrm{r} G ⟶ ∞$ when $Q⟶∞,ξ⟶max$ (vertical slope asymptotic to a line at ξ = max).
Can anyone reconcile this graph with the equation to calculate $\Delta_\mathrm{r} G$?

G is a finite quantity for ξ = 0 or max. The slope of the graph is vertical on the extremes, though. This is because the chemical potential for reagents approaches negative infinity on one side, and that of the product positive infinity on the other side. (A vertical or infinite slope does not mean that the function value has to be infinite - a half-circle or the letter U is an example to illustrate).
The interpretation is that (unless $\Delta_r G^\circ$ is of extremely large magnitude), there are always a least a couple molecules of solute or gas species at equilibrium.
The sketch of the graph could be made more accurate by adjusting the slope at the ends accordingly. For a graph of $\Delta_r G$, see here: https://chemistry.stackexchange.com/a/115544

Thanks for the link, I have seen that graph, it is a good one (thanks to @Night Writer for posting that) and it makes sense. It seems that the reason for there being an absolute value for G (rather than infinite) for a pure species is that the species needs to be in equilibrium with something (even a few molecules of products will do) in order for the the equations we are using to actually hold. So the graph essentially fails at the extreme values of ξ as this is no longer true (either no reactant, or no products exist). Would you agree?

I would also like to thank @orthocresol for answering my initial comments. Hopefully they will get a chance to comment on the answers.

@Withnail - The graph does not fail;  x ln x approaches zero when x approaches zero. This is because x approaches zero "faster" than ln x approaches infinity. So there is no discontinuity (except for the usual one that substances are made of particles, and you go abruptly from zero to one particle, i.e. the x-axis is not continuous if you look too closely).

We have solved the equation now for the limits of ξ. I will write up the work and post when I have time as an edit to the answer. It turns out that G ⟶ RT (C - 1) when ξ ⟶ 0 (from the + side) and G ⟶ ΔrG∘ + RT (C - 1) when ξ ⟶ 1 (from the - side). Where C is the constant of integration. I knew that ΔrG∘ was the slope at ξ = 0.5 (Q = 1), but i didn't realize that the difference in G between pure reactants and products was the same thing. @ Night Writer

@Withnail Thanks for sharing your insights! They are the same because of $\Delta_r G^\circ$s definition. We are imagining that the reaction proceeds while the concentrations remain at standard state, and it is per mole. So you can either see it as differential, or in this somewhat artificial reaction where you keep $\Delta_r G$ constant, either by having large vats of reactants, or by having very little react. I recommend Treptow (1996) J Chem Ed 79(1) 51-54.

@Withnail It looks like we've attempted roughly the same thing. I also edited my answer and added what seems a satisfying if somewhat anticlimactic math explanation. It's been an interesting set of detailed observations you've made.

I have added edits to my answer, involving limits to the function of G vs ξ. Thanks for your help. Feel free to edit/improve/clarify any of the answer. @Night Writer

After a lot of help, I have the following to suggest as an answer:
Imagine a reaction with $\Delta_\mathrm{r} G^{\circ} = -1000Jmol^{-1}$ at 298K. Using the following equations: $$\Delta_\mathrm{r} G^{\circ} = - RT\ln K$$ $$e^{-\frac{\Delta_\mathrm{r} G^{\circ}}{RT}} = K$$
This would give a K value of 1.50, which indicates the Q value on the curve where G is at a minimum (favoring products slightly in the equilibrium mixture). Using another equation (valid e.g. for a reaction $\ce{A(aq) <=> B(aq)}$): $$ξ = \frac{Q}{1+Q}$$
Gives the ξ value (1.50/2.50 = 0.600) which is the x coordinate of the minimum on the graph, where the x axis runs from ξ = 0 (pure reactants) to ξ = 1 (pure products). This graph is pretty similar to the one posted in the question:

Using the equation:$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^{\circ} + RT\ln Q$$
And substituting in:$$Q = \frac{ξ}{1-ξ}$$
Gives us:$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^{\circ} + RT\ln \frac{ξ}{1-ξ} $$
or:
$$\frac{\delta G}{\delta \xi} = \Delta_\mathrm{r} G^{\circ} + RT\ln \frac{ξ}{1-ξ} $$
$\Delta_\mathrm{r} G^{\circ}$ is a constant even as G changes as the value of ξ changes, so are R and T. This differential equation can be separated and integrated to gives us the original function of how G depends on ξ:
$$\begin{align} \int \delta G = \Delta_\mathrm{r} G^{\circ}ξ + RT \int \ln \frac{ξ}{1-ξ}\delta \xi \end{align}$$
$$\begin{align} G = \Delta_\mathrm{r} G^{\circ}ξ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1 + C)\end{align}$$
Graphing this on desmos gives us the correct graph for how G varies with ξ.

The graph shows the correct ξ value for the equilibrium point and also the difference in G between the "pure" reactants and products. All of this is as predicted by the theory.
If you trace the line along the domain of x towards zero (or 1) the y values increase until the value becomes undefined, but I have zoomed in until my brain hurt and the value never seems to shoot upwards (as I thought the maths predicted). L'Hopital's rule seems to prevent this from happening.
I can't pretend that I fully understand the maths (I don't), maybe others' comments will help here. If anyone wants to use the desmos calculator to produce their own plots (you can change T or $\Delta_\mathrm{r} G^{\circ}$ to other values to see the effect) please do so, I may use this in teaching the material in the future.
I credit @orthocresol @Night Writer and @Karsten Theis with helping me. Also Darrell Bach and my students Nick Kleiderer and Sylver Riddell.
****** Addition to deal with limits ******
The limits of G at each end of the domain of ξ are below.
(For ease of reading I will use x = ξ)
$$\begin{align} G = \Delta_\mathrm{r} G^{\circ}ξ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1 + C)\end{align}$$
$$ G(x) = \Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)$$
Limit of G when $x \to 0 ^+$ :
$$ \lim_{x \to 0 ^+} G(x) = \lim_{x \to 0 ^+} [\Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)] $$
$$\lim_{x \to 0 ^+} G(x) = \lim_{x \to 0 ^+} [\Delta_\mathrm{r} G^{\circ}x - RT + RTC] + \lim_{x \to 0 ^+} [RT(x \ln (x) + (1-x) \ln (1-x))] $$
Note (1):   $ \lim_{x \to 0 ^+}[(1-x) \ln (1-x)] = 0 $
Note (2):   $\lim_{x \to 0 ^+} [x \ln (x)] = 0$   (by l'Hopital's rule)
$$\lim_{x \to 0 ^+} G(x) = - RT + RTC + RT(0+0) $$
$$\lim_{x \to 0 ^+} G(x) = - RT + RTC $$
$$\lim_{x \to 0 ^+} G(x) = RT(C-1) $$
Limit of G when $x \to 1 ^-$ :
$$ \lim_{x \to 1 ^-} G(x) = \lim_{x \to 1 ^-} [\Delta_\mathrm{r} G^{\circ}x + RT~(x \ln (x) + (1-x) \ln (1-x) - 1 + C)] $$
$$\lim_{x \to 1 ^-} G(x) = \lim_{x \to 1 ^-} [\Delta_\mathrm{r} G^{\circ}x - RT + RTC] + \lim_{x \to 1 ^-} [RT(x \ln (x) + (1-x) \ln (1-x))] $$
Note (1):   $ \lim_{x \to 1 ^-}[(1-x) \ln (1-x)] = 0 $ (by l'Hopital's rule)
Note (2):   $\lim_{x \to 1 ^-} [x \ln (x)] = 0$
$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} - RT + RTC + RT(0+0) $$
$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} - RT + RTC $$
$$\lim_{x \to 1 ^-} G(x) = \Delta_\mathrm{r} G^{\circ} + RT(C-1) $$
These limits to G show that there is a value to G as you approach pure reactants and products along the curve of G vs ξ. The algebra shows some interesting points, which agree with other parts of the theory of equilibria and state functions:

The curve of G vs ξ can be thought of as the sum of a simple line between the G of reactants and the G of products $$G = \Delta_\mathrm{r} G^{\circ}ξ + RT(C-1)$$

for $C = 0$
$$G = \Delta_\mathrm{r} G^{\circ}ξ - RT$$
and a second function (which is always negative) that deals with the Gibbs free energy of mixing $$ + RT~(ξ \ln (ξ) + (1-ξ) \ln (1-ξ) - 1)$$
This is seen as a dip from the line, reducing the Gibbs free energy in the mixed system. This disappears as ξ approaches 0 or 1, as the system is becoming pure reactants or products:

For a reaction where K > 1 and $\Delta_\mathrm{r} G^{\circ} <0$ (such as the graphs in this example) the pure products are lower in G than the reactants, but the equilibrium mixture is lower still
$\Delta_\mathrm{r} G^{\circ}$ refers to the tangent of the curve at standard conditions ($Q=1$ and $ξ = 0.5$) and also to the slope of the straight line that connects $G_{reactants}$ and $G_{products}$. As ξ changes by 1 from reactants to products, $\Delta_\mathrm{r} G^{\circ}$ is also the absolute $\Delta G$ between pure reactants and products

Two short comments, since I did not inspect your math thoroughly yet. (1) Again, it's the slope that diverges at 0 and 1, not $\Delta G$ which you obtained by integration. (2) There is a small error which I spot right away in the $\Delta G^\circ$ term: should be $\times (\xi_f - \xi_i)$ but this gets buried in the integration constant.

@Night Writer. Thanks, I'll inspect and make changes.

I also zoomed and zoomed. The trick is to seeing the increasing slope is keep the aspect ratio constant, otherwise you don't see the larger and larger slope as you zoom in. If you express ξ as a negative power of 10, the slope is linear with the exponent as ξ approaches zero, so it is subtle. Another way of saying it is that in order to see extreme slopes, you have to zoom in $10^\text{extremely}$ and keep the aspect ratio or, when letting the graphing program choose the aspect ratio, pay attention to the distinct scale of the two axes.

Note that from your own analysis you can conclude that the initial value of the (integral) Gibbs free energy is $G_i=RT(C−1)$

Yes. I would guess that there is little significance to this absolute value (as it is a state function) beyond the fact that it differs from final G by Delta G std. Maybe this is also inherent in the constant of integration bring there too?

Nice! Of course, now your answer is much more complete than mine...

Logically, that would mean that absolutely pure reactants have infinite G (which doesn't seem right).

It's the slope (rate of change) that is infinite, not the actual property (Gibbs free energy).
$\Delta_r G$ (expressed sometimes as $\Delta G'$, sometimes you'll see just $\Delta G_m$ without subscripts indicating explicitly that this is the molar Gibbs free energy change for a reaction) is a partial molar quantity and describes the change in the Gibbs free energy per mole unit change in the reaction progress coordinate. It is a differential quantity (a slope).
If you want to know the value of the Gibbs free energy at any point during the reaction, what you actually want $^\dagger$ to compute at constant $T$ and $p$ is
$$G=\sum_i n_i \mu_i \tag{1}$$
where $n_i$ and $\mu_i$ are the number of moles and chemical potential of substance $i$.
In order to arrive at the expression for $\Delta_r G$ you need to take the derivative of equation (1) with respect to the reaction progress coordinate ($d\xi=\frac{dn_i}{\nu_i}$):
$$\Delta_r G =\frac{dG}{d\xi}=\sum_i \mu_i \left(\frac{\partial n_i}{\partial\xi}\right)_{n_j}=\sum_i \nu_i \mu_i$$
(where use was made of the Gibbs-Duhem equation).
With appropriate substitutions this last expression can be reformulated into the more familiar expression
$$\Delta_r G =\Delta_r G^\circ + RT \log(Q)$$
where
$$\Delta_r G^\circ = \sum_i \nu_i \mu_i^\circ $$
and
$$Q = \prod_{i} a_{i} ^{\nu_i}$$

$\dagger$ Want but can't as written. You instead compute $\Delta G$, the value as a difference relative to a reference state.

Mathematical aside

if the slope is infinite as ξ approaches zero, the line will never hit the y axis, so the y intercept should be infinite too

Consider the inverse problem, determining the change in G from its slope. This involves evaluating integrals of the sort $\int_{c_i}^{c_f}\log(c)dc$:
$$\begin{align} \int_{c_i}^{c_f}\log(c)dc = \left[c\log(c)-c \right]^{c_f}_{c_i} \end{align}$$
Now if one of the limits ($c_i$ or $c_f$) equals zero, it would appear at first glance to result in a singularity, but it doesn't. To see why we can evaluate $0 \times \log(0)$ by invoking L'H$\hat{\text{o}}$pital's rule:
$$\begin{align} \lim_{c\rightarrow 0} [c\log(c)] &=\lim_{c\rightarrow 0} \left[\frac{\log(c)}{1/c}\right] \\ &= \lim_{c\rightarrow 0} \left[\frac{1/c}{-1/c^2}\right]\\ &= \lim_{c\rightarrow 0} \left[-c\right]\\&=0 \end{align}$$

In the poster's answer an expression analogous to the following is provided for the integral of $\Delta_r G$:
$$G = G_i + \Delta_r G^\circ \xi + RT[\xi\log(\xi) + (1-\xi)\log(1-\xi)] $$
If plotted and expanded around $\xi=0$, this provides little evidence of the presumed infinite slope, thanks to the log dependence on $\xi$. The slope (free energy of reaction) near $\xi=0$ can be written
$$ \Delta_r G =  \Delta_r G^\circ + RT\log(\xi)$$
The logarithmic dependence on $\xi$ guarantees that the slope goes to $-\infty$, but that it does so slowly, as seen in the following plots (same parameters as in the poster's answer):

Thanks for the detailed reply. I know that the equation gives us change in G per mol of reaction proceeding (ξ) - i.e. the slope of the graph, but if the slope is infinite as ξ approaches zero, the line will never hit the y axis, so the y intercept should be infinite too. Please let me know what I am missing.

@Withnail - I edited my answer to give examples of finite functions with infinite slope - a half circle or U shape.

I appreciate the work (and from @Karsten Theis) in helping me understand. Looks like I need to study some more Maths! I will work on integrating the differential equation ΔrG=ΔrG∘+RTlnQ with respect to G and ξ to see if I can visualize the y intercept. This is a great forum and you both have been invaluable.

I searched for images that show the correct slope for no reactants or no products, and most "schematic" drawings have it wrong. Two exceptions: https://media.cheggcdn.com/media%2F76e%2F76e64fec-0d7d-4ac6-b35e-c3f47d7a90a6%2Fimage and https://commons.wikimedia.org/wiki/File:Entropy_of_Mixing.jpg @Withnail

I am working on a maths solution, I have a colleague helping with the calculus, and will post as an answer when we are done. I appreciate your help, the function of G vs ξ is quite interesting as ξ approaches 0 or 1 @Karsten Theis

Thanks for the graphs of G (and delta G) dependence on ξ. I will post the work on the limits when I get a sec.

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What is the most stable pear-shaped nucleus?
I have heard on the Internet that there are at least three pear-shaped nuclei: radium-224, radon-220 (a direct product of the previous isotope), and barium-144. CERN has a webpage about this discovery https://home.cern/news/news/experiments/first-observations-short-lived-pear-shaped-atomic-nuclei and it calls them "short-lived". Their half-lives are all less than one week and I wonder if radium-224 is the most stable pear-shaped nucleus or if there are more stable ones that have that property in common.

Probably better on Physics SE. Even as a long-time supporter of Glen Seaborg's nuclear chemistry contributions, this one really is nuclear physics.

I'm voting to close this question as off-topic because it's not about chemistry.

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Difference of Ascorbic acid dissolved or precipitated in Glycerol?
Difference of Ascorbic acid (AA) dissolved or precipitated in Glycerol?
Referring to pat US20110008445A1
Solution 1
At 25C, AA was dissolved in glycerol and alcohol (subsequently removed by distillation).
The AA was measured using scattered light (JIS K0101 using formazine as ref).
5 µm sized particles of AA were found to be dissolved in the glycerol.

Solution 1a
Solution 1 was aged at 28C for 10 days.
10% AA remained dissolved
90% AA precipitated
25 µm sized particles of AA were found to be suspended in the glycerol.

Solution 1b
Solution 1 was aged at 15C for 10 days.
10% AA remained dissolved
90% precipitated
80 µm sized particles of AA were found to be suspended in the glycerol.

Analysis
A key aspect of the experiment focusses on AA being in two different states:
Dissolved and Precipitated
The implication is that the AA is somehow different, otherwise why mention dissolved and precipitation?
Also, the AA was initially dissolved in glycerol and alcohol.
Therefore the AA was

Dissolved in Glycerol and Alcohol
Dissolved in Glycerol
Precipitated and suspended in Glycerol

Question
Does Ascorbic Acid remain in its usable form, in all three scenarios?
('Usable' relates to applying this concept, of dissolving the AA in Glycerol and Alcohol, as a precursor to AA encapsulation in lecithin emulsion).

My short answer is yes. L-Ascorbic acid is stable, in mostly neutral and basic conditions. And, I assure that the particle sizes are given in $\pu{\mu m}$ (See abstract of the patent application and SEM figures).

Thanks for that @Mathew Mahindaratne. With the 1000's of words, confirmation of units was missed :( I've ordered glycerine, a high shear (rotor stator) homogeniser & am now tracking down a high HLB ester supplier - not easy in small qty. Will be using de-oiled sunflower lecithin powder & fine milled ascorbic acid. Will prototype a contact sonificator, as bath method is poor - I have a spare head :) Probably a month to get together. Do you want to put your answer up confirming that the ascorbic acid remains usable when dissolved in glycerine - and await potential disagreement :) Thanks again...

Could you separate the patent text and your commentary in the question? Otherwise, it is hard to read, and I don't expect anyone to go and read the actual patent.

@Karsten Theis - Thanks for taking the time on my question. Hopefully it is now easier to understand. I think that we already have the answer, but it was worth the effort to present the question with greater clarity. This experience will no doubt help with future questions :)

To acknowledge OP's request, I am putting this as my answer:
What OP's initially referring to (US20110008445A1) was a patent application. Since it was a 2011 application, I searched for the patent and found it (Ref.1). It's abstract states that:

A Suspension of ascorbic acid in glycerol or in glycerol comprising diglycerol, in which the content of ascorbic acid is 13% by mass or greater, and further in which 8 to 12% by mass of the ascorbic acid is dissolved in glycerol or in glycerol comprising diglycerol, and the rest of ascorbic acid is precipitated in the form of microcrystals having a particle diameter of $\pu{25 \mu m}$ or smaller and is uniformly dispersed in the suspension. The suspension of ascorbic acid in glycerol is useful as a base material for cosmetics containing ascorbic acid which exhibits excellent feel in the use (spreadability and Smooth feel on application to the skin).

The patent described the preparation of ascorbic acid suspentions in alcohol or glycerol. Based on my knowledge in ascorbic acid, all forms of suspensions contain ascorbic acid in  in its usable form. However, one should be aware that ascorbic acid tends to have browning effect due to slow formation of related furan derivative on long standing (Ref.2).
References:

Mikiharu Okumura, Takanori Ohta, “Suspension of Ascorbic Acid in Glycerin and Process for Production Thereof,” United States Patent 2012, 8,211,472 B2 (https://patentimages.storage.googleapis.com/62/29/bf/f95fde91d869cd/US8211472.pdf).
T. Lalikainen, M. A. Joslyn, C. O. Chichester, “Browning Reactions, Mechanism of Browning of Ascorbic Acid-Citric Acid-Glycine Systems,” J. Agric. Food Chem. 1958, 6(2), 135–139 (DOI: 10.1021/jf60084a011).

Thanks for that answer & additional research. I will set aside some samples in a fridge and cupboard to monitor any colour change over time. I found a supply of Sucrose Stearate that can be tested as an aid to emulsifying lecithin. In the intervening period, other options will be procured. Thanks again for your time :)

Good. I'm sure it stays same for long lime. Good luck on your work though. :-)


text:
Isopropyl Alcohol leaves a white haze on plastic, is there anything I can use to clean it off?
Recently I wanted to clean up some lubricant on some keyboard switches. On some forums, an Isopropyl alcohol bath was recommended so I decided to give that a try and went to purchase some 91% Isopropyl from Target(purest I could find near me and I admit, I am a little impatient).
Good news - it worked in getting the lubricant off the switch! Bad news, on some of the switches, it left this white haze.
I'm not sure what it is, though from what I've been reading, it sounds like this is a common issue when using Isopropyl, my best guess is that it's some mix of the lubricant mixing with the alcohol and reacting poorly with the switch housing plastic given that I'm only seeing this haze on the bottom half of the switches which were the only parts of the housing that were lubed, the other parts are clean. Another reason for my guess is that this haze isn't appearing in the same amounts on all the switches if at all, some switch bottoms are basically clear.
It does appear to scrape off but looks like it will require a fairly strong brush which I don't have at the moment and I'm not sure if there's one available for purchase that could be used in this situation.
Is there any plastic-friendly household cleaner available that might be able to help speed up the process?

Try 1/2 and 1/2 vinegar with water and soak for 20 min. Then rub (soft cloth) rinse and dry. Should work.

Olive oil will take it all off with a dry cloth or q-tip after a few applications. It worked perfect for me on a dark marble designed plastic camper sink after 3 lighy wipes in one afternoon.

Even the common alcohol ( ethanol, ethylalcohol) is known to corrupt surface of some plastics, and rubbing alcohol ( isopropanol, isopropylalcohol) is even better in that, as it contains bigger organic chain and is less polar.
These alcohols do not dissolve the plastics in large extent, like acetone often does, but they may corrupt surfaces, especially if they are shiny or polished.
The typical case is compact polystyren or polymethylmethacrylate, that are often clear and shiny, where the white milky haze is very well visible. The opposite case is filled black shiny plastic where the haze and lack of shining  is notable.
For similar reason, lukewarm water with detergents is preferred for plastics cleaning.
The only way I see to get rid of the haze is polishing, or applying a selfpolishing  coating, if applicable and worthy. Perhaps polishing of wax coating.

Heat polishing with a heat gun works too.

Yes, I agree, I forgot about that. But it must be done very carefully, unless the heat gun is dedicated for such  polishing.

I did this by accident just today. Had gotten an almost new black PA sound system from Freecycle and wanted to wipe it down for germs and used alcohol (90% isopropyl) and it left lots of white streaks. I tried to clean it with soapy water, glass plus, detergent, all to no avail. Then tried comet and that may have helped a little and then I saw someone mention toothpaste and the earlier answer here speaking to polishing and I put the two together and recalled that's an approach for car lights. In any case, it worked like a charm. A couple rounds of scrubbing with a white toothpaste/toothbrush (Colgate) with a rinse in-between and then some buffing with a towel and the PA system is back to normal...so toothpaste is a good abraiter I suppose to assist in polishing.

A reason for using isopropanol to clean greasy surfaces is that it has a lower surface tension (21 dyne/cm) than water (72 dyne/cm). The same effect (reduced surface tension) can be achieved by adding 0.1% of surfactant to water.
The difference is that 0.1% of additive in a non-solvent hardly dissolves anything, ever, while 91% of a solvent is almost as good as the pure solvent.
A possible way to use isopropanol, which is a really good cleaner, is to add non-solvent (water) to decrease its solvent power, while hopefully not diminishing its surfactant action (reducing the surface tension of the water). In fact, adding just 20% of isopropanol to water will reduce its surface tension to 30 dyne/cm, which is a big reduction and will help cleaning a lot, but this will probably reduce the solvent power of the mixture by at least 80%. A quick wipe with a  cloth wetted with a 20% solution of isopropanol could remove grease, leaving only a little of the solvent mixture, which would quickly lose more isopropanol by evaporation.
Soaking in the mixture would tend to allow isopropanol penetration of plastic; a quick wipe minimizes the exposure. The isopropanol mixture does have an advantage over detergents in that detergents may leave some involatile residue if you use too much (remember 0.1%!), but isopropanol and water are completely volatile.
A very very light rub with a cloth very very lightly saturated with a mineral oil, petroleum jelly or silicone oil could fill in the voids created by partial solution or disruption of the polymer matrix and diminish or eliminate the haze.
Beautiful tables of surface tensions of mixtures of water and methanol, ethanol, and propanols at various temperatures are shown at https://pubs.acs.org/doi/pdf/10.1021/je00019a016.


text:
Henry's law p = kx vs m = kp
What exactly is the difference between the following equations?
$$p = kx \qquad\text{and}\qquad m = kp$$
Is the $k$ in both same, because I have seen same $k$ in textbooks for both equations?

Both are different Henry's law constants. See: https://en.wikipedia.org/wiki/Henry%27s_law

Henry's law constants also reflect an implicit scale: molality, mole fraction, concentration, etc. (with a given reference state defined as such).


text:
Why isn't 'chemically-strengthened glass' made with potassium carbonate to begin with?
Instead of making potassium-strengthened glass by creating ordinary soda-lime glass first, then replacing the sodium atoms/ions with potassium by putting the glass in a bath/solution of potassium nitrate, why not replace the 'soda' (sodium carbonate) in the initial process(es) with potassium carbonate?

Related: https://chemistry.stackexchange.com/q/77249/23561

@A.K. That question (and your answer to it) is actually much more interesting than this one.

The potassium is not added first because the potassium does not intrinsically make stronger glass, it is the substitution of a larger ion for a smaller one at the surface that does.
To understand why ion exchange strengthens glass, you have to understand why the ion exchange makes the glass harder. The process of ion exchange hardening is done at a temperature that allows ion diffusion, but disallows a reconfiguration of the glass structure or relaxation of bonds (i.e. below $T_{g}$, the glass transition temperature). When sodium atoms are replaced by potassium in glass, the potassium ions occupy a site that is sized for sodium. This substitution of ions creates a compressive force in the sites where it occurs. Since the process is diffusive, the compressive force is produced on the surface of the glass and the interior is put in tension. The compressive force at the surface causes any surface defects, namely scratches, to be in compression. These surface defects being in compression prevents their ability to grow or cause failure of the glass, thus strengthening the glass.
If you make a glass with potassium, you have potassium ions occupying potassium sites, thus no strengthening occurs. Thus the only way to chemically harden the glass is to substitute larger ions into the glass.

Preemptive Question: What about internal defects causing failure?

Gorilla Glass is made via the fusion draw process where molten glass flow over two sides of a platinum trough and meets at the bottom to be rejoined.  This process prevents many internal defects to the point that they are very rare and quality control can remove the few pieces with defects that are made.

Interesting, is there a way you could make softer glass by replacing potassium with sodium instead? I'm not sure what use that would have, but is it chemically possible?

Or would that actually also make the glass stronger by adding different kind of stress?

@KevinWells It would not.

Do you mean it wouldn't make it stronger or that it would not be possible?

@KevinWells potassium does not cause stress by being in the glass as any stresses present will relax above the annealing (or melting) temperature. It is only by ion substitution that strengthening will occur.

But just to be clear on the question here since the answer given seems to be talking about total substitution of K for Na:-  If this kind of glass needs x wt% substitution of K ions for Na then why not add a suitable amount of the K ions (or some oxide or carbonate of them) into the mix before melting ? Is there some chemical, structural, vitrification or other reason why GG glass can't be processed directly in this way rather than by a secondary post-melting process to infuse the K ions ?

@Trunk I don't understand, is this not in the answer?

If you make a glass with potassium, you have potassium ions occupying potassium sites, thus no strengthening occurs. This doesn't make much sense to me, an ex-materials scientist. If you mean to say that intimately mixing crushed solid $K_2O$ and $Na_2O$ and melting them together will simply produce a 2 phase oxide mixture on vitrification, then it's a poor way of saying it. There can't be "potassium sites" unless there is a putative phase based on a definite concentration of potassium ions after melting and solidifying.

The answer by DrMoishe Pippik (see below) makes much more sense. He's clearly saying that the diffusion process results in only a surface layer of the glass being infiltrated by potassium ions and this produces a compressive stress within that depth which naturally increases the tensile load needed to cause fracture in it. Perhaps your familiarity with this field made you assume that we all knew that the diffusion treatment would be limited in depth.

@Trunk I have edited the question for completeness. The answer by rMoishe Pippik makes incomplete sense as it does not discuss any mechanisms.

@Kevin If you replace by smaller ions, the glass will not become soft but very likely just brittle. Probably shatters even while you do it.

Chemically-strengthened glass is similar tempered glass in that the outside of the glass is under compression, while the inside is not compressed. If all the sodium in chemically-strengthened glass were to be replaced by potassium, there would be no difference in stress between the interior and the exterior layer.
How does this difference between layers help make the glass stronger? As in prestressed concrete, this makes the substance stronger under tension because the initial stress counteracts that force.

As A.K. notes, it is the substitution of larger atoms for smaller atoms that adds stress to the glass, making it stronger.  So you can't increase the stress by starting with a larger metal, but you could on the other hand start with a smaller metal:
Chemically strengthened lithium aluminosilicate glass (Int. Patent Application)

All credit is due to A.K., I just made some minor edits


text:
Alkyl insertion with cobalt and CO

Reaction of $\ce{(Cp)Co(Me)2(CO)}$ in the presence of excess $\ce{CO}$ yields two different organic products, $\ce{A}$ and $\ce{B}.$ The only metal-containing product is $\ce{(Cp)Co(CO)2}.$ The IR spectrum of $\ce{A}$ contains a peak at $\pu{1731 cm-1}$ whereas that of $\ce{B}$ has no peak in the carbonyl region. Write down mechanisms that explain the formation of $\ce{A}$ and $\ce{B}$.

From the IR, I think it is likely that an aldehyde is being produced, and looking at the reaction I assume it is the alkyl insertion of the methyl into the $\ce{CO}$ forming ethanal $(\ce{A}).$
$\ce{B}$ is confusing me however as the only other reactants are $\ce{Me}$ and $\ce{CO}$ (as the $\ce{Cp}$ remains in the metal product). Is some other reaction occurring (is $\ce{B}$ simply the enolised form of $\ce{A}$ formed from a β-hydride shift)?

I would think that at least one of the products would be acetone due to $\alpha$ migratory insertion and reductive elimination.

I concur with @Zhe. The other one, I would guess, is ethane. To form acetaldehyde you need a proton from somewhere to make the aldehyde C–H bond, and this system doesn't have it.


text:
Why is the nucleophilic substitution of OH- with 1-bromopentane slower than with 2-bromo-2-methylbutane?

Which statement about the reaction of a hydroxide ion with the organic reagent is correct?
A. 1-bromopentane predominantly follows an $\mathrm{S_N1}$ mechanism.
B. 2-bromo-2-methylbutane predominantly follows an $\mathrm{S_N2}$ mechanism.
C. Reaction with 1-bromopentane occurs at a slower rate than with 1-chloropentane.
D. Reaction with 1-bromopentane occurs at a slower rate than with 2-bromo-2-methylbutane.

I got the answer correct (D), but only by eliminating the other three options. However, why is D true? Does this mean that $\mathrm{S_N1}$ reactions are always faster than $\mathrm{S_N2}$ reactions when they are reacting with the same nucleophile?
I'd really appreciate some help.

There is no SN1 in D. The tertiary bromide in the presence of hydroxide will give E2 elimination. Whether the SN2 rxn. on the primary bromide is slower than the E2 is at issue.


text:
What is "detwinning" as a 2D crystallography technique?
The Phys.org article Iron selenide revealed as 'garden-variety iron-based superconductor' is quoted below. Normally I would hesitate to quote so much but in this case it all seems germane.
Question: What is "detwinning" as a crystallography technique? It seems a bit complicated for 2D crystals, with phrases like "... reasoning that the pressure needed to align the larger sample would also cause the layers of iron selenide to snap into alignment." it sounds like manual re-arrangement at almost the atomic scale.
The article links to Anisotropic spin fluctuations in detwinned FeSe in Nature Materials. (also in arXiv and Researchgate)

Tong Chen, a Rice PhD student "detwinned" iron selenide crystals by gluing them atop much larger crystals of barium iron arsenide. Using a 2014 method developed at Rice, the larger crystals are placed under pressure and detwinned, causing the smaller iron selenide crystals to also snap into alignment. Credit: Jeff Fitlow/Rice University

Physicists refer to this directionally dependent behavior as anisotropy or nematicity, and while structural nematicity is known to occur in iron selenide, Dai said it has been impossible to measure the exact electronic and magnetic order of the material because of a property known as twinning. Twinning occurs when layers of randomly oriented 2-D crystals are stacked. Imagine 100 baseball diamonds stacked one atop the other, with the line between home plate and second base varying randomly for each.

"Even if there is directionally dependent electronic order in a twinned sample, you cannot measure it because those differences average out and you wind up measuring a net effect of zero," Dai said. "We had to detwin samples of iron selenide to see if there was nematic electronic order."

Study lead author Tong Chen, a third-year Ph.D. student in Dai's research group, solved the twinning problem by cleverly piggybacking on a 2014 study in which Dai and colleagues applied pressure to detwin crystals of barium iron arsenide. It was impossible to apply the same method to iron selenide because the crystals were 100 times smaller, so Chen glued the smaller crystals atop the larger ones, reasoning that the pressure needed to align the larger sample would also cause the layers of iron selenide to snap into alignment.

Chen spent weeks creating several samples to test in neutron scattering beams. About 20 to 30 1-millimeter squares of iron selenide had to be aligned and placed atop each crystal of barium iron arsenide. And applying each of the tiny squares was painstaking work that involved a microscope, tweezers and special, hydrogen-free glue that cost almost $1,000 per ounce.

The work paid off when Chen tested the samples and found the iron selenide was detwinned. Those tests with neutron scattering beams at Oak Ridge National Laboratory, the National Institute of Standards and Technology, the Technical University of Munich and U.K.'s Rutherford-Appleton Laboratory also showed iron selenide's electronic behavior is very similar to that of other iron superconductors.

there is no neutron-diffraction tag, so I've added x-ray-diffraction and neutrons.

and so I've asked Tagging questions on diffraction of things other than X-rays?

Detwinning is not a novel technique; it's been around since 1940s. These days it's more often done computationally rather than treating crystals "mechanically".

@andselisk Thanks, that will be good to know once I understand what it is!

Well, the first question would be, do you know what a twin in a crystal is?

@JonCuster not in a 2D crystal I don't. I think an answer will have to at least touch on how twinning arises in 2D crystals.

@uhoh - if you know what it is in a 3D crystal, the extension (contraction?) to 2D is pretty straightforward and what you would expect. The twin is a line defect in the 2D structure with mirror symmetry between the two sides.

@JonCuster I am not sure if the paper is talking about that within the plane, or normal to it, (i.e. plane-to-plane) in the stacking of the planes.

@uhoh - good point. Although, if a "2D' crystal is thick enough to have twin planes I'm not sure I'd call it "2D" anymore.

@JonCuster and If they can be moved around relative to each other just by pushing on them, I'm not sure I'd call it "3D" yet either, thus I've asked this question to find out more about what's actually going on here.

@uhoh - well, that kind of is how twins in real 3D crystals are eliminated as well - you 'just' have to push one layer one spot over and now the layers all stack up 'properly'.

Given the lesser number of observables (compared to walking around a 3D crystal, even if it were one of elder Stoe imageplates offering only a $\phi$-scan) discerning merohedral and non-merohedral twinning in 2D crystallography still is as routinely possible by means of x-ray diffraction as for 3D samples?

TL,DR: Detwinning splits twins.
A crystal may consist of multiple domains, i.e. areas (2D crystal) or
volumes (3D crystal) sharing the same chemical composition (what atoms and
molecules occur), and crystal phase (think polymorphism as e.g. introduced
for anatase vs rutil).  Per sé, each of these domains can be seen as a small
crystal, i.e. with isotropic and anisotropic properties.  The spatial alignment
of these sub-units can be described relative to the coordinate system of
the large crystal, as well among each other.
In the model of mosaic crystals, the crystal consists of small grains;
their orientation (in respect to the coordinate system of the large
crystal) typically varies only a little, and the variation is random.
This contrasts to twins; here the orientation of the domains relative
to each other can be described by symmetry operations like a mirror
plane (leading to mirror twins), an axis of rotation (rotation
twins), a centre of inversion (inversion twins).  That is, you can
describe domains relative to each other and their atomic positions by
a transformation matrix, a twin law.  (An important tool solving and
refining a crystallographic models by X-ray diffraction.)  The mere
description of a crystal as twinned, however, does not describe (yet)
the number of of individual twin domains in the crystal, nor the ratio
of the volumes concerned:

(source: Britannica)
Detwinning then is a procedure to split the domains apart from
each other.  This can be attempted numerically at an analytical level
(at various stages analyzing diffraction patterns, for example), or
mechanically.  Depending on the case, this can be simple, difficult, or
impossible.  Diffraction intensities of different domains can overlap, or be separate of each other (merohedral / nonmerohedral twinning)**. The
mechanical approach can fail for (a non-exhaustive list):

sample crystal overall is too small to be manipulated
domains are too small to be easily recognized, e.g. by polarization
microscopy monitoring manipulation, or/and manipulated
twin domains can contact each others (easier case), or intersect and
mutually penetrate each other
Cutting crystals introduces mechanical stress.  A split along the
cleavage planes is an approach more gentle and hence less harmful for the
sample quality.  Yet the application of mechanical stress can introduce
additional domains, increase mosaicity, or render the fragments too small
for the X-ray single crystal diffraction equipment at disposition.
(However, diffraction analysis is not constrained to X-rays, (cf. vide
supra).  Crystals suitable for electron diffraction (not imaging*)
can be considerably smaller than the ones for in-house X-ray diffraction
analysis [down to µm scale].)

Finally the discern of 2D/3D crystallography.  Apparently, there is no
generally accepted definition how many µm along $(x,y,z)$ represent the
critical threshold between a film/a surface 2D crystal, and a volumetric
3D one. Probably better defined with the number of translations of the
unit cell along vectors $(a,b,c)$, I guess.
* Gruene, T.; Holstein, J. J.; Clever, G. H.; Keppler, G. Establishing
electron diffraction in chemical crystallography. Nat Rev Chem 2021, 5,
660–668. https://doi.org/10.1038/s41570-021-00302-4
** Herbst-Irmer, R. Twinning in chemical crystallography – a practical guide. Zeitschrift für Krist. Cryst. Mater. 2016, 231, 573-581. https://doi.org/10.1515/zkri-2016-1947

Instead of "separate of each other, or overlap (merohedral / nonmerohedral twinning" I think it should be "overlap, or separate of each other (merohedral / nonmerohedral twinning. Also, I think after detwinning, the crystals are 3D.

Interesting and helpful, thank you! The fourth sentence of the Nature Materials paper (I added more links to it) begins as follows: "By applying a uniaxial pressure along one-axis of the orthorhombic lattice to detwin the sample..." which is confusing, but I've asked "What is" and not "How does it work" so that will have to be asked separately (along with "What (the heck) is 'uniaxial pressure' in the context of detwinning?") Thanks for the idea of https://en.wikipedia.org/wiki/Mosaicity now I'm going to see if there's a "Difference between polycrystallinity and mosaicity?" question possible.

update: Thank you for "mosaicity" - now I'm finding all kinds of interesting material to read. This might be exactly what I need to read about in the real world (outside of SE).

@Karsten Re merohedral/nonmerohedral twinning: I agree, there was an edit.  And with a reference to one or Regine Herbst-Irmer's publications, there now is one of the more accessible introductions to the topic added, too. I equally agree on «if an intentional split a crystal along cleavage planes is possible, and one subsequently can stack fragments on top of each other, it is a 3D crystal»

@uhoh There has been an edit to the answer about twins/detwinning.  Former member of Sheldrick's group, much of Mrs Herbst-Irmer's work (both journal publications, as well as presentations and tutorials) dealt with the topic. Hence one of the references was added, too.

@uhoh "Uniaxial pressure" There are these jacks to lift a car (e.g. when replacing a flat tire). Column-like in shape, put orthogonal to the street's surface, there is only one vector to describe the force. Or a cell where the sample is compressed like in a vise.  If the crystal sample however were floating freely in a liquid, the application of pressure on the liquid would then exercise force on the crystal, now from all directions (top, bottom, left, right, you name it) from around the sample. Uniaxial in respect to the crystal's axes, too.

Thanks! I'll take a look in the morning. Okay, one can take a macroscopic force, divide it by the area of a surface over which it is applied, and call it uniaxial pressure at the surface. Inside that material though, the (what I want to call actual) pressure would be isotropic, no? Uniaxial pressure is a concept taught in school and used by jack sales people to sell more jacks, but "actual pressure" is still a scalar, isn't it?

@uhoh In absence of a surface / phase boundary, is there an internal pressure for a homogeneous sample? Preparing mashed potatoes, for example, I apply force in one direction on them ($p = \frac{\vec{F}}{A}$). Yet simultaneously to the compression, there is expansion of potatoes as a conservation of volume below the masher (let's assume they are not a gas, not a foam, nor a sponge and hence, compression is less likely). Based on this, "uniaxial pressure" better should be read as "uniaxial force" - don't you think so, too? And second, isn't a vector like force always uniaxial/unidirectional?

I think there is only pressure while the mashed potato "fluid" is moving - while it's a dynamical system. Since they are not confined to a volume, in a static situation they will always expand to relieve any internal pressure (except of course the tiny gradient due to gravity). So in the case of mashed potatoes, there is a pressure defined at the masher-potato interface specified by the force per unit area at that point, with the axis being the normal to the interface. After all this, I now understand what they mean; they apply a force.

I'll try to check tomorrow, but I don't think we can assume that this force does not induce any strain in the other two directions however. I don't know about mashed potatoes, but for this solid material there must be some kind of a Poisson's ratio...

I've added a 2nd bounty to I can award both answers.

@uhoh The article about Poisson's ratio wasn't anticipated.  Its implications (incl. orthotropic materials, and auxetic materials) provide an additional perspective on many topics, both on macroscopic scale (support structures architecture, construction, mechanical engineering), as well as on microscopic scale (I think piezoelectric crystals were an example).

The work is about the magnetic structure of the material, probed with neutron diffraction. Thus, the twinning disorder is not only in the positions of the atoms but also in the orientation of the nuclear spins.
Wikipedia states in an article on magnetic structure:

Such ordering can be studied by observing the magnetic susceptibility as a function of temperature and/or the size of the applied magnetic field, but a truly three-dimensional picture of the arrangement of the spins is best obtained by means of neutron diffraction. Neutrons are primarily scattered by the nuclei of the atoms in the structure. At a temperature above the ordering point of the magnetic moments, where the material behaves as a paramagnetic one, neutron diffraction will therefore give a picture of the crystallographic structure only. Below the ordering point, e.g. the Néel temperature of an antiferromagnet or the Curie-point of a ferromagnet the neutrons will also experience scattering from the magnetic moments because they themselves possess spin. The intensities of the Bragg reflections will therefore change. In fact in some cases entirely new Bragg-reflections will occur if the unit cell of the ordering is larger than that of the crystallographic structure. This is a form of superstructure formation. Thus the symmetry of the total structure may well differ from the crystallographic substructure. It needs to be described by one of the 1651 magnetic (Shubnikov) groups rather than one of the non-magnetic space groups.

Here is a conceptual sketch of the expected diffraction image (reciprocal space or Q-space) from a paper on a different material, NaFeAs (with the same senior author):

The green dots are the signals they are interested in. The red dots are from parts of the crystal where the spins have a different pattern. The uniaxial strain (squeezing in one direction rather than placing in a pressure chamber exerting pressure in all directions) suppresses the signals shown with the red dots, but in this case not completely.
For a general introduction about twinning (merohedral and non-merohedral), see e.g. this paper. What the Phys.org article is trying to say is that the 2D order is perfect, but in order to get 3D order, they applied pressure along one crystal axis. You can see this nicely in a perfect crystal of a barium-iron-arsenide crystal that undergoes a twinning transition at low temperature. Here is the description from a video showing the transition, with a screenshot below:

Structural "twin" domains forming in a barium-iron-arsenide crystal as it is cooled down to 5 K (- 451 °F). This video, using polarized optical microscopy, shows the formation of these domains, which are caused by a slight change of the positions of the atoms within the crystal.

As you can see from the titles of the journals where this work is published, this would probably get a better answer from a physicist familiar with solid state methods, magnetic properties and superconductivity.


text:
How does hyperconjugation explain stability of partial positive charges?
Recently I was reading this answer to the question Why does boron add to the less substituted carbon in the hydroboration of an alkene?.
See the transition state given in the above answer, and the reason given to explain it's stability.
As far as I know, hyperconjugation is a result of σ-p interaction in carbocations.
If partial charges are developed(as in the case of linked answer), there is no empty p orbital in them.
So how can hyperconjugation be used to explain their stability?

The solution is referring to the partial positive charge on the carbon opposite the developing C-B bond. It is stabilized by substituents similar to how a full carbocation is stabilized by substituents, e.g., in the tert-butyl carbocation. This doesn't need to be a completely empty p-orbital—electron density will be dynamically influenced by any imbalance in charge distribution—but if it's easier to conceptualize, you could imagine it as the pi-bond electrons being fully transferred to the C-B bond, leaving an empty p-orbital on the opposite carbon, which is stabilized by hyperconjugation.
The reality should be closer to a dynamic process at both sites, though. The pi-bond transfers electron density into the C-B bond, while the substituents stabilize the rising positive charge on the other carbon. As the C-B bond develops, the B-H bond simultaneously weakens, and begins moving its electron density toward the partial-positive carbon.


text:
Disproportionation of 2 KO₂ + 2 H₂O → 2 KOH + H₂O₂ + O₂
How is the following reaction a disproportionation reaction?
$$\ce{2 K\overset{-1/2}{O}_2 + 2 H2\overset{-2}{O} ->2 KOH + H2\overset{-1}{O}_2 + \overset{0}{O}_2}$$
In this, the OS are $-1/2$, $-2$, $-1$, $0$, respectively, but in disproportionation  the element$ O $both oxidizes and reduces. How do we know that this happens here? $\ce{KOH}$ may get oxidized, and $\ce{H2O}$ get reduced. How do we know the same compound has undergone both?

Compounds don't get oxidized or reduced; elements  do.

Note that it is adviced against MathJax or other intensive formatting in the title. Rather, plain text titles are preferred, if applicable. E.g."Disproportionation at potassium superoxide hydrolysis"

@Poutnik I will take a note of that

The disproportionation happens to elements, not to compounds.
In this particular case, to the part of oxygen,
$$  \overset{\ce{2 KO2}}{ 4 \times -0.5}=  \overset{\ce{H2O2}} {2\times -1} +\overset{\ce{O2}}{ 2\times 0}$$
while the other part keeps the oxidation state constant:
$$  \overset{\ce{2 H2O}} {2 \times -2}=  \overset{\ce{2 KOH}} {2\times -2}$$
$\ce{H2O}$ does not get directly involved in disproportionation.
$\ce{KO2}$ oxidizes and at the same time reduces  itself.
This is the nature of isproportionation, which happens , if the particular compound of the element in the intermediate oxidation state has higher Gibbs energy then particular compounds of this element in higher and lower oxidation states. Therefore redistribution of oxidation states is thermodynamically preferred.
Typical cases are $\ce{2 Cu^{I}->Cu^{II} + Cu^{0}}$ and $\ce{2 Hg^{I}->Hg^{II} + Hg^{0}}$

Thanks , but I got confuse with if $\ce{K\overset{-1/2}{O}_2}$ is getting oxidised to $\ce{O2}$ and $\ce{H2\overset{-2}{O}}$ to $\ce{H2O2}.$

No, it does not. H2O does not get directly involved in disproportionation. KO2 oxidizes and at the same time reduce  itself. This is the nature of isproportionation, which happens , if the compounds with the element in intermediate oxidation state have higher Gibbs energy then compounds in higher and lower oxidation states. Therefore redistribution of oxidation states is thermodynamically preferred.


text:
Number of hyperconjugative structures in aromatic systems
I know that to find the number of hyperconjugative structures in aliphatic systems we use the formula $a + 1$ (where $a$ is the number of α-hydrogens in that compound)
However, my teacher answered to the above question as:

No, this formula is not valid for aromatic systems.

Why is it so? What is the exact formula? Is it that we don't count its original structure as a hyperconjugative structure?

Because of the aromaticity which occurs in cyclic system. It means that you have already two limiting structures in which hydrogen are alpha-hydrogens (respect to the double bound which is once at left and once at the right).

@Alchimista alright, but then what's the exact formula?

What such a formula should be useful for? Which already is not that clear for aliphatic molecules, at least to me. One should analyse case by case and keep in mind is thinking of limiting structure, the weight of which might be even unrelated to their number. Understand what mesomers are and forget formulae to count things existing only within a precise frame. Anyway, I would say that for toluene one can draw 2 x the three structures, so 2n.

@Alchimista okay, I will keep that in mind, thankyou :)


text:
Glucose spatial arrangement
It was written in my textbook that the spatial arrangement of hydroxy groups was determined by Fischer by studying certain properties. But I wanted to know to how the relative configurations of hydroxy groups were determined.
Glucose is dextrorotatary but that could be possible with many arrangements.

The full description is probably too long for an answer, and you should look it up. The basic idea is that one can build bigger sugars from smaller sugars. Once you establish the stereochemistry of the smaller sugar, you can use it to deduce the stereochemistry of the more complex sugars by examining what larger sugars can built from smaller ones via simple reactions.

It's not certain properties, but a very careful observation if and with what products glucose reacts to other chemicals. Took many years. And even then Fischer did not know the absolute configuration, because at his time nobody knew the absolute configuration of anything.
Only when the first absolute configurations were known from X-ray diffraction experiments, one could start comparing.
Today we can calculate the absolute optical activity via quantum mechanics, but Fischer solved the Glucose structure in 1891, that's a quarter century before quantum mechanics.


text:
Ozonolysis of 1,2,3,4,5,8‐hexahydronaphthalene
Consider 1,2,3,4,5,8‐hexahydronaphthalene:

If one equivalent of ozone is added, then which double bond will undergo ozonolysis?
If it attacks the more electron rich one, then unstable 10 membered ring will form, and if it attacks the other,less sterically hindered one, stable 6 membered ring having 2 branched aldehydes will form.
What will actually happen? What factors decide this?

I would be interested to know how you propose to add 1eq of ozone. I suspect the less hindered double bond is likely to be attacked preferentially, but with such a reactive species as ozone I think you'll get a mixture as there is insufficient difference between the two double bonds.

@Waylander Since addition of ozone is electrophilic addition, shouldn't it be added to the more electron rich one? Also, the carbon under consideration is sp2 Hybridised, so attack from above and below is not hindered much.

There is not enough difference between the electron densities of the two bonds IMHO,

@Waylander Then will hyperconjugation play a role? The left bond on ozonolysis produces an alkene having more hyperconjugating structures.

A minor role, none of it adds up to a large enough difference to give preference.

@Waylander another question, what will happen when MCPBA or any other peroxy acid is added to this compound? Where will epoxidation take place?

The rate of epoxidation with MCPBA increases with the degree of alkyl substitution so the central double bond will react preferentially. http://ifasonline.com/chemicalScience_topic15.jsp   MCPBA is a much more controllable reagent than ozone.


text:
What happens if an orbital has no entry in a point group table?
A while ago I cited a few passages from a book on $f$-orbital contributions in metal complexes.
And although usually the $4f$-orbitals are too isolated to really do chemistry and the elctron density is mostly on the f-orbitals there should still
be at least some interactions with the ligands to a point where the degenerecy is destroyed.
What I'm currently facing is a mixed-anionic Ce(III) complex as capped square antiprism. I know this would be way better dealt with by using band structure calculations.
So this not going to be a question about the use of this method but rather a problem I faced when I tried to treat the complex like a $d$-orbital complex.
So for this let us consider the folling conditions:

The mixed-anionic situation would automatically create a non-symmetric polyhedron so we simply reduce it to 9 equal ligands $L$.
For Ce(III) we have a small gap between $4f$- and $5d$-orbitals. That is why for example Ce(III) or Eu(II) doped luminescent materials are so
dependend on their host materials.
Therefore we expect quite some correlation between the $f$- and $d$-orbitals. And according to what I read effects like first and second order Jahn-Teller
exist for Ce(III) complexes as well, but it becomes quite difficult to calculate them.

So we are talking about a qualitative picture here:
When I looked for examples for ligand field splitting in Ce(III) I found the other $CN$ = 9 isomer as tricapped trigonal prism:
That means that we are looking for a ${D_{3h}}$ point group:
$$\begin{array}{c|cccccc|cc} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 & x(x^2-3y^2) \\
\mathrm{A_2'} & 1 & 1 & -1 & 1 & 1 & -1 & R_z & & y(3x^2-y^2)\\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0 & (x,y) & (x^2-y^2,xy)& (xz^2, yz^2), x(x^2+y^2), y(x^2+y^2) \\
\mathrm{A_1''} & 1 & 1 & 1 & -1 & -1 & -1 & & \\
\mathrm{A_2''} & 1 & 1 & -1 & -1 & -1 & 1 & z & &   z^3, z(x^2+y^2) \\
\mathrm{E''} & 2 & -1 & 0 & -2 & 1 & 0 & (R_x,R_y) & (xz,yz) & xyz, z(x^2-y^2) \\ \hline
\end{array}$$
That means for the $d$-orbitals we expect a splitting into $A'_1$ + $E'$ + $E''$ and for the $f$-orbitals into $A'_1$ + $A'_2$ + $E'$+ $A''_2$ + $E''$
And when we look into the group's computational results that matches the point group:

Source: M. Kotzian, N. Rösch, J. Phys. Chem. 1992, 96, 7288-7293.
So what about my case? I am looking for the point group ${C_{4v}}$.
$$\begin{array}{c|ccccc|cc} \hline
C_\mathrm{4v} & E & 2C_4 & C_2 & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & 1 & 1 & z & x^2+y^2, z^2 & z^3, z(x^2+y^2) \\
\mathrm{A_2} & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_1} & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 & z(x^2-y^2) \\
\mathrm{B_2} & 1 & -1 & 1 & -1 & 1 & & xy & xyz \\
\mathrm{E} & 2 & 0 & -2 & 0 & 0 & (x,y),(R_x,R_y) & (xz,yz) & (xz^2, yz^2), (xy^2, x^2y),(x^3, y^3) \\ \hline
\end{array}$$
The $f$-orbitals can be described with two different sets, the general set and the cubic set leading to:
General Set: $f_{z^3}$, $f_{xz^2}$, $f_{yz^2}$, $f_{y(3x^2-y^2)}$, $f_{x(x^2-3y^2)}$, $f_{xyz}$, $f_{z(x^2-y^2)}$
Cubic Set: $f_{x^3}$, $f_{y^3}$, $f_{z^3}$, $f_{x(z^2-y^2)}$, $f_{y(z^2-x^2)}$, $f_{z(x^2-y^2)}$ and $f_{xyz}$
Now if I try to match them to the point group I won't be able to include the $f_{y(3x^2-y^2)}$ and the $f_{x(x^2-3y^2)}$ for the general set and for the cubic set.
What does that mean now? The first similar example I came up with is the $t_{2g}$ problem for σ-only $d$-metal complexes. As long as we don't have a π-bonded ligand that will give us the $t_{2g}$ symmetry
three d-metal orbitals will be unpaired and simply won't change in energy. But in that example they are still there. They have a representation in the point group but just no matching ligand.
Here we are talking about the opposite case, they would be there in general but they have no representation. So are they just left at the $f$-orbital energy level then? And how would I give them a Mulliken symbol in the end?
Sidenote:
I also tried a simple computational calculation using ORCA but I guess that I cannot simply print the orbitals for systems that involve $f$-orbitals and further more strong correlations. I tried to increase the base to a point where it also considered the $g$-orbitals to make sure that it would certainly include the $f$-orbitals but it also went up to the point where it calculated the $f$-orbitals for the oxide-ligand (I chose oxide as example). And in the end I ended up with an MO-diagram that had 300+ energy levels. So perhaps should be discussed much more qualitatively at this point.

The $x((x^2-3y^2)$ and $y(3x^2-y^2)$ both belong to E. The direct product of $B_1 \otimes  E = E$ in $C_{4v}$ point group.

Could you please elaborate how you knew that both were E? I guess since they are somehow degenerate this will give you E but why did you then multiply it with B1 here?

because the squared terms are $B_1$ in $C_{4V}$.


text:
Remove ethanol from gasoline by mixing it with water?
I have read on forums that ethanol can be removed from commercial gasoline by mixing it with water, then draining water from the bottom to remove the water-ethanol solution (for example, with a separatory funnel).
Does this really work? I would assume the ethanol is homogenized with the gasoline somehow, so would that interfere with it binding to the water?

The "homogenization" will cause a percentage of the ethanol to remain in the gasoline, but the rest will migrate to the water phase. You can repeat the extraction (separation) multiple times to reduce this percentage remaining. I don't know anything about what the resulting solutions will be usable for though.

Yes, it would work, but would not be very useful: the ethanol would be very dilute, and some water droplets would remain in the gasoline, making it a problematic engine fuel.

Please don't just create tags to summarise your question. Tags should be used to categorise the questions. There will not be enough questions (probably ever) to sustain an ethanol tag.

@Martin-マーチン There are hundreds of questions dealing directly with ethanol and have "ethanol" in their title. Maybe instead of deleting my tag, you could have tagged the 100 or so major questions dealing with ethanol with my tag.

Yes I could have done that, but this would have undermined the process in which we create and use tags. Just because ethanol is a very common chemical, it doesn't mean that the uses are similar enough to group them apart from what is already accomplished by an ordinary search. Tags should be applied to things that cannot be searched easily.


text:
Why does moving from left to right on the periodic table result in higher conductivity?
Question is in the title. I understand that the metals towards the left form stronger bonds with other chemicals compared to metals towards the right, and that the ones with weaker bonds break down easier and pass more current. (if that's wrong, please correct me) Why is this the case? Could someone please explain this concept to me more?

Al -> Si -> P -> S seems problematic for the assertion.

Actually, this only applies for metals. This trend does not apply for non-metals: non-metals form covalent bonds, which are due to the sharing of electrons - thus, there are no free electrons that can carry charge. Additionally, atoms are held tightly by covalent bonds and thus cannot dissociate into ions that carry electric charge.
With regards to metals, metallic bonding consists of metal ions surrounded by a sea of delocalized electrons - and it is these delocalized electrons that can carry charge. Each atom in a metal has the same number of electrons as the charge on the ion. For example, Na+ has one delocalized electron for one atom, Mg2+ has two, Al3+ has three, and so forth. So as the charge increases, the number of delocalized electrons per atom increases - and so the number of delocalized electrons in the metal overall also increases. This allows it to carry more charge per unit time, leading to a higher rate of conductivity.
Similarly with regards to ionic lattices, a greater charge means there will be more ions formed per molecule that is made molten. So one mol of NaCl will have 2 x 6.022 x 10^23 ions total, while 1 mole of MgCl2 will have 3 x 6.022 x 10^23 ions. So, MgCl2 has more ions to conduct electricity and thus shows greater conductivity.
I'm not sure if it has much to do with the actual 'ease' of breaking bonds. Both NaCl and MgCl2 would have to be in a molten state to be able to conduct electricity anyway, so I don't think it plays a role for ionic compounds. For metallic compounds, again, as they are simply ions surrounded by electrons there isn't any bond-breaking to be done.


text:
Why do Cu⁺ ions spontaneously form copper metal and Cu²⁺ ions in solution?

The standard electrode potentials for three reactions involving copper and copper ions are:
$$
\begin{align}
\ce{Cu^2+(aq) + e- &-> Cu+(aq)} &\quad  E^\circ &= \pu{+0.15 V} \\
\ce{Cu^2+(aq) + 2e- &-> Cu(s)} &\quad   E^\circ &= \pu{+0.34 V} \\
\ce{Cu+(aq) + e- &-> Cu(s)} &\quad   E^\circ &= \pu{+0.52 V} \\
\end{align}
$$
Which statement is correct?
A. $\ce{Cu^2+}$ ions are a better oxidizing agent than $\ce{Cu+}$ ions.
B. Copper metal is a better reducing agent than $\ce{Cu+}$ ions.
C. $\ce{Cu+}$ ions will spontaneously form copper metal and $\ce{Cu^2+}$ ions in solution.
D. Copper metal can be spontaneously oxidized by $\ce{Cu^2+}$ ions to form $\ce{Cu+}$ ions.

I thought that because copper is more likely to get oxidised than $\ce{Cu+},$ it is a better reducing agent and thus B is the answer. But the given answer is C. Why?

See related https://chemistry.stackexchange.com/questions/115660/disproportionation-of-2-ko%e2%82%82-2-h%e2%82%82o-%e2%86%92-2-koh-h%e2%82%82o%e2%82%82-o%e2%82%82/115662#comment218026_115662

@Poutnik - thanks, yes I did. But that does not explain why in terms of the standard electrode potentials involved. I'm having a hard time understanding how you could infer that from the SEPs given. Do you have any idea? Thanks in advance.!

Did you compare standard potentials of all 3 half reactions ?

as they are directly related to delta G via nF factor.

Strongly related; https://chemistry.stackexchange.com/questions/42855/why-cu-is-unstable-in-aqueous-medium

Take it as if Cu(I) had been 2 different compounds in one. One as oxidizing agent, the other as reduction agent. Compare potentials for their half reaction.

@Poutnik I got it!!! Thank you!!! Essentially, both reactions involving Cu+ have the lowest and highest SEPs respectively - so they have to be the strongest oxidizing/reducing agents - thus undergoing disproportionation. :)))

Both reactions involving $\ce{Cu+}$ have the lowest and highest standard electrode potentials, respectively. Thus, $\ce{Cu+}$ has to be the strongest oxidizing/reducing agents, and therefore it undergoes disproportionation.
Because the formation of $\ce{Cu}$ has a higher SEP than $\ce{Cu+},$ it is more likely to get reduced and therefore is a better oxidizing agent — not a better reducing agent — than $\ce{Cu+}$ (which is more likely to get oxidized as it has a higher SEP).


text:
How do you get contents of large flask into a smaller separatory funnel?
Separatory flasks that I use have a ground glass joint, but it is not clear how that joint is to be used. Videos I have seen just show people pouring stuff into the throat of the funnel.
Normally the situation I have is that the solution is in a large flask, maybe a 1000ml flask, and I need to separate a thin layer of the solution that is at the bottom of the flask using a 500ml separatory funnel. The question is how to get the bottom of the flask to the funnel.
Saying "oh just use a bigger funnel" is not a solution, because the flasks, reactors and carboys I have tend to be significantly larger than any normal separatory funnel. For example, if I have a 3000ml reactor or 5 gallon carboy with a solution in it, I am not going to go looking around to spend hundreds of dollars on a gigantic separatory funnel.

I'm not sure I understand the question, but why not carry out separation in batches? Or, if the liquids are highly immiscible, why not to try to get rid of some volume of the upper layer by decantation? // Also, I'm not sure how to interpret the part about the ground glass joint — are you asking what it is for in general, or whether it can be helpful for processing separation of large quantities of liquids?

Have you considered syphoning? It can be very useful at times, especially in large scales.

@NicolauSakerNeto Right, I would think the natural use of the ground joint of the separatory funnel is to attach it to some kind of siphoning system, but I don't know the proper way to set that up, so that is why I am asking.

@andselisk Part of my question is to ask what the purpose of the ground glass joint is on a separatory funnel. Every single separation setup I have seen, either in books or on YouTube makes no use of this joint (except for attaching a wide mouth funnel), so what is its purpose?

@ShakaBoom From what I know, the ones with the joint can not only be used as separatory funnels, but are also usually used as dropping funnels, e.g. by attaching it to the round-bottom flasks directly or via Claisen adapter to carry out reflux or inert atmosphere synthesis.

Could you want to remove most of the upper liquid by syphoning or pumping? Or do you want to transfer the bottom liquid without removing the top solution?

You should proceed as follows:

Charge the funnel with the mixture until it is almost full.
Let the content settle.
Separate the lower phase.
Separate the upper phase.
Repeat until you have processed all the mixture.

The ground joint is to close the funnel hermetically.
This is necessary when you need to perform an extraction.
A liquid with a mixture of components is charged in the funnel and an immiscible liquid is added.
The funnel is closed with a stopper and it is vigorously shaken.
BE VERY CAREFUL BECAUSE PRESSURE CAN DEVELOP INSIDE THE FUNNEL.
It is better to shake a bit first, take off the pressure, shake again, etc.
To release the pressure, put the funnel upside down and open the valve.
Put the funnel upright and let settle.
When there are two clean phases, release the stopper and take them out separately.
The components will be distributed between the phases in different quantities, so it is a way to purify those substances.

does not answer the question

It answers some of your questions. I read it and learned something.

I did too. I would instinctively have removed the plug at the top joint to let out some pressure, thought after reading this I'll always use the valve, which is more practical and almost seems self-evident now.

The ground glass joint at the top of the separatory funnel is for a ground glass stopper. Occasions arise where you have a homogeneous liquid which you wish to extract. You put this liquid into the funnel, add the extraction liquid, stopper the funnel, shake (carefully...), then separate the layers.
The pressure development is frequently a result of using a volatile solvent like ether to extract organics from an aqueous emulsion or possibly of gas production (like CO2).
In your case, where you already have two layers (let's say, 800 mL over 100 mL in a 1 L flask), it should be relatively easy to slowly pour off 400-500 mL into another vessel, leaving much less liquid to put in the separatory funnel. This is comparable to siphoning, but may be easier. If any of your desired underlayer gets carried over, you can do the separation on this portion of the batch also.


text:
Reaction between an acid and acid
My initial thought was that the stronger acid will act as an acid and the weaker acid as base like the reaction between $\ce{H2SO4}$ and $\ce{HNO3}$ but in this reaction there is a lone pair available for protonation, what if the lone pair was not present , will the reaction still occur? Is the answer dependent on what the acids exactly are?

Can you give an example of acid where the lone pair is not present?

I meant an acid in which lone pair is unavailable like in Ammonium ion(NH4+).

OK, then there will be no reaction of this type.

Being an acid and ability of acting as base are 2 independent properties. The   same for bases and their ability to act as acids.

So a stronger acid cannot protonate a weaker one(in general)?

You can, but you will need one of the acids as pure solvent; a solvent like water will just absorb the protons. For example, sulfuric acid or hydrochloric acid in glacial acetic acid. If you only want to consider protonation of acids bearing a + charge like ammonium, then this will not exist.

related https://chemistry.stackexchange.com/questions/112087/why-doesnt-h%e2%82%84o%c2%b2%e2%81%ba-exist

@IvanNeretin There could be if strong enough acid is used (and the other "acid" is good enough base).

In the case of $\ce{HNO3}$ and $\ce{H2SO4}$, the proton from sulphuric acid protonates the $\ce{-OH}$ group of $\ce{HNO3}$ , which gives $\ce{NO2+}$ , $\ce{HSO4-}$ and water.
This mixture of nitric acid and sulphuric acid is widely used as a nitrating mixture for nitration of aromatic compounds.
In case of other acid mixtures, no reaction happens, but there is an equilibrium. The stronger acids dissociates almost completely , decreasing the dissociation of the weaker one by common ion effect.


text:
Why is the pH level of carbonation loss decreasing then increases afterwards?
We have done an experiment of the titration of unsealed sprite soda to find the amount of carbonic acid being loss on various duration (0 to 1 hour), due to decrease pressure.
The calculation was to determine the pH of carbonic acid overtime, thus the number of concentration of flat and unsealed soda are subtracted and is converted to pH level using -log[concentration of flat soda - unsealed soda].This was done multiple times on multiple duration until 1 hour. Now the results has demonstrated that carbonic acid pH level was decreasing in the first 30 minutes but then afterwards it was increasing. This contradicts on what my hypothesis is as it states that due to pressure being decreased (by unsealing the soda) the pH will increase turning more basic. Is this due to the factors of pressure and temperature? Can you please describe this thoroughly? Thank you for your time on reading this.

Was that pH observation before the titration or after it ?

The pH observation was after the titration sir.

Is not the pH computation ignoring citric acid content ?

Plus, is not it ignoring the strength of the carbonic acid ?

Is it a question of kinetics ?

At the beginning, there was titrated mostly just the citric acid and the equilibrium amount of carbonic acid.
After that, carbon dioxide slowly converted to carbonic acid and $\mathrm{pH}$ has been decreasing.
Finally, the long term process took dominance as dissolved carbon dioxide together with carbonic acid were  leaving solution as gaseous carbon dioxide and $\mathrm{pH}$ have been raising.

Sorry I haven't specified the description clearly, we have titrated both the flat soda (where all carbonation is loss) and the soda that is titrated in various duration. The amount of moles has been determined for both of the flat soda and the 'duration soda' and was subtracted, resulting the moles of only carbonic acid. This was then converted to pH level and was done multiple times for various duration. I am very sorry that I wasn't clear on this.

You know, the prerequisite for good answers is a good question. If others have to guess, imagine or interpret what is the real question, answers are rarely what was wanted and clarification wastes resources on all sides. :-)

Converted to pH level ? I thought it was measured... Do you mean estimation of pH of sprite soda drink before and after titration ? The best would be, if you rewrite the question as a detailed description of experiment, with  raising the clear questions about experiment details in a particular experiment stage.

I will do that. Thank you for your time


text:
Chemical Structures as Graph Labels in R
I asked the following question here https://stackoverflow.com/questions/56236135/chemical-structures-as-graph-labels-in-r but I ask it here for a different perspective.
I want to make a plot that labels a point with the chemical structure represented by the value at that point.
In the example below I have a plot of melting points which are labelled by the smiles strings of the molecules.
chem_data <- c(40.5, -6.3, -44)

names(chem_data) <- c("c1ccc(cc1)O", "Nc1ccccc1", "Fc1ccccc1" )

plot(chem_data, main="Example Plot\n Labelled with Smiles String", ylim=c(-50,50), xlim=c(0,4),
     ylab="mp /°C")

text(chem_data, labels=names(chem_data), cex= 0.7, pos=1)

How do I label the points with the chemical structures rather than the smiles strings?
Can R generate the image of the chemical structure from the smiles string?

It won't get your data points "labeled" by SMILES-based structure pictures yet by it seems CheminerR either alone, or working together with openabel is able to convert similes into pictures as by this post: https://www.bioconductor.org/packages/release/bioc/vignettes/ChemmineR/inst/doc/ChemmineR.html.  An other post suggests  rcdk and rpubchem (http://www.cureffi.org/2013/09/23/a-quick-intro-to-chemical-informatics-in-r/). (I have no experience at all with R so far.)

thanks so far ... I was thinking of rcdk but I was hoping that others would point me in the right direction.


text:
Implication of negative order on kinetics
In the reaction
$$\ce{2 O3 -> 3 O2}$$
the order of reaction that I have been taught is
$$\ce{O_3 \underset{$k_{-1}$}{\overset{$k_{1}$}{<=>}} O_2 + O}$$
$$\ce{O_3 + O \overset{$k_{2}$}{->} 2O_2}$$
$$\mathrm{rate} = k_{2}[\ce{O3}][\ce{O}]$$
$$ k_{1}[\ce{O3}]= k_{-1}[\ce{O2}][\ce{O}]$$
So $$\mathrm{rate} = \frac{k_{1}k_{2}}{k_{-1}} [\ce{O3}]^2[\ce{O2}]^{-1}$$
What does this $[\ce{O2}]^{-1}$ part mean? For example, initially when there is no $\ce{O2}$, the rate should be $\infty$; how is that possible? Or is it normal and I should accept it?

There is no "there is no". In reality, the concentration of anything anywhere is never exactly 0, so don't bother about the possible singularity.

It means that the reaction you quote is the overall reaction, a summary if you like, but in reality there are other reactions occurring, perhaps  $\ce{O3 \to O2 + O^.}$, and when the full scheme is worked out the rate law is more complex.

"The order of reaction that I have been taught" This is a statement that worries me a lot. The two ways I interpret this are (1) you were taught to derive the rate law in a certain way or (2) you were taught this specific rate law. The former statement has problems because the derivation would require knowledge of elementary steps, which are not provided here. The latter doesn't make sense because you can't teach a rate law for a reaction; you can only posit a mechanism consistent with experimental data and derive the associated rate law.

@Zhe The way we derived it was , the teacher gave us the mechanism. Told us what the Slowest step is and derives it from that

Arrhenius-type rate equations are never valid at extremely low concentrations or extremely small samples, because they implicitly assume number of molecules is a continuous quantity. This is true even with idealised single-step processes. The discrete nature of matter requires statistical corrections when there aren't billions and billions of entities in your sample. This has the side-effect of removing any mathematical singularities, like the one you've encountered.

What does this $[\ce{O2}]^{-1}$ part mean?

It means the higher the concentration of oxygen, the lower the rate. The two-step mechanism you show relies on oxygen atoms (radicals) to react with ozone in the second step. Dioxygen acts as a radical scavenger (first reaction, reverse direction). So if dioxygen concentrations are high, the reaction will not proceed via this mechanism.
When ozone is produced as part of smog in the lower atmosphere (i.e. the air we breathe), this mechanism is not expected to help lower the ozone concentration. The equilibrium of the first reaction would be too far on the side of the reactants.
In the higher layers of the atmosphere, however, this sequence is more feasible. Concentrations of all species are extremely low, so bimolecular reactions are slow as well. In fact, in the image linked below, concentrations are given as molecule per mL (as apposed to mol / L) and rate a similarly expressed as molecules reacting per time unit.
https://www.ems.psu.edu/~brune/m532/meteo532_ch7_stratospheric_chemistry_files/image018.jpg
What actually happens in the ozone layer is a bit more complicated than the two step process you show, see for example https://www.ems.psu.edu/~brune/m532/meteo532_ch7_stratospheric_chemistry.htm

For example, initially when there is no O2, the rate should be ∞; how is that possible? Or is it normal and I should accept it?

See the excellent comment by Nicolau Saker Neto. Another way to say this is "the molecules don't know the concentration of species, but they still do their thing without worrying about division by zero".

The scheme gives $rate=k_2\ce{[O3][O]}$ and $\displaystyle \frac{d\mathrm{[O]}}{dt} =k_1\mathrm{[O_3]}-k_{-1}\mathrm{[O_2][O] } -k_2\mathrm{[O_3][O] }=0$ for steady state on the oxygen atom.
This gives the rate expression as:  rate=$\displaystyle \frac{k_1k_2\mathrm{[O_3]^2} }{k_{-1}\mathrm{[O_2]}+k_2\mathrm{[O_3]}}$.
Thus the problem when the oxygen concentration is zero does not arise, which must be the case as initially pure ozone exists and then its rate of decomposition is $k_1\mathrm{[O_3]}$. At enormously high oxygen molecule concentration the dissociation is restricted because of the equilibrium step.


text:
How do α,β-unsaturated acids undergo decarboxylation?
I know the concerted mechanism for β-keto acids, but neither could I figure out, nor was I able to find out the mechanism for α,β-unsaturated acids.
Any help is appreciated.

Unanswered duplicate: Decarboxylation mechanism and stability of α,β-unsaturated and α-hydroxy acids.

Albeit carboxylic acids and their derivatives lose carbon dioxide under a variety of experimental conditions, the literature on the decarboxylation of $\alpha,\beta$-unsaturated carboxylic acids is not common. The reason for this is all $\alpha,\beta$-unsaturated carboxylic acids seemingly do not undergo decaboxylation under mild conditions, except for those containing substitutions at $\beta$-$\ce{C}$ position, which stabilize the intermediate carbocation.
A wide variety of experimental conditions undoubtedly make the reaction to undergo by many different mechanisms. For example, the role of acid catalysis in the decomposition of cinnamic acid (and related compounds) has been clearly demonstrated in Ref.1.
The mechanism for this acid catalyzed decarboxylation is illustrated in the following figure (see I in the diagram). The structures of some example of these susceptible substrates are also included (Ref.1) in the diagram:

Some carboxylic acids undergo decarboxylation by pyrolysis. It is an experimental fact that the olefins resulting from pyrolysis of simple $\alpha,\beta$-unsaturated acids (note that simple means at least no substitution on $\alpha$-$\ce{C}$) are homogeneous and contain terminal unsaturation (Ref.2). Two different mechanisms leading to the same olefin are formulated in Ref.2. The first one is essentially the same mechanism suggested for the decarboxylation of cinnamic acid and related compounds (see I in the diagram).
Before discuss about the second mechanism, let’s look at the proposed mechanism for $\beta,\gamma$-unsaturated carboxylic acid compounds (see II in the diagram). The decarboxylation mechanism of $\beta,\gamma$-unsaturated carboxylic acid compounds undergo the reaction through a cyclic transition complex to give alkene products. This mechanism mimics the one proposed for the thermal decomposition of $\beta$-keto acids.
As a consequence, an alternative mechanism for thermal decomposition of $\alpha,\beta$-unsaturated carboxylic acid compounds was also proposed if the substrate contains allylic $\mathrm{sp^3}$ carbon to extend the double bond (see III in the diagram). It was assumed that under thermal energy, substrate first isomerizes to the $\beta,\gamma$-isomer, which then decarboxylates through a cyclic transition complex to give the terminal alkene product as given by mechanism I.

References:

W. S. Johnson, W. E. Heinz, “The Acid-Catalyzed Decarboxylation of Cinnamic Acids,” J. Am. Chem. Soc. 1949, 71(8), 2913–2918 (DOI: 10.1021/ja01176a098).
R. T. Arnold, O. C. Elmer, R. M. Dodson, “Thermal Decarboxylation of Unsaturated Acids,” J. Am. Chem. Soc. 1950, 72(10), 4359–4361 (DOI: 10.1021/ja01166a007).

The decarboxylation of cinnamic acid 1 and related structures has been studied by Johnson and Heinz. The reasonable mechanism below was suggested.

W. S. Johnson and W. E. Heinz, J. Am. Chem. Soc., 1949, 71, 2913.


text:
Why does the following "Battery" experiment work?
I did the following experiment:
Experimental setup:
Two electrodes (one made out of Iron,the other made out of Copper)
are put into a (bluish) Copper(ii) sulfate lotion.

Then, one connect a voltmeter to them to measure the potential.

Finally, one disconnect the voltmeter and connect an Amperemeter to measure the electrical current.

Observations:
-Copper accumulates around the iron electrode,forming a thick copper coloured slick around it.
-there is a potential of about 0.5 V between the electrodes, which stays for several minutes.
-when one connect the amperemeter, one can measure a current which stays at least for a few seconds. (didn't measure further)
-there is no (visible) activity around the copper electrode.
-the solution doesn't change its colour notably.
-especially, one could not observe any evolution of gas. (although this does not mean that there is absolutely no gas evolution)
How is the measured tension and current caused?

What's unclear about it? This is a simple galvanic cell.

@Ivan Neretin : There is just one chamber! AFAIK galvanic cells have two chambers.

All right, this is a bad  galvanic cell. So what? The reactions are the same.

Could you give a detailed description of the reactions going on?

Could you  do the same for a normal galvanic cell with two chambers?

@Ivan Neretin : Yes. I could do that for a normal galvanic cell. But this bad one is hard to understand.

Now imagine another situation: no wires, just an iron rod in the CuSO4 solution. Surely you must have heard about the substitution reaction which will take place here.

Let us continue this discussion in chat.


text:
Is there dye that attaches only to human DNA?
I am looking for simple way to tell if blood stain is from human blood or some other animals. I know there are fluorescent DNA dyes, but is there dye that only attaches to human DNA?

I highly doubt it: human DNA and DNA from other species are very similar in structure, and in some cases (eg: primates) are very similar as sequences as well.
I think that "complex" analytical techniques are necessary in order to tell human DNA from animal DNA.

It would have to be a superintelligent dye.


text:
Acid-Base Reaction in MacGyver Episode "Countdown"
This question is about the old MacGyver show, Season 1, Episode 14, "Countdown." Could someone please explain how/why MacGyver neutralizes the acid in the bomb? He claims it is an electrolyte switch, and proceeds to neutralize it with oven cleaner and milk before draining it out.

wouldn't the milk act as a buffer?
wouldn't the neutralized solution also contain electrolytes?

Thanks!

Can you present the relevant dialog from the episode? There isn't much to go on here. MacGyver sometimes had accurate science, but they never let accuracy get in the way of drama.

Some oven cleaners contain sodium hydroxide, which is a base

It is determined that the bomb has an "unstable electolyte fluid trigger," consisting of an acid solution. MacGyver: "The fluid carries a current. Now our problem is to neutralize it."

It is determined that the bomb has an "unstable electolyte fluid trigger," consisting of an acid solution, that if released, enters some wiring, creates a short, and sets off the bomb. MacGyver: "The fluid carries a current. Now our problem is to neutralize it."  He mixes the oven cleaner and milk and injects it into the acid solution, which turns cloudy at first. Eventually the mixture turns clear, he drains it out, and shuts down power to the bomb.

Does he take a chug of milk first? Knowing McGyver he probably does. It's good for you: http://ansci.illinois.edu/static/ansc438/Milkcompsynth/milkcomp_minerals.html

The answer to your first question is yes, to the second maybe.
You are right that milk can act as a buffer, as explained for instance by this page from the Food Science Department of the University of Guelph:

This leads to the concept of buffer capacity, which is an important principle in cheese making. The effect of protein removal on the TA of whey, is related to the ability of protein to 'buffer' the milk against changes in pH. That same buffer property is the reason it helps to take acidic medication, like aspirin, with milk.
  [...]
  The two most important buffer components of milk are caseins (buffer maximum near pH 4.6) and phosphate (buffer maxima near pH 7.0). The buffer maximum near pH 5.0 is extremely important to cheese manufacture because the optimum pH for most cheese is in the range of 5.0 - 5.2. As the pH of cheese is reduced towards pH 5.0 by lactic acid fermentation, the buffer capacity is increasing (i.e., each incremental decrease in pH requires more lactic acid). The effect is to give the cheese maker considerable room for variation in the rate and amount of acid production. Without milk's built in buffers it would be impossible to produce cheese in the optimum pH range.

On the other hand denaturation of the milk proteins might act to form a gel that could perhaps help stabilize the trigger fluid, but milk gelation appears to display a complex pH dependence and a mechanism of stabilization is unclear.
The electrolyte concentration might be diluted by milk, since its ionic strength is surprisingly moderate, see Ref (1) for instance:

Taking into account these different associations, the diffusible fraction of milk at pH 6.6–6.7 appears to be supersaturated in calcium phosphate (Tab. V), and to have an ionic strength of about 80 mM.

Much of the salt is bound to protein which forms part of the buffering system. So much depends on the composition of the sinister trigger fluid.
The substantial concentration of divalent calcium (~1 g/L or ~50 mM) might be useful, but how will presumably also remain a well-kept secret of the writers privy to the composition of the very unstable trigger electrolyte fluid.
Reference

Frédéric Gaucheron. The minerals of milk. Reproduction Nutrition Development, EDP Sciences,2005, 45 (4), pp.473-483.


text:
Why does the rate of the reaction decrease with an increase in temperature?
Why does the following reaction
$$\ce{2NO + O_2 -> 2NO_2}$$
have negative reaction coefficient, i.e., why does the rate of the reaction decrease with an increase in temperature ?
I have been taught that this is the only reaction that has this property.

I find it extremely unlikely that this is not only reaction in the universe with this property. Looking at the free energy of the reaction, and using $G = H - TS$, contemplate the effect of $T$ on the reaction.

@JonCuster How does the thermodynamic equation of G equate to kinetics

When this occurs it is because the reaction is more complex than the stoichiometric equation suggests. If there is a ratio of rate constants each of the Arrhenius form rate coefficient = $A_1e^{-E_1/RT}/A_2e^{-E_2/RT}  \sim e^{-(E_1-E_2)/RT}$ then the measured value can increase of decrease with temperature depending on the relative size of the two energies $E_{1,2}$

@Harsh Wasnik, the question refers to the rate constant not the equilibrium constant or reaction rate.

@porphyrin exactly

I believe my answer is very relevant to you: https://chemistry.stackexchange.com/q/95343/44877

It's strange, but there are mechanisms whereby this type of thing can happen.  Basically you have a rate determining step that involves an intermediate species, and that rate determining step must have the usual positive temperature dependence.  But the preceding equilibrium that produces the intermediate species could have a negative temperature dependence that dominates.
Let's translate this into concrete terms for the reaction given.  A quite plausible mechanism would look like this:
$\ce{NO + O2 <=> ONOO}$ Fast, reversible
$\ce{ONOO + NO -> 2NO2}$ Rate determining step
The equilibrium in the first step is given by
$K_\mathrm{p1} =\dfrac{p^\mathrm{eq}(\ce{ONOO})}{p^\mathrm{eq}(\ce{NO})p^\mathrm{eq}(\ce{O2})}$
Assuming $\ce{ONOO}$ is a minor species, it's formation hardly affects the equilibrium partial pressures of the other species at any time, so we may just use $p(\ce{NO})$ as the nominal nitric oxide partial pressure and similarly for the molecular oxygen.  Then we can solve for the partial pressure of the $\ce{ONOO}$ intermediate:
${p^\mathrm{eq}(\ce{ONOO})}=K_\mathrm{p1}(p(\ce{NO}))(p(\ce{O2})$
Now put this result into the rate determining step with rate
$r_2=k_2{p^\mathrm{eq}(\ce{ONOO})p^\mathrm{eq}(\ce{NO})}=k_2K_\mathrm{p1}(p(\ce{NO})^2)(p(\ce{O2})$
Now for the eye-opener:  put in the temperature dependence of the equilibrium and rate constants
$K_\mathrm{p1}=K_\mathrm{p1}^0\exp(-\Delta H_1/RT)$
$k_2=k_2^0\exp(-E_{a2}/RT)$
So then
$r_2=k_2^0K_\mathrm{p1}^0\exp(-\dfrac{\color{blue}{E_{a2}+\Delta H_1}}{RT})p(\ce{NO})^2p(\ce{O2})$
The activation energy then contains two terms, one from the inherent rate in Step 2 and the other from the equilibrium in Step 1.  The latter is negative if Step 1 is exothermic, as it might well be if a bond is being (temporarily) formed between the reactants.  Then if this enthalpy term is the absolutely larger part of the sum ... we see a reversed temperature dependence!


text:
Addition to conjugated dienes
$$\ce{CH2=CH-CH=CH2 ->[\ce{H2/Pt}][\pu{1 equiv}] ?}$$
I thought $\ce{H2}$ should add to one of the π bonds and form but-1-ene. But the answer is but-2-ene.
I couldn't understand this.

Just read about $1,2$-addition vs. $1,4$-addition.

Okay. But isn't 1,4-addition only for electrophilic addition to conjugated dienes? But addition of H2 is by metal adsorption.

Yes, it is by metal adsorption. So, take a look at the simplest mechanism given here https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Catalytic_Hydrogenation . Now at higher temperature there can be rearrangements in the binding of the alkene to the metal surface after addition of one hydrogen. That means the double bond can rearrange to form an intermediate in which the binding to metal surface is from the different carbon after rearrangement. And hence that can take hydrogen to form rearranged product.