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Negative Molar Specific Heat of a Gas
It is known that the specific heat of a gas is process dependent. So it must be theoretically possible to have a negative value for a gas according to the following equation (for polytropic process, $PV^n = \mathrm{const}$):
$$C = \frac R{γ - 1} + \frac R{1 - n}$$
where $C$ is molar specific heat and $γ$ is adiabatic exponent.
Supposing $γ$ is $\frac{5}{3}$ and $n$ is $\frac{4}{3}$, $C$ comes out to be negative. Is it practically possible and if so what would it signify? As you provide more heat to a gas in such a process, would it lose temperature?
Please clarify.

As Cv=x.R and Cp=Cv+R, I do not think it is possible.

But putting the values of monotomic gas and n as 4/3 (both being valid and within domain) seem to give a negative value.

While it is possible to add heat to gas and get the final temperature lower, one has to count done work and also possible change of C.

Monoatomic gas has Cv=3/2R, Cp=5/2R, gamma 5/3.

It is because I have no idea whether this phenomenon can happen or not that I've posted this question. I don't understand by what you mean by considering the work done here.

also the variable n in the equation can be any rational number if I'm right

It is a very bad idea to consider to consider heat capacity (a function of state) to be a function of process path (such as dQ/dT).  Doing this will drive you nuts in the future.  In thermodynamics, we purposely define heat capacity as a function of either enthalpy or internal energy, both of with are functions of state (and not functions of path).  Problems like your polytropic gas example force you to think of heat capacity in the wrong (and non-productive) way.

I second Chet. It is an interesting question though, and stimulating so +1

@ChetMiller does that mean C is process independent ?

In Thermodynamics (as opposed to Freshman Physics), we define heat capacity in terms of the partial derivatives of state functions with respect to temperature, not in terms of heat Q (a non-state function).  Anyway, your equation is incorrect and shouldn't have an R in the denominator.

Yes, that means that C is process independent.

Sorry, I'm still in high school, so we've not come across this definition of C

I'm going to continue to resist calling C the heat capacity because, as used here, it is not path-independent.  Instead, I will take it to merely represent the derivative of the heat added per mole of ideal gas with respect to temperature for a polytropic process path.
With this caveat, here is my interpretation of what is happening mathematically when C takes on negative values:  If the gas is expanding, a negative value of C means that we added heat to the gas during the expansion, but not enough to offset the expansion cooling that would occur in an adiabatic reversible expansion, so its temperature decreases while we add heat.  If the gas is being compressed, a negative value of C means that we removed heat from the gas during the compression, but not enough to offset the compression heating that would occur in an adiabatic reversible compression, so its temperature increases while we remove heat.  It's as simple as that.


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PLGA - poly(lactic-co-glycolic acid) Synthesis
Does anyone have experience with the synthesis of PLGA?  PLGA is synthesized by means of ring-opening co-polymerization of two different monomers, the cyclic dimers (1,4-dioxane-2,5-diones) of glycolic acid and lactic acid. PLGA is very expensive, due to its production costs. Can someone explain why is this synthesis process so expensive?

I assume it's simply a low-volume speciality. Have you got any more specifics about the synthesis? Also it's afaik mainly sold for medical applications, which makes the production expensive due to regulatory demands.

Most popular catalyst seems to be tin(II) octanoate.

source: Stack Exchange

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Atomic number ranges in "integral units"?
My textbook says the following:

For an electrically neutral or complete atom, the atomic number also equals the number of electrons. This atomic number ranges in integral units from 1 for hydrogen to 92 for uranium, the highest of the naturally occurring elements.

I'm wondering what is meant by "integral units" in this context?
I would greatly appreciate it if people could please clarify this.

It means the said number can be 1, or 10, or 25, but never 1.5.

@IvanNeretin In that case, I think the word they're look for is "discrete", not "integral".

No, it should be just integers. "Integral number" looks like literal translation from another language.

@Mithoron Yes, the word "integral" doesn't even have a grammatical connection to the word "integers", which makes its usage very unusual.

If the author has used "integral units of charge" it would have been much better. Definition 1b(1) at Merriam-Webster.

@ThePointer Integral can be used to mean "of or relating to integers". An example of this usage in mathematics is the concept of an integral domain

@ThePointer "discrete" is not the same thing as "integers"; 1, 1.1, 1.2, 1.3 ... is discrete. I also second what Tyberius has said.

This is only a way of saying that the atomic number will never be a decimal number like 1.5 or 16.8
One consequence is that if you carry out an experiment which gives you a decimal number as an atomic number, then you must have a mixture or something is wrong.

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Gas evolution in reaction between zinc and HCl
So I was doing a chemistry lab for school, and we were analyzing how the concentration of a solution affects the rate of a reaction. In order to do this, we used the inverse burette method for the reaction between zinc and HCl, which would produce hydrogen gas. We had an Erlenmeyer flask connected to the burette by a red tube, and the reaction would occur in the flask. I noticed that the flask stayed blurred, meaning that the hydrogen gas would be trapped inside of it. I asked my teacher if that meant that the pressure was not enough for the hydrogen to displace the water, but he said that was wrong. I cannot think, however, another reason for that. I would appreciate any help.

Ok, we're getting closer... --- What did you mean by "the flask stayed blurred"? Did you mean that the solution was not clear?

I also assume that you weighed the zinc and didn't get as much hydrogen gas as you expected?

@MaxW I mean that the Erlenmeyer flask was foggy inside. As if there was some kind of gas inside (I suppose it was hydrogen gas). Unfortunately, I did not weight the zinc. All I can say is that it was excess (although it would still be nice to do that).

So the gas above the solution in the erlenmeyer was foggy? That's the problem?

@MaxW Yes, I would like to know why the Hydrogen did not go to the burette (did not travel through the tube) but instead stayed in the Erlenmeyer flask.

Did you see bubbling around the zinc metal in the acid solution?

@MaxW I did see some bubbles in the Erlenmeyer flask.

Why do you think you can see hydrogen gas? Can you see air, or propane from a torch? What else might create fog?

@DrMoishePippik Before the reaction started, there was no "fog", the "fog" could only be seen after the reaction started. As the only gas the reaction produces is hydrogen gas, that must be the cause of the "fog".

Again, read the question: why do you think a gas might look foggy? What makes fog in air??

Probably acid spray?

When HCl reacts with zinc or any other active metal such as aluminium, hydrogen is produced. This is what your textbook tells you. Whats is perplexing is the appearance of "fog" in the Erlenmeyer. In the beginning there is air above the liquid. When hydrogen evolution starts, the flask begins to fill with $\ce{H2}$ gas and creates enough pressure to displace the column of water in the buret. So that part of your observation is correct and you see a colorless gas displacing water column in the buret. What is in the conical flask that is causing the fog? If you were to touch the Erlenmeyer you could have noticed that it was hot, secondly there was a very vigorous evolution of hydrogen bubbles. There are several reasons, each contributing to the fogginess in the flask. That "fog" is just not hydrogen, air but vapors and very fine droplets of acid.
(i) creation of very fine droplets of acid by vigorous bubbling process (major contribution)
(ii) steam generated by the heat of the reaction
(iii) Hydrochloric acid has the tendency to produce fumes in moist air (seen with very concentrated acid). You did not mention the concentration of the acid used.

I'd think that it is mostly steam. The bubbles of gas are equilibrated water at the temperature of the hot solution. Glass doesn't conduct heat well, so the bottom of Erlenmeyer with the solution is hot, but the top of the Erlenmeyer is cold. So when the hot gas bubble breaks the surface, the gas above the solution is cooled by the relatively colder glass above the liquid level. This creates water condensation in the form of fog. // A important point is that the gas doesn't have much heat capacity, the glass has a lot.


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Saturation dissolved oxygen level at pressure
The saturation level of dissolved oxygen in contact with a moist atmosphere is
$$c = 0.21(p_\mathrm{atm} - p_\mathrm{vap}) K_\mathrm{H} M$$
where the $p_\mathrm{atm}$ and $p_\mathrm{vap}$ are the atmospheric and vapour pressures, respectively; $K_\mathrm{H}$ is the Henry's law constant, and $M$ is the molar mass.
If I take this volume of water away from the atmospheric interface, and reduce the pressure, does the saturation level equation stay the same as above, except swapping out $p_\mathrm{atm}$ for $p_\mathrm{local}$?

Yes, if p-local is made up of air and water vapor and the water vapor partial pressure at local is included.  But this will be when the system re-equilibrates.


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Cyclohexane derivative formation of carambolaflavone using a Lewis acid catalyst
The scheme below summarizes the results from a study into the optimum conditions needed for a key step in the synthesis of the proposed structure of carambolaflavone, a natural product that has antidiabetic properties. Note that $\ce{Sc^3+}$ acts as a Lewis acid in these reactions.

I'm very confused about how the acid is coordinating to the Benzene derivative here?

The Sc3+ may bind to the benzene derivative but that is not important. The binding to the oxygen of the tetrahydropyran is what matters.

Hi thanks for your answer, but I'm afraid I'm not quite understanding what you mean?

The key intermediate in the reaction step is an oxonium species R-O=R+ created by the Scandium ion binding to the acetal oxygens on the tetrahydropyan ring. In the first step it binds to the carbonyl of OAc gruop, in the second step to the oxygen of the THP-OPh

@Waylander provided a correct written explanation for your question. I'll expand on his comments by considering the effect of temperature and provide diagrams. Using M+ = Sc+3 to save some space, the Lewis acid  coordinates with the more basic of the two acetate oxygens in 1 to provide the reactive oxonium species 2. [The oxonium species is planar about the double bond but I have left it in the chair conformation as a space-saving device.] This step may be reversible. At room temperature there is not sufficient activation energy to permit substitution of the aromatic ring. Following the red arrows of phenolic species 3, axial attack of the phenol oxygen occurs on the oxonium ion 2 to form, kinetically, compound 4. At 50 oC this reaction is reversible catalyzed by Sc+3 to reform 2 and 3 via the blue arrows. Following the green arrows, now there is sufficient energy to disrupt the aromaticity of 3 and allow for axial attack by carbon on the oxonium ion 2 to form, kinetically, structure 5. An increase of temperature to 70 oC permits coordination of the tetrahydropyran oxygen with the Lewis acid permitting ring opening, bond rotation about the red bond in 6 and ring closure of conformation 7 forming the thermodynamically stable, all equatorial tetrahydropyran ring of 8. [Structures 6 and 7 may also be drawn as benzyl cations with the carbonyl oxygen protonated.]


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What is the balanced chemical equation for a redox reaction between a lithium-cobalt battery and a graphite electrode?
I'm doing a project for my last year of secondary school on lithium cobalt batteries and got confused about the equation.
This is the only equation I can find (source: www.science.org.au):
\begin{align}
\ce{LiC6 &→ $x$ Li+ + $x$ e- + C6}\tag{anode}\\
\ce{Li_{(1−x)}CoO2 + $x$ Li+ + $x$ e- &→ LiCoO2}\tag{cathode}\\
\hline
\ce{C6 + LiCoO2 &→ Li_xC6 + Li_{(1−x)}CoO2}\\
\end{align}
But I don't understand it at all! I figured it must be balanced, but the '$x$' part of the equation is beyond my level of understanding. Is there a simpler way of writing this formula for someone at my level? If not, can someone please explain what this equation means?

Related, if not a duplicate: What does x mean in (1-x)A – xB composite materials, molar ratio, weight ratio or volume fraction?

Below we have a simpler way of the $\ce{Li-Co}$ cell reactions.
Cathode half-reaction:
$$\ce{CoO2 + Li+ + e- <=> LiCoO2}$$
Anode half-reaction:
$$\ce{LiC6 <=> C6 + Li+ + e-}$$
Net-cell reaction:
$$\ce{LiC6 + CoO2 <=> C6 + LiCoO2}$$
Over-discharging supersaturates lithium cobalt oxide, leading to the production of lithium oxide.
$$\ce{Li+ + e- + LiCoO2 -> Li2O + CoO}$$
Similarly overcharging (up to $\pu{5.2 V}$) leads to following reaction:
$$\ce{LiCoO2 -> Li+ + CoO2 + e-}$$

At overcharging, there also occurs deposition of metallic lithium and its reaction with the dialkylcarbonate solvent. https://pubs.acs.org/doi/abs/10.1021/la980454y


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Why do Hydrogen Halides(with the exception of HF) form mostly strong acids?
Why do Hydrogen Halides(with the exception of HF) form mostly strong acids?
My guess is that the negatively charged Halide part of the molecule is attracted by the partially positive part of the water molecule. Thus it is separated more easily. Although because it has one more electron it is more stable once it has dissociated

This question has already been answered:https://chemistry.stackexchange.com/questions/34818/why-is-hf-the-least-acidic-hydrogen-halide

I would like to correct a few points firstly that the halide part is not negatively charged. Also the electronegativity of the halide is the dominating effect in this case.
Hydrofluoric Acid as we know is not a very strong acid as compared to other hydrogen halide acids, had it been the case of electronegativity this would not have been the case.
The dominating factor here is the sizes of the atoms forming the molecule. As the size of halide increases, the bond strength of the H-X bond decreases, making it easier to break the Hydrogen halide bond and act as a stronger acid.
So the acidic strength comes out to be:
              HF < HCl < HBr < HI

Agree mostly but why "the halide part is not negatively charged"? What about $F^-$, $Cl^-$, $Br^-$ ,$I^-$ ...?

The question stated that the negative charged fluoride existed before dissociation, which wont be the case

Ah, I interpreted that to mean the halogens are electronegative, but ok.


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Why doesn't solvent form the major product in SN1 reactions?
If in a reaction SN1 pathway is followed then after the formation of carbocation the reagent or the solvent molecule can attack to form the product. As the number of solvent molecules are significantly higher than the reagent molecules the solvolysis product should be the major one, but then why is the reagent substitution product given as the major one?
Is is just because we have to 'use' the reagent since it is given to us?

Depends on what the solvent is. Which solvent are you asking about?

The solvent will react IF it can react irreversibly with a carbocation.

https://en.wikipedia.org/wiki/Solvolysis

So, how to decide which nucleophile will form the major product the reagent or the solvent?

@Waylander so is my assumption that the solvent (as a nucleophile) should form the major product because the number of solvent molecules are significantly higher wrong?

Depends entirely on the solvent. For EtOH this is a reasonable argument. It is a terrible argument for hexane.


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What is the range of values of n in [Cu(NH3)4(H2O)n]SO4?
My textbook represents the copper ion as "tetraaminecopper(II) sulfate-1-water", so I assumed that this complex ion has only 1 water of crystallization.
Then, in the Wikipedia article, they represent the ion as $\ce{[Cu(NH3)4(H2O)_n]SO4}$. The same formula is used but the number of water of crystallization is variable, denoted by $n$. So $n$ must have a range of values.
My question is what determines the value of $n$? What condition would cause $n$ to be larger or lower?

From the top of my head I only recollect tetraamminecopper(II) sulfate monohydrate. Probably that's the way Wikipeda article suggesting that this compound is hygroscopic and water content may deviate (e.g. $n≥1$).

Tetraamminecopper(II) Sulfate or cuprammonium sulfate can be synthesized as its monohydrate (Ref.1):

A solution of $\pu{50 g}$ of finely divided $\ce{CuSO4 . 5 H2O}$ in $\pu{75 mL}$ of conc. ammonia and $\pu{50 mL}$ of water is filtered and precipitated by slow addition of $\pu{75 mL}$ of alcohol. After standing for several hours in the cold, the crystals are filtered on a Büchner funnel, washed with a mixture of alcohol and conc. ammonia (1:1) and then with alcohol and ether, and dried by suction.

$$\ce{CuSO4.5H2O + 4NH3 -> [Cu(NH3)4]SO4.H2O + 4H2O}$$
The properties of tetraamminecopper(II) Sulfate monohydrate is listed as follows (Ref.1):

Deep blue crystals, $\mathrm{d}^{20}_4$: 1.81. Solubility ($\pu{21.5 ^{\circ}C}$): $\pu{18.5 g} / \pu{100 g}\: \ce{H2O}$. Decomposes in air. Loses $\ce{H2O}$ and $\ce{2NH3}$ on heating to $\pu{120 ^{\circ}C}$; the remaining ammonia is evolved at $\pu{160 ^{\circ}C}$.

The crystallographic study of tetraamminecopper(II) Sulfate monohydrate was done several times, and most recently by Mazzi (Ref.2). In the abstract of Ref.1, the author states that:

$\ce{Cu(NH3)4SO4.H2O}$ is orthorhombie, $Pmcn-D^{16}_{2h}$, with four molecules in the unit cell. The determination of the structure was carried out by means of Patterson and Fourier projections on $(001)$ and $(100)$. The structure consists in layers parallel to $(010)$ formed by $\ce{Cu(NH3)4}$ squares and $\ce{SO4}$ tetrahedra held together by $\ce{H2O}$ molecules. Copper atoms, which lie on symmetry planes, have four nearest $\ce{NH3}$ neighbours ($\ce{Cu-N} = \pu{2.04-2.06 Å}$) and two more $\ce{H2O}$ neighbours ($\ce{Cu-O} = \pu{2.59-3.37 Å}$), which complete the distorted octahedral co-ordination group. Each $\ce{SO4}$  tetrahedron ($\ce{S-O} = \pu{1.40-1.43-1.46 Å}$) has two oxygen atoms linked with hydrogen bonds to adjacent $\ce{H2O}$ molecules ($\ce{OH...O} = \pu{2.67 Å}$).

All of these data directed to the comment of @andselisk who states that tetraamminecopper(II) sulfate is a monohydrate but, Probably why Wikipeda article suggesting it as  $\ce{[Cu(NH3)4(H2O)n]SO4}$ that this compound may be hygroscopic, and thus, its water content may deviate (or vary). The crystal structure has confirmed it is a monohydrate.
References:

Georg Brauer, Ed., Handbook of Preparative Inorganic Chemistry – Volume 2; 2nd Edn, P. G. Stecher, Translation Ed., Translated by Scripta Technica, Inc.; Academic Press: New York, NY, 1965, “Section 19: Copper, Silver, Gold, by O. Glemser & H. Sauer,” p. 1021.
F. Mazzi, “The crystal structure of cupric tetrammine sulfate monohydrate, $\ce{Cu(NH3)4SO4.H2O}$,” Acta Cryst. 1955, 8, 137–141 (https://doi.org/10.1107/S0365110X55000571).


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Rate of cathodic reaction vs Tafel slope
One of my reference engineering books says that for the cathodic protection process (in seawater):
Al $\rightarrow$ Al$^{3+}$ + $3e^-$ (anode, setting the protection potential at about -900 mV SCE)
2H$_2$O + 2$e^- \ \rightarrow$ H$_2$ + 2OH$^-$
Book says that the rate of hydrogen evolution grows roughly one order of magnitude every -120mV of over-polarization (more negative potential).
I found online that the Tafel slope for the hydrogen evolution process is roughly $\beta$=200 mV/dec in seawater (should be $\beta$=120mV for uni-molecular mechanisms).
Following this reasoning, I get (using the Tafel equation):
ln$\frac{i*}{i}$=ln10=-$\frac{1}{\beta} \left ( \eta^*-\eta \right )$
therefore the rate should rise one order of magnitude every $\Delta\eta$=460 mV ($\beta=200mV$, and $\Delta\eta$=276 mV if $\beta=120 mV$).
Does anybody see any mistake in my reasoning? If this is correct either my Tafel constant is off (should be about 55 mV) or the book has a typo...
Thanks in advance.


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Any examples of liquids volatile at room temp but non-flammable?
I'm thinking of something similar to the liquid used in the classic dipping birds. Temperature differential in two "bulbs" at each end of a tube should cause vapor to expand and push liquid to the top, shifting the weight and tipping it. The closer to room temperature the boiling point is, the better. Thanks.

Dichloromethane

Dichloromethane is a carcinogen and liver toxin (if this is for schools). If you want to be green, try "A dunking bird of the second kind". It just uses water and patience. Published in American Journal of Physics: https://aapt.scitation.org/doi/10.1119/1.1703543

@M.Farooq It's suspected to be carcenogenic. And not very badly, because with its widespread use, one would know, otherwise.

Benzene is present in fuels, and it is a known human carcinogen. Widespread use does not guarantee safety (standards change all the time). If you talk to an old timer, they always say that they used benzene freely in labs. Not anymore. Organic chemists use DCM all the time and work with worse substances. Still they live very long or sometimes quite short. I would avoid volatile halogenated materials in general because they are quite bad for the liver even if they are suspected carcinogens.

Dichloromethane is surely toxic. While it can be acceptable for our columns I would not use it for demonstration aimed to general audience nor in a general class activity.

Toxicity is definitely a concern, as it should be. Plus it's a decent song. But long term my application may be industrial in nature, so certain safety precaution could be enforced.

@AChem Standards do not guarantee safety either.  Ideally standards and safety point in somewhat similar directions, but there are plenty of times they are not the same at all.

Yes, there are lots. Start with Wikipedia's list of refrigerants; while it doesn't call out flammability, any fully-halogenated compounds will be effectively non-flammable. Trichlorofluoromethane has a standard boiling point just above standard room temperature.
You didn't say anything about pressure in your question. If you don't mind pressures higher than one standard atmosphere, and don't mind toxicity, candidates like ammonia, sulfur dioxide, and chlorine come into play. If you don't mind low pressure, you could even use water, although it might freeze as it boils off.

Don't forget bromine

@A.K. He's the rugged guy in Lord of the Rings who gets shot up with arrows, right?

I suppose the higher the pressure the faster the whole sequence would progress, so lets says as high a pressure as is reasonable.

Can you tell us more about your safety constraints (flammability, pressure limits)? HFC-134a (1,1,1,2-Tetrafluoroethane) has a room-temperature vapor pressure between 4 and 5 atmospheres, maybe 25%-60% higher than the pressure in a full two-liter soda bottle.

The higher the pressure the better. A higher pressure differential at as little of a temperature difference would be best. It should be able to hold at least 10 atm.  Flammability isn't a huge concern. It will all be inside a sealed system. It's just a bit easier and safer to work with in general.

If you're looking for "general" safety, then a household/automotive refrigerant is probably your best bet. It looks like I was slightly off-base on 134a, its vapor pressure at room temp is about 6 atm. If this is a one-off and you aren't concerned about expense, other halocarbons let you dial in just about any boiling point or pressure you'd like.

Probably a lot of general anesthetic gases like desflurane, isoflurane, and sevoflurane would also be worth a look.

Most commercial refrigerants would work. There is a caveat, which is that they are usually also ozone depleting and are also greenhouse gasses with high , and the EPA will ream your bunghole if you handlGlobal Warming Potential (GWP)e them improperly. Almost all of these refrigerants are halogenated hydrocarbons (usually HFCs or CFCs).
One of the more modern refeigerants is R-1233zd (1-Chloro-3,3,3-trifluoropropene). R-1233zd is neither an HFC or a CFC, it is an HFO (hydrofluoro-olefin), It is safer to handle and more environmentally friendly than the former categories (zero ozone depletion and low GWP). However, it's also very expensive (~$50/lb), and you'll likely need an EPA 608 license to purchase it (though I may be mistaken here).
Update R-1233zd is actually flammable but only slightly, and it also has a very slow combustion rate of <10cm/s.


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Removing PTFE marks on glass bottles after using magnetic stir bars
I am using magnetic stir bars to agitate solutions for 4 weeks at a time.  I have noticed at the bottom of the glass bottles that white marks have accumulated where the stir bar spins.  Even with soaking, thorough washing and scrubbing, acid washing, and sonicating, the marks remain.  I suspect these marks are white PTFE that has eroded from the stir bars then stuck to the glass.
I have 2 questions:

Are the marks removable?  I know PTFE is very resistant to most solvents, which makes removal seem difficult.
Is it necessary to remove these marks?  Perhaps because PTFE is so resistant to solvents, it will not affect my analysis.

I am extracting ammonium from soil extracts.  The soil extract is an acidified water solution, which I immediately add MgO to, in order to increase the pH to above 9.  This pushes the NH4+ into the headspace of the bottle.  Floating on the solution is a glass filter in a sealed Teflon packet.  The ammonium ions diffuse through the Teflon, and react with citric acid on the filter to make an NH4Cl salt.  This is then analyzed on an EA-MS for d15N.

Are you absolutely sure it's traces of PTFE and not just scratched/etched glass? Such matte coating is likely formed due to mechanical friction between the glass surface and some particles or stirring bar itself in alkaline medium. This would also explain why you cannot remove those marks.

Agreed with andselisk.
Glass, and expecially borosilicate glass, is attacked by bases. For sure the long exposure time and the alkaline pH are the causes.
No worries: it's the normal destiny of glassware in such reaction conditions. Just be aware, in the future, because eroded glass could trap particles, so it might not be suitable for every use, and it might require cleaning with "aggressive" agents in order to remove them. Also, it's a good habit not to use eroded glassware under vacuum or strong pressure, because it might lose mechanical resistance and fail.

Stir bar has a very rough grinding action. It is quite well known among silica manufacturers that if you stir micron sized silica with a stir bar for a long time in a reaction mixture, the spherical particles are damaged to some extent. In such cases, overhead stirring is preferred. This is not applicable to your case , I guess. If you had soil as well, the scenario is no different. It is a grinding action of the stir bar on glass. The "frosting" of glass is like rubbing a sand paper on glass for a long time. Nothing can heal this, except a glass-blower's flame treatment. Of course this is not worth it, as it is not going to affect the analysis of ammonia. I don't think that this is PTFE on glass. It will not stick to glass and secondly it is so hydrophobic that it would be floating on your solution rather than firmly sticking to glass.


text:
Band of stability
Here’s the band of stability as it appears in my textbook:

1- My textbook states that the last stable atom is Pb-206 and every element that has an atomic number greater than 82 is radioactive. But if every dot resembles a stable atom, why aren’t the dots after Pb-206 in the graph considered stable?
2- Why is the $\frac{n}{p}$ ratio be $\frac{1.5}{1}$? Doesn’t that mean that for every proton there’s one and a half neutron?

The image is mislabeled: that red dot is 204 Hg, not 206 Pb.
Plus, the text is wrong: the "last" (heaviest) stable isotope is not lead-206, but lead-208!
The elements on the same vertical line of 206-Pb (which is 3 dots on the right of the mislabeled red dot) are 207-Pb and 208-Pb.
The last dot on the top right is 209-Bi. Bismuth was found in 2003 to be metastable (radioactive, with an extremely long half-life), so that chart might be older than that discovery.
A similar (and more accurate?) table, with colors and n and Z shown, can be found here:
https://en.wikipedia.org/wiki/Table_of_nuclides_(combined)#Isotopes_for_elements_75-89
The "1.5" ratio you mention comes from a linear fit. It means that not necessarily every single data fits it, but it represents well a general trend of the data. For instance, the image shows you that the ratio for 206-Pb "looks like" the 1.5 ratio, being 1.51.
Some other ratios are: 1.52 for 207-Pb, 1.55 for 204-Hg, etc...


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Nitrobenzene - Aromatic Substitution Reactions
Which EAS (electrophilic aromatic substitution) reactions does nitrobenzene show?
We know that nitrobenzene is particularly deactivated towards most such reactions, as the nitro group is strongly electron withdrawing via resonance and inductive effects. However, I wish to know exactly where to draw the line - which reactions nitrobenzene undergoes, and which ones it doesn't.
Of course, there's no clear boundary as to whether or not a compound undergoes a particular EAS, but can't we at least comment on the yield obtained in such reactions?
I haven't been able to find much about EAS reactions of nitrobenzene online, hence I come to Chem SE, seeking data and sources.
The reactions of interest are:
1.Friedel Crafts Alkylation, Acylation
2.Nitration (concentrated sulphuric and nitric acid)
3.Sulphonation (fuming sulphuric acid)
4.Halogenation in presence of Lewis acids
etc.
Please help! Thanks a lot!

1,3 Dinitrobenzene is commercially available so that should give you an idea.

chlorination of nitrobenzene  https://www.sciencedirect.com/science/article/abs/pii/S0926860X1200614X

Sulfonation of nitrobenzene  https://pubs.rsc.org/en/content/articlehtml/2012/ra/c2ra20406a

Of course, there must be some boundary as to whether or not nitrobenzene undergoes a particular electrophilic aromatic substitution (EAS) since nitro-group is very deactivating. I leave it to you to decide that fact on the yield obtained in following reactions, because there are some data you may fail to find but, nonetheless, exist in literature.
Chlorination of nitrobenzene at $\pu{35\!-\! 45 ^{\circ}C}$ in the presence of iron (III) chloride ($\ce{FeCl3}$) gives an isomer mixture, which contains 86% 3-chloronitrobenzene, 10% 2-chloronitrobenzene and 4% 4-chloronitrobenzene (Ref.1). Even if nitro group is $m$-directing, this reaction has given significant amount of $o$-prodouct. You may compare this reaction with nitration reaction of chlorobenzene at $\pu{40\!-\! 70 ^{\circ}C}$ with mixed acids (30% $\ce{HNO3}$ : 56% $\ce{H2SO4}$ : 14% $\ce{H2O}$), which typically gives a 98% yield of an isomer mixture consisting of 34-36% 2-chloronitrobenzene, 63-65% 4-chloronitrobenzene, and about 1% 3-chloronitrobenzene (Ref.1).
A simple and efficient method for the sulfonation of nitrobenzene using liquid $\ce{SO3}$ as the sulfonating agent in a microreactor at $\pu{40\!-\! 110 ^{\circ}C}$ was developed and conducted recently, and 88% yield of 3-nitrobenzenesulfonic acid (m-NBSA) has been obtained (Ref.2).  The main by-product is bis(3-nitrophenyl)sulfone (BNPS) and proposed mechanism was depicted in the following Figure:

On the other hand, the sulfonation of toluene has given $o:m:p$ ratio of  21.2% : 2.1% : 76.7% in 77.8% $\ce{H2SO4}$ (Ref.3). The $o:m:p$ ratio has become 38.8% : 2.6% : 58.6% when solvent has changed to 84.3% $\ce{H2SO4}$ (significant decrease of $p$-product in increasing acidity; Ref.3). Also, note that the sulfonation by sulfuric acid of nitrobenzene itself and of various aromatic compounds in nitrobenzene solution has been studied at a series of temperatures (Ref.4). Yet, no concomitant sulfonation of nitrobenzene in the presence of other activated substrates such as toluene and xylene has been reported.
Using $\ce{NO2+PF6-}$ as the nitrating reagent, Olah had studied nitration of nitrobenzene in two different solvents: He achieved $o:m:p$ ratio of  7.1% : 91.5% : 7.4% in 96% $\ce{H2SO4}$  and that of 10% : 88.5% : 1.5% in $\ce{CH3NO2}$ (Ref.3).
For comparison, same nitrating agent ($\ce{NO2+PF6-}$) has given $o:m:p$ ratio of  68.2% : 2% : 29.8% in $\ce{CH3NO2}$ when subjected to nitrate toluene (Ref.3).
To my knowledge, Friedel-Crafts reaction of nitrobenzene (alkylation or acylation) is yet to be known. I'd love to hear if somebody say otherwise.

References:

International Agency for Research on Cancer (World Health Organization), lARC Monographs on the Evaluation of the Carcinogenic Risk of Chemicals to Humans, Vol. 65, 1996, “2-Chloronitrobenzene, 3-chloronitrobenzene and 4-chloronitrobenzene,” pp. 263-296.
Y. Chen, Y. Su, F. Jiao, G. Chen, “A simple and efficient synthesis protocol for sulfonation of nitrobenzene under solvent-free conditions via a microreactor,” RSC Adv. 2012, 2, 5637–5644 (DOI: 10.1039/C2RA20406A).
George A. Olah, “Aromatic substitution. XXVIII. Mechanism of electrophilic aromatic substitutions,” Acc. Chem. Res. 1971, 4(7), 240–248 (DOI: 10.1021/ar50043a002).
F. J. Stubbs, C. D. Williams, C. N. Hinshelwood, “The Kinetics of Aromatic Sulphonation Reactions. Sulphonation by Sulphuric Acid in Nitrobenzene Solution,” J. Chem. Soc. 1948, 1065–1078 (DOI: 10.1039/JR9480001065).

What about nitration of nitrobenzene using HNO3 + H2SO4?

Reference 3 has that values as well.


text:
What is the precise definition of Ramachandran angles?
The Wikipedia article Ramachandran plot is very schematic and doesn't even say when these angles are considered to be zero. It seems to assume that the peptide chain has no end, and always shows the next carbon atom. What happens when the amino acid is in the terminal position? It also asserts that the $\omega$ angle is always 180°, while in the practical cases this isn't exactly true.
The definitions seem to rely on the adjacent bonds. Two on one side, and one on the other side. They rely on the fact that two bonds on one side are in the same plane as the rotating bond. How are these angles defined when relevant bonds don't form a plane?
What is the precise definition of $\psi-\phi$ angles? Are there formulas for them that take atom positions as arguments with amino acids being in both terminal and non-terminal positions? Are they defined for amino acids in terminal positions?
Additionally, an article by Zhou et al. [1] also talks about the bond angle $\tau$. $\tau$ seems to be a logical addition because it corresponds to an additional degree of freedom, not covered by $\psi-\phi$ angles.
References

Zhou, A. Q.; O’Hern, C. S.; Regan, L. Revisiting the Ramachandran Plot from a New Angle. Protein Science 2011, 20 (7), 1166–1171. https://doi.org/10.1002/pro.644.

I updated the figure in the Wikipedia article to indicate when the angles are considered to be 180. The article on dihedrals has the information on where zero is as well: https://en.wikipedia.org/wiki/Dihedral_angle#Stereochemistry

Definition

Protein backbone dihedral angles are called φ (phi), involving the backbone atoms COn-1 - Nn - Cαn - COn , ψ (psi) involving the backbone atoms Nn - Cαn - COn - Nn+1) and ω (omega) involving the backbone atoms Cαn - COn - Nn+1 - Cαn+1). Thus, φ controls the COn-1 - COn distance, ψ controls the Nn - Nn+1 distance and ω controls the Cαn - Cαn+1 distance. Where n is the current atom for which you want to calculate the dihedral angles. Dihedral angle varies between -180 to +180. It is positive if rotation between atoms is clockwise and negative if rotation is anticlockwise.

Source: http://mebinfotalk.blogspot.com/2013/06/how-to-calculate-dihedral-angle-using.html
And here is an illustration from the same source:

To calculate, use the cross product to get vectors normal to the (v2, v1) and the (v2, v3) planes, and calculate angles between them (it is easy to get the direction wrong, so I would test with known cases).
In the question, the OP states:

They rely on the fact that two bonds on one side are in the same plane as the rotating bond.

No, the plane is between the rotating bond and the adjacent bond. Two bonds of one atom always form a plane. Here is an illustration: Dihedral angle of gaseous and crystalline HOOH
3D demonstration
A nice interactive tutorial can be found at Proteopedia: https://proteopedia.org/wiki/index.php/Tutorial:Ramachandran_principle_and_phi_psi_angles
Omega
Omega is the torsion angle of the peptide plane. It is most often near 180 (trans peptide) but sometimes near zero (cis peptide).
In a PNAS paper from 2012 (https://doi.org/10.1073/pnas.1107115108), the authors state in the abstract:

Analyses as a function of the φ,ψ-backbone dihedral angles show that the expected value deviates by ± 8° from planar as a systematic function of conformation, but that the large majority of variation in planarity depends on tertiary effects.

In protein structure determinations, the omega angle is often restrained, especially in low resolution structures. Looking at structures deposited in the protein data bank, it might therefore seem that the distribution of the omega torsion around 180 is more tight (less varied) than it is in reality.
First and last dihedral
The first amino acid has no phi angle and the last has no psi angle. Exceptions are when the N-terminus or C-terminus are modified.
Bond angle tau
Different from the torsion angles phi and psi, the angle tau is a bond angle (depends on three atom position, not four). It is the angle at the C-alpha carbon between the nitrogen and the carbonyl carbon. As an $sp^3$ carbon, the C-alpha carbon is expected to have an angle around $109^\circ$. The larger the angle, the more (phi, psi) combinations are possible without steric clashes, i.e. the larger the "allowed" regions in the Ramachandran plot.

Peptide bonds are rigid and planar in nature and Ramachandran plots are used to analyze protein structures and predict stable amino acid conformations in the protein.
A Ramachandran plot is a method to visualize energetically stable regions for polypeptide torsion angles psi (ψ ) against (phi) φ of amino acid residues present in a  protein structure.
A polypeptide chain consists of a regularly repeating part, the main chain or backbone, and a variable part comprising the distinctive side chains. The three dimensional spatial arrangement of the peptide backbone is determined by the relative orientation of the groups connected by these repeating bonds in the peptide backbone.
The relative orientation of groups attached via a series of bonds can be described by dihedral or torsion angles. Therefore, the conformation of the peptide backbone (of a polypeptide) can be described by (a series of) three dihedral (or torsion) angles that describe the rotation about each of the three repeating bonds in the peptide backbone.
In a protein chain three diehedral angles are defined as phi, psi and omega. The planarity of the peptide bond usually restricts ω to be 180° (the typical trans case) or 0° in the rare cis case.

What if many peptides are connected, and there is strain in the system. Would all angles still be strictly 180° and 0°?

Under stress, angle distortions can occur. This term 'disallowed Ramachandran conformations' has been applied where local constraints impose conformational distortions.
Specific phi, psi angle distortions are preferred under stress and high resolution analysis of protein structures show that short contacts are eliminated in most cases by local distortions of bond angles.
For example see Gunasekaran, Kannan & Ramakrishnan, C & Balaram, Padmanabhan. (1996). Disallowed Ramachandran Conformations of Amino Acid Residues in Protein Structures. Journal of molecular biology. 264. 191-8. 10.1006

The figure shows the $\psi,\, \phi$ angles on the peptide backbone.


text:
pKa value of an indicator
Can we deduce the actual value of $\mathrm{p}K_\mathrm{a}$ of an indicator if the $\mathrm{pH}$ range has been given?
I know that $\mathrm{pH}$ range is physically the range of values over which an indicator changes its colour from its acid form to its base form, but can we judge the $\mathrm{p}K_\mathrm{a}$ value from the data?
For example: if $\mathrm{pH}$ range of $3.4$ to $4.6$ has been provided, then I know for certain that the $\mathrm{p}K_\mathrm{a}$ value of the indicator lies in between, but can we deduce the actual value?

Visually, we can only estimate the  $\mathrm{p}K_\mathrm{a}$ value to be near the middle of the indicator range, shifted to more intense colour.
The $\mathrm{p}K_\mathrm{a}$ of an indicator could be determined more precisely by  combining $\mathrm{pH}$ meter with photometry, where we would get the well known " round step function" of absorbance versus $\mathrm{pH}$.

E.g phenolphthalein has indicating range $\mathrm{pH}=8.2 - 10.0$ for reaction
$$\ce{HInd(clear) <=> H+ + Ind^-(violet)}$$
(I will not fight over the colour name. It may be magenta, carmine, purpur, dark pink -  you know the 16 colours of Windows jokes. )
Let suppose the phenolphthalein $\mathrm{p}K_\mathrm{a}$ is in the middle of the range, i.e. $\mathrm{p}K_\mathrm{a}=9.1$.
Let suppose the ratio of concentrations of clear and violet forms at $\mathrm{pH}=8.2$ is $\ce{[Ind-]/[HInd]}=X$.
The equation for indicates are the same as for any other weak acid:

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind-]}}{\ce{[HInd]}}$$
If we consider the mentioned border $\mathrm{pH}$ values,
then
$$\begin{align}
8.2 &=9.1+\log X \\
10.0 &=9.1+\log \frac 1X \\
X&=0.126 \\
\end{align}$$
The problem is the relative eye sensibility to small additions  of one form of an indicator to the other form.
 We notice much easier a small amount of violet substance in clear substance, than a small amount of clear substance in violet substance.
$$\begin{align}
\mathrm{pH_{low}}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind^-_{noticeble}]}}{\ce{[HInd_{abundance}]}} \\
\mathrm{pH_{high}}=\mathrm{p}K_\mathrm{a} + \log \frac {\ce{[Ind^-_{abundance}]}}{\ce{[HInd_{noticeble}]}} \\
\end{align}$$
As the consequence, the $\mathrm{pH}=10.0$ is significantly closer to $\mathrm{p}K_\mathrm{a}$ than $\mathrm{pH}=8.2$, therefore
$$\mathrm{p}K_\mathrm{a} \gt \frac {8.2+10.0}{2}$$
Let suppose for now the real $\mathrm{p}K_\mathrm{a}=9.7$.
Then $\ce{[Ind-]/[HInd]}$ is

2 for pH 10.0, i.e full violet weakened by 1/3
0.03 for pH 8.2, i.e 3% of the full violet.

The similar visual shifts happen to other indicators as well.

Could you please elaborate this , "Let suppose pKa=9.1 and the border values are reciprocal values [Ind−]/[HInd]"?

"still think insisting on writing pKa as \mathrm{p}K_\mathrm{a} is kind of perversion." Please avoid leaving irrelevant complaints in the answers and keep it for Meta. Yes, MathJax is cumbersome, but deal with it. Set a hotkey or a snippet for \mathrm{ if it's such  big issue for you. Also note that in "pH" both letters are upright (I'm not sure why you reverted my edit).

Well, you did whether you were aware of that or not. As for pH, the rule has been established in 1960 (https://doi.org/10.1351/pac196001010163) and hasn't changed since – $\mathrm{pH}$ (\mathrm{pH}) should be written in roman upright type.

@andselisk  I have not reverted your edit, I was not aware of it. You may have edit it in parallel. I did ask myself 25 min ago as well, why is not pH written upright ? The issue is not with writing MathJax but with staying MathJax :-) I am sorry, I could not resist.

Yeah, the editing system is unfortunately not tolerant towards simultaneous edits. Well, at least now you have an IUPAC article to back up the claim of correct typography:)

In fact I knew it I may have been corrupted by MathJax formatting terrorism :-)

I was writing my graduation thesis in era, where it was still usual to write it on a typewriter, and my one was one of the first, printed on 9pin Epson printer. LaTeX was for gloves in that time.


text:
Do the mechanism (SN1 or SN2 ) which reaction follow depends more upon intermediate stability than on substrate and nucleophilic?
For eg : would the reaction of 3-methylbut-2-en-1-ol with HBr follow SN1 mechanism due to stability of your intermediate carbocation stability  or SN2 due to the compound being primary alcohol and good nucleophilic. Also please correct me if I got something wrong.

It will follow SN1 as the intermediate carbocation formed is resonance stabilised.

So should we consider that the stability of intermediate is key factor in deciding mechanism.

It's one of the factors. For example : 2° alkyl halide can undergo solvolysis by SN1 ( in aqueous medium) as well as SN2 pathway (with alcohol and amines)


text:
Which of the following options are correct?

The equilibrium constant $K$ for the reaction $\ce{2HI <=> H2 + I2}$ at room temperature is $2.85$ and at $\pu{698 K}$ is $0.014$. This implies that:
a) $\ce{HI}$ is exothermic compound
b) $\ce{HI}$ is very stable at room temperature
c) $\ce{HI}$ is relatively less stable than $\ce{H2}$ and $\ce{I2}$
d) $\ce{HI}$ is resonance stabilized

I was getting both a and c (only one of the options is supposed to be correct). Which is the most appropriate answer?

For an exothermic reaction $\Delta H \lt 0$ and $K_p$ must decrease as the temperature increases. (It is the opposite way round for an endothermic reaction, i.e. more dissociation at higher temperature which makes sense if the reaction is endothermic: 'more energy more product'.)
Using the integrated Van Hoff equation you  can calculate what happens;  $\displaystyle \ln\left(\frac{K_2}{K_1} \right) = -\frac{\Delta H}{R}\left( \frac{1}{T_2}-\frac{1}{T_1} \right)$ the slope is $-\Delta H/R$.  If you plot your data at $298$ and $698$ K vs $1/T$ the slope is positive so $\Delta H$ is negative and the reaction exothermic.

You can understand this as follows : Le Chatelier's principle says that on changing some factor ( like temp.,pressure,find.,etc ) the equilibrium will shift in a direction so as to undo the effect of change . Now in this case on increasing the temperature the reaction equilibrium is shifting in backward direction (descrease in value of K ) so reverse reaction must lower the temp. i.e.,the reaction must consume heat which means endothermic process and hence 'option a' is correct.

Option (a) is more appropriate as in case of (c) the temperature or the condition is not mentioned. As in case of 698 K, since $K_\mathrm{eq}$ is reasonably less, the equilibrium will shift in the backward direction, so HI will be stable.


text:
When there is an increase in temperature, how is the pH of the ocean affected? (Question is related to ocean acidification)
This is my current understanding:
Ocean acidification is the result of atmospheric $\ce{CO2}$  dissolving in the ocean's water. When this occurs the $\mathrm{pH}$ of the ocean decreases (from the increase in $\ce{H3O+}$ ions produced from the dissociation of $\ce{H2CO3}$).
When there is an increase in temperature, the solubility of carbon dioxide decreases which results in a lower concentration of carbonic acid (the formation of carbonic acid is from the reaction of $\ce{CO2}$ with $\ce{H2O}$ which is exothermic, so the reverse reaction is favoured).
So since the concentration of carbonic acid is lower, the concentration of $\ce{H3O+}$ ions decreases (lower concentration of acid dissociates in the water).
Therefore, the $\mathrm{pH}$ of the ocean increases as there are fewer $\ce{H3O+}$ ions present in the water.
Is my understanding correct, or have I made a mistake in my explanation?
If you can help me out I would be very grateful :)

Related: https://chemistry.stackexchange.com/q/15939/30836

You have to compare the size of these counteracting effects.
The temperature increase slightly decreases the acidification caused by increased carbon dioxide partial pressure.
The shift of reaction constant is minimal.
From the Wikipedia carbon dioxide data page, Water solubility section, we can see for temperature  15 resp 17 $^\circ$C V/V solubility 1.019 resp. 0.956. That is 6.2% lower solubility.
Increased partial $\ce{CO2}$ pressure from 0.03 to 0.04% gives 33%, what is significantly more than by 6.2% decreased solubility.
Aside of that, higher $\ce{CO2}$ concentration and lower sea water $\mathrm{pH}$ are already observed.
This is directly affecting carbonate related equilibriums, particularly in context of forming of mollusk shells.
Some mollusks are already observed  having troubles in forming aragonite - the more soluble form of calcium carbonate.

@Night Writer Well, in my understanding, in given context, the raise of temperature, increase of CO2 And sea acidification cannot be separated.

Ah, yes, I see my mistake!


text:
Adjustment of calcium, magnesium and alkalinity in drinking water?
It's been a long time since I did any molar chemistry in school and despite doing a lot of reading on the subject this week, I am making a mess of some basic calculations. I hope that someone who understands water chemistry & molar equations a lot better than I do can point me in the right direction.
I am trying to add mineral salts to soft water to adjust the calcium, magnesium and alkalinity in a predictable manner. But I can't get past the basic equations so far.
The source water I am working with is:
12.6 mg/L Hardness as CaCO3 equivalent
11.3 mg/L Alkalinity as CaCO3 equivalent
  4.8 mg/L Calcium
  0.2 mg/L Magnesium
  1.5 mg/L Sodium
And the desired target water is:
70    mg/L Hardness as CaCO3 equivalent
40    mg/L Alkalinity as CaCO3 equivalent
20    mg/L Calcium
  4.9 mg/L Magnesium
  8   mg/L Sodium
I would prefer to have most of the hardness derive from calcium, rather than magnesium if possible. I plan to try

adding sufficient baking soda, NaHCO3 to the source water to obtain 8 mg/L sodium, then
add sufficient calcium carbonate (chalk), CaCO3 to the source water to make 50/50 GH/KH water (general hardness/carbonate hardness) i.e. 50 mg/L hardness and 50 mg/L alkalinity, both expressed as CaCO3 equivalents
add sufficient lactic acid (88%  CH3CH(OH)COOH) to neutralize some carbonate and (I hope) get to 50/40 GH/KH
add sufficient epsom salt, MgSO4*7H2O to obtain a final result of 70/40 GH/KH.
This assumes that adding lactic acid won't remove (precipitate) calcium hardness.
The solubility of calcium carbonate in water shouldn't be an issue, I can put it under pressure with carbon dioxide, C02 if necessary to dissolve it.

If the above is not feasible, then alternately

add sufficient baking soda, NaHCO3 to the source water to obtain 8 mg/L Sodium
add sufficient calcium carbonate, CaCO3 to reach the alkalinity target of 40 KH (40/40 GH/KH) i.e.. 40 mg/L hardness and 40 mg/L alkalinity, both expressed as CaCO3 equivalents
Add non-carbonate hardness by adding sufficient epsom salt, MgSO4*7H2O to obtain a final result of 70/40 GH/KH

Any help, guidance as to molar calculations and best practices in water chemistry or well honed levity welcome
Edit to add molar amounts
Source water
12.6 mg/L Hardness as CaCO3 equivalent
11.3 mg/L Alkalinity as CaCO3 equivalent
Ca 2.4 x 10-4 M
Mg 1.7 x 10-5 M
Na 1.4 x 10-4 M
Target water
70 mg/L Hardness as CaCO3 equivalent
40 mg/L Alkalinity as CaCO3 equivalent
Ca 1.0 x 10-3 M
Mg 4.1 x 10-4 M
Na 7.3 x 10-4 M

I advice to recalculate all masses to molar amounts according to the molar mass tables. You will find the ion matching easier.

Is not better to get a mineral water of suitable composition and dilute it with this water ? As drinking water with added chemicals can have unpredictable taste and smell.

BTW, 1 dGH/dKH are equivalent of 10 mg CaO of general/carbonate hardness, the latter known also as alkalinity.

I have also consider the final pH of water, and that by adding of lactic acid, you create a double oH buffer of lactic acid/lactate and carbon dioxide/bicarbonate. Additionally, you will get into computation troubles in determination of the right amount of acid for dissolving CaCO3.

Be also aware to use the right optical isomer of the acid.

The parallel thoughts:
The source water I am working with is:
12.6 mg/L or 0.126 mmol/L GH as CaCO3 equivalent
11.3 mg/L or 0.113 mmol/L KH as CaCO3 equivalent
  4.8 mg/L or 0.12 mmol/L Ca
  0.2 mg/L or 0.0082 mmol/L Mg
  1.5 mg/L or 0.065 mmol/L Na
And the desired target water is:
70    mg/L or 0.7 mmol/L GH
40    mg/L or 0.4 mmol/L KH
20    mg/L or 0.5 mmol/L Ca
  4.9 mg/L or 0.2 mmol/L Mg
  8    mg/L or  0.35 mmol/L Na
What is needed to add:
0.574 mmol/L GH, as Ca+Mg
0.287 mmol/L KH  as Ca+Mg as bicarbonates, I.e. 2x0.287=0.574 mmol/L of bicarbonate
0.38 mmol/L Ca
0.192 mmol/L Mg
0.285 mmol/L Na
Reaction of 0.285 mmol/L sodium bicarbonate
with 0.285 mmol/L lactic acid
releases 0.285 mmol/L carbon dioxide,
dissolving 0.285 mmol/L calcium carbonate
forming 0.285 mmol/L calcium bicarbonate.
( Not exact due equilibrium reactions but good enough )
What is needed to add:
0.287 mmol/L GH, i.e Ca+Mg
0.093 mmol/L Ca
0.192 mmol/L Mg
Adding 0.192 mmol/L MgSO4.
What is needed to add:
0.093 mmol/L GH, i.e Ca only
0.093 mmol/L Ca
This can to be added as 0.093 mmol/L calcium carbonate(optionally supported by pressured carbon dioxide) + 0.186 mmol/L lactic acid.


text:
Find the Total Pressure When the Faucet is Released

Two chambers are connected as shown in the figure. The total pressure of the first chamber that contains Ar and He gas is 98.4 atm. If the faucet is released, find the total pressure. (There's 1 mol of Nitrogen gas on the second chamber)
(A) 52.2 atm (B) 61.5 atm (C)98.4 atm (D)123 atm (E)172.2 atm
My attempt:
Because of Ar and He are inert gas I only calculate the second chamber's partial pressure.
$$\begin{aligned}
P_2&=\frac{1\cdot0.082\cdot300}{1}\\
&=24.6\mbox{ atm}\\\\
P_{tot}&=P_1+P_2\\&=123\mbox{ atm}\\\\
\end{aligned}$$
My other attempt:
$$\begin{aligned}
n_2&=\frac{1\cdot98.4}{0.082\cdot300}\\
&=4\mbox{ mol}\\\\
n_{tot}&=n_1+n_2\\&=5\mbox{ mol}\\\\
P_{total}&=\frac{5\cdot0.082\cdot300}{2}\\&=61.5\mbox{ atm}
\end{aligned}$$
My question is which one is the correct method and why.

Suppose you open a door from one room of your house to another. Do the pressures add up?

Always write the values with the correct units and carry the units through the calculation. Do not omit the units while performing intermediate steps and do not just reintroduce units at the end of the calculation.


text:
When it rains, it puddles. Spilled salt cycles between wet and dry with humidity. Is this akin to a phase change?
I have some spilled un-iodized table salt (NaCl). It is very humid where I live, and when it gets very humid for a few days, the salt absorbs so much water that it becomes a puddle of (probably) saturated salt water. When the humidity drops for a few days, the water evaporates and I have dry salt crystals again. This is repeatable.
My questions about this process are:

What is this process called exactly? As far as I understand "deliquescence" is a property and not a process. Does the exchange of water between my pile of salt and the air have a specific terminology? Something that would go along with terms like evaporation, sublimation, condensation, etc.? Perhaps some mash-up of adsorption and dissolution?
Is this process akin to a phase change in some way? For example, if I had a sealed box of air with a dish of salt and a beaker of water, and I could ramp the  temperature slowly enough, would I see a plateau at a certain humidity as the salt hit a threshold and started absorbing water until saturated?

Some may be familliar with the Morton Salt logo and slogan "When it rains, it pours" meaning that other moisture-absorbing materials in the package will keep the sodium chloride as table salt from absorbing too much water and becoming stuck and un-pourable:
Source

Well, hygroscopic comes to mind, and condensation. It is a phase transition. Initially, it may be considered a condensation reaction (a process on the surface of salt crystals, possibly with water migrating into the crystal). At some point, it is transtion from water in the gas phase to water in a saturated NaCl solution. In other words, this involves multiple stages as you move from hydrated crystals to solvated salt ions.

@NightWriter "hygroscopicity" is a property, not a process, and condensation takes place without any hygroscopic materials being present. Also I don't think NaCl has a hydrated form. update: wait, maybe it does! https://en.wikipedia.org/wiki/Water_of_crystallization

See also https://chemistry.stackexchange.com/a/112915/35806

NaCl does not have a hydrated form. It simply adsorbs water at its surface, which starts dissolving.
When this surface phase becomes thick enough, it coalesces into one continous, NaCl saturated liquid phase, with residual salt particles still in it.

also See also https://chemistry.stackexchange.com/a/63901/16035

Well, hygroscopic does capture the initial property of a salt attracting humidity, so it seems relevant: en.wikipedia.org/wiki/Hygroscopy#Deliquescence . As for the lack of hydrated forms, well, that's interesting in its own right don't you think?

What is this process called exactly? As far as I understand
"deliquescence" is a property and not a process. Does the exchange of
water between my pile of salt and the air have a specific terminology?
Something that would go along with terms like evaporation,
sublimation, condensation, etc.? Perhaps some mash-up of adsorption
and dissolution?

The property of salts to become wet as I mentioned in another post Scientific Reason for Salt Solution Gaining Volume is deliquescence. The corresponding verb is to deliquesce. It is a pretty old word. From OED (not open access):

To melt or become liquid by absorbing moisture from the air, as certain salts.
1756   C. Lucas Ess. Waters i. 14   They attract the humidity of the
air, and deliquesce, or run liquid.
1781   Philos. Trans. 1780 (Royal Soc.) 70 349   This
pot-ash..deliquesces a little in moist air.
1876   D. Page Adv. Text-bk. Geol. (ed. 6) xvi. 299   Pure chloride of
sodium is not liable to deliquesce.*

Regarding phase changes, there were many phases changes involved in that process. The water as gas became liquid. The solid NaCl phase become one phase with a liquid. Once it evaporated, you got the solid again.

Thank you for the historical references! For item #2, would you expect such a plateau?

@uhoh The salty solution with excess of solid salt, or solid salt alone, would try to keep the relative humidity at 75%, until all salts dissolves. Then the humidity will raise, until the saturated vapour pressure above the diluted salt solution equals the vapour pressure. Note that relative humidity above saturated solution is temperature dependent, see the provided link.

@Poutnik If you would like to add an answer here, not just linking-to, but actually citing from and explaining Humidity Fixed Points that would be excellent!

It was not nice to provide a big PDF link without warning to expiring mobile data plan. :-)

@Poutnik sorry, I don't even have a mobile data plan, lucky you! :-) I'll try to remember to flag pdfs in the future.

fyi more water: Does water really have strong EM absorption at 3 kHz in solid and 2 GHz in liquid? Why the huge shift?


text:
Halogen bonding in organic reactions
Recently, a group of researchers from the Nanyang Technological University in Singapore have investigated the novel $\ce {S_N2X}$ mechanism which involves a "frontside attack" on the substrate, rather than the usual "backside attack"$^1$. This may seem peculiar at first. However, the fact that there is favourable halogen bond formation between the electrophilic halogen atom and the incoming nucleophile certainly does help to explain why this unusual "frontside attack" would occur.
There are quite a few other similarly peculiar reactions in organic chemistry in which the nucleophilic species first attacks a halogen atom. These include the Appel reaction and the Corey-Fuchs reaction. I was wondering if the formation of an intially halogen-bonded complex the reason why these reactions can proceed.
Reference

Zhang, X.; Ren, J.; Tan, S. M.; Tan, D.; Lee, R.; Tan, C.-H. An enantioconvergent halogenophilic nucleophilic substitution ($\ce {S_N2X}$) reaction.  Science 2019, 363, 400-404.


text:
Home-grown diamond using CVD
I want to make a diamond at home using chemical vapor deposition (CVD).

If I use a metal chamber filled with methane and hydrogen gas and heated with a microwave oven oscillator element, will the gas explode?
What is the ratio of methane and hydrogen for the diamond to be able grow?

You can't do that at home. Think of something more realistic, like a nuclear reactor.

The easiest way to grow diamond through chemical vapor deposition (CVD) is on a heated filament, which just takes a few watts.
You'd need a low-pressure chamber (~30 torr, or ~0.04 atm) with gas-tight fittings for electricity, feed-gas of ~1% $\ce{CH4}$ and ~99% $\ce{H2}$ and $\ce{Cu}$, $\ce{Ti}$, $\ce{W}$ or $\ce{Ta}$ filaments. Since this is a flow-though reactor, you'll also need to dispose of the flammable feed-gas without letting it accumulate near the vacuum pump.
N.B. This is not a simple project one could do at home, since you'd need vacuum equipment, pressure regulators, high-pressure bottles* of these highly flammable gases, and a vacuum-vessel which could implode. If you can get a knowledgeable instructor involved who has access to laboratory facilities and a thorough understanding of safety procedures, though, it's not impossible to accomplish.
* One of the greatest dangers would be a gas bottle falling, taking off like a rocket and leaving explosive gas in its wake!

Logically for a 2 gram diamond I need the equivalent weight of carboon in the methane gas, theoretically I don't have to feed the gas constantly into the chamber, only to feed it once along with the hydrogen gas and let it react with the graphite seed, is this true?

Probably not true, because, apparently, the process requires a constant ratio of methane to hydrogen. Of course, this could be subject to experiment...

What is the purpose of hydrogen in the reaction, why can't I use only methane, or instead of methane only carbon dioxide witch doesn't explode, as far as I know real diamonds are made of CO2

There are many allotropes of carbon, e.g. graphite and diamond. See https://en.wikipedia.org/wiki/Allotropes_of_carbon .Did the reference above give any information why H2 helps diamond form rather than something else? This may have been determined empirically or from some theoretical basis.


text:
How to chelate magnesium oxide and l-lysine
Given magnesium oxide and l-lysine dissolved in water. What needs to happen so that both form a chelate inside the solution.

Not a complete answer, but too long for a comment.
First recognize what is really in solution.  Magnesium oxide does not dissolve in water.  It reacts to give magnesium hydroxide, which in turn is only sparingly soluble in water alone but could react with the acidic lysine.  You should proceed as if your solution had magnesium ions and lysine anion -- because after the reactions involving the magnesium oxide that is what you end up with!


text:
What makes an isotope stable?
I have heard many times that when there is the “right balance” of protons and neutrons the atom will be stable, and therefore it will decay to achieve this stability. But what about this supposed balance of protons and neutrons makes it stable? Does it have to do with quarks? Density? I am quite curious as to what causes this.

To be honest, no matter what we come up with, in the end it is just the way the universe works:-)

Useful reading: https://en.wikipedia.org/wiki/Valley_of_stability

Simply put, stable isotopes are particular combinations of neutrons and protons which minimise the total energy of the system, which is the sum of kinetic energy, electric potential energy, and strong nuclear potential energy. Unfortunately, that last term is fiendishly difficult to calculate/understand theoretically. Furthermore, local energy minima can lead to stable isotopes, depending on the kinetics of weak nuclear force interactions. We understand the weak nuclear force quite well, but kinetics is naturally a harder topic.

There is a shell structure within the nucleus also. So the neutron/proton ratio varies from 1:1 for the light elements to 1.5 for the heavy elements.

Note: I ripped off the image from another website. Not sure where they got it from...

Looks like it's Figure 21.2 Belt of Stability that's been replicated with pgfplots from Brown's Chemistry: The Central Science 8th Ed. (2000). A more recent 14th Ed. (2018) has an updated plot (for the reference it's Figure 21.1 on p. 950).


text:
Organic conversion involving Grignard reagent and alkyl group with a carbonyl carbon on an benzene ring
I have been asked to synthesize the organic compound shown below using the following reagents and solvents.
$\ce{Mg, Zn, Hg, FeBr3, AlCl3, Br2, LiAlH4, KMnO4}$, dil. $\ce{H2SO4}$, conc. $\ce{NH3}$, conc. $\ce{HCl, H2O, D2O}$, ether, ethanol, $\ce{CH3C(O)Cl, CH3C(O)CH3, C6H6}$.
I have done it as shown below.

But in the answer given that the -$\ce{C(O)CH3}$ group attached to the benzene ring is reduced to -$\ce{CH2CH3}$ before introducing $\ce{Mg}$/dry ether.
Then the $\ce{Mg}$/dry ether is introduced followed by $\ce{D2O}$. Then the alkyl group is oxidized and so on...
Is there is specific reason for doing that or is my method correct?

Cannot form a Grignard in the presence of a ketone. Form an acetal of the ketone then do the Grignard and your scheme will work

I would also be inclined to reduce the amide with LiAlH4

HBr is formed during the ring bromination of acetophenone. Be aware that acid catalyzed bromination of the methyl group may occur.


text:
Why the concentration with respect to time equation of a first-order reaction does not depend on the stoichiometric coefficient of reactant?
According to my Chemistry textbook (Chemistry by Raymond Chang 10th edition, pg 570), for a first-order reaction of the type $$\ce{A->product}$$
the rate is $$rate=-\frac{[\Delta A]}{\Delta t}$$
From the rate law we also know that $$rate=k[A]..... (13.2)$$
Combining the two equations for the rate we write $$-\frac{[\Delta A]}{\Delta t}=k[A]$$
And finally, using calculus, $$\ln\frac{[A]_t}{[A]_0}=-kt...... (13.3)$$
All of this feels okay to me. However, later on, the book uses the same equation (13.3) to solve a first-order reaction of the type $$\ce{2A->product}$$ and I don't understand why. Because the rate of reaction would now be
$$rate=-\frac12\frac{[\Delta A]}{\Delta t}$$ and equating this relation with (13.2) gives $$-\frac12\frac{[\Delta A]}{\Delta t}=k[A]$$ and finally $$\ln\frac{[A]_t}{[A]_0}=-2kt$$ I know I'm probably misunderstanding something. Please can someone explain this to me. I've been trying to figure it out for days now.
[Edit] The problem that I'm referring to is a Practice Exercise, the book only gives the problem and the answer, not the solution. However, based on the answer is obvious that equation 13.3 was used to solve it.
Problem: The reaction $\ce{2A->B}$ is first order in A with a rate constant of $\pu{2.8 \times 10^{-2}s^{-1}}$ at $\pu{80 ^{\circ}C}$. How long (in seconds) will it take for A to decrease from $\pu{0.88M}$ to $\pu{0.14M}$?
Answer: $\pu{66s}$
If equation $$\ln\frac{[A]_t}{[A]_0}=-2kt$$ were to be used, the answer would ended up being $\pu{33s}$.

It could be that the 2k here be changed to a, which one could designate as the rate constant. As far as I can tell, it just seems as a change of variable.

Hi! I added (as an edit) the practice excercise where the book uses equation 13.3 to solve a reaction of the type 2A --> B

@user79161 But in the case of the practice excercise above, how would you know that the rate constant that was given is 2k or a?

The genuine of your question based on the idea which confused a lot of chemistry teacher may be this link help you:  https://chemistry.stackexchange.com/questions/38167/do-coefficients-matter-in-rate-law/38169#38169

Thanks @AdnanAL-Amleh !! but I unable to see how that thread answers my doubt

The key point which you will study in advanced classes, is the concept of elementary reactions. When a given reaction is an elementary reaction, the stoichiometric coefficients become the power of the rate law or in better words, the reaction order corresponds to the stoichiometric coefficients.
Definition of elementary reaction: An elementary reaction is a chemical reaction in which one or more chemical species react directly to form products in a single reaction step and with a single transition state (more details on elementary reactions).
Rate of reactions are determined by experiments. We never write the rate law based on the balanced equation. Compare this with equilibrium constant where coefficients are important. Even $\ce{2 A -> B}$ may be proceeding in several elementary steps. The observed rate may be the effect of several combined elementary reactions.
In those kinetic problems like yours, the writer will tell you the order of the reaction. Don't proceed to write the rate law on the basis of balanced equation.
Reaction rate and writing the rate law are two different things. For a reaction
$$\ce{a A + b B → p P + q Q}$$
Reaction is written as
$$r = -\frac{1}{a}\frac{\mathrm d[\ce{A}]}{\mathrm dt} = -\frac{1}{b}\frac{\mathrm d[\ce{B}]}{\mathrm dt} = \frac{1}{p}\frac{\mathrm d[\ce{P}]}{\mathrm dt} = \frac{1}{q}\frac{\mathrm d[\ce{Q}]}{\mathrm dt}$$
but we cannot write the rate law i.e. you do not know what would be the order of the reaction in general.
The relation above only shows how the rate of disappearance of reactants and rate of appearance of products are related. It should not be used anywhere else. Your book answer is correct.

If given a problem with no other restrictions, then an elementary reaction should be assumed. However for the particular problem under consideration  it was explicitly stated that the reaction was first order.

Exactly! @MaxW If it is first ordered and the stoichiometric coefficient of A is 2, aren't I supposed to include the number 2 in there? Since, according to the IUPAC definition, the reaction rate depends on the stoichiometric coefficients  https://goldbook.iupac.org/html/R/R05156.html

@Manuel1299 - First order $r = k\ce{[A]}$ is first order. $r = k\ce{[A]^2}$ would be a second order reaction.

The goal is not to write the rate law. The rate law is already given since it is stated in the problem that is a first-ordered reaction. Hence the rate is $r=k[A]$. Also, based on IUPAC definition of rate, $r=-\frac12\frac{[\Delta A]}{\Delta t}$. Aren't these two expressions equal to each other?

You are missing the punchline of the answer above. The IUPAC definition of the rate just shows the inter-relationship of rates among reactants and products. That is all. It is not the definition of rate. Rate is just [conc]/time- no coefficients. IUPAC relation only helps when you know the rate of a reactant but now you wish to know rate of appearance of a product. The expression r=k[A] is not equivalent to the relationship shown by IUPAC.

Oh ok, now I see your point. But, if that's so, explain to me please why my book writes $-\frac{[\Delta A]}{\Delta t}=k[A]$. I mean, if rate is just [conc]/time- no coefficients, would it be also ok to write  $-\frac{[\Delta B]}{\Delta t}=k[A]$?

No matter what the reaction is, rate = [concentration of a reactant or product]/ time. With reactants we attach a negative sign and with products positive. Again no coefficients are attached. They are only good for comparing rates. This explains the left hand side. Now the book said that this is a first order reaction for A. It means rate = k * [A]. This part must have been determined experimentally. If someone asks what is the rate of appearance of B if the rate of disappearance of A is x? Now you use the IUPAC relationship shown in my answer above.

But by stating "rate = [concentration of a reactant or product]/ time" you are saying that   - [concentration of a reactant]/ time =  [concentration of a product]/ time for any given reaction which is a contradiction don't you think?

Nope this is not a contradiction. We can choose only one species at a time for determining a rate law either a reactant or a product. The rate of disappearance of A is different from the rate appearance of B. They are related by the equation shown by IUPAC.

If equation $$\ln\frac{[A]_t}{[A]_0}=-2kt$$ were to be used, the answer would ended up being $\pu{33s}$.

You cannot substitute $k=2.8\cdot{10^{-8}}$  in the above equation because of the value of $k$ in the last equation of your question equal half the value of k in equation(3.2).
Why the values of k in the two equations are different?
Because they are different in defining the rate:

The equation(3.2) defines the rate as the rate of disappearance of the reactant A:
$$r_1 = -\frac{\Delta[A]}{\Delta t}= k_1[A]$$
The last equation in the question defines the rate as the rate of appearance of the product P:
$$r_2 = \frac{\Delta[P]}{\Delta t}= k_2[A]$$
But , the rate of appearance of the product $P=\frac{1}{2}$disappearance of the reactant$A$
$$\frac{\Delta[P]}{\Delta t}= \frac{1}{2}\frac{\Delta[A]}{\Delta t}$$
$$  k_2[A] = \frac{1}{2} k_1[A]$$
So : $$  k_2= \frac{1}{2} k_1$$


text:
Which "exotic salt" can lower water's freezing point by 70 °C?
The Medium.com article Mars Phoenix Lander, 10 Years Later shows several remarkable images and discoveries on Mars by the Mars Phoenix Lander circa 2008.
One image (shown below) shows what looks like droplets of liquid water, condensed on the surface of one of the lander's legs.
The article says (emphasis mine):

Shortly after landing, the camera on Phoenix’s robotic arm captured views of blobs of material on one of the landing struts. Over time, these blobs moved, darkened, and coalesced, behaving like droplets of liquid water. The hypothesis here was that these blobs “splashed up” on the struts when the descent thrusters melted the ice exposed upon landing mentioned above.
But if liquid water isn’t stable on the martian surface, how did Phoenix observe liquid water on Mars? The key here lies in salt. If you live anywhere that gets snow, you’re probably familiar with salt as a de-icer for roads, sidewalks, etc. Salt lowers the freezing point of water, allowing it to remain liquid at temperatures lower than that of non-salty water. For example, pure water freezes at 0 °C/32 °F, but ocean saltwater freezes around −2 °C/28.4 °F. While the de-icing salts you get at the hardware store lower the freezing point by a few degrees, more exotic salts can lower the freezing point as much as −70 °C/−89 °F! Phoenix discovered some of these exotic salts in the soil around the lander—in particular, magnesium perchlorate. (note, minor editorial changes have been made)

Question: Which "exotic salt" can lower water's freezing point by 70 °C?
Is it in fact magnesium perchlorate (which was found on Mars) or is it a different salt?

Blobs of possible brine (really salty water) imaged on one of Phoenix’s landing struts shortly after arriving on Mars. Credit: NASA/JPL-Caltech/University of Arizona/Max Planck Institute

You know why freezing point of water is decreased by using salt? It is because of relative lowering in vapour pressure which depends on number of particles present in solution, it doesn't matter whatever sizes are or whatever the salt is, the necessary conditions are that the substance you are mixing must be non volatile and it must be a solution. So basically the more salt you mix the lower the freezing point is but there is a limit on how much you can mix salt in water. And freezing point is decreased by a few degrees only. -70° is like a dream. the text you are reading might be wrong.

Also -70 C seems awful cold for a super cooled liquid.

@SauravSingh, vapor pressure has nothing to do with freezing point, though it does affect the boiling point.

@DrMoishePippik as an aside, vapor pressure may have a role in the appearance of the droplets nonetheless. Assuming this is indeed water, it may have been the heat of the spacecraft that drove water vapor out of the Martian soil, allowing some of it to "deliquesce" on the salt-encrusted leg of the spacecraft.

@DrMoishePippik I too realized that I used the wrong term 'vapour' but it was too late I cannot edit. But still freezing point dependent on pressure not vapour pressure.

If "lower water's freezing point by 70 °C" makes its freezing point -70 °C, then "lower water's freezing point by −70 °C", lowering it by a negative amount should raise its freezing point to 70 °C.  I think you should remove that "-" from the title and question itself.

@MontyHarder Done! Yes, thank you for that. I remember feeling a little weird about it at the time, wasn't sure if the change should be "to -70" or "by 70", waffled, then probably got distracted. Interestingly after over 6,000 views you're the first person to comment on it.

Vapor pressure has Everything to do with the freezing point [if head space is present].  At the freezing point the vapor pressure of the solution is equal to the vapor pressure of the solid

@jimchmst it's hard for me to get up to speed here after 2 years; my coffee hasn't kicked in yet and I'm not a chemist. 1) Is your comment directed at me, or the first comment, or to future readers? 2) does "if head space is present" mean there's a sealed container with the "exotic" salt solution and some (albeit low pressure) air above it, rather than being exposed to "open air" on Mars? 3) Doesn't vapor pressure simply correspond to the pressure of H2O in the atmosphere? How could the atmosphere's H2O vapor pressure "of the solution" ever be different than H2O vapor pressure "of the solid"?

I recently got a chance to attend a talk by someone who was working on developing analytical instrumentation on Mars. The interesting story is that the initial results by ion-selective electrode was that Mars soil is full of nitrates. Nobody knew on Earth that the nitrate ion selective electrode is far more responsive to perchlorate than nitrate. After learning this, it was an eye opener for analytical chemists! Now they wish to use chromatography rather than electrochemistry. So this was a good lesson for us on Earth.
The perchlorate ion was discovered in 2008 by the nitrate selective electrode. No specific electrode was attached to detect perchlorate, it was rather an accidental discovery. The Science Report makes a footnote "Detection of Perchlorate and the Soluble Chemistry of Martian Soil at the Phoenix Lander Site" (paper: Science 2009, 325 (5936), 64–67)
A Hofmeister anion ISE was intended to monitor nitrate from a $\ce{LiNO3}$ reference electrolyte that was part of the leaching solution, but was ultimately used for perchlorate detection

[Footnote] The relative sensitivity of the Hofmeister series ISE to perchlorate
  over nitrate is 1000:1, and substantial quantities of perchlorate will
  overwhelm any other signal. If, as was observed, >1 mM perchlorate
  accounts for the observed signal, it would require >1000 mM nitrate to
  produce the same response. This would correspond to more than the mass
  of the entire sample.

Now that they know it is a perchlorate ion, people did some studies on supercooled brines. See this paper: Toner, J.; Catling, D.; Light, B. The formation of supercooled brines, viscous liquids, and low-temperature perchlorate glasses in aqueous solutions relevant to Mars. Icarus 2014, 233, 36–47 (also available here). They clearly show that if calcium or magnesium perchlorates are slowly cooled, one can get supercooled brines up to -120 Celcius. This is a rather amazing finding. They call it a glassy state.

"Nobody knew on Earth" (!!) Let's add this to the list of "Reasons to go to Mars" ;-)

Yes it was a surprising finding because everyone used nitrate selective electrodes without ever realizing that it also responds to perchlorate far more effectively.

Nobody on earth knew? Seems a little over the top. For instance Rocheleau and Purdy "Ion-selective electrode with fixed quaternary phosphonium ion-sensing species", Analyst (1992) Vol 117(2), p 177-179 specifically discuss perchlorate interference with a nitrate ion electrode.

Your Question: Which "exotic salt" can lower water's freezing point by $\pu{-70 ^\circ C}$?

Here is your "exotic compound" although it is not a salt by definition. It is a base: Aqua ammonia, also called ammoniacal liquor, ammonia liquor, or ammonia water, is produced by dissolving ammonia gas ($\ce{NH3}$) in water. The proper chemical name of aqua ammonia is ammonium hydroxide ($\ce{NH4OH}$), which is in the following equilibrium with water:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
Ammonia is very soluble in water: According to Wikipedia, its solubility in water is $47\% (w/w)$ at $\pu{0  ^\circ C}$, $31\% (w/w)$ at $\pu{25  ^\circ C}$, and $18\% (w/w)$ at $\pu{50  ^\circ C}$. Therefore it is ideal to cause large freezing point depression since its solubility increases with decreasing temperature. Now, let's see how are the freezing points of aqua ammonia solutions behave with increasing concentrations. The large scale manufacturer of aqua ammonia, Tanner Industries, listed following values of boiling and freezing points of various solutions in its Customer Manual:
$$
\begin{array}{ccc} \\\hline
\% \ce{NH3} \text{ (by weight)} & \text{Approx. Boiling point} & \text{ Approx. Freezing point} \\\hline
23.52 & \pu{103 ^\circ F}\: (\pu{39.4 ^\circ C}) & \pu{-56  ^\circ F}\: ( \pu{-48.9 ^\circ C})\\
25.48 & \pu{95 ^\circ F}\:  (\pu{35.0 ^\circ C}) & \pu{-69  ^\circ F} \: ( \pu{-56.1 ^\circ C})\\
27.44 & \pu{88 ^\circ F} \: (\pu{31.1 ^\circ C}) & \pu{-89  ^\circ F}\:  ( \pu{-67.2 ^\circ C})\\
29.40 & \pu{85 ^\circ F}\:  (\pu{29.4 ^\circ C}) & \pu{-110  ^\circ F} \: ( \pu{-78.9 ^\circ C})\\
31.36 & \pu{73 ^\circ F}\:  (\pu{22.8 ^\circ C}) & \pu{-123  ^\circ F} \: ( \pu{-86.1 ^\circ C})\\
33.32 & \pu{66 ^\circ F} \: (\pu{18.9 ^\circ C}) & \pu{-148  ^\circ F}\:  ( \pu{-100 ^\circ C})\\\hline
\end{array}
$$
Accordingly, anything between $29-33\%$ of aqua ammonia solution would do the job.
On the other hand, if you are looking for only a "exotic magic salt," then, don't look too far: Ammonium fluoride ($\ce{NH4F}$) would do the job. The solubility of $\ce{NH4F}$ in $\pu{100 mL}$ of water is listed as $\pu{100 g}$ at $\pu{0  ^\circ C}$ (Wikipedia). That's give you a $\pu{27.0 m}$ solution at $\pu{0  ^\circ C}$. Theoretically, that would freeze at
$\pu{-100.4  ^\circ C}$! ($\Delta T = \pu{27.0 m} \times 2 \times \pu{-1.86 ^\circ Cm^{-1}} = \pu{-100.4  ^\circ C}$)

@uhoh I'd ask the definition of a salt, if $\ce{NH3}$ or any of the $\ce{NH_4X}$'s don't fit the bill...

@StianYttervik I'm just going by what the author says in the first sentence:  "...although it is not a salt by definition." If you feel that it can be a salt, then that would be great! If so, perhaps you can propose an edit to adjust the wording?

@MathewMahindaratne Excellent, thank you for that! I've made a formatting change to start a new paragraph to improve the visibility.

Ixnay on the ammonium fluoride calculation.  Solutions, especially of ionic compounds, are far from ideal when they are as concentrated as quoted in the edit; therefore the linear dependence of freezing point depression on on concentration does not apply.

@Oscar Lanzi: I agree. It is an important point. Similar to Beer's law. That's why I stated as "theoretically"! :-) Also, thank you uhoh for the edit.

"ammonium hydroxide" isn't a proper name for anything...

Magnesium perchlorate is far from unique.  In fact, if you're willing to spend a little money at that hardware store you could pick up some calcium chloride, whose eutectic reaches about -50°C, not quite as low as magnesium perchlorate but still good enough to cover much of the temperature range on Mars.
Hydrogen chloride, which becomes ionic upon reacting with water, gives a liquid down below -70°C according to a German reference:  "Systemnummer 6 Chlor, Ergänzungsband Teil B – Lieferung 1". Gmelins Handbuch der Anorganischen Chemie. Chemie Weinheim. 1968.

Hydrogen chloride however has a vapor pressure (normal BP is $-85 ^\circ C$)


text:
What are the other ways (if they exist) that atoms can gain or lose electrons that aren't bonding?
I know that electrons can be isolated through human intervention from their atoms. My question rather concerns if electrons are ever lost or gained naturally. I recognize that this is a stupid question but I couldn't find the answer on the Internet.

Transferring electrons between atoms is quite normal, but creating or destroying electrons outright is a different matter. That depends on what you count as natural. Does radioactive decay count?

See electron capture, beta decay, and electron-positron pair production

At high temperature everything loses electron and cannot stay in the bonded state. Check the term plasma.

Also x-ray fluorescence, and Auger effect

And photoionization -- in the beginning, all was ionized...https://en.wikipedia.org/wiki/Photoionization

My question rather concerns if electrons are ever lost or gained naturally.

Yes, they can!
Here are two examples of natural processes by which atoms can gain or lose electrons "naturally":
Electron Capture
In the natural radioactive decay process known as electron capture the nucleus captures an inner-shell electron, K- or sometimes L-shell. This results in a proton being converted to a neutron, and the emission of a neutrino. The atom has lost an electron, but conveniently a proton as well, so it remains neutral but with a deep vacancy.
Auger Process
If the vacancy is filled by a higher shell electron falling into it this can result in the emission of a photon, or in the emission of yet another electron. That emission is referred to as the Auger process.

See Electron capture; Reaction_details for examples.
 Source


text:
When a solution is diluted, the number of moles of the solute decreases. what would happen to the molarity?
When a solution is diluted, the number of moles of the solute decreases, so the molarity decreases??

moles = molarity X volume

Nunber of moles is still the same but the volume of the solution increases

The number of moles of solute does not decrease on dilution.
For example suppose you add 2 moles of NaOH to 100cm3.
Now if you dilute this solution to 1000cm3 you  are still going to have 2 moles of NaOH.
But now the total volume has changed ,hence the concentration also changes.


text:
Electrolytes are ionic compounds that conduct electricity
Electrolytes are ionic compounds that conduct electricity, or the flow of charged particles, when dissolved in water. This occurs via the separated ions acting as those charged particles, or charge carriers, when a voltage is applied. Which compound would not create an electrolyte solution when dissolved in water?
A) ammonium acetate, $\ce{NH4C2H3O2}$
B) sucrose, $\ce{C12H22O11}$
c) sodium hydroxide, $\ce{NaOH}$
D) potassium nitrate, $\ce{KNO3}$

You have to improve your questions because it looks like homework.Most likely your future questions will get flagged.Write the problem you are having when you are asking a question.

Welcome to Chem.SE. For these questions, can you show us and edit in your way of approaching the question so we can help you better? Because what you said sems like you just copy from a textbook and not understanding what it means

Sucrose is an covalent compound and it does not ionize.
$\ce{KNO3}$, $\ce{NaOH}$ are good ionic compounds hence the can form electolytic solutions.
Ammonium acetate also can ionize.
Hence the answer is 2

Hey! It's not a hw question, its asking about Which compound would not create an electrolyte solution when dissolved in water

We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you.
Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework.
Don't worry, they're not banned. But, we require a minimal effort.
Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.
See homework


text:
The products of a chain reaction beginning with acetone and sodium amide
I encountered the following question in an exam:

$\ce{CH3COCH3 + NaNH2}$  giving out $\ce{A}$.
$\ce{A}$ then reacts with $\ce{C2H2}$ giving out $\ce{B}$.
$\ce{B}$ then reacts with $\ce{H+}$ to give $\ce{C}$.
$\ce{C}$ then reacts with 1 equivalent of $\ce{H2}$ in presence of $\ce{Pd}$ to give $\ce{D}$.
$\ce{D}$ then is heated to $\pu{400 ^{\circ}C}$ in presence of $\ce{Al2O3}$ to give $\ce{E}$.

I reasoned it as the first reaction could be an acid-base reaction, but that didn't help much. Then I tried a nucleophilic substitution, but there aren't any good leaving groups and still we get stuck after the first step.
One of the teachers reasoned it as the first reaction being reversible and it goes back to $\ce{CH3COCH3}$ which then reacts with all other and it ends up at isoprene which is the correct answer according to the answer key, but this explanation seems very flawed to me as then what was the purpose of the first step? Can someone please check and write the steps and products in each reaction?

Acetone and sodamide give acetone sodium enolate A, nothing further happens until acetylene is introduced. The pKas of acetone and acetylene are both around 25 so an exchange occurs reforming acetone and creating sodium acetylide. This seems an unusual way of making sodium acetylide, but I don't think this is a real world example (unless someone can find a reference).
Sodium acetylide adds to acetone giving the sodium alkoxide of tertiary alcohol B
Work up with dilute acid gives the tertiary alcohol C.
Hydrogenation over Pd catalyst reduces the triple bond to a double bond giving D.
Passing D over alumina at 400C dehydrates the tertiary alcohol to give isoprene.

Nice answer to a poorly formulated reaction sequence.  Given the pKa's of acetone (~20) and acetylene (~25), a practicing chemist would prepare sodium acetylide first and then add acetone. One equivalent of hydrogen in the presence of "Pd" will not be selective, also giving over reduction of the initially formed alkene. That's why we have Lindlar's catalyst. Alas!

Is it possible for an aldol self-condensation to happen in the first step? I went along a different line of thought. The product then undergoes DA reaction with acetylene.

An aldol, as the alkoxide, is likely reversible. Aldol, itself,  is difficult to prepare in the presence of base IN the solution of acetone. A Soxhlet technique is used. You are presuming that mesityl oxide is formed. Less likely. Unsure what you had in mind for a Diels-Alder between mesityl oxide and acetylene.


text:
Plotting the Arrhenius curve without the rate constant
I was carrying out an experiment about finding the activation energy of the Maillard reaction using the Arrhenius equation. Since I didn't know the rate constant for the reaction due to the complex reaction mechanism of the Maillard Reaction, I tweaked the rate reaction to fit into the Arrhenius reaction.
$$\text{Rate}=k[\text{glucose}]^m[\text{asparagine}]^n \\\implies k=\frac{\text{Rate}}{[\text{glucose}]^m[\text{asparagine}]^n } \\\implies k=([\text{glucose}]^m[\text{asparagine}]^n)\,\frac{\mathrm d[\text{product}]}{\mathrm dt} \\\implies k=\frac{c}{t}$$ where $c=([\text{glucose}]^m[\text{asparagine}]^n)\,\mathrm d[\text{product}]$. By ln both sides, I got $\ln k=\ln c-\ln t$ which I then subbed into the Arrhenius equation to get $$\ln c-\ln t=-\frac{E_\mathrm a}{R}\times\frac{1}{T}+\ln A$$ By manipulating, I got $$\ln t=\frac{E_\mathrm a}{R}\times\frac{1}{T}-\ln A+\ln c$$ Which then produce the below graph based on the experimental data
However, the gradient is positive which technically should be negative according to other Arrhenius plots I have seen. Does anyone know if this is correct or is there something I missed out?

Ignoring whether the manipulation you give is correct, experimentally it is possible to have a positive slope if the reaction is complex. For example if the rate expression has the form $k_{overall} \sim k_1/k_2$ where both rate constants have an Arrhenious form, the result is proportional to $e^{-(E_1-E_2)/RT}$ so the effective activation energy $E_1-E_2$ can be positive or negative.

I implicitly assume the reaction is stopped at time t at the same concentration of the product, so c is constant. otherwise the slope Ea/R is affected by additional slope d(ln c)/d(1/T)

But as X axis variable is 1/T,  the ln t as Y should raise with raising X. as rates increase with temperature. Is possible you have seen charts ln rate versus ln T ?

So this graph is actually possible? The other graphs showed lnk

Please remember that descriptive terms or names of quantities (such as $\text{Rate}$) shall not be arranged in the form of an equation and shall not be used in the place of proper quantity symbols.

Could Rate be the change in product over time? Since the change in product is always constant since I was measuring the time taken for completion, then Rate=Total product/Time where time is the variable?

@Matthew Tan Yes, the rate is the 1st derivative of the product concentration over the time, what is the same as the product change over time, if it is constant over time. But why do you ask, when you have already correctly stated it in the question ? You can easily replace ln t by ln rate as Y variable.


text:
Could this solution be resolved by titration?
I have a mixture of $\ce{NH_4NO_3}$ and $\ce{NH_3}$ at $\ce{pH}$ of about 13.
It is possible to determine the concentration of both on them by acid-base titration?
I was told it's possible, and that you need first titrate with $\ce{HCl}$ and then with $\ce{NaOH}$.
Thanks.
Edit: In the lab were I was working, there were an auto titrator, that had a program that could tell the concentracion of $\ce{NH_4NO_3}$ and $\ce{NH_3}$ in a solution, just by titrating it and controlling the $\ce{pH}$ with a pH meter. I simply don't understand how it does it, because depending of the $\ce{pH}$, $\ce{NH_4+}$ and $\ce{NH_3}$ would be in the same form and I don't know how you could tell them apart.

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Note that at pH=13, NH4+ ions are practically all converted to NH3 by OH- ions. Notice the pKb constant for ammonia and what NH4+/NH3 concentration ratio exists at pH 13.

The ammonium nitrate dissociate in water:
$$\ce{NH4NO3 -> NH4+ + NO3-}$$
The $\ce{NH4+}$ and $\ce{NH3}$ form together the ammonia $\mathrm{pH}$ buffer with $\ce{NH4+}$ acidity constant $\mathrm{p}K_\mathrm{a}=9.25$:
$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+ \log { \frac {c_{\ce{NH3}}}{c_{\ce{NH4+}}}}$$
The titration curve as $\mathrm{pH}=f(V)$ is the most flat at $\mathrm{pH}=\mathrm{p}K_\mathrm{a}=9.25$, where the buffer has the biggest capacity.
For a given solution, containing $\ce{NH4+}$ and $\ce{NH3}$, titration with $\ce{HCl}$ or $\ce{NaOH}$ causes  respective acid-base reactions:
$$\begin{align}
\ce{NH3 + H+ &-> NH4+ \\
NH4+ + OH- &-> NH3 + H2O} \\
\end{align}$$
The ammonia buffer causes   $\mathrm{pH}$ being quite stable during titration, until one of the 2 ammonia forms is practically spent.
At that point, the $\mathrm{pH}$ titration curve gets very progressively bent up or down, respectively and the titrator stops the titration.
The content of $\ce{NH4NO3}$ and  $\ce{NH3}$ is then obtained by obvious routine calculation.
$$\begin{align}
m_{\ce{NH3}}&= M_{\ce{NH3}}\cdot c_{\ce{HCl}}\cdot V_{\ce{HCl}}\\
m_{\ce{NH4NO3}}&= M_{\ce{NH4NO3}}\cdot c_{\ce{NaOH}}\cdot V_{\ce{NaOH}}\\
\end{align}$$
Note that at $\mathrm{pH}=13$, $\ce{NH4+}$ ions practically do not exist, as $ \frac {c_{\ce{NH3}}}{c_{\ce{NH4+}}}=5620$
If we consider $\mathrm{pH}$ of ammonia solution itself, it is
$$\begin{align}
\mathrm{pH}=14 - 0.5 \cdot ( \mathrm{p}K_\mathrm{b}-\log c) \\
13=14 - 0.5 \cdot ( 4.75 -\log c) \\
\end{align}$$
$c_{\ce{NH3}}$ would be 563 mol/L, what is nonsense. Such $\mathrm{pH}$ is not achievable by ammonia alone, but a mineral hydroxide would be present.
If there was initially any ammonium nitrate, it would be converted to ammonia:
$$\ce{NH4NO3 + NaOH -> NaNO3 + NH3 + H2O}$$
In such a case, determination of both compounds by titration is not possible, as there are no ammonium ions and ammonia would be titrated together with hydroxide.

Ok, I get it, thanks. I don't know why, but such a simple thing and I was completely stuck. And sorry, $\mathrm{pH = 13}$ was just an example, but now I see that it would make no sense with ammonia alone.
So the titrator would first titrate with $\ce{HCl}$ and then with $\ce{NaOH}$, having to titrate first the $\ce{HCl}$ that it used.


text:
Aromaticity of Tropone
When we are determining if a given compound is aromatic using Hückel's rule, we count the number of delocalized $\pi$ electrons.
In tropone for example, if I wanted to count the number of delocalized $\pi$ electrons, should I count the $\pi$ electrons of the $\ce{C=O}$ bond?
Or is the aromaticity of tropone because we can draw a more stable aromatic resonating structure in which the $\ce{C}$ acquires a positive charge and the $\ce{O}$ a negative charge?

See also: Do the π-electrons from the C=O bond in tropone contribute to Hückel 's rule?