text:
Does pure HCl (without water) is an acid?
Hydrochloric acid is always discussed as dissolved in water, and acidity is measured in terms of pH by the concentration of the hydronium ion.
Is pure HCl still an acid, and if so, how can we measure its acidity?

HCl without water is gas. It is called hydrogen chloride and it is not an acid.

@Nilay Ghosh but HCl recives lone pair from oxygen in water when dissolves in water ( lewis acids recieve lone pair no?)

Does this answer your question? Is HCl(l) acidic?

Pure HCl is a gas that does not react with $\ce{CaCO3}$

If we look at the classic case of ammonia and hydrogen chloride in the gas phase, the answer is yes -- but it's more complicated than one might think.  The salt does form, but it requires either a proton-transfer medium such as water vapor or the formation of particles out of the gas phase to actually get the ions.
The acid-base reaction is formally,
$$\ce{HCl + NH3->(NH4)^+Cl^-}$$
but according to an an ab initio calculation by Cazar et al. [1] a single pair of molecules do not actually form the indicated ions.  They form instead a hydrogen-bonded species in which the would-be transferred proton remains primarily bonded to the chlorine atom.  Only with the introduction of water molecules do we find the formation of ammonium and chloride ions in the gas phase from a single molecule of ammonia and one of hydrogen chloride.
An empirical study by Countess et al. [2], going beyond the work if Cazar et al., considered both the initial formation of $\ce{NH4Cl}$ monomers and their subsequent polymerization to form solid particles. They demonstrate the formation of ammonium chloride salt as condensed particles in anhydrous nitrogen and give a rate constant in terms of molecules per unit volume for formation of the monomeric gas-phase species, which is polar and attracts other monomers or reactant molecules to nucleate the particles. Then, during this polymerization phase, the ionic lattice is definitively formed Chaban et al. [3] compute that even two ammonium chloride entities are sufficient to form an ionic structure.
References

Robert A. Cazar, Alan J. Jamka, and Fu-Ming Tao,
"Ab Initio Investigation of Proton Transfer in Ammonia−Hydrogen Chloride and the Effect of Water Molecules in the Gas Phase", J. Phys. Chem. A., 1998, 102, 26, 5117–5123.

Richard J. Countess and Julian Heicklen, "Kinetics of particle growth. II. Kinetics of the reaction of ammonia with hydrogen chloride and the growth of pariculate ammonium chloride", J. Phys. Chem. ,1973, 77, 4, 444–447.

Galina M. Chaban, R. Benny Gerber, and Kenneth C. Janda, "Transition from Hydrogen Bonding to Ionization in (HCl)n(NH3)n and (HCl)n(H2O)n Clusters:  Consequences for Anharmonic Vibrational Spectroscopy", J. Phys. Chem. A 2001, 105, 36, 8323–8332.

So Cazar et al. are wrong?

Not wrong, but limited in their calculations. They only considered the structure of the initial monomeric NH4Cl. Ref. 3 shows you have to grow the particle to at least two NH4Cl units to begin proton transfer.


text:
What is the reason that ice melts with increasing pressure at constant temperature (revelation)?
I've seen many explanations regarding this topic where they explain it in the sense of the anomalous expansion of water, but I actually want an answer where I can understand the different forces at play in the molecular level.
I've seen a paper where it is said that when the pressure is applied the h-bonds contract thus making them stronger but this also makes the covalent bond weaker which results in higher probability of the hydrogen to get separated from one of the two oxygen atoms which in turn decreases the energy required to break the bonds therefore lowering the melting point of ice.
Is this explanation correct?

https://chemistry.stackexchange.com/questions/44011/how-does-ice-melt

The term is regelation not revelation.

Sorry auto correct changed it

Try this https://chemistry.stackexchange.com/questions/94117/why-does-increasing-the-pressure-cause-ice-to-melt-more?r=SearchResults

The potential energy between the O atoms in the H bond has the form of an anharmonic potential an example of the shape is the Morse potential or a Lennard-Jones 6-12. The energy is zero at large O-O separation and moves to a minimum as this distance is reduced (the point at which a the H bond exists with the H atom between the two O atoms, typically 0.15-0.25 nm). As the O-O distance is further reduced the potential rises again to zero then positive potential energy. The H bond thus forms in a lop-sided potential well, with the H atoms between the two O atoms. (I imagine the effect on the covalent bonds would be less as they are stronger and shorter than the H bond.)
Suppose that the distance between the O-O atoms is reduced then the potential rises and becomes closer to zero or repulsive (positive) potential. Suppose that the pressure causes this to happen just a little and is such that the H bond energy is now only slightly below zero. At this point the thermal energy (as molecular vibrations) could in places be enough to increase the energy above zero and so break the H bond, at which point the solid would start to melt. The structure beginning to break down could further collapse rather like a building collapses by bending and twisting if it structural girders are removed.

But if I apply this to CO2 the consequence should much more drastic as the molecules are attracted by London forces(which are weaker than h-bonds). But the phase diagram of CO2 contradicts with this fact

Yes, this is a good point, but the vdw are (probably) at shorter range so the effect if any would be limited to far higher pressures. Hopefully someone will know of some experimental data to sort out the actual mechanism.

The key is that when water freezes increase its volume. Only substances that increase their volume when solid should melt if pressure is applied.

@blu potatos can you list some experimental data to support your suggestion ?

If a substance increase its volume when it freezes that means that within the structure there are gaps between molecules (see Cristal structure of ice) and when pressure is applied you are forcing molecules to be close to each other something that is detrimental for keeping their solid structure that involves more spacing between the molecules.

@blu potatos , this is what I have tried to explain in my answer.

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text:
How Do I Determine The Strength Of Two Ammonia Solutions Using Titration Methods?

Two housewives disagreed on the strength of an ammonium cleaning agent. Housewife A thought that her brand was more effective than the brand used by housewife B. Plan and design an experiment that could be used to determine the more effective brand of cleaner.

I know that the aim is to determine which if the ammonium solutions are stronger, more concentrated (as the stronger ammonium solution would therefore be the more effective brand). So, a back titration must be used. $\ce{HCl}$ (hydrochloric acid) could be used in the titration.
However, I am not sure how the experiment would be carried out.

Assume concentration of solution A and B are $a\%$ and $b\%$, respectively (both are $\%(w/v)$). Suppose $a\% \approx b\% \approx 5\%$. Molar mass of ammonia ($\ce{NH3}$) is $\pu{17.03 gmol-1}$. Thus, if molarity of A and B are $M_A$ and $M_B$, respectively, then:
$$M_A = \frac{a \ \pu{g} \ \ce{NH3}}{\pu{100 mL}} \times \frac{\pu{1 mol}}{\pu{17.03 g} \ \ce{NH3}} \times \frac{\pu{1000 mL}}{\pu{1.0 L}} = \frac{10a}{17.03}\pu{molL-1}$$
Similarly, $M_B = \dfrac{10b}{17.03}\pu{molL-1}$.
Let's take $\pu{100 mL}$ of each solution in separate containers. Add to each $\pu{100 mL}$ of $\pu{1.0 M}$ $\ce{HCl}$ (assume volumes are additive). The following reaction would happen in each container:
$$
\begin{array}{lccc}
\ce{& NH3 &+ & HCl & -> & NH4Cl} \\
\text{I} & \frac{10a}{17.03}\times 0.10 && 1.0 \times 0.10 && 0 \\
\text{C} & -1.0 \times 0.10 && -1.0 \times 0.10 && +1.0 \times 0.10 \\
\text{E} & \frac{10a}{17.03}\times 0.10 - 1.0 \times 0.10 && 0 && +1.0 \times 0.10 \\
\end{array}
$$
Therefore, at equilibrium in solution A, $[\ce{NH3}] = \frac{\frac{10a}{17.03}\times 0.10 - 1.0 \times 0.10}{0.20} \ \pu{M}=\frac{10a-17.03}{34.06}  \ \pu{M}$.
And, $[\ce{NH4Cl}] = \frac{1.0 \times 0.10}{0.20} \ \pu{M}= \pu{0.50 M}$.
Similarly, at equilibrium in solution B, $[\ce{NH3}] = \frac{\frac{10b}{17.03}\times 0.10 - 1.0 \times 0.10}{0.20} \ \pu{M}=\frac{10b-17.03}{34.06}  \ \pu{M}$.
And, $[\ce{NH4Cl}] = \frac{1.0 \times 0.10}{0.20} \ \pu{M}= \pu{0.50 M}$.
As a result, Solution A and B become buffer solutions after addition of $\ce{HCl}$. Now, we can measure $\mathrm{pH}$ of each solution. If both are in same strength, both buffer solutions give same $\mathrm{pH}$ value.
Suppose they have given different values. Suppose the values given are: $\mathrm{pH}_{(A)} = 9.55$ and $\mathrm{pH}_{(B)} = 9.45$. Assume for $\ce{NH4+/NH3}$ system, $\mathrm{p}K_\mathrm{a} = 9.25$
Now we can use the Henderson–Hasselbalch equation, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{\text{[base]}}{\text{[acid]}}\right)$, to calculate each concentration (acid and base here are $\ce{NH4Cl}$ and $\ce{NH3}$, respectively):
$$9.55 = 9.25 + \log \left(\frac{\frac{10a-17.03}{34.06}}{0.50}\right) = 9.25 + \log \left(\frac{10a-17.03}{17.03}\right)\\ = 9.25 - 1.231 + \log (10a-17.03) $$
$$\log (10a-17.03) = 9.55 - 9.25 + 1.231 = 1.531 \\ \Rightarrow \ \therefore \ a = \frac{1}{10}(33.96 + 17.03) = 5.099$$
Similarly, for Buffer B:
$$9.45 = 9.25 + \log \left(\frac{\frac{10b-17.03}{34.06}}{0.50}\right) = 9.25 + \log \left(\frac{10b-17.03}{17.03}\right)\\ = 9.25 - 1.231 + \log (10b-17.03) $$
$$\log (10b-17.03) = 9.45 - 9.25 + 1.231 = 1.431 \\ \Rightarrow \ \therefore \ b = \frac{1}{10}(26.98 + 17.03) = 4.401$$
Thus, strength of the solution A is $5.1\%$ while that of the solution B is $4.4\%$. Clearly, solution A is stronger.

For housewives, a solution is simply to add Epsom salt in equal doses to each mix and let the $\ce{Mg(OH)2}$ settle out of solution.
Decant and repeat the experiment until one solution no longer produces any results of a white precipitate. The latter brand is weaker in its aqueous ammonia concentration based on the reaction:
$$\ce{\overset{Epsom salt}{MgSO4 · 7 H2O} + 2 NH3(aq) -> Mg(OH)2(s) + (NH4)2SO4(aq)}$$
To speed up the comparative test, one could experiment with a sample of each brand of ammonia which has been equally diluted with distilled water (record the respective volumes used for future tests).
For calculation purposes, I would start with an assumption that each solution is approximately $5\%$ aqueous ammonia.

What was the point of switching from \underset to \overset in your edit? All style guides I know recommend to utilize space below for naming since the above area is reserved for charges, oxidation numbers and arrow pushing. If you edit the question solely for bumping it, note that this might be considered as abusive activity.

I agree with @andselisk. Writing names above compounds looks so wrong and off-balance.

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text:
On heating in the Earth's atmosphere, can magnesium react with nitrogen to form magnesium nitride?
This question is derived from a question asked in my school test.

What happens when a magnesium ribbon is heated in air?

My first response was the formation of magnesium oxide $(\ce{MgO})$ when oxygen in air reacts with magnesium at a high temperature which can be expressed in the form of a chemical equation like this:
$$\ce{2 Mg(s) + O2(g) ->[\Delta] 2 MgO(s)},$$
but I was wondering if magnesium could react with any other gas in the air to form a compound with that gas and I found out that magnesium does react with nitrogen in the air to form magnesium nitride too:
$$\ce{3 Mg(s) + N2(g) ->[\Delta] Mg3N2(s)}.$$
What determines whether the heated magnesium ribbon will react with the oxygen in the atmosphere or the nitrogen in the atmosphere?
Two possibilities that I can think of are:

composition of the air;
temperature.

I don't think that composition is the answer because on average the atmosphere of Earth has more nitrogen than oxygen, so I think that the answer may be temperature.
I'd also like to know how the factor affects the chemical reaction on an atomic level.

Magnesium Is known to burn in oxygen, nitrogen and carbon dioxide, if heated enough

I think this is one of the first methods of obtaining argon from air.

A large pile of grey magnesium powder, when lit in air, produces a smouldering pile which cools down to reveal a crusty white solid of magnesium oxide. However, if you break apart the mound, you can find something quite strange in the middle - a clearly brownish powder that wasn't there before.
Seeing is believing! The author of the video also has a clever idea to identify the brown solid. By adding water and placing some moist pH paper above the puddle, it clearly shows the transfer of some alkaline substance across the gap. This is ammonia gas, $\ce{NH3}$, whose presence is explained by the hydrolysis of magnesium nitride:
$$\ce{Mg3N2(s) + 6H2O(l) -> 3 Mg(OH)2(aq) + 2 NH3(g)}$$
It is important that the pH paper not come in direct contact with the water used to hydrolyze the magnesium oxide, as $\ce{Mg(OH)2}$ is itself also basic, and could also be formed by reaction with either $\ce{MgO}$ or $\ce{Mg}$ directly. Only $\ce{Mg3N2}$ produces a basic gas which forms an alkaline solution in water.
As you can see, magnesium metal does react directly with molecular nitrogen ($\ce{N2}$) when burned in air. However, the reaction is thermodynamically and kinetically less favourable than the reaction with molecular oxygen ($\ce{O2}$). This is almost certainly due to the extreme strength of the bond between nitrogen atoms in molecular $\ce{N2}$, whose bond dissociation energy of $\mathrm{945\ kJ\ mol^{-1}}$ is one of the strongest in all of chemistry, second only to the bond in carbon monoxide. For comparison, the bond dissociation energy of molecular $\ce{O2}$ is drastically lower, at $\mathrm{498\ kJ\ mol^{-1}}$.
So why did the Chem13 magazine article referenced in Aniruddha Deb's answer not find any magnesium nitride? It is likely that 1 g of magnesium metal is far too little for the experiment run under their conditions. It takes a significant amount of "sacrificial" magnesium to completely consume the oxygen in its surroundings. Only once practically all the oxygen is consumed (and while the pile of magnesium is still hot enough from the reaction between magnesium and oxygen) will the remaining magnesium metal react with the nitrogen in air. Alternatively, the reaction would have to be performed in an oxygen-free environment. Magnesium metal is such a strong reductant that many substances can act as an oxidant for it, including pure $\ce{CO2}$ (also shown in the video above) and water (never put out a magnesium fire with water!).

Re "the reaction is thermodynamically and kinetically less favourable": Is the reaction with magnesium and nitrogen actually exergonic? It could be endergonic and the energy is taken from the reaction with oxygen going on at the same time. What are the actual numbers for the two reactions?

@PeterMortensen That's a good question. We can be certain that the reaction causes a reduction in entropy (a gas is turned into a solid), so now we can just check the enthalpy of formation. According to Heat Contents and Heat of Formation of Magnesium Nitride, Ind. Eng. Chem. 1949, 41, 9, 2027–2031, the enthalpy of formation of $\ce{Mg3N2}$ is $\mathrm{-110.2\ kcal\ mol^{-1}}$ in ambient conditions. So the reaction is decidedly exergonic even at low temperatures, and if anything becomes less favourable at high temperatures.

For comparison, the heat of formation of $\ce{MgO}$ appears to be approximately $\mathrm{-143\ kcal\ mol^{-1}}$ in ambient conditions. The difference is surprisingly smaller than I would have expected, even taking into account that the $\ce{Mg3N2}$ lattice is more strongly bound than the $\ce{MgO}$ lattice.

Actually, the article linked above also explicitly discusses the free energy of formation of $\ce{Mg3N2}$. The author estimates an entropy of formation using a semi-empirical equation, and then provides an approximate function of free energy of formation (in $\mathrm{cal\ mol^{-1}}$) versus temperature (in $\mathrm{K}$): $\mathrm{\Delta G(T)=-224,240+189.67T+5.00\times10^{-4}T^2+8.13T\ lnT}$. At ambient conditions, this corresponds to $\mathrm{\Delta G=-153.8\ kcal\ mol^{-1}}$. Therefore, the reaction is clearly exergonic. The formula also estimates the decomposition temperature of 1500 °C

I don't think that composition is the answer because on an average,
the atmosphere of Earth has more Nitrogen than Oxygen, so I think that
the answer may be temperature.

Indeed, temperature is an important factor for this reaction and the reaction carries at a specific temperature. It was intensively studied in the late 19th century$\ce{^{[2]}}$. It was predicted that the reaction begins at $\pu{450 ^\circ C}$ and proceeds most intensely at $\ce{600-700 ^\circ C}$ at atmospheric pressure of ammonia($\pu{1003 kPa}$ at $\pu{25 ^\circ C}$). The temperature dependence of the reaction was found to be parabolic in nature. Later, it was  established that magnesium nitride can be formed by heating magnesium in air by means of a gas burner. Researchers at that time suggested that a relatively high temperature was needed for the reaction to get going specifically in the range of $\pu{700-900 ^\circ C}$. Interaction of magnesium with nitrogen started at $\pu{780-800 ^\circ C}$ and within 4-5 hrs at a temperature of $\pu{800-850^\circ C}$, the nitride is formed with nitrogen content of $\ce{27.3-27.6 {%}}$ which corresponded to theoretical nitrogen content in $\ce{Mg3N2}$ i.e $\ce{27.4{%}}$.
Notes and references

Encyclopedia of the Alkaline Earth Compounds by Richard C. Ropp
Discovered in 1854 by Saint-Claire Deville during a study of sublimation of magnesium in air. In 1885, it was synthesized by heating magnesium in the atmosphere of ammonia.

Ammonia? Do you mean nitrogen? If not, how does ammonia enter the picture?

@PeterMortensen it was indeed ammonia. Magnesium nitride was first made by burning magnesium in ammonia atmosphere. Later, it was found out that magnesium nitride can also be made by burning magnesium in air at high temperature.

@peter magnesium could displace hydrogen from ammonia. Then you have magnesium nitride plus the displaced hydrogen.

Nicolau Saker Neto's answer does a more accurate job of answering the question. Do give that answer a read as well.
Exactly the same question was pubilshed in University of Waterloo's Chem13 magazine. More details can be found in the link but the conclusion was:

Given that no evidence of $\ce{Mg3N2}$ formation could be found, it appears that the hydration step is not necessary and only makes the experiment more difficult. Not only could no ammonia be detected by smell; within the precision of the electronic scale ($\pu{0.01 g}$) the results were consistent with pure $\ce{MgO}$ being the product.

NOTE: Wikipedia mentions a contrary result

In fact, when magnesium is burned in air, some magnesium nitride is formed in addition to the principal product, magnesium oxide.

However, Wikipedia does not seem to provide a citation for the same. In this case, I would believe in the first reference more than the Wiki article.

On 'heating' in the 'earth atmosphere' elemental magnesium in what form?
This is, in my assessment, actually quite a good chemistry essay type question, because required clarification on the chemistry is likely needed. This arises from noted ambiguities, as for example, with respect to the intensity of the heat applied, the form of Mg present (as a powder or sheet metal), and even the composition of the experiment's atmosphere.
In the case of minor heating in the presence of an 'earth atmosphere' per Wikipedia, to quote:

Magnesium occurs naturally only in combination with other elements, where it invariably has a +2 oxidation state. The free element (metal) can be produced artificially, and is highly reactive (though in the atmosphere, it is soon coated in a thin layer of oxide that partly inhibits reactivity – see passivation).

So one quick answer with mild conditions with bulk magnesium is simply a protective coating of MgO.
However, as we are dealing with an 'earth atmosphere', which can contain water vapor, and per the same source:

Magnesium reacts with water at room temperature, though it reacts much more slowly than calcium, a similar group 2 metal. When submerged in water, hydrogen bubbles form slowly on the surface of the metal – though, if powdered, it reacts much more rapidly.

So, especially powdered Magnesium in the presence of water vapor at room temperature could create some Magnesium hydroxide and hydrogen gas also per the reaction:
$\ce{Mg (s) + 2 H2O (l) -> Mg(OH)2 (s) + H2 (g)}$
Further, assuming sufficient heat and amenability of the form of Mg to ignition, as per the same source, to quote:

Magnesium is highly flammable, especially when powdered or shaved into thin strips, though it is difficult to ignite in mass or bulk. Flame temperatures of magnesium and magnesium alloys can reach 3,100 °C (5,610 °F),[16] although flame height above the burning metal is usually less than 300 mm (12 in).[17] Once ignited, such fires are difficult to extinguish, because combustion continues in nitrogen (forming magnesium nitride), carbon dioxide (forming magnesium oxide and carbon), and water (forming magnesium oxide and hydrogen, which also combusts due to heat in the presence of additional oxygen). This property was used in incendiary weapons...

which apparently adds to the list carbon and magnesium nitride (Mg3N2).
However, again in the presence of water vapor, the nitride is not stable possibly reacting as follows:
$\ce{Mg3N2(s) + 6 H2O(l) -> 3 Mg(OH)2(aq) + 2 NH3(g)}$
So, lastly add ammonia to the product list (for a grade of 'A').

Forming magnesium nitride comes with some subtleties. Consider these two points:

Magnesium in air burns hot. Really hot, as in roughly 2000°C adiabatic flame temperature.

Magnesium nitride is stable under ambient conditions, but "thermal decomposition of magnesium nitride gives magnesium and nitrogen gas (at 700-1500 °C)." The 1500°C figure is the estimate suggested by Nicolau in a further comment to his answer; decomposition could be observed at a lower temperature if the nitrogen were contaminated with oxygen or moisture.

Yet experimental evidence for magnesium nitride is seen in Nicolau's answer.
The solution to this paradox is the quoted adiabatic temperature assumes a stoichiometric mixture. Deep in a large mound it is entirely possible that magnesium is in excess, thus acting as a heat sink, as well as the oxygen being burned out by the outer part of the mound. That is where you get the right temperature range and nitrogen-rich composition to get the nitride identified in the video.
Magnesium has also been reacted directly with nitrogen to form the nitride [1].
Reference
1.
F. Zong, C.Meng, Z. Guo, F. Ji, H. Xiao, X. Zhang, J. Ma, H. Ma, "Synthesis and characterization of magnesium nitride powder formed by Mg direct reaction with N2",
Journal of Alloys and Compounds
508 (1), 15 October 2010, Pages 172-176.


text:
What's the correct definition of activity?
I learned activity equals 'activity coefficient' times 'mole fraction' but i also learned it equals 'activity coefficient' times 'partial pressure / standard pressure (1bar)'.
But there's something really confusing me that if activity coefficient for mole fraction and 'partial pressure / standard pressure (1bar)' is equal, then mole fraction and 'partial pressure / standard pressure (1bar)' should be the same.
But from the definition of partial pressure, which is 'partial pressure = mole fraction times total pressure', 'mole fraction and 'partial pressure / standard pressure (1bar)' should be the same.' doesn't seem to make sense.
So is the activity coefficient for mole fraction and activity coefficient for 'partial pressure/standard pressure' the same?

First of all, these are not definitions (where would you derive eg the activity coefficients?). Second, depending of the physical form we may measure concentration for the activity of solutions, partial pressure for gases, etc. So the role/meaning of activity coefficient is the same, but the corresponding activity coefficient would have the corresponding numerical value and unit/dimension.

A good definition per Wikipedia, to quote:

An activity coefficient is a factor used in thermodynamics to account for deviations from ideal behaviour in a mixture of chemical substances.[1] In an ideal mixture, the microscopic interactions between each pair of chemical species are the same (or macroscopically equivalent, the enthalpy change of solution and volume variation in mixing is zero) and, as a result, properties of the mixtures can be expressed directly in terms of simple concentrations or partial pressures of the substances present e.g. Raoult's law. Deviations from ideality are accommodated by modifying the concentration by an activity coefficient. Analogously, expressions involving gases can be adjusted for non-ideality by scaling partial pressures by a fugacity coefficient.

So, to answer the question: "So is the activity coefficient for mole fraction and activity coefficient for 'partial pressure/standard pressure' the same?", my take on the explanation above is that while one can apply a scale correction factor to partial pressures as opposed to concentration to address deviations from ideal behavior, technically, the latter adjustment factor is referred to as the fugacity coefficient.


text:
What is the reaction mechanism for this reaction?
Is it Polar unsymmetrical bond breaking (heterolytic) mechanism?

Google "free radical halogenation".

It is halogenation by free radical mechanism. In UV radiations halogen gas produces two halogen free radical which can cause allylic substitutions
!

Thank you. Can you please draw the entire reaction mechanism of this by hand ? Its hard to understand without equation and arrow heads

(@Abhishek) This mechanism is not correct. The C–X bond is (primarily) formed via a propagation step $\ce{R^. + X–X -> R–X + X^.}$, not a termination step $\ce{R^. + X^. -> RX}$. The "influence of $\ce{X^.}$" is also not an appropriate way to describe the hydrogen atom abstraction $\ce{R–H + X^. -> R^. + H–X}$.


text:
Identify the structure of compound with C₉H₁₀ formula using IR and NMR data

Use IR and NMR data to define the structure of the molecule, the formula of the given structure $\ce{C9H10}.$

On what basis do we conclude that its structure is that (the answer) only? On what points can we identify the structure of this molecule?
I have found out that since its frequency is just less than $\pu{3000 cm-1}$ it is an alkane. Also, the approximate value of the chemical shift is near $7.2,$ hence it is aromatic.
But what I did not get was the meaning of two peaks at $7.2$ and the shifts at $2.9$ and $2.1.$

There are peaks beyond $\pu{3000 cm-1}$ in FT-IR. What's that means?

Good. Actually they are $\mathrm{sp^2}$-$\ce{C-H}$ stretching. You have an aromatic ring so that's it for IR. Now count how many unsaturation in your compound. You can count out 4 for aromatic ring.

@Andrews: I have given the explanation for peaks in NMR for you to learn.

The molecular formula of the compound is $\ce{C9H10}$. The number of unsaturations in the compound of $\ce{C_nH_m}$ is $\frac{2n+1-n}{2}$. Thus, the number of unsaturations in $\ce{C9H10}$ is $\frac{2\times 9+1-10}{2}=5$. You already found out that compound is aromatic by FT-IR, and hence $\ce{C6}$ of the formula counted out with 4 unsaturaations (3 double bond and a ring for aromatic nucleus).The remaining number of unsaturarion is $5-4=1$, and remaining number of $\ce{C}$ in the formula is 3. The remaining unsaturation could be a double bond or a ring.
When look at $\ce{^1H}$-$\mathrm{NMR}$ spectrum, chemical shifts of all but aromatic peaks are less than $\pu{3.00 ppm}$. Therefore, it is safe to conclude that the compound is a aromatic bicyclic rather than aromatic alkene. Therefore, the compound is indane:

The protons on 1-$\ce{C}$ and 3-$\ce{C}$ are chemically and magnetically equivalent, and benzylic, hence the their relevant signal would be down-field to that of protons on 2-$\ce{C}$, which is usual acyclic aliphatic carbon. The benzylic protons have two vicinal protons, thus the corresponding peak is a triplet. Meanwhile, protons on 2-$\ce{C}$ have four vicinal protons (benzylic protons), thus the corresponding peak is a quintet (5 peaks).
Note: The integration given is a ratio, not correct number of protons.


text:
Why do isotopes have different chemical reaction rates?

Isotopes of a given element differ in the rates at which they undergo chemical reaction.

Why is this so? The only explanation I could think of is that the gravitational force on the electrons increases and hence the electrons are better bound to he nucleus. But this force is very negligible.
Is there perhaps another contributing factor?

Gravity is not relevant, inertia is.

Isotopes differ in mass.
Otherwise the same molecules but with the different mass have the different mean speed.
The different speed means the different molecular collision rate.
The different collision rate means the different kinetic constants.
There are some indirect effects of different isotop mass on the kinetics, but these are much smaller than the direct kinematic effect.


text:
How much time for a water molecule in pure water to reach the top of a glass of water?
I have a curiosity to know how long it will take for a water molecule to move a certain distance in pure water. I can put it simply as how much time for a water molecule to travel the whole length of a typical glass of water if no collective motion is present (therefore very still water with no convective currents).

To simplify things, this is the model:

Suppose I have a completely still and ideally insulated glass of water where the temperature in each part of the fluid is exactly the same (therefore no convective motion).      In addition the top of the glass container is closed and no water air interface is present to avoid evaporation that if present changes the dynamic.    In addition the glass of water is big enough so that the dynamic of the water in the core is not influenced by the the limited range of motion of the water near the borders of the glass.
Since it is a liquid, the water molecules move around each others until the original water molecule is at distance r from where it started(A). In absence of gravity the movement is equally probable in all directions, while with gravity I suppose it may be more likely a layer layout of the water molecules making horizontal movements more likely (see elipse shape of the trace).
The questions are (no need to reply to both):
1) How much time to be at distance r in both situations (gravity or not)? (it will depend on temperature, so let's say at room temperature)
2) How much time roughly will it take for a water molecule to reach reach almost the top of a typical glass of water starting from the bottom?    (in absence of convective motion and evaporation)

Just keep in mind that the gravitational influence on a single water molecule must be negligible as compared to thermal motion.

I am aware that gravity force on the single isolated molecule is negligible compared to its thermal motion but if applied on many molecules it may cause water molecules to form layers sliding on top of each others  and in an enough long distance to reach(like bottom to top of a glass) this may be relevant

Eukaryotic cells have transport systems because they can't wait for materials to get from one to the other side by diffusion. From a certain size up, animals have circulatory systems because oxygen would not diffuse quickly enough to reach the cells far away from the surface. The time of random walks scales with distance square. This calculator says 60 days: https://www.physiologyweb.com/calculators/diffusion_time_calculator.html

I was expecting something that is many hours but actually seems days.

I am posting here the answer since it has been quickly found thanks to the useful comments. To give an estimate the best way is to look at the self-diffusion coefficient.
This is defined as the diffusion coefficient when the chemical potential gradient equals zero, or more simply, it is the diffusion coefficient in a medium of the same molecules.
According to Ref.1, the self-diffusion coefficient of water at $\pu{25^\circ C}$ is $D = \pu{2.3 \times 10^{−9} m^2 s^{−1}}$.  The diffusion time can be calculated as: $$t \approx \frac{x^2}{2D}   $$
As pointed out by @Karsten Theis, you can use the calculator and insert the water diffusion coefficient D.
For example, if the distance x is $\pu{8cm}$ like a typical glass of water and we are at $\pu{25 ^\circ C}$, then $t \approx \pu{16 days}$!
Instead if the temperature is $\pu{90 ^\circ C}$ then it's around $\pu{5 days}$.
I have drawn a graph that shows how much time to reach the top of the glass depending on temperature:

References:

Manfred Holz, Stefan R. Heila, Antonio Sacco, "Temperature-dependent self-diffusion coefficients of water and six selected molecular liquids for calibration in accurate $\ce{^1H}$ NMRPFG measurements," Phys. Chem. Chem. Phys. 2000, 2(20), 4740-4742 (https://doi.org/10.1039/B005319H).

It's actually a bit longer because this is the time to go in any direction, but you have to go up. Also, it is a bit more complicated because any point on the surface will do, and they have slightly different distances from a given point on the bottom of the glass. Finally, it would be great to put error bars because this is just the average in a stochastic process, as anyone going on a random search to find their wedding band on a sandy beach will know.

Also, random walks, varying path lengths,...not exactly a precise answer, more like an approximate statistical expected average (see, for example, the illustration here and also next page https://www.mit.edu/~kardar/teaching/projects/chemotaxis(AndreaSchmidt)/random.htm ).

Yes, I know it's not a precise answer but a very rough estimate just to have the sense of the timescale. However if someone wants to write a detailed answer this is very welcomed.


text:
Hydrogen peroxide and ThickenUp (potassium chroride)
Retr0bright is used to whiten yellow plastic.
The original recipe contains xantham rubber that is a bit hard to find.
I'm thinking about using Nestlé ThickenUp that contains xantham rubber, but it also contains potassium chloride. I know you should be careful with mixing hydrogen peroxide with other chemicals.
Does anyone here know if mixing 12% hydrogen peroxide with ThickenUp should be avoided or is considered safe?
Content of ThickenUp is corn and potato maltodextrin, xanthan gum, potassium chloride.

Hydrogen peroxide is an oxidizer and unstable.  It can react with a variety of common materials to exothermically release oxygen gas, exothermically oxidize the material, or both.  Additionally it can damage skin.  Fortunately these safety concerns are only really a major consideration at higher concentrations, where the amount of oxygen released can be problematic, and the heat is sufficient to pose a burning or fire hazard, made worse by the oxygen.  At lower concentrations (such as the 12% you're talking about), it poses mild hazard to skin, and significant hazard to clothes and eyes, but shouldn't present any major safety concerns.
The potassium chloride shouldn't do anything, but the hydrogen peroxide may react with the maltodextrin, or other unlisted ingredients.  To test for this, you can dilute down your hydrogen peroxide (or buy some medical hydrogen peroxide that is already lower concentration), and mix it with the TickenUp.  If bubbles or foam forms, that's an indicator that the hydrogen peroxide is reacting with the maltodextrin or some other ingredient in the TickenUp, and you shouldn't use it, because it will destroy your peroxide before it can bleach your plastic.
Of note, additional care should be taken to wash the plastic after using this form of Retrobright, because the maltodextrin and potassium chloride may leave an undesirable residue otherwise.  Additionally, the xantham gum is only required if you wish to apply the Retrobright as a gel, if you are planning on submerging the plastic, the xantham gum is unnecessary.


text:
Organic Chemistry Reactions
Would any of you mind sharing the named reactions associated with the conversion of alkanes to haloalkanes.
Thank you.

Have a look at this and the thionyl chloride reaction: https://www.chemguide.co.uk/organicprops/haloalkanes/making.html

There is no direct process for this conversion other than nucleophilic substitution (I am assuming all types are not allowed eg. SN1 SN2 SNi ) but can be done by first elimination leading to alkene and next reacting the alkene with a halogen hydracid (HCl;HBr) .
Another process can be only for primary alcohols to oxidised them to carboxylic acid and then to the salt of silver .next perform the hunsdicker reaction with Br2 (It occurs through free radical mechanism )but in this clreaction haloalkane product is one carbon less than reactant .


text:
Permeability of Nafion membrane
Nafion membrane is used as a proton exchange membrane (PEM) in fuel cells due to its selectively high permeability to hydronium ions $(\ce{H3O+})$ and water.
A permeability study was done on small organic polar molecules like 1-propanol and d1-propanol and dimethyl methylphosphonate [1] showing that those small organic polar molecules can pass through the membrane, however, I can't find anything about bigger molecules like the examples below.

Questions

Is Nafion permeable to polar organic molecules with bigger dimensions than typical solvents?
Is Nafion permeable to non-polar organic molecules?

References

Rivin, D.; Meermeier, G.; Schneider, N. S.; Vishnyakov, A.; Neimark, A. V. Simultaneous Transport of Water and Organic Molecules through Polyelectrolyte Membranes. J. Phys. Chem. B 2004, 108 (26), 8900–8909. DOI: 10.1021/jp037448h.


text:
Why is H5PO5 not called as phosphoric acid?
Recently, my course of inorganic chemistry has begun, and in the nomenclature of inorganic acids, our professor told that phosphoric acid is $\ce{H3PO4}$ and phosphorous acid (orthophosphorous acid) is $\ce{H3PO3}$.
He explained to us that usually, the oxyacid of the highest oxidation state end with the suffix -ic.
So my question is, even in $\ce{H5PO5}$, phosphorous exhibits its highest oxidation state, so why isn't it called phosphoric acid, instead of $\ce{H3PO4}$?

So, the highest oxidation state acids arent supposed to end with -ic?

what are the rules for nomenclature then?

Because there's no such thing as H5PO5!

$\ce{H3PO4}$ is called orthophosphoric acid$\ce{^{[3]}}$. The prefix ortho- generally applies to the highest hydroxylated acid know either in free state or in salts or esters etc. pyro- and meta- denotes successive stages of dehydration. No provision is made in I.U.C(International Union of Chemistry) rules for naming the highest possible hydroxylated acid if they are hypothetical in any way but they are designated by a prefix holo- proposed by the German commission. So, in this case $\ce{H5PO5}$ is called holophosphoric acid.
But this type of nomenclature has been declared obsolete (see: What do prefix ortho, meta and pyro mean in chemistry) and shouldn't be used. Also see: When to use -ous and -ic while naming acids.
Notes and References

Inorganic Syntheses, Volume 2
The nomenclature of mineral acids
The name "orthophosphoric acid" can be used to distinguish $\ce{H3PO4}$ from other "phosphoric acids (phosphorus acid series)". Nevertheless, the term "phosphoric acid" often means $\ce{H3PO4}$; and that is the current IUPAC nomenclature. (Wikipedia)


text:
In what way trans alkenes are more symmetric than cis alkenes?
While providing the reason for the higher melting point of trans molecules, everyone says that trans molecules are more symmetric than cis molecules. An example can be found over here, where the OP states that:

trans-2-Butene has more symmetry than its cis isomer which results in better packing and hence higher melting point.

But to my (uneducated) eyes there doesn't seem any more symmetry (if any) in trans form than cis form of but-2-ene. I mean cis form is vertically and horizontally symmetric, whereas trans form is not.
Why do we call trans form more symmetric than cis form?

When comparing the melting points, the most important point is not the symmetry of the molecule. It is the symmetry of the spatial arrangement of the molecules in the solid state. l

"More symmetric" is not a thing at all. However, trans isomers tend to have central symmetry, which is not the case with cis.

Trans alkenes have a $C_\mathrm{2h} $ symmetry, identified as such because it has a $C_2$ rotational axis (you need to rotate 180° to have an identical molecule) and a mirror plane perpendicular to that axis $\sigma_\mathrm h$. You need to follow the symmetry flowchart to assign symmetry groups.
A $C_\mathrm{2h}$ group is characterised by having:

$E$: the identity operation
$C_2$: a twofold rotational symmetry axis
$i$: a center of inversion
$σ_\mathrm h$: a horizontal mirror plane

Cis alkanes have a $C_\mathrm{2v}$ symmetry; therefore they have:

$E$: the identity operation
$C_2$ :  a twofold rotational symmetry axis
$2σ_\mathrm v$: two mirror planes that include the rotation axis

As you see a $C_\mathrm{2h}$ group has more symmetry elements and therefore sometimes people say "more symmetry" but this is not a technical way of describing it. To say which is more symmetric you need to consider the possible symmetry operations you can do.
Easier answer:
A symmetric object is an object that looks the same if you rotate, reflect etc. The more operations you can do keeping the object identical, the higher the degree of symmetry. For example a sphere is "more symmetric" than a cube because you have much more symmetry operations you can do (you can have infinite rotation axis and infinite mirror planes etc.). In the case of alkenes the main difference between the trans and cis ones is the so-called inversion point (blue drawing in first figure) that only the trans alkene has and therefore is "more symmetric".

Please don't mix bold symbols with MatJax. When people see bold italic serif, they immediately recognize it as a vector. Also, <br> are unnecessary if you are writing in Markdown and pollute the code. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
Order of boiling point of group 14 halides

Find out the species having lowest boiling point:
(A) $\ce{SnCl4}$
(B) $\ce{GeCl4}$
(C) $\ce{SiCl4}$
(D) $\ce{CCl4}$

I have considered the inert pair effect of $\ce{Sn}$ here which will make the compound $\ce{SnCl4}$ least stable (and perhaps the one with lowest b.p.) and hence came up with the answer being option A. But, when I checked the answer, it was option C.
Why was my assumption incorrect and what should be the way of approach here?

https://chemistry.stackexchange.com/questions/134794/why-is-boiling-point-of-carbon-tetrachloride-greater-than-silicon-tetrachloride

@Mithoron, the question is not answered there as well. And this one is quite different since I have asked about 4 of the elements of group 14 whereas the OP in the question that you linked asked for only two.

It's unanswered, but not different. The crucial point is that that for some reason CCl4 has higher boiling point then SiCl4. Unfortunately, you'd have to simply know it. While there have to be "reasons" for that, it's still an anomaly and no rule would let you guess it.

Thank you @Mithoron. I have to memorize it then.

@Sanu_012 You don't necessarily have to memorize it I guess. C has got a much greater capacity for catenation as compared to Si, and so the number of molecules weakly bonded together in CCl4 in the liquid state should be more than SiCl4,and hence,the former should have a higher boiling point.


text:
Calculating the molar volume of a perfect gas under standard ambient temperature and pressure
I am trying to calculate the molar volume of a perfect gas under standard ambient temperature and pressure, but I'm having problems with the units.
$$\begin{align} V_m = \dfrac{RT}{p} &= \dfrac{(8.314 \ \text{J K$^{-1}$ mol$^{-1}$})(298.15  \ \text{K})}{10^5 \text{Pa}} \\ &= 0.024789 \ \text{N$^{-1}$ m$^2$ mol$^{-1}$ Nm} \\ &= 0.024789 \ \text{m$^{3}$ mol$^{-1}$} \\ &= 0.24789 \ \text{dm$^{3}$ mol$^{-1}$}  \ \ \ (\text{1 metre $=$ 10 decimetres)} \end{align}$$
I am told that it should be $24.789 \ \text{dm$^3$ mol$^{-1}$}$.
I would greatly appreciate it if someone would please take the time to help me understand what I'm doing incorrectly.

1 m^N = 10^N dm^N, so 1 m = 10 dm, 1m^2 = 100 dm^2, 1 m^3 = 1000 dm^3

$$\begin{align} V_m = \dfrac{RT}{p} &= \dfrac{(8.314 \ \text{J K$^{-1}$ mol$^{-1}$})(298.15  \ \text{K})}{10^5 \text{Pa}} \\ &= 0.024789 \ \text{N$^{-1}$ m$^2$ mol$^{-1}$ Nm} \\ &= 0.024789 \ \text{m$^{3}$ mol$^{-1}$} \end{align}$$
Here's where you're going wrong.
$1\ m = 10\ dm$. So, $$1\ m^3 = 1000\ dm^3\ !!!$$
That automatically makes it $24.789 \ \text{dm$^3$ mol$^{-1}$}$
Cheers!

Your calculation is correct, except the last line. One meter is $10$ decimeter. But one cubic meter is $1000$ cubic decimeter. So $0.0248$ m$^3$ = $24.8$ dm$^3$. That's all.


text:
What is an ambident substrate?
I had to identify whether a few compounds are ambident substrates or not.
However, as I did not know the term, I took to googling it, but I could not find any explanation for this group anywhere.
I know about ambident nucleophiles and electrophiles, yet I had never heard of ambident substrates.
Please explain what they are.

Where did you come across the term? There are also ambident dienophiles.

@KarstenTheis I was solving a couple of questions from general organic chemistry from a book by the company megacosm.

There is a subsection in March’s Advanced Organic Chemistry [1, pp. 450–451] that provides a straightforward definition and a couple of examples:

10.G.viii. Ambident Substrates
Some substrates (e.g., 1,3-dichlorobutane) can be attacked at two or more positions, and these may be called ambident substrates.
In the example given, there happen to be two leaving groups in the molecule.
Apart from dichlorobutane, and in general, there are two kinds of substrates that are inherently ambident (unless symmetrical).
One of these, the allylic type, has already been discussed (Sec. 10.E).
The other is the epoxy (or the similar aziridine524 or episulfide) substrate.525
Selectivity for one or the other position is usually called regioselectivity.

Substitution of the free epoxide, which generally occurs under basic or neutral
conditions, usually involves an SN2 mechanism.
Since primary substrates undergo SN2 attack more readily than secondary, unsymmetrical epoxides are attacked in neutral or basic solution at the less highly substituted carbon, and stereospeciﬁcally, with inversion at that carbon.
Under acidic conditions, it is the protonated epoxide that undergoes the
reaction. Under these conditions the mechanism can be either SN1 or SN2.
In SN1 mechanisms, which favor tertiary carbons, attack may be expected be at the more highly substituted carbon, and this is indeed the case.
However, even when protonated epoxides react by what is expected to be an SN2 mechanism, attack is usually at the more highly substituted position.526
This result probably indicates signiﬁcant carbocation character at the carbon (ion pairing, for example).
Thus, it is often possible to change the direction of ring opening by changing the conditions from basic to acidic or vice versa.
In the ring opening of 2,3-epoxy alcohols, the presence of $\ce{Ti(O\textit iPr)4}$ increases both the rate and the regioselectivity, favoring attack at C-3 rather than C-2.527
When an epoxide ring is fused to a cyclohexane ring, SN2 ring opening invariably gives diaxial rather than diequatorial ring opening.528
Cyclic sulfates (108), prepared from 1,2-diols, react in the same manner as epoxides, but usually more rapidly:529

524 Chechik, V.O.; Bobylev, V.A. Acta Chem. Scand. B, 1994, 48, 837.
525 Rao, A.S.; Paknikar, S.K.; Kirtane, J.G. Tetrahedron 1983, 39, 2323; Behrens, C.H.; Sharpless, K.B. Aldrichimica Acta 1983, 16, 67; Enikolopiyan, N.S. Pure Appl. Chem. 1976, 48, 317; Dermer, O.C.; Ham, G.E. Ethylenimine and Other Aziridines, Academic Press, NY, 1969, pp. 206–273.
526 Biggs, J.; Chapman, N.B.; Finch, A.F.; Wray, V. J. Chem. Soc. B 1971, 55.
527 Caron M.; Sharpless, K.B. J. Org. Chem. 1985, 50, 1557. See also, Chong, J.M.; Sharpless, K.B. J. Org. Chem. 1985, 50, 1560; Behrens, C.H.; Sharpless, K.B. J. Org. Chem. 1985, 50, 5696.
528 Murphy, D.K.; Alumbaugh, R.L.; Rickborn, B. J. Am. Chem. Soc. 1969, 91, 2649. For a method of overriding this preference, see McKittrick, B.A.; Ganem, B. J. Org. Chem. 1985, 50, 5897.
529 Gao, Y.; Sharpless, K.B. J. Am. Chem. Soc. 1988, 110, 7538; Kim, B.M.; Sharpless, K.B. Tetrahedron Lett. 1989, 30, 655.

Reference

Smith, M. March’s Advanced Organic Chemistry: Reactions, Mechanisms, and Structure, 7th ed.; Wiley: Hoboken, New Jersey, 2013. ISBN 978-0-470-46259-1.

I agreed with andselisk's answer. Yet, Ambident Substrates are not limited to March’s given examples in that answer. A substrate, which can be consecutively acted either as an electrophile of a nucleophile is also called ambident substrate (Ref.1). One of the best example is acetaldehyde or propanaldehyde in aldol condensation:

In 2002, Barbas III & coworkers have reported that L-proline ($\bf{\text{cat1}}$) could catalyze the twofold condensation of propionaldehyde ($\bf{1}$), acting consecutively as a pro-nucleophile and an electrophile, to other aliphatic aldehydes providing lactols, $\bf{3}$, in a moderate yield after intramolecular hemiacetalization (Scheme B)(Ref.2).

References:

Céline Sperandio, Jean Rodriguez, Adrien Quintard, "Catalytic strategies towards 1,3-polyol synthesis by enantioselective cascades creating multiple alcohol functions," Org. Biomol. Chem. 2020, 18, 1025-1035 (DOI: 10.1039/C9OB02675D).
Naidu S. Chowdari, D. B. Ramachary, Armando Córdova, Carlos F. Barbas III, "Proline-catalyzed asymmetric assembly reactions: enzyme-like assembly of carbohydrates and polyketides from three aldehyde substrates," Tetrahedron Lett. 2002, 43(52), 9591-9595 (https://doi.org/10.1016/S0040-4039(02)02412-7).


text:
Is there any easy trick to know if a substance is a Brønsted–Lowry acid or base?
I know that Brønsted–Lowry acids can donate a proton and a Brønsted–Lowry base can receive a proton. I have also memorized the strong acids and bases, which makes it easier to figure out. However, if I was given a substance that I didn't memorize, I would struggle to tell if it is an acid or base. Are there any quick methods to know if a substance is an acid or base? Or do I have to memorize as many as I can?

Has H - can act as acid. Virtually anything can act as base.

Not perfect, but here are a couple rules of thumb to start with for the most common cases.  The key with both rules is to look at the molecular structure.

A Bronsted acid typically has hydrogen attached to oxygen, sulfur or a halogen, all these being relatively electronegative elements.  The hydrogen could also be attached to another nonmetal, especially nitrogen or phosphorous, if the atom is a positive charge center like the nitrogen in ammonium ion.  If the atom is both a positive charge center and oxygen, sulfur or a halogen, you likely have a strong acid.

A Bronsted-Lowry base typically has a nonbonding electron pair that can combine with a proton, thus resembling a Lewis base.  In stable compounds the nonbonding pair us usually on the same kind of atom as one of the elements mentioned above for acids.  This is one reason Bronsted-Lowry acid-base reactions are typically fast and reversible.


text:
In the gold foil experiment why didn't thallium form?
When alpha particles hit the gold nucleus shouldn't they form thallium.
More over shouldn't it also ionize the gold or any atom it hits how does it remain unchanged unionized?

How does titanium not form in Rutherford gold foil experiment?

https://en.m.wikipedia.org/wiki/Geiger%E2%80%93Marsden_experiment

Update your question rather than add comments. Use the edit option. Also be specific in your question to get quality answers otherwise.

Think simple, 2+2= 4, not 3. If you add gold nucleus and helium nucleus would you get a lower mass element such as titanium?

It will be thallium

Most of the alpha particles passed through the gold foil. Negligible amount hit the nucleus and so it didn't change to thallium.

Why do you think a nuclear reaction should happen? The alpha particle is deflected by the gold nucleus. Nobody said it would hit it.

Both the nuclei and $\alpha$ particles are positively charged so repulsion is expected; if the collision energy were utterly vast then it might be possible to fuse the two, but I expect that such energies are not experimentally possible even nowadays.

@porfyrins Available energies are able to break kernels into pieces. Juliet Curie discovered induced radioactivity by alpha particle bombardment of light elements. New super heavy elements are discovered by bombardment of heavy kernels ( like 206Pb) by medium mass/charge kernels. As there is the only stable isotope of gold 197Au, there could be formed traces of 201Tl, that decays with halflife about 3 days by an electron capture to stable 201Hg. But very most of alpha particles would just pass, and from the rest, very most would just get deflected.

@Poutnik, that is interesting, I did'n't know about the Curie experiment. In this case some Tl should be formed by your argument and I greatly overestimated the energy needed because I was thinking of laser fusion experiments.

@porphyrin Needed energy is not problem in various fussion attempts. The problem is energy efficiency, the net energy outcome and sustainable output power. Generally, more energy is spent than produced.

We don't care if a few gold atoms turned into thallium. Ditto for ionization. This is not what we measure here.

@Poutnik - you need another particle (likely a neutron) departing the compound nucleus, so the likely isotope formed is 200Tl. In fact, the reverse reaction, TL-200(N,A)AU-197, can be found in the Evaluated Nuclear Data files, although one needs a fairly energetic (> 3MeV) neutron to make it happen.

@Joe Custer. You are probably right. Definitely for light kernels. I was not so sure for heavy kernels, where the fusion is endotermic.

The Rutherford formula, as derived, assumes purely elastic scattering from the Coulomb force. No formation of a compound nucleus is considered. Generally, for most of the initial experiments, the available alpha particle energies from various decays (in the range of a few MeV) were not high enough for large deviations from pure Rutherford scattering. Of course, folks rapidly wanted both other energies (or controlled energies) or other charged particles (such as the protons used by Cockroft and Walton for their first accelerator measurements).
These days, much higher energy particles are available, and are commonly used in nuclear physics. A nice recent paper on the topic is in Physics Review C, "Cross section of $\alpha$-induced reactions on $^{197}$Au at sub-Coulomb energies". Here, the authors measure the reaction cross-sections for $^{197}$Au($\alpha$,$\gamma$)$^{201}$Tl, $^{197}$Au($\alpha$,n)$^{200}$Tl, and $^{197}$Au($\alpha$,2n)$^{199}$Tl. Although mainly of interest to astrophysical heavy-element nucleosynthesis, the results bear on this question.
The bottom line is that high (> 13MeV) $\alpha$ energies are needed to drive these reactions. At the lowest energy used (~ 13.7 MeV), the cross-section for the $^{197}$Au($\alpha$,n)$^{200}$Tl reaction is 0.67 $\mu$b (micro barns). At 20 MeV it has increased to 37 mb (milli barns).
Anyway, it is quite clear that these reactions do not occur to any measurable extent in the few MeV regime.


text:
Chiral carbons in glucose
In the open chain structure of glucose, one can identify 4 chiral carbon atoms but some sources consider the ring structure and then say that glucose has 5 chiral carbon atoms. If asked to state the number of chiral carbon atoms in glucose, which case should I consider?

Well, it seems you have answered yourself. It depends.

On a mcq test (single option correct type) I was asked this question. It simply said " How many chiral carbons are present in beta D-(+)-glucose? " and there were options for both 4 and 5. What to choose now?

That Greek letter descriptor, "beta" (should be written as: β) is a game changer. It indicates that in fact it's an anomer, i.e. one of cyclic forms, and has 5 chiral carbon atoms.

To elaborate on @mykhal 's comment, the $\beta$ symbol indicates that we are talking about the Haworth Structure of glucose, which is commonly called the the cyclic structure of glucose.

@Firefox1921 To be precise, there's not "Haworth structure", it's a type of projection formula, β is related to the stereochemical configuration, and the "ordinary" structure formula is maybe somewhat more helpful, in the context of natural compounds nomenclature α/β stereodescriptors meaning.

Yes @mykhal, I meant that. My mistake! Sorry!

@Firefox1921 up vote for mature response to critique


text:
Are glycosides stable to acidic hydrolysis?
I recently read a fact about glycosides:

Strong reagents like $\ce{C6H5NHNH2}$ or $\ce{NH2OH}$ are required to decompose it and acidic hydrolysis fails to do so.

Why is it so? What makes it different from other hemiacetals which can be decomposed by both acidic and alkaline hydrolysis?

You are missing the context there...

This information is wrong. Glycosides do undergo acidic hydrolysis. The accepted mechanism of this hydrolysis is as follows1:

References:

Vernon, C. A. “The Mechanisms of Hydrolysis of Glycosides and Their Revelance to Enzyme-Catalysed Reactions.” Proceedings of the Royal Society of London. Series B, Biological Sciences, vol. 167, no. 1009, 1967, pp. 389–401. JSTOR.
Timell, T. E. “The Acid Hydrolysis of Glycosides: I. General conditions and the effect of the nature of the aglycone.” Canadian Journal of Chemistry, vol. 42, no. 6, 1964, pp. 1456–72. doi:10.1139/v64-221.


text:
How is the Born-Lande equation modified when the structure is not NaCl?
Often the Born-Lande equation is quoted (alongwith the calculation of the Madelung constant and Born exponent) with reference to rock-salt structure. But what if we take some other crystal, like rutile $\textrm{TiO}_2$. The Coulomb interaction energy for a reference $\text{Ti(IV)}$ ion is now:
$$
U_{E} = \frac {e^2}{4π\epsilon_0} \sum_{\text{all surrounding ions}} \left( \frac {16}{r_{\text{Ti}^{4+}}} -  \frac {8}{r_{\text{O}^{2-}}} \right)
$$
I'm unable to figure out how does one get the Born-Lande equation using this expression for the electrostatic energy, that for the Pauli repulsion remaining as it is (that is, $U_P = \sum A/r_k^n$). If it were a rock-salt structure, we would say, lattice energy is,
$$
U_{\text{lattice}} = \sum_{\text{entire lattice}} \left( U_E + U_P \right)
$$
And after a little mathematical manipulation we would get,
$$
U_{\text{lattice}} = -\frac{N_A Mz^2e^2}{4πR_0\epsilon_0} \left(1 - \frac1n\right)
$$
However this doesn't seem proper for the case of rutile (and other cases too, like perovskite). Any suggestions are welcome.


text:
In which condition CO uses the oxygen end for bonding?
Generally Metal Carbonyl complex are formed from Carbon end. Is it possible under some specific condition that the Oxygen end is used?
(PS -I know about the structures of M-CO bonding)

Not quite what you are looking for: there probably are some Frustrated Lewis Pair systems with $\ce{CO}$ wherein the oxygen is coordinated. Technically, those are complexes.

Of course it's possible - there's impressive variety of modes of coordination of CO.

@Mithoron I mean under what conditions will CO be O-Coordinated


text:
Alkylation of benzene
I was solving a question where I had to determine a way to convert benzene into n-propyl benzene.
I figured out a way by first performing Friedel-Craft acylation with propanoyl chloride and then performing Clemmensen reduction to form the desired compound.
On checking the solution,  there were 2 alternative ways portrayed:
One was the same method I had used.
The other involved reaction of benzene with n-chloropropane in the presence of $\ce{FeBr3}$.
However wouldn't there be carbocationic rearrangements taking place that convert the carbocation to isopropyl carbocation?
Please tell me what prevents the rearrangements from happening in the second case.

IMO carbocationic rearrangements will occur and cumene will be the major product in the second reaction scheme.

Yes that's what I think. But are you sure that this specific case will also result in the same thing? Because I believe that the alternative solution was presented because of some special property here

https://chemistry.stackexchange.com/questions/83518/friedel-crafts-alkylation-of-benzene-with-propyl-chloride

Initially one would assume that cumene (isopropyl benzene) would be the major product. However, kinetic control does have a say in this reaction, according to Gilman and Means.1

At - 2°, n-propyl bromide with benzene and aluminum chloride gave n-propylbenzene, identified as its sulfonamide. Genvresse (6) obtained both n-propylbenzene and isopropylbenzene by conducting the reaction at reflux temperature. Konowalow (7) found that below $0^\circ$, n-propyl chloride gave n-propylbenzene, while from $0^\circ$ to reflux temperature it gave mixtures of n-propylbenzene and isopropylbenzene. More recently it has been shown that at $-6^\circ$, 60% of the monopropylbenzene was n-propylbenzene, and 40% was isopropylbenzene. At $35^\circ$, this ratio was reversed.

a low temperature would facilitate the production of larger amounts of n-propylbenzene due to kinetic control. However, this is still not a very efficient method of preparing n-propylbenzene, and acylation followed by Clemmensen/Wolff-Kishner reduction would be more suitable here.
References:

Gilman, Henry, R. N. Meals, “Rearrangements in the Friedel-Crafts Alkylation of Benzene” Journal of Organic Chemistry 1943, 08(2), 126–146. doi:10.1021/jo01190a003.

Aniruddha Deb gave an excellent answer to your question. Yet, there is another important point of differences on two Friedel-Craft processes: Alkylation vs acylation. I admit that rearrangement is the major drawback on alkylation process (using alkyl halides with a catalyst). However, another fact is you cannot stop the reaction after monoalkylation. The reaction would proceed to give di- and trialkylated products if steric effect did not stop giving more substitution. The reason for this is because alkylation activates the benzene nucleus to electrophilic substitution. For example, monoalkyl benzene is more active than benzene.
In contrary to direct alkylation, the alkylation by the acylation first followed by reduction method gives you much cleaner product. The reason for this is monoacylation deactivate the ring (recall o,p-directive and m-directive groups). As a result the reaction stop after monoacylation. You got only one product.
Finally, the difference between two reduction conditions: 1) Clemmensen Reduction: This is done in acidic conditions, hence should avoid if your compound contains acid-sensitive groups. 2) Wolff-Kishner Reduction: This is done in basic conditions, hence should avoid if your compound contains base-sensitive groups.


text:
What's the minimum amount of energy needed to remove CO2 from the atmosphere?
I've been learning about CDR (Carbon Dioxide Removal) techniques to combat climate change. These all seem to involve extracting $\ce{CO2}$ from the atmosphere and converting to a mineral, which is then buried.
I am not a chemist, but something seems wrong about this to me.  Wouldn't the decrease in entropy need to balanced by an increase of entropy somewhere else, e.g., by using up a low entropy source of energy?
Is it possible to come up with a theoretical lower bound for the amount of energy needed to extract $\ce{CO2}$ as a proportion of the energy obtained by burning the fuel that released it in the first place?
If this is more than 100% then CDR schemes cannot possibly be an alternative to decarbonising our energy sector asap.  Even if it's less than 100% the calculation is still useful as it allows us to compare the true cost of fossil fuels with renewables, e.g., a value of 33% would indicate that the cost of fossil fuels is 50% higher than advertised (assuming we remove the $\ce{CO2}$, otherwise it's infinitely more expensive!)
(Edit)
Assume the original CO2 came from coal at 24MJ/Kg.
I’m not going to specify the mineral to which this is converted because I’m looking for a theoretical lower bound across all (scalable) technologies.

Suggestion - remove the "proportion of energy obtained" part of the question, since that is dependent on the fuel source and so is highly variable. You can decide yourself what fuel source(s) you want to do the proportion calculation for. Also, please specify the final form of the CO2. If mineral, what mineral and how do you account for the energy cost of supplying the other reactant (eg CaO). Maybe pure room temp CO2 gas would be a better final state? That can be pumped into underground caverns or reacted with something to mineralize.

How about we stop cutting trees first?

I wasn't kidding about trees: To achieve negative emissions, the study recommends using forestry and agricultural practices that increase the carbon-sucking ability of soil and trees, with the goal of pulling 100 gigatons of carbon out of the atmosphere.

Per a source: "Like any other plant, algae, when grown using sunlight, consume (or absorb) carbon dioxide (CO2) as they grow, releasing oxygen (O2) for the rest of us to breathe. For high productivity, algae require more CO2, which can be supplied by emissions sources such as power plants, ethanol facilities, and other sources." Link: http://allaboutalgae.com/benefits/#:~:text=Like%20any%20other%20plant%2C%20algae,ethanol%20facilities%2C%20and%20other%20sources. So, not only remove CO2, make O2 and some algae are even a fuel source. Green technology that is itself green.

Yes, the energetic cost of the carbon capture can't exceed the energy produced in forming the CO2. And, in general, it doesn't -- the net energy production, even with carbon capture technologies, is positive (they wouldn't make any sense otherwise).  So the question becomes:  What are the energetic and economic costs, compared to other alternatives?  This article, from 2019, provides a comprehensive analysis:  https://www.nature.com/articles/s41560-019-0365-7?platform=hootsuite&platform=hootsuite&dom=prime&src=syn

Solar, wind and wave are  low entropy sources of energy so only by using these makes sense for CCS/CDR. (Cutting down forests in the US, shipping wood to the UK and burning this in power stations to provide energy (for CCS) makes no sense). A general thermodynamic analysis of chemical storage is described by G. Porter, J.Chem. Soc. Faraday Transactions 2, 1983, v 79 p 473-482.

@theorist It sounds like you are able to answer the question.  If you can put numbers to the claim that net energy production is positive, would you be able to put that in an answer. Then I can upvote it!
PS  assume the CO2 came from coal at 24 MJ / Kg

@theorist Sorry I didn’t spot you were discussing carbon capture (CCS). My question is about Carbon Dioxide Removal from normal air (CDR). CDR is much less efficient than CCS.

The OP is specifically interested in the efficiency of removing $\ce{CO_2}$ from ambient air, rather than from point sources like power plants.  The former is obviously less efficient because the $\ce{CO_2}$ is far less concentrated.
According to p. 50 of the the SAPEA* Evidence Review Report No. 2
(Schlögl, Robert, et al. Novel carbon capture and utilisation technologies: research and climate aspects. 2018.  Available at : https://www.sapea.info/wp-content/uploads/CCU-report-proof3-for-23-May.pdf),
"the thermodynamic minimum energy required to extract CO2 from ambient air is about 250 kWh/(ton $\ce{CO_2}$)."  This equals 0.99 MJ/(kg $\ce{CO_2}$).
The OP asks that we assume the $\ce{CO_2}$ came from a coal-fired plant that produces 24 MJ/(kg $\ce{C}$) or 6.55 MJ/(kg $\ce{CO_2}$).
Hence (assuming SAPEA's number, and the OP's, are correct), at at the theoretical maximum thermodynamic efficiency, the energetic cost of capturing $\ce{CO_2}$ from ambient air is 0.99/6.55 x 100 = 15% of the cost of the energy generated in producing that $\ce{CO_2}$ from a coal-fired plant.
Of course, this is not a lifecycle analysis.  One needs not only to determine the actual thermodynamic efficiency of the process, one also needs to account for the energetic costs of putting the infrastructure in place for the carbon capture, and for maintaining that infrastructure.
*For more information on SAPEA, see:
https://en.wikipedia.org/wiki/All_European_Academies

Hi I'm a bit confused about that report.  It says "Removing and concentrating CO2 from air (390 ppm) to a pure stream (> 90%) implies a greater energy input and treated gas volume than CO2capture from concentrated point sources.  For example, the thermodynamic minimum energy required to extract CO2 from ambient air is about 250 kWh/tonCO2 [...] A  thermal  energy consumption around 2000 kWh/tonCO2 is considered".   This only applies to purifying CO2 - how would storing as a mineral change the figure of 0.99MJ/Kg?  Also, why do they jump from 250kWh/Ton to considering 2000kWh/Ton?  Thx.

From https://www.ipcc.ch/site/assets/uploads/2018/03/srccs_chapter7-1.pdf: "The  best  case  studied  so far is the wet carbonation of the natural silicate olivine, which costs between 50   and    100    US$/tCO2 stored and translates into a  30-50% energy penalty on   the   original power plant. When accounting for   the   10-40% energy penalty in  the   capture plant as  well,  a  full  CCS  system  with  mineral  carbonation  would need    60-180% more energy than    a  power plant     with    equivalent output without CCS".  NB this isn't a theoretical minimum, and its CCS not CDR.

I've accepted the answer, but now realise that my question was wrong. "Extracting" CO2 can mean just generating a pure gas, which is not particularly useful. It seems this can be done relatively cheaply (4% of original energy extracted from the fuel) but the full cost of converting it to a form that can be buried is (as the IPCC point out) vastly more expensive.

There are free, natural, methos of removal of CO2 from the air, rainfall that dissolves CO2 that eventually resides in the Ocean as bicarbonate, and biological sequestration by plants etc. that give the added bonus of replenishing our O2 supply. A third method in use is landfills [and Ocean dumping] that temporarily reduce CO2 emissions [an argument against biodegradable materials made from fossil fuels]. Rather than mining or manufacturing minerals to react with CO2 [and not replenishing O2] a better approach is to severely reduce energy usage and the burning of fossil fuels, use biomass, and couple biological sequestration to combustion energy generation.  This approach should, slowly of course, slow the increase of CO2 and lower it to controllable levels.


text:
Reaction of Alkyl Halide with Ag2S / Na2S
What does alkyl halides give on reaction with $\ce{Ag2S / Na2S}$?  In my book, it is given as $\ce{R-S-R}$ but I am not quite sure as alkyl halide react with moist $\ce{Ag2O}$ to gives $\ce{R-OH}$ and thus I am expecting an analogous product.

RSH is a better nucleophile than ROH


text:
The most convenient process to purify Oxalic Acid up to a high level of purity
I am set out to purify a batch of commercial oxalic acid with the aim in mind to end up with a highly pure product (starting from $\pu{4,000 ppm}$ total impurities and doing so with basic lab equipment and no advanced analytical capacity).
Looking at the distribution of solubility vs. temperature in aqueous medium (which - despite the widely discrepancy in data from different sources - seems to generally adopt a somewhat exponential shape), I was thinking that the best may be to carry out dissolution/filtration at high temperature, followed by fractional recrystallization.
A few questions remain though:

I have no idea how I could discriminate against Vanadium impurities with the basic equipment I have at my disposal.
According to the data I've found $\ce{V2O5}$ has a fairly low solubility at ~$\pu{10^{-1} g}/\pu{100ml}$ at $\pu{25 ^\circ C}$, a level which makes it much too soluble for proper filtration, and too little for the recystalization to efficiently remove it.
I found data for the solubility of Vanadyl Oxalate to be of ~$\pu{22 g}/\pu{100ml}$ at $\pu{20 ^\circ C}$, and that of vanadium oxalate complexes of ~$\pu{14.5 g}/\pu{100ml}$ at $\pu{25 ^\circ C}$, which makes it very close to the solubility of oxalic acid itself (though, unlike for oxalic acid, I don't have data for how the solubility of Vanadium evolves with temperature).
I had the notion that slowly recrystalizing the oxalic acid would yield better results, but I found an old paper* on its purification which found/states exactly the opposite, for a reason I don't very well understand. If someone had enough understanding to explain to me why a faster recrystalization would yield better results and/or to confirm/refute the paper's statement on the subject, I would greatly appreciate.
I read upon a preliminary search here on Chemistry Stack Exchange that upon excessive heating, oxalic acid decomposes into $\ce{CO2 + CO + H2O}$ (in the ~$\pu{103-185 ^\circ C}$ range), thus I was wondering if it would be worth it to attempt to further exploit the exponential shape of its water solubility vs temperature by carrying out the dissolution/filtration in a $\ce{CO2}$ atmosphere, maybe under a slightly increased pressure. Would that significantly decrease the rate of its decomposition or simply complicate things without significant advantage?

https://core.ac.uk/download/pdf/29156537.pdf

Do you have any methods available to analyze the oxalic acid? If so, what?

Convert all oxalate salts to calcium oxalate first, which is extremely insoluble in water.  Then dissolve in dilute $\ce{HCl}$ to get back oxalic acid form and recrystallize.

If you want to, you can also buy food-grade 99.6% pure oxalic acid, for almost $\pu{2 g}$ per cent (USD).

Actually, likely impure oxalic acid hydrate I once purchased online just decomposed in my basement during a warm summer. As such, don't invest in a lot and keep it cool. Also, do ample research pre-experimentation, employ lots of safety equipment/precautions as it can rapidly decompose releasing gases and ejecting reagents.

@MaxW Hello, as written I don't have any advanced analytical capacity (nor equipment). But you're welcome to suggest a method if you have a good one in mind.

@Mathew_Mahindaratne Thank you for the suggestion. I'd be left with having to purify a source of Calcium then. Is it much of an advantage over purifying the oxalic acid directly?

The purity of 99.6% (based on my comment about buying) means $\frac{40}{100} \times 1000 = \pu{400 ppm}$ of impurity. How pure you want your oxalic acid?

You can get $\ce{CaCl2}$ in high purity and all soluble in water. No need to purify. If you don't have equipment to check the purity, then how you are going to know how pure your crystals, any way?

What Methew asked, and also: Why? (I´m sure there could be good reasons, no sweat, but what are they?)

My oxalic acid is already at 99.6%. Isn't 0.4% 4,000 ppm?

I cannot find any CaCl2 anywhere near as pure for an equivalent price.

Here it is 99% pure $\ce{CaCl2}$.

@Hans: You are correct: $$\frac{\pu{0.4 g} \ \text{Imp}}{\pu{100 g} \ \ce{CaCl2}} \times \frac{\pu{10^3 mg} \ \text{Imp}}{\pu{1.0 g} \ \text{Imp}} \times \frac{\pu{10^3 g} \ \ce{CaCl2}}{\pu{10^6 mg} \ \ce{CaCl2}} = \frac{\pu{4000 mg} \ \text{Imp}}{\pu{10^6 mg} \ \ce{CaCl2}} = \pu{4000 ppm},$$ I was just rushing on calculations.

@Mathew Mahindaratne Thank you for the link. Can you explain to me how the CaCl2 route would be an advantage in the purification of oxalic acid? I find it easier to purify large amounts of a product with high solubility rather than one with low solubility, also, I'd risk to collect a lot more impurities on the way since the CaCl2 is of a lower grade than my Oxalic Acid. Are you suggesting this to address the Vanadium issue specifically? i.e. that I start this way to get rid of the vanadium impurities and only then fractionally crystalize the oxalic acid?


text:
Mass percent and mole percent of a composition
The Problem
A gas mixture contains hydrogen, carbon monoxide, and carbon dioxide. Calculate:
a) The mass% hydrogen if the mixture contains 15 mol% $\ce{CO}$ and no $\ce{CO2}$;
b) The molar composition if the mass fraction of $\ce{CO}$ and $\ce{CO2}$ are both 0.35.
My Attempt
I first calculated the molar masses:
molar_mass$\ce{H2}$ = 2.01588 g
molar_mass$\ce{CO2}$ = 44.0095 g
molar_mass$\ce{CO}$ = 28.0101 g

a) The mass% hydrogen if the mixture contains 15 mol% $\ce{CO}$ and no $\ce{CO2}$;

Since the mixture contains 15 mol% $\ce{CO}$ and no $\ce{CO2}$, then the mol% of $\ce{H2}$ must be 85.
I set the mixture equal to 10 mols. This means that there are 1.5 moles $\ce{CO}$ and 8.5 moles $\ce{H2}$. I then converted these values to grams.
1.5 mol $\ce{CO}$ = 42.015 g $\ce{CO}$
8.5 mol $\ce{H2}$ = 17.135 g $\ce{H2}$
The total grams is 59.15 g.
The percent of $\ce{H2}$ would then be 29% since 17.135 / 59.15 = 0.29.

b) The molar composition if the mass fraction of $\ce{CO}$ and $\ce{CO2}$ are both 0.35.

Since the mass fractions of $\ce{CO}$ and $\ce{CO2}$ are both 0.35, the mass fraction of $\ce{H2}$ must be 0.3.
I set the mixture equal to 100 g. This means that there would be 35 g $\ce{CO}$, 35 g $\ce{CO2}$, and 30 g $\ce{H2}$.
Converting these to mols:
35 g $\ce{CO}$ = 1.25 mol $\ce{CO}$
35 g $\ce{CO2}$ = 0.795 mol $\ce{CO2}$
30 g $\ce{H2}$ = 14.88 mol $\ce{H2}$
This gives us a molar composition of
0.88 $\ce{H2}$;
0.07 $\ce{CO}$;
and 0.05 $\ce{CO2}$.
My answers matched up with the solution key, but did I solve this question in the 'right' way? Or is there a better/quicker alternative?

Your approach in general looks OK except for

terminology (e.g. "total grams" should be total mass);
units (e.g. values provided for molar masses should have dimensions $\pu{g mol-1},$ not $\pu{g});$
excessive precision (you are given fractions with two significant figures, but there are mass values reported in your intermediate calculations having five significant digits; also, there is no need to calculate molar masses up to six significant figures — two would suffice);
math notations ("textual formulas" including chunks such as "molar_mass" or "1.5 mol CO" are not acceptable; use standardized symbols instead, e.g. $M$ or $n(\ce{CO}) = \pu{1.5 mol},$ correspondingly).

I was, however, taught to solve problems differently.
Solve the problem algebraically first, then plug in the physical quantities.
This way your solution becomes modular, generalized, and errors are easier to notice.
No imaginary "set" or "fixed" masses, volumes or other quantities introduced out of nowhere.
There is no point in intermediate calculations and they make tracking significant symbols harder than it should be.
Finally, assigning numerical indices to the compounds in the mixture saves you some time and paper, so let's do it now:
$$\underset{1}{\ce{H2}}\quad\underset{2}{\ce{CO}_\vphantom{}}\quad\underset{3}{\ce{CO2}}$$
and move on to the questions:

a) Calculate the mass% hydrogen if the mixture contains 15 mol% $\ce{CO}$ and no $\ce{CO2}.$

By definition, mass fraction of $i$-th component in the mixture is
$$ω_i = \frac{m_i}{\sum_j m_j}.\tag{1.1}$$
For a binary system with given mole fraction of the second component $x_2:$
$$ω_1 = \frac{m_1}{m_1 + m_2} = \frac{1}{1 + \frac{m_2}{m_1}} = \frac{1}{1 + \frac{x_2M_2}{(1 - x_2)M_1}} = \frac{1}{1 + \frac{0.15 × \pu{44 g mol-1}}{(1 - 0.15) × \pu{2.0 g mol-1}}} = 0.29.\tag{1.2}$$

b) Calculate the molar composition if the mass fraction of $\ce{CO}$ and $\ce{CO2}$ are both $0.35.$

By definition, mole fraction of $i$-th component in the mixture is
$$x_i = \frac{n_i}{\sum_j n_j} = \frac{m_i}{M_i\sum_j\frac{m_j}{M_j}} = \frac{ω_im}{M_i\sum_j\frac{ω_jm}{M_j}} = \frac{ω_i}{M_i\sum_j\frac{ω_j}{M_j}}.\tag{2.1}$$
Plugging in the corresponding quantities:
$$x_1 = \frac{0.30}{\pu{2.0 g mol-1} × \left(\frac{0.30}{\pu{2.0 g mol-1}} + \frac{0.35}{\pu{28 g mol-1}} + \frac{0.35}{\pu{44 g mol-1}}\right)} = 0.88 \tag{2.2.1};$$
$$x_2 = \frac{0.35}{\pu{28 g mol-1} × \left(\frac{0.30}{\pu{2.0 g mol-1}} + \frac{0.35}{\pu{28 g mol-1}} + \frac{0.35}{\pu{44 g mol-1}}\right)} = 0.07 \tag{2.2.2};$$
$$x_3 = \frac{0.35}{\pu{44 g mol-1} × \left(\frac{0.30}{\pu{2.0 g mol-1}} + \frac{0.35}{\pu{28 g mol-1}} + \frac{0.35}{\pu{44 g mol-1}}\right)} = 0.05 \tag{2.2.3}.$$
As you can see, using this approach takes less space and in my opinion is much more presentable.
Given you know elementary algebra and use standardized notations, you can greatly improve literacy of your answers and make them more concise.


text:
Why does HF generate a carbocation with an alkene
I was reading of different ways to perform Friedel Craft alkylations.
One of the methods I came across involved an alkene in the presence of H+ or a mineral acid like $\ce{HF}$. Here $\ce{H+}$ or $\ce{HF}$ caused an electromeric shift of the pi electrons and the $\ce{H+}$ bonded there resulting in a carbocation.
However I felt like the bond strength of $\ce{HF}$ is so high that it couldn't be possible to cleave it for the donation of $\ce{H+}$ (If that's what happens. Correct me if I'm wrong).
Why does it still generate a carbocation?

Well, it sure can do that if it's concentrated enough. Somewhat depending on alkene and temperature, though, I guess.

You felt wrong. HF is a typical acid, behaves like an acid, and (most likely) tastes like an acid.

@IvanNeretin But HF is a really weak acid, owing to its extremely strong bond with hydrogen

@Mithoron But what could cause the HF bond cleavage? Surely the pi electrons wouldnt suffice?

https://chemistry.stackexchange.com/questions/34818/is-hf-the-least-acidic-hydrogen-halide/34829#34829

It's neither truly weak, nor typical. Diluted aq HF wouldn't do it, but pure could work wonders.

However I felt like the bond strength of $\ce{HF}$ is so high that it couldn't be possible to cleave it for the donation of $\ce{H+}$

HF has the highest single bond energy and hence you were right in expecting that it may not dissociate.However in concentrated solutions, due to the formation of stable hydrogen bonded complex $\ce{[HF_2]-}$, the reaction is pushed in the forward direction making it  a resonably strong acid (in fact strongest among corresponding concentrated solutions of all the hydrogen halides).So, you could generate a carbocation with that.However, as generally dilute solutions are used which are economically feasible, more stronger mineral acids in dilute solutions  like $\ce{HCl, HBr}$ and $\ce{HI}$ are used.
EDIT:
As suggested by Oscar Lanzi in the comments, acids like $\ce{HCl, HBr}$ and $\ce{HI}$ are safer to handle when compared to $\ce{HF}$ as the reactions of $\ce{HF}$ are associated with highly negative enthalpy changes [highly exothermic]

Not just an economic advantage.  HF is more dangerous to work with than HCl or HI, especially if you need a strong acid and thus would have to concentrate the HF.

@OscarLanzi Thanks for adding value.I will update my answer to include this as well.

Also more dangerous than HBr.  Left that out!


text:
Why isn't the chemistry of a zinc-air battery (meaningfully) hindered by an activation energy barrier?
Virtually every source I've read pertaining to zinc-air batteries explain the chemistry of it either using these formulae or a simplified version of them:
$$
\begin{align}
&\text{Anode:} &\quad \ce{Zn + 4 K(OH) &-> [Zn(OH)4]^{2-} + 4 K+ + 2 e-} \\
&\text{Electrolyte:} &\quad \ce{[Zn(OH)4]^{2-} + 2 K+ &-> ZnO + 2 K(OH) + H2O} \\
&\text{Cathode:} &\quad \ce{1/2 O2 + 2 K+ + H2O + 2 e- &-> 2 K(OH)}
\end{align}
$$
For all intents and purposes, I believe it is acceptable to simplify the formulae (correct me if I'm wrong).
How does the anode reaction occur in the first place without some sort of catalyst? It momentarily pushes the system out of equilibrium, at least until the cathode reaction can occur, so I would think that the activation barrier would be some significant obstacle. What am I missing?
Apologies if this is a secretly a simple question, I'm not very adept in chemistry.

The first equation as written is a spontaneous reaction. Zinc readily dissolves in KOH soln.

A couple of questions : 1) Why don't you erase the potassium atoms and ions in all these equations. They are always spectator. 2) What the use of the so-called Electrolyte reaction. 3) Why do you think the first equation needs a catalyst ? 4) Where do you see an activation barrier ? 5) Which equilibrium are you speaking of ?


text:
How do I calculate the humidity inside sealed containers containing mixing salts?
At $\pu{25 ^\circ C}$:

saturated Magnesium chloride $\ce{MgCl2}$ keeps humidity 33% inside sealed containers.
saturated Sodium chloride $\ce{NaCl}$ keeps humidity 75% inside sealed containers.

I want to keep humidity at 60%. Which ratio should I mix this two chemicals?
How we can calculate humidity for mixing salts? Is there a formula or table list for this?
At $\pu{25 ^\circ C}$ temperature, $\ce{NaCl}$ sets 75% humidity and $\ce{KCl}$ sets 85% humidity. NaCl mixed 1:1 with $\ce{KCl}$ will give an RH% around 70–71 between $20$ and $\pu{30 ^\circ C}$.
What is the formula to calculate 70% with using 75% and 85%?

There is no formula. It is pretty complicated.

thanks for answer, i'll buy salts and do mixing experiments.

As Ivan Neretin pointed out in his comment, there is no simple formula on this matter. Howevr, if you have access to ammonium nitrate ($\ce{NH4NO3}$) or sodium nitrite ($\ce{NaNO2}$), saturated solution of both give 64% relative humidity (RH) at $\pu{25 ^\circ C}$ (Engineering Toolbox).
Interestingly, magnesium nitrate ($\ce{Mg(NO3)2}$) gives exactly 60% RH at $\pu{0 ^\circ C}$, but it was only 54% at $\pu{25 ^\circ C}$. A few salts and their measured RH values in different temperatures are given in the following Table for readers' convenience:
$$
\bf{\text{Relative Humidity over the Saturation Salt Solution (%)}}\\
\begin{array}{c|ccc}
\hline
\text{Salt} & \pu{0 ^\circ C} & \pu{10 ^\circ C} & \pu{20 ^\circ C} & \pu{25 ^\circ C} & \pu{30 ^\circ C} & \pu{40 ^\circ C} & \pu{50 ^\circ C} & \pu{60 ^\circ C}\\
\hline
\ce{NH4NO3} & - & 75 & 67 & 60 & - & - & - & - \\
\ce{(NH4)2SO4} & 82 & 82 & 81 & 81 & 81 & 80 & 79 & - \\
\ce{MgCl2} & 34 & 33 & 33 & 33 & 32 & 32 & 31 & 29 \\
\ce{Mg(NO3)2} &99 & 98 & 98 & 97 & 97 & 96 & 96 & - \\
\ce{KNO3} & 96 & 96 & 95 & 94 & 92 & 89 & 85 & - \\
\ce{KCl} & 89 & 87 & 85 & 84 & 84 & 82 & 81 & 80 \\
\ce{NaCl} & 76 & 76 & 75 & 75 & 75 & 75 & 74 & 75 \\
\ce{NaNO2} & - & - & 65 & 64 & 63 & 61 & - & - \\
\hline
\end{array}
$$


text:
Is there any database, or reference, for retardation factors (Rf)?
I'll like to compare the Rf of myrcene, obtained in Thin Layer Chromathography, for different species.  Toluene:Ethyl Acetate (97:3) would be fine, but is not a necessity.

There are just too many different solvents, and even more mixtures.

Is it not possible to create a database like this?

No there is not database for retardation factors, because Rf values are useless outside your lab. It is not a fundamental property like melting point or boiling point. One batch of silica is slightly different from another. It also depends on exact solvent composition and history of the treatment of the stationary phase.

Even though the values will never match perfectly, I can't help but imagine there is some kind of information which could be extracted by performing a wide-ranging analysis on a large database, and even just having an approximate notion could be useful. I don't think this would ever get off the ground, though, people are already stingy about placing Rf information in suppinfo

@NicolauSakerNeto, There is a reason why there is no such database despite the fact that chromatography is a century old technique. Silica and alumina behavior is heavily dependent on their history of manufacture and moisture content. It is useful to state the typical mobile phase and stationary phase on TLC. There are thick books on TLC theory and practice.


text:
¹H NMR and MS analysis of products obtained by Claisen condensation reaction between methyl 2‐methylpropanoate and sodium methoxide

A student attempted to perform the Claisen condensation below and  analyzed their isolated product by 1H NMR and MS.
Propose structures for  both the expected Claisen product and the observed product of this  reaction.
What could the student change to obtain their desired product?

This question was asking me to predict the product, both expected and observed.

The premise here is weird. This substrate cannot undergo a Claisen condensation at all. There is no expected product. If you don't know why, it will be a good lesson to figure out why that is.

I understand that it won't undergo a Claisen condensation because there aren't two alpha protons (the structure I got from the proton NMR basically just gives me the reactant so that's probably it) but yah I was just confused what the desired product would be then and what the student should change. Like he should start with a different reactant?

Consider starting with propionate

Start with methyl isobutyrate (your starting compound) and methyl propionate (mixed Claisen). React resulting product with methyl iodide/NaOEt. You get desired product.

@MathewMahindaratne I think that will give a mixed product in ways you don't want, unless you're using a large excess of the isobutyrate.

Looking at your comment, I believe the main source of confusion may be understanding the underlying reason why the Claisen condensation requires the two α protons.
The first one, obviously, is required for the enolate to form. The second one is arguably equally important because it is the removal of this proton in the condensed product that prevents a retro-Claisen. The Claisen condensation is equilibrium driven to the product by that last deprotonation. Under the conditions listed here, you cannot deprotonate the product to generate the "real" product, so a retro-Claisen will quickly take place to provide the original starting material.
The "desired" product is the bis-methylated β-keto ester. You cannot use the conditions for a Claisen, but, as mentioned in comments, alkylation of the unmethylated β-keto ester should work. If you use a strong electrophile, like methyl iodide, you shouldn't need any harsh basic conditions, especially given the fact that a β-keto ester is quite acidic.

@andselisk The main reason is that I don't have a meta key on this machine, and I don't know the alt codes without looking them up. LaTeX is fast and easy to remember.

I don't know how it's on macOS, but on Windows and Linux Meta key can be assigned to pretty much any key on the keyboard (or a key sequence). Plus upright Greek is the proper typesetting in this case, not slanted version that MathJax renders. Anyway, this is a minor thing.

Well, you motivated me to finally bite the bullet and look into it. I've using this: https://github.com/samhocevar/wincompose

@andselisk I posted this on meta: https://chemistry.meta.stackexchange.com/questions/4766/whats-the-best-alternative-to-mathjax-in-titles-and-for-one-off-symbols/4767#4767 Hopefully, it'll be helpful to someone else.


text:
Relationship between the partial pressure $p_J = x_J p$ and the perfect gas law $pV = nRT$
I am currently studying Atkins' Physical Chemistry, 11th edition, by Peter Atkins, Julio de Paula, and James Keeler. Chapter 1A.2 Equations of state says the following:

When dealing with gaseous mixtures, it is often necessary to know the contribution that each component makes to the total pressure of the sample. The partial pressure, $p_J$, of a gas $J$ in a mixture (any gas, not just a perfect gas), is defined
$$p_J = x_J p \tag{1A.6}$$
where $x_J$ is the mole fraction of the component $J$, the amount of $J$ expressed as a fraction of the total amount of molecules, $n$, in the sample:
$$x_J = \dfrac{n_J}{n} \ \ \ n = n_A + n_B + \dots \tag{1A.7}$$
When no $J$ molecules are present, $x_J = 0$; when only $J$ molecules are present, $x_J = 1$. It follows from the definition of $x_J$ that, whatever the composition of the mixture, $x_A + x_B + \dots = 1$ and therefore that the sum of the partial pressures is equal to the total pressure:
$$p_A + p_B + \dots = (x_A + x_B + \dots)p = p \tag{1A.8}$$
This relation is true for both real and perfect gases.

When all the gases are perfect, the partial pressure as defined in eqn 1A.6 is also the pressure that each gas would exert if it occupied the same container alone at the same temperature.

It is this last part that I am unsure about:

When all the gases are perfect, the partial pressure as defined in eqn 1A.6 is also the pressure that each gas would exert if it occupied the same container alone at the same temperature.

I don't understand why this is true. Furthermore, is this not in reference to the perfect gas law, $pV = nRT$? Then how does this relate to $pV = nRT$?
I would greatly appreciate it if people would please take the time to clarify this.

Well, if $p_{total}V=n_{total}RT$, and $n_A=n_{total}\cdot x_A$, then what is $p_A$?

@IvanNeretin So $p_A = x_A p_{\text{total}} \Rightarrow p_{\text{total}} = \dfrac{p_A}{x_A}$ by 1A.6, and so we have that $\dfrac{p_A}{x_A} V = n_{\text{total}} RT$?

@IvanNeretin And what about the other part of the question? Why is it that, when all the gases are perfect, the partial pressure as defined in eqn 1A.6 is also the pressure that each gas would exert if it occupied the same container alone at the same temperature?

Why is it that if $2a=4$ then $a=2$? That's how algebra works.

@IvanNeretin I mean the qualitative explanation.

Qualitative? Well, let's try. Say you have a gas, which in fact is a mixture. Half of it is A, and the other half is B. Gas exerts pressure. Half the gas exerts half the pressure. It is as simple as that.

@IvanNeretin No, I mean, what is it about the properties of perfect gases so that the partial pressure as defined in eqn 1A.6 is also the pressure that each gas would exert if it occupied the same container alone at the same temperature? It seems that the authors are implying that there is something about the properties that allows this to occur.

What is it about the properties of numbers so that when you divide a number by 2, you get exactly the half of it?

@IvanNeretin You're not even using the definition of "perfect gas" or mentioning temperature here.

Well, we have our pV=nRT thanks to the fact that the gas is perfect.

I think that the qualitative point is that when you refer to a 'perfect gas' you can not distinguish the identity of your atoms. There are equal and each one of them contributes equally to the total pressure. The rest is algebra.

Partial pressure is defined as the pressure a component of a gaseous mixture would exert if it alone were to occupy the container at the same temperature as that of the mixture.
If the gases in the mixture are ideal, you can apply the ideal gas equation for it $$p_A V = n_ART$$where $p_A$ and $n_A$ represents the partial pressure and number of moles of a component A. Also $$p_tV = nRT$$ where $p_t$ and $n$ represents the total pressure and number of moles of the gaseous mixture. Dividing these equations you will get $$\frac{p_A}{p_t}=x_A$$ This equation will only be true for ideal gas mixture.
Also, partial pressure is a notional concept. It is not a the actual pressure exerted by the components. It was introduced to explain and form physical properties of gaseous mixtures.

If we define the partial pressure of a gas (real or ideal) as
$$p_J = p_\mathrm{tot}x_J \tag{1}$$
as is done in Atkins, then the ratio $p_J/p_\mathrm{tot} = x_J$ is a definition and can't be wrong. The confusion comes from two different ways of defining partial pressure. Dalton's law uses a definition based on ideal gases, whereas the definition in Eq. 1 is based on actual parameters of the system. See this question for more on the topic.