text:
What is the interaction between dissolved ions of opposite valence in a solution at rest?
NaCl is dissolved in water. Ions sodium and chloride are sufficiently free from each other so that they may occupy different regions of the solution after an active transport process through a membrane, or a mere electric field is in place.
They still act upon each other, and a membrane voltage is set by oppositely charged ions lining its interior and exterior surfaces.
So my question is this: how free are those ions in a uniform solution at equilibrium? Aren't they actually forming pairs of watered ions, although weakly bound? Can they form even bigger clusters, be them of short lifetime?
Note: I originally asked this on Physics SE but after several weeks I have received no answer nor any other kind of reaction, so I am trying Chemistry SE now.

The answer is: it depends. Dissolution of a salt implies that the entropy gained exceeds the cost of breaking lattice interactions (the solution enthalpy, assuming it is positive). Electrostatic interactions compete with kT (thermal jostling). Under physiological conditions, long range interactions are strongly screened by intervening solvent molecules and other ions. At best ions are subject to an effective potential due to distant ions, in particular between segregated charges as in the cases you present (e.g. on opposite sides of a membrane).
However, ionic solutions are generally regarded as "non-ideal". This means that interactions between the ions and with the solvent cannot be dismissed. At high salt concentration ion pairing will be encouraged. Above the solubility limit salt precipitation will occur, preceded by formation of clusters (seeds). If an ion has multiple charges, then electrostatic interactions will be stronger and pairing will be encouraged. Some metal ions form permanent ligand complexes that can attenuate the overall charge or distribute it over the complex, and the ligands can form a "cage" about the central ion, altering the interaction potential with solvent and other ions. The effect of temperature is complex. The entropic cost of association increases with T, but the dielectric constant of the solvent tends to decrease also.
If you are interested in reading on ion-pairing in NaCl solutions you may want to start with Ref. 1. It explains that association is negligible in dilute solutions:

For dilute NaCl, we predict a coordination number of 5.3 with only waters in the solvation sphere for Na+ at 25 °C, 1 bar. With increasing NaCl molality, we find increased Cl complexation of Na. At 25 °C and 1 bar the Na+ ions in an 8 m solution of NaCl have 4.3 waters and 1.0 Cl ligands in their first coordination shell. There is some evidence of solvent-separated Na–Cl ion pairs from the peak in the Na–Cl pair distribution function at 5 Å. However, this disappears with increasing temperature.

There is more on formation of NaCl complexes and clusters in that article.
References

Sherman, DM Collings, MD. Geochem. Trans., 2002, 3(11), 102–107

I am reading the article which contains a lot of interesting references, thanks. It seems this is still a matter of active research which is in line with the feeling I had that the topic was swept under the carpet by textbooks I came to read so far.


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Why are double bonds shorter than single bonds?
We're told in chemistry class that the order of bond length is:
triple < double < single
Can anyone explain WHY this is the case? Is it because multiple bonds involve stronger attraction between the two atoms or something else?

Yes, stronger attraction is quite enough of a reason.

Do you know why the atomic distance is without any bond, i.e. the van der Waals distance, is larger than the atomic distance of covalently bonded atoms? This would be a starting point in discussing the more subtle question of single, double and triple bonds. There are two things that change when going from single to triple bond - the strength of the bond, but also the coordination geometry around the bonded atoms (for carbons, it has to be linear for triple bonds, whereas for a single bond, it could be tetrahedral).

Besides the screening effect you can also see that considering the orbital involved (s character).

In general, in a single bond, there is involvement of sigma bonds only. In double bonds, there is also overlap between pi orbitals. More is the overlap shorter will be the distance between the atoms.

That's the start of an answer, but it still begs the question: why does more overlap imply shorter distance?

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text:
Forming Basis Functions from 6-31G Basis Set for Carbon Atom
I am a computer science grad and I am working to write an electronic structure calculation program and I am stuck at forming basis functions using 6-31G Basis set for atoms having higher atomic numbers (p,d,f shells).
For example, consider a gaussian g as
g(alpha) = (2*alpha/pi)^(3/4) * exp(-alpha*r^2).

Looking at the Carbon (1s2 2s2 2p2) basis set data for 6-31G basis set
C     0
S   6   1.00
      0.3047524880D+04       0.1834737132D-02
      0.4573695180D+03       0.1403732281D-01
      0.1039486850D+03       0.6884262226D-01
      0.2921015530D+02       0.2321844432D+00
      0.9286662960D+01       0.4679413484D+00
      0.3163926960D+01       0.3623119853D+00
SP   3   1.00
      0.7868272350D+01      -0.1193324198D+00       0.6899906659D-01
      0.1881288540D+01      -0.1608541517D+00       0.3164239610D+00
      0.5442492580D+00       0.1143456438D+01       0.7443082909D+00
SP   1   1.00
      0.1687144782D+00       0.1000000000D+01       0.1000000000D+01

How many basis functions will carbon have in this case? I have been following a lecture series which does HF SCF calculation for a simple H2 molecule where each Hydrogen atom has two basis functions as per 6-31G basis set.
H     0
S   3   1.00
      0.1873113696D+02       0.3349460434D-01
      0.2825394365D+01       0.2347269535D+00
      0.6401216923D+00       0.8137573261D+00
S   1   1.00
      0.1612777588D+00       1.0000000

I can write the two basis functions in the form
b1= 0.0334*g(18.73) + 0.2347*g(2.825) + 0.8137*g(0.64);
b2= g(0.161);

In a similar manner, how would I go about writing the basis functions for Carbon atom. Any advise would be very helpful as I am stuck in this for couple of days and haven't been able to find any substantial resource for this.
I did stumble upon this previous post but having a hard time figuring out how did 9 basis functions come for Carbon.

Maybe this would be a better fit at [scicomp.se], but maybe not. I am not saying it is off-topic here.

It may be useful to look at the different formats available on https://www.basissetexchange.org In this case, exponents are shared between s and p functions, which are themselves entirely separate. This was done for efficiency reasons in some codes.

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Inorganic foreign author books for odd electron species
I need a foreign author book which explains odd electron species in great detail ,(explaining dimerization , energetic) telling hybridization of central atom with one unpaired electron on it pls tell me the book. I need to understand this topic and I don't have any idea of foreign author books pls

What do you mean by foreign author books? Do you mean foreign language (non-English) books?

I need books name which covers the above topic ( English)

Holleman and Wiberg 2000 paged book on Inorganic Chemistry.

@M.Farooq, The OP means non-Indian books. It's a frequent irrelevant colloquialism used in India, which actually refers to good quality books. To the OP: using the phrase foreign author books makes no sense in an international community.

I had guessed that from OPs previous and current post ithat he is implying foreign books=good quality books. Holleman & Wiberg is originally written in German but current editions are in English as well. I have not seen a thicker single volume book on Inorganic Chemistry. It should be available in a good university library in India and/or from Amazon. Some scammer on Amazon is selling it for $5000. Amazing! The actual price should be much much lower.


text:
Calculating partial pressure of carbon monoxide

$\ce{ZnO}$ is exposed to pure $\ce{CO}$ at $\pu{1300 K}$ and the equilibrium
$$\ce{ZnO(s) + CO(g) <=> Zn(g) + CO2(g)}$$
is then established at $\pu{1 atm}$ pressure. The density of the gas mixture is $\pu{0.344 g L-1}$ at the same temperature. Calculate the partial pressure of $\ce{CO(g)}$ at equilibrium (atomic weight of $\ce{Zn}$ is $65.4).$

My attempt
Since at equilibrium the amounts of $\ce{CO}$, $\ce{Zn}$ and $\ce{CO2}$ are equal (as no information about initial amounts are given), mole fraction of $\ce{CO}$ is 1/3.
Since pressure is given as $\pu{1 atm},$ the partial pressure of $\ce{CO}$ is 1/3 atm. Where I am wrong?

You might want to revise your idea of equilibrium.

@GRSousaJr: What about $\ce{Zn}$ vapors? At $\pu{1600 K}$, $\ce{Zn}$ stays as vapors (or gas).

@MathewMahindaratne Sorry! My mistake!

For simplicity, let's assign numerical indices to the compounds of interest — all gaseous products participating in equilibrium:
$$\ce{ZnO(s) + \underset{1}{CO(g)} <=> \underset{2}{Zn(g)} + \underset{3}{CO2(g)}}$$
Partial pressure of carbon monoxide can be found via its mole fraction $x_1$ and given total pressure $p$:
$$p_1 = x_1p\tag{1}$$
To find unknown mole fraction, two facts can be taken into consideration:

As Karsten also pointed out, the molar ratio between the products is $1:1$ so that $x_2 = x_3$. Which gives
$$x_1 + x_2 + x_3 = 1 \quad\implies\quad x_2 = x_3 = \frac{1 - x_1}{2}\label{eqn:2}\tag{2}$$

All three mole fractions are linked together in the expression for the average molar mass:
$$\bar{M} = x_1M_1 + x_2M_2 + x_3M_3\label{eqn:3}\tag{3}$$

The average molar mass can be found via average density of the gaseous mix $\bar{ρ}$ by applying ideal gas law:
$$pV = nRT$$
$$p = \frac{nRT}{V} = \frac{mRT}{\bar{M}V} = \frac{\bar{ρ}RT}{\bar{M}}$$
$$
\begin{align}
\bar{M} &= \frac{\bar{ρ}RT}{p} \\
  &= \frac{\pu{0.344 g L-1} × \pu{0.082 L atm K-1 mol-1} × \pu{1300 K}}{\pu{1 atm}} \\
  &\approx \pu{36.67 g mol-1}\tag{4}
\end{align}
$$
Now, knowing numerical value for $\bar{M},$ we can combine \eqref{eqn:2} and \eqref{eqn:3}:
$$\bar{M} = x_1M_1 + \frac{1 - x_1}{2}M_2 + \frac{1 - x_1}{2}M_3$$
Grouping the like terms and expressing $x_1$:
$$
\begin{align}
x_1 &= \frac{M_2 + M_3 - 2\bar{M}}{M_2 + M_3 - 2M_1} \\
  &= \frac{\pu{44.01 g mol-1} + \pu{65.38 g mol-1} -2×\pu{36.67 g mol-1}}{\pu{44.01 g mol-1} + \pu{65.38 g mol-1} -2×\pu{28.01 g mol-1}} \\
  &\approx 0.67\tag{5}
\end{align}
$$
Finally, the partial pressure of carbon monoxide:
$$p_1 = 0.67×\pu{1 atm} = \pu{0.67 atm}$$

Since at equilibrium, Moles of CO= Moles of Zn= moles of CO2 (as no information about initial moles are given)

If no information is given, you should just give the three amounts names and treat them as unknowns.

ZnO is exposed to pure CO at 1300 K

This means that initially there is no carbon dioxide and no elemental zinc. Carbon dioxide and elemental zinc will be present in a 1:1 ratio because that is the ratio they are produced at.

The density of the gas mixture is 0.344 g/L at the same temperature.

Let's find the density of pure CO at 1300 K and 1 atm pressure, assuming it behaves as ideal gas.
$$\rho_1 = m/V = M n / V = M P / (R T) = $$ $$\frac{\pu{28.01 g/mol} \cdot \pu{1 atm}}{\pu{0.08206 L atm K-1 mol-1}\cdot \pu{1300 K}} = \pu{0.2626 g/L}$$
And now the density of a 1:1 mixture of carbon dioxide and zinc under the same conditions. The average molar mass of the two species is 0.5 (65.4 + 44.01) g/mol.
$$\rho_2 = m/V = M_\mathrm{ave} n / V = M P / (R T) = $$ $$\frac{\pu{54.71 g/mol} \cdot \pu{1 atm}}{\pu{8.314 J K-1 mol-1}\cdot \pu{1300 K}} = \pu{0.5129 gm3 g/L}$$
So the density of the gas for pure reactants is lower than the observed density, and the density for pure products is higher. The equilibrium mixture contains some reactant and some product, and the density is a weighted average of the two densities just calculated (with $X_\ce{CO}$ the mole fraction of carbon monoxide in the gas phase of the reaction mixture):
$$\rho = X_\ce{CO} \rho_1 + (1-X_\ce{CO}) \rho_2$$
We can solve for $X_\ce{CO}$ and calculate its value:
$$\rho - \rho_2 = X_\ce{CO} (\rho_1 - \rho_2)$$
$$ X_\ce{CO} = \frac{\rho - \rho_2}{\rho_1 - \rho_2} = 0.675 $$
So the partial pressure of carbon monoxide is 0.675 atm.

How to do this without percent completion of reaction as we aren't taught that. Moreover, I think R should be taken as 0.082 L atm K-1 to match the units. And the answer key given to this question was 2/3.

I'm not sure it's correct to write "$\rho = \xi \rho_2 + (1-\xi) \rho_1$" for density.

@andselisk - There were two problems in my answer. I mislabeled what I called $\xi$ as extent of reaction (it is just a mole fraction), and then I gave the partial pressure of the (combined) products rather than the reactants. I obviously confused myself with my own choice of variables, so in the corrected answer I simply used the mole fraction of carbon monoxide to express the density of the gas mixture. Thanks for pointing it out.

@dhaneshvijay I fixed my answer after looking at andselisk's, who got the correct result that matches what you saw in the answer key. I also swapped the units for R so it is easier to follow the arithmetic.


text:
Reduction potential as a function of pH
I have read that Latimer diagrams can be represented both in an acidic medium (pH = 0) and in a basic medium (pH = 14).
I would like to know how to calculate the reduction potentials as a function of pH. That is, if I know the reduction potential for the pair $X^{+}/X$ at pH = 0, calculate it for another pH.
I thought the Nerst equation could be applied, but I'm not sure about that.

Nernst equation is the only way of solving this problem. Let's consider the example of the reduction of permanganate. The half-reaction is:
$$\ce{MnO4-  +  8 H+  +  5 e-  ->  Mn^2+  +  4 H2O}$$
The potential can be written as:
$$E  =  E^\circ  +  \frac{0.059}{5}\log \left(\frac{\ce{[MnO4-][H+]^8}}{\ce{[Mn^2+]}} \right)$$
If you want to see explicitly the $\mathrm{pH}$, you can rewrite the preceding expression, which gives:
$$E  =  E^\circ  +  0.012 \log \left(\frac{\ce{[MnO4^-]}}{\ce{[Mn^2+]}}\right) - \left(\frac{8\times 0.059}{5} \right) \mathrm{pH}$$
or :
$$E  =  E^\circ  +  0.012 \log \left(\frac{\ce{[MnO4^-]}}{\ce{[Mn^2+]}}\right) - 0.094 \: \mathrm{pH}$$

So, for example, what would be the reduction potential at pH = 0? And, another question, if I want to determine the potential for calcium reduction is $\mathrm{Eº(Ca^{2+}/Ca)} = -2.760 V$ under normal conditions. What would it be at pH = 10? Thanks


text:
Thermodynamic statements
Question 13 from [1, p. 4.2]:

Which of the following statements is true?
(a) An ideal gas always obeys the equation: $PV^γ = \mathrm{constant}$ in adiabatic process.
(b) An ideal gas always obeys the equation: $PV^x = \mathrm{constant}$ in
  polytropic process.
(c) In a polytropic process, the heat capacity of the system remains same
  throughout.
(d) In all the cyclic process, $w_\mathrm{net}$ by the system is non-zero.

To me (a), (b) and (c) appear to be correct, but the solution manual says only the (c) is right.
References

Neeraj, K. Advanced Problem in Physical Chemistry for Competitive Examinations, 2nd ed.; Pearson India Education Services Pvt. Ltd: Dehli, Chennai, 2015. ISBN 978-93-325-4373-7.

Please avoid posting textual information as a screenshot. Also, it would be nice if you cite the source of the question (e.g. a textbook reference), otherwise it may be considered a plagiarism. As for the question, neither (a) or (b) mention that the process is reversible.

Yes I will make sure of it from the next time. It was my first post. And yes I totally forgot  about the reversible part. Thanks!

Thank you for adding the reference, I corrected the styling and publisher's info.

Understood the format now . Will take care of it from the next posts . Thank you .

(a) is not true, because, in addition to being adiabatic, the process must be reversible.
(b) is not true, because the gas must be assumed to have constant heat capacity for this equation to apply


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Classic/killer/easy application for the deuterium kinetic isotope effect?
I'm a physicist. While visiting someone's o-chem class I learned about the existence of the deuterium kinetic isotope effect. This seems like an extremely sweet example of zero-point energy, and I would like to use it as an example when teaching the basic quantum mechanics that we teach in freshman physics at a community college.
In this context, I would like to be able to give a good example where we actually care about the deuterium kinetic isotope effect. Ideally this would be an example that would be at least somewhat understandable to people who (like me) have no clue about organic chemistry. Something like:

Without the deuterium kinetic isotope effect, we wouldn't be able to synthesize x (where x is nail polish, fullerenes, methamphetamine, ...)
Historically, there was intense controversy about how this mysterious chemical process in x worked (where x is learning in slime molds, work-hardening of bronze, ...), but then they injected their (slime mold, ...) with deuterium and used the deuterium kinetic isotope effect to prove that ...

Related: Why is a C–D bond stronger than a C–H bond?

Involvement of water in enzyme-based catalysis, I'm sure there are examples.

The effect can be smaller that you imagine as the energy at the transition state may also be lowered so the barrier height need not change that much. In reactions involving many steps, particularly with enzymes a small, say 10%, change at each step becomes important and results in a big overall effect.

You might read about heavy water toxicity to find an overall impressive one. For instance an aquarium of heavy water can host no fishes, at least after a certain time.

@porphyrin "Big" is relative. I dont think there is a natural reaction pathway that can actually accumulate deuterium in a specific chemical function to twice the concentration in the surrounding water or so. Much less, usually.

The site specific hydrogen (and carbon) isotope ratios tell you if the alcohol in your wine was actually made by fermentation of fructose (instead of saccharose, maltose, ...), and even if the fructose was actually in grapes and not apples. https://en.wikipedia.org/wiki/Isotopic_analysis_by_nuclear_magnetic_resonance.

@Karl, Sure, 'big' is relative, but I meant only that if hydrogen was replaced with deuterium then the effect would be accumulative and so much slower.

@porphyrin A hydrogen atom can only be replaced by deuterium once. ;) Where du you see an accumulative slowing effect?

@Karl, quite so but in a sequence $A \to B \to C\cdots$  each step is slower so slower overall.

@Alchimista: Your comment sounds like it would make a good answer.

An expensive but possible demo would be to make two candles, one out of perdeuterated wax, and another out of regular wax, and to burn them both at the same time.  The rate at which the candles burn should be significantly different.


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Clarification on the Clausius inequality
In Atkins' Physical Chemistry the criteria for spontaneity using Gibbs energy was calculated using the Clausius inequality:
$$\mathrm dS ≥ \frac{\mathrm dq}{T},$$
and at constant pressure, $\mathrm dq = \mathrm dH.$ Thus, $\mathrm dH - T\,\mathrm dS$ is negative.
So, if we define a function $G = H - TS$ at constant pressure and temperature, we get
$$\mathrm dG = \mathrm dH - T\,\mathrm dS ≤ 0,$$
which we use as a criteria for spontaneity.
My issue with this argument is that it implies that the Clausius inequality is for spontaneous processes only, otherwise we would end up with $\mathrm dG ≤ 0$ for all processes.
However, the derivation of the inequality simply used the fact that
$$|\mathrm dW_\mathrm{reversible}| > |\mathrm dW_\mathrm{irreversible}|$$
when the system does work.
So, where is it implied that the derivation is for spontaneous processes only?

It's not the case that $\mathrm dS ≥ \frac{\mathrm dq}{T}$ is the criterion for spontaneity.  Furthermore, $\mathrm dS ≥ \frac{\mathrm dq}{T}$ is not a statement of the Clausius inequality.   The Clausius inequality states that $\mathrm dS > \frac{\mathrm dq}{T}$ for spontaneous processes.  I.e., in the expression $\mathrm dS ≥ \frac{\mathrm dq}{T}$, the inequality sign applies to spontaneous processes, and the equals sign applies only for reversible processes.

Denbigh, in his book the Principles of Chemical Equilibrium, considers the case where "(a)the only heat transferred to the system is from a reservoir which remains at the constant temperature T; (b) the initial and final temperatures of the system are equal, and are equal to the temperature T of the reservoir; (c) apart from the system, the only other body with has undergone a change of volume at the end of the process is at a constant pressure p (e.g. a surrounding fluid such as the atmosphere); (d) the initial and final pressures are equal, and are equal to p."
For this situation, from the first law, we have
$$\Delta U=q-w$$ so that, from the definition of Gibbs free energy, it follows that:  $$\Delta G=q-w+p\Delta V-T\Delta S$$
Applying the Clausius inequality to this situation yields:
$$\Delta S=\frac{q}{T}+\sigma$$where $\sigma\geq0$ is the entropy generated, with the = sign applying if the process is reversible and the > sign applying if the process is irreversible or spontaneous.  If we combine the previous two equations, we obtain:$$w=-\Delta G+p\Delta V-T\sigma$$ So, for any process path between two specified end states of a system under these pressure and temperature constraints, the maximum work over and above p-V work is $-\Delta G$, and all other irreversible cases will involve less work (over and above $p\Delta V$).

Your question is how does the inequality $|\mathrm dW_\mathrm{reversible}| > |\mathrm dW_\mathrm{irreversible}|$ follow from more general statements of the Second Law (a "corollary" being the Clausius inequality). Or, how is that inequality a definition of spontaneity?
There are various established formulations of the Second Law. Clausius' statement is that a pump cannot transfer heat from a lower T into a higher T reservoir without external work being supplied. Such a process, in other words, is not spontaneous. While Clausius' statement seems obvious, it can be shown to be consistent with other statements of the Second Law, including Clausius' inequality: $|\mathrm dW_\mathrm{reversible}| > |\mathrm dW_\mathrm{irreversible}|$. The inequality means that the amount of work that can be performed by a heat engine from a heat input $dq_{rev}$ never exceeds  the reversible work $|\mathrm dW_\mathrm{reversible}|$. A violation of this condition would mean that an irreversible engine requiring only a fraction of the reversible heat could be used to power the reversible engine as a heat pump, driving more heat from the cold into the warm reservoir than was sent in the opposite direction by the irreversible engine, and this would violate the Clausius statement, and thereby the Second Law.
You can thus see that Clausius' statement and inequality are tied together and represent different expressions of the Second Law, and so of the condition of spontaneity.


text:
How do I figure out how many hydrogens my compound actually has using a mass and NMR spectrum?
Question 3:

It said $m/z = 122,$ and $m/z = 124$ is in a $3:1$ ratio, so I figured that meant that chlorine is present. Then I thought $m/z$ was the actual compound's molecular mass.
So I used the rule of 13, and did:
Chlorine's molar mass = 35
122 - 35 = 87
Using the rule of 13
87/13 = 6C + 9H/13
So the molecular formula is $\ce{C6H9Cl}.$ But the integral values I rounded are: 2,2,2,2 and 3 adding up to 11. Did I use the wrong mass to charge ratio? What am I missing?

I would check for consistency with the proton spectrum: that suggests you have 11 H or a multiple thereof.

How do I know for sure if there are exactly 11 H?

Well, I posted an explanation as an answer...

Being an NMR fan myself I would inspect that NMR spectrum:

The integrals suggest you have 11 $\ce{^1H}$ or a multiple thereof (the number under each peak is the normalized integral, which is proportional to the number of protons represented by the multiplet). That leaves you with $\pu{87 Da -11 Da}=\pu{76 Da}$ to explain. If you throw in an oxygen you get $\pu{60 Da}$, which is neatly divisible by 12, for a formula $\ce{C5H11OCl}$. Next use the NMR spectrum again to infer the connectivity. The rightmost integrated multiplets (multiplet and triplet) are typical of $\ce{-CH2CH3}$ with $\ce{CH2}$ split by methyl and something else. From the integrals we see clearly there is only one methyl group, and there are no lone H which seems to eliminate the possibility of carbonyl or hydroxyl groups, suggesting an ether.
In the end I just "cheated" (it's 2019 after all) and used the online NMR simulator predictor:

This nice piece of software lets you tinker quickly with different structures and inspect assignments interactively, helping you develop your intuition.
Next you need to verify the MS and IR spectra, but the NMR spectrum is a giveaway that this is the correct hit.

‘verify the IR spectrum’? I thought IR spectra nowadays were only recorded because some journal editors complain if they are not there; then a couple of wavenumbers are written down and the spectrum filed.

@Jan Yes, well, in that case consider it "fine print". Evidently this is homework and the OP should incorporate all available info into the answer :-)

I would probably also use the method Buck has suggested, but let’s say the NMR broke down or somebody is measuring a $\ce{^13C}$ of $\pu{2.5mg}$ meaning it will be blocked until tomorrow; in this case, we can still extract more information from the mass spectrum.
In addition to the molecule peak at 122, you have:

a chlorine-containing fragment $m/z=93$
a chlorine-containing fragment $m/z=63$
a chlorine-free fragment $m/z=73$
and some more stuff at lower masses that doesn’t analyse itself quite as easily.

Those three peaks should derive from fragmentation processes, so we can get an idea of what groups we have from looking at the mass difference (i.e. the bit that fragmented away).
For the peak at 93: $122-93=29$. The chlorine atom is retained. The most common groups to leave are methyl, ethyl etc., and 29 happens to be exactly the mass of $\ce{CH3CH2}$ (methyl is 15, methylene is 14; you internalise those numbers pretty quickly). Therefore, your molecule should have an ethyl group somewhere.
The chlorine-free peak at 73: $122-73=49$. That’s an uncommon number at first sight but remember we lost a chlorine. Taking that into account, $49-35=14$ and we find another methylene group. So we can say that there is a terminal $\ce{CH2Cl-{}}$ group.
Finally, the peak at 63: $122-63=59$. That again doesn’t say everything at once, but what happens if we remove the 29 we found earlier for the ethyl group? $59-29=30$. 30 could mean two methyl groups but that’s somewhat unlikely given how few fragments we have overall—especially since we would also expect a peak with a mass difference of 15 for a single methyl group. But $30=14+16$, meaning there could be a $\ce{-CH2-O-{}}$ group attached to the ethyl group. Other possibilities for 16 might be $\ce{NH2}$, but the molecule disobeys the odd-nitrogen rule and two amino groups doesn’t look good with the data.
If we take what we have, we have a strong suspicion that there might be a $\ce{CH3-CH2-CH2-O-{}}$ group and a $\ce{CH2Cl-{}}$ group. Adding these together gives an intermediate mass of $59+49=108$ or 14 missing from the complete molecule. It seems reasonable to assume a $\ce{CH2}$ group connecting the fragments.
We should consider this all a working hypothesis until we can then take a look at the NMR spectrum which beautifully confirms that molecular structure.

I like both answers provided before me where one has used exclusive use of internet to suggest structure by NMR spectrum, and the other has used thorough analysis of mass spectrum. Although these two are valuable techniques, I feel OP needs to analyse step by step analysis of spectra given to predict the structure since he/her seemingly in a graduate course, Spectroscopic Analysis of Organic Compound, or one similar to that. Hence OP needs knowledge of how to analyse given spectra to determine relevant structure.
Thus, I'd like to give some insight to provide that knowledge. I usually start with $\ce{^1H}$-NMR since it gives the most information. The given NMR is showing 5 resonances indicating 5 different protons in 2:2:2:2:3 ratio, according to the corresponding integral values given. By that information, you can guess there are at least 11 protons and 5 carbons are in the molecule to be analyzed. Therefore, the molecule possibly contains $\ce{C5H11}$- portion, which counts for $\pu{71D}$. There are three -$\ce{CH2}$- resonances present in the downfield of the NMR spectrum, each having chemical shifts greater than $\pu{3 ppm}$. This clearly suggest that at least one side of each of these  -$\ce{CH2}$- groups attached to a electron withdrawing group (such as oxygen or halide) making the corresponding proton deshielded than ordinary alkyl group.
Mass spectra of that compound has shown that it must contain one $\ce{Cl}$ atom, as OP correctly guessed (peaks with 3:1 ration intensities with $2 \: m/z$ difference). It also  has shown molecular peak is $122 \: m/z$ and $124 \: m/z$ for $\ce{^{35}Cl}$ and  $\ce{^{37}Cl}$ fragments. Therefore, $\ce{C5H11^{35}Cl}$- fragment gives $\pu{(71+35) D}=\pu{106D}$. Thus, the rest is $\pu{(122-106) D}=\pu{16 D}$, which indicates the presence of $\ce{O}$ in the molecule. Therefore, molecular formula of the molecule would be  $\ce{C5H11ClO}$, which accounts for $\pu{122 D}$ (for $\ce{C5H11^{35}ClO}$). Based on these values, you can predict the NMR resonance as follows:

Since two electron withdrawing groups in the molecule are $\ce{O}$ and $\ce{Cl}$, we can predict that $\mathrm{EWG^{I}}$ is $\ce{Cl}$ and $\mathrm{EWG^{II}}$ is $\ce{O}$,  based on the chemical shift values of peaks at ~$\pu{3.9 ppm}$ and ~$\pu{3.6 ppm}$. Therefore, it can be suggested that $\ce{Cl-CH2-CH2-O}$- group is present in the molecule. The third deshelded peak at ~$\pu{3.3 ppm}$ suggests the presence of $\ce{R-CH2-CH2-O}$- group in the molecule as well. Thus, conventional wisdom would suggest the molecular structure  $\ce{Cl-CH2-CH2-O-CH2-CH2-R}$ for the molecule. Based on two shielded resonances at ~$(\pu{1.5 ppm},\: sextet)$ and ~$(\pu{0.9 ppm},\: t)$ suggest that $\ce{R}$- group would be  $\ce{CH3}$-, based on chemical shifts of $1.5$ and $\pu{0.9 ppm}$, and integral ($3.07$) and splitting pattern of resonance at $\pu{0.9 ppm}$. Therefore, the tentative structure of the molecule is $\ce{Cl-CH2-CH2-O-CH2-CH2-CH3}$ as given in the diagram above. The given mass spectrum confirms this structure as shown in fragmentation illustrated in following diagram and elsewhere in one of the answers:


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UV-Vis absorbtion spectra of a radical cation/anion. Is it different compared to its even counterpart?
I am studying gas-phase Uv-Vis spectra of radical molecules. If one electron is removed/captured from the neutral even electron molecule, what can I expect from the absorption spectra? Will new peaks appear? Will they be blue/red-shifted?

A cheeky answer is 'yes and yes '. Why not look in the literature for  example spectra, there are bound to be many examples.

I have not found a general answer to how the ground state absorption would shift. Here are two publications 1, 2 that observe shifts, but both times a redshift, be it anion or cation. So as of now I dont know how to explain that.
But, I can certainly say that new peaks will apear and they will most likely be in the red part of the spectrum. In the image below you see that the HOMO/LUMO gap is the biggest. Assume now an radical anion, i.e. an additional electron which occupies the LUMO, then transitions to LUMO+i (i=1,2,3,...) are possible which likely have lower energies than the HOMO/LUMO gap.

Image taken from here.


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What is the right color of calcium copper titanate?
I'm browsing Alibaba for calcium copper titanium oxide $\ce{CaCu3Ti4O12}$ (CCTO) and I find it variety of colors: pure brown, pale brown, reddish brown or grayish dark brown. Wikipedia does not tell the right color either.
What is the right color for CCTO?
All of these below are sold as CCTO on Alibaba. Can it be that they have different color but are equally valid products?

Check the MSDS of calcium copper titanate: https://www.ltschem.com/msds/CaCu3Ti4O12.pdf

@NilayGhosh Thank you. Although this card does not answer my answer I still find a valuable source of information for this compound. Yet, if it say brown without any other remarks then I suspect that the first example is the closest to expected.


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Chemichal difference between liquid sucralose and powder sucralose
In my country (Brazil), sucralose comes in this two forms, liquid and powder.
Looking on the nutritional table, liquid sucralose have no significant amount of carbohydrates or kcal; On the powder sucralose says that have about 4kcal/g and about ~1g of carbohydrates(Basically equal to sugar) per g of serving.
There is any chemical difference in the powdered form that makes it been broken down to be used as energy? Or powdered sucralose is basically sugar with a little amount of actual sucralose? Or the chemical substance is the same but described differently
Side note: I've seen the same pattern in other types of recommended (to high blood sugar patients) powdered sweetener with 4kcal/g and ~1g of carbohydrates per g of serving. With the argument that's better than actual sugar.

It's a question of formulation, the liquid contains the sweetener dissolved in water, eg here is an example of ingredients for a version I found online: Purified Water, Sucralose, Potassium Sorbate, Citric Acid; the sweetener represents 25% of the weight.

I didn't understand what you meant. It means that powdered form should have kcal?

I cannot speak for the manufacturer, you should check the label of contents. It should say if it has anything other than sucralose as a sweetener. IF it only has sucralose and is a liquid, then it is simply a solution (water+sucralose) and the same weight of solution will have less calories (kcal) than the powder (assuming they didnt add stuff  to the powder). The liquid also has "kcal", but if it is a 25% solution for example then you have to drink 100 mL to get the same kcal as 25 g of powder.

The side note sweeteners have maltodextrin (basically sugar), so it's the same as sugar. The main sucralose powder have: lactose, silicon dioxide, sucralose and acesulfame-k. As I see, the powdered version always have some type of carbohydrate (maltodextrin, lactose). So it's reasonable to say, (given the almost 1:1 carbohydrate to serving) that those powdered sweeteners are equal to sugar?

Without having the exact % composition to inspect I can't tell you exactly, but it sounds like they have sugar. I don't know how it is in Brazil but in other countries they list ingredients in order of abundance. However the sweetness of the artificial sweeteners is so high that you are only supposed to use a very small amount. The wikipedia says: "Sucralose is about 320 to 1,000 times sweeter than sucrose"

In the US, if there is less than 0.5 g of something in a serving, the FDA specifies rounding down to zero. Cooking oil spray is labeled as 0 g fat even though it’s 100% fat.

In the United States, sucralose is sold under the brand Splenda. There is a liquid product with the food label shown below:

The low calorie sweeteners in solid (powder) form containing sucralose come in different form, and they often contain sugars with calories. Here is one example:

It contains glucose as the main ingredient. Nevertheless, they are allowed to say it is suitable for diabetics. Also, they are allowed to say zero calories. From the FDA rules:

(b) Calorie content claims. (1) The terms "calorie free," "free of calories," "no calories," "zero calories," "without calories," "trivial source of calories," "negligible source of calories," or "dietarily insignificant source of calories" may be used on the label or in the labeling of foods, provided that:
(i) The food contains less than 5 calories per reference amount customarily consumed and per labeled serving.
(ii) As required in 101.13(e)(2), if the food meets this condition without the benefit of special processing, alteration, formulation, or reformulation to lower the caloric content, it is labeled to disclose that calories are not usually present in the food (e.g., "cider vinegar, a calorie free food").

Source: https://www.accessdata.fda.gov/scripts/cdrh/cfdocs/cfcfr/cfrsearch.cfm?fr=101.60

Is the second label a sweetener in a powdered form?
As in Brazil,it's fair to say that there is no apparently difference between sucralose in liquid or powdered form? The problem is the "hidden" sugar in the tabel of contents(dextrose, maltodextrin) that give the non zero carbs and non zero kcal to the product.

@KaiqueGomes Yes, the second label is from a solid sweetener; I added it to my answer. The molecule sucralose is the same, no matter whether in a solid form or dissolved in water (notice the first ingredient in the liquid is water), also in terms of carbohydrate content and calories. From my remarks about how food labels are written in the US, you can see that to understand them fully you also need to know the regulations about rounding down and "negligible amounts".

@KaiqueGomes The solid sweetener label says "'Free food' = up to 4 packets". This refers the "exchange list" which has foods with less than 20 calories per serving (or no calories at all) labeled as 'free'. For the sweetener, this means that 4 packets have less than 20 calories, and 5 packets already have more. So this is a backdoor way to figure how many calories per packet (all we knew was less than 5 calories because it says zero).


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Number of Stereoisomers for Terpin
Terpin (4-(2-hydroxypropan-2-yl)-1-methylcyclohexan-1-ol) has two $\mathrm{sp^3}$ carbons with different substituents on a cycloalkane, so it has two cis-/trans-isomers, and from what I was taught, is a stereoisomer.

The way I was taught to find stereoisomers was to count the number of double bonds that can be E/Z and the stereocentres, and do $2^n.$ There are no stereocentres, and no double bonds, so the number of stereoisomers would only be 1 (only one molecule). This disagrees with the cis-/trans-isomers number which would be 2.
I can't find anything online about these kinds of cases. Is this general rule of counting stereocentres and double bonds with E/Z not apply in every case? And is this a case that happens a lot that when I'm solving problems, I should always keep this in mind?

I don't get your point - there are 2 geometric isomers, neither of them is optically active, that's it...

The simple rules don't work with complicated examples. There are non-chiral molecules with "stereocenters", and chiral molecules lacking "stereocenters". The textbook example for your question is 1,4 dichloro cyclohexane (or 1-chloro 4-bromo cyclohexane to get even closer to your example):

William Reusch explains this and other cyclohexane cases here: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Stereoisomerism_in_Disubstituted_Cyclohexanes

The way I was taught to find stereoisomers was to count the number of double bonds that can be E/Z and the stereocentres, and do $2^n$.

You established that this rule does not work here. If you added an oxo group to one of the ring carbons (turning it into a ketone and making the two branches of the cyclohexane different from one another), you would get the expected pattern: two stereocenters, four stereoisomers.
I found a more exact formulation of the rule on another one of William Reusch's pages:

[William Reusch] As a general rule, a structure having $n$ chiral centers will have $2^n$ possible combinations of these centers. Depending on the overall symmetry of the molecular structure, some of these combinations may be identical, but in the absence of such identity, we would expect to find $2^n$ stereoisomers.

Source: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Compounds_with_Several_Stereogenic_Centers
So what about 1-chloro 4-bromo cyclohexane or terpin? If you use the following definition of stereocenter, both have two stereocenters:

[Gamini Gunawardena] If the interchange of two ligands on an atom in a molecule results in a stereoisomer of the molecule, the atom is called a stereocenter or stereogenic center.

Source: https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Stereocenter
You have 4 combinations, but there are pairs of identical molecules when you consider the two chair conformations inter-converting.
Other examples where the simple $2^n$ rule does not work are the following:

Meso compounds: Stereochemistry of meso compounds
Axial chirality: M/P nomenclature for axial-chirality
Planar chirality: Why is trans-cyclooctene chiral?
Molecular knots: Molecular knots absolute configuration


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Winkler's method in the absence of iodide
I'm currently doing an IB chemistry individual investigation on the effect of adding bleach to water on the water's dissolved oxygen content. I've used Winkler's method for this and I'm just wondering whether it is possible to do Winkler's method without the presence of iodide ions $(\ce{I-})$, i.e. without adding chemicals like potassium iodide?
I used $\ce{NaOH},$ $\ce{MnSO4},$ $\ce{H2SO4}$ in this experiment (and $\ce{Na2S2O3}$ as the titrant) and I'm not too sure whether I used potassium iodide. However, I've obtained a result and found out a relationship where the dissolved oxygen content increases as I add more bleach into the water.
So my question is: is it possible to do a Winkler's method experiment without the presence of iodide ions? If so, could I know how my experiment worked?

You shouldn't have used the winkler method for bleach analysis. Why not just stick to the iodine-thiosulfate method given in the textbook? Pg 421 in the Pearson HL book.

Bad news is that Winkler's method will fail in the presence of bleach or any other oxidizing agent besides oxygen. Neither sodium thiosulfate nor potassium iodide are compatible with bleach, whether it is a oxygen bleach or a chlorine bleach.

I used $\ce{NaOH},$ $\ce{MnSO4},$ $\ce{H2SO4}$ in this experiment (and $\ce{Na2S2O3}$ as the titrant) and I'm not too sure whether I used potassium iodide. However, I've obtained a result and found out a relationship where the dissolved oxygen content increases as I add more bleach into the water.

As a good scientist of future, you always write whatever you are doing at the time of experiment in your notes.
Your results are skewed towards higher numbers because sodium thiosulfate is being consumed by released iodine not only from the oxygen reaction but also from the bleach.


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Resonance with different size orbitals
The equilibrium for this reaction is on the left side.

Why is this? I thought since the sulphur in molecule X has bigger p-orbitals than the oxygen in molecule Z, resonance would be a less important factor for stabilization, meaning Z is more stable, and the equilibrium should go to the right.
What part of the concept am I misunderstanding?

You should compare second mesomeric structures of anions - they're more important here.

The fundamental question is: which is more acidic, 2-pyrrolidone or 2-pyrrolidinethione?

If you took a 50-50 mixture of the two and added enough alkali to neutralize half of the mix, what would be the composition of the solution? The mismatch in p orbitals between C and S leads to more C=N bonding, rather than C=S, with a charge separation, leading to more negative charge on sulfur in Y and more positive charge on N, so this N will be more acidic than the N in W. The charge separation by oxygen, leading to more double bond character between positions 1 and 2 in 2-pyrrolidone is an effect also, but apparently not as large as the p-p mismatch between C and S.
The Lewis diagrams don't show clearly enough the actual charge distribution in the anions.
And I thank Jan for pointing out that I completely flipped my actual explanation when I thought I understood what two effects were happening. Although this is not a trick question, I think it needs more context to be able to be of predictive value. Maybe just a way to get you thinking: what could be going on here, and where else might it explain.

This is just repeating OP’s statement.


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Why doesn't hydrazine contain a double bond between two nitrogen atoms?
I know that the Lewis structure of hydrazine contains $\mathrm{sp^3}$ hybridized nitrogen atom but on analyzing it's structure I realized there are many bond pair and lone pair repulsions. Instead of a single bond between the N atoms they could also have a double bond and a dative bond with h atoms.
When I asked my professor about this he said that the formal charge imparts much more instability. I was not satisfied with the argument. So, can anyone please provide a rigorous explanation of why I am wrong?

You can't have a dative bond with H atoms. How many electrons will an H atom have, if you would?

Please edit to show the two Lewis structures you are comparing (the conventional one, and your alternate one).

Very interesting question! Let's start with the nitrogens in hydrazine being approximately sp2-p hybridized, flat. The p orbitals on each nitrogen would overlap in a double bond configuration - but the "double bond" has four electrons, two from each nitrogen. If you consider a molecular orbital, the bonding AND the antibonding orbitals are filled, so the net bonding is zero. Actually worse than that, because the filled p orbitals on each nitrogen are aligned in the worst way - maximum repulsion.
The clever little nitrogens rehybridize to become pyramidal - perhaps not completely sp3, but not at all planar. This rearrangement hardly affects the N-H bonds, but allows the p orbitals to accept some s-character, which increases one lobe at the expense of the other. And of course, the prominent lobes find a way to point away from each other and reduce electrostatic repulsion of the electrons. You could say that the double bond character is reduced/eliminated, but the bonding loss is more than balanced by reducing the antibonding.
A useful comparison can be made with ethylene: planar, sp2-p, because there are only 2 electrons in the double bond.

Thank you for such a beautiful explanation!


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Heat involved in isenthalpic process
Question 28 from [1, p. 4.4]:

An ideal gas is expanded irreversibly from $\pu{5 L}$ to $\pu{10 L}$ against a constant external pressure of $\pu{1 bar}.$ The value of heat involved $(q)$ in this isenthalpic process is
(a) $0$
(b) $\pu{+500 J}$
(c) $\pu{+5 J}$
(d) $\pu{-500 J}$

It is mentioned in the problem that the process is isenthalpic, which means the change in enthalpy must be zero. So, my conclusion was there is no heat involved here. But the answer given is (b).
References

Neeraj, K. Advanced Problem in Physical Chemistry for Competitive Examinations, 2nd ed.; Pearson India Education Services Pvt. Ltd: Dehli, Chennai, 2015. ISBN 978-93-325-4373-7.

After further reflection I realized an initial comment I made was flawed*. To quote the Wikipedia:

Isenthalpic processes on an ideal gas follow isotherms, since $\mathrm dh = 0 = c_p\,\mathrm dT$.

That means that for an ideal gas isenthalpic is synonymous with isothermal, and this implies $\Delta U = 0$ and so $q = -w = P_\mathrm{ext}\Delta V$.
For the isobaric process described in the problem, the work done by the gas is $\pu{500 J}$, and therefore the heat expected for an isothermal process is also $\pu{500 J}$.

* The source of my error is simple: for an irreversible process the heat is not equal to $\Delta H$. In fact, since the external pressure is not balanced by that of the system, $P_\mathrm{ini}\neq P_\mathrm{fin}$ for the system (the process is not strictly isobaric for the system). For a reversible isobaric process $q_p = \Delta H$.


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Real Gas Behaviour

For a fixed amount of a real gas when a graph of $Z$ vs $P$ was plotted, then at a very high pressure slope was observed to be $\pu{0.01 atm^-1}.$ At the same temperature and pressure if a graph is plotted between $PV$ vs $P,$ then for $\pu{2 mol}$ of the gas the $y$ intercept is found to be $\pu{40 atm L}.$ Calculate excluded volume in litres for $\pu{20 mol}$ of the real gas.

Kindly note that Z is the compressibility factor of the gas.
I am not sure whether or not this problem involves differentiation in order to find the slope. Maybe the excluded volume is actually the product of the van der Waals constant $b$ times the amount of substance (in this case, $\pu{20 mol}).$
So, how can the value of $b$ be determined? Also, why is the slope between $PV$ and $P$ has its units in terms of atm-litre? Shouldn't it be just in terms of litres?
The answer was given to be 4. But I wish to know how it was calculated.

Can you define $Z$? It's not obvious from the statement of the problem what that is.

Z = PV/nRT, is the compressibility factor of the gas under the given conditions.

The van der Waals equation is
$$\left(p+\frac{a}{V_m^2}\right)\left(V_m-b\right)=RT$$
which for $p \gg a/V_m^2$ $^\ast$ can be rewritten as
$$pV_m=RT+bp$$
or
$$pV=nRT+nbp \tag{1}$$
The problem says that the tangent of this curve evaluated at the same $T$ has an intercept (the value of $pV$ when $p$ goes to zero) of $\pu{40 Latm}$ when $n=2$, which means that $RT=\pu{20 Latm/mol}$. But equation (1) can be written as
$$\frac{pV}{nRT}=\frac{pV_m}{RT}=Z=1+\frac{bp}{RT}$$
which means that
$$\left(\frac{\partial Z}{\partial p}\right)_T=\frac{b}{RT}$$
The problem states that this slope is $\pu{0.01atm^{-1}}$ so that $b=\pu{0.2 L/mol}$ and so that when $n=20$, the excluded volume is $nb=\pu{4 L}$.

$^\ast$ More precisely, assuming ideal gas behavior, the inequality is satisfied when $$\frac{(RT)^2}{a}\gg p$$ Looking up typical values of a (typically $\pu{1-20 atmL^2/mol^2}$) means that for the given problem this equality holds if $p\ll\pu{20 atm}$.

Hmm, But for the pressure term of the van Der Waals equation, for small and medium pressures, a/Vm^2 grows faster then pressure. The simpification is applicable, when gas is not well compressible anymore and the  equation is not well applicable.

@Poutnik good point, I will add a comment to my answer.

Just complementing the first answer of @Buck Thorn :
Full rewriting of the van Der Waals equation in terms of $Z=f(p,V_\mathrm{m})$:
$$\left(p+\frac{a}{{V_\mathrm{m}}^2}\right)\left(V_\mathrm{m}-b\right)=RT$$
$$pV_\mathrm{m} - pb + a/V_\mathrm{m} - ab/{V_\mathrm{m}}^2=RT$$
$$pV_\mathrm{m} \left( 1 - \frac{b}{V_\mathrm{m}} + \frac{ a}{ p{V_\mathrm{m}}^2} - \frac{ab}{p{V_\mathrm{m}}^3}\right)=RT$$
$$Z=\frac{pV_\mathrm{m}}{RT}\\ =\frac{1}{1 - \frac{b}{V_\mathrm{m}} + \frac{ a}{ p{V_\mathrm{m}}^2} - \frac{ab}{p{V_\mathrm{m}}^3}}\\=\frac{p{V_\mathrm{m}}^3}{p{V_\mathrm{m}}^3 - pb{V_\mathrm{m}}^2 + aV_\mathrm{m} - ab}\\
=\frac{p{V_\mathrm{m}}^3}{(pV_\mathrm{m}^2 + a)(  V_\mathrm{m} - b)}\\
=\frac{1}{(1 + \frac{a}{pV_\mathrm{m}^2})(  1 - \frac{b}{V_\mathrm{m}})}\\
\overset{p\gg \frac{a}{ {V_\mathrm{m}}^2}}
=\frac{V_\mathrm{m}}{V_\mathrm{m}-b}$$
$$\overset{b \ll V_\mathrm{m}}=1 + \frac{b}{
V_\mathrm{m}}$$

There is an error in equation 4, if I understand correctly what you are doing. When you correct that you get $Z=1+b/V_m$. Also, I am not sure the series in Z is useful.

@Buck Thorn Well,I guess you are right, it should be reciprocal. More eyes see more. :-)

I think you'd agree that in general $b<V_m$ and at lower pressures, $b \ll V_m$, meaning that $(1-b/V_m)^{-1} \approx 1+b/V_m$

Yes, sure. :-) But the point is, for low pressures is Z often < 1, as the cohesive pressure often overtakes non ideal compressibility.

But your inequality is OK.


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What is the stability order of the carbocation and carboanion?
So I came up with examples as below:
1
2
Now I am confused about the +I and -I effect in these structures since to my knowledge it can only be applied when there is a substituent (I might be wrong). I understand the role of resonance but do not get how to apply the +I and -I effect in these so I need help. I have basic knowledge about the substituents which give +I and -I effect but still can't get it.

Alkanes or $\mathrm{sp^3}$ hybridized carbons show +I effect.
Alkenes contain $\mathrm{sp^2}$ hybridized carbons. Due to greater $\mathrm{s}$  character they have higher electron withdrawing tendency. So they show -I effect.
Coming to the first question:
In the given figures we have $\mathrm{sp^3}$ and $\mathrm{sp^2}$ hybridized carbons in the given rings. Notice that in the second figure there are greater number of $\mathrm{sp^2}$ hybridized carbons. Hence, the net withdrawal of electron density is more, and accordingly, stability goes down. In the third figure there is no resonance, which is the more dominating effect than the inductive effect. So it is the least stable.
In the second question:
In figures 1 and 2, we have equal number of $\mathrm{sp^2}$ hybridized carbons. But notice in the first figure, the negative charge is on the ring and is closer to the $\mathrm{sp^3}$ carbons which are electron donating (Remember, inductive effects are distance dependent). In the third figure greater +I is being imposed on the anion, hence it has the least stability.


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Cases when Raoult's law is unsolvable
Setup
Consider a closed binary mixture of known total molar composition $n_i$, held at volume $V$ and temperature $T$.  The equilibrium phase composition is determined by $\mu_i^{vap} = \mu_i^{liq}$. Assume the vapor is an ideal gas mixture, the liquid is an ideal mixture, Poynting's correction is negligible, and $V_{liq} \ll V$. Then Raoult's law holds:
$$y_i P = x_i P_i^{sat}(T)\quad \tag{1}\label{1}$$
Simplify
For convenience, denote vapor quantity = $v_i$, liquid quantity = $l_i$, total quantity = $n_i=v_i+l_i$, and "saturation quantity" = $s_i=\frac{P_i^{sat}(T)\,V}{R T}$.  A closed system means $n_i$ are constant, and $s_i$ are constant at constant $T$.  For simplicity, choose units such that $R T = V = \sum_i n_i=1$, so that Raoult's law becomes
$$ n_i - l_i = s_i \frac {l_i}{\sum_i l_i} \tag{2}\label{2}$$
where $l_i$ are the only independent variables.  (The solution/result is independent of these choices. It holds for any Raoult's law system at constant $N, V, T$. It generalizes to infinite systems with fixed $\sum n_i/V$.)
Question
Eq. ($\ref{2}$) can be algebraically solved.  But it has no solution when $s_1>1$ and $s_1 - n_1 ≥ n_2 \frac{s_1}{s_2}$.  For example, if $s_1=2$ and $s_2=1/2$, then no solution exists when $n_2≤1/3$ ($x_2=y_1$ and $y_2=x_1$ at $\lim n_2\to \frac{1}{3}^+$).
This seems to predict that any generic species can form an uncondensable mixture if one is uncondensable (as a pure component), and there are few enough of the condensable species.  Is this correct?

Hi there, Alex, and welcome to the Chemistry StackExchange. It seems that you are trying to solve Raoult's Law in a manner that I have never seen before, and with quite a few non-standard variables. Might I suggest that you look into using the Antoine Equation to solve for the Saturation Pressure, of each species, as a function of Temperature. With this you can solve the Raoult's law equation as a function of Temperature, Pressure and either the gas or liquid fraction of species i. As for the prediction, I would not draw any conclusion about species based off of so many assumptions.

Thanks.  My only assumptions are ideality and constant T, V, and N.  That's a vessel with no reactions.  Everything else is just a choice of units to simplify Eq.1 into Eq.2.

I use an Antoine-like equation to obtain values for $P_i^{sat}(T)$.  But I'm still left to solve for the equilibrium values of $x_i$ and $y_i$.  That's nontrivial, because condensing any of species j changes the total amount of vapor & liquid, and thus affects every single $x_i$ and $y_i$.  Above I'm describing the algebraic solution, using different letters because standard ones overcomplicate the problem.

How do you justify this inequality: $s_1 - n_1 ≥ n_2 \frac{s_1}{s_2}$? What does it mean in plain words?

That inequality is the conditions for which simultaneously solving $n_1 - l_1 = s_1 l_1 / (l_1 + l_2)$ and $n_2 - l_2 = s_2 l_2 / (l_1 + l_2)$ has no solution for positive $l_1$ or $l_2$.  In plain words, Raoult's law cannot be satisfied when that inequality is true.

@alexchandel, it is most typical in a bicomponent system to first find the $x_1$, deduce $x_2$ as $x_2 = 1-x_1$, then apply Rault's Law to find the gas phase concentrations. A popular expansion of this is to condense the gas phase, collect it and measure the concentration of 1 and 2 in the collected liquid to find $y_1$ and $y_2$ and apply the activity modified Rault's law $y_iP=\gamma_ix_iP_i^{sat}(T)$. This renders activity coefficents which can then be compared to those predicted by tabulated van Laar coefficents.

So to say that you still have to solve for $x_i$ or $y_i$ of your system is entirely correct - this is the intended use of Rault's Law.

Perhaps towards this point it would be better to define your non-standard variables of $l_i$ and $v_i$ in more explicit terms. As you have it written, the only sensible unit for these would be in terms of moles, and they could be re-written as $l_i = n_{liq}x_i = n_{liq,i}$, and $v_i = n_{vap}y_i = n_{vap,i}$, where $\sum_i x_i = 1$, $\sum_i y_i = 1$, and $n_{total} = n_{vap} + n_{liq}$. Using this method, you can keep track of both the phase and the component of each variable

I think that a large amount of my confusion about your approach is in how and why you are defining a "saturation quantity". As per your definition from the ideal gas law, and Rault's law, you can find that $s_i=\frac{n_{vap}y_i}{x_i}=\frac{n_{vap,i}}{x_i}$. This has no intuitive meaning to me, and as I have never seen such an approach, I am unsure as to why you are using it.

That said, if you consider $s_i=\frac{n_{vap}y_i}{x_i}$, given the system you described a solution exists at $n_{vap}=0.682$, $x_1 = 0.088$ and $y_1 = 0.279$. Note that $n_{vap}$ does not change between the two calculations, as it represents the total quantity of gas phase molecules. If you need a derivation of this, I an provide one.

After defining fugacity and activity coefficients, to predict a mixture's equilibrium I still have to simultaneously solve $\mu_i^L=\mu_i^V$ (i.e. Raoult's law for ideal gases + ideal mixtures) either numerically or algebraically.  I changed from ${x_i, y_i, P_i^{sat}, R, T, V, n_{liq}, n_{vap}}$ to ${n_i, l_i, s_i}$ to eliminate redundant terms in Eq (1) for the latter purpose.  (Intuitively, $s_i$ is to a species' vapor quantity what $P_i^{sat}$ is to its vapor pressure.)

The algebraic solution to Eq (2) was bound by that inequality.  $n_{vap}=0.682$, $x_1 = 0.088$, $y_1 = 0.279$ does solve the system, but it doesn't satisfy the inequality ($s_1-n_1 ≈ 1.94$, but $n_2\frac{s_1}{s_2} ≈ 3.14$), so no surprises there.

Try to find a solution to Raoult's law when $s_1=2$, $s_2=0.5$, and $n_2 = 1/4$.  (If units are a must, assume $\frac{P_1^{sat}(T) V}{R T} = 2 \mathrm{mol}$ and $\frac{P_2^{sat}(T) V}{R T} = 0.5 \mathrm{mol}$, and introduce $0.25 \mathrm{mol}$ of species #2 and $0.75 \mathrm{mol}$ of species #1 into an empty, rigid vessel held at constant T.)  None exists.

@alexchandel With this setup, you are only defining component 1 has 4x the saturation pressure of component 2, and that component 1 is present in 3x the quantity. There is no reason that Raoult's law is cannot be solved for such a system. I think you went wrong with an assumption somewhere.

As you have not provided a source for your definition of $s_i$, I shall redefine it in terms of Raoult's Law and the Ideal Gas Law. To do this, we shall first take Raoult's Law and solve for $P$.
$$\begin{align}y_iP&=x_iP_i^{sat}(T)\\P&=\frac{x_iP_i^{sat}(T)}{y_i}\end{align}$$
we shall then take this form, and plug it into the ideal gas law, considering only the gas phase of the equilibrium system. Note that because the system is at equilibrium, $P_{bulk}=P_{liq}=P_{vap}$ and $T_{bulk}=T_{liq}=T_{vap}$. If the volume of the liquid phase is assumed to be significantly lower than that of the vapor, then $V_{bluk}=V_{vap}$. After plugging in this, we will then work towards the form of $s_i$ as shown.
$$\begin{align}P_{vap}V_{vap}&=n_{vap}RT_{vap}\\PV&=n_{vap}RT\\n_{vap}&=\frac{PV}{RT}\\n_{vap}&=\frac{x_iP_i^{sat}(T)}{y_i}\frac{V}{RT}\\\frac{n_{vap}y_i}{x_i}&=\frac{P_i^{sat}(T)V}{RT}\end{align}$$
From this, and the definition of $s_i$ as you stated, we can show that
$$s_i=\frac{P_i^{sat}(T)V}{RT}=\frac{n_{vap}y_i}{x_i}$$
which is subject only to the reasonable assumptions of ideality as stated above, and the bounds per the definition of each term. Namely, $\left\{P_i^{sat}(T),V,T\right\}$ must be positive, real numbers and that $\left\{x_i,y_i\right\}$ must be positive, real numbers with bounds of $\left\lbrack0,1\right\rbrack$.
No further inequalities are required to describe the system.
To solve such a system, you would then set up a system of equations and numerically backsolve for an answer, in terms $n_{vap},x_{i},y_{i}$. For example you might feed a set of equations the seed values of $n_{vap}=1$ and $x_i=.5=y_i$ such as seen below, then optimize your values to minimize the error between the calculated and target values.

The above seed values could then optimize to a particular solution that looks like. Note that this does not represent the whole of the solution space.

If you wish to further control where your particular solution ends up, you can restrict the solution space by hold one of the values to be a known constant. Below is one such example as described per your comments.

In this case, it is not possible to find a reasonable solution, as $n_{vap}=0.25~\text{mol}$ lies outside the solution space for this system of equations. Without going too heavily into the derivation, it can generally be stated that this value is a bit too low to garuntee that you will be able to find a solution. This more says that if you had this system, it would likely evolve off more of its liquid into the gas phase at in order to reach equilibrium than it says the system could not exist, though yours is not an unreasonable reading of the result.
A good rule of thumb for this would be that you can expect to find a reasonable solution for $\left\{x_i,y_i\right\}$ for any system in which $s_1 \leq n_{vap} \leq s_2$ holds, however this is by far the most complicated way to solve out this system of equations. A simpler approach to this would be to hold either $x_i$ or $y_i$ constant to restrict the solution space, as any constant value on the bounds of $\left\lbrack0,1\right\rbrack$, as already assumed will garuntee a unique solution to the system of equations.

My question isn't how to solve Raoult's law; for < 5 components, at constant $V, T$, & $V_{liq} \ll V$, Raoult's law has (if any exists) an exact algebraic solution for a given $N_i$, which numerical methods approximate.  I'm talking about equilibrium behavior when Raoult's law is unsolvable.  "This more says that if you had this system, it would likely evolve off more of its liquid into the gas phase at equilibrium" — then that isn't equilibrium.

"…it can generally be stated that this value is a bit too low to garuntee that you will be able to find a solution."  — The proof of that is the premise of the question.  "…you can expect to find a reasonable solution for $\left{x_i,y_i\right}$ for any system in which $s_1\leq n{vap}\leq s_2$ holds"_ — Not so: take $s_1=0.5$ mol, $s_2=2$ mol, & begin with $n_1 = 0.25$ mol, $n_2 = 0.75$ mol ($\to n_{vap}≤1$ mol).  No solution exists for any $n_{liq}>0$, and limiting $f_i^{liq}>f_i^{vap}$ imply $n_{liq}=0,n_{vap}=1$ mol at equilibrium.  So $s_1\leq n_{vap}\leq s_2$ with no solution.

"this is by far the most complicated way to solve out this system of equations."  — How else would you determine the equilibrium phase compositions of a given closed mixture at a constant $V, T$?  "hold either $x_i$ or $y_i$ constant to restrict the solution space…, as already assumed will garuntee a unique solution to the system of equations." — Fixing $y_1$ and $y_2$ (or $x_1$ and $x_2$) is incompatible with finding the equilibrium of a given closed binary mixture at constant $V, T$, as solution space is already zero-dimensional.

In regards to the second comment, note that $n_{vap} \neq n_{vap,i}$, and is the total number of moles in the gas phase from all $i$ species, and therefore must be the same of both species. If you hold $n_{vap}=0.75$ a solution exists at $x_1=0.167$ and $y_1=0.444$.

In regards to the third comment - I was assuming that $n_{vap}$ would be able to be modulated by changing the pressure of the system, but after looking over your most recent revision, I see that this is not the case, which means that $n_{vap}$ would be held constant under the ideal gas law and the definition of the system.

With that said, yes, it would be fair to say that if you are holding, $P, V, T$ and therefore $n_{vap}$ as being constant, then you will not always be able to design a system such that a target $s_1$ and $s_2$ values are met.


text:
Determine structure from NMR
So I've come across this lovely spectrum of a compund where only its chemical formula is given with C5H9NO.
I have made the pictures of the spectrum available below.
obviously there is some pretty hefty coupling going on but I cant really figure out what it could be.
So far I think we definitely have a CH3 group that could be attached to either heteratom. On the other side, the single proton at 6.66 ppm could also be an H attached directly to the N-atom (maybe?).
Anyways, I hope you guys can help me here and maybe also explain me the couplings that are going on.

EDIT: to not get marked as homework, because it isnt, I suggesting that it should be some sort of pyrrole or pyrrolidine structure type. Further I think that the CH3 group must be shielded from other protons and is either near or is sitting at a heteroatom.
Next I have narrowed down two structural possibilities: either we have a proton sitting at the nitrogen and the other sitting on a double bound, or we have diastereotropic protons.
My best candidate so far is the picture below (please let me know if it would be possible):

Sorry, but if you dont want this closed as "homework", at least try to identify the couplings ("ddt" etc.), and calculate the coupling constants. Btw. marking the individual peaks of a splittings in ppm is just wrong. And one more hint, your normalisation of the integrals is rather weird. The methyl group consists of three protons, so it should be 3, not "100".

I admit the one at 5.2pmm is kind of weird. There is a clear roof with the one at 5.8, however. Forget figuring out the exact splittings there. Thats a real higer order spectrum.

@Karl I´ve now thought about it for quite a while, please look at my edited version. I hope it makes it clear, that its not homework

Methoxy resonates at ~3.1 ppm so its not consistent.

You've kind of already done it, but it's good to explicitly determine the degrees of unsaturation so you know what you have to work with...

The signal at 5.2 ppm is not a multiplet, but two signals. Thus, the suggested structure with two CH2 groups is not consistent with the spectrum. https://www.chem.wisc.edu/areas/reich/nmr/Notes-05-HMR-v26-part2.pdf

Is this a 600 MHz spectrum? That would make the largest couplings on the order of 20 Hz and the smallest couplings on the order of 1.8 Hz. How about an allyl system to explain the signals between 3 and 6 ppm, e.g. compare to the spectrum of allyl chloride https://www.ebiochemicals.com/Wiki/QcEB000030949_HNMR_1.html.

Is there a mention of solvent for this spectrum? I was able to find a the free supplemental information for a paper where they made N-allyl acetamide (which has been proposed below) and the chemical shifts don't quite match up in chloroform-D.

no unfortunately no mention of solvent, but suspect that it should be chloroform-d.

Finding a structure with the correct formula that matches the chemical shifts reasonably well is not so difficult.
The multiplet integrals, normalized such that the area corresponding to one $\ce{H}$ is $\approx 33$ suggests the following number of $\ce{H}$ (moving upfield): 1,1,2,2,3.
The upfield singlet suggests an uncoupled terminal methyl group, so we begin with the simplest assumption, that we have a molecule with a linear alkane.
For a linear alkane, the 4 remaining $\ce{C}$ require 7 bonds in addition to $\ce{3 C-C}$, $\ce{1 C-N}$, and $\ce{1 C-O}$ single bonds. The $\ce{N}$ requires 2 additional bonds, the O 1. Since there are $\ce{6 H}$ remaining, either unsaturation or multiple bonds such as $\ce{C=N}$ or $\ce{C=O}$ bonds are required (we dismiss ring formation for the time being).
The high chemical shift and simple coupling pattern suggests the methyl group sits next to an electron withdrawing group without protons, carbonyl would fit.
If we include a carbonyl, we now have remaining $\ce{3 C}$ requiring 6 bonds in addition to $\ce{2 C-C}$ and $\ce{1 C-N}$ single bonds. The $\ce{N}$ requires 2 additional bonds. Since there are $\ce{6 H}$ remaining, either unsaturation or a multiple bond such as either $\ce{C=N}$ or $\ce{C=O}$ bond is still necessary. The broad singlet at $\pu{6.7 ppm}$ suggests a single $\ce{N-H}$ proton, so we assume the $\ce{N}$ is part of the chain. Let's try an amide.
If we include the amide, we now still have 3 remaining $\ce{C}$ requiring 6 bonds in addition to $\ce{2 C-C}$ and $\ce{1 C-N}$ single bonds. Since there are $\ce{5 H}$ remaining, unsaturation is necessary. We assume an  ethylene group is present with an intervening methylene, we arrive at the following proposed structure with the right formula, which I checked with an online simulator:

The match of the chemical shifts is pretty good, but the detailed splittings are not explained. Hans Reich provides tutorials on his site showing you how to decompose complex multiplets, you may want to look at that. Other simulation software are also available.

Thank you so much. I knew that the structure cant be too difficult...
very nice explanation of how one should think when analyzing these structures

This is an interesting question. However, disadvantage is having only $\ce{^1H}$-$\mathrm{NMR}$ to deal with. Yet, it has pretty good resolution (probably using $\pu{400 MHz}$ machine) so we can resolve the splitting pattern very easily and predict the structure.
First, OP has correctly assigned two degree of unsaturation. For molecular formula, $\ce{C5H9NO}$:
$$u=\frac{5\cdot2+2+1-9}{2}=2$$
However, OP's suggested structure was incorrect since chemical shift of -$\ce{OCH3}$ is usually greater than $\pu{3 ppm}$. Because of the chemical shift of -$\ce{CH3}$ group in the given $\ce{^1H}$-$\mathrm{NMR}$ is about $\pu{2.0 ppm}$, one can conclude that it is either olefinic or one $\alpha$ to a cabonyl group. Nonetheless, OP's suggestion of proton resonance around $\pu{6.66 ppm}$ is due to $-\ce{NH}-$ group is correct, but due to its deshielding nature ($\pu{6.66 ppm}$), one can suggest that it is an amido $-\ce{NH}-$ group. However, OP correctly assign it next to a $-\ce{CH2}-$ group.
Since I suspect non cyclic nature of the unknown, I have assigned splitting patterns and calculate relevant coupling constants assuming the spectrum is taken in $\pu{400 MHz}$ machine. All splitting patterns of remaining resonances are $ddt$ as depicted in fallowing diagrams:

Accordingly, these patterns and chemical shifts well match with N-allyl acetamide (see insert in first diagram).

thank you as well for your answer! Its very much appreciated and now I understand the coupling as well. good to see, that I wasnt completely wrong. the -OCH3- I should have caught to be at 3ppm, that was a rookie mistake

One question though: what is this formula for the degree of unsaturation?

It is a little complected to write in details in a comment. But, you can find it in any textbook used in spectroscopic analysis of organic compounds course or just google it. For a simple hydrocarbon, $\ce{C_nH_a}$: $u = \frac{n \times 2+2-a}{2}$.

Also, for a simple hydrocarbon with halogen, $\ce{C_nH_aX_b}$: $u = \frac{n \times 2+2-a-b}{2}$

The triplet in the methylene group  must be a $dd$ of 1 Hz each (fitting to the two 1 Hz $t$ around 5.2

@Karl: I agreed.

Well, do you still agree? Because your answer is still the same as it was this morning. ;)

@Karl: It's going to be so messy if I do so. Besides, $dd$ with identical coupling constants becomes a $triplet$, so I leave it alone. Thanks for the comment.

@Karl: My point above is fulfilled with my answer here.

Thanks for showing the details of the doublet of triplets at 5.21 ppm that looks like a quartet because the coupling constants are related by a factor of 2. Is there any way to tell them apart?

@ Karsten Theis: Not in this case (with $\pu{400 MHz}$). However, if we can get a spectrum using a $\pu{600 MHz}$ machine, we'd see further splitting of middle two peaks showing clear $dt$, or even $ddd$ as @Carl suggested.


text:
Will disodium EDTA effectively destroy Tollens' reagent?
As I understand it, if I mix $\ce{[Ag(NH3)2]NO3}$ and $\ce{Na2EDTA}$ the silver will complex much more strongly with the EDTA than the ammonia, and the hydrogen ions will protonate the ammonia, leaving $\ce{AgEDTA^{3-} + 2NH4+ +NO3- + 2Na+}$, preventing the formation of silver nitride. Does that make sense? I'm concerned that the nitrate is incompatible with the EDTA somehow (that is, that it will oxidize something and break it apart).

Reaction is OK, but I don't get what you mean by "incompatible".

He may mean mutual reaction, that would occur in concentrated nitric acid only.

There would be needed review of complex stability constants. I am afraid for M+ ions and EDTA it would not be great, compared to ammonia.I would prefer acidification and silver recycling, e.g. by metallic zinc or solution based reduction agens.

Silver has been found to form complex with EDTA: https://www.researchgate.net/publication/222356275_Thermal_stability_and_bonding_in_the_silver_complexes_of_ethylenediaminetetraacetic_acid

@Nilay Ghosh I rather mean, if it's stability constant is high enough to avoid forming diamminsilver complex.


text:
How to find osmolarity of a solution contatining salt and glucose?

What is the osmolarity of a solution that contains $4.00\%$ (m/v) $\ce{NaCl}$ $(M = \pu{58.44 g mol-1})$ and $3.00\%$ (m/v) glucose $(M = \pu{180.18 g mol-1})?$

I know you have to convert percentages to mass soultion/Liter solution and multiply by number of moles in $\ce{NaCl},$ which is 2 moles:
$\pu{2 osmol}$ $\ce{NaCl}$ / $\pu{1 mol}$ $\ce{NaCl}$
but I'm being thrown off by being given molar mass.

Hint: Osmolarity is osmotic molarity. Molarity is ...  The adjective osmotic means...

Let's begin by finding the molarity of each solute in that solution. We'll get to osmolarity later.
NaCl
The concentration of NaCl given in the problem is $0.04 \frac{\text{g}}{\text{mL}}=40 \frac{\text{g}}{\text{L}}$. We can divide by the molar mass, getting $\frac{40\text{g}}{\text{L}}\cdot\frac{\text{mol}}{58.44\text{g}}\approx0.6845\text{ M.}$ (M represents molar, or mol/L.)
Glucose
The concentration of glucose given in the problem is $0.03 \frac{\text{g}}{\text{mL}}=30 \frac{\text{g}}{\text{L}}$. We can divide by the molar mass, getting $\frac{30\text{g}}{\text{L}}\cdot\frac{\text{mol}}{180.18\text{g}}\approx0.1665\text{ M.}$
It is at this point that we consider the distinction between osmolarity and molarity.
According to Wikipedia,

$\text{osmolarity}=\displaystyle\sum_{i}\phi_in_iC_i$
where

$\phi$ is the osmotic coefficient, which accounts for the degree of non-ideality of >the >solution. In the simplest case it is the degree of dissociation of the solute. >Then, $\phi$ is between 0 and 1 where 1 indicates 100% dissociation. However, $\phi$ can >also be larger than 1 (e.g. for sucrose). For salts, electrostatic effects cause $\phi$ >to be smaller than 1 even if 100% dissociation occurs (see Debye–Hückel equation);
n is the number of particles (e.g. ions) into which a molecule dissociates.
C is the molar concentration of the solute;
the index i represents the identity of a particular solute.

For the moment, we are going to ignore $\phi$ and assume that that everything dissociates perfectly. We can make this assumption because glucose and NaCl generally dissolve near-completely in water.
From that, we get $\text{osmolarity}=\displaystyle\sum_{i}n_iC_i=n_\text{NaCl}C_\text{NaCl}+n_\text{glucose}C_\text{glucose}$
We know that NaCl dissociates into two ions: Na$^+$ and Cl$^-$, so $n_\text{NaCl}=2.$ Glucose, however, does not dissociate, but rather stays as a single molecule. Therefore, $n_\text{glucose}=1.$
We now have $\text{osmolarity}=2\cdot0.6845+1\cdot0.1665=\boxed{1.5355 \text{ osmolar}}$


text:
Why do significant figure rules contradict each other?
So I came across a problem in my chemistry textbook that brought up another question. The problem had to do with calculating the mass% composition of chlorine in a certain molecule.
So to go about solving this, you could calculate the mass in grams of 1 mole of the molecule and divide the mass in grams of the corresponding amount of moles of chlorine by that number and multiply that by 100 to get a percent.
The molecule had 3 Cl molecules, so I would look and see that the molar mass of 1 mole of Cl is 35.45g. So I would multiply that by 3 to get 106.35g of Cl, but I would have to round that to 4 significant figures since 3 has infinite significant figures, and the mass of Cl was given in 4. Which means the mass of 3 moles of Cl is 106.4g.
Which poses my question... Isn’t multiplying the molar mass of Cl by 3 the same as adding the molar mass of Cl 3 times? And if so, wouldn’t that mean that if you were to add the mass 3 times, then the significant figure rules for adding would tell us to round to the number in the problem with the least decimal places, which is 2 decimal places? So when you add the mass of Cl 3 times you get 106.35g so then you round to 2 decimal places and get 106.35g.
Why does this give a different answer? How come when you multiply by a number with infinite significant figures, the number of significant figures in your rounded answer can be different than if you were to add that something to itself that same amount of times?
Can’t you be certain to the tenths place in this example, and have the hundredths place by your uncertain digit since when you add, you don’t have to round any numbers off? If not, why not?

TLDR - With significant figures you should only round the final answer. For intermediate calculations you should carry extra figures to minimize round-off errors. I'd recommend two extra figures if you need to round intermediate results.

I suggest to make the question better structured for easier reading.

If you multiply by an integer, it is clear that you are talking about the same quantity. When you are adding, it could be three independent measurements that happen to have the same value. In the latter case, some of the errors might cancel out (the combined error can be taken as root mean square). Thus, even in a more sophisticated error propagation treatment, the results might be different.

Bear in mind the significant digit rules are just  simplifications of the rules of absolute (+, -) and relative (*,/) variance propagation.
The rule of significant digit numbers for summing is for 2 statistically independent variables. Their  correlation coefficient = 0 and the error is $$  s_\mathrm{C}= \sqrt{(s_\mathrm{A})^2+(s_\mathrm{B})^2}$$
and therefore about at the same order as for less precisely known variable.
Multiplying a variable by number means summing the same variable $N$ times, with the correlation coefficient 1 and the error is growing linearly by $N$. As here applies the multiplicative  rule for propagation of relative variance.
$$ s_\mathrm{C,r}= \sqrt{(s_\mathrm{A,r})^2+(s_\mathrm{B,r})^2}\\=\sqrt{(s_\mathrm{A,r})^2+(s_\mathrm{3})^2}=s_\mathrm{A,r}$$
I agree with the comment to follow the rule of significant digits on the final result only, to avoid cummulative rounding errors.
On the other side, it can be hard to track the digits but for simple formulas. Applying the variance propagation rules may be easier way to go.

Yes, there is an inconsistency in how significant figures work. As you noted when using significant figures
$$35.45 \times 3 = 106.4$$
but
$$35.45 + 35.45 + 35.45 = 106.35$$
As a final answer, I'd go with 106.4 but for intermediate calculations I'd use 106.35.
Significant figures are really a sloppy way of doing error propagation. The gist is that significant figures prevents stupid mistakes like equating 10.0/3 with 3.3333333333... which has infinite exactness.
To go in somewhat more detail, consider that $35.45$ is taken to mean $35.45 \pm 0.005$. In percent this is $0.005/35.45 = 0.014\% $.
$106.4 \pm 0.05 \ce{->} 0.047\%$
$106.35 \pm 0.005 \ce{->} 0.0047\%$
So an error of 0.047% is a little worse than the "real" error of 0.014%, but 0.0047% is an error which is better than the "real" error of 0.014%.


text:
How do Precipitation Reactions behave in the Absence of Gravity?
How do precipitation reactions behave in the absence of gravity, say on the International Space Station (ISS)? I have seen water taking the shape of a sphere and not that of the container in space due to its surface tension. It takes a spherical shape in order to minimize the energy due to forces of surface tension. In space all directions are equivalent and hence there is no notion of up or down.
In precipitation reactions, generally, the precipitate formed either floats on the surface or settles down. I know the solid particles will be formed in space due to the chemical reaction between the reactants. What will happen to the solid particles (precipitate), will they concentrate at the centre of the sphere of the reaction mixture (solution) or will they come to the surface, or will they remain suspended, or is there any other possibility? Has such an experiment been done on the ISS to date?
For reference, this is how water looks in the absence of gravity:

Please note: Water is coloured to enhance visibility

@Poutnik, I thought of that only for particles which form at the centre of the sphere where all the forces balance each other to give no net force. But, I think if they were formed somewhere between then they must come to the surface. So, I think some particles may be at the centre and most on the surface. But, I don't know what will happen in reality. And that's why I posted the question here :)

@Poutnik, Thank you. So are you trying to say all the particles will be uniform throughout? If so will the surface be entirely water molecules or does have some precipitate?

OK, I have summarised the comments in the answer.

In gravityless environment, all directions are equivalent save the very vicinity of the surface(*), where just few water molecules are toward the surface.
Therefore, in the bulk volume, the precipitate would stay where it is, unless some currents exist for whatever  reason.
At the very surface, it depends on nature of precipitate.
If it is rather hydrophilic, water molecules would have tendency to surround it, so it would be rather pulled from surface inwards.
If it is rather hydrophobic, water molecules would have tendency to expelled it, so it would be rather pushed outwards to the surface.
If it is far enough ( tens molecules ) from surface, all effects will get spherically symmetric, but one....
(*)  .....  as in a very very long term, precipitate denser than water would be more concentrated near the centre by the own gravity of the "bubble" and vice versa.
I am not aware of such an ISS experiment, but it makes little sense to do it for me. In the microgravity context, it is very very long term experiment.

"In very very long term, precipitate denser than water would be more concentrated near the centre by the own gravity of the "bubble" and vice versa." But the force of gravity according to my physics knowledge says the force of gravity at the centre of sphere is $0$.

Sure, but just in the center. :-) There will be  microgravity force toward the centre, strongest at the surface, zero at the centre.

Ok, then we can think it of like - a particle at the centre forms at some point of time, due to the particle's very weak gravitational force, some more particles might migrate towards the centre, and so on and so forth.

Rather, they will, not might, but very very slow.

Do you know any such experiments conducted on the ISS?

I am not, but it makes little sense to do it. It is very very long term experiment.

When it comes to the self-gravity of the drop vs. Brownian motion, I suspect that the gravity will only ever become apparent statistically - the particles of precipitate won't hang around near the centre for very long.  This random motion may also mean that over time particles will get close enough to the surface for them to stay there if hydrophobic.  I wonder how we could model this in the lab - acoustic levitation probably creates internal forces in the drop; electrostatic perhaps less - A nice undergrad project (if only I had the kit)

@Intellex Precipitation of supersaturated solutions has been studied on the ISS and Space Shuttle.   Protein crystals for X-ray diffraction are the goal.  https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5515504/

Inspired by @Jan's Answer.
Following image considers the case when the liquid drop is under rotation (shown by blue curved arrows). Let us consider the frame of reference as one which is rotating with the same angular velocity as that of the droplet. Since we are in a non-inertial frame, we need to consider the pseudo forces - and here it is the centrifugal force (acts away from the axis or centre of rotation). Due to this, there is some kind of artificial gravity inside the sphere as mentioned in @Jan's answer. This force is represented by the orange arrow in the diagram. Due to this, and due to the light nature of bubbles of dissolved gas, the bubbles form in the central region as denoted by the dotted circle region. Here the precipitate will collect on the inner surface of the sphere if it is denser than water (shown by the light green portion along the circumference in this cross-sectional view).

The above image is a view from the top of the axis. The following image is a side view (cross-sectional view again!) of the rotating system. See the sphere is slightly compressed at the poles and elongated at the equator. The precipitates accumulate near the equatorial region of the sphere.

For Both Images:
Image Source: My own work :) (Hope you like it!)

In addition to Poutnik’s answer, it is possible to manipulate a drop of water in space to generate a force similar to gravity within it: by applying gentle airflow to make the drop of water rotate.
Water in a rotating drop is subject to centrifugal forces, meaning that there is now an ‘along the force’ and an ‘opposite to the force’ direction which act in a very similar manner as ‘up’ and ‘down’ on Earth. Therefore, a precipitate that would collect at the bottom of the flask on Earth collects on the outside of a rotating drop in space.
I read that an experiment of this type was performed with either carbonated soft drinks or beer in space. The amusing part is that carbon dioxide is, of course, lighter than water and thus collects in the middle of a rotating drop.

Now, the drop will be compressed at the poles (or along the axis) and elongated at the equator if you consider it to be rotating.

With reference to the previous comment kindly see this answer.


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Gibbs free energy of phosphorus pentachloride decomposition reaction

The equilibrium constant at $\pu{227 °C}$ for the equation
$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$
is $K_p = \pu{4.50E3 bar}.$ Calculate the value of $Δ_\mathrm{rxn}G^\circ$ at  $\pu{227 °C}$.
With the temperature at $\pu{227 °C}$ and the partial pressures of the gases of  $p(\ce{PCl3}) = \pu{0.286 bar},$ $p(\ce{Cl2}) = \pu{0.713 bar},$ and $p(\ce{PCl5}) = \pu{3.06E-6 bar},$ calculate the value of $Δ_\mathrm{rxn}G.$

I calculated $Δ_\mathrm{rxn}G^\circ$ to be $\pu{-34.97 kJ mol-1}.$ Wouldn't it just be
$$Δ_\mathrm{rxn}G^\circ = -\pu{0.0083145 kJ mol-1 K-1} × \pu{500 K} × \ln(\pu{4.5E3})?$$
For the second part of the question I got $\pu{11.20 kJ mol-1},$ but apparently this is also wrong, no doubt because I got the first part wrong.
Would someone care to explain to me?

What makes you think the first answer is wrong?

The website we submit the answer to (Sapling) claims it is wrong.

I have no idea what Sapling is, but probably they expect another format, number of sig figs or different units? Do they provide an answer?

It's like a HW platform. I've tried different sig figs and units are already provided. They don't provide an answer unless you either solve it correctly or press "Give Up".

So I ended up pressing "Give Up". Apparently it wanted only 1 Sig Fig, but no where in the question did it ask for it.

Well, that's one of the reasons I don't like online tests. Good job on your side though:)


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Does the explosion of TNT include oxidation?
This comment below the question Rocket explosion compared to kT of TNT; has one ever knocked something over at a distance? suggests that

...TNT includes it's own oxidizer...

Explosion isn't the same as combustion and you don't need to add a separate oxidizer to an explosive for it to do its thing.
But technically speaking does the explosion of TNT include any steps that could be considered as oxidative? Can some intermediate products be considered to be acting as oxidizers?

currently unanswered Rocket-related questions: What reactions involving hydrazine, dinitrogen tetroxide and possibly a contaminant can produce a salt? and also Approximate magnetic susceptibility of these liquid propellants?  Still not sure I have a complete answer to Why would red fuming nitric acid become white fuming nitric acid if left out at low temperature? either.

It is not about intermediate products at all. Yes, TNT itself acts as an oxidizer. And as a reducer too.

That's just disproportionation, nothing special, really.

3 nitrogroups of TNT are rich of oxygen, but the aromatic ring has only carbon and hydrogen atoms.
Therefore oxygen of nitrogroups oxidizes carbon and hydrogen atoms.
There is no principle difference between oxidation and reduction agens being separate molecules, or being different parts of the same molecule.
Note that TNT, in contrary to e.g nitroglycerine, is oxygen deficient, so the amount of available oxygen is not enough for complete oxidation to water and carbon dioxide.
TNT:
$$\ce{2 C7H5N3O6 -> 3 N2 + 5 H2O + 7 CO + 7 C}$$
or
$$\ce{2 C7H5N3O6 -> 3 N2 + 5 H2 + 12 CO + 2 C}$$
Nitroglycerine
$$\ce{4 C3H5N3O9 -> 6 N2 + 10 H2O + 12 CO2 +  O2 }$$

Second equation for TNT should rather be with CO2 and C - probably most accurate. Definitely no H2 there, sooner there would be only C then any gaseous hydrogen.

@Mithoron That may be questionable, as a mix of CO and H2 is/was industrially produced by reaction of gaseous H2O with hot coke. Independently, both reactions are mentioned in the Wikipedia TNT page https://en.wikipedia.org/wiki/TNT#Explosive_character

https://pubs.acs.org/doi/abs/10.1021/ja410020f may shed some light on this, but is paywalled :(

Note I am not an expert in chemistry of explosives. See also https://en.wikipedia.org/wiki/Chemical_explosive particularly section Balancing chemical explosion equations.

Thing is all this looks like equations/ calculations wrote from top of a head. I guess just mentioning that these are simplified/borderline examples would be okayish as question doesn't ask about it.

See also https://fas.org/man/dod-101/navy/docs/es310/chemstry/chemstry.htm

https://link.springer.com/chapter/10.1007%2F978-3-642-86682-1_2 Thermodynamics of Explosions :  .....Direct products from a chemical explosive include the usual combustion products of carbon dioxide (CO2), carbon monoxide (CO), and water vapor (H2O), and molecular nitrogen (N2). They may also include molecular hydrogen (H2) and molecular oxygen (O2), along with nitric oxide (NO).....


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Do I understand correctly that simulation volume influences relative permittivity calculations? If so, how is the result useful?
I am having a lot of trouble understanding the influence of volume on relative permittivity/static dielectric constant.  I'm not a chemist, but am using molecular dynamics to calculate relative permittivity for use in other work.
A predecessor used this equation for relative permittivity $ε_\mathrm{r}$:
$$ε_\mathrm{r} = 1 + \frac{4π\sigma_M}{3k_\mathrm{B}TV}$$
where $V$ is the volume of the simulation, $T$ is temperature, $\sigma_M$ is the variance of the total dipole moment, and $k_\mathrm{B}$ is the Boltzmann constant.
Here is what I do not understand:  If I use a different $V $(i.e., a different number of atoms) for my simulation compared to my predecessor, then my calculated relative permittivity will be different from his.  I have done this to make sure, and indeed, the difference is quite significant.
Is there something silly I am misunderstanding here?  Like "$V$" is not the average box/cell volume of the simulation, but something that is not size-dependent?

Your equation looks right. It is the same as can be found in this paper, https://arxiv.org/ftp/arxiv/papers/1009/1009.5958.pdf.
I am not certain of this, but I would guess that the variance of the total dipole moment would change when you change the total volume, and thus your result should be more or less volume independent.

Thanks very much, Erik.  I wondered if that were meant to be the case.  So then does it seem likely to you that when two calculations (each with a different number of atoms) lead to different calculated values, that the system volume is too small for one of the calculations?

And a follow-up question to that for anyone who is interested: would it make sense to run the smaller system for longer to achieve the same relative permittivity value as the larger system?  Or is there something more fundamental that's a problem about using a smaller system?

It's not quite clear how you determine the variance of total dipole moment. Is it just a mean square? Could you probably elaborate on this a bit further and add the exact math expression you are using in the equation?

Thank you very much for asking, andselisk, and thank you for your edits.  The molecular dynamics software I'm using outputs the total dipole moment and I'm using Python/numpty's "variance" function (e.g., "np.var") to calculate the variance.  Please feel free to ask if you would like me to add any more detail about this.

@ErikKjellgren Thank you for your edit. I couldn't figure out how to denote the  variance of the dipole moment in that formula, but you saved the day:)


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Why don't we use silver hydroxide instead of Tollen's reagent for test of aldehydes?
AgNO3+NH4OH=AgOH+NH4NO3
AgOH is used for the reaction with aldehyde. But why can't we use silver hydroxide directly?

Silver hydroxide is not really a thing.

Silver hydroxide does not exist. Silver salts react with hydroxide ion to produce silver (I) oxide. See: https://chemistry.stackexchange.com/questions/75705/finding-ions-in-water-samples/75728?r=SearchResults#75728

Silver hydroxide does not exist.
Hydroxides of alkalic metals form insoluble silver oxide:
$$\ce{ 2 Ag+ + 2 OH- -> Ag2O + H2O}$$
Ammonia  forms with silver ions the linear diammin complex, the nature of the Tollens reagent:
$$\ce{ Ag+ + 2 NH3 -> [Ag(NH3)2]+}$$