text:
Nitrogen inversion barriers of trimethylamine and N,N-diisopropylmethylamine

Which would have a smaller nitrogen inversion barrier: $\ce{NMe3}$ or $\ce{N(i-Pr)2Me}?$

I think that $\ce{NMe3}$ should have a smaller inversion barrier as it is less bulky, but data shows otherwise. Are there some other effects that contribute to the inversion barrier that I might not know of?

You says "data shows otherwise," so would you kindly show that data?

@Mathew Mahindaratne: This very Question is given in MS chauhan organic chemistry book, so OP probably would have got it from there

So I thought about this for a bit and I think I have an answer. Be warned, I might be wrong.
The inversion process involves a conversion from $\mathrm{sp^2}$ to $\mathrm{sp^3}$. So $\mathrm{sp^2}$ hybridization increases angle between these bulky groups, which would be in fact favorable. Hence the low activation energy.

Yes, your thinking is correct.  However, since the roughly tetrahedral ground state is ~$\ce{sp^3}$ hybridized while the planar TS for inversion is ~$\ce{sp^2}$ hybridized one would more likely say that you convert $\ce{sp^3}$ to $\ce{sp^2}$.  But your argument that the steric destabilization is greater in the ground state than the TS thereby reducing the activation energy more in the di-isopropyl case is correct.

Barriers to intramolecular motions, which lead to inversion, are intimately related to the configurational identity of a molecule ($\ce{XY3}$). These barriers turn out to be most sensitive to changes in $\nu_2$ (the frequency of symmetrical deformation vibration) and $\ce{Y-X-Y}$ angle ($\alpha$) of $\ce{XY3}$ (Ref.1). The inversion is assumed to take place by a gradual increase in the amplitude of the $\nu_2$, until the molecule passes through plainer configuration (Ref.2). The values of the barriers have mostly been calculated mathematically. In fact, one of the long-standing problems of stereochemistry is that of determining barriers of nitrogen in amines, which has been assumed at time that barriers must be too small to be determined experimentally. No quantitative estimates were reported prior to the discovery that the barriers in aziridines are large enough to study by the dynamic NMR methods (Ref.3), which lying in the range $\pu{10-15 kcal/mol}$. For example, the activation energy for N-methylaziridine (1-methylaziridine) is found to be $\pu{19 \pm 3 kcal/mol}$ (Ref.4 & 5) while that of N-methyl-2,2-dimethylaziridine (1-methyl-2,2-dimethylaziridine) is found to be $\pu{10 kcal/mol}$ (pure compound; Ref.6). Further, when $\Delta E$ is measured for 1-methyl-2,2-dimethylaziridine in methanol and $\ce{CCl4}$, the values obtained are $\pu{6.8 kcal/mol}$ and $\pu{7.8 kcal/mol}$, respectively. The authors attributed this decrease of activation energy for both polar and non-polar nature of relevant solvents. The also explained the extra reduction of activation energy in methanol due to H-bonding of hydroxyl group of methanol with nitrogen lone pair in aziridine. However, the authors have admitted that they cannot explain the large difference in $\Delta E$ between 1-methylaziridine and 1-methyl-2,2-dimethylaziridine.
The mathematical estimated activation barrier of trimethylamine was found to be $\pu{7.5 kcal/mol}$ ($\pu{31.4 kJ/mol}$). The experimental value was found to be $\pu{10-11 kcal/mol}$, which was determined by indirect NMR methods (Ref.7). The first direct experimental measurement to barrier to nitrogen inversion in a simple acyclic amine, dibenzylmethylamine has been made using dynamic NMR techniques. The barrier was reported as $\pu{6.7\pm 0.2 kcal/mol}$  in vinyl chloride solution at $\pu{-135 ^\circ C}$ , which has agreed well with that of the calculated theoretical value (Ref.8).
As explained before, these barriers turn out to be most sensitive to changes in $\ce{Y-X-Y}$ angle ($\alpha$) of $\ce{NY3}$. Using any molecular modeling software, one can show that when bulkiness of $\ce{Y}$ increase the  $\alpha$ increases. For example, using online software provided by ChemEd DL, one can calculate $\ce{H-N-H}$ angle of $\ce{NH3}$ as $107.2^\circ$ while $\ce{C-N-C}$ angle of $\ce{N(CH3)3}$ is $110.9^\circ$. Further, $\ce{C-N-C}$ angle of $\ce{N(CH2CH3)3}$ is $111.9^\circ$. Therefore, it is safe to assume that activation barrier of diisopropylamine is smaller than that of trimethylamine similar to competitive values of trimethylamine and dibenzylmethylamine.

References:

Gerald W. Koeppl, Dalius S. Sagatys, G. S. Krishnamurthy, Sidney I. Miller, “Inversion barriers of pyramidal ($\ce{XY3}$) and related planar (($\ce{=XY}$) species,” J*. Am. Chem. Soc.* 1967, 89(14), 3396-3405 (https://doi.org/10.1021/ja00990a004).
Ralph E. Weston Jr., “Vibrational Energy Level Splitting and Optical Isomerism in Pyramidal Molecules of the Type $\ce{XY3}$,” J. Am. Chem. Soc. 1954, 76(10), 2645-2648 (https://doi.org/10.1021/ja01639a012).
Albert T. Bottini, John D. Roberts, “The Nitrogen Inverson Frequency in Cyclic Imines,” J. Am. Chem. Soc. 1956, 78(19), 5126-5126 (https://doi.org/10.1021/ja01600a083).
H. S. Gutowsky, “Chemical Shifts and Electron-Coupled Spin-Spin Interactions,” Ann. N.Y. Acad. Sci. 1958, 70(4), 786-805 (https://doi.org/10.1111/j.1749-6632.1958.tb35431.x).
J. P. Heeschen, Ph.D. Dissertation, University of Illinois, ILL, 1959.
A. Loewenstein, John F. Neumer, John D. Roberts, “The Activation Energy of Inversion in Substituted 1-Methylaziridines (N-Methylethylenimines) Measured by the Nuclear Magnetic Resonance Technique,” J. Am. Chem. Soc. 1960, 82(14), 3599-3601 (https://doi.org/10.1021/ja01499a030).
Martin Saunders, Fukiko Yamada, “Measurement of an Amine Inversion Rate Using Nuclear Magnetic Resonance,” J. Am. Chem. Soc. 1963, 85(12), 1882-1882 (https://doi.org/10.1021/ja00895a049).
Michael J. S. Dewar, W. Brian Jennings, “Barrier to pyramidal inversion of nitrogen in dibenzylmethylamine,” J. Am. Chem. Soc. 1971, 93(2), 401-403 (https://doi.org/10.1021/ja00731a016).


text:
Cis-Trans Isomers Heat of Combustion

Consider the 4 cis-trans isomers for the above compound. In what way
  would the cis-trans isomers affect their heat of combustion.

I think that if the substituents are on alternating sides of the ring, then there will be minimum steric crowding. But is there anything else that would affect their heat of combustions?
UPDATE: Here is as far as I have gotten. Now I think for the methyl group on carbon 4, either conformation would be stable. I am guessing that an axial methyl group would be more stable, because the axial position is free from all types of steric repulsion (This is just a guess). Is this correct?

You need to consider the conformation of the ring.

The t-butyl group will be equatorial then draw the possible diastereomers. Read this: https://chemistry.stackexchange.com/questions/118093/low-or-high-enthalpy-of-combustion-in-relation-to-stability/118130#118130

@user55119 is it true that whenever we have a single group attached on a cyclohexane, it will take an equatorial position? And if we have more than one group, the most bulky group takes the equatorial position

No. For a single group on cyclohexane it is an equilibrium as a function of its effective size. The larger the group, the more equatorial. t-Butyl is essentially always equatorial. This might help: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS i/Avalues.htm

@user55119 OK, so after that what do we do?

Do you realize that all 4 isomers give the same amount of CO2 and H2O on combustion? The more stable the isomer, the less the heat of combustion. Your job, not WE, is to draw all four and make a conclusion.

@user55119 I drew all four of them, but I still don't know what to do with one of the methyl groups. See my update in my question.

Is there a penalty for an axial methyl group? Are there any gauche t-butyl/methyl interactions? Make an effort to solve this qualitatively and amend your post before I would provide an answer.

Let us continue this discussion in chat.

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text:
Laplacian of the Hartree Fock second-order reduced density matrix
I am attempting to replicate and understand the Colle-Salvetti method for approximating the electron correlation energy.[1] I have successfully replicated the method, but there is one part of the derivation that I cannot correctly derive, the Laplacian of the Hartree–Fock, second-order reduced density matrix.
With some trial and error I was able to modify my derived form for the Laplacian operator to then exactly match the correlation energies calculated by Colle and Salvetti for the helium atom. My derivation of the following term is as follows
$$
\nabla^2_\mathbf{r} P_{2 \text{HF}}\left(\mathbf{R} - \frac{\mathbf{r}}{2}, \mathbf{R} + \frac{\mathbf{r}}{2}\right)_{r=0},
\tag1
$$
where $\mathbf{R} = \frac{1}{2}(\mathbf{r}_1 + \mathbf{r}_2)$ and $\mathbf{r}=|\mathbf{r}_1 - \mathbf{r}_2|$. I found a useful article written by Colle and Salvetti in 1983,[2] which states a form for $\nabla^2_\mathbf{r}$ in terms of the inter-particle coordinates $\mathbf{r_1}$ and $\mathbf{r_2}$
$$
\nabla_\mathbf{r}^2 = \frac{1}{4}\left(\nabla_{\mathbf{r_1}}^2 + \nabla_{\mathbf{\mathbf{r_2}}}^2 - 2 \nabla_{\mathbf{r_1}} \cdot \nabla_{\mathbf{\mathbf{r_2}}}\right).
\tag2\label{eq:two}
$$
This form is useful as the Hartree Fock wavefunction for helium I am using is in terms of $\mathbf{r_1}$ and $\mathbf{r_2}$. I apply the chain rule to the derivatives in terms of $\mathbf{r_1}$ and $\mathbf{r_2}$
\begin{aligned}
  \frac{\partial}{\partial \mathbf{r_1}}
    &= \frac{\partial \mathbf{r}}{\partial \mathbf{r_1}}
       \frac{\partial }{\partial \mathbf{r}}
     + \frac{\partial \mathbf{R}}{\partial \mathbf{r_1}}
       \frac{\partial }{\partial \mathbf{R}} \\
    &= \frac{\partial}{\partial \mathbf{r}}
     + \frac{1}{2}\frac{\partial}{\partial \mathbf{R}},
\end{aligned}
and
\begin{aligned}
  \frac{\partial}{\partial \mathbf{r_2}}
    &= \frac{\partial \mathbf{r}}{\partial \mathbf{r_2}}
       \frac{\partial }{\partial \mathbf{r}}
     + \frac{\partial \mathbf{R}}{\partial \mathbf{r_2}}
       \frac{\partial }{\partial \mathbf{R}} \\
    &= -\frac{\partial}{\partial \mathbf{r}}
      + \frac{1}{2}\frac{\partial}{\partial \mathbf{R}}.
\end{aligned}
The second order derivatives take the form
\begin{aligned}
  \frac{\partial^2}{\partial \mathbf{r_1}^2}
    &= \left(\frac{\partial}{\partial \mathbf{r}}
     + \frac{1}{2}\frac{\partial}{\partial \mathbf{R}}\right)
       \left(\frac{\partial}{\partial \mathbf{r}}
       + \frac{1}{2}\frac{\partial}{\partial \mathbf{R}}\right) \\
    &= \frac{\partial^2}{\partial \mathbf{r}^2}
     + \frac{\partial^2}{\partial \mathbf{r} \partial \mathbf{R}}
     + \frac{1}{4}\frac{\partial^2}{\partial \mathbf{R}^2},
\end{aligned}
and
\begin{aligned}
  \frac{\partial^2}{\partial \mathbf{r_2}^2}
    &= \left(-\frac{\partial}{\partial \mathbf{r}}
       + \frac{1}{2}\frac{\partial}{\partial \mathbf{R}}\right)
       \left(-\frac{\partial}{\partial \mathbf{r}}
       + \frac{1}{2}\frac{\partial}{\partial \mathbf{R}}\right) \\
    &= \frac{\partial^2}{\partial \mathbf{r}^2}
     - \frac{\partial^2}{\partial \mathbf{r} \partial \mathbf{R}}
     + \frac{1}{4}\frac{\partial^2}{\partial \mathbf{R}^2}.
\end{aligned}
The Laplacian is evaluated when the inter-electronic distance, $\mathbf{r}$ is $0$, thus all these derivatives simplify
$$
\frac{\partial}{\partial \mathbf{r_1}}
  = \frac{\partial}{\partial \mathbf{r_2}}
  = \frac{1}{2}\frac{\partial}{\partial \mathbf{R}},
$$
and
$$
\frac{\partial^2}{\partial \mathbf{r_1}^2}
  = \frac{\partial^2}{\partial \mathbf{r_2}^2}
  = \frac{1}{4}\frac{\partial^2}{\partial \mathbf{R}^2},
$$
Substituting these into equation \eqref{eq:two} gives
\begin{aligned}
  \nabla_\mathbf{r}^2
    &= \frac{1}{4}\left[
         \frac{1}{4}\frac{\partial^2}{\partial \mathbf{R}^2}
       + \frac{1}{4}\frac{\partial^2}{\partial \mathbf{R}^2}
       - 2\left(\frac{1}{2}\frac{\partial}{\partial \mathbf{R}}\right)
         \cdot \left(
           \frac{1}{2}\frac{\partial}{\partial \mathbf{R}}
         \right)
       \right] \\
    &= \frac{1}{4}\left[
         \frac{1}{2}\frac{\partial^2}{\partial \mathbf{R}^2}
       - \frac{1}{2}\left(
           \frac{\partial}{\partial \mathbf{R}}
         \right)^2
       \right]
\end{aligned}
Now act this operator on the two-electron reduced density matrix
$$
\nabla^2_\mathbf{r} P_{2 \text{HF}}\left(\mathbf{R}
  - \frac{\mathbf{r}}{2}, \mathbf{R} + \frac{\mathbf{r}}{2}\right)_{r=0}
= \frac{1}{4}\left[
    \frac{1}{2}\frac{\partial^2}{\partial \mathbf{R}^2}
  - \frac{1}{2}\left(
      \frac{\partial}{\partial \mathbf{R}}
    \right)^2
  \right]
  \rho_2^\text{HF} (\mathbf{r_1}, \mathbf{r_2})\rvert_{
    \mathbf{r_1} = \mathbf{r_2} = \mathbf{R}
  },
$$
where I assume the two-electron density is a product of the two, one-electron densities
$$
\rho_2^\text{HF} (\mathbf{r_1}, \mathbf{r_2})\rvert_{\mathbf{r_1}
= \mathbf{r_2}
= \mathbf{R}}
= \frac{1}{2}\rho(\mathbf{r_1})\rho(\mathbf{r_2})\rvert_{\mathbf{r_1}
= \mathbf{r_2}
= \mathbf{R}}
= \frac{1}{2}\rho(\mathbf{R})\rho(\mathbf{R})
$$
We now have
$$
\nabla^2_\mathbf{r} P_{2 \text{HF}}\left(\mathbf{R}, \mathbf{R}\right)
= \frac{1}{8}\left[\frac{1}{2}
    \frac{\partial^2 \rho(\mathbf{R})\rho(\mathbf{R})}{\partial \mathbf{R}^2}
  - \frac{1}{2}\left(\frac{\partial \rho(\mathbf{R})\rho(\mathbf{R})}{\partial \mathbf{R}}\right)^2
  \right]
$$
Apply the product rule and factor out $\rho(\mathbf{R})$
$$
\nabla^2_\mathbf{r} P_{2 \text{HF}}\left(\mathbf{R}, \mathbf{R}\right)
= \frac{1}{8}\rho(\mathbf{R})\left[
    \frac{\partial^2 \rho(\mathbf{R})}{\partial \mathbf{R}^2}
  + \frac{1}{\rho(\mathbf{R})}\left(
      \frac{\partial \rho(\mathbf{R})}{\partial \mathbf{R}}
    \right)^2
  - 2\rho(\mathbf{R})\left(
      \frac{\partial \rho(\mathbf{R})}{\partial \mathbf{R}}
    \right)^2
  \right]
$$
This gives the incorrect answer. However by dividing the last term, $2\rho(\mathbf{R})\left( \frac{\partial \rho(\mathbf{R})}{\partial \mathbf{R}}\right)^2$ by $\rho(\mathbf{R})^2$ allows for Colle and Salvetti's correlation energy to be replicated exactly; and you can combine the $\left( \frac{\partial \rho(\mathbf{R})}{\partial \mathbf{R}}\right)^2$ terms. As this example is for the helium atom there is a spherical symmetry to exploit so the vector notation is not strictly required.
I cannot see what I have done wrong in this derivation. It is very close to being correct with a seemingly rogue $\rho(\mathbf{R})^2$ in the final term. Any assistance is greatly appreciated in finding what I have done wrong.
References:

Colle, R.; Salvetti, O. Approximate calculation of the correlation energy for the closed shells. Theor. Chim. Acta 1975, 37 (4), 329–334. DOI: 10.1007/BF01028401.
Colle, R.; Salvetti, O. A general method for approximating the electronic correlation energy in molecules and solids. J. Chem. Phys. 1983, 79 (3), 1404–1407. DOI: 10.1063/1.445899.

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text:
Where to put the number in organic nomenclature?
Should one write 4-hydroxypentan-2-one or 4-hydroxy-2-pentanone?
It has been seen that both have been used frequently.

Check PubChem. As I understand it that site uses a computer program LexiChem for naming which is setup to follow all the current rules.

The IUPAC recommendations on this matter have been changed.
According to the 1979 recommendations, the correct name was still 4-hydroxy-2-pentanone.
According to the 1993 recommendations, however, locants were placed immediately before the part of the name to which they relate. This rule is still included in the 2013 version. Therefore, the correct name is now 4-hydroxypentan-2-one.


text:
How to convert benzene to N,N-dimethylbenzamide?

This is my solution to convert benzene to N,N-dimethylbenzamide. If you think is the right way, even just saying the equations are correct would help me! Thanks in advance.

What is exactly that you're unsure of?

How many steps are you proposing doing this in?

I'm unsure of which inorganic reagent that can make this reaction happen @Mithoron

And you know what? At least "organic" part? Unless you do know something and edit it in, the question is gonna get closed.

This can be done in one step, but it is pretty sophisicated chemistry and probably not what your teacher is thinking of. It is more easily done in 2 steps which I'm happy to tell you about, but you need to make more of an effort first.

I would recommend changing the Friedel-Crafts alkylation into a Vilsmeier-Haack formylation. Because the alkyl groups are activating the ring more, this will lead to poly-substitution. If, on the other hand, you use a Vilsmeier-Haack formylation, followed by oxidation of the resulting formyl group, you should be able to avoid the issue of poly-alkylation (If you are curious, a Vilsmier-Haack formylation uses N,N-dimethylformamide and phosphoryl chloride to form a reactive complex that will formylate your benzene ring)

@Eli Jones: The Vilsmeier-Haack formylation requires activated benzene rings (anilines, phenols, etc). Benzene probably doesn't have enough nucleophilicity.

@user55119: Yes, I ran a SciFinder search and could not find any examples of this being used on benzene. You are probably right. Thanks for pointing this out!

The scheme you have written will satisfy your teacher in my opinion but it can be shortened. The suggestion of @Eli Jones of the V-H formylation to give benzaldehyde is a good one. However, it can be done in 2 steps using reasonably well-known chemistry:
Step 1 - Brominate with $\ce{FeBr3/Br2}$ to give bromobenzene
Step 2 - Either lithiate by Li-Halogen exchange using $\ce{n-BuLi}$, or form the Grignard (directly with Mg or by Knochel exchange with $\ce{i-PrMgBr}$) then react the resulting benzene organometallic with dimethylcarbamoyl chloride ($\ce{(Me)2NCOCl}$) to give the benzamide.
There are a couple of ways of doing this in one step from benzene by the use of biscarbamoyl diselenides/Lewis acids$\ce{^{[1]}}$ and by $\ce{(Me)3SiOTf}$ activated carbamoyl chloride$\ce{^{[2]}}$.
References

Biscarbamoyl diselenides as new carbamoylating reagents. Lewis acid promoted carbamoylation of aromatic compounds by Shin-Ichi Fujiwara, Akiya Ogawa, Nobuaki Kambe, Ilhyong Ryu, Noboru Sonoda, Volume 29, Issue 47, 1988, Pages 6121-6124, DOI: https://doi.org/10.1016/S0040-4039(00)82282-0
Jin-Wei Yuan, Qian Chen, Chuang Li, Jun-Liang Zhu, Liang-Ru Yang, Shou-Ren Zhang, Pu Mao, Yong-Mei Xiao and Ling-Bo Qu, Silver-catalyzed direct C–H oxidative carbamoylation of quinolines with oxamic acids, Organic & Biomolecular Chemistry, 10.1039/D0OB00358A, (2020).

Thank you for this answer! I have to perform an acylation in my research and I have the issue of regioisomer formation (which makes purification very difficult); however, I have not considered performing a lithiation followed by treatment with an acyl chloride. This would definitely solve the regioisomer problem. It is Interesting how it it hard to consider different synthetic routes when you are so set on using a particular reaction!

Aryl lithiums and Grignards do not react well with acyl chlorides. You'll get a higher yield if you transmetallate to the arylmanganese with MnCl2  https://onlinelibrary.wiley.com/doi/abs/10.1002/adsc.201400654

If you have to use an acylation in your scheme, form acetophenone from benzene via an FC reaction. Then do a haloform reaction to get benzoic acid and continue as you planned.


text:
Boiling point of oxygen
I am studying chemistry from the book titled "Fundamental Chemistry for Cambridge O Level" written by Rosemarie Gallagher and Paul Ingram.
On page 17, it states that

oxygen boils at −219 °C and freezes at −183 °C.

Isn't this incorrect? Should it not be the other way round?

Yeah that looks like they have it mixed around. The boiling point will always be higher than the melting point measured at the same pressure. Wikipedia gives the same values, but in the correct order.

BTW, write (or have your instructor write) to the publisher, citing the error. You might get a "freebie" (e.g. a copy of the next edition, with corrections) and the satisfaction of being acknowledged, as well as performing a service for those who would otherwise have to "unlearn" the error.

I tried to find some sort of errata for the book and couldn't. According to Amazon the paperback  was published January 1, 2018 so it isn't an old book.

According to NIST Chemistry Webbook, for oxygen:
$$T_\mathrm{boil} = \pu{90.2\pm0.2 K}$$
$$T_\mathrm{fus} = \pu{54.8\pm0.2 K}$$
$$T_\mathrm{triple} = \pu{54.33\pm0.6 K}$$
The $T_\mathrm{boil}$ and $T_\mathrm{fus}$ are from Ref.1 while Brower and Thodos reported $T_\mathrm{triple}$ as $\pu{54.363 K}$ at pressure of $\pu{1.125 mmHg}$ (Ref.2). The uncertainties were given by Thermodynamics Research Center, NIST Boulder Laboratories, CO.
Accordingly, $T_\mathrm{boil} = \pu{-183 ^\circ C}$, $T_\mathrm{fus} = \pu{-218.4 ^\circ C}$, and $T_\mathrm{triple} = \pu{-218.8 ^\circ C}$. Therefore, as you correctly suspected, the values in the textbook is a misprint.
References:

Alex G. Streng, “Miscibility and compatibility of some liquefied and solidified gases at low temperatures,” J. Chem. Eng. Data 1971, 16(3), 357-359 (https://doi.org/10.1021/je60050a024).
Gerald T. Brower, George. Thodos, “Vapor pressures of liquid oxygen between the triple point and critical point,” J. Chem. Eng. Data 1968, 13(2), 262-264 (https://doi.org/10.1021/je60037a038).


text:
is H2O in gas form in poly styrene combustion?
original question:
if we burn 390g of poly styrene (STP), how much gas produces?
me: 1008L
my teacher: 672L
he says h2o is not in gas form!(because of STP)
who is wrong?

Neither of you is correct. The problem is indeterminate assuming that after the reaction the system was brought back to STP because the initial volume of the gas was not specified. If say initially a million liters of oxygen was used then all the water would be present as the vapor. However if a stoichiometric amount of oxygen was used then most, but not all, of the water would be in liquid form

You can't burn polystyrene at STP. You didn't state the entire problem so the final conditions are a bit of a mystery too. But it would seem that the system (products after combustion) was brought back to STP after the reaction.

@MaxW why we can't burn?

You can't burn polystyrene at STP. i.e. constantly at $\pu{O ^\circ C}$. When you burn polystyrene it gets hot. So you then have to cool the products down.

@MaxW so we couldn't burn things while we are near sea and temp is O°C?

Find a walk in freezer. Strike a match and hold it in your fingers till it burns completely. Did the burning match stay at $\pu{0 ^\circ C}$?

Neither of you is correct. The problem is indeterminate assuming that after the reaction the system was brought back to STP because the initial volume of the gas was not specified. If say initially a million liters of oxygen was used then all the water would be present as the vapor. However if a stoichiometric amount of oxygen was used then most, but not all, of the water would be in liquid form
$$\ce{\pu{M.W. }\ C8H8 = \pu{104.15 g/mol}}\tag{1}$$
$$\ce{\pu{mol }\ C8H8 =\dfrac{\pu{390 g}}{\pu{104.15 g/mol}}} = \pu{3.7446 mol} \tag{2}$$
$$\ce{C8H8 ->[O2] 8CO2 + 4H2O}\tag{3}$$
1.000 mole of a gas at STP ( $\pu{0 °C}$ and $\pu{100 kPa}$) has a volume of $\pu{22.711 L/mol}$.
Let's suppose that a huge excess of oxygen gas was used for the combustion. After the combustion the system is brought back to STP, and all the water is present as vapor. So the upper limit of the gaseous products is:
$$V_\mathrm{gas, UL} =3.7446 \times 12\times 22.711 = \pu{1020.52 L} \ce{->[round to 3] \pu{1.20\times 10^3 L}}\tag{4}$$
Now suppose that a stoichometric amount of oxygen gas was used for the combustion. After the combustion the system is brought back to STP, and some of the water is present as vapor, but most is present as the liquid. The vapor pressure of water at $\pu{0 °C}$ is $\pu{0.6113 kPa}$. So the lower limit of the gaseous products is:
\begin{align}
V_\mathrm{gas, LL} &= (3.7446 \times 8\times 22.711) \times \left(\dfrac{\pu{100 kPa}}{\pu{100 kPa} - \pu{0.6113 kPa}}\right) \\
&= (680.35)\times (1.00615) = \pu{684.53 L} \ce{->[round to 3]} \pu{685 L}\\
\end{align}
Old definition of STP
Before 1983 the definition of STP was $\pu{0 ^\circ C}$ and $\pu{101.325 kPa}$ for which 1.000 mole of a gas has a volume of 22.414 L/mol. Using the old value:
$$V_\mathrm{gas, UL} = \dfrac{22.414}{22.711}\times \pu{1020.52 L}= \pu{1007 L}$$
$$V_\mathrm{gas, LL} = \dfrac{22.414}{22.711}\times \pu{684.53 L}= \pu{676 L}$$

The OP finally made me realize that the above answer for $V_\mathrm{gas, LL}$ was not correct for the old STP.
The problem lies with the fact that the pressure correction for the vapor pressure of water involves subtraction and not just multiplication. As the OP points out:
$$\dfrac{\pu{100 kPa}}{\pu{100 kPa} - \pu{0.6113 kPa}} \ne \dfrac{\pu{101.325 kPa}}{\pu{101.325 kPa} - \pu{0.6113 kPa}}$$
So my calculation yielded
$$V_\mathrm{gas, LL} = \dfrac{22.414}{22.711}\times \pu{684.53 L}= 675.58 \ce{->[Round(3)]}\pu{676 L}$$
But the correct calculation is:
\begin{align}
V_\mathrm{gas, LL} &= (3.7446 \times 8\times 22.414) \times \left(\dfrac{\pu{101.325 kPa}}{\pu{101.325 kPa} - \pu{0.6113 kPa}}\right) \\
&= (671.45)\times (1.00607) = \pu{675.52 L} \ce{->[round to 3]} \pu{676 L}\\
\end{align}

could you explain about 100kPa/(100kPa - 0.6113kPa) ?

@aStudent - The current specification for STP defines the pressure as 100 kPa. If a stoichiometric amount of oxygen was used then the gas would contain $\ce{CO2}$ and $\ce{H2O}$. at $0 ^\circ \mathrm{C}$ the vapor pressure of water alone is  0.6113kPa. The vapor pressure of $\ce{CO2}$ alone would be $\pu{100kPa - 0.6113kPa}$.

oh excuse me! i made a small mistake. i will delete my comment and i will tell you after checking.

Let's get back to basics. Was the volume of the products measured at STP, ($\pu{0 ^\circ C}$ and  $\ce{100 kPa}$) or not? If not what T and P?

now I understood. thank you a lot! PS: in Iran, at least in schools we are using old definition of STP. (you should also consider this at the end. I will edit it)

I DID consider an older value of STP at the end.

you did this at original formula -> *100kPa/(100kPa - 0.6113kPa) so at the end you should also do this: * ((100kPa - 0.6113kPa)/100kPa) * (101.325kPa/(101.325kPa - 0.6113kPa)

Did I get the wrong answer? The point is to do simple calculations not something complicated that is hard to follow.

you did this : (3.7446×8×22.711)×(100 kPa/(100 kPa−0.6113 kPa)) in old STP we have 22.414 L/mol and 101.325 kPa you only considered 22.414 L/mol.

I don't know why I did the calc that way. It just seemed the right way at the time. Note that it is just a pressure correction: $$\dfrac{100}{101.325} = \dfrac{22.414}{22.711}$$

we should do three corrections : (3.7446×8× 22.414 )×( 100 kPa /( 100 kPa −0.6113 kPa)) .(we should also change pressure of CO2 and total gases) i.e. : ( 100 kPa /( 100 kPa −0.6113 kPa)) does not exactly equal to ( 101.325 kPa /( 101.325 kPa −0.6113 kPa))

Please post your own answer. I'll give you my opinion if it is right or wrong.

LOL - My high school chemistry class was in 1969. I automatically think of 22.414 liters as being the molar volume at STP. I have to remind myself that the molar volume is now 22.711 at STP. I also think in torr, not kPa. My brain is also wired for calories for heat. There are certain numbers that you will just remember after working a lot of problems.

:) look at this picture. if you correct red ones, final answer decreases by 0.05.

I finally understand your point. You are correct.

we didn't have any pressure correction for the vapor pressure of water! (in your last edit : "... that the pressure correction for the vapor pressure of water involves ..." )


text:
Which is more effective at raising pH, Tums or baking soda?
Chemical buffers like Tums (calcium carbonate) and bases like baking soda (sodium bicarbonate) are both used medically to reduce acidity in the stomach to ease heartburn symptoms. I'm curious which is more effective in vitro.
I'm interested specifically in the comparison in effectiveness of bases and buffers, which act in slightly different ways to raise pH. I'd also love to learn the math behind figuring this out so next time I can answer such a question myself :)

Useful terms to search are Henderson-Hasselbalch equations in buffers. Also note that Na and Ca are spectator ions, but they have different charges. Which one will be more effect in terms of mass for neutralizing 1 mol of HCl ?

@M.Farooq All else being equal, baking soda has a molar mass 16% lower than that of calcium carbonate making baking soda naively 16% more effective by mass, but the "all else being equal" assumption is what I'm really trying to correct with this question. Neither reaction will proceed 100% to completion, so I'm trying to figure out which does a better job with the HCl

The completeness of the reaction will depend on the limiting reactants. If HCl is in excess, the reaction will proceed to completion. However if you don't have enough, a buffer system will be formed. First of all, fix the amounts of baking soda and calcium carbonate. It is up to you. How many Tums are recommended in a single dose. What is their mass? Find the concentration of HCl in stomach acid and also find how many moles of HCl you are supposed to neutralize.

@M.Farooq How is this math done? What numbers am I looking for and how are they used? Also, this is starting to sound a lot like an answer :)

Did you search the equation I mentioned above?

@M.Farooq Ah that's great. So all you need is the dissociation constant and the ratio of acid to conjugate base concentration?


text:
Can you call H_2 an Atom?

I am currently studying to complete the Chemistry CLEP exam, I ran into this question and was thoroughly confused by the phrase

12.04x10^23 atoms of H_2

According to the question, this corresponds to 12.04X10^23 molecules (2 moles) of H_2, or 24.08x10^23 atoms (4 moles) of Hydrogen. The use of the word "atom" over "molecule" seems useless, confusing, and possibly wrong.
I'm not sure if this question is written poorly or if I have made a mistake in my grammar either way, any help is appreciated. Thank you!

No you cannot call them an atom, you are right to be confused. The most natural interpretation is a mistaken attempt to say x molecules of H2.

There is no doubt that the statement should have been $\pu{12.04 \times 10^{23}}$ molecules of $\ce{H_2}$ given that the given answer is $\pu{(E) 44.8 L}$. // PS - Since 1983 a mole of gas at STP occupies 22.711 Liters.

An online practice test which contained this question had more, um, questionable questions. E.g. asking which of Na, Zn, Hg, N, He "exists as a liquid in its elemental form". Well, they all do. The question fails to mention "STP". The singular "exists in its" is a giveaway, but we are taking a chemistry test, not linguistics.

The question is indeed poorly worded. What the question seems to be saying is that there are $12.04\times10^{23}$ molecules of hydrogen gas/dihydrogen ($\ce{H2}$).
I'm also noting that the correct answer ends with "... $\pu{44.8L}$ of $\ce{H2O}$ present", while there was absolutely no mention of $\ce{H2O}$. The question is quite misleading.

The question and answer are not just misleading, they are abysmally bad. ;-)

It's badly proof-read, or probably not at all. Curious about the other questions.

C'mon @Raphaël. You've been around long enough to know basic mhchem formatting! Please try to use $\LaTeX$ appropriately as much as possible.


text:
Naming terminal monosubstituted alkenes conformations
Alkenes can be described by E/Z nomenclature and cis/trans nomenclature.
However, is there a way to name alkene conformations which are monosubstituted? For example, how could one describe the difference between the following conformations?
Any suggestions are welcome, even if they aren't IUPAC accepted -- I would just like to hear what the community thinks would be a good way to differentiate these type of structures.

Huh, you mean like s-cis / s-trans?

Awesome thanks! It slipped my mind. Should I delete the question?

Only if you want, perhaps there more to be added ,or this can work as sort of redirect. Duplicates are often useful.


text:
Deriving the equation for reaction spontaneity
I'm trying to make sense of the derivation of the equation $$\Delta G = \Delta G^0_r + RT\ln(Q)$$
While searching for a derivation of this equation that made sense to me I came across a publication in the Journal of Chemical Education that suggests it is incorrectly applied if not completely wrong:
"Following this development, the paper investigates the source of the mathematical
and logical errors contained in textbooks ca. 1950 to the present that led to the
fallacious statements still present in most introductory chemistry textbooks and some
more advanced texts that the conditions for spontaneity are ΔG < 0 at constant T and
p and ΔA < 0 at constant T and V, whereas the corresponding conditions for
equilibrium are ΔG = 0 or ΔA = 0. This investigation shows the principal errors to be
(i) incorrect evaluation of definite integrals; (ii) failure to determine whether the
results of such integrations produce criteria for spontaneity and equilibrium that are
necessary conditions, sufficient conditions, both, or neither; and (iii) incorrect logical arguments related to equilibrium. (Spontaneity and Equilibrium III: A History of Misinformation)"
https://pubs.acs.org/doi/full/10.1021/ed500253e
Also, by the same author: https://pubs.acs.org/doi/10.1021/ed400453s
Here is the standard derivation:
For a given chemical reaction, let $v_i$ be the signed stoichiometric coefficient of reagent $i$ (That is, $v_i$ is negative for reactants and positive for products, with $|v_i| = $ the stoichiometric coefficient of reagent $i$).
Define the extent of the reaction, $\xi,$ as: $$\xi := \frac{n_i - n_{i,o}}{v_i} =  \frac{\Delta n_i}{v_i}$$
where $n_i$ and $n_{i,o}$ are the instantaneous number of moles and initial number of moles, (of reagent $i$) respectively. This quantity will have the same value regardless of which reagent is chosen to compute it and is therefore well-defined. It is effectively a measure of the reaction coordinate.
Note that: $$n_i = n_{i,o} + v_i\xi \to dn_i = v_i d \xi $$
$ \\ $
$$dG = Vdp - SdT + \sum_i \mu_i dn_i \to \left(dG\right)_{P,T} = \sum_i \mu_i dn_i = \sum_i \mu_i v_i d \xi$$
In particular, $$\left( dG/d \xi \right)_{P,T} = \sum_i \mu_iv_i = \sum_i \left(\mu^o_i + RT\ln(a_i)\right)v_i = \sum_i \mu^o_iv_i + RT \sum_i \ln(a_i)v_i = \Delta G^o_r + RT \sum_i \ln(a_i^{v_i}) = \Delta G^o_r + RT \ln\left(\prod_i a_i^{v_i}\right) = \Delta G^o_r + RT \ln(Q)$$
In particular, $$\left( dG/d \xi \right)_{P,T} = \Delta G^o_r + RT \ln(Q)$$
I assert this is where it should stop. It tells you everything you need to know about the direction of spontaneity and the condition for equilibrium. For instance, the condition for equilibrium is that the free energy of the system is at a (local) minimum. To find the minimum of a differentiable function, you set its derivative equal to zero. Thus at equilibrium you have $Q = K$ and $dG/d \xi = 0$. Plugging this in gives $$K = e^{- \Delta G^o_r / RT}$$What I don't understand is how the left side gets equated with $\Delta G$. The left side is $\sum_i \mu_i v_i$. This is not the "instantaneous difference in free energy between the reactants and products". That would be $\sum_i \mu_i n_i$ (assuming such a notion even makes physical sense). I suppose you could call it $\Delta G_r$, but that perspective is confusing and would have no physical significance in the context of this equation (as you are viewing the reagents' concentrations as the variable that is moving as opposed to pressure or temperature).
I'm wondering if I am conceptually missing something?

What are the dimensions of $\Delta G$ and $\Delta_r G^\circ$ in your question?

Related: https://chemistry.stackexchange.com/questions/39641/difference-between-chemical-potential-and-gibbs-free-energy

Note that it is discouraged to use LaTeX or MathJax formatting of question titles in Chemistry SE site for searching reasons.

@Poutnik Ok. Illl edit

You do realize that there are two somewhat different derivations of the relationships you are talking about, right?

@ChetMiller I came across a couple when looking for one I liked. If you have a link I'll check them out.

The method you described for deriving the desired relationship is one of two that I am aware of.  In my judgment, it is important to mention that this approach looks at a reaction in progress,  determining the free energy of the mixture of reactants and products as a function of the reaction conversion.  As such, it makes the tacit assumption that it is valid for this irreversible process of a reaction occurring spontaneously to determine the free energy at each conversion as if one is considering a thermodynamic equilibrium state of an unreacting mixture.  In my judgment, this is an excellent approximation, but some purists might argue that such an approach is not valid.
The other approach I am aware of is illustrated for the case of a reaction involving an ideal gas reaction.  It evaluates the change in Gibbs free energy between the following pair of initial and final thermodynamic equilibrium states:

Stoichiometric proportions of pure reactants in separate  containers at temperature T and 1 bar pressure
Corresponding stoichiometric proportions of pure products in separate containers at temperature T and 1 bar pressure

The process for evaluating the change from state 1 to state 2 is reversible, and involves successive steps involving isothermal gas expansions and compressions within cylinders in series with a van't Hopf equilibrium box.  The net result is the standard free energy change of the reaction $\Delta G^0$.
If one then modifies the process slightly, by starting with pure gases in separate containers at pressures different from 1 bar, this merely changes the compression and expansion steps in the reversible process, and results in the value of the Gibbs free energy change altering from $\Delta G^0$ to $$\Delta G=\Delta G^0+RT\ln{Q}$$Moreover, if the initial and final pressures in the separate containers are chosen to exactly match the equilibrium partial pressure of the various species in the equilibrium box, then the compression and expansion steps are not necessary, and the $\Delta G$ for this process is zero.  Under these circumstances the reaction parameter Q is equal to the equilibrium constant:  $$0=\Delta G^0+RT\ln{K}$$

As you've already outlined in your derivation: the change in Gibbs Free Energy ($\Delta G$) at constant T & P, is the sum of changes in the chemical potentials ($\mu_i$) of a given reaction system.
$$dG_{T,P} = \sum_{i=1}^n\mu_idn_i$$
If you look at the definition of $\mu$, you will see that this is just the partial derivative of Gibbs Free Energy with respect to a component $n_i$, such that,
$$\mu_i = (\frac{\partial G}{\partial n_i})_{T,P}$$
Part of the derivation you've done is to look at changes in Gibbs Free Energy ($\Delta G$) assuming changes in the composition of a reaction system. When you substitute in the equation for the change in a chemical species with respect to conversion or extent of reaction ($\xi$), and you then simplify the expression to,
$$(\frac{dG}{d\xi})_{T,P} = \sum_{i=1}^n\mu_i\nu_i$$
For the righter-most expression, you are essentially looking at the overall change in the mixture's inherent chemical potential (at a given T & P), for a given unit of converted reagent.

@DavidReed was this comment meant for my take on your question? The units of dG/dxi are in fact in units of kJ/mol—I agree with what you’re saying about these expressions being normalized to 1mol... that is the purpose of having a differential for conversion. You’ve shown this in your derivation that the change in mols is proportional to the stoichiometric coefficient and the differential change in conversion.

@DavidReed for the reaction system you've shown above, $\ce{aA + bB -> cC + dD}$, yes the heat of reaction $Q_{rxn}$ would be in units of kJ, however this is because the extent of reaction, $\xi$ (on a 1 mol basis), like you've stated, is assumed to be 1. $$Q_{rxn}  = \xi \sum \nu_iH_{fi}^o = \sum \nu_iH_{fi}^o$$For the chemical potentials, $\sum{\nu_i\mu_i}$, this quantity will be in units of kJ/mol, you can see this through the derivation you've provided, i.e $\frac{dG}{d\xi}$ or $\frac{Energy}{mol}$. Going back to units of energy, $$d\xi \frac{dG}{d\xi} = dG = \sum{\nu_i\mu_i}d\xi$$

I'll have to chew on that. You would still have equilibrium corresponding to the value of $\xi$ that minimizes the free energy of the system.  A better expression would likely be $ \Delta \mu_r$.

nvm. We are in total agreement. I got caught up in a weird thought process. So the argument that a reaction is spontaneous but "slow" until it reaches its activation energy is misleading. It doesn't become spontaneous until it reaches its transition state, at which point it becomes spontaneous in both directions. The point of maximum energy corresponds to $dG < 0$.


text:
Unit derivation concerning Nernst equation
We know that Nernst equation is given by:
$E=E^{\circ}+\frac{RT}{nF} \ln[\ce{M^{n+}}]$.
Now, A question asks me to derive the unit of  $\frac{RT}{nF}$.
Which I think can easily be derived this way:
$\frac{RT}{nF}=\frac{E-E^{\circ}}{\ln[\ce{M^{n+}}]}$,
Now, the unit of $E-E^{\circ}$ is volts (since it is emf), and $\ln[\ce{M^{n+}}]$ is just a constant numerical value.
Hence, I think the unit of $\frac{RT}{nF}$ is Volts.
This was my approach. Can anyone tell me if it is a correct approach?

M. Farooq have given good description of how to analyse units in a equation. Therefore, your assumption of the unit of $\dfrac{RT}{nF}$ should be Volts is correct. However, your question is not to assume but to derive. What's that means is unit conversion. Let's see the units of all four variables:

$R$ is a constant: $\pu{8.314 J mol-1 K-1}$
$F$ is also a constant: $\approx \pu{96485  C mol-1}$
$T$ is a variable: $\pu{K}$
$n$ is number of total electrons in the redox reaction involved, and hence a variable but unit less.

Now lets see the unit conversion:
$$\frac{RT}{nF} = \frac{\pu{J mol-1 K-1}\times \pu{K}}{\pu{C mol-1}} = \frac{\pu{J}}{\pu{C}} $$
By definition (Ohm's law):
$$ V = \frac{\text{potential energy}}{\text{Charge}} = \frac{\pu{J}}{\pu{C}} $$
$$\therefore \:\; \frac{RT}{nF} = \frac{\pu{J}}{\pu{C}}= \pu{V} $$

You should read the article "Can One Take the Logarithm or the Sine of a Dimensioned Quantity or a Unit? Dimensional Analysis Involving Transcendental Functions" Here
The concepts mentioned there will help for life even if you don't pursue chemistry. In short, natural log factor ($\ln$) should be dimensionless. Now your equation is
$E=E^{\circ}+\frac{RT}{nF} \ln[\ce{M^{n+}}]$.
Now do a unit analysis:
The left hand side has units of volts $\pu{V}$.
$[\pu{V}] = [\pu{V}] + \frac{RT}{nF}$ times (dimensionless natural log)
So you cannot add apples to oranges. It implies that $\frac{RT}{nF}$ must have units of Volts.
Now do a proper dimensional analysis of the equation in terms of SI base units of [M], [L], [n], [T], [I] etc. and see if it is all consistent.


text:
Determining Half-lifes from a graph
So I was studying for an exam I have thursday when I came across a question regarding half-life. I had previously thought that the definition of a half-life is the time it takes for the amount of material to half in its decay process. However, this question had a y-axis of Disintegration Rate, and used the same process.

They gave an answer of 8 days for the half life of this radioactive element(so they treated it as normal)
So my question is, what exactly does the half in half life represent? A halving of rate or an actual halving of decay of the nuclei?

the y axis in the graph given is just decayconstant*N(where N is number of nuclei of radioactive element half life represents actual halving of decay of the nuclei(as muchas i know)

At any time, the number of decays per reasonable$^1$ time unit is proportional to the amount of the isotope present. (This is expressed by the first-order rate equation for a one-decay process.) Therefore, it does not matter whether one looks at the amount or the number of decays. It may be easier to measure the number of decays by e.g. a Geiger counter and prefer it for that reason.
It now matters what you understand to be the rate: the number of decays per second or the percentage of nuclei decaying at any given time. The first becomes less over time, the second one is a constant for each isotope.
$^1$ You must have a reasonable number of decays within your time unit. If the half-life is long, it makes little sense to look at millisecond slices.

It is a bit quirky, but the reasonable time unit should be much smaller than the half life, but long enough to get a decent number of counts. Radioactive decay follows the Poisson distribution. For more than a 100 counts the square of the number of counts is a very good estimator of the standard deviation. So $\sqrt{8000} = 89$ and the %RSD = 1.1%.

My explanation relates to the underlying math upon examination of the exponential decay graph.
I start with the cited statistical model given by:
$ y = \alpha \ e^{\beta t} $
Taking the natural log of both sides of the equation, we have the following equivalent equation:
$ \ln y = \ln{\alpha} + {\beta t} $
This equation has the form of a linear regression model where the rate of change in a natural log scale is a constant rate of ${\beta} $ per unit of time. In the current test question, no need for a regression, just checking some ${\Delta ln y}$ and ${\Delta t}$ appears to suggest a constant slope of ${(ln(4,000)-ln(2,000))/(24-16)}$, which equals ${ln(2)/8}$ as the value for the slope ${\beta} $.
Next, we have to know for the Exponential distribution, an expression for the half-life relating to ${\beta} $, namely, ${T = \ln (2)/\beta} $ (see the derivation at this reference). So, the respective half-life per the chart is 8 days.
As such, to answer the question, "what exactly does the half in half life represent", mathematically due to the employment of a natural log transform, its represents a ${\ln (2)}$ change in natural log of decay between two points, which when scaled by the rate of decay (represented by ${\beta}$), produces the half-life, as was demonstrated in the cited reference.

While another answer nicely explains in words what follows, and as an additional bonus points out some subtleties related to the condition of statistical sampling, for some reason it doesn't explicitly show equations, and sometimes it's nice to just look at equations, in the same way that a picture can be worth many words, so at the risk of redundancy I post this answer.
If the amount or population of something (call it $y$) decays exponentially in time, then we can write
$$y(t)=y(0)\exp(-kt)$$
where we define $y(0)$ as the initial concentration (at time $0$).
Then from the property of the exponential
$$\frac{dy}{dt}=-ky(0)\exp(-kt)$$
It follows, defining the decay rate as $\rho=|\frac{dy}{dt}|$, that
$$\rho(t)=ky(0)\exp(-kt)=\rho(0)\exp(-kt)$$
As explained in another answer, it follows that the decay rate also decreases exponentially and according to the same rate constant $k$. The time dependencies of the normalized functions $y(t)/y(0)$ and $\rho(t)/\rho(0)$ - described by $\exp(-kt)$ - are therefore identical. Since the half-life depends only on $k$ (as explained in another answer), it follows that both the original property and its rate of change havethe same half-life.

The others have explained the decay process of radioactive material very well. Therefore, I'm not going to elaborate the same thing again, but want to point out certain thing you seemingly do not understand clearly. In your question, you states that:

I had previously thought that the definition of a half-life is the time it takes for the amount of material to half in its decay process.

That statement is not quite to the point. The decay doesn't mean it vanish (or disappear) to air. It is not mass decay (kind of, theoretically but some mass remains, e.g., as $\ce{^{206}Pb}$, which is stable and not radioactive). The process is complicated one. For instance, see the total decay process for $\ce{^{238}_{92}U -> ^{206}_{82}Pb}$:
$$\ce{^{238}U ->[t_{1/2} = 4.4 \cdot 10^9 y] ^{234}Th ->[t_{1/2} = 24.1  d] ^{234}Pa ->[t_{1/2} = 46.69 h] ^{234}U ->[t_{1/2} = 2.455 \cdot 10^5 y] ^{230}Th \\ ->[t_{1/2} = 7.54 \cdot 10^4 y] ^{226}Ra ->[t_{1/2} = 1599 y] ^{222}Rn ->[t_{1/2} = 3.82 d] ^{218}Po ->[t_{1/2} = 3.04 min] ^{214}Pb ->[t_{1/2} = 27 min] ^{214}Bi\\ ->[t_{1/2} = 19.9 min] ^{210}Po ->[t_{1/2} = 160 \mu s] ^{206}Pb}$$
Therefore, for novices, what half-life simply means is the original radioactivity of given material to become half of its initial value (Refer to TAR86's answer). Thus, I decide to explain this process using your graph:

Radioactive decay of any active material is a spontaneous process, which follows first order kinetics:
$$\alpha = \alpha_\circ e^{-\beta t} \tag{1}$$
where $\alpha$ is activity of the material at any time $t$ and $\alpha_\circ$ is activity of the material at time you started to measure, $t=0$. The constant $\beta$ depends on several factors including decay process (e.g., $\beta$ is not same for $\ce{U}$ and $\ce{Po}$). We can simplify this as:
$$\frac{\alpha}{\alpha_\circ } = e^{-\beta t} \Rightarrow  \ln \left(\frac{\alpha}{\alpha_\circ }\right) = -\beta t \Rightarrow  \ln \alpha = \ln \alpha_\circ -\beta t \tag{2}$$
This is an equation for straight line, the slope of which is equal to $\beta$ and $y$-interception is $\ln \alpha_\circ$. By definition, $t_{1/2}$ is the time when $\alpha = \frac{1}{2} \alpha_\circ$. Applying this to equation $(2)$ gives:
$$\ln \frac{\alpha_\circ}{2} = \ln \alpha_\circ -\beta t_{1/2} \quad  \Rightarrow  \quad \therefore \; t_{1/2} = \frac{\ln 2}{\beta} \tag{3}$$
Thus, you can find $t_{1/2}$ by just getting $\beta$ from above straight line (Note that $t_{1/2}$  is independent of $\alpha_\circ$). Unfortunately, you don't have that straight line here. But still, you can find $t_{1/2}$ by analysing the given graph.
The equation of your graph is equation $(1)$. According to your graph, at $t=0$, activity has been measured as $\pu{16000 decays/min}$, which is your $\alpha_\circ$. Thus, $\frac{1}{2} \alpha_\circ$ should be $\pu{8000 decays/min}$ (see above graph). Accordingly, time taken to decay $\pu{16000 decays/min \rightarrow 8000 decays/min}$ is apparently $\pu{8 d}$. Therefore, $t_{1/2}$ is $\pu{8 d}$. If you uncertain of the value, you can check the next half-time by finding how much time it takes to decay $\pu{8000 decays/min \rightarrow 4000 decays/min}$. Not surprisingly, it is also $\pu{8 d}$ and so on (Note: If you choose $\alpha_\circ = \pu{12000 decays/min}$, you'd see time taken to decay $\pu{12000 decays/min \rightarrow 6000 decays/min}$ is still $\pu{8 d}$).
To go to an extra mile, now you can calculate the constant $\beta$ for this process. From the eqtation $(2)$:
$$\beta = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{\pu{8 d}} = \pu{0.087 d-1}$$


text:
What is the correct pH value of weak and strong alkalis?
In the book titled fundamental chemistry for Cambridge O level, it states at page 149,

Ammonia solution with a pH of 11.1 is a weak alkali and that's why it shows much lower conductivity of 0.50 US/cm.

However at p. 147 it's written that

a solution with a pH of 10 is a strong alkali.

So either the correlation of pH with conductivity is incorrect  or the statement that ammonia solution with a pH of 11.1 is a weak Alkali is wrong.   What are the pH values for a weak and strong Alkali?
Is there anything wrong with the two diametrically opposite sentences/ statements given in the book?

Proper word is base and it's strength has nothing to do with pH!

There might be confusion as below:
strong/weak "Alkali"(1) - or rather an alkaline solution, which is related to the $\mathrm{pH}$ value. The conductivity and the strength increases with $\ce{pH}$, but conductivity can be obviously affected by presence of salts.
There is not a strict threshold between strong or weak alkaline $\ce{pH}$. It is often considered relatively and/or it depends on context. As a guide the threshold can be between 10-12, context dependent.
strong/weak "Alkali"(2) - or rather a base ( as an opposite to an acid ), which is related to the base dissociation constant. IF a base has too low dissociation constant, so there is significant fraction of not dissociated base ( like ammonia ), the base is considered a weak base. There is ongoing equilibrium $\ce{ BOH <=> B+ + OH-}$, resp. $\ce{ B + H2O <=> BH+ + OH-}$

E.g.
$\ce{NaOH}$ solution with $\mathrm{pH} = 10$ is weak "alkali"(1), but strong "alkali"(2).
$\ce{NH3}$ solution with $\mathrm{pH} = 11$ is stronger "alkali"(1) then above, but weak "alkali"(2).
Note: For historical reasons, "alkali" is reserved to hydroxides of alkalic metals, so it should not be used for ammonia. As the general term, a base is preferred, with alkalic/alkaline ( along with basic ) as adjective being accepted.

In the 19th century, ammonia solutions were called "volatile alkali" by druggists and keepers of hardware shops. And sometimes the word "volatile" was omitted by subsidiary people. This information was given to me a long time ago by my father who was a druggist in a hardware shop in the first half of the 20th century (I am an old man, born 1937). I am surprised to see that this word "alkali" is still used for ammonia solutions in a serious document... It should not.

I recently worked at a company with people who had been with the company 50 years(!), and when it started, the chemist was already old. The names toluol and xylol kept being used. BTW, 83 isn't old; you are just up to bismuth. You have another 35 to go.

I wouldn't tell my age (or other data) on the net without good reason. Also this is more of a comment on archaic wording then full answer.


text:
Does a [B4O8]4- anion really exist?
I am currently collecting information for an abstract and therefore I wanted to discuss different $\ce{[BO3]^3-}$ and $\ce{[BO4]^5-}$ connectivities. I also found a book on crystallography and the following passage
(Summarized translation of the original German text in [1])

When a finite connection of $\ce{[BO3]^3-}$ units is considered group borates form. [...] $\ce{[B2O5]^4-}$ (via vertices), $\ce{[B2O4]^2-}$ (via edges) or for example $\ce{[B4O8]^{4-}}$ (four-membered ring). [...]

It's about the last example that I cannot really find any other mention in literature. Although it looks okay if you consider vertex-connected oxoborate units forming a $\ce{[B4O8]^{4-}}$ anion, I cannot seem to find any example in literature. Even the book doesn't give any example.
I found some minerals for example $\ce{Ca2B4O8}$ but it's also described as $\ce{CaB2(OH)8}$. There are examples with $\ce{[B4O8]^{4-}}$ but they always contain additional $\ce{(OH)-}$ ligands.
The any other explanation I could think of would be $\ce{[B4O7]^{2-}}$. That would have a different structure (they also sketched their molecule anion and the got the charges correct so I don't think $\ce{[B4O8]^{4-}}$ is typing mistake). So, did anyone else find any example on a $\ce{[B4O8]^{4-}}$? I would usually check the Gmelin but due to the virus spread our library is currently closed.
[1] W. Kleber, H.-J. Bautsch, J. Bohm, Einführung in die Kristallographie, 19th edition, Oldenbourg Verlag, Munich, 2010.
Edit: I also did a quick DFT, hybrid functional B3LYP with Def2-SVP on the anion and according to my bond analysis (don't know if you can trust that for a tetra-anion here) all $\ce{B-O}$ bonds inside the ring have a bond order of 1.x while the external $\ce{B-O}$ bonds have a double bond character (I guess it's treating them as $\ce{B=O}$ rather than with charges here). But in theory that thing should be stable to me.


text:
Why are sulphide salts of p and d block coloured?
I have studied that sulphide salts of s block are colourless while p and d block salts are coloured.
But colour intensity of a salt is directly proportional to polarisation in salt.
And s block cations being smaller offer more polarisation. According to this logic, s block salts should be coloured.
Is my knowledge incomplete? Or am I using wrong logic?

Polarisation and colour are not related. Well, at least not directly.

sodium sulfide is yellow.

Pure sodium sulfide is colorless. I do have a bottle of $\ce{Na2S·9H2O}$ in my collection. Technical sodium sulfide is yellow because of the presence of sulfur as impurities. And this impurity of often produced by atmospheric oxidation releasing $\ce{Na2O}$ and free sulfur. Then free sulfur reacts with $\ce{Na2S}$ to produce disulfide $\ce{Na2S2}$ which contains the ion $\ce{S^{2-}}$ which is yellow, the same yellow color as in pyrite $\ce{FeS2}$
–

"But colour intensity of a salt is directly proportional to polarisation in salt." This is a  false premise.

Color of a salt is mainly due electronic transitions, MLCT, LMCT in transition metal complexes. Also convergence of HOMO-LUMO gap leads to easier excitations, small HOMO-LUMO gap make π electrons mobile, since it is easy for the electron to jump to a higher energy level that is close in energy. "Higher polarisation" in salts definitely lead to decrease in HOMO-LUMO gap. But polarization in group 1 salts is not so well due to large size of cations.

See the answer of this question: https://chemistry.stackexchange.com/questions/138309/the-color-of-the-most-sulfides-of-p-and-d-block-elements


text:
Why is the energy needed to transform PbO+CO into Pb+O2+C equal to the sum of breaking down PbO and CO?
Since PbO to Pb+(1/2)O2 releases 217.3kJ and CO to C+(1/2)O2 releases 110.5kJ, what I think is that PbO+CO to Pb+(1/2)O2+C+(1/2)O2 releases 327.8kJ. Now, the book says PbO+CO to Pb+C+O2 releases 327.8kJ.
My question is, why does is there no energy involved making Pb+(1/2)O2+C+(1/2)O2 into Pb+C+O2? Wouldn't there be some energy released joining the two halfs of oxygen?

why did the O become (1/2)O2?

It's like how half a dozen eggs + half a dozen eggs = one dozen eggs. It's not a dozen of half-eggs (that would be a tragedy!).

oh ok, thanks! thats a funny analogy!

There are no two halfs of oxygen. The half denotes that there is $\frac{1}{2}$ a mole of oxygen produced for every mole of $\ce{PbO}$ (or $\ce{CO}$) reacting. So there is no such "joining of oxygen halves" going on.

ok thanks, I get it! the 1/2 is just a simplification so we dont have to multiply everything by 2.


text:
Isotopes and electronegativity

Is there any difference between electronegativities of isotopes of the same element. For example consider the above compound. Will all the resonance structures be equivalent?
Also if I place the above compound in water (which will act as a Bronsted acid), which carbon will get the H atom?

An equally intriguing question could be how the resonance hybrid might be represented in this case.

@orthocresol Can you please explain why a negative charge on a $\ce{{14} C}$ would be different then a negative charge on a $\ce{{12} C}$. Is there a difference of electronegativities (hence the question)?

It's really just a wording nitpick. 14C isn't the same thing as 12C; thus a negative charge on 14C isn't the same thing as a negative charge on 12C. Asking whether the resonance structure are "equivalent" has the answer "no", because they aren't the same thing. Chemically speaking, it may of course be that the difference is so small so as to be completely negligible. I am guessing that is the point of your question. But we don't describe that as "equivalence". We'd describe it by saying that they contribute (almost) equally.

@orthocresol , the background of this question was what happens when we place this compound in water. Where would the hydrogen get placed in the acid-base reaction? This would be based on which resonating structure would be more stable, in turn giving different carbons a different negative charge. My guess was that maybe it was due to electronegativities, but I may be wrong.

(Star represents $\ce{^14C}$ atom)
This figure represents three out of the five possible resonance structures. Of these five structures, four have a negative charge on a $\ce{^12C}$ atom whereas one has negative charge on $\ce{^14C}$ atom.
As the first premise, let's consider stability of the two kinds of resonating structures. The more stable the resonating compound, the more likely it is to be protonated.

We can compare stability using the premise of electronegativity (or rather charge density). According to this property, the more heavy an atom, the larger is its volume. If volume increases, the charge density decreases making it less likely to be able to stabilize the negative charge (becomes less electronegative). Due to this we can say that the resonance structure with $\ce{^14C}$ having the negative charge is less stable and so is less likely to be protonated.
Now, moving onto the second premise, the number of resonating structures containing the negative charge on $\ce{^14C}$ and $\ce{^12C}$ respectively.
If you notice only one resonating structure has a $\ce{^14C}$ with the negative charge, whereas there are $4$ resonating structures containing $\ce{^12C}$ atom with the negative charge. This would mean that probabilistically (considering stability to be equal) The resonance structure having a $\ce{^12C}$ with the negative charge would be more likely to be attacked.
Now, if we combine the two premises, we see that both stability and frequency of the resonance structure containing $\ce{^12C}$ with negative charge is greater its $\ce{^14C}$ counterpart. Concluding, we can say that the resonating structure with $\ce{^12C}$ stabilizing the negative charge would be more likely to be protonated.
Further Reading on how stability of isotopic compounds can be determined :  Inductive effect of hydrogen isotopes

A tentative reasoning could be made on the basis of charge density (on nucleus).
More number of neutrons means larger size of nucleus, but with the same number of protons. Hence, positive charge density (on nucleus) will get reduced, and hence the electronegativity.
But these comparisons can be made on numbers only, as discussed in the comments.

EDIT: (after I read the marked answer carefully, I found that I'd not answered the second part of question. So, here it goes.)
As the $\ce{^{14}C}$ isotope is less EN, therefore it is relatively less stable while carrying a negative charge on it. Hence, protonation would happen on a stable intermediate, which is $\ce{^{12}C}$.

Intuitively it's not clear how extra neutral will impact electronegativity. Will it make protons more sparse in the nucleus? Or vice versa? I assume denser protons would decrease electronegativity while sparser protons would increase it?

@StanislavBashkyrtsev, I guess I thought it the other way. But, here's what I think: more number of neutrons will decrease charge density and hence the $\ce{Z_{eff}}$, which in turn makes it easy for the atom to lose electrons, and therefore make it, relative to its isotope, a good e-donor. Hence, reduces its electronegativity. Am I missing something?

@StanislavBashkyrtsev This may help

Yes, it is there due to the increased size of nucleus. But, the difference in the electronegativity of the two isotopes would be so small that you can easily neglect it.
The resonating structure will be different in $\ce{^14C}$ and $\ce{^12C}$. The reason is $\ce{^14C}$ is radioactive (a very small effect) so it may cause  a difference between the chemical properties of the two resonating structures.

so where will hydrogen be added if we react this with water

An isotope is defined as an element that has similar chemical properties but varied physical properties due to a different number of neutrons in the nucleus. So your premise feels flawed.


text:
From diagram, assuming that the reaction is 40% complete in the vessel: a. The weight of the methanol added per day b. The amount of Hl recycled

Here is my solution on the problem and the diagram below is the actual given of the problem. I hope you could help me correct it.

What's your question?

(1) I assume that this is a check my work question? (2) Were you given any information other than what is in the diagram?

Note the posting a picture of written text instead of typing that text is discouraged for several reasons. 1/It cannot be searched for. 2/It cannot be reused as reference or quoting, putting extra effort on shoulders of responders. 3/It can be challenging to read and interpret. 4/ Low effort of question preparation leads often to low effort of question answering, even to question closure.

Edit 5/16 3pm Another equation just occurred to me...
Edit 5/16 6pm fixed a math mistake in equation (2)

This isn't an answer yet, I'm trying to clarify the question.
(1) I assume that this is a check my work question?
(2) Were you given any information other than what is in the diagram below?

So using the variables indicated in the diagram:
R = recycled HI, units = lb/day
X = feed of $\ce{CH3OH}$, units = lb/day
W = waste stream, units = lb/day
Y = product stream, units = lb/day
Additional points:

The waste and product outputs give % values. Weight % or Mole fraction %?
Weight % would seem to be the right choice to me.
The diagram would seem to indicate that the total input to the reactor is $\pu{2000 + R lbs/day}$ of $\ce{HI}$.

(A) - 2000 lb/day of HI feed
Assume 2000 lb/day of $\ce{HI}$ is input into the reactor plus whatever is recycled. We can only assume that the recycling has reached a steady state, so 2000 lbs/day of HI, or its equivalent, must be in the input stream to the separator. So summing the two steams from the separator:
\begin{align}
\pu{1 lb \ce{HI}} &= \pu{1 lb \ce{HI}} \times \pu{2200 g/lb} \times \dfrac{\pu{141.939 g \ce{CH3I}/mol}}{\pu{127.911 g \ce{HI}/mol}}\times  \dfrac{1}{\pu{2200 g/lb}} \\
&= \pu{1.1097 lb \ce{CH3I}} \\
\end{align}
$\therefore 2000 = 0.8265\times W  + \dfrac{0.816}{1.1097}\times Y = 0.8265\times W + 0.73533\times Y\tag{1} $
(B) Now for $\ce{CH3OH}$ feed
$$\pu{1 lb \ce{CH3OH}} = \dfrac{141.939}{32.04} = \pu{4.4301 lb \ce{CH3I}}$$
$\therefore X = \dfrac{0.816}{4.4301}\times Y + 0.184 \times Y = 0.36819 \times Y\tag{2}$
(C) Balance water and $\ce{CH3I}$
The moles of water in the waste stream has to equal the moles of $\ce{CH3I}$ in the product stream.
$$\dfrac{0.1735\times W}{18.015} = \dfrac{0.816\times Y}{141.94}$$
$$\therefore W = 0.59693 \times Y\tag{3}$$
Substituting (3) into (1)
$$2000 = 0.8265\times (0.59693 \times Y) + 0.73533\times Y\tag{1a} $$
or $Y = 1,627.75 \approx \pu{1628 lbs/day}$
and $W = 0.59693\times 1627.75 = 971.65 \approx \pu{972 lbs/day}$
and $X = 0.36819 \times 1,627.75 = 599.32 \approx \pu{599 lbs/day}$
(D) For R -- "40% Complete" of what?

40% of HI is reacted?
40% of $\ce{CH3OH}$ is converted to $\ce{CH3I}$?

If the 40% percent is for the fraction of $\ce{CH3OH}$ which reacted, then  81.6:18.4 is not 60:40 in weight % or mole per cent for $\ce{CH3I:CH3OH}$.
\begin{array}{|c|c|c|c|c|}\hline
& \pu{M.W.} &\pu{mol \%} & \pu{\dfrac{\pu{g}}{\pu{mol}}} & \pu{wt \%} \\ \hline
\ce{CH3I} & 141.939 & 82.65 & 117.31 & 95.47 \\ \hline
\ce{CH3OH} & 32.04 & 17.35 & 5.56 & 4.53 \\ \hline
\end{array}
\begin{array}{|c|c|c|c|c|}\hline
& \pu{M.W.} &\pu{wt \%} & \pu{\dfrac{\pu{mol}}{\pu{100 g}}} & \pu{mol \%} \\ \hline
\ce{CH3I} & 141.939 & 82.65 & 0.58229 & 51.814 \\ \hline
\ce{CH3OH} & 32.04 & 17.35 & 0.54151 & 48.186 \\ \hline
\end{array}
The only way to use the information seems to be to assume that 40% of the total $\ce{HI}$ feed, 2000 + R, is converted to $\ce{CH3I}$.
$\therefore (2000 +R)\times 0.4 = \dfrac{0.816}{1.1097}\times Y = 0.73533\times Y\tag{4}$
Using Y from above, $R = 992.33 \approx \pu{992 lbs/day}$

Note - In comments below user Chet Miller makes a convincing argument about two points.

The 40% reacted would refer to the % of the limiting reactant which reacted. This would mean that the 40% referred to $\ce{CH3OH}$.
The OP's diagram is incorrectly drawn. Being a chemical engineer he proposes a diagram which makes more sense. There has to be a second separator of some sort.

I doubt that you could get all the HI out of water stream, but concentrated HI is about 57% HI by weight. So the reactor output and the waste stream, W, would have to be above atmospheric pressure.

Given Chet's proposed diagram below $R = 0.8265\times W = 803.06 \approx \pu{803 lbs/day}$

Shouldn't the recycle stream be coming from the separator, and have the same concentration as in the waste stream?  For 40% complete, they mean that the limiting reactant is 40% consumed.

@ChetMiller - RE: "recycle stream be coming from the separator" ---I don't know. I took the diagram literately, without making any assumptions about the actual process. Obviously some kind of separation would have to be happening if just HI was recycled from the reactor output. That being said I don't think  you'd recycle waste from the separator shown in the diagram. You wouldn't want to put the water back into the reactor.

RE: 40% complete, they mean that the limiting reactant is 40% consumed. --- Could be, I'm not extremely familiar with how chem engineering problems are setup. I also wonder how accurate the OP's drawing is.

It's much easier to work in terms of lb-mole flow rates than in terms of lb or in terms of metric units.  So I will express X,Y, W, and R in terms of lb-moles/day.  2000 lb/day of HI is 15.6 lb-moles/day.  Since the stoichiometry is such the number of moles of product from the reaction are equal to the number of moles of reactant, the molar flow must be constant.  Therefore we can use molar balances on the system.  The overall molar balance on the system is $$15.6+X=Y+W$$ In addition, a carbon balance on the system is $$X=Y$$ and thus $$W = 15.6\ lb-moles/day$$.
The mole fractions in stream Y are:$$CH_3I=0.4998$$ and $$CH_3OH=0.5002$$
The mole fractions in stream W are:  $$HI=0.4012$$and$$H_2O=0.5998$$
So an overall balance on iodine yields: $$15.6=0.4012W+0.4988Y=(0.4012)(15.6)+0.4998W$$From this, it follows that $$Y=18.76\ lb-moles/day=X$$
These values agree with the mass flow rates determined by @MaxW.


text:
Determining symmetry correction by looking at rotational quantum states
I'm trying to understand the relationship between symmetry correction and rotational quantum states, particularly in the case of dipoles with identical atoms.
For an angular momentum quantum number $J=1$ there are 3 quantum states according to $g(J)=2J+1$. Each quantum state determines the z-vector of the angular momentum $M$. The spacial direction of the angular momentum $M$ for each quantum state is then:

The orbital plane of a dipole is perpendicular to $M$.
I can see that I can re-orentiate scenario a) in space to make it look like scenario c).

Thus for $J=1$ there are actually 2 quantum states instead of 3.
I noticed that, at any value of $J$, this can be done for each pair of quantum states that have the same z-vector momentum $M_z$ but in opposite directions. The only quantum state at any $J$ that isn't part of a pair is when $M_z=0$
This would make me deduce that the number of quantum states for a dipole after symmetrical correction $\sigma$ is equal to $\frac{2J}{\sigma=2}+1$
However, here's a snippet of a source

Apart from this discussing the density of states, why does it show that after symmetrical correction the number of quantum states is $\frac{2J+1}{\sigma}$ instead of $\frac{2J}{\sigma}+1$?

The density of states calculation is correct, the multiplicity is $2J+1$ and these levels are degenerate. I don't follow why you want to ignore the  $m_z=0$ case. It should be treated in the same way as the others.

@porphyrin I am not ignoring $m_z=0$, I am considering $m_z=1$ and $m_z=-1$ to be the same quantum state in space. So any non-zero value of $m_z$ and its negative counterpart should be treated as 1 quantum state. The number of quantum states that have a negative counterpart is $2J$ so this should be divided by $\sigma$. The $+1$ part is reserved for $m_z=0$ which does not have a negative counterpart and thus is a quantum state by itself. Furthermore, how can a dipole at $J=2$ have $\frac{2J+1}{\sigma} = 2,5$ quantum states according to the source while it should be integers?


text:
Origin in a crystal
I am trying to understand the International Tables for Crystallography.
How can I identify the origin in the image on the left?
Without knowing the origin, I cannot identify the position of the symmetry elements. In the International Tables for Crystallography (Vol. A) it appears that the origin is at the "$(\overline{3}$ m 1) at $\overline{3}$ 2/m c" position. What does the $(\overline{3}$ m 1) element and the $\overline{3}$ 2/m c element mean? Where is the reference center really located?

You picked a fairly difficult one. I'd suggest starting with some of the more symmetric ones and parsing from there. The times I've used the tables it took a few exercises to build up from some of the simple cubic-type lattices to the ones I wanted.

@JonCuster Thanks for the reply. I've seen some simpler ones and I wanted to know the origin of this to confirm my guess. I think the origin is the $6_3$ axis that is in the upper left corner, at a vertex. That's right? Why do two positions appear: one with parentheses and one without parentheses?

For centro-symmetric space groups such as this one, the origin coincides with the center of inversion. You can see this specifically for this group if you look at symmetry operation (13): $$\overline{1} \mathrm{\ at\ } 0,0,0$$ Of course, there an an infinite number of origins in an infinite lattice because of the translational symmetry. In the tables, the origin is in the plane of the paper on any of the unit cell vertices shown, so upper left corner would be fine.
$(\overline{3}\ m\ 1)$ is a complicated way of saying inversion center. "$\overline{3}$" points at us, "$m$" is along the axes b or c, and "1" is the third direction (in the a b plane, 30 degrees rotated away from a or b or a+b). The next set of three symmetry elements $\overline{3}\ 2/m\ c$ are also in those three directions, and tell us how they are oriented with respect to the origin (they all go through it). The table listing all symmetry operations (and the list of transformations of a coordinate $x, y, z$ given here) confirms this.

There are a few centro-symmetric space groups with multiple settings; for those, the second setting has the inversion center at the origin. The space group in question is not one of those, though.

Thanks for the reply. Although you have explained, I did not understand why two directions appear (one with parentheses and one without parentheses). I didn't understand what you meant by "how they are oriented with respect to the origin". Could you explain otherwise please? When you say that 1 in $\overline{3},m,1$ refers to the diagonal, is the diagonal you consider to be [1 1 0] or are you considering the bisector: [1 -1 0]? The direction $\overline{3},m,1$ belongs to the trigonal group, so I am a little confused with the choice of axes for each index.

@JoseMarin "how they are oriented with respect to the origin": A two-fold axis along c could go through the origin (i.e. 0 0 z) or not (i.e. 1/2 0 z or 1/2 1/2 z). Same for a mirror plane (e.g. 0 -y -z would be through the origin, and 1/4 -y -z would not). As for the exact formatting on the statement about the origin (i.e. with/without parentheses), I don't know. The third direction (I misleadingly called it diagonal) is 30 degr rotated from a, b or the bisector. The directions a, b, and bisector are all equivalent because of the 6-fold screw axis, so the third direction is not any of those.

So in $\overline{3},m,1$, $\overline{3}$ is parallel to z, m is perpendicular to a, b, or diagonal: a+b ([1 1 0]) and 1 (identity element) is the direction that makes 30º with a, b or diagonal (a+b) axis, right?
I don't understand the part of indexes with or without parentheses. In the case of the axis of rotation, you are right, it is necessary to say the axis for which it is parallel and the point of application. But in inversion center, it is usually indicated the position where it is, because in this case, as it is not an axis, I don't understand why 2 sets of indexes are needed.

@JoseMarin I don't think 2 sets of indices are needed. It would suffice to say that the origin is at the center of symmetry (but this does not appear in the short or long form of the space group symbol). I suspect this is a quick way of making a statement about the relative location of the symmetry elements, but as I said, just a guess.

Why is the last index $1$ (in $\overline{3},m,1$), if in the upper left corner, in the bisector (a, b, or diagonal: a+b, each rotated by 30º) is there a glide plane perpendicular to the bisector? Why do you say $\overline{3},m,1$ represents the inversion center?


text:
Fundamentals of mathematics and physics for physical chemistry
Before I start studying chemistry, I want to gain some knowledge about the fundamentals of mathematics and physics that are important to better understand the principles of physical chemistry, especially since I haven't taken any maths or physics courses yet. Which fields of maths and physics should I take a look at? Suggestions for literature are welcome.

Quora has a really good answer to your question. https://www.quora.com/What-are-the-best-physical-chemistry-textbooks. I would check this out if you have not already.

We have a list of many textbooks from beginner to advanced, see Resources for learning Chemistry. If you get stuck or have additional questions while studying these texts, you are very welcome to ask for clarification here.

I would like to advise not dive into math textbooks from cover to cover, esp. when time is limited. Rather get some essentials:

the Rule of Three frequently is a life saver.
linear algebra and basic matrix stuff is important
differential and integral calculus as well

Unfortunately the question remains quite broad and also opinion-based, and the rest is probably covered by the resources post. I'm not so sure this question will be reopened, and there is nothing from a moderator's point of view to be done. If you want to have a more detailed explanation, please ask it on [meta].

There are three levels of scholarship. An analogy is that someone has looked at the ocean from far away and he feels that he has seen it all (standard undergrad mindset). Next stage, is when you walk on the beach and your feet become wet and dirty with sand. You have a feeling how ocean water and the beach feels like (fresh MS/PhD). The higher level of scholarship is when you are deep diving into the ocean and realize there is too much to learn and the knowledge you have is quite limited (a mature scientist who has spent his life in research).
Since you are saying that you have not taken mathematics and physics courses, it may not be the right time to begin a proper physical chemistry course.
Couple of nice books:
Mathematics in Chemistry An Introduction to Modern Methods by Hecht
Mathematics for Physical Chemistry by Robert G. Mortimer
Get hold of them from a library and get a feel for it (the first level of scholarship). You don't need to master all to open an undergraduate physical chemistry book. You rather learn it along the way. See if you like the subject in general.


text:
Confusion about Titration Formula
When performing a titration of a monoprotic acid and a monoprotic base, water can not be added to the titrant but water can be added to the analyte. To prove this statement to myself, I decided to analyze the equation for a hypothetical titration where the acid is the analyte and the base is the titrant:
[Acid] = (Volume of Base) $\cdot$ [Base] / (Volume of Acid)
So, if water is added to the base in the burette, the volume of base would increase but the molarity of the base would decrease and the effects seem to cancel out, leading to the same [Base] calculation. However, when water is added to the acid in the flask, the volume of acid increases which would lead to a decrease in the [Acid] calculation.
This contradicts the common protocol and knowledge when adding water to the burette or flask. Many answers I have seen state that the moles of acid and base stay constant if water is added to the analyte; however, wouldn't this also increase the volume of analyte and skew the calculation of [Acid]?
It would be awesome if someone could explain the effects of adding water to the burette and analyte on the equation above.

Your title don't match the body. Avoid asking multiple questions once

RE: "water can be added to the analyte" - Yes and no. Yes small quantities can be usually be added without a problem. However if you wanted to add a million liters of water to the analyte solution then that would be a problem. // Let's say that you need to add 500 ml of water to get the solid analyte to dissolve. What you'd do is run a blank. No analyte, just 500 ml of water. The blank value would be how much titrant it takes to reach the titration end point with just 500 ml of water. That volume would be subtracted from all the analyte titration volumes.

Your title and the main body of the post are two different questions. Anyway, it is important to understand the difference in each case.
Let us talk about the analyte. Your analyte can come in two forms: (i) either as a solid or (ii) as a solution.
In the case of a solid analyte you would quantitatively transfer a known exact weight of the sample to a flask, let us say an Erlenmeyer flask. You can now add some water to dissolve it and you do not need to worry because you have recorded the actual weight of the sample. [yes the conc. will change, depending on how much water you are adding].
In the case of a analyte solution you would quantitatively transfer a known volume with the help of a pipette to an Erlenmeyer flask. You can still add some water and need not worry because you know the initial volume.
[again the conc. will change with the addition of water].
*What is important in titration calculations are absolute moles of the analyte not its total volume in which it is being titrated*.
Now why can't we add water to a buret? The reason is that by adding water to buret, it will change the concentration of the titrant. As I said earlier, only moles are important in the calculations, and moles of titrant can only be calculated by
[Volume of titrant][Conc. of titrant] = moles of titrant
if the [Conc. of titrant] is an incorrect number, which has changed by adding water, the moles of titrant will be a wrong number.
Eventually, we calculate the concentrations with the initial volume we added with the help of pipette or a known mass.
*The problem is arising because you are using a titration formula taught by those who shouldn't be teaching science. This formula is not only wrong (prone to error), but it will lead to all type of problems in future with redox calculations. I strongly discourage using the formulae given in your query.
[Acid] = (Volume of Base)*[Base] / (Volume of Acid)
This formula is valid for all acid-base and redox titrations iff (if and only if) concentrations are provided in normality. Normality is an ancient 18th century concept. It has stuck in certain education systems like cancer. You quash it here but it appears somewhere else.*


text:
Entropy change of surroundings and system
The entropy change of the surroundings can be calculated by the equation $dS_{sur}=\frac{dq}{T_{sur}}$ regardless of the path (irreversible or reversible).
The argument is that because the surroundings may be approximated as either constant volume or constant pressure, the heat absorbed/released by it is equal to the internal energy or enthalpy change, respectively. As both are state functions, it does not matter if the path taken was reversible or not, the changes would be the same for both, so $dU_{sur,rev}=dU_{sur,irrev}$ and $dH_{sur,rev}=dH_{sur,irrev}$. However, $dH=dq$ for constant pressure, and as per definition $dS=\frac{dq_{rev}}{T}$, the initial equation follows.
This is somewhat the way my book derives the formula. It seems to be inconsistent to me because it equals $\frac{dq_{rev}}{T}=\frac{dq_{irrev}}{T}$, which some pages after the book says that is not the case and uses the fact that $dq_{rev}\geq dq_{irrev}$ to derive the Clausius inequality.
I do understand the Clausius inequality and its derivation, however I do not understand why is it that it is not used to calculate the entropy change of the surrounds, only the system. The arguments for its derivation seem to apply to the first case as well.
Similar question that has no answer: Entropy change of surroundings

In evaluating the entropy change of the surroundings, you need to first look exclusively at its initial state and its final state, before and following the irreversible process.  The only difference between the initial state and the final state is a difference in enthalpy of the surroundings, equal to amount of irreversible heat that flowed to the surroundings.  To get the entropy change for the surroundings, you next need to separate it entirely from the original system, and subject it to a new reversible process while in contact with its own new surroundings (i.e., a second set of surroundings).  No matter what new (alternative) reversible process you devise, you will always end up with the same enthalpy change and same beginning and ending pressures, and its entropy change will always be the same, and equal to the irreversible heat it received in the original process divided by its absolute temperature.
On the other hand, for the original system, if you follow the same procedure (separating it from the original surroundings and putting it into contact with a third surroundings), the reversible process you devise will give you a different value for the integral of $dq/T_{boundary}$ than the integral of $dq/T_{boundary}$ for the original irreversible process.

Could you elaborate on that separation part? I've never seen that.

That's too bad.  It has annoyed me for some time that the text books out there do such a bad job of explaining how to determine the entropy change for an irreversible process.  So I wrote a short primer of how to do it, which many people have found very useful:  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/ Please feel free to ask follow-up questions.

Regarding separating them:  If the process was irreversible, to get the entropy change of the system and the entropy change of the surroundings, we need to devise a path that takes each of them from their initial thermodynamic equilibrium state to their final thermodynamic equilibrium state reversibly.  If the original process was irreversible, there is no single path that will take them both between their initial and final states (of the irreversible process) simultaneously.  So we need to separate them, and subject each of them separately to a reversible path between these states.

The most important concept to understand is that entropy is a state function.
Best explained by this image:

The change in entropy is just based on the final state and the initial state. Therefore we model all processes as a reversible process because it has a simple equation. The reversible process is the one in blue above. That allows us to use the equation for the change in entropy for reversible process:
$$\mathrm dS =\frac{\mathrm dq_\mathrm{rev}}{T}.\tag{1}$$
Hint: use the first law to get the equation for $q.$
So if I want to calculate change in entropy for the system I integrate from state 1 to state 2. It will give you the same answer if you want to calculate $ds$ for any process.
Nuance of Surroundings:
If I want to calculate the change in entropy of the surroundings it changes, slightly, because the temperature of the surroundings is constant. That is because the surrounding is a reservoir. So we end up with
$$\int^{2}_{1}\mathrm dS_\mathrm{surr} = \int^{2}_{1}\frac{\mathrm dq_\mathrm{rev}}{T_\mathrm{surr}},\tag{2}$$
which will simplify to
$$S_2 - S_1 = \frac{q_\mathrm{rev}}{T_\mathrm{surr}}.\tag{3}$$

"The nuance of surroundings" part is incorrect as the final result must be $\Delta S=\frac{ -q_{\text{irrevsible}}}{T}$

If what you're trying to point to is that a heat transfer to the surroundings is irreversible then yes. However I wanted to explain in a way where the reader could see that this is the same q.  Then they will see sign convention correctly. My point being it's the same base equation. Same heat transfer, always starting tith definition. Keeps reader focused on that then they can decide what the "system" is and sign vectors accordingly.  In other words... My solution is right, your just being it from a different basis system.


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Basicity order of 2nd group oxides
It's a general trend that basicity of oxides decreases down the group.
But recently I came across a statement "the basicity of second group oxide decreases down the group and decrease in basic nature of oxide from Be to Ba is due to decrease in polarizing power with increase in ionic size"
I'm not able to understand the relationship between ionic nature and basicity.
Also I feel that more the ionic nature, more the basicity, (as oxygen easily dissociates )
This is question is related to  this , but generalised explaination is given, I'm looking for specific explaination to this this particular case

I have never heard until now that BeO is more basic than BaO...  :-) BeO is in fact amphoteric oxide, while CaO,SrO,BaO are strongly basic.

So the statement is wrong ?

Search for chemical properties of elements of the group 2 as broader context.

@poutnik Wikipedia also identifies magnesium as forming a strongly basic oxide.

@Oscar Lanzi  I do not object. But it is strongly basic wrt BeO, not wrt CaO-BaO.

Ah the dangers of classifying things in a bipolar manner when the properties are really a continuum.  See this answer on magnesia, or even lime, being an "alkali" or not.


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Does it make sense to have a control group for a school chemistry experiment?
I'm wondering if it makes sense to have a control group for the following experiment:

measure the speed of gas produced  when hydrochloric acid and calcium carbonate react, as a function of surface area of the calcium carbonate.
independent variable: coarseness of calcium carbonate ($\ce{CaCO3}$) as a way to control the surface area of the $\ce{CaCO3}$ from large chips down to powder.
dependent variable: speed of glass production measured using water displacement in a calibrated volume.

Usually I think of a control group as being used to control uncontrollable variations in the experiment, especially in biology or psychology experiments. But in this case, assuming that the volumes of $\ce{HCl}$ and $\ce{CaCO3}$ are measured carefully, and the gas is accurately measured, are there other variables that a control group would assist with?
We know that there will be an effect of mixing $\ce{HCl}$ and $\ce{CaCO3}$. We can hypothesize that increased surface area will speed the reaction. But does it make any sense to have a control where we mix $\ce{CaCO3}$ and water? Or $\ce{HCl}$ and pieces of unreactive glass?

A control where you only mix HCl and water will show you the baseline: You should observe some volume expansion as the mixture get warm, due to the heat liberated on dilution of the acid.
Otherwise, I'd opt for replicates, as you'll see some variation and it will nicely show that measured data is not absolute. Plus, you can do some statistics.

@M.Farooq if you copy your comments into an answer I would be happy to edit them.

@imalipusram if you copy your comments into an answer I would be happy to edit them.

@nilay-ghosh thank your for your edits

In your demo, you don't need any control experiment. A control experiment would have been needed if there were a competing side reaction producing a gas and affecting your gas measurement.

@JasonHarrison, I deleted my comment and copied it as an answer upon your request.

Unfortunately this isn't a demo experiment. If you modify your answer to include the other information that are included in the comments it would make this answer complete


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Are there any organic compounds that do NOT contain hydrogen ? And if so, what are they?
Organic compounds are typically defined as compounds containing carbon and nearly all contain hydrogen as well. Which organic compounds do not contain hydrogen?

e. g. $$\ce{ CCl4}$$

@Karsten Theis: And $\ce{CBr4}$. How about diamond and graphite? :-)

$\ce{ CS2}$, $\ce{ CO2}$, $\ce{ CO3^2-}$, $\ce{(CN)2}$,

@KarstenTheis https://chemistry.stackexchange.com/questions/49704/should-carbon-tetrachloride-really-be-considered-an-organic-compound

@imalipusram https://chemistry.stackexchange.com/questions/22195/is-carbon-dioxide-organic-or-inorganic

what about $C_{60}$ and other fullerenes?

@porphyrin I guess there is a thin ice here. Are graphite and diamond organic polymers ? Sometimes, distinguishing organic and inorganic substances is very formal. As we say CX4 are organic compounds, we should also say carbon does not form carbon halogenides, which are inorganic compounds.

I agree that the distinction organic/inorganic is somewhat arbitrary and is rather pointless anyway. The compartmentalisation of chemistry has had the unfortunate  effect of holding the subject back.

Does organometallic compounds count? eg. Ni(CO)4?

I’m voting to close this question because it is too broad.

Another definition I've seen of organic is that it contains BOTH hydrogen AND carbon. In that case not containing H would render a compound inorganic. But as others point out, the usefulness of such definitions is largely arbitrary and of questionable utility.

Distinguishing organic and inorganic substances is often very formal. We say CX4 are organic compounds, we should also say carbon does not form carbon halogenides, which are inorganic compounds.

One of more known halogen-hydrogen-free examples is carbon suboxide $\ce{C3O2}$,  $\ce{O=C=C=C=O}$, that is somehere on inorganic-organic border, being often considered as the second anhydride of malonic acide $\ce{HOOC-CH2-COOH}$, as it is prepared from it by $\ce{P4O10}$ aggressive dehydratation.
But the true malonic anhydride has 4-atom cycle.

One of less known examples is the anhydride of the mellitic acid, which could be formally considered as a carbon oxide with the summary formula $\ce{C12O9}$

... [mellitic anhydride] is one of the only four [oxides] that are reasonably stable under standard conditions.  It is a white sublimable solid, apparently obtained by Justus Liebig and Friedrich Wöhler in 1830 in their study of Mellite

As an interesting side info:

Mellite, also called honeystone, is an unusual mineral being also an organic chemical.   Chemically identified as an aluminium salt of mellitic acid; that is, aluminium benzene hexacarboxylate hydrate, with the chemical formula $\ce{Al2C6(COO)6 \cdot 16H2O}$

Mellite possess water of crystallization. So, it can be considered to have hydrogen atoms. Doesn't it disqualify it from the set of "organic compounds not having hydrogen"?

@Nilay Ghosh You have misinterpreted my answer, probably due my wrong answer layout. Mellite mentioning was intended as an interesting side information related to the anhydride, not as an another example. :-)

Oh I see, there are also some other minerals(Whewellite, Zhemchuzhnikovite) of organic origin but both of them possess water of crystallization :(

@Nilay Ghosh  But the question is, if this hydrogen counts.

Yes, OP's question sounds vague to me. Is OP asking for organic compounds not having hydrogen atoms or organic compounds not having hydrogen bond (C-H bond)? If OP is asking for the 1st question, then CX4 (X=halogen) counts and if OP is asking for 2nd question, then oxalic acid and benzenehexol counts.

It depends, of course, on what "organic" means.  If we define "organic compounds" as "carbon compounds generated by biological processes", then carbon dioxide is an obvious choice.  But, of course, most chemists consider that an inorganic compound.  A choice more likely to be favored by chemists is carbon disulfide, which Wikipedia reports is produced in trace quantities from marshes.


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How did the temperature in the solution with a lower concentration of CuCl2 increased in a faster rate?
My lab assignment is about the reaction between $\ce{CuCl2}$ and Aluminium and we tracked the temperature of the solution every 15 seconds and we needed to make our own experiment to check how different things affect the temperature of the solution every 15 seconds. So I chose the concentration of $\ce{CuCl2}$ and thus we had four cups, one with $\ce{0.1 M}$, one with $\ce{0.2 M}$, one with $\ce{0.5 M}$ and one with $\ce{0.8 M}$. Obviously the temperature in the cups with a lower concentration of $\ce{CuCl2}$ barely changed and throughout most of the experiment stayed at the exact same state.
Now, the cup with the $\ce{0.5 M}$ concentration of $\ce{CuCl2}$ changed temperature in a faster rate than the cup with the $\ce{0.8 M}$ concentration. Not only that, at the end of the experiment, the temperature in the cup with $\ce{0.8 M}$ was 28° C meanwhile the temperature of the cup with the $\ce{0.5 M}$ concentration of $\ce{CuCl2}$ was 29° C .
Can somebody explain to me how that happened?

The reaction $\ce{Cu^2+ + 4 Cl- <=> CuCl4^2-}$ may lead to slowing down reaction with aluminium, decreasing the equilibrium concentration of $\ce{Cu^2+}$ in $\pu{0.8 M}$ $\ce{CuCl2}$. That all would lead to lower peak temperature due bigger heat dissipation before reaching maximal temperature.
Contributing factors may be

decreasing of chemical activity of water for the electrode surface reaction $\ce{Al(s) + 6 H2O -> [Al(H2O)6]^3+(aq) + 3 e-}$. These  electrons, charging the electrode, then reduce copper: $\ce{[Cu(H2O)n]^2+(aq) + 2 e- -> Cu(s) + n H2O}$
increasing of viscosity, therefore decreasing of diffusion coefficients and another decreasing of the reaction rate.