text:
Why proton concentration is divided by 10⁻⁷?
I am reviewing the book Biochemistry Concepts and Connections by Appling, Cahill, and Mathews and I cannot understand why they divide by the hydrogen concentration by $10^{-7}$.  Why not just leave it at the antilog(-8.1) over $\pu{1 M}$ like the other concentrations? I literally have see nothing else on the internet like this.

The textbook is precisely correct.
The equilibrium constant $K$ which the logarithm is taken of is dimensionless, and includes activities or fugacities, and not concentrations and pressures.
In practice this is achieved by using standard states which refer to the pure materials: standard concentration $c^⦵$ and standard pressure $p^⦵$.
One must be very fastidious with units when finding the equilibrium constant.
For example, the reaction
$$\ce{aA + bB <=> cC + dD}$$
equilibrium constant $K_c$ is exactly
$$K_c = \frac{([\ce{C}]/c^⦵)^c\cdot ([\ce{D}]/c^⦵)^d}{([\ce{A}]/c^⦵)^a\cdot ([\ce{B}]/c^⦵)^b}$$
For pure water in its standard state $c^⦵ = [\ce{H+}] = \pu{1e-7 M}$.
It also correlates with so-called biological standard state of $\mathrm{pH} = 7$.
You probably haven't seen it before because many authors use sloppy notations omitting mentioning standard states since they can often be cancelled out.
In this case those cannot be cancelled out, and must be written explicitly.

In fact, your own textbook contains extensive explanation [1, p. 91]:

For chemical reactions the standard state for solutes is defined as $\pu{1 M}$; however, in living cells the concentration of $[\ce{H+}]$ is roughly $10^{-7}~\mathrm M$, much lower than the standard value of $\pu{1 M}$.
  It is therefore appropriate to define the reference concentration of $\ce{H+}$ in biochemical reactions relative to the $\ce{H+}$ concentration found in the living state (i.e., $10^{-7}~\mathrm M$), rather than the value $\pu{1 M}$ defined by the chemical standard state.
  Recall that when a solute in a dilute solution has a concentration of $\pu{1 M}$, the activity of that solute is unity.
  For the biochemical standard state we define the activity of $\ce{H+}$ to be unity when $[\ce{H+}] = 10^{-7}~\mathrm M$.
[...]

The mass action expression $Q$ is unitless.
  We strip the units from each concentration term in $Q$ by dividing each by its
  proper standard concentration (e.g., $\pu{1 M}$ for all solutes
  except $\ce{H+}$; $10^{-7}~\mathrm M$ for $\ce{H+}$; $\pu{1 bar}$ for gases, etc.).

Refrences

Appling, D. R.; Anthony-Cahill, S. J.; Mathews, C. K. Biochemistry: Concepts and Connections (Global Edition); Pearson: Boston, 2015. ISBN 978-1-292-11210-7.

There are two standard states for [H+]. In the absence of a prime after the $^\circ$, it is 1 M, and with the prime (biochemical standard state), it is $10^{-7}$ M.

Thanks,  I have the second edition and this explanation is not on this page 91.  Is it from Chapter 3?  If so I will find it.

Nevermind, it is on page 59 of the second edition for those with the second edition.

For an explanation you may want to inspect Recommendations for Biochemical Equilibrium Data 1, which states:

Buffer and pH. If only  a limited number of measurements are to be
  made, they  should be carried out at pH =  7.0 and, if possible, also
  at a  pH value at which the apparent equilibrium constant $K_c^\prime$, has
  little or no dependence on pH. ($K_c^\prime$ is defined in a  later
  section.) If direct measurements at pH = 7.0 are not practicable, the
  calculated values for this pH should be reported. The procedure used
  in making these calculations must be carefully  described. Care should
  be taken that the solution is adequately  buffered so that the pH is
  well  defined throughout the experiment. It is desirable to determine
  the effect of varying the nature and concentration of the buffer in
  order to identify  buffer effects. Buffers that are known to interact
  with reactants (including macromolecules) or salts, such as phosphate
  or pyrophosphate in the presence of divalent metal ions, should be
  avoided.

The highlighted portion means that reported values for biochemical reactions are (or should be) referenced to pH 7.0. The data in Table 3.6 refer to this biochemical standard state.
Another question on the subject of equilibrium constants also addresses the importance of properly considering reference states.
Reference
1 The Journal of Biological Chemistry (1976), Vol. 261,  No. 22, pp.  6859-6885.


text:
How the unreactivity of lithium hydride makes it useful in synthesis of other useful hydrides?
I have seen the following chemical property of lithium hydride in the textbook which seems to be incomplete:

Lithium Hydride is rather unreactive at moderate temperatures with ${O_2}$ or
${Cl_2}$.It is therefore, used in the synthesis of other useful hydrides.
$\ce{8LiH + Al2Cl6 -> 2LiAlH4 + 6LiCl}$
$\ce{2LiH + B2H6 -> 2LiBH4 }$

But why the unreactivity of lithium hydride makes it useful in synthesis of other useful hydrides?

"Usefulness" is a subjective term and cannot be judged outside the context. What textbook you've seen it in?

here is the link for the page(http://ncert.nic.in/textbook/textbook.htm?kech2=2-7) (page 15 pdf )(288)

The equations make no sense. Not only are they unbalanced, but they have different elements in and out.

Sorry for the mistake,it was just an error.

Lithium hydride is by no means unreactive towards air.  In powdered form it can react with air at low humidity forming the oxide, hydride and carbonate; if the air has higher humidity lithium hydride can combust spontaneously.  Pellets are less violent but still not fully stable in air containing water vapor (as air typically does).  See the Wikipedia article.
Lithium hydride is less reactive than heavier alkali metal hydrides.  The textbook may be referring to chemists being better able to minimize reactions with the surroundings using lithium hydride; thus better able to target the intended synthesis reaction.

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text:
charge transfer between layers of graphite with stacking fault
As a general rule, we only expect two parts of a system to have the exact same amount of charge if they are symmetry-related to each other. How much charge transfer would occur in graphite with a stacking fault (e.g., with stacking ABABABCBCBC...), due to the layers no longer being all symmetry-equivalent as in pure Bernal graphite?

Well, that's interesting point, but I wouldn't expect much CT.

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text:
What does thermal stability of compounds depend on?
Does it somehow depend upon the lattice energy of the compound?
My textbook says that Lithium carbonate is not so stable to heat and forms more stable $\ce{Li2O}$ and $\ce{CO2}$.
Could it depend on the electropositive character? Because my textbook further sates that the stability of carbonates increases down the group.

thermal stability depends on the structure, and bond lenghts, bond angle etc. it also depends on the difference in electronegativity of the compounds.

Does this answer your question? Difference between lattice energy and thermal stability

Lattice energy might be important if the decomposition took place in the solid phase.
But Li2CO3 melts at 723 C and boils (with decomposition) at 1310 C. Na2CO3 melts at 851 C and decomposes before boiling, according to an older CRC Handbook and Wikipedia, which seems to be pretty current.  According to those numbers, thermal stability of sodium and lithium carbonates seems fairly similar, and decomposition does not take place because of solid issues.
K2CO3 melts at 891 C, decomposes before boiling. Rb2CO3 melts at 837 C, decomposes at 900 C. Cs2CO3 melts/decomposes at 610 C. Lattice energy might be important with Rb and Cs, but thermal stability seems to become less stable as you go down the group, not more so.
The first three seem fairly stable for several degrees above the melting point. We could consider the melt to be a mix of CO3-- anions with the metal cations, all oxygen-bonded (similar to hydrogen bonding!), but much looser than in a crystal. The thermal stability of the CO3-- anion does not seem to be very dependent on the ionic size of these three metals (Li+, Na+, K+).
So, I would ask the textbook author to substantiate his claims.

As we move down the alkali metal group, the electropositive character increases. This causes an increase in the stability of alkali carbonates. However, lithium carbonate is not so stable to heat. This is because lithium carbonate is covalent. Lithium ion, being very small in size, polarizes a large carbonate ion, leading to the formation of more stable lithium oxide.
Therefore, lithium carbonate decomposes at a low temperature while a stable sodium carbonate decomposes at a high temperature.
The smaller the size of the ion, the higher the lattice energy and the greater the extent of the polarization. Li+ has a smaller size compared to $\ce{Na+}$, hence polarizes the $\ce{CO3-}$ ion at a greater extent compared to $\ce{Na+}$. $\ce{Li2CO3}$ therefore requires less energy to break while $\ce{Na2CO3}$ needs a higher energy making it more stable than $\ce{Li2CO3}$.


text:
Developing a lab to identify the concentration of casein in organic milk
I am currently attempting to create a feasible lab design that will enable me to identify the concentration of casein protein in organic milk without having to use spectroscopy. Thus far I have thought of separating the casein from the milk by exploiting its pH of isoelectric point of 4.6 to form curdles and filter the curdles out using a filter. Furthermore, I would rinse and leave the curdles to dry till I am left with casein powder. Following this I would compare the weight of the casein powder to the weight of the original sample of milk I extract it from. Is this a feasible method or are the any other alternative ways also not using any major equipment?

That sounds pretty reasonable. You can search for "casein isolation method".  I found this for instance: facstaff.bloomu.edu/mpugh/Experiment%2011.pdf

@NightWriter The link you posted has been very helpful so far, thank you.

This book describes in great detail an experimental procedure for what are you looking for.

And what's an organic milk? Milk with alcohol instead of (quite inorganic) water? BTW you just described a method of making cottage cheese, not isolating casein (more useful and tasty though).

@Mithoron I can't tell if your response is tongue-in-cheek or not, but https://www.ams.usda.gov/grades-standards/organic-standards.


text:
Surface Chemistry - Dipole Moment of Surfactant with Water and Phases of a Monomolecular Layer
Recently, I have been trying to study surface chemistry and I have come across questions that I do not understand how to answer. I understand what a dipole moment is, defined as a charge separation over a defined distance, and I have a fair understanding of surfactants and other areas of surface chemistry such as Langmuir Isotherms. But I am struggling to answer such questions:

How would surfactants with a dipole moment in contact with water be characterised?

Would this mean that most likely the ionic head of a surfactant interacts with either the partially-positive hydrogen or partially-negative oxygen moiety of a water molecule? If so, what would be the proper way to characterise this phenomenon?

In what phase of the monomolecular layer are the sizes of surfactants determined?

I am puzzled by this question, as I do not understand what exactly is meant by phase. I have spent relentless time trying to search for monomolecular layers and their phases but I have come out empty-handed. I see that phase can refer to distinct boundaries of physical states, such as those between gasses and liquids, but I do not know how this could be used to find the size of molecules. I know that in order to find the area that each surfactant molecule covers at an interface the Gibbs Adsorption Equation is utilised, however it uses concentration of surfactant molecules plotted against surface tension.
I would be much obliged if anyone could provide assistance to aid me in understanding this subject of surface chemistry better.

Wow, that is a question!

Surfactants (aka tensioactive compound) have a dipolar moment at the air-water interface but that is not enough to characterize their behavior. There is a cohesive force between their lipophilic tails (due to Van der Walls force) and there is also a cohesive force between their polar head but only because water of metal ions can make a bridge between identically charged moieties.
I will address only your main question which seems to be "how to know the size of molecules in a monolayer"? Well, a surfactant floats at the surface of water. So you can make an apparatus filled with water and with a movable barrier which will move across the surface to slowly reduce the overall surface available for a tensioactive compound. We also know how to measure variations in the "tensioactive power" (whatever its name is) which depends on the number of molecules at the surface.

Using such a device, we can create a monolayer of a lipophilic compound, reduce the whole area (which we call a "compression") and detect the resulting effect (which correlates to the number of molecules at the surface). Whenever the available surface becomes too small for all the molecules to be nicely arranged at the surface (i.e. the "collapse").

Regarding question (2), the wikipedia page describing Langmuir-Blodgett films may help.
The following figure shows a pressure-area curve (top) and a diagram of a Langmuir-Blodgett trough (bottom, adapted from a description at Ameslab) which is used to modify the area while a Wilhelmy plate balance measures the surface pressure:

When a condensed phase is formed, the surface pressure-area curve goes through a discontinuity. At the point where the liquid expanded phase transitions into the condensed solid phase (indicated by the arrow), the area divided by the amount of surfactant provides a measure of the (molar) headgroup area.


text:
Distillation: temperature not reaching azeotrope boiling point
I ran a simple distillation of $\ce{HBr}$/water mixture because I wanted to purify it. Boiling point of azeotropic hydrobromic acid is approx. $\pu{124 ^\circ C}$, however my distillation ran almost all the way through at vapor temperature of $\pu{112\! -\! 113 ^\circ C}$. Atmospheric pressure is normal. It is around $\pu{8 ^\circ C}$ where I'm working though. What could be the reason the temperature never reached $\pu{124 ^\circ C}$?

See, the concentration of HBr is somewhat important. If it would be 0%, your distillation will run at 100°C, after all.

It probably means you have too much water. You can either dehydrate your original solution with activated molecular sieve, or you can re-distill it... though it would be better after drying it with the same activated molecular sieve.

Have you calibrated your thermometer? What is your distillation setup? Did you use a fractional distillation column? // See video

Initial concentration was around 30%. I used simple distillation setup.

I think your major mistake is using a simple distillation method, which may not give accurate reading for azeotropic distillation. The early distillate may not have enough cooling to get to expected azeotrope temperature of aqueous $\ce{HBr}$. If you have followed the work of Carriére et Cerveau (Ref.1), you may able to assume approximate composition of your fractions. Although their work cannot be regarded as of high precision as the composition was determined by titration (Ref.2), they have given an insight to $\ce{HBr}$ distillation. Their work has clearly showed aqueous $\ce{HBr}$ has been yielding a "high-boiling" azeotrope (boiling temperature of azeotrope is higher than boiling points of either of pure components). For example, the $\ce{HBr}$ composition of the early distillate collected at $\pu{110 ^\circ C}$ was reported as 50.8%. The temperature had increased to $\pu{126 ^\circ C}$ maximum (47.5% $\ce{HBr}$), and then declined passed  $\pu{111.5 ^\circ C}$, at which, the $\ce{HBr}$ composition was found to be 30.8%. Since your initial $\ce{HBr}$ concentration was around 30%, I’d say your distillate is already at downward slope and never have able to reach constant boiling azeotrope temperature. If you had used fractional distillation setup with long enough column (and measured atmospheric pressure accurately), you may have able to collect some azeotrope. For your convenience, I listed published data of azeotrope boiling point, density, and composition at different pressure (measured by pressure control apparatus):
$$
\begin{array}{ccc}
\text{Reference} &\text{Pressure} & \text{Boiling Point} & \text{Density at}\: \pu{25 ^\circ C} & \text{Composition, % HBr}\\\hline
\text{Carriére et Cerveau}^1 & \pu{760 mmHg} & \pu{126 ^\circ C} & - & 47.5\\
\text{Ewing and Shadduck}^3 & \pu{760 mmHg} & \pu{125 ^\circ C} & - & 47.795 \pm0.033\\
\text{Bonner et al.}^2 & \pu{760\pm0.5 mmHg} & \pu{124.3\pm0.02 ^\circ C} & - & 47.63\pm0.01\\
\text{Bonner et al.}^2 & \pu{100 mmHg} & \pu{74.12 ^\circ C} & 1.5116 & 49.80\\
\text{Bonner et al.}^2 & \pu{200 mmHg} & \pu{90.35 ^\circ C} & 1.5030 & 49.28\\
\text{Bonner et al.}^2 & \pu{300 mmHg} & \pu{99.91 ^\circ C} & 1.4961 & 48.83\\
\text{Bonner et al.}^2 & \pu{400 mmHg} & \pu{107.00 ^\circ C} & 1.4908 & 48.47\\
\text{Bonner et al.}^2 & \pu{500 mmHg} & \pu{112.94 ^\circ C} & 1.4866 & 48.19\\
\text{Bonner et al.}^2 & \pu{600 mmHg} & \pu{117.82 ^\circ C} & 1.4832 & 47.95\\
\text{Bonner et al.}^2 & \pu{700 mmHg} & \pu{122.00 ^\circ C} & 1.4802 & 47.74\\
\text{Bonner et al.}^2 & \pu{800 mmHg} & \pu{125.79 ^\circ C} & 1.4775 & 47.56\\
\text{Bonner et al.}^2 & \pu{900 mmHg} & \pu{129.13 ^\circ C} & 1.4752 & 47.40\\
\text{Bonner et al.}^2 & \pu{1000 mmHg} & \pu{132.12 ^\circ C} & 1.4733 & 47.27\\
\text{Bonner et al.}^2 & \pu{1100 mmHg} & \pu{134.80 ^\circ C} & 1.4716 & 47.14\\
\text{Bonner et al.}^2 & \pu{1200 mmHg} & \pu{137.34 ^\circ C} & 1.4700 & 47.03\\
\end{array}
$$

References:

MM. E. Carriére and Cerveau, Compt. rend. Acad. Sci. 1923, 177, 46-48 (https://gallica.bnf.fr/ark:/12148/bpt6k3130n/f50.image).
W. D. Bonner, L. G. Bonner, F. J. Gurney, “Azeotropic Hydrobromic Acid Solutions at Pressures of 100 mm to 1200 mm,” J. Am. Chem. Soc. 1933, 55(4), 1406-1409 (DOI: 10.1021/ja01331a012).
D. T. Ewing, H. A. Shadduck, “The Composition of a Constant-Boiling Solution of Hydrogen Bromide in Water,” J. Am. Chem. Soc. 1925, 47(7), 1901–1904 (DOI: 10.1021/ja01684a017).


text:
Why is 1‐tert‐butylcyclopentan‐1‐ylium more stable than 1‐(2‐methylbutan‐2‐yl)cyclopentan‐1‐ylium?

(1) should be more stable as the $\ce{CH3}$ group is electron donating.
But apparently (2) is more stable.
Is there another phenomenon taking place here?

As formulated one should expect 2 to be more stable. Point is eventually why is the ethyl fragment not stabilising more than a methyl, as usually found and as resumed here https://chemistry.stackexchange.com/questions/47008/why-do-larger-alkyl-groups-show-a-greater-inductive-i-effect

This is getting absurd. I quite wonder how detached from reality these textbook exercises would get. For all practical purposes there no difference here. You can argue about "inductive effect" vs "hyperconjugation" however long you wish...

On the same line of @Mithoron though the example might be of interest in a proper contest (e.g. computational approach) it seems to stretched simple rules to the limit.

So does that mean these structures simply cannot be compared?

No - you YES can. But they should be rather the same and whatever difference should be experimentally seen than explained, not the vice versa. It is the weight of subtle hyperconjugation that might differentiate between.

@Alchimista can you not argue using the Boyer's strain?, The pentagonal geometry increases the s character on the carbocation...which further develops it's way asking for the $e^-$ density from 3 hydrogens using hyperconjugation (methyl) or 2 (for ethyl)

@Alchimista Indeed the s char is high, I checked and found it to be 35% s and remaining  64%


text:
What are some ways to measure sugar concentration of fruits?
I've tried to do titrations using the quantitative Benedict's Solution and it was difficult to use as I do not have a way of keeping a constant temperature. I am currently using a Polarimeter to find the angle of the various concentrations of sugar in my solutions but it does not seem very clear or efficient.

Are you looking for the total sugar content or for a certain carbohydrate, such as fructose, glucose or saccharose, for example?

I'm looking specifically for the D-fructose sugar concentration, but I am able to switch to total sugar concentration if needed.

The solute will alter the density of the solvent, which is used for example to determine the concentration of sulfuric acids by a hydrometer and in particular by help of a saccharometer.  Depending on your equipment you will need at least about 20 mL of analyte.
A popular alternative among winemakers is to rely on the change of the refractive index of a solution (refractometry) instead.  With portable devices as on the picture, you basically determine the critical angle of total reflection of the juice of few grapes placed on a prism.  The analysis is quick (seconds) and needs little (mL or less), the reading typically yields refractive index and content of sugars present as expressed for example in Oechsle, Baumé, Brix, or mass%.

(source)

(source)
Note, however, that these techniques are not specific to sugar(s).  The presence of other compounds soluted equally may affect the refractive index, as well as the wavelength of light used and temperature during the recording.  To prevent adulteration of wine, for example, some more advanced refractometers already determine the refractive index at multiple wavelengths simultaneously, too.


text:
Question about reactions that are always spontaneous and non spontaneous
I wanted to ask for examples of spontaneous chemical reactions meaning that their enthalpy would be negative (exothermic) and their variation of entropy would increase, therefore the Gibbs energy would be also negative and the reactions would be spontaneous at all temperatures (?).
I was thinking about the boiling of water as the reaction feels hot and the entropy change is positive as it goes from liquid to gas (entropy increases) but then I thought that to do that, we need an external energy and so the reaction would be non-spontaneous?
Another one could be dropping mentos into a carbonated drink. I can be sure that the reaction doesn't requiere an external force to happen BUT the reaction itself doesn't feel hot (I think I saw it once when they did it in primary school but I'm assuming it isn't releasing any heat?)
I know that when enthalpy is negative and entropy is negative, it would only be spontaneous if the temperature was low enough.
For the non spontaneous reactions, I can't really think about one that has positive enthalpy and negative entropy. I CAN think of a reaction that needs a push to "occur" like taking off a stain of a shirt or even a reaction of decomposition of water into h2 and o2.
I think I'm getting something wrong.

I was thinking about the boiling of water

That is a good example of a process that goes forward at higher temperature (water boils when heated above the normal boiling point) and goes in the opposite direction at lower temperature (steam condenses below the normal boiling point). This process also depends on the pressure, but my statements are valid at atmospheric pressure.
For this process, the entropy change is positive (water has more freedom to move in the gas phase compared to the liquid phase) and the enthalpy change is positive (endothermic, it takes energy to break the hydrogen bonds between water molecules).

Reactions that are always spontaneous

How about combustion of cyclohexane? The entropy change is positive (lots of gas molecules are formed - six carbon dioxide molecules per hexane) and the enthalpy change is negative (exothermic).

Reactions that are always non-spontaneous

Once you are certain that a reaction is always spontaneous, it is easy to find one that is always non-spontaneous - just swap reactants and products. So according to this argument, cyclohexane and oxygen will never form from carbon dioxide and water spontaneously. That sounds reasonable.

+1 for realizing VLE is just a type of reaction!

VLE: vapor liquid equilibrium... I had to look it up

Well, if we were all engineers, the world would be a worse place.


text:
Boiling methanol in a microwave
Some protocols (metabolomics) use boiling methanol for metabolites extraction and cleaning glassware. Is it safe to boil methanol in a microwave?

Sure if you have an explosion proof microwave. ;-)

It is possible and is not unheard to perform an extraction assisted by microwaves (even "dry ones" e.g.).  Either you seal the reaction vessel with a special cap to clamp, and monitor temperature and pressure evolution in function of time and power applied by the dedicated microwave oven.  This has the potential advantage of both rapid heating, and "pressure cooking".  (Depending on the material, using the reaction tube sealed a twice may be discouraged, though.)
Or you opt-in for a system where the the reaction flask is in the microwave oven, and the reflux condenser passes across the walls of the oven.  While in principle ovens designed for use in a kitchen may be used, e.g.,

(source)
it is both safer and more reproducible to use ovens dedicated to this (especially: focussing on the spot where your reactor is situated) and providing proper shielding and monitoring of the process parameters.

(source)

Methanol flashpoint is listed as 11 °C to 12 °C (52 °F to 54 °F),
upper flammable limit is listed as 36%, so it seems likely that it could enter into a flammable limit.
Also you may need to understand heat input/vaporization rate vs vent path through your condenser.
I too think a “home built” microwave setup does not sound safe.


text:
How to use Kohlrausch's law with ionic equivalent conductances?
My text says that for an ionic compound like AxByif ionic equivalent conductance of Ay+ is a and that of   Bx- ion is b, then the equivalent conductance for AxBy would be a/y + b/x
Now my question is:
as the unit of equivalent conductance is S cm2 (gm equivalent)-1 so if i multiply it by a quantity x with unit (gm equivalent) mol-1, then i will get a quantity with unit S cm2 mol-1. This should be the Molar conductance of the ion granted that the value of x was correct. Now for example in Al2(SO4)3 for Al3+ ion the value of x would be 3 because one mole of Al3+ ions would contains 3 gram equivalents. So if equivalent conductance of Al3+ ion was a, then its molar conductance would be 3a and similarly for SO42- ion with equivalent conductance b, the molar conductance would be 2b. Now from all that i can tell that the molar conductance of Al2(SO4)3 would be, 2(3a) + 3(2b) = 6a + 6b
Now from the formula of equivalent conductance and molar conductance
(Molar conductance)/n = Equivalent conductance
where n is the valence factor for desired compound.
So in our case n=6 for Al2(SO4)3
∴ equivalent conductance = (6a + 6b)/6 = a + b
but according to my text it would be a/3 + b/2
Tell me where am i going wrong.
And the result I ended up on works for some questions and the thing that my text provided works for some questions, and then there are some questions in which nothing works. I am really confused... And sorry for any grammatical atrocities. EDIT: I am just trying to ask that when equivalent conductance of two ions are given to me as a and b, then what would be the equivalent conductance of the compound to which these ions belong? Would it be a + b or would it be a/(charge on first ion) + b/(charge on second ion) or is it something totally else?

Can i get the hold removed? I edited it.

You textbook writer has explained it in a very complicated way. I think you are confused by the concept of equivalent. Gram equivalent is an obsolete unit now but it is still used in conductance measurements.
First understand the definition of gram equivalent weight for a salt (and it applies to salts only). Acids/ Bases/Redox agents have different definitions.
Gram Equivalent weight of a salt or an ion = gram molecular weight/charge on the ion
This is the definition of gram equivalent.
From the context of your book, ionic equivalent conductance is same as "molar conductance" of aluminum ion. Here a and b refer to molar conductance. This is the mistake you are making. Thus, a 1 Molar Al3+ solution has 3 gram equivalents of Al/ 1 L or it is a 3 N solution (N = Normal).
So if you are given ionic equivalent conductance of Al2(SO4)3 (at infinite dilution) then your textbook is right = (ionic equiv. conductance Al)/ charge on Al + (ionic equiv. conductance sulfate)/ charge on sulfate = a/3 + b/2
By the way, equivalent conductance can only be added at infinite dilution. Whatever you see in tables are usually given at "infinite dilution".

But aren't molar conductances and equivalent conductances different?  Because it is clearly mentioned in my book that a and b are the equivalent conductances. And yes I am really confused about this whole equivalent thing. Thanks for replying!

Molar conductance and equivalent conductance are same for monovalent ions but not for mutivalent ions, acids or bases. I think it will be easier for you if you realize that molar conductance = ionic equivalent conductance in the context of your textbook. "Equivalent" is a way of expressing amount like mole. Just like 1 M solution  of X contains 1 mol/L, a 1 Normal solution of X contains 1 gram equivalent/ 1 L. The author has done a poor job in explaining it and using poor terminologies. https://en.wikipedia.org/wiki/Molar_conductivity.

Ok thanks! I guess i will try another book.


text:
When surface area of electrodes is increased, why does the current in the galvanic cell increase?
I did a lab in galvanic cells where I changed the surface area. As a result, the current increased, but I don't really understand why. I guessed it was because more are means that more ions from the anode are oxidized and more ions from the cathode reduced, more electrons are transferred and taking into account that current is the flow of electrons per unit time, the current increases.
Is that even correct?
Thank you all

You nailed it...

what I guessed is actually right?

Yeah, you're on the right track. In any battery reaction there is some rate-limiting step that determines the maximum power the battery can supply. Most of the possible rate-limiting steps involve the surface of the electrode.
As an example, imagine that the rate-limiting step is the diffusion of new ions from the electrolyte bulk to the cathode (or diffusion of ions away from the surface at the anode. The physics of diffusion in this case is governed by Nernst-Planck equation, where the quantity of interest is the ion flux (number of ions of unit area). However, the total current is determined by the total number of ions. To get from flux to total number of ions you have to integrate over the area of the electrode: if you double the area of the electrode you'll double the total number of ions (ie. current) if all else stays equal.
That's just one example that depends on the specifics of your cell, but many of the surface-related limited steps in a battery follow this type of behavior.

One of my electrochemistry professors used to give a nice example. Why can't you start a car with a 9 V Everyday Alkaline Batteries? It is all about the rate of current which can drawn from a battery.


text:
Why does the 31P{1H} NMR spectrum of cis-[Mo(CO)2(dppe)2] show two signals?
I've tried searching for literature references that explain this, finding only a single reference which refers to the two signals as originating from phosphine environments cis/trans to carbonyls, however I'm having trouble rationalising why this occurs.

I just realised why, I'm a total idiot  - I blame this on my lack of sleep. To ask a question that requires an actual answer @orthocresol I see you added a far better representation of the complex, do you have any tips on how to draw better looking chelating ligands?

Not really! I just drew a regular octahedral complex in ChemDraw (from Templates > Stereocentres), made it a bit bigger (or else the PPh2 clashes into the CO), filled in the PPh2's and CO's, then drew a few bonds between phosphorus. Yours was perfectly fine, just a bit large, and I was drawing it for my answer anyway so it was trivial to add it into the question itself.

Are you using ChemDraw's default document settings? If I try to follow a similar process the ligand bonds end up being too long so I end up with disproportionately large ligands.

Nope, for all the diagrams on SE, I use a version of the Trauner group ChemDraw template where everything is scaled by 0.75x. If you use the ACS Document 1996 template in ChemDraw it gives stuff which look pretty similar, so you could try that out.

In this complex there are two different 31P environments which are not related by symmetry:

The two green phosphorus nuclei can be interconverted by a $C_2$ rotation (the rotation axis bisects the OC–Mo–CO angle), and so can the purple ones, but green and purple cannot be interconverted.
As extra proof, consider that the green P is cis to both carbonyl ligands whereas the purple P is cis to one and trans to the other.
They therefore have different chemical shifts and in the spectrum you would expect to see two different peaks. Presumably they would show coupling to each other, so I would expect two triplets, if we ignore satellites from all other nuclei (e.g. 13C).


text:
Subscript on Orbitals
What do the subscripts on the orbitals (such as px, py, pzetc) mean and how can one derive these or at least gain a fundamental understanding of each? Do they describe the geometry of the shape? And if so- how do derive them?
Please describe in an advanced high school level.

There's no "suborbitals" - these are just orbitals and subscript describes their orientation in space.

Atomic orbitals are actually functions with a well defined (and rather complicated) mathematical form. We often have those functions coming in groups, where the shape is the same but the orientation is different. Those indexes help to identify which one we are talking about, and also help figure out symmetry properties.

If you don't like the question, don't just downvoat it: help improve it. In my experience this question reflects a common confusion among beginners, and beginers, by definition, are not the best in formulating their questions.

"Advanced high school level" is ambiguous. It would be helpful to know your math and/or physics background to help formulate an answer. Greg's comment is a helpful start down the right path. However, if you're looking for "fundamental understanding," you're only going to get there by studying quantum mechanics.

Depending on your level, you may think that they are kind of components like in a vector. NO. They are single orbitals and the label describes their orientation (it is that the axes are chosen in a convenient way). In thist case their longitudinal axes lies right on x or y or z. Each would form, if it contains a valance electron, a sigma bond in that direction - unless  hybridisation must be used. Anyway do not think of them as components, they can mix but not as components of a vector.

@Zhe Atomic orbitals, orbital interaction etc are often taught in high school without any relevant context or quantum mechanical background.

@Greg That's true, but I think OP is looking for more something deeper than that.

As a high school student, here’s how I considered it:
The wavefunction of an orbital describes the region of space that an electron residing in that orbital would be in about 90% of the time.
The wavefunctions would have a radial component (depending on period), which is why the 3p orbitals Arena larger than the 2p etc.
The wavefunction would also have an angular component, which would describe its shape.
In the case of the p orbitals, the x, y and z labels simply means that their angular components are x/r, y/r and z/r respectively (hence the subscript). Do note that we usually use the radial forms of the wavefunctions though.
Hope this helps.

Unfortunately, the $x$, $y$, $z$ have more to do with the angular part of the wave function than the radial part.

I actually messed up- I meant the angular components were x/r, y/r and z/r


text:
Cerium (CeO) behavior in Glass Polishing?
I am basically working in semiconductor industry and I specifically taking care of CMP (Chemical Mechanical Polishing). We did polish glass (borosilicate type of glass) and we are using slurry (which is cerium-based) as the chemical compound. Our polishing time was quite long (~40 min) and without chilling the slurry, it could deteriorate the polishing pad, material removal etc. I have decided to chill the slurry (CeO) at temp ~20 °C prior supply to my polishing machine.
How does this Ce chemically and mechanically impact my CMP performance? What is the agglomeration index of this Ce with the changes of temperature?
Any good article to share?

CeO is not a thing at all.

@IvanNeretin https://stellafane.org/tm/atm/mirror-refs/grit.html#Polishing%20Compounds

@Sansiro there are some amateur telescope makers active in Astronomy SE and mirror polishing is discussed there from time to time. You might consider asking there instead.

@uhoh You don't learn chemistry by pointing at random sites out there. I mean, the site might have been maintained by knowledgeable and respectable professionals in their field; it is just that chemistry is not their field. Like I said, CeO is not a thing. It is CeO2 they polish their glass with.

@IvanNeretin I just thought I'd let you know that the OP may have gotten "CeO" from a seemingly authoritative source and repeated it in good faith. Instead of writing "I think you mean Cerium oxide or "CeO2 is what they polish their glass with", or even making a helpful edit to improve a new user's first question, you wrote something terse and discouraging. I'm all for helping people learn chemistry, but a ruler across the knuckles isn't (always) the best way.

@uhoh Discouraging, if you ask me, is to have the door slammed shut in your face. I mean all those close votes - where are they from? The question is totally legitimate, even if I have nothing to say on it other than pointing out one minor error (which is not even the OP's fault, nor did I say it is).

As a fun fact, ceria containing ores also commonly contain radium, so your polishing compound is likely radioactive and needs proper disposal.

Be careful with how the words end - cerium is a metal or a chemical element, whereas ceria is a cerium oxide.

My guess is that for the reason why your slurry deteriorates without chilling you should look the other way. Maybe the other slurry components are to blame? I do not see how ceria could impact the slurry performance in your case. Ceria (and yes, it's $\ce{CeO2}$ not $\ce{CeO}$) and some other oxides like zirconia are used in milling and polishing chiefly because of their physical hardness and chemical inertness. In other word, $\ce{CeO2}$ should probably be regarded as purely abrasive component. You need extreme conditions to chemically alter ceria, such as concentrated acids and a lot of prolonged heating, and the chemical interaction between ceria and the other solid materials is highly unlikely unless the temperatures are so high that the whole system starts to glow.


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What is the nomenclature of this alkane?

In this alkane, the blue line is parent chain. It is 12 carbon based.
I wonder why red line cannot be parent chain. It does not compete with blue line about the position of substitution.
And also, i understand 5-isopropyl part but i cannot understand octan-4-yl part. As i learned, when we take substituent part of already substituted part, we take carbon number one on the branch point. So i think this molecule has nomenclature about 6-(1-propylpentyl) but it is wrong. Is there any nomenclature rule that i have missed?

Related questions: Choosing Parent Chain, Complexity of Side Chains (similar compound) and How should I name this branched alkyl substituent? (about the branched substituent)

The nomenclature rule that you have missed is given in Section P-46 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) (explaining that compound acyclic substituents consisting of a principal chain and one or more acyclic substituents can be named in two ways which can be found in Subsection P-29.2) and in Subsection P-29.3.2.2, which clarifies that the preferred names for simple substituent prefixes derived from saturated parent hydrides with a single free valence that is not at the end of the longest chain are formed by using alkanyl substituents according to Method (2) of Subsection P-29.2:

(2) The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ are added to the name of the parent hydride with elision of the terminal letter ‘e’, if present, when followed immediately by the letter ‘y’. The locants for the atoms of free valences are as low as is consistent with any established numbering of the parent hydride and, except for mononuclear parent hydrides or the suffix ‘ylidyne’, the locant ‘1’ must be cited. This method is used to generate names of ‘alkanyl-type substituent groups’ that are simple substituent groups with free valences at positions other than ‘1’.

For the substituent that is mentioned in the question, the preferred alkanyl-type prefix according to Method (2) is ‘octan-4-yl’.
Note that, according to current IUPAC nomenclature, the prefix ‘isopropyl’ is only retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’, which is also formed according to Method (2).
Therefore, the complete name for the compound that is given in the question is 6-(octan-4-yl)-5-(propan-2-yl)dodecane.


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Canonical Restricted Open-shell Hartree–Fock equation
In Frank Jensen's Introduction to Computational Chemistry ($\mathrm{2^{nd}}$ edition), on page 99, it says:

For an ROHF wave function, it is not possible to choose a unitary
  transformation that makes the matrix of Lagrange multipliers in eq.
  (3.41) diagonal, and orbital energies from an ROHF (Restricted
  Open-shell Hartree–Fock) wave function are consequently not uniquely
  defined and cannot be equated to ionization potentials by a
  Koopmans-type argument.

Here ROHF eq. (3.41) is the non canonical Hartree-Fock equation:
$$F_i\phi_i=\sum\lambda_{ij}\phi_j$$
But in ATTILA SZABO, NEIL S. OSTLUND's Modern Quantum Chemestry, they prove that the Hartree-Fock equation can always be put into the canonical form:
$$F_i\phi_i=\lambda_{i}\phi_i$$
What am I misunderstanding here?

I'm guessing that "restricted open-shell" part is important.

ATTILA SZABO, NEIL S. OSTLUND showed that there is a basis where the hartree-fock
equation has the form $F_i\phi_i=\lambda_{i}\phi_i$ that is, it does not work for every basis , only for a kind special of basis , so there is no issue in the two statements

The treatment found in Szabo-Ostlund, although perhaps it is not explicitly stated, is appropriate only for non-degenerate spin-singlet electronic states, which in the Hartree-Fock method correspond to closed-shell systems.
For open-shell systems one may use the restricted open-shell Hartree-Fock method (ROHF), which is somewhat more complicated. In particular, in the ROHF method six arbitrary constants have to be introduced. Many choices for these parameters have been suggested.
The total ROHF energy, as well as the overall ROHF wave function and therefore the expectation value of any operator, are independent of these six parameters, so that all these ROHF sub-variants are largely equivalent.
However, the six parameters affect the Hartree-Fock orbitals and their orbital energies. Also, approximate post-HF methods such as MP2 or CISD will depend on the particular choice of those six parameters.
Knowles et al suggested in 1991 a particular choice leading to called "semi-canonical orbitals" whose orbital energies do obey Koopmans theorem and which are also a good choice for MP2 calculations.
I suggest the following paper for an introduction to the problem (the paper by Knowles metioned above is ref. 18):
T. Tsuchimochi and G. E. Scuseria, ROHF Theory Made Simple, J. Chem. Phys. 133, 141102 (2010) https://arxiv.org/abs/1008.1607
The paper above shows that one can obtain semi-canonical ROHF orbitals by a minor modification of the unrestricted Hartree-Fock method and that this approach is numerically very stable.


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How to determine Acid-Base Indicator
How do we determine which is the best indicator to use when given the initial pH’s of both the acids and bases and the endpoint value of our base?

The first row in the worksheet is unconventional because it represents a titration of a weak acid with a weak base. The change in pH after reaching the equivalence point will be much smaller than in cases where the titration is with a strong acid or base, but when carefully done, these titrations do yield results (source).

I have a poor impression of the question since the pH values are given for the titrant and the analyte. In order to properly judge the equivalence point it would be better to know the pKa values of the two.

All acid-base indicators are weak acids or bases themselves. They have a pKa value associated with them. The pH at which they change color "quickly" is
pH of abrupt color change = pKa-1 to pKa+1;
For example phenolphthalein has a pKa of 9.3, so it will quickly change colors in the pH range of 8.3 to 10.3, whichever side your are approaching this range. This is how those numbers are listed for your indicators.
So how do we choose an indicator? The punchline is that your end-point* must lie close the pH range of abrupt color change. Take the titration of HCl (in a flask)and NaOH (in a burette). Now you see that pH has reached 7; in order to ensure that there is no trace of acid remaining, you would add a fraction of a drop of a base to the flask; the pH will quickly jump to an alkaline pH because there is no acid anymore. Which indicator would abruptly change color as it sees a slightly excess of base? Phenolphthalein (8.2 to 10.2 as per your text)
You can do the rest using similar logic.

* Finer details: Your text is wrong in the sense of using the word endpoint. End point in a classical sense means that you have added a slight excess (fraction of a drop) to the titration flask. This indicates end of the titration. The correct term in your scanned image should be "equivalence point" where the titration has just been completed (equivalent amounts of base and acids have reacted); so the end point of a strong acid- base titration is not at pH of 7.0, it is slightly basic if the base is being added from the burette. Ask your teacher to clarify further.

You're confused about the terms end point and equivalence point. The end point is the pH which you are trying to reach in the titration, ie when the titration "ends" due to the color change of the indicator. Assuming you're starting with an acidic solution, then the equivalence point is when you have added as many moles of base as there were of acid.

RE: "For example phenolphthalein has a pKa of 9.3, so it will be useful in changing colors in the pH range of 8.3 to 10.3. // Not true. When using  phenolphthalein you're looking for the color to appear around 8 or disappear at 8. It is impossible to tell by eye when all the  phenolphthalein has been converted to the colored form. In other words you can't tell the difference between 90% converted and 99% converted by eye.

Also the pH does not have to be 7 at either the end point or the equivalence point.

@ MaxW, regarding equivalence and end point, both of us saying the same thing. The scanned images uses the wrong term. It should be equivalence point. I have further clarified the statement regarding phenolphthalein for the student.

The moderator removed some comments. // M Farooq - We're not saying the same thing and the OP's scanned image uses the term end point correctly. One can only titrate to the end point where the indicator color changes.

The moderator removed comments upon my request including yours. The forums are meant for discussion in a constructive way without denigrating each other. If you don't like someone's response, feel free to write another answer. My opinion is that the way it is written is misleading for students. Showing that the end point pH of NaOH/HCl titration is at a pH of 7.0 is misleading. Yes, the equivalence point pH = 7.0. It is a classical volumetric titration not a potentiometric one. In potentiometric titration we get the equivalence point because it is free from the bias of an indicator.

Assuming relatively equal concentrations of acid and base, then the equivalence point will be half way between the two pKa values, not at pH =7. The pH value of 7 has nothing to do with either the end point or the equivalence point. // Also as I have repeatedly stated a titration can only be done to the end point, not the equivalence point. // I can't add another answer, the question has been locked.

Please look up the definition of equivalence point and end point from any standard analytical chemistry textbook (Vogel's or even Kolthoff are classical ones). Remove this misconception from your mind that equivalence point cannot be located. Potentiometric titrations can do it (in principle). When we use indicators, we locate the end point. Who said that pH at the end point must be 7.0? It could just be a high school boy's misconception which has nothing to do with real life analytical chemistry. https://en.m.wikipedia.org/wiki/Equivalence_point


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How to identify the structure of an organic acid based on its reactivity, mass and NMR spectrum?
Taken from OCR A Chemistry Unified Chemistry Sample Paper.

Information about a monobasic organic acid D is shown below.

D reacts by both electrophilic substitution and electrophilic addition.
The molecular formula of D is $\ce{C_xH_yO2}$.
The mass spectrum of D has a molecular ion peak at $m/z = 148$.
The $\ce{^13C}$ NMR spectrum of D contains seven peaks.

Determine and draw a possible structure for D. Explain your reasoning from the evidence provided.

I understand that D must contain a benzene ring (electrophilic substitution) and a double carbon bond (electrophilic addition). However, I'm at a loss about how to deduce the molecular formula. Taking away $M_r(\ce{O2})$ from $\pu{148 g mol^-1}$, gives $M_r(\ce{C_xH_y})= \pu{116 g mol^-1}$. The mark scheme gives the molecular formula as $\ce{C9H8O2}$. I can't see how the values of $x$ and $y$ were deduced. All you know is that D contains a $\ce{C=C}$, benzene ring and most likely a $\ce{-COOH}$ group (due to monobasic nature).

Divide 116 by 12  (with integer then modulo division) which gives 9 with a remainder of 8. You know the molecule contains only C, H and O, so this gives your answer.

If you were a mathematician, you would say why $\ce{C9H8}$ and not $\ce{C8H20}$ or even $\ce{H116}$. Studying chemistry, you can exclude those "solutions" because even in an alkane, you never have more than 2n+2 hydrogen atoms in a hydrocarbon with n carbon atoms.

If you were an engineer, you might say why $\ce{C9H8}$ and not $\mathrm{C_{9.5}H_2}$. Studying chemistry, you can exclude a formula with non-integers if you want to describe a molecule with distinct structure.

It's nothing other than Cinnamic acid.

@Soumik Das You might be right, but that was not the question. The OP is asking how to deduce the chemical formula from the molar mass.

There is no contradiction on that question. Because it is an A-Level question (assuming you are high school student), you should be able to solve with your basic knowledge:
You have correctly suggested that D contains a $\ce{C=C}$, a benzene ring, and most likely a $\ce{-COOH}$ group, based on the data given (at least $4+1+1=6$ number of unsaturations).
The molecular formula of D is $\ce{C_xH_yO2}$ and molecular ion peak of its mass spectrum is at $148\:m/z$, thus, $12\times x + 1\times y + 16\times 2 = 148$ or $12 x + y = 148-32= 116$.
Since $\ce{^13C}$ NMR contains 7 peaks, the molecule must have at least 7 $\ce{C}$ atoms so you can deduce that amount from your  $\ce{C_xH_y}$ portion. That leaves you with: $\pu{(116-12\times7) g/mol} = \pu{32 g/mol}$ to resove ($x-7\lt3$), which can be either $\ce{CH20}$ ($= \pu{32 g/mol}$) or $\ce{C2H8}$ ($= \pu{32 g/mol}$). That leave you with the molecular formula of your compound been either $\ce{C8H20O2}$ ($\mathrm{FM}= \pu{148 g/mol}$) or $\ce{C9H8O2}$ ($\mathrm{FM}= \pu{148 g/mol}$). Since you know that compound has at least $6$ unsaturations, molecular formula $\ce{C8H20O2}$ can be eliminated. Therefore the molecular formula of your compound should be $\ce{C9H8O2}$.

I'm at a loss about how to deduce the molecular formula.

Compounds containing only C, H, and O have a certain number of hydrogens (h) compared to the number of carbons (c).
If there are no double bonds and no rings, h = 2c + 2. This is true for linear chain alkanes. Adding oxygens to those (hydroxyl groups, ethers) does not change the number of hydrogens. Turning the linear chain to a branched structure does not change the number of hydrogens.
For every double bond you add, you lose two hydrogens (think of it as turning sigma bonds to hydrogen into pi bonds between the heavier atoms). For every ring you close, you lose two hydrogens (one additional C-C bond instead of each carbon bound to a hydrogen, for example).
So for molecules with double bonds (d) and rings (r), you get:
h = 2c + 2 - 2d - 2r

All you know is that D contains a C=C, benzene ring and most likely a -COOH group

So the benzene ring accounts for three double bonds and a ring, the C=C for a double bond and the -COOH for a double bond (C=O). At this point, we don't know if there are even more double bonds or rings, so we have d >= 5 and r >= 1 and the following inequality
h = 2c + 2 - 2d - 2r < 2c - 10
So the molecule will not have the maximum possible number of hydrogen atoms (2c + 2) but much less.

$M_r(\ce{C_xH_y})= \pu{116 g mol^{-1}}$

Just like porphyrin said in his comment, there is only one solution to x and y that makes chemical sense: 9 carbons and 8 hydrogens. You don't even have to think about the rings and double bonds like I discussed above, but it is a nice check. For a hydrocarbon with 9 carbon atoms we would expect 20 hydrogen atoms in the absence of rings and double bonds. We have 12 less than that, so the number of rings and double bonds in the molecules is 6. That fits with the expectation of a benzene ring (accounting for 4), another C=C double bond (1 more) and a carboxylate group (1 more).


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How does existence of alpha and beta form of glucose prove that it exists as a cyclic structure
My book says that

Glucose is found to exist in two different crystalline forms which are
  named as $\alpha$ and $\beta$.

Next it says that

This behaviour could not be explained by the open chain structure for
  glucose.

I am not able figure out how does existence of two crystalline forms prove that glucose exists as a cyclic structure? Are there no noncyclic compounds which exist in more than once crystalline forms?
And why can't it be explained by the open chain structure?
I think I am missing something, but what?
Thanks!
Edit:
If we take the example of glycine, it exists in two different crystalline forms, but doesn't have a cyclic structure like that of glucose. So why does existence of two crystalline forms for glucose seen as a supporting evidence of its ring form?

Look at @MathewMahindaratne's answer. The two forms of glucose exist in solution where the only additional structure would be what the molecule adopts on its own. You're confused because you're confusing crystalline polymorphs with anomers. They're just not the same thing.

@Zhe Oh okay. I think I got it now.

In old days, polymorphs (different spatial arrangement of the same molecule in a crystal) were labeled $\alpha$, $\beta$, $\gamma$, etc. as e.g., still seen for sulfur, iron, and other elements.  But Today, polymorphs are labeled by Roman numbers I, II, III (e.g., carbamazepine).  Thus, the pitfall to confuse it with anomers labelled with $\alpha$, or $\beta$ should be a thing of the past.

According to Wikipedia:

Glucose is usually present in solid form as a monohydrate with a closed pyran ring (dextrose hydrate). In aqueous solution, on the other hand, it is an open-chain to a small extent and is present predominantly as α- or β-pyranose, which partially mutually merge by mutarotation.

Glucose predominantly occurs in nature in the form of the D‐enantiomer, which is generally believed to exist in three crystalline forms: $\alpha$‐D‐glucose monohydrate (Figure 1A)(Ref.2), anhydrous $\alpha$‐D‐glucose (Figure 1B)(Ref.2), and anhydrous $\beta$‐D‐glucose (Figure 1C)(Ref.1,3). Both anhydrous $\alpha$‐D‐ and $\beta$‐D‐glucose crystals are orthorhombic while $\alpha$‐D‐glucose monohydrate crystals are monoclinic (see Fig. 1A-C). However, a fourth form, which is metastable in solution phase at $\pu{38\!-\! 50 ^{\circ}C}$ and thought to be a hydrated form of $\beta$‐D‐glucose, has been reported as well (Ref.1,4).
The crystal structure of $\beta$-D-glucose published in 1960 (Ref.5) clearly showed the exsistence of pyranose ring system. As in the inserted box in Figure 1 state that, in aqueous solutions, 99% of D‐glucose exists as a mixture of the $\alpha$- and $\beta$-forms (approximately 62%  $\beta$ and 38% $\alpha$ when equilibrated at $\pu{31 ^{\circ}C}$ (Ref.1). Recent NMR study using fully $\ce{^13C}$ labelled glucose (Ref.6) clearly showed $\alpha/\beta$ ratio of $37/63$, which is almost identical to this literature value (Figure 2):

References:

F. W. Schenck, "Glucose and Glucose-Containing Syrups," In Ullmann's Encyclopedia of Industrial Chemistry: Ullmann's Food and Feed, Vol. 2; B; Elvers, Ed.; Wiley‐VCH Verlag GmbH & Co. KGaA: Weinheim, Germany, 2017, pp. 781-802, doi 10.1002/14356007.a12_457.pub2.
W. B. Newkirk, “Manufacture and Uses of Refined Dextrose,” Ind. Eng. Chem. 1924, 16(11), 1173-1175, doi 10.1021/ie50179a028.
G. R. Dean, “Optical-Crystallographic Properties of $\beta$-D-Glucose,” Anal. Chem. 1973, 45(14), 2440–2441, doi 10.1021/ac60336a005.
G. R. Dean, “An unstable crystalline phase in the D-glucose-water system,” Carbohydr. Res. 1974, 34(2), 315–322, doi 10.1016/S0008-6215(00)82906-7.
W. G. Ferrier, “The crystal structure of $\beta$-D-glucose,” Acta Cryst. 1960, 13, 678-679, doi 10.1107/S0365110X60001588.
T. Richter, S. Berger, “A NMR method to determine the anomeric specificity of glucose phosphorylation,” Bioorg. Med. Chem. 2013, 21(10), 2710–2714, doi 10.1016/j.bmc.2013.03.008.

The open chain is floppy. The relative stereochemistry between two centers is fixed and unchanging.
But the existence of geometric isomers implies additional structure imposed on the molecule. In this case, in 6-member ring, one can have axial or equatorial substitution which are distinct. The relative stereochemistry is the same but the relative substitution patterns are now different.
Take a look at the following structure (source).

In the $\alpha$-form, the hydroxyl group at C-1 is axial, and in the $\beta$-form, it is equatorial. However, this change does not affect anything else in the molecule of glucose, so the relative positions of substituents becomes different.
The tricky part is that this doesn't necessarily imply a cyclic structure, but a cyclic structure is probably the easiest way to impose structure on the molecule that will rationalize two forms. (And I am hard pressed at the moment to actually come up with an alternative.)

But a compound like glycine doesn't have a cyclic structure, but still exhibits polymorphism. Can't glucose exist like that?

@Natasha For glycine, I believe those polymorphs are in the crystalline phase, which is structure that has been imposed on the system (and conveniently provides the example my answer was missing). For glucose, the isomerism is present in solution.

Oh okay. Can you please explain you last comment by comparing polymorphism of glycine, and that of glucose (if that's also known as polymorphism).

@Natasha Those are two different concepts. Based on your last comment, I am missing something in interpreting your question. There is some bit of confusion that you have, but I haven't yet figured out what that is.

@Zhe From what I understood by reading your answer,the energy barrier for axial-equatorial transformation in the cyclic structure is less than the barrier to actually make geometric isomers in the open-chain structure. So to rationalize the existence of 2 isomers,the conformational(kind of)isomerism in ring structure was preferred for the solution phase?

@Zhe I have edited the question. I think that might help in clarification of my confusion.


text:
NMR Prediction for compounds at variable pH values
Like the title says, I'm looking for any software package that may be able to simulate a variety of NMR at a variety of different pHs. The majority of the software programs that I've found do not have an option that also incorporates a pH specification.
Any advice is appreciated, I found a few prior threads on the topic, however they were all 2+ years old so I was curious if there were any advances.
Thanks in advance for any advice.

Please link to those earlier threads, esp. as you already looked them up!

Quick answer: Impossible for any general purpose NMR predictor. You need to know not only the electronic structure but also the respective lifetimes of the states of all exchangeable protons. Then you can exactly predict how their NMR lines coalesce. But that's a multi-scale simulation project, from ab-initio up to molecular dynamics, and I wouldn't put large stakes on the outcome before I also have corroborating experimental evidence, i.e. measured a series of spectra.

What compounds are you interested in? And by simulation, you mean $\mathrm{^1H}$ chemical shifts?

If you still have the elder publications (2 years old), a search in data bases like Web of Science (Clarivate), Scopus (Elsevier) or even to some extent free google scholar (Google) may indicate who cited these, hence pointing toward more recent publications in the fiel of interest.

Hi all, thanks for the info. I'll try your suggestions! Sorry for the late replies!

I don't know of a good, "easy" way to do this. A bad, but easier way than that outlined above is to use a Louiville-von-Neumann simulation package, such as spinach [http://www.spindynamics.org/wiki/index.php?title=Simple_liquid_state_NMR_simulations] which is able to take the scalar couplings and estimate T1s directly in a variety of relaxation frameworks (e.g. redfield). You can get these couplings from a quantum chemistry package, eg Gaussian. You'd then do this for your protonated /deprotonated form and perform the averaging based on their relative mole fractions in the high T limit.


text:
Measuring bioanode potential against Ag/AgCl reference electrode
I'm trying to make a bioanode for use in a microbial fuel cell.  I have a piece of carbon cloth in an acetate solution with river sediment.  I'd like to know the potential of the anode, so I connected it to an Ag/AgCl reference electrode also in the sediment/acetate solution (note that ref electrode has 3M KCl, about 0.2V relative to SHE).  Between the reference electrode and anode, I have a multimeter connected with the positive going to the Ag/AgCl reference electrode and negative going to the carbon cloth anode.
The multimeter reads +0.2V.  Does this mean my anode potential is 0.4V (0.2V from multimeter reading + 0.2V for reference electrode correction)?  OR, does this mean that my anode potential is 0V (-0.2V multimeter reading + 0.2V for reference electrode correction). I'm not clear if the positive side of the multimeter and negative side are connected correctly to the electrodes.  If they are connected incorrectly, then the voltage sign should be flipped.
Thank you!

Coming to your experiment, there is a nice description along with sufficient details which shows you how to make connections.
Microbial Fuel Cells & Bacterial Power
https://agsci.oregonstate.edu/sites/agscid7/files/bioenergy/fuel_cell_interactive.pdf
Coming to the confusion: Electrochemists do not measure electrode potentials by using a multimeter as shown in general chemistry textbooks. The numbers shown on a multimeter display only qualitatively showing us that there is a potential difference between these two electrodes. Secondly, carbon cloth is the least reproducible electrodes in terms of its potential (heavily dependent on the history of how it was made). You won't find its values in standard tables. Therefore we cannot compare your experimental values with standard conditions (in the tables, the values are calculated under controlled conditions) because your system is nowhere close to standard conditions. It is like comparing apples and oranges.
An interesting part for you is, how to decide which one is cathode and which one is anode of your cell? Connect the electrodes in such a way that you get a positive potential difference on the multimeter. If you happen to get it, you have made the right connections i.e., negative terminal of your cell is connected to the negative terminal of your multimeter and so on.


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What does the dissociative mechanism for cis/trans interconversion of [Mo(CO)4(PPh3)2] look like?
All I can find in the literature is that the cis→trans interconversion proceeds via dissociation of one of the PPh3 ligands resulting in a square-pyramidal Cs intermediate, with the remaining PPh3 in an axial position.
I assume it proceeds via either dissociation of the CO trans to the PPh3, followed by PPh3 and CO binding again to form the trans isomer; or via a Berry pseudorotation-like conversion of the square pyramidal intermediate, however considering that the former seems unlikely due to the CO being a more strongly binding ligand than PPh3 and the latter would require quite a bit of energy, neither of these make sense to me.

More specifics are needed in the question, but it's possible that the square pyramid is in equilibrium with a trigonal pyramid (simple isomerization). That allows blurring of what it means to be cis and trans.

I'm not sure what how to be more specific, any suggestions welcomed. Also I assume you meant trigonal bipyramid?

Oh, the complex is given in the title... Yes, I meant bipyramid. There aren't too many viable options here. I don't have a reference right now, but I remember seeing something similar before.


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Why is change in entropy negative and change in enthalpy negative for the reaction of Magnesium and Hydrochloric acid?
So
$$\ce{Mg(s) + 2 HCl(aq) -> H2(g) + MgCl2(aq) + Heat}$$

From a conceptual standpoint, shouldn't the change in entropy be
positive for this reaction because there are more moles of gas in
the products vs the reactants? Why is this not the case? Can this
be justified through a conceptual view of the problem rather than
a pure mathematical approach(simply calculating delta s as
negative)?
From a conceptual standpoint, why is the change in entropy negative for this reaction?

Screenshot of reference data (source):

What's your source for the entropy data? @ShuryuKisuke

http://chemistry-reference.com/reaction.asp?rxnnum=697

Can you show me the work for the entropy calculations because when I asked my Chemistry teacher about it she also said it was negative?

For starters, you should be suspicious when the reference page is formatted as it is.
The real issue I see is that the page lists $\Delta S_{f}$. This is where you should immediately call BS.
We don't use this quantity because entropy is not an energy, therefore, you cannot set the standard state entropy to 0. That means that entropy of formation is pretty useless because you need to know the actual entropy of the standard state of components to know what the entropy of the species is. For elements, the standard molar entropy is positive. However, for solutions (and this is what I got wrong in my now-deleted comment), the standard molar entropy can be negative thanks to solvation (specifically, solvation shells of water can be highly "ordered").
In order to compute the entropy of this reaction, you need to know the standard molar entropy of each species. The easiest thing to do is to look it up in a trustworthy reference source, for example a NIST database or CRC Handbook of Chemistry and Physics.
$$
\begin{array}{|c||c|}
\hline\hline
\text{substance} & S^{\circ} (\mathrm{J}\ \mathrm{mol}^{-1}\ \mathrm{K}^{-1}) \\
\hline
\ce{H2(g)} & 130.68 \\
\ce{Mg(s)} & 32.67 \\
\ce{HCl(aq, 1 M)} & 56.6 \\
\ce{MgCl2(aq, 1 M)} & -25.1 \\
\hline
\end{array}
$$
Sources: hydrogen, magnesium, hydrochloric acid CRC handbook under "Thermodynamic Properties of Aqueous Ions", magensium chloride CRC handbook under "Thermodynamic Properties of Aqueous Ions"
These are the same values in your reference, but I had trouble believing those data. So, basically, the answer is correct. This reaction has negative entropy change.
Why? Because solutions are kind of evil. Solubility and other solution phenomena can be really hard to predict because they depend so much on how the solvent interacts with the solute. In this case, it's probably driven by some idiosyncratic size of the chloride anion, which somehow allows water molecules to form very nice solvation spheres around the anion. I'm rationalizing this because the standard molar entropy for solutions of magnesium iodide (84.5) and magnesium bromide (26.8) are all positive. These ions are probably of a size that prevents the water molecular from fitting together nicely into a single shell. The imperfections in the solvation shell increase entropy.
That means that the large entropy of hydrogen gas is balanced by some stupid solvent effect with negative entropy, so the overall change in entropy is negative.

Still is difficult to relate the effect to solvation of chloride ions as for formally nothing change between the two sides of the reaction.  It should be a fine balance involving Mg++ too. Puzzling. Isn't? And, again formally, we attain two "entities" starting  from three, which can also contribute at least in the specific case.

@Alchimista I did actually contemplate this last night when I wrote out the answer. My guess is that hydration of the proton is very favorable from an entropic standpoint. I probably shouldn't have put away my CRC handbook in the most hard to reach place in the basement after using it last night... Anyway, without looking it up, I would guess that the ability to easily shift a proton around in a hydrogen bonded network of water might actually provide the high entropy I'm looking for. Let me know if you have additional thoughts. This is the most interesting question I've answered in a while.

Zhe's answer is good and I could have added a comment but think it is worth a separate answer to ellaborate on his. It's true that it is difficult to develop an intuition around entropy changes associated with solutions and gases, because their entropy is a strong function of concentration.
The solutions in this case contain dissociated ionic species, respectively $\ce{2HCl}$ (dissociated into $\ce{2H+}$ and $\ce{2Cl-}$) and $\ce{MgCl2}$ (dissociated into $\ce{Mg^{2+}}$ and $\ce{2 Cl-}$), so the effective change in moles of ions per mole of reaction is -1. This counters the entropically favorable release of $\ce{H2}$ gas.
In addition, the data in the table refers to formation of 1 M solutions or gas at STP (1 bar, 298.15 K). But what if we'd worked with different concentrations or pressures?
Say the concentrations are 0.1 M. Then the entropy of reagents would be increased more than that of the products, by $$ \mathrm{\Delta S = Rlog(10)} =  \pu{19.14 J/molK}$$ so that the total change in entropy change would be $ \pu{-59.31 J/molK}$.
Instead if the concentrations were 5.5 M (saturated $\ce{MgCl2}$), then assuming again ideality $$ \mathrm{\Delta S = Rlog\left(\frac{1}{5.5}\right)} =  \pu{-14.17 J/molK}$$ so that the total change in entropy change would be $ \pu{-26.00 J/molK}$.
If the product $\ce{H2}$ gas had been very dilute, say 0.01 bar, then the entropy of the gas would have been greater by $$ \mathrm{\Delta S = Rlog(100)} =  \pu{38.29 J/molK}$$ and the reaction entropy would be  $ \pu{-1.88 J/molK}$, a much smaller change. Another order of magnitude reduction in the pressure of the gas and the entropy change would be favorable by $ \pu{17.26 J/molK}$.

A possibly useful analysis is to compute the enthalpy change and think about how much the entropic change actually even matters at ambient temperatures. Put it another way, what's the dominate contribution to the favorable free energy change, entropic or enthalpic?


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Organic chemistry, hybridization concept
Hybridization of phenyl cation and venyl[sic] vinyl cation and acetylene cation ,I am totally confused , I think that the hybridization of both phenyl and the venyl cation is SP , I think that in both cases the positive charge remains at pz orbital , but my tuition teacher said me that the hybridization of phenyl cation is SP2 but the hybridization of venyl cation is SP now please give me a proper explanation of hybridization of those cations

This may help: keep in mind what shape a phenyl ring has. Your $sp$ hybridization would correspond to one of those carbons in the hexagon having an 180-degree angle instead of a 120-degree angle.

Putting it another way, in the formula cosθ=(p−1)/p ,where p stands for % p character and θ stands for the bond angle, see what amount of approximate p-character you are obtaining if you put θ as 120°(which is the bond angle of phenyl cation) and then try to see what type of s-p mixing would justify this result. Remember, hybridization is just a mathematical construct,and it can be moulded so as to justify a certain geometry

@YUSUFHASAN , please tell the orbital in which the +ve charge resides in every cation

A positive charge isn't like a lone pair which would reside in a particular orbital; it is the absence of electrons in orbitals that gives a positive charge

So please say the vacant orbitals of the concern carbocations

It is becoming customary to write literally nonsensical sentence. Unless you resumed the question in your own way. There is not hybridisation of pheyl cation etc. Unless hybridisation refers to another level of specificity. This made clear, tgen the other comments gives you the way of thinking.

https://chemistry.stackexchange.com/questions/10509/why-are-vinylic-and-arylic-carbocations-highly-unstable https://chemistry.stackexchange.com/questions/38093/do-vinyl-cations-adopt-a-classical-or-non-classical-structure

https://chemistry.stackexchange.com/questions/49648/is-the-phenyl-cation-or-ethynylium-more-stable


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Separating NiCl2 and CoCl2
My undergraduate lab is currently doing a project in which we have to separate inorganic compounds from organic compounds, identify each compound, and quantify how much of each compound there is. We've identified that the inorganic compounds are NiCl2 and CoCl2, and quantified them using Beer's Law. However, our project specifications say we need to come up with two ways of quantifying them, and we're stumped on finding a second way. Is there a reaction that can possibly titrate one of the compounds without having the other compound interfere, or perhaps separate them and then weigh them using a gravimetric approach? The problem is that the two compounds seem to have the same solubility properties so we don't quite know how to proceed. Thanks!

Read about atomic absorption spectrophotometry; Also read this classical method....https://pubs.acs.org/doi/abs/10.1021/ac60094a033?src=recsys&journalCode=ancham

Using dimethylglyoxime (DMG) to precipitate nickle is a classic. At one time the pipe dream was to find reagents that would specific for each element, or a very limited sets of elements.


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Explain why body-centered tetragonal is a Bravais Lattice, but based-centered tetragonal is not?
Can you explain it using schematics? I think the reason is the symmetry that doesn't allowed to unique lattice points.

Let's assume you stacked two face-centered (base-centered) tetragonal lattices. At the side where they meet the atom, on the center of a surface, turns into an atom that is in the center of the combined volume of the two cells (just draw two face-centered tetragonal cells). That means by stacking you also get atoms inside the volume and not only on it's surface (face-centered). And in this case a much smaller body-centered tetragonal cell can be constructed from this arrangement.

@Justanotherchemist You seem to have proved that face-centered and base-centered cells are the same (which they are not), and moreover, that they all are impossible. I'm not ready to agree with this.

Well if been calculating this through with all possible orientations right now. In the cubic case, wouldn't the body centered lattice become tetragonal and thus have a lower symmetry? Because no matter how I draw it on side will be 'a' and the other two will be smaller as they are diagonal. While for the tetragonal face-centered case the resulting body-centered lattice remains tetragonal in symmetry.

Face-centered case (that is, with positions in the centers of all faces) is not even mentioned in the question, so I see no reason to bring it up. Base-centered tetragonal cell will be equivalent to a smaller primitive cell which is still tetragonal. And of course the body-centered cubic cell is equivalent to some primitive cell which is not cubic. So you are essentially right, but I'd rather put it in somewhat different words. Lattice, for one, does not care about our choice of unit cell; lattice symmetry does not depend on it.

@ivan sounds like an answer to me.

Related: https://chemistry.stackexchange.com/questions/47834/why-can-a-face-centered-cubic-lattice-not-be-redrawn-as-a-body-centered-tetragon/47868#47868

Related: https://www.quora.com/Why-is-there-is-no-base-centered-cubic


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Packing fractions of HCP and FCC
The packing fractions of hexagonal close packed unit cell and face centred cubic unit cell come out to be same when calculated geometrically.
Is it just a coincidence? Differently put, could it be predicted that the packing fractions of HCP and FCC will be same without calculating them geometrically?

Look at them hexagonal layers. They are the same in both structures. Now stack one layer on top of the other. This is also done in exactly the same way in both structures. How could the volumes be any different?

@IvanNeretin Can you explain more? From what I understand, HCP has AB type packing, and FCC has ABC type packing, and hence are different. Or are they ultimately the same thing?

True, they are different, but in a way that's of no importance to us when we consider density. Look again at the layers: they are identical. Look at the way one layer is placed on top of the other, and consider the distances between layers: they are identical. What else do we need? Nothing.

The C layer is just the A layer placed differently and causes no difference in the density. Is that right?

Yes, but that's not quite my point. Take just two adjacent layers out of the structure. Look at them as hard as you can. What are  they? Maybe A and B? Or B and C? Or C and A? Where are they from, HCP or FCC? You won't be able to tell. There is simply no way of knowing. Any pair is precisely identical to any other. The distance between the layers is the same, and hence so is the packing density.

Thanks a lot, I understand it completely now.


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XAS spectroscopy - Differences in magnitude of 2p SOC and 3d SOC?
Within L-edge x-ray absorption spectroscopy, the spin-orbit coupling within the 2p shell results in the splitting of the L edge into L3 and L2 edges.As given in the following paper http://dx.doi.org/10.1063/1.4896373, the 2p SOC is 8ev whereas the 3d SOC is 0.05eV. Could some please explain why the 3d SOC constant is much smaller than that for 2p SOC?

The energy of a spin orbit level for a '1 electron' or 'hydrogenic' atom such as sodium is given by
$$E_{so}=\frac{AZ^4}{n^3(l(l+1)(l+1/2))}\frac{j(j+1-l(l+1)-s(s+1)}{2}$$
where $A$ is a collection of constants and $j,\,l,\,s$ quantum numbers derived from the term symbols for the levels involved.
The second fraction in the equation is the angular momentum term, the first due to $\langle1/r^3 \rangle$ where $r$ is the distance from the nucleus.
For a D electron you might have $j$ quantum numbers $j$=5/2 and 3/2 with $l$=2 and $s$=1/2 and for a P,  $j$=3/2 and 1/2 with $l$=1 and $s$=1/2 thus the separation of the D level pair is less than that of the P.
Most good textbooks will have a derivation of this equation and other corrections such as due to relativity etc.


text:
What are the limitations of a waters of hydration experiment?
When a hydrated salt such as $\ce{CuSO4.5H2O}$ is subjected to heat, an anhydrous salt ($\ce{CuSO4}$) is produced. However the theoretical molar ratio of $\ce{H2O:CuSO4}$ and experimental molar ratio of $\ce{H2O:CuSO4}$ are not the same. Instead there is a percentage error, why does this occur? And how can we minimise this in this experiment?
The hydrated salt was left on hot plate at 70 degrees celsius for 25 mins.

Why, all experiments contain errors. If you want to know more, you have to tell us more. What is your experimental molar ratio, to begin with?

My experimental molar ratio of H2O: CuSO4 is 4:1

Then perhaps your experiment resulted in a monohydrate, rather than an anhydrous salt?

The hydrate contained 4 moles of water for every 1 mole of CuSO4. The expected was 5 moles of water for every 1 mole of CuSO4. It was not a monohydrate as there was more than 1 mole of water per mole of compound.

That's not what I was saying. See, what is it that's left after the heating? You seem to believe this is anhydrous CuSO4. What if this is not so?

Are you saying that maybe the hydrated salt was not heated enough, therefore not all water turned into steam therefore we were left with CuSO4.H2O + 4H2O (gas) instead of the intended CuSO4 + 5H2O (gas)

Yes, that seems plausible.

How was the amount of water evaporated measured?

You can mention experimental details and how you achieved 1:4 ratio, which may explained if we know the maximum temperature of heating.  CuSO4.5H2O loses water of hydration in steps as a function of temperature showing that those five water molecules are not equivalent. In this figure from German Wiki, follow the green line (ignore blue). The y-axis is % loss in weight and x-axis is the temperature. It clearly shows that water is lost in steps. The labels read Mass change (loss of crystal water). If your heating temperature is not enough you are left with CuSO4.H2O.

Also I would mention that it might be difficult to keep the sample anhydrous both in the • sense and in a more trivial one. As the answer explain this matter is really linked to experimental details not to mention even humidity in the lab.


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Protons on aromatic rings in NMR
Recently, I have started to learn about nuclear magnetic resonance (NMR) in school and something which I cannot seem to reconcile is the fact that all aromatic protons on any substituted benzene ring would give the same chemical shift. This is very, very strange... Clearly, the aromatic protons are in different chemical environments due to the fact that they are different distances away from the substituents from the benzene ring. I understand that the strong delocalisation in the aromatic ring may tend to "even out" the electron distribution across all the aromatic $\ce {C-H}$ bonds. However, I would not think that this effect would result in all the aromatic protons having the same chemical shift. Could anyone propose a explanation to this peculiarity?
The following image was taken from p. 279 of Clayden, Warren & Greeves (2012):

It also seems to suggest the equivalence of aromatic protons despite them being of different physical distance from the substituent. Here's another one from p. 278:

If strong delocalisation and aromaticity really is the effect responsible for this apparent chemical equivalence of the aromatic protons, then would we also observe such equivalent signals in other aromatic heterocycles or even in polyaromatic hydrocarbons?
Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

They don't have the same cs.

That’s not a fact, it’s totally untrue.

This is totally contrary to experiment and theory. Please provide concrete examples.

@orthocresol Please refer to the image in the updated post.

@NightWriter Please see the update.

@NightWriter I am critically doubting these examples too. I believe that they are extremely misleading and are likely wrong. My school teacher's best rationalisation is that the aromatic multiplets "clump" together on the spectra, giving a single "big multiplet" signal. However, that is not a very convincing reply.

That just means they're quite close to each other and cannot be adequately resolved when taking into account the multiplet shape (which can be very complicated for molecules like these).

Even if that spectrum was acquired at a low field (say 100 MHz) the brown window is ~30 Hz wide, which is ~4 x 8 Hz, or 4 x value of J-coupling between vicinal (ortho) protons on the ring. In fact, you expect two triplets with max splitting in each triplet of ~16 Hz (keeping things simple). In other words, you have overlap of multiplets from different protons separated by ~10 Hz. Assuming 100 MHz.

After editing is much clearer. It is just a problem of resolution rather than an absolute fact. And this is indeed exacerbated by the protons being all aromatic, in the sense that they should be expected at similar cs, plus eventually symmetry at works as usual, but this last is trivial. All comments, including that of the teacher (he said they clump not the same) should answer your question.

This is true that all the aromatic protons come near the region having $\delta$: $\pu{7\!-\! 8 ppm}$. But you can always easily find examples where all the aromatic protons doesn't have the same chemical shift (which definitely include your examples). In fact the cases where all the aromatic protons have exactly similar chemical shift is hardly found.
The main idea is the difference between the terms Chemical Equivalence & Magnetic Equivalence, which is often misinterpreted. In the first example you have given, all the protons don't have the same chemical shifts and this is seen from the multiplet pattern as well. What you can obviously predict is that the two ortho protons are chemically equivalent, and the two meta protons are also chemically equivalent, and the para proton is a separate different type of proton. For observing these differences, what you need to do is to record the NMR at high frequency (e.g., $500$ or $\pu{600 MHz}$). There you can see the three different peaks clearly. Now in this case only, the two ortho protons are also magnetically equivalent as they couple with all the other protons equally, i.e. the $^3\!J_{H-H}$ and $^4\!J_{H-H}$  values are same for the two ortho protons, but in general this may not be the case at all (which is simply to consider another different para substitution at the ring, then the two ortho protons will become magnetically non-equivalent and therefore will have different chemical shift which is observable). Similar is the case with meta potons i.e. they are also magnetically equivalent. So those two protons will have same chemical shift. But ortho, meta, and para protons will definitely have different chemical shifts in your first example. If you zoom in or record the spectra in a high frequency machine, you should see a doublet of doublet peak (due to $^3\!J_{H-H}$ and $^4\!J_{H-H}$) for both ortho and meta protons, and a triplet of triplet for the para proton.
In the second example also, the two protons ortho to $\ce{-N(CH3)2}$ are chemically as well as magnetically equivalent as they both couple with the remaining proton in the same way. So, applying the similar logic you should see a doublet peak (due to $^4\!J_{H-H}$ only) for those two protons and a triplet peak for the remaining one and these two peaks should be at two different chemical shifts.
To encourage you more here are some spectra from the book Introduction to Spectroscopy which emphasize on aromatic regions and consider different spin-spin splitting and show how complicate a structure can be:

The above spectra depicts the differences in the spectra between ortho, meta, and para substitution in the benzene ring. The below two are some typical examples of complicated spectra due to spin-spin splitting as well as magnetic non-equivalence.

From these examples you can see that the aromatic protons can definitely come at distinct regions and what you need is to just record them in high field NMR machine and zoom into that region.

I don't agree that the ortho protons in the OPs first example are magnetically equivalent, since they have different couplings to other protons on the ring, but they duplicate the coupling pattern and long distance couplings are negligible. In the second case I do agree, with the caveat that any coupling between the methyl group and the aromatic protons is very small.

I think that the key here is to understand that stereo-electronic effects are limited because of the ring.

Consider the case where a hydrogen atom is 2 bonds away of another atom $\ce{X}$ like in $\ce{H–C(=X) \!–}$ or $\ce{H2C–X\! –}$ (or something like that). There will be a huge difference in chemical shift between $\ce{X=C}$ and $\ce{X=O}$, typically $\pu{2\!-\! 3 ppm}$.
Then consider the case where there are 3 bonds between a hydrogen atom and $\ce{X}$: The effect on the chemical shift of $\ce{H}$ will be much less because it decreases exponentially with the distance/number of bonds.
Now let's consider the structure you gave "from p. 278" (but it is true in every case): the minimum number of bonds you can have between a proton attached to the benzene ring and its closest neighbor (N) on the ring is 3 bonds, i.e. $\ce{H–C–C–N}$. You cannot compare this case with another case like $\ce{H–C–N}$ (only 2 bonds apart).

Overall, the main factor for the chemical shift of aromatic protons is the aromatic ring itself with a base chemical shift of $\pu{7.27 ppm}$. Some powerful substituents can modify electronic distribution in the benzene ring but many of them don't so aromatic protons usually appear very near to there base chemical shift of $\pu{7.27 ppm}$.


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Lanthanoid Contraction
Why is the radius of Europium so unusually high and out of the general trend ? Moreover, at different sources I am getting different values of radii. In some (e.g., Wikipedia) the radii are following the regular trend and in some Europium has been shown to be bigger. (Source : NCERT Class 12 Chemistry Part-1)

What do the other sources say?

Well spotted! I dare say, the book appears to have used a covalent radius for Europium, while using atomic radii for the other lanthanides.
See "atomic radius" vs "covalent radius" values: for Europium, Samarium, and Gadolinium.
Also see this article if you don't like my referencing Wikipedia :P

For what it's worth, a similar trend is given in Housecroft and Sharpe's Inorganic Chemistry, 4th ed. In there, both Eu and Yb have anomalously large "atomic radii" (199 and 194 pm respectively), which could be related to this: https://chemistry.stackexchange.com/questions/53960/why-are-the-densities-of-europium-and-ytterbium-anomalously-low I assume that these radii relate to the interatomic separation in the metal.


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Calculate pKa for lipids
I would like to calculate $\mathrm{p}K_\mathrm{a}$ of lipid molecules using some software tools. I have tried ChemAxon and compared with ChemDraw professional 16.0 version they are mostly similar. I think both are in gaseous phase.
Do I have any chance to calculate the $\mathrm{p}K_\mathrm{a}$'s in aqueous phase using solvent parameters? Any software tool suggestion are highly appreciated.

The pKas are not gas phase, they are in aqueous solution, but they are for the individual lipid, and predicted based on similar molecules.

What kinds of lipids are we talking about here? The set of functional groups is pretty limited, and you can probably extract the pKa from the most significant functional group from an acidicity/basicity standpoint, at least to first order.

Alishetty, did this answer your question?

See H++ if you want a rough prediction of $\mathrm{p}K_\mathrm{a}$.
If you want to more accurately calculate the $\mathrm{p}K_\mathrm{a}$ of your molecules, you should run a quantum mechanics calculation with a continuum solvent model (example calculation: [1]).
The general outline is as follows. Run a geometry optimization and vibrational mode analysis on your system in basic and conjugate acidic forms. So, $\ce{BH+}$ and $\ce{B}$, or $\ce{BH}$ and $\ce{B-}$, depending on what your system looks like. This will allow you to extract a free energy of reaction for each of your lipid molecules.
Importantly, these calculations will not give accurate single point free energies, but the relative values ($∆G$) will be. As such, you will need a reference $∆G$ from literature of a similar reaction to "set the 0" of your values. Finally, you can get $\mathrm{p}K_\mathrm{a}$ from $∆G$ using
$$\mathrm{p}K_\mathrm{a} = \frac{∆G^\circ}{2.303RT}$$
Refrences

Liptak, M. D.; Shields, G. C. Accurate $\mathrm{p}K_\mathrm{a}$ Calculations for Carboxylic Acids Using Complete Basis Set and Gaussian-n Models Combined with CPCM Continuum Solvation Methods. Journal of the American Chemical Society 2001, 123 (30), 7314–7319. https://doi.org/10.1021/ja010534f.