Are these alternate names for long chain hydrocarbons with complex substituents accepted? Let's consider the following compound:

$\ce{CH3-CH2-CH(CH3)-CH2-CH2-CH(CH(CH3)-CH(CH3)-CH2-CH(CH3)3)}$

Its IUPAC name is 2,2,7-trimethyl-4-(1-methyl-propyl)-nonane

$\ce{CH3-CH2-CH(C2H5)-C(CH3)(C(CH3)3)-CH2-CH2-CH(CH3)-CH2-CH2-CH3}$

Its IUPAC name is 3-ethyl-4,7-dimethyl-4-(1,1-dimethyl-ethyl)-decane

I wish to ask if the following names are acceptable (respectively):

4-(1-methyl-propyl)-2,2,7-trimethyl-nonane
4,7-dimethyl-4-(1,1-dimethyl-ethyl)-3-ethyl-decane

Do complex substances follow a rule regarding where they should be placed in the IUPAC name or can they be placed anywhere?
Do prefixes like bis, tris, tetrakis, etc. need to be considered while naming?

Okay so here is the situation: You have two subsitutions, methyl and 1-methylpropyl as the two substituents here. You wish to know in which order the substituents must be written in the name. You knew that substituents are arranged alphabetically, and both methyl and 1-methylpropyl start with the same letter m.

Consider a dictionary. It has over thousands of words that start with the letter C. But we can still arrange the words in order by looking at the second, third, fourth, fifth, etc. letters of the names. Similarly, if the first letters are the same, look at the second, third and fourth letters.

Here's where we arrive at the conclusion. methylpropyl will be ranked below methyl as per dictionary order. Thus the name of the compound would, in fact, be 2,2,7-trimethyl-4-(1-methyl-propyl)-nonane and the other name is not accepted.

I believe you can continue from here to name the second compound.

Important notes:

The inclusion of di, tri in the name must only be applied for complex subsituents such as in the case of 5-(dimethylethyl)decane. In cases such as 4-ethyl-5,5-dimethyldecane, do not use d as the starting letter. This is becase d is not the part of the name of the subsitituent (methyl) and only indicates how many times this subsitituent occurs in the compound.
Do not add hyphens between methyl and decane such as "methyl-decane". Use "methyldecane" Hyphens must only be used between numbers and subsituents for which the number represents its position in the name.


Which of the following is the most basic compound? Which of the following is the most basic compound?

$\ce{HS-}$
$\ce{H3Si-}$
$\ce{H2P-}$
$\ce{Br-}$

I can't figure out which of these is the strongest base. Would it be $\ce{H3Si-}$, since $\ce{Si}$ is the least electronegative of all these central elements?

I agree with Doctor Insult's argument that "the stronger the base, weaker is the conjugate acid." In that sense, higher the $\mathrm{p}K_\mathrm{a}$ of acid, weaker is the acid, and hence, stronger is the conjugate base. The $\mathrm{p}K_\mathrm{a1}$ values of $\ce{HBr}$, $\ce{H2S}$, $\ce{H3P}$, and $\ce{H4Si}$ are -8.7, 6.89, 27, and ~35, respectively (Ref.1). Therefore, the order of the strength of their conjugate bases should be:
$$\ce{H3Si- \gt H2P- \gt HS- \gt Br-}$$
Hence, strongest base is $\ce{H3Si-}$.

Note: Ironically, this is the same decreasing order of electronegativity of center atoms $\ce{Si, P, S, Br}$: $1.90 \lt 2.19 \lt 2.58 \lt 2.96$, respectively.

References:

William L. Jolly, In Modern Inorganic Chemistry; 1st Edn.; McGraw-Hill, Inc.: New York, NY, 1984, p 177.


Other molecules like fullerenes? When I first heard about buckminsterfullerene, I thought it was a very fascinating molecule, with its interesting complex ring structure and pseudo-symmetry.  I wonder though, can you get similar structures with other atoms?  Maybe neighbors of similar size, like boron, nitrogen or oxygen?  Or maybe silicon, one row down?   What would their chemistry be like?

Cool question! Always stay curious!

A while ago, I had a similar question that started it's journey with silicon, being of similar bonding capability to carbon. However, over the course of my research, I realized that silicon was just too low on the periodic table (and therefore much heavier) to be so extremely stable in such a wide variety of forms such as fullerenes and long, complex hydrocarbons.

However, my search was not in vain. I stumbled upon Boron Nitride, an isoelectronic analog to carbon graphite. Many different compounds have been synthesized from boron nitride, consisting of a wide array of fullerene analogs, such as Boron Nitride Nanotubes, or BNNTs, which by themselves have plenty of applications. For example, paper products made from these nanotubes are highly flame retardant, as is shown in the image below:

The trick is that $\ce{B-N}$ has very similar properties to $\ce{C=C}$, with very little difference in molecular weight, and so many analogous compounds can be built using either monomer. However, boron nitride is different enough to have unique applications in certain areas. Cool, right?


What are the products of the reaction between fluorine and ammonia? What are the products of the reaction between fluorine and ammonia? Is it the same as with chlorine and ammonia, which depend upon which reactant is taken in excess?

Fluorine reacts with ammonia to give dinitrogen and hydrogen fluoride as the major product.

The Formation of Dinitrogen Tetrafluoride in the Reaction of Fluorine and Ammonia - J. Am. Chem. Soc. 1959, 81 (23), 6338-6339 - states that dinitrogentetrafluoride accounts for 6% of the yield while the major reaction is:

  $\ce{3F2 + 2NH3 → N2 + 6HF}$

The same article quotes in its first paragraph that the result remained the same irrespective of the concentration of the reactants. (In the case of chlorine it matters which reactant is taken in excess)


How to assess sunscreen effective via blueprinting? How do I assess the effectiveness of sunscreen creams by using the blueprinting process? I know blueprinting is done when something is sensitive to light and this may cause reactions to produce a product that causes Prussian Blue to appear. However I am unsure how to connect sunscreen and blueprinting - does it have to do anything with the colour in relation to its wavelengths and how it is connected to UV light?

From Wikipedia page on the blueprint process:

  The blueprint process is based on a photosensitive ferric compound. The best known is a process using ammonium ferric citrate and potassium ferricyanide. The paper is impregnated with a solution of ammonium ferric citrate and dried. When the paper is illuminated, a photoreaction turns the trivalent ferric iron into divalent ferrous iron. The image is then developed using a solution of potassium ferricyanide forming insoluble ferroferricyanide (Prussian blue or Turnbell's blue) with the divalent iron. Excess ammonium ferric citrate and potassium ferricyanide are then washed away. The process is also known as cyanotype.

Exposing blue print paper to UV light will cause it to develop blue. Sunscreen is meant to block UV light so applying sunscreen over blueprint paper then exposing to UV light and developing will give you a semi-quantitative way to evaluate the sunscreen based on the intensity of the blue.

See: Colorimetry


How do people check the type of defect in a crystal? There are many types of defects like

Vacancy
Interstitial
Schottky
Frenkel
Non-Stoichiometric
Alien Impurities

But given a sample, how do scientists know which kind of defect is present in it [practically]?

As you said there are many different kind of defects in crystals. There are basically two different methods to find the Crystallographic defect.

Mathematical classification methods
Computer simulation methods

Mathematical classification methods:

A successful mathematical classification method for physical lattice defects, which works not only with the theory of dislocations and other defects in crystals but also, e.g., for disclinations in liquid crystals and for excitations in superfluid $\ce{^{3}He}$, is the topological homotopy theory.

Computer simulation methods:

Density-functional theory, classical molecular dynamics and kinetic Monte Carlo simulations are widely used to study the properties of defects in solids with computer simulations. Simulating jamming of hard spheres of different sizes and/or in containers with non-commeasurable sizes using the Lubachevsky-Stillinger algorithm can be an effective technique for demonstrating some types of crystallographic defects.

Source: Wikipedia


Does entropy depends on temperature? Does the entropy of a system depends on the system's temperature?

I know that $\Delta S_{ext}=\frac{q_{rev}}{T}$ ($q_{rev}$ is the heat in a transformation if it is reversible).

My textbook says that the entropy of the exterior always depends on that formula (so it depends on the temperature).

But it says the entropy doesn't always depends on the formula, and we can only say the that in a transformation the entropy of a system can increase or decrease. So does it depend on the temperature?

Thanks!

Entropy is a physical property of a material that, for a single phase substance of constant composition, can be expressed as a function of temperature and one other intensive physical property such as pressure or specific volume.  The equation that you wrote down is a relationship that we can use to determine the change in entropy of a substance from one thermodynamic equilibrium state to another, by devising a reversible process from the first state to the second state, and calculating the value of the integral of dq/T for that process.  All reversible processes between these same two states will give exactly the same value for the integral.


Prove that internal energy is a state function Why is it that heat and work are path functions but their sum i.e. internal energy is a state function? How is it even possible?
$ΔU=q+W$

The three laws of thermodynamics (or four, depending on the source) are postulates that scientists have assembled in an attempt to describe the way the universe works. The First Law is a reformulation of the principle of conservation of energy. It says very simply that while energy is indestructible it can be exchanged in various ways. It also implies that a system can exist in different states, each with a set of clearly identifiable properties including a fixed amount of energy. Declaring the state a system is in also stipulates the energy of the system. We can write this mathematically as $U=f(\mathrm{state})$.
One also presumes that if the energy of a system changes, then energy has been transmitted in some measurable form by an identifiable mechanism. In other words, for some process you can always write
$$\Delta U = \sum_x \mathrm{energy~in~transit~via~mechanism~}x$$
where the sum goes over all the ways we imagine energy might be transferred.
Classical thermodynamics suggests partitioning the energy in transit into two basic categories, work and heat. Heat is transferred when objects are at different temperatures (this forms the basis for the Zeroth Law of thermodynamics). If you insulate a system, heat is no longer a viable way to transfer energy. That leaves all other possible ways of transferring energy, which we call work. Alternately, if all identifiable means to do work are inaccessible, then the system can exchange energy only as heat.
The point is that if the initial and final states are known then $\Delta U$ is fixed, but there might be different ways (paths) that change the energy of the system by that amount. For instance, work can be used to change the temperature of a substance just as putting the substance in contact with an external heat bath can.


Influence of protic and aprotic solvents on SN reactions How does protic and aprotic solvents influence reaction mechanism considering SN1 vs SN2?

As you know, $\mathrm{S_N1}$ and $\mathrm{S_N2}$ reactions have different mechanisms. $\mathrm{S_N1}$ occurs in 2 steps: formation of a carbocation and addition of the nucleophile. However, $\mathrm{S_N2}$ has a one-step mechanism where the addition of the nucleophile and the elimination of the leaving group happen simultaneously.

A protic solvent would participate in hydrogen bonding with the nucleophile since it can easily share one proton. A solvent shell is thus formed around the nucleophile, making him less able to undergo a $\mathrm{S_N2}$ reaction.


Why is the carbonyl IR frequency for a carboxylic acid lower than that of a ketone whilst an ester is higher than a ketone The approximate stretching frequencies are as follows:

  Acid: $1710$, Ketone: $1715$, Ester: $1730$ (all $\mathrm{cm^{-1}}$)

This would seem to suggest that the acid decreases the carbonyl bond strength and the ester weakens it. This must be as a result of the fact that in the ester inductive withdrawal is more effective than resonance and the converse is true in the acid. However, I am finding it hard to reasonably justify why. Presumably this is also linked to the relative electrophilicity of the ester and the acid.

First things first, it's important to realise that functional groups may have IR frequencies over quite a broad range. I've provided the table below, from which you can see that actually there is a great deal of overlap depending upon the actual 'environment' of the functional group in question. Ketones and acids, for example, cover a very similar portion of the spectrum.

  Stretching frequencies of common carbonyl groups (Taken from Introduction to Spectroscopy, Pavia and Lampman.

That said, the difference between the frequency range of an acid vs an ester is an interesting one, as on the surface of it they are electronically quite similar (in both cases the carbonyl is next to an electronegative oxygen).

  @RobChem: This would seem to suggest that the acid decreases the carbonyl bond strength and the ester weakens it. This must be as a result of the fact that in the ester inductive withdrawal is more effective than resonance and the converse is true in the acid.

In your statement, you may have gotten a little confused, remember that the stretching frequency increases with bond strength. Therefore using your data we can say that the ester is stronger than the ketone whilst the acid is weaker.

We can explain the ester being stronger by considering the effect of the electronegative oxygen withdrawing electron density:

This argument however, fails to explain why the carboxylic acid has a stretching frequency below that of the ketone, since, as was already said, they both have the oxygen adjacent to the carbonyl.

One rationalisation for this is that the carboxylic acid group doesn't exist in isolation but rather interacts with other carboxylic acids in a hydrogen bonding interaction, which weakens the C=O bond:

This hydrogen bonding concept is universal, for instance in methyl salicylate, the C=O bond has dropped to well below that expected for an ester:


Salt analysis using spectroscopy A few days ago, I learned the lengthy process of salt analysis during my practicals. The process is based on deductive reasoning but is too lengthy.

What I am thinking is, can't it be done using the fact that all molecules have unique emission spectra? I searched for it and found that it's the same thing as the flame test which was covered in my class.

But a flame test is only used for metal ions. So, my question is: Is there a way in which I can determine the metal ion as well as the radical part(anion part) of the salt using spectroscopic methods? What equipment would I need for it?

Two of the methods are:

Atomic Absorption Spectroscopy - similar to the flame test, but is usually quantitative. The sample is heated (in a flame or furnace) and the absorbance at a defined wavelength is measured against standardized solutions. It will work for any atom if you use a tunable laser.

ICP/MS (Inductively Coupled Plasma/Mass Spectrometry) - a solution of the compound is ionized and passed into a mass spectrometer. Some atoms (especially halogens) cannot be properly ionized and passed into the mass spectrometer. Oxygen and nitrogen (and a few others) would be difficult to quantify since it may come from impurities in the sample.


What are the major environmental drawbacks with using methane as a combustable fuel? What are the biggest environmental issues with using methane, $\ce{CH4}$, (which is found as the major energy source in natural gas, shale gas, biogas, and synthetic natural gas), as a combustible fuel?

Methane has many lovely properties. It's a very dense source of energy. It is plentiful, it seemed to be cheap to extract (until we worked out the real cost), and similarly seemed to be cheap to transport long distances. It can be stored for many months in vast quantities.

But.

It's a greenhouse gas - a powerful one at that. On a 100-year horizon, it has about 20 times the global warming potential of $\ce{CO2}$. That needn't be a problem, if natural-gas extraction had no leakages, and distribution systems had no leakages.  But they do have leakages, which means the supply chain worsens global warming.

Furthermore, when methane is burnt, $\ce{CO2}$ is released into the atmosphere, worsening global warming even further.

We've learnt to handle methane's other main problems: that it's colourless and odourless, so can just accumulate unnoticed waiting to explode, if you let it; and that it's also dangerous if handled carelessly in any other way.

It also generates $NO_x$ when burnt in air, and they are local as well as global pollutants. It is possible to have flue-scrubbers to remove the $NO_x$.

That doesn't mean that that methane is permanently bad.  It can be created in a carbon-neutral fashion (from biological sources, or synthesised from atmospheric  $\ce{CO2}$), and if leakages are minimised, offers a potential useful form of energy storage. (note that in these forms, it would be an energy & storage vector, not a net energy source, so its economic value would be very different - but that's economics, not chemistry)


Organic chemistry textbook for self-learning? I am interested in learning organic chemistry in order to solidify my knowledge of molecular biology.

However, there are a lot of options for organic chemistry textbooks, and I plan on simply sampling a few in order to figure out which ones are best, but I thought I would also ask for some advice.

Since I am self-learning, I have some specific items on my textbook wishlist:

The textbook should be insightful - I don't simply want facts presented, I want to see how the facts connect together into a larger whole. Usually, teachers fulfill this task, but in this case, I will need some help from the textbook.
The textbook should have a good section on learning IUPAC nomenclature.
I don't want organic chemistry to be an experience in advanced memorization for me. The previous point helps in addressing this issue, but I would like to reiterate it. I would like to have a textbook that helps me understand why each class of reactions is important (what can one do, that the other can't?) in a chemist (or an organism's) toolbox - I don't want to be memorizing reactions, I want to be putting them into my toolbox.

With such requirements in mind, which textbooks would you recommend?

I'll provide a rundown of the general organic chemistry textbooks I've studied in detail:

Organic Chemistry, David R. Klein: A very solid introductory text. Emphasis is on expediency and development of intuition over rigorous theoretical correctness. The coverage of molecular orbital theory is extremely scant, and classic valence bond theory (with associated concept of orbital hybridization) is the major explanatory paradigm. This is probably the most accessible text for students with a minimal background, but it's also somewhat superficial.
Organic Chemistry, Jonathan Clayden, et al.: This is a very thorough text, with a slightly wider array of reactions discussed than Klein's textbook (most notably in the coverage of organometallic reagents). Again, the MO treatment is lacking, but it's invoked more frequently than in Klein. This book tends to delve quite deeply into the mechanisms underlying specific reactions, making it rather lengthy. There's also some superficial discussion of HSAB theory, which is mostly (if not completely) absent in Klein's book.
Advanced Organic Chemistry, Carey & Sundberg. This is something of a seminal classic, but it's not for the faint of heart or for those uninitiated in the fundamentals. Mechanisms are elucidated in great detail (primarily via MO theory, with results from both ab initio and semi-empirical methods), and the exposition of pericyclic reactions is notably excellent. It also contains a great deal of illuminating content on conformational analysis. This is definitely not appropriate for a first textbook, but I would consider it essential reading for advanced undergrads and above.

Ultimately, either of Klein's or Clayden's textbooks should satisfy all of your criteria, but I strongly recommend Carey & Sundberg for subsequent study, if possible.

Beyond those, I've found Fleming's Molecular Orbitals and Organic Chemical Reactions to be a very good introduction to qualitative MO theory, but it's not a general-purpose text. March's Advanced Organic Chemistry is the ultimate, all-encompassing reference manual, and I'd consider it the sine qua non of any good organic chemistry library. As an aside, McQuarrie and Atkins have both published very good physical chemistry textbooks, which in many ways have been more enlightening than any textbook specific to organic chemistry. I'm also a fan of Oxtoby's Principles of Modern Chemistry as a general chemistry textbook; its mathematical and physical rigor is incredibly admirable, even if it does sometimes lack severely in concision. I've also heard unanimous praise for Housecroft & Sharpe's Inorganic Chemistry, but I've yet to read more than a few sections of it.


Elimination reaction with 1,2-dibromo-4-methylcyclohexane I was doing a question from a practice exercise on the topic of elimination reactions and alkene synthesis. It is shown below:

I predicted that the product would be 1,2-diiodo-4-methylcyclohexane because I thought the reaction conditions favoured substition with iodide via the $\ce {S_N2}$ mechanism. Firstly, iodide is a rather good nucleophile and secondly, the substrate is a secondary alkyl halide. Even if it was not $\ce {S_N2}$, I would expect perhaps $\ce {S_N1}$ since iodide is a rather weak base and the conditions do not seem to favour elimination. However, the answer provided is very interestingly, 4-methylcyclohexene. Would someone kindly enlighten me on where I went wrong?

Dehalogenation of vicinal dihalides by iodide ion occurs stereospecifically and
  concertedly via an $\mathrm{E2}$ mechanism. An anti-coplanar arrangement of both bromine atoms is required (see diagram below):

Since, the given compound is a 1,2-trans-dibromocyclohexane, two bromo groups  can arrange in 1,2-anti-diaxial positions in one of chair confirmations, fulfilling the requirement. It won't be possible if it'd have been a 1,2-cis-dibromocyclohexane.

Reference: Structure & Synthesis of Alkenes (Chapter 7): Leroy G. Wade, Organic Chemistry; 8th Edn, Pearson Education Limited: Harlow, Essex, UK, 2014.


Enthalpy of Formation problems using Hess's Law How do I solve problems where I'm asked for $\Delta H^\circ_f$ and given $\Delta H^\circ_f$ values for equations with the involved elements? For example:

  Find the $\Delta H^\circ_f$ of the following reaction:
  $\ce{C + O2 -> CO2}$, given:

  $\ce{SrO + CO2 -> SrCO3}$ ; $\Delta H^\circ_f = \pu{-234 kJ}$

  $\ce{2SrO -> 2Sr + O2}$ ; $\Delta H^\circ_f =  \pu{1184 kJ}$

  $\ce{2SrCO3 -> 2Sr + 2C + 3O2}$ ; $\Delta H^\circ_f = \pu{2440 kJ}$

I know that I'm somehow supposed to manipulate the equations to make them "add up" (adding straight down) to the target equation, but I'm not sure how, exactly, to achieve this.

Also, is there a different and/or better method for doing these?

I've made some progress on my own, using this About.com page. What I'm wondering now, is if elements are locked into compounds, can I split them up, as long as I keep them on the same side? In the example above, I've manipulated it to where I have $\ce{2CO2 + O2 -> 2C + 3O2}$; am I allowed to split the $\ce{2CO2}$ into $\ce{2C + 2O2}$?

I've made it as far as getting to $\ce{2C + 2O2 -> 2CO2}$. If you take these as molar ratios, it's the same as $\ce{C + O2 -> CO2}$, which is the target equation. What do I have to do to the enthalpies to go from the first to the second?

Find the $\Delta H^\circ_f$ of the following reaction:
$$\ce{C + O2 -> CO2}$$

We have our list of reactions with known enthalpies:
$\ce{SrO + CO2 -> SrCO3} \quad \Delta H^\circ_f = \pu{−234 kJ}$
$\ce{2SrO -> 2Sr + O2} \quad \Delta H^\circ_f = \pu{1184 kJ} $
$\ce{2SrCO3 -> 2Sr + 2C + 3O2} \quad \Delta H^\circ_f = \pu{2440 kJ} $
Hess's Law takes advantage of the fact that enthalpy is a state function, looking at our reactions, we need to find a way to derive what we want:
Carbon dioxide is on the reactants side, however, we want it on the products, so we can run that reaction reverse; however we need to change the sign of enthalpy if we do that:
$\ce{SrCO3 -> SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$
We need to get rid of $\ce{SrO}$, thus, we use our second equation:
$\ce{2SrO -> 2Sr + O2 } \quad \Delta H^\circ_f = \pu{1184 kJ} $
Since, there's only one strontium oxide in the first reaction, we divide by the integer 2 across the chemical equation including the enthalpy of formation:
$\ce{SrO -> Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $
Now, with equation 3, we need carbon and oxygen on the reactants side, and need to remove $\ce{SrCO3}$ and also divide again by 2, since we only have one strontium.
$\ce{Sr + C + 3/2O2 -> SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $
Now we add up all our reactions and enthalpies, and cancel out common elements on both sides:
$\ce{SrCO3 -> SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$
$\ce{SrO -> Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $
$\ce{Sr + C + 3/2O2 -> SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $

$\ce{C + O2 -> CO2} \quad \Delta H^\circ_f= (\pu{234 kJ}) + (\pu{592 kJ}) + (\pu{-1220 kJ}) = \pu{-394 kJ}$


What exactly is the 'shuttle effect' in sulfur batteries? Ref :

https://www.electropages.com/blog/2020/01/why-lithium-sulphur-batteries-are-taking-so-long-be-used-commercially

  Because they continually shuttle between the electrodes, they never change back into a neutral state (which the surface polysulphide ions do)

I don't think this explains the 'shuttle effect' clearly. I still don't understand what it is.

During discharging, lithium ions and sulfur form Li2S8 at the cathode, then it becomes Li2S4, Li2S2, Li2S with continued discharging. During charging, the Li2S move towards the cathode, now, why can't it be reversely reduced to Li2S4, Li2S8. Why can't they ever change back into a neutral state?

OP's title indicated OP is looking for the real mechanism of  the Shuttle Effect in $\ce{Li-S}$ batteries. Yet, OP is asking different questions in the text body. I'd try to address as much as possible in this answer.

A most recent review article (Ref.1) states that:

  Lithium-sulfur ($\ce{Li-S}$) batteries are one of the most promising batteries in the future due to its high theoretical specific capacity ($\pu{1675 mAh g−1}$) and energy density ($\pu{2600 Wh kg−1}$). However, the severe capacity fading caused by shuttle effect of polysulfide needs to be addressed before the practical application of $\ce{Li-S}$ batteries.

To understand what is the Shuttle Effect, OP need to first understand the working chemical mechanism of $\ce{Li-S}$ batteries, and can take a head stat in that direction with this article about Lithium–Sulfur Battery, which gives a fair description of the chemistry involve in the battery. The same Wikipedia article simply described the Shuttle Effect as follows:

  The lithium polysulfide $\ce{Li2S}_x$ ($6 < x < 8$) is highly soluble in the common electrolytes used for $\ce{Li-S}$  batteries. They are formed and leaked from the cathode and they diffuse to the anode, where they are reduced to short-chain polysulfide and diffuse back to the cathode where long-chain polysulfide are formed again. This process results in the continuous leakage of active material from the cathode, lithium corrosion, low coulombic efficiency, and low battery life.

The formation of polysulfide shuttle effect is classified into five steps in Ref.1:

  Formation of long-chain polysulfide,
  Detaching of polysulfide from sulfur host,
  Dissolution of polysulfide into electrolyte,
  Migration of polysulfide toward lithium anode side, and
  Reaction between polysulfide and lithium anode.

If interested, OP may read Ref.1 and Ref.2, which thoroughly discussed the mechanism and possibilities of how to prevent the Shuttle Effect in depth. For example, one such techniques is modifying the separator, which has been done successfully recently (Ref.3):

References:

Wenchen Ren, Wei Ma, Shufen Zhang, Bingtao Tang, "Recent advances in shuttle effect inhibition for lithium sulfur batteries," Energy Storage Materials 2019, 23, 707-732 (https://doi.org/10.1016/j.ensm.2019.02.022).
Chao Deng, Zhuowen Wang, Shengping Wang, Jingxian Yu, "Inhibition of polysulfide diffusion in lithium–sulfur batteries: mechanism and improvement strategies," J. Mater. Chem. A 2019, 7(20), 12381-12413 (https://doi.org/10.1039/C9TA00535H).
Yanpeng Fan, Zhihui Niu, Fei Zhang, Rui Zhang, Yu Zhao, Guang Lu, "Suppressing the Shuttle Effect in Lithium–Sulfur Batteries by a UiO-66-Modified Polypropylene Separator," ACS Omega 2019, 4(6), 10328-10335 (https://doi.org/10.1021/acsomega.9b00884).


How to measure UV index? I am a high school student so please apologise if I say something very dumb or irrelevant. I am conducting an experiment in which I want to find UV index at different points. Such as under shade, inside house, near beach etc. Anything will be greatly appreciated, also even if there is a device thats very expensive please tell me about that as well.

Finding 'UV index' at a point might be difficult. UV index is an index created by I think was WHO, and is calculated using
 - Thickness of ozone above
 - Cloud density of the area
 - Altitude of location
 - Sun's height
more about it here - https://www.who.int/uv/publications/en/UVIGuide.pdf
So I think you are better off finding the UV light intensity instead, using some measurement tools. There's few different options like photo reactive dyes, and digital scales, but probably the most cheap and quantitative data can be obtained using arduino or other similar development board and creating a small module yourself. Good Luck!


Logic behind this statement (On Transcription) There's this little paragraph in my textbook, parts of which I find redundant. My teacher is absolutely fine with this ("Aaron, you've got to write down what's in the textbook if you want to score in exams, nothing extra...."), but all the same.....

It's regarding 'Why does transcription occur at one DNA strand at a time?'

According to their first point, forming two complementary RNA strands would result in the production of two different proteins for the same character.

Now according to their second point, formation of two complementary strands of RNA side-by-side would just result in them pairing to form dsRNA, which would be transcriptionally inactive and therefore wouldn't take part in the translation process.

The thing is, they've mentioned the second point after the first point, which I find confusing. For if the second point were correct, the first point wouldn't even exist.

Is there a sound reason for the points being stated this way? Am I missing something here?

This textbook would never have made it into my classroom. The second "answer" is an arguably correct concept but is specious as a reason. The first "point" has a silly justification.

The real reason is an evolutionary one. In polypeptides, particular sequences of amino acids fold in ways that provide function. Tiny variations in this sequence can render the resultant polypeptide non-functional. With this sort of sensitivity what are the odds that the non-coding strand, whose sequence is dictated by the coding strand, could actually possess a viable gene? Furthermore, genes evolve over time. Any changes in the sequence of one would necessitate a change in the other. What are the odds of that being efficatious?

That said, I believe there are some cases where a gene on one strand has some overlap with a gene on the sister strand. I doubt these are lengthy or common, possibly occuring with the help of introns, portions of the DNA message that are "spliced out" of the final RNA transcript.


Optical activity of cis coordination compounds I came across a statement saying,

  "Consider a Octahedral coordination compound of type [MA2X2]. Then,
  if the compound is in a cis form, it shows optical activity as there
  is no plane of symmetry and hence is non-superimposable on its mirror
  image. On the other hand, a trans form is symmetric and superimposable
  on its mirror image and so is optically inactive."

I do not see why the cis form is not superimposable on its mirror image. If all the ligands lie on the same plane, then isn't there a line of symmetry passing diagonally? Does that not make it superimposable on its mirror image?

It is not always as easy to see from a standard octahedral depiction. Optical activity in octahedral complexes can be understood better, if you draw the octahedron as looking onto a triangular face rather than along an edge; this results in a hexagonal arrangement of ligands around the central atom with ligands alternately pointing into and out of the paper plane.

If we do that with $\ce{[Co(en)2Cl2]}$, we get the following two enantiomers which you cannot convert into each other:

The complex is chiral because the $\ce{en}$ ligands necessarily cross the paper plane and can do so either in a right-turn screw motion or in a left-turn screw motion.


Approximate time for pure water pH to equilibrate in open air? I'm a complete non-chemist, so I don't know how to calculate this,  or even what to google for (my searches yielded esoteric research papers).
I'm doing some hobby experiments, and I'd  like to know what approximate timeframe to expect for relatively pure (distilled or double distilled) water left in open air at around 24 °C/75F to reach whatever the pH equilibrium point is (I read around 5.5 somewhere).

I'm not looking for a giant equation please,  just a quick approximate number.
Thanks

Assume you have poured your distilled water in a 1 liter container.

We write the flux of $n_{\ce{CO2}}$ between air towards your container, as
$$
J_{\ce{CO2}}=\frac{dn_{\ce{CO2}}}{dt}\times\frac{1}{A}=\frac{d[\ce{H2CO3}]^{*}}{dt}\times\frac{V}{A}
$$
where $n_{\ce{CO2}}$ is the number of ${\ce{CO2}}$ moles in water, $A$ the water surface and $V$ the water volume.
Assuming $\ce{CO2}$ atmospheric pressure equal to $10^{-3.5}\,atm$, we obtain the concentration $[\ce{CO2}]_{air}$ approximately equal to $10^{-5}\,M$ (the latter value obtained via $PV=nRT$).

We call $[\ce{H2CO3}]^{*}$ the sum of $[{\ce{CO2}}]_{water}$ and $[\ce{H2CO3}]_{water}$ concentrations in water: at the beginning, we set it equal to $0$. Note that $[\ce{H2CO3}]^{*}$ could be reasonably written as:

$$
[\ce{H2CO3}]^{*}\approx[{\ce{CO2}}]_{water}
$$

indicating that the majority of carbon dioxide in water is not converted into carbonic acid, being the kinetics for such a conversion very slow, as pointed out by @Nicolau Saker Neto, .

By expressing $J_{\ce{CO2}}$ as:

$$
J_{\ce{CO2}}=\frac{D_{{\ce{CO2}}}}{Z}\times ([\ce{CO2}]_{air}-[\ce{H2CO3}]^{*})
$$

with $D_{\ce{CO2}}$ the diffusion coefficient of ${\ce{CO2}}$ equal to $7.2\times10^{-4}\,dm^{2}h^{-1}$and $Z$ the thickness of superficial layer through which the exchange occurs. For the latter, we take $40\times10^{-5}\,dm$.

Putting previous equations together (with $A/V=1\,dm^{-1}$), the following is obtained:

$$
\frac{d[\ce{H2CO3}]^{*}}{dt}=1.8\times10^{-5}-1.8\times[\ce{H2CO3}]^{*}
$$
which yields:

$$
[\ce{H2CO3}]^{*}=1\times10^{-5}(1-e^{-1.8t})
$$

The graph below shows that about $2$ hours later, a plateau is established, for $[\ce{H2CO3}]^{*}= 1\times10^{-5} M$.

By considering the acid-base equilibrium ($pK_{a}=6.3$):
$$
\text{H}_{2}\text{C}\text{O}_{3}^{*} \ce{<=> HCO3- + H+}
$$
we obtain a $\text{pH}=5.7$, at the plateau.

UPDATE 1: the value for $Z$ has of course a big impact on time. A larger value will increase the time needed to reach the plateau. I have used $40\times10^{-5}\,dm$, considered a "typical value" by the authors of this book (sorry in French, page 181, 3rd edition). Any more accurately derived value for $Z$ is welcome !

UPDATE 2: I used the notation $\text{H}_{2}\text{C}\text{O}_{3}^{*}$ to represent the two species $\ce{CO2}$ and $\ce{H2CO3}$, in water.


Chondroitin Sulfate (greater volume in solution than in crystal form) I've read in a textbook on biophysics that chondroitin sulfate occupies a greater volume in a solution than in crystalline form. Why is that?

Is it because of hydrogen bonds in solution that the volume is larger?

$\hspace{4.5cm}$

In solution, the negative charges on the chondroitin sulfate repel each other and force the molecule into an extended conformation. The polar molecule attracts many water molecules, which increases its molecular volume. In the dehydrated solid, each negative charge is counterbalanced by a positive ion, and the molecule is condensed.


How to get the coordinates or .cif file of β  and χ ? Can someone tell me how to draw the β  and χ  in Materials Studio or any other software? I've just started my project work, and I'm a beginner in this field.
I have searched for the coordinates or .cif file in the literature, but I could not find any. If anyone can provide me with those, it'll also be helpful for me.

Jiang et al. 2018 (DOI: 10.1016/j.jechem.2018.01.026) may be relevant literature? Here's a quote:

These include three allotropes named borophene, beta-12(β12) and chi-3(χ3), which
have been successfully synthesized by physical methods [26,27].
[26] A. J. Mannix, X. F. Zhou, B. Kiraly, J. D. Wood, D. Alducin, B. D. Myers, X. L. Liu, B. L. Fisher, U. Santiago, J. R. Guest, M. J. Yacaman, A. Ponce, A. R. Oganv, M. C. Hersam, N. P. Guisinger, Science 350 (2015) 1513–1516.
[27] B. J. Feng. J. Zhang, Q. Zhong, W. B. Li, S. Li, H, Li, P. Cheng, S. Meng, L. Chen, K. H. Wu, Nat. Chem. 8 (2016) 563-568.

I also found this one:
Izadi Vishkayi, S., Bagheri Tagani, M. Edge-Dependent Electronic and Magnetic Characteristics of Freestanding β 12-Borophene Nanoribbons. Nano-Micro Lett. 10, 14 (2018). DOI: 10.1007/s40820-017-0167-z
Perhaps you may find the link to the .cif from all of the above?


How can compounds be called reduced if nothing is known about the reaction they formed in? I'm reading a research article about a model trying to explain the elemental compositions of meteorites.[1]

What confuses me, are statements of the following type (here in chapter 2.1, second paragraph about enstatite chondrites, a type of meteor):

  [...] Enstatite condrites contain minerals that are remarkably reduced chemically, including abundant sulfides like oldhamite (CaS).

In my understanding reduction or oxidization only refers to a specific element in a specific reaction, but here they talk about a compound  with unknown reaction path to form it.

How can CaS be considered to be (remarkably) reduced, if it is not an element in a reaction?
How can CaS be considered to be (remarkably) reduced, if we don't know the reaction path that it formed in?

Steven J. Desch, Anusha Kalyaan, Conel M. O'D. Alexander. The Effect of Jupiter's Formation on the Distribution of Refractory Elements and Inclusions in Meteorites. arXiv:1710.03809 [astro-ph.EP]

While we don’t know exactly how minerals on meteorites were formed, we can make very educated guesses on the conditions they were formed in. Most notably, outer space only contains very low concentrations of the compound most associated with oxidation on Earth: gaseous oxygen.

But this is not the main reason why the label ‘remarkably reduced’ is accurate here. We will immediately recognise the mineral in question as an ionic compound and given that it is made up of a metal and a nonmetal, we can equally immediately assign oxidation states as shown below.

$$\ce{\overset{+II}{Ca}\overset{-II}{S}}$$

There is no surprise about calcium which generally is found in its $\mathrm{+II}$ oxidation state. The surprise here is that sulphide is the anion which is a rare occurance. On Earth, most mineralic sulphur is oxidised and mainly to sulphate ($\mathrm{+VI}$). This is basically the extreme opposite and caused among others by the high atmospheric oxygen concentration.

Of course, the label ‘remarkably reduced’ can only be applied to some of the elements of the compound. Language-wise, we are looking at the oxidation state and merely stating that a state this low is uncommon.


Imiscible fluids for demonstration I have been considering building a fluid dynamics demonstration.  Basically, it would be a thin plexiglass box, with some turbulence inducing elements attached to the back face, and a pump to move the fluid around.

Can anybody recommend a pair of fluids that will:

not mix/dissolve permanently
can be dyed different colors, while respecting (1)
won't react with plexiglas (or, alternatively, polycarbonate)
isn't toxic or particularly flammable
can be pumped with a "very" low power pump at temperatures suitable for humans (say, 60-100F)

The point of the demonstration is to create a "sustainable" chaotic system.  I might even point an HD camera at it, and extract random numbers from the video stream (maybe).

Have a look at this paper

Summarizing:

put food coloring to 35% isopropyl alcohol in water.
add ammonium sulfate

What happens ? Once ammonium sulfate added and the solution mixed, the ammonium sulfate dissolves, pushing the isopropyl alcohol out of the aqueous solution (and the food coloring as well).

A separate layer thus forms, that is clearly visible thanks to the food coloring.

The weak point of the whole story is that isopropyl alcohol is flammable...

PS: I would have preferred to reply to @Michiel's answer and @Nicolau Saker Neto's comment: but I do not have enough reputation to do that...


Stability of F- Cl- Br- I-? In aqueous state If we talk about hydrolysis F- will be more heavily hydrolysed so it should be more stable in aqueous state.
But I know that HI is a stronger acid than HF. So the stability of conjugate acid ie I- should be more than F-. As dissociation takes place in water.

If we consider $\ce{HX}$ as the common formula for hydrogen halogenides, then there is an equilibrium
$$\ce{HX(aq) + H2O <=> X-(aq) + H3O+(aq)}$$
that can be expressed also as:
$$\ce{acid1 + base2 <=> base1 + acid2}$$
The equilibrium is competition of 2 acids, $\ce{HX}$ ansd $\ce{H3O+}$. $\ce{H3O+}$ is the strongest acid stable in aquaeous environment. Acids we consider as strong acids ($\ce{HCl}$, $\ce{H2SO4}$, $\ce{HNO3}$ ) are much stronger acids than $\ce{H3O+}$. Therefore for $\ce{HCl}$, $\ce{HBr}$, $\ce{HI}$, the equilibrium is practicaly fully pushed toward right:
$$\ce{HX(aq) + H2O -> X-(aq) + H3O+(aq)}$$
On the other hand, $\ce{HF(aq)}$ is a weak acid, weaker than $\ce{H3O+}$, with dissociation constant $\mathrm{p}K_\ce{a}=3.17$, for which the equilibrium is rather on the left side:
$$\ce{HF(aq) + H2O <<=> F-(aq) + H3O+(aq)}$$
Highly concentrated $\ce{HF(aq)}$ is rather a strong acid:
$$\ce{2 HF(aq) + H2O <=>> HF2-(aq) + H3O+(aq)}$$
and liquid HF is very strong acid ( H0 = −11 ) :
$$\ce{3 HF <=>> HF2- + H2F+}$$


Relative stability of cis and trans cycloalkenes Why is the stability of (Z)-cyclodecene 2 greater than that of (E)-cyclodecene 1?

I'd think that since 1 contains a trans double bond instead of a cis double bond, it should experience less steric repulsion and hence be more stable.

However, I've read that the cis isomer 2 is more stable than the trans isomer 1 if the ring size is 11 or smaller. For ring sizes larger than 11, the opposite is true.

Why is this so? Also, what would happen if there were more double bonds?

Usually trans-olefins are more stable than their cis isomers for steric reasons, like you suggested.  However in small and medium size rings this is not the case; here the cis-cycloalkene is more stable than the corresponding  trans isomer.

trans-Cyclooctene is the smallest trans-cycloalkene that is stable at room temperature (trans-cyclohexene and trans-cycloheptene have been detected as short-lived intermediates).  In your drawing up above of trans-cyclodecene (a) draw in the two hydrogens on the trans double bond.  One of them is "inside" the ring and causes steric problems with the carbons on the other side of the ring as pictured below.  Because of this destabilizing transannular interaction trans-cycloalkenes with up to around 11 carbons do not exist as planar molecules.  They exist in a conformation as pictured below for trans-cyclooctene.

(image source)

If you can, build a model of trans-cyclooctene.  Notice how you have to pull and stretch the methylene chain in order to attach that last methylene back onto the other end of the double bond - there is a lot of strain in this molecule. Now build a model of cis-cyclooctene; it is easy to loop the methylene chain from one end of the double bond to the other end.

Not until you get to around trans-cyclododecene (12 carbons) do you have enough carbons to easily connect from one end of the trans double bond back to the other end.  trans-Cycloalkenes with less than 12 carbons have additional strain because of this difficulty in spanning the double bond. Once you get to around 12 carbons there is no difficulty spanning the double bond and so once again trans-cycloalkenes become more stable than their cis counterparts.

Since double bond carbons are $\mathrm{sp^2}$ hybridized with 120° angles, if we add more double bonds, the 120° angles make it more difficult to loop the more flexible $\mathrm{sp^3}$ methylene (with their 109.5° angles) chain around the double bonds.


Optical activity of (1r,3R,5S)-3,5-dimethyl-4-methylidenecyclohexan-1-ol Could anyone explain why compound L is not optically active?

The given molecule,(1r,3R,5S)-3,5-dimethyl-4-methylenecyclohexanol , is optically inactive .This molecule has a plane of symmetry (as shown below) and, therefore, is achiral.


Balancing ionic equations and determining number of ions There were two questions that I've listed below that I've had troubles on. I got an answer in both, but both of them weren't correct.

For the first question, they list the answer to be D and then for the second they list the answer as A.

  Determine the net ionic equation for the reaction between $\ce{HNO3}$ and $\ce{Mg(OH)2}$.
   a. $\ce{2H+(aq) + (OH)2^{2-}(aq) -> H2(OH)2(aq) }$

  b. $\ce{2HNO3(aq) + Mg(OH)2(aq) -> Mg(NO3)2(aq) + 2H2O(l) }$

  c. $\ce{Mg2+(aq) + 2NO3 (aq) -> Mg(NO3)2(s) }$

  d. $\ce{H+(aq) + OH(aq) -> H2O(l) }$

  e. $\ce{Mg+(aq) + NO3 (aq) -> MgNO3(s) }$
  Determine the number of ions that are present in $38.8~\mathrm{mL}$ of $0.113~\mathrm{M}$ $\ce{(NH4)2CO3. }$

  a. $7.92\cdot 10^{21}~\text{ions }$

  b. $2.64\cdot 10^{21}~\text{ions }$

  c. $2.07\cdot 10^{23}~\text{ions }$

  d. $6.20\cdot 10^{23}~\text{ions }$

  e. $1.32\cdot 10^{23}~\text{ions }$

For the first question, the work that I did was $\ce{2 H+ + 2NO3- +Mg(OH)2 -> 2H2O + Mg+2 + 2NO3-}$. The problem that I have in this problem is why the $\ce{Mg(OH)2}$ splits up. I don't get it because the things you split up are the things that are aqueous or soluble. $\ce{Mg(OH)2}$ isn't soluble so I listed it as a solid and thought that you wouldn't break it down. All I need to know is why I have to break the $\ce{Mg}$ and the $\ce{OH2}$ down because it isn't soluble.

For the second, I got an answer of $2.64\cdot 10^{21}~\text{ions}$ but the review said it was A. I set it up as $.113~\mathrm{M} = \text{moles} / 0.0388~\mathrm{L}$ and then multiplied $.113$ by $0.0388$ and then multiplied that amount by $6.022\cdot 10^{23}~\text{ions}$ which got me the answer B.

Actually $Mg(OH)_2$ has (low) solubility ($K_{sp}=1.5\cdot 10^{-11}$). Since it is asking for the net ionic equation, the only ions reacting (since $NO_3^-$ stays in solution) are $H^+$ and $OH^-$.

You forgot that each molecule of ammonium carbonate produces 3 ions, so multiply your answer by 3.


A quick method for separating liquids? I'm looking for a way to take 3 separate liquids (liquids which function as paint -- they will all be different colors), put them together into a container, and then rapidly separate them back to distinct colors.

I can handle all the electrical engineering aspects of my project, but this portion is beyond me. Does anyone have any ideas?

When I say "a quick method", I mean something that can be done easily without an excess of time or equipment. Ideally an way to make the paints intrinsically immiscible, but I need to be able to separate them entirely, so that when they are poured together into a container, I can then separate and reuse them.

A stable 3-phase liquid system can be created, as Spießbürger mentioned with three mutually immiscible liquids. The best one that I know of is the organic-aqueous-fluorous 3-phase system.

Fluorocarbons (hydrocarbons in which every hydrogen atom is replaced by a fluorine atom), are interestingly immiscible in both water and common organic solvents. This property has added a dimension to chemical separations using fluorous phase extractions.

If you have three paints, one oil-based, one water-based, and one fluorous-based, then you could separate them easily using a separatory funnel or similar device. You would have to make sure that the pigments in each paint was only significantly soluble in that solvent, which would take experimentation.


How to trace an energy versus time diagram? I'm looking for data to create a simple graph of an exothermic or endothermic reaction but can't find any.
Could someone help me with some enough data to create a simple graph displaying energy as function of time.

Assuming you want to plot heat exchanged by reaction versus time, you need at least to know what reaction you want to study. Then you will have to assess two different quantities: enthalpy if the process is isobaric (thermodynamic) and reaction rate (kinetic). This kind of data can be found on NIST Database.

If you do not find reaction enthalpy $\Delta_{R} H$, you can assess it by using enthalpies of formation:

$$ \Delta_{R} H = \sum\limits_i \nu_i h_{\mathrm{f},i} $$

Where $i$ indexes reactants and products, $\nu_i$ are stoichiometric coefficients (note: they are negative for reactant) and $h_{\mathrm{f},i}$ are enthalpies of formation.

Then you need to find out which is the kinetic $v$ of your reaction and solve the related differential equation. If you do not find any information you might try the activated complex model:

$$ v = \frac{1}{V}\frac{\mathrm{d}\xi}{\mathrm{d}t} = k(T)\prod_j x_j^{\alpha_j} $$

Where $\xi$ is the molar coordinate of your reaction (De Donder variable), $k(T)$ the kinetic constant, $x_j$ are reactant concentrations and $\alpha_j$ are partial orders of reaction (they can be taken equals to $|\nu_j|$). Solving this differential equation - for a given set of initial conditions - will provide you a relation, maybe explicit, between $\xi$ and $t$.

Merging both step results will lead to enthalpy-time expression which can be plotted. Of course those calculation might be perform numerically.

If you were looking for an hand-made and pedagogic Energy Plot like this, you have not searched so far.

Method exposed above does not create this kind of plot which shows internal energy of sum of compounds versus coordinate of a molecular transformation. Gap between the initial and final state is the energy relaxed/absorbed by the reaction (often pure heat). Such plot are arbitrarily plotted because the activated state is often unknown and definitely non measurable. Furthermore everything between initial and final state is also not-well-known because we just have access on macroscopic data so we ignore microscopic details. Last hint: activation energy (maximum) is always higher than the higher state, you need extra energy to initiate the transformation.


Is the reaction thermodynamically favored or not? $\ce{CO(g) + 2H2(g) +O2(g) <=> 2CH3OH(g)}$,

$\Delta H=-128\ \mathrm{kJ/mol}$
$\Delta S=-409.2\ \mathrm{J/(mol \cdot K)}$

Determine if the reaction if thermodynamically favored.

I just wanted someone to see if my logic is right. Since there are 4 moles on the reactants side and and 2 on the products side the reaction would not be thermodynamically favored because the reaction decreases the amount of entities, which decreases the entropy.

Your reasoning is correct but there is more to it.

Look at it from the point of view of Gibbs Free Energy,

$\Delta G = \Delta H - T \Delta S$

For a given reaction if:

$\Delta G >0$, the reaction is nonspontaneous in the forward direction, not thermodynamically favourable

$\Delta G < 0 $, the reaction is spontaneous in the forward direction, thermodynamically favourable

If we substitute the values for $\Delta H$ and $\Delta S$:

$\Delta G=(-128\ \mathrm{kJ/mol}) - T\left(-0.409\ \mathrm{kJ/(mol \cdot K)}\right)$

Mathematically, $\Delta G$ will become positive only when $T$ is greater than 313K. Because of this,  the reaction is thermodynamically favorable at any temperature less than 313K (since $\Delta G$ will be negative), but is not thermodynamically favorable at any temperature greater than 313K (since $\Delta G$ will be positive).


Hybridization of heteroatom Why is the hybridization of oxygen in the seven membered ring pictured below $sp^3$, while the oxygen has a hybridization of $sp^2$ in the 5 membered ring? This also may be an error in the answer key as well.

It relates to the Hückel rule for aromatics. With the electron pair of the oxygen participating in the $\pi$ system of the oxazol (5-membered ring)  we get 6 $\pi$-$\ce{e-}$. Doing the same for the oxazepin would get 8 $\pi$-$\ce{e-}$.

Hückels rule states that planar, circular $\pi$ systems with $4n+2$ electrons are aromatic, resulting in a lower overall energy if the system is conjugated. (with $n$ being a positive integer or zero)
The other statement that Hückels rule does is that $\pi$ systems with $4n$ electrons are anti-aromatic, which avoid conjugation as it would increase the overall energy of the system.

Therefore it is favoured for your oxazol to conjugate, which results in a sp$^2$-hybridization for the oyxgen, while the oyxgen in the oxazepin avoids conjugation, resulting in a sp$^3$-hybridization.

The Hückel rule in turn can be quantized by a Hückel method computation or  graphically utilizing a Frost-Musulin-diagram.


Does the main washing effect of soap come from foam? I am curious if the main cleaning effect of soap comes from its foam (or, maybe, it is just a by-product).
Here are some  points, that might clarify what I mean:

Does foam sort of "carry" dirt and germs away from the surface?
If it is not the "main motor", which makes the soap function: is it possible to make a soap which will still be as effective without it producing any foam?
What percent of the cleaning work, if any, does foam undertakes (i.e. how effective would soap be against germs and dirt without its foaming ability)?

Foam is a side effect of using a tensio-active agents (though some tensio-active agents are engineered to produce as little foam as possible, e.g. in washing machines). In fact, you could consider that, if soap has nothing better to do, it will form foam. This answers some of your questions: foam does not (really) carry away dirt and its percentage of efficiency in the whole process is low.

Molecules of soap are made of 2 parts: a hydrophobic tail (which loves fat, grease, lipids, whatever you call them) and a hydrophilic head which loves water, and hate lipids.

When the hydrophobic tails are busy linking to a greasy stain or grease on your hands, they no longer participate into forming bubbles/foam.

The easiest way to verify that is to pour a little bit of oil in your hands and rub them. Wash your hands with soap a first time: there will be almost no foam. Now rinse your hands and wash them again: there should be much more foam!

Note that it does not mean that your hands are cleaner, regarding bacteria, but just that they are less greasy!

If you want to remove bacteria, rub your hands with soap for at least 30 seconds. So-called "anti-bacterial" soaps do not achieve better and promote bacterial resistance.


Simplest distillation setup I would like to start learning practical chemistry. I have not done chemistry before and am seeking help assembling a simple distillation setup.

I do not want to use the cold water tubes etc for the time being, I am trying to keep it as simple as possible.

I have a few questions which I am hoping someone can answer:

What are the plastic clips for?
How do I vacuum seal all the joints as simply as possible (preferably without using grease). Would tape work?
How do I seal the small side arms? (In the liebig condenser tube this isn't a problem because they don't lead to the inner distillation tube.)

That's a nice bit of kit to get started. I've annotated the picture a little to aid discussion:

I've labelled the "male" and "female" joints that should be connected with the same letter (e.g. B inserts into B'). The thermometer goes in the top of the blue plastic screw joint.

Joint A serves the purpose of connecting your system to a vacuum source. If you are doing vacuum distillation, you should connect the vacuum tubing at this point. However, if you are just distilling at atmospheric pressure you should leave this open. You should NOT seal this joint, as this would mean there is no pressure relief in the system - when you start to heat your sample, gases expand and they need somewhere to go, otherwise you will at best pop a joint, and at worst cause an explosion.

As you say, the joints on the liebig condenser (G) can be left open since they aren't open to the interior of the apparatus. However, you will get better results if you do hook up the condenser to water using the tubing provided, especially if you are distilling at relatively low temperatures.

The clips are there to secure ground glass joints so they don't fall out. You only really need them on junctions that arent held in place by gravity (e.g. you need one on D/D' since the joint could fall apart under gravity, but not on B/B' since gravity holds the male joint in place).

You shouldn't need to vacuum seal your joints - ground glass quick-fit joints should be manufactured to a reasonably high tolerance so the leakage is minimal. It would only become a problem if you needed very strong vacuuum.

Note: if you are going to use vacuum for your distillation you should carefully inspect every piece of glassware for cracks, imperfections etc. - if the glassware is damaged then when you apply the vacuum it can implode. You should always wear eye protection when doing any chemistry at all.

Hope this helps, please ask questions if you aren't sure of anything - people on here will be happy to help a new recruit to develop their chemistry skills!


Why did a former emulsion of oil and water precipitate after heating and cooling when heat made it originally? I added 100 g coconut oil, 100 g distilled water and 500 mg lecithin to an 1800 watt blender.  The motor from the blender has has a passive heating action.  I blended it at full for four minutes.  The result seemed to be a hot, uniform substance with a thin layer of foam on top.

I poured this mix into a smaller container and placed it in the fridge.  After a while it set into a solid, uniform substance with no apparent precipitation.

I then heated the substance in the microwave  for thirty seconds.  When I removed it, the components had separated and precipitated.  Cooling it in the fridge again resulted in a delicate solid sheet at the top which broke through to liquid underneath.

Why did the mix precipitate (presumably de-emulsify) after being microwaved but not occur when blended?

The simple explanation is that the speed of de-emulsification depends strongly on physical state and temperature.

If you have created a solid emulsion by cooling the initial blend, the de-emulsification will be very slow as there is little scope for the components to move around at all never mind separate. warming this solid up so it liquefies creates two factors that make separation easier: the mixture liquefies and the temperature rises, creating many easier ways for separation to occur. Once separation has occurred only re-blending will restore the emulsion.

Emulsions often separate slowly (milk, for example, slowly forms a layer of creamier fat on top but it doesn't happen quickly after thorough mixing). In your case the initial cooling "froze" the emulsion but the microwave heating enabled de-emulsification to occur by heating the mix. Cooling the result without blending is not going to restore the emulsion without further blending.


Strength of oxoacids of Chlorine I have seen various different contradicting statements about oxoacids of chlorine regarding oxidising strength & reducing strength of these.

I understand the trend for acidity that HClO4 is strongest and HClO is weakest , but what about oxidisind and reducing character,

Some day that HClO4 is strong reducing as well as strongest oxidising agent as well,

I understand that it will perform reduction of itself and oxidise other element but what about reducing strength?

When you are considering redox reactions, you should get familiar with electrochemical series (Ref.1). Oxidizing and reducing characters of certain reagent also depends on the conditions  used as well. For example, according to Ref.1, $\ce{HClO}$ in acidic medium:
$$\ce{HClO + H+ + e- <=>  1/2Cl2 + H2O} ~~~~~~~~ \mathrm{E}^\circ = \pu{+1.594 V} $$
On the other hand, $\ce{ClO-}$ in basic or neutral medium:
$$\ce{ClO- + H2O + 2e- <=>  Cl- + 2OH-} ~~~~~~~~ \mathrm{E}^\circ = \pu{+0.890 V} $$

Consider the following half reaction:
$$\ce{Br2 + 2e- <=>  Br- } ~~~~~~~~ \mathrm{E}^\circ = \pu{+1.098 V} $$

Now, you can say that $\ce{HClO}$ in acidic medium spontaneously oxidizes $\ce{Br-}$ to $\ce{Br2}$, but $\ce{ClO-}$ in basic or neutral medium cannot spontaneously oxidize $\ce{Br-}$ to $\ce{Br2}$.

In similar way, you may also say that $\ce{HClO}$ in acidic medium is a better oxidizing agent than that of $\ce{ClO4-}$ in acidic medium, because:
$$\ce{ClO4- + 8H+ + 7e- <=>  1/2Cl2 + 4H2O} ~~~~~~~~ \mathrm{E}^\circ = \pu{+1.392 V} $$

Reference:

Steven G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at 298.15 K,” Journal of Physical and Chemical Reference Data 1989, 18(1), 1–21 (https://doi.org/10.1063/1.555839).


How to calculate the amount of substance of nitrogen that dissolves in water? The Henry's law constant for the solubility of $\ce{N2}$ gas in water at $298~\mathrm{K}$ is $10^5~\mathrm{atm}$. The mole fraction of $\ce{N2}$ in air is $0.8$. The amount of substance of $\ce{N2}$ for air dissolved in 10 moles of water at $298~\mathrm{K}$ and $5~\mathrm{atm}$ pressure is ___.

I think the mole fraction of nitrogen will be calculated using the formula $$x(\ce{N2})=P/K_h.$$ But the answer is wrong.
In the solution for the question, the mole fraction $x(\ce{N2})$ is multiplied further by $0.8$ which seems to the mole fraction of nitrogen in air.

Unfortunately, various definitions of Henry’s law and the corresponding proportionality constant $H$ or $K_\mathrm H$ exist. (For several definitions and corresponding parameter values, see Sander, R. Compilation of Henry’s law constants (version 4.0) for water as solvent. Atmos. Chem. Phys. 2015, 15, 4399–4981.) Therefore, it is important to identify the dimensions of the given data. In the question, the proportionality constant is given as Henry volatility $K_\mathrm H$ with

$$K_\mathrm H = 10^5\ \mathrm{atm}$$

Apparently,* since the given Henry volatility $K_\mathrm H$ is expressed in terms of pressure (the unit symbol “atm” stands for “standard atmosphere”, which is an obsolete unit of pressure; the use of this unit is actually deprecated), the used definition is

$$K_\mathrm H=\frac{p_{\ce{N2}}}{x_{\ce{N2}}}$$

where $p_{\ce{N2}}$ is partial pressure of nitrogen and $x_{\ce{N2}}$ is amount-of-substance fraction (the use of the unsystematic name “mole fraction” is deprecated) of nitrogen in the aqueous phase.

The amount-of-substance fraction $x_{\ce{N2}}$ is defined as

$$x_{\ce{N2}}=\frac{n_{\ce{N2}}}{n}$$

where $n_{\ce{N2}}$ is the amount of substance of nitrogen and $n$ is the total amount of substance. For dilute aqueous solutions, the total amount of substance is approximately equal to the amount of water

$$n\approx n_{\ce{H2O}}$$

which is given as $n_{\ce{H2O}}=10\ \mathrm{mol}$.

Thus, the amount of nitrogen $n_{\ce{N2}}$ can be calculated from the partial pressure of nitrogen $p_{\ce{N2}}$ as

$$\begin{aligned}
n_{\ce{N2}}&=n\cdot x_{\ce{N2}}\\
&=n\cdot \frac{p_{\ce{N2}}}{K_\mathrm H}\\
&\approx n_{\ce{H2O}}\cdot \frac{p_{\ce{N2}}}{K_\mathrm H}
\end{aligned}$$

For a mixture of gases, the partial pressure of nitrogen $p_{\ce{N2}}$ is defined as

$$p_{\ce{N2}}=x_{\ce{N2}}\cdot p$$

where $x_{\ce{N2}}$ is the amount-of-substance fraction of nitrogen in the gaseous phase (as apposed to the above-mentioned parameter in Henry’s law) and p is the total pressure.

Since the total pressure is given as $p=5\ \mathrm{atm}$ and the amount-of-substance fraction of nitrogen in air is given as $x_{\ce{N2}}=0.8$, the partial pressure of nitrogen in air is

$$\begin{aligned}
p_{\ce{N2}}&=x_{\ce{N2}}\cdot p\\
&=0.8\times5\ \mathrm{atm}\\
&=4\ \mathrm{atm}
\end{aligned}$$

* This use of the unit is in the question is actually not permissible. The unit symbol should not be used to provide specific information about the quantity and should never be the sole source of information on the quantity.


Fentons Reagent tips: H₂O₂ + FeSO₄ I've been having trouble with reducing COD (Chemical Oxygen Demand) below approx. 500 ppm by using Fentons Reagent.

For 100 gm of the sample (initial COD 3000 ppm) I use 5 gm of 30% $\ce{H2O2}$ & 0.1 gm of $\ce{FeSO4}$. pH is adjusted to between 3 to 5. Temp. is approx. Room Temp. & I've tried reaction times up to 6 hrs.

Do these sound reasonable choices?

Any advice?

I'm not an expert in Fenton chemistry but it appears that difficulty in getting the Fenton reagent to reduce COD to below 500 ppm is not unusual.

Does your particular wastewater have  chelants such as phosphates, EDTA, formaldehyde, and/or citric/oxalic acids?
Perhaps at long reaction times all the iron you add is being precipitated as iron (hydr)oxides.  Could you lower the pH further?

  Reaction rates with Fenton’s Reagent are generally limited by the rate
  of $\ce{.OH}$ generation (i.e., concentration of iron catalyst) and less so
  by the specific wastewater being treated. Typical $\ce{Fe:H2O2}$ ratios are
  1:5-10 wt/wt, though iron levels < 25-50 mg/L can require excessive
  reaction times (10-24 hours). This is particularly true where the
  oxidation products (organic acids) sequester the iron and remove it
  from the catalytic cycle. Fenton’s Reagent is most effective as a
  pretreatment tool, where CODs are >500 mg/L. This is due to the loss
  in selectivity as pollutant levels decrease...In addition to free radical scavengers,
  the process is inhibited by (iron) chelants such as phosphates, EDTA,
  formaldehyde, and citric/oxalic acids. Because of the sensitivity of
  Fenton’s Reagent to different wastewaters, it is recommended that the
  reaction always be characterized through laboratory treatability tests
  before proceeding to plant scale.

My tip would be to increase the concentration of both $\ce{H2O2}$ and $\ce{FeSO4}$, and, if possible, to make sure the pH is 3 (or maybe even lower).  If that doesn't work, do you think iron chelators could be present in your particular wastewater sample(s)?


How to know properties of ion Aluminum -27? How do I know the charge/properties of this Ion Aluminum -27? All the information given is that Al has an atomic number of 13. I want to know how to know that it has an electric charge of +3 from the given information.

Aluminum-27 refers to an isotope of aluminum. For almost all chemical reactions the isotope makes no difference in the chemistry. You need "exotic" (meaning very unusual) conditions for the isotope to matter.

Unusual conditions would mostly happen in the gas state. For example a lighter gas would diffuse faster. This was the basis for the separation of uranium isotopes at Oak Ridge during WWII to make the first atomic bomb.


Does DIBAL-H reduce carboxylic acids? Diisobutylaluminium hydride(DIBAL-H) reduces only carboxylic acid, ester or both?

The answer given was that it only reduces ester.

Is it correct that under any condition carboxylic acids are not reduced by DIBAL-H?

Does diisobutylaluminium hydride (DIBAL-H) reduce only carboxylic acids? or  only carboxylic acid esters? or reduce both both types of compounds?

The answer is yes, it does reduce both. Not only them, it also reduces nitriles to aldehydes, and is a more selective reagent than lithium aluminum hydride (LAH) in the reduction of nitriles (Ref.1). About reduction of carbozylic acid, Miller et al. (Ref.1) found that DIBAL-H can be used to reduce benzoic acid to give 72% yields of benzyl alcohol in 1959. In same publication, they have also shown that esters can be reduced to corresponding alcohols as well.

In reduction of nitriles to aldehydes, following results were achieved and compared with the result obtained by Amundsen and Nelson (Ref.2) using $\ce{LiAlH4}$:

The summery of reduction of carboxylic acids and their esters are illustrated in following scheme, where two different groups (Ref.1 & Ref.3) achieved two different results for ester reduction under different conditions:

References:

A. E. G. Miller, J. W. Biss, and L. H. Schwartzman, “Reductions with Dialkylaluminum Hydrides,” J. Org. Chem. 1959, 24(5), 627–630 (DOI: 10.1021/jo01087a013).
L. H. Amundsen, L. S. Nelson, “Reduction of Nitriles to Primary Amines with Lithium Aluminum Hydride,” J. Am. Chem. Soc. 1951, 73(1), 242–244 (DOI: 10.1021/ja01145a082).
L. I. Zakharkin, I. M. Khorlina, “Reduction of esters of carboxylic acids directly into aldehydes with diisobutylaluminum hydride,” Tetrahedron Letters 1962, (14), 619-620 (https://doi.org/10.1016/S0040-4039(00)70918-X).


Is it possible to make an anticatalyst? I'm wondering if it is possible, theoretically, to create compounds which perform the opposite function of a catalyst (thus an anticatalyst). That is to say, could a compound be made which raises the activation barrier of a reaction rather than lowers it?

I know enzyme inhibitors exist, but I would argue these merely disable or slow the catalyzed reaction rather than slowing the reaction down relative to its baseline speed. (If I'm mistaken and there are enzyme inhibitors that actually lead to a slower reaction than if the enzyme weren't there at all, I would also consider that an answer).

I would think an anticatalyst could be made by having an enzyme which binds a molecule in a way that reduces the favorability of the transition state (or alternatively, stabilizes the reactants, as mentioned by Nicolau Saker Neto below). The binding to the anticatalyst enzyme would also have to be favorable enough where the reactants would not just react and ignore the slower anticatalyzed pathway (one could also assume, for the sake of argument, that enough of the anticatalyst is added where the majority of the reactant molecules are bound).

Some substances do exist that slow down reactions and they're different from catalytic inhibitors/poisons. Such substances are called negative catalysts. Here is one example I can think of right now:
$$\ce{2H2O2 \xrightarrow{\large\mathrm{glycerol}} 2H2O + O2}$$

Glycerol, in this case behaves as a negative catalyst and slows down the reaction.

Another notable example you might have heard is tetraethyllead (TEL) which was used in petrol to prevent knocking.

It may be tempting to think that a negative catalyst will raise the energy of activation for a given reaction, however this isn't true. It takes away the internal energy of reactants and deactivates them to reduce the rate of a reaction.


Effect of dopant size on XRD peak shift If we add an impurity atom (doping) to the crystalline struture of another material and take the XRD of the doped material what kind of peak shift can we expect. For the better description of my question i have added an image below

In the image black bold line ($2\theta $) is the standard XRD peak of a material; blue ($2\theta ^{L}$) and green ($2\theta ^{R}$) dotted lines are the peak shift caused due to the doping of impurity atom. Can you tell me if there is any relationship between dopent size and peak shift (right/left dotted lines).

Doping, not to confuse with alloying, introduces a low amount of impurities or more generally point defects in a crystal. The overall periodicity of the system is thus not changed and no shift should be expected in XRD. However doping introduces some disorder that makes XRD peaks broader.


Cross-linking in Latex My understanding was that cross-linking would generally make a material harder and less flexible, but if latex does have cross-linking, how would this affect its shape and hence its properties, especially as a material with elasticity?

In addition, is there a method of testing if a substance has crosslinks?

I have googled to no avail.

"Raw" latex consists of long hydrocarbon chains. The chains themselves have covalent bonds and are therefore very strong; imagine them like hemp threads. A volume of latex is basically a huge mess ("ball of wool") of those chains, which are weakly held toghether by van der waals forces.
This raw latex is therefore a thermoplast: It is soft and sticky when hot and rather brittle when cold. This must have been rather annoying when only this form was known. It is also not very elastic but malleable,  (is that the word?)

But now it was discovered that sulfur likes to form cross-connections between said chains. This basically makes the latex a single, huge molecule (hence a duroplast). But there are not too many of those cross connections, and the material looks rather like a loose net. If this vulcanized latex is streched, it can regain its form rather well; it is therefore called an elastomer. The more sulfur-bridges there are, the harder is the material; i.e. a hair band has less bridges than an ice hockey puck.

I think this view should explain the behaviour of latex quite well and on an intuitive level.

Testing for crosslinks seems rather difficult in general; but in the case of latex you can test for sulfur however. If you only have an emulsion this is very easy, if it smells repugnant there is sulfur in it. If you have solid material you might as well just test for elasticity.

In general it seems a save bet that elastic materials have crosslinks.


Which bond - C-C or C-H - is stronger? I've read that nonpolar bonds are stronger than polar bonds. If it is true, why $\ce{C-H}$ bond entalpy ($\pu{413 kJ/mol}$) is more than $\ce{C-C}$ ($\pu{348 kJ/mol}$)? I'm very confused at the moment.

The "strength" of a bond can't be evaluated just based on its enthalpy.

According to this relation

$$ \Delta H = \sum \Delta H_\text{bonds broken} - \sum \Delta H_\text{bonds created} $$
the strength of your bond will depend on the reactive nearby the bond. If the molecule is likely to create bonds with higher enthalpy after the bond's breakdown, then the enthalpy variation will be negative. The process will then be spontaneous and the initial bond will breaks "easily".
There is also some steric consideration, the hydrogen atom is smaller, the bond is then more easy to be "attacked".
Finally, for a polar bond, the electronic cloud is delocalised closer to one atom rather than the other (according to their electronegativity) so you have "less perturbation to add" to fully remove the covalent electron from the less electronegative atom.


Why does water evaporate spontaneously at room temperature despite ΔG > 0? Standard Gibbs free energy of formation of liquid water at $\pu{298 K}$ is $\pu{−237.17 kJ mol-1}$ and that of water vapour is $\pu{−228.57 kJ mol-1}$ therefore,

$$\ce{H2O (l) -> H2O (g)}\qquad\Delta G =\pu{8.43{kJ mol-1}}$$

Since $\Delta G>0$, it should not be a spontaneous process but from common observation, water does turn into vapour from liquid over time without any apparent interference. Why does this happen?

Standard Gibbs free energy of formation of liquid water at 298 K is −237.17 kJ/mol and that of water vapour is −228.57 kJ/mol. Therefore, $$\ce{H2O(l)->H2O(g)}~~\Delta G=8.43~\mathrm{kJ/mol}$$
Since $\Delta G>0$, it should not be a spontaneous process but from common observation, water does turn into vapour from liquid over time without any apparent interference.

Your math is correct but you left out a very important symbol from your equations.  There is a big difference between $\Delta G$ and $\Delta G^\circ$.  Only $\Delta G^\circ$ means the Gibbs energy change under standard conditions, and as you noted in the question, the free energy values you quoted are the standard gibbs free energy of water and water vapor.
Whether or not something is spontaneous under standard conditions is determined by $\Delta G^\circ$.  Whether something is spontaneous under other conditions is determined by $\Delta G$.  To find $\Delta G$ for real conditions, we need to know how they differ from standard conditions.
Usually "standard" conditions for gases correspond to one bar of partial pressure for that gas.  But the partial pressure of water in our atmosphere is usually much lower than this.  Assuming water vapor is an ideal gas, then the free energy change as a function of partial pressure is given by $G = G^\circ + RT \ln{\frac{p}{p^\circ}}$.  If the atmosphere were perfectly 100% dry, then the water vapor partial pressure would be 0, so $\ln{\frac{p}{p^\circ}}$ would be negative infinity.  That would translate to an infinitely negative -- i.e. highly spontaneous -- $\Delta G$ for the water evaporation reaction.
Small but not-quite zero dryness in the atmosphere would still lead to the $\Delta G$ of water vapor that is more negative than liquid water.  So water evaporation is still spontaneous.
Extra credit: given the standard formation energies you found, and assuming water is an ideal gas, you could calculate the partial pressure of water vapor at which $\Delta G = 0$ for water evaporation.  And the answer had better be the vapor pressure of water, or else there is a thermodynamic inconsistency in your data set!


Collision rates between two molecules in a liquid How would I calculate the collision rate between kaolinite and amylopectin particles in a liquid with a density of ($\pu{1.01g/m^3}$) and a viscosity of ($\pu{0.000434 Pa s}$)? I know this is related to orthokinetic flocculation, however, I do not know the exact equation. If anyone knows the equation, can you please inform me of it?

If you assume that the reaction is diffusion controlled  then this is the fastest that a bimolecular solution phase reaction can be and the second order rate constant is $k=8000RT/(3\eta)$ (dm$^3$/mol/s) where $\eta$ is the solution viscosity and assuming that the molecules are approximately the same size and even if they are not its still a good estimation. In mobile solvents (hexane, water etc)  $k\approx 10^{10} \to 10^{11}$ dm$^3$/mol/s.

If the species a, b are ions (charge $Z_{a,b}$) then the rate constant is multiplied by $\delta/(e^{\delta}-1)$ where  $\displaystyle \delta = \frac{Z_aZ_bq^2}{(4\pi\epsilon_0)\epsilon k_BTd_{ab}}$ where $d_{ab}$ is the centre to centre molecules size, $q$ electronic charge, $k_B$ the Boltzmann constant, $\epsilon$ the solvent dielectric constant and $\epsilon_0$ the permittivity of free space.

[If the species are not the same size radius $r$ then $\displaystyle k= \frac{2000RT}{\eta}\frac{(r_a+r_b)^2}{r_ar_b}$.  ]


Why is it O-Cl-O and not O-O-Cl? I'm currently preparing my exam and while I did never think about this question during the course, it begins to confuse me right now...

Why is it the structure of $\ce {ClO_{2}-}$ like $\ce {O-Cl-O}$ and not like $\ce {O-O-Cl}$?

I know how to place the covalent bonds and the electrons, but I don't know how to figure out the basic structure of the molecule. Is there a relatively simple rule to solve (and understand) such relatively simple molecules?

Is there a relatively simple rule to solve (and understand) such
  relatively simple molecules?

The most electropositive atom is usually at the center of a molecule because

a "central atom" implies that there are multiple (more than 1) bonds to the central atom. If the central atom is relatively electropositive, then it will be better able to share its electrons and form bonds with other atoms, at least more so than an electronegative central atom would.
electronegative atoms tend to carry multiple lone pairs of electrons. If this electronegative atom and all of its lone pairs were at the center of the molecular structure, then we would have many more destabilizing (lone-pair - bonding pair) electron-electron repulsions, then if all of these lone pairs were on the periphery of the molecule.


Tallow and polymer pellets Bank of England has issued a new £5 note which contains Animal Tallow. I was wondering if you can explain to me as to how the animal tallow is used in this process. Also would you be able to speculate as to tallow of which animal could potentially be used in this production process?

Does the animal tallow completely transform or are its original characteristics retained?

Bank of England official Press Release is as follows:

We are aware of some people’s concerns about traces of tallow in our new five pound note. We respect those concerns and are treating them with the utmost seriousness.

This issue has only just come to light, and the Bank did not know about it when the contract was signed.

Information recently provided by our supplier, Innovia, and its supply chain shows that an extremely small amount of tallow is used in an early stage of the production process of polymer pellets, which are then used to create the base substrate for the five pound note.

Innovia is now working intensively with its supply chain and will keep the Bank informed on progress towards potential solutions.

According to the UK news outlet The Sun:

  The new polymer note uses beef tallow made from suet, which is hard fat found around the animal’s kidneys, stomach and other organs.
  [...]
  The animal product is used as a source of stearic acid.
  This allows the note to be lubricated and to ensure it is smooth inside your wallet.

Suet is mainly used to make tallow, though it is also a widely used cooking ingredient.

Stearic acid is a waxy material who's ionic form (stearate salts like sodium stearate) is widely used in detergents. This property gives it the desired smooth, slippery property desired in the manufacture of the polymer notes.


Isomerism and namegiving Is this the right chemical formula of the $\ce{Rb[Pt(NH3)Br2Cl]}$ complex: rubidium amminedibromidochloridoplatinate(II)?

And is it correct that it has geometric isomerism, being square planar with a cis- and trans-isomers, leaving the rubidium out?

Your formula, $\ce{Rb[Pt(NH3)Br2Cl]}$ is correct for the named complex, rubidium amminedibromochloroplatinate(II) (For clear understanding of naming, you may read Naming Coordination Compounds-1 and Naming Coordination Compounds-2).

Many coordination compounds have distinct geometric structures. Two common forms are the square planar, in which four ligands are arranged at the corners of a hypothetical square around the central metal atom, and the octahedral, in which six ligands are arranged, four in a plane and one each above and below the plane. Altering the position of the ligands relative to one another can produce different compounds with the same chemical formula.
Coordination compounds often exist as isomers—i.e., as compounds with the same chemical composition but different structural formulas. Many different kinds of isomerism occur among coordination compounds (Source).

Cis-trans-isomerism  is one of the more common types. Cis-trans (geometric) isomers of coordination compounds differ from one another only in the manner, in which the ligands are distributed spatially. Most of platinum(II) complexes are square planner. Most common example is Cisplatin ($\ce{Pt(NH3)2Cl2]}$) where the each of two ammonia molecules and the two chlorine atoms are situated next to one another (see the diagram in the blue box below), and hence isomer is called the cis (Latin for “on this side”). Thus, coordination name for Cisplatin is  cis-diamminedichloroplatinum(II) (Wikipedia).

When the identical groups are in opposite sides, it is called the trans-isomer (Latin for “on the other side”). To emphasize this nomenclature, I sketched cis- and trans-isomers of diamminedibromoroplatinum(II) (above) and your compound (below) in the maroon box below for better understanding (cis-isomers are in LHS and trans-isomers in RHS).


Does hydrogen really have seven isotopes? I know that there are 3 isotopes of Hydrogen. But according to this website it is showing 7 http://periodictable.com/Properties/A/KnownIsotopes.htmlWebsite

Yes, there are seven known isotopes of hydrogen, though only two ($\ce{^1H}$ and $\ce{^2H}$) are stable with respect to nuclear decay, and only three ($\ce{^1H}$, $\ce{^2H}$ and $\ce{^3H}$) exist/can be made in enough quantities to be relevant outside of nuclear physics. All other hydrogen isotopes have extremely small half-lives. The next most stable hydrogen isotope after $\ce{^3H}$ ($t_{1/2}=12.3\ yr$) seems to be $\ce{^5H}$ ($t_{1/2} \sim 10^{-21}\ s$). Given that even the fastest chemical reactions happen in timespans of the order of $10^{-15}\ s$, we can safely discard any influence of hydrogen isotopes heavier than $\ce{^3H}$ in chemistry.

Out of the currently 115 elements which have been confirmed to exist, there are approximately 254 isotopes which are considered to be indefinitely stable, or have half-lives so enormous (over trillions of times greater than the age of the Universe, 13.8 billion years) that no value could yet be determined. Theoretically, only 90 of those 254 isotopes are expected to be truly indefinitely stable (proton decay aside), but there are great difficulties in setting up experiments capable of measuring such rare decay events.

On top of the previous 254, another 34 isotopes have half-lives which have actually been measured in laboratory, but are still large enough (>50 million years) so that a small fraction of the isotope could conceivably have survived since the creation of the Earth 4.6 billion years ago, and may be of interest in geological timescales. There are also isotopes with smaller half-lives, but which are continuously replenished by the decay of heavier atoms or by the action of cosmic rays, so they can still be found on Earth today. In all, about 339 different isotopes can be found naturally on our planet.

However, as many as 3100-3300 different isotopes of the first 118 elements are claimed to have been detected in laboratories, most of them with very small half-lives (seconds or less). Nobody knows for sure how many elements and how many isotopes can exist, though the number is finite.


p-value for ions in a solution? The question in my class notes read as:

  Calculate the p-value for each ion in a solution that is $2.00 \times 10^{-7}\ \mathrm{M}$ in $\ce{NaCl}$ and $5.4 \times 10^{-4}\ \mathrm{M}$ in $\ce{HCl}$.

I was confused how to approach it, but my teacher did the solution on the board with the work done out and obtained the values of $\mathrm p\ce{H}$, $\mathrm p\ce{Na}$ and then used the totals of those values to get the concentration of [$\ce{Cl-}$]. From that, she then took the negative log of the concentration of [$\ce{Cl-}$].

For reference, she obtained the following values:
 $$\begin{align}
\mathrm p\ce{H} &= 3.27, \\
 \mathrm p\ce{Na} &= 2.699, \\
 [\ce{Cl-}] &= 2.54 \times 10^{-3}, \\
 \mathrm p\ce{Cl} &= 2.595
\end{align}$$

I understand how one has to take the negative log of a given concentration to obtain the p-value of it in a "rote" type of sense.

But what I misunderstood was the log rules and the calculation of the first $\mathrm p\ce{H}$ value.

Her work for the calculation of $\mathrm p\ce{H}$ was as followed:

(1st step)

$\mathrm p\ce{H} = -\log[\ce{H3O+}] = -\log[5.4 \times 10^{-4}]$

(use the log rule, $\log(a \cdot b) = \log(a) + \log(b)$)

(2nd step)

$\mathrm p\ce{H} = -\log 5.4 - \log 10^{4}$

(3rd step)

$\mathrm p\ce{H} = 3.27$

But why in the 2nd step above is the $\log 10^{4}$ not have a negative value, such as $\log 10^{-4}$?

In her solution of p$\ce{Na}$ it was the case that the second additive had a  negative exponent, which can be seen below in the second step.

(1st step)

$\mathrm p\ce{Na} = -\log[2.00 \times 10^{-3}]$

(2nd step)

$\mathrm p\ce{Na} = -\log2.00 + \log10^{-3}$

(3rd step)

p$\ce{Na}$ $= -0.301 -(-3.00) = 2.699$

Also if anyone could explain to be why we sum the values of $\mathrm p\ce{H}$ and $\mathrm p\ce{Na}$ to obtain $[\ce{Cl-}]$? Does this have to do with the "stoichiometry" of something? My chemistry is not great, nor is my math. Thank you.

It seems like the calculation done or copied from the board by you has gone a little wrong.

In your first step when taking $-\log$ of the given concentration you have ignored the signs on the lower of the argument of the log function.
$$-\log 10^{-3} = \log 10^3 = 3 \log 10 = 3$$
since the base of log here is 10. And same correction for next steps.

And to your other question ... We add up the concentrations (adding up p values is irrevalent) because here NaCl is a strong electrolyte and HCl is a strong acid. They both dissociate completely and individually contribute to the concentration of chlorine anion.


Why is an energy cost associated with splitting the HOMO spins in singlet oxygen? The molecular orbital schemes for both forms of singlet oxygen ($\mathrm{^1\Delta_g}$ and $\mathrm{^1\Sigma_g^+}$) and triplet oxygen ($\mathrm{^3\Sigma_g^-}$) are typically given as shown in the image below.

Figure 1: Molecular orbital schemes of two types of singlet oxygen and triplet oxygen with the highest energy electrons highlighted in red.

Hund’s rule sufficiently predicts that the triplet $\mathrm{^3\Sigma_g^-}$ state is the most stable, i.e. lowest in energy. When one speaks of singlet oxygen, one typically has the $\mathrm{^1\Delta_g}$ form in mind which is $\pu{0.98eV}$ higher in energy than triplet oxygen according to a group member’s recent literature seminar. This is often explained on this site by the fact that spin pairing puts another electron into the space that is already populated by one electron, causing electron-electron repulsion that must be overcome by adding energy.

Extending this explanation, I am inclined to think that the $\mathrm{^1\Sigma_g^+}$ form of singlet oxygen, whose MO scheme is depicted in the centre of figure 1, should have a lower energy than the $\mathrm{^1\Delta_g}$ form because the spins are occupying spacially different (and orthogonal) orbitals. However, the same seminar talk also included the energy difference of $\pu{0.65eV}$ between $\mathrm{^1\Delta_g}$ and $\mathrm{^1\Sigma_g^+}$ with the $\mathrm{^1\Sigma_g^+}$ being higher in energy.

Why is this so?

Short story: The comment by orthocresol is correct, based on Hund's rules. Given the same spin multiplicity (i.e., the two singlet states), the state with higher orbital angular momentum ($\mathrm{^1\Delta_g}$) wins.

Loosely speaking, one can imagine that comparing the two singlet states, the higher angular momentum implies that the electrons are a bit further out, and suffer less electron-electron repulsion.

Longer story...

The "unified atom" model can apply to simple diatomics - that is, many of the same considerations for the electronic structure can be used for multi-electron atoms. There's discussion in both Mulliken (Rev. Mod. Phys. 1932, 4, 1. and Herzberg's Spectra of Diatomic Molecules).

The vibrational-electronic spectra of $\ce{O2}$ was determined fairly readily:
 (Figure from Chem. Rev., 2003, 103 (5), pp 1685–1758, adapted from Herzberg).

We can assign the lines as discussed by Mulliken in 1932:

  The $\mathrm{^1\Sigma_g^+}$, is 1.62 volts above the $\mathrm{^3\Sigma_g^-}$ according to theoretical considerations,[118] the $\mathrm{^1\Delta_g}$, is probably about halfway between $\mathrm{^1\Sigma_g^+}$, and $\mathrm{^3\Sigma_g^-}$ (cf. Fig 48).

Reference 118 is to Hund and Hückel:

F. Hund, Zeits. f. Physik 63, 726 (1930).
E. Huckel, Zeits. f. Physik 60, 442-3 (1930).

Both Mulliken and Herzberg discuss the Hund's rules in regards to many diatomics - that the triplet will lie lower than the singlet because of higher S, but then the two singlet states will differ based on angular momentum (here L).


How does resonance fail in approximating chemical structures? In the book "Concise Inorganic Chemistry" by Prof. JD Lee, it says here:

  These contributing structures do not actually exist. The $\ce{CO3^2-}$ does not consist of a mixture of these structures, nor is there an equilibrium between them. The true structure is somewhere in between and is called a resonance hybrid. Resonance was widely accepted in the 1950s but is now regarded at best as clumsy and inadequate, and at worst as misleading or wrong!

I want to know where exactly does resonance fail in approximating the structures?

Resonance structures are one model that explains bonding, but not a very good one

Resonance structures are widely misunderstood. Individual resonance structures are not real compounds that exist, but an approximation that assumes all structures are made from bonds consisting of two electrons. The true structure of a compound is thought to be an average mixture of the various possible resonance structures consisting of simple bonds.

The archetypal example is benzene. If we model benzene as a structure consisting of "normal" 2-electron bonds, we get a ring consisting of alternating single and double bonds. This contradicts the observed structure where all the bonds are the same (and are somewhere between the character of single and double carbon-carbon bonds). The resonance model of benzene consists of two structures with the normal double and single bonds swapped, giving an average overall structure where all the bonds are an average of single and double bonds.

For structures with more complexity than benzene, the resonance structures are more complex and frequently don't make equal contributions, nor do they do a good job of predicting the real structures. Molecular orbital based approaches (which recognise the difference between pi-bonds and sigma bonds and do a much better job of recognising the fractional character of many real bonds) are far better.

That is why resonance structures "fail". They make drawing simple structures easier, but they don't do a good job of predicting real structures.


Doubt in a question related to electronic transitions in Bohr's orbits Question
In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inner-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?
(A) $3\to2$
(B) $5\to2$
(C) $4\to1$
(D) $2\to5$
Only one option is correct.

My approach
I just drew the electron transition diagram like this:

Then, I counted the third line from the red end, i.e., from the left side starting from the first line of the particular series. I then counter checked with the given options. Surprisingly two options matched, i.e., option B as well as C. I don't know further how to decide which is the correct answer out of B and C, since only one answer is correct. There is no other details mentioned in the question and thus I am confused.
Please clarify my doubt.

Since only one answer is correct, we must consider the transitions in the visible region only i.e., the Balmer Series. If multiple answers are correct then it is wise to choose both the options B and C.

The transition $n_2 \to n_1$ can be easily calculated by using the following formula and it is not necessary to draw the transitions, which might be difficult for higher transitions where drawing the transitions would be time-consuming.

$$n_2=n_1 + n$$
where $n$ represents the $n^{th}$ line in a particular series. Since here we are talking about Balmer series, $n_1=2$ and for the third line $n=3$. Substituting the values
$$n_2=2+3=5$$

Hence the required transition is given by option B.


What does the hydrolysis of a salt tell us about its basic/acidic character? I'm curious to know how basicity and acidity arises when a salt or covalent compound is dissolved in water.

Think of the salt in terms of its components: for example, sodium sulfide can be thought of as derived from sodium hydroxide (a strong base) and hydrogen sulfide (a weak acid) - resulting in a net basic character. When you dissolve sodium sulfide in water, the sulfide ion partially reacts with water to form hydrosulfide ion and hydrogen sulfide, plus hydroxide ion. Aluminum chloride, on the other hand, is derived from aluminum hydroxide (a weak base) and hydrogen chloride (a strong acid), so it is an acidic salt. Sodium chloride, from sodium hydroxide and hydrogen chloride, is a neutral salt (fully ionized). Knowing the dissociation constants of the corresponding acid and base of a salt, one can calculate how acidic or basic the salt is.


How does the treatment of resin R-SO3H with sodium chloride lead to formation of RNa? I am studying the methods of removal of permanent hardness of water, and came across the "synthetic resins method". From my textbook,

Cation exchange resins contain large organic molecules with $-\text{SO}_3\text{H}$ groups and are water insoluble. Ion exchange resin $\text{RSO}_3\text{H}$ is changed to $\text{RNa}$ by treating with $\text{NaCl}$. The resin then exchanges $\text{Na}^+$ ions with $\text{Ca}^{2+}$ and $\text{Mg}^{2+}$ ions present in hard water to make it soft.

I don't understand how the treatment of $\text{RSO}_3\text{H}$ with $\text{NaCl}$ to form $\text{RNa}$ works. For one thing, in the former compound, since $-\text{SO}_3\text{H}$ is an anion, $\text{R}$ is a cation, but in $\text{RNa}$, $\text{Na}^+$ being a cation, $\text{R}$ is an anion. What is the redox reaction that is responsible for the conversion here?

It is a typing mistake. Cation exchange process can be written as
$$\ce{R-SO3^-H^+ (s) + NaCl (aq) -> R-SO3^-Na+(s) + H^+(aq) +Cl^-(aq)}$$
You can immediately see why the resin must be in sodium form for water softening. If the acid form of the resin is used for softening water, what would you get? Extremely strong acids!!
For a divalent ion, the sulfonic acid groups should be in close proximity to "catch" ions such as $\ce{Ca^{2+}}$. Each calcium ion would quantitatively release two hydrogen ions from the resin.