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Heat capacity during state change
I read that the heat capacity of water when it is at 100 °C or 0 °C is infinite because all the heat given to it will not be used to increase temperature but to change the state.
I am unable to understand this. Say, we have water at 100 °C. If we continuously provide heat, water will change to steam and then it's temperature will increase, right?
Sure, we'll have to provide more heat than before to break the bonds. But it will definitely become a 101 °C steam after a while.
So why is heat capacity infinite at state change?

The temperature of the steam stays at 100°c unless it is superheated. The extra heat/energy being added (latent heat) is consumed by the water changing phase. So the more heat you add the faster all the water will convert to steam when it reaches 100°C.

Say, you have 1 kilogram of water at 20°C. You add 42kJ of heat, and your water gets warmer by 10°C. You divide 42000 by 10, and that's the heat capacity. Now you take 1 kilogram of water at 100°C. You add the same amount of heat, and... what is the change of temperature?

Both the comments are pointing at the fact that we need to add a lot of heat to change temperature. So I'm guessing infinite heat here refers to a lot of heat, not to the fact that it is impossible to increase temperature.

You got it wrong. A lot of heat does not matter that much (also, it is not such an awful lot, if we are talking about the boiling of liquid helium, for example; but the infinite heat capacity would still be there even in this extreme case). What matters is that $\rm{c={Q\over\Delta T}}$, and $\Delta\rm T$ is zero. Not just very small, but zero. What happens when you divide by zero?

Oh! OK. See if this is right:  when you heat water at 100°c  by a certain amount (not a VERY large amount) and divide by the temperature change which is zero, you get infinity, hence the heat capacity. Is that correct?

Now that's right.

Guys, you know that such calculation doesn't really mean this?  It's completely idealised situation with magically constant pressure


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Existence of orbitals
Do orbitals exist even when they are not occupied?
For example: $\ce{Cr^{+3}}$ has the configuration $\ce{[Ar]}\mathrm{3d^3}$ with the other two $\mathrm{3d}$ orbitals empty. We know the  other two orbitals exist, since the metal "uses its empty $\mathrm{d}$ orbitals" to form complexes.
But, for $\ce{Na+}$ with configuration $\ce{[Ne]}$, do the $\mathrm{3s,3p}$ and $\mathrm{3d}$ orbitals exist  too?

click on the edited $x$ ago to see that I did not touch anything that was not MathJax, thank you. (Brian is the culprit.)

@Jan Apologies.

Do orbitals exist even when they are occupied? Orbitals are mathematical constructs and as such they exist in a mathematical sense regardless of whether they are occupied or not. Whether you can actually say that orbitals exist in reality is a bit dubious. They certainly provide a good description of the observed electron density and behaviour of atoms but they are still only a model. Like everything in science, we develop a model to describe what we observe but that model is not perfect and does not necessarily represent what actually exists. Particularly on a quantum scale, the notion of well defined objects that exist in space starts to become a bit fuzzy.

In that case  every atom can be thought of having  empty  orbitals.Then why do we argue this happens because this particular element has empty orbitals to accommodate the electron pair?
Empty orbitals always exist as per your answer.

@yasir Yes, they do; but what is meant is usually accessable empty orbitals. Example: phosphorus and sulphur have unaccessable $\mathrm{3d}$ orbitals (they are high in energy and cannot really participate in hybridisation). A carbon-bromine bond has an accessable $\unicode[Times]{x3c3}^*$ orbital which can be attacked by the lone pair of a nucleophile because it is relatively low in energy.

@yasir By calculating (or measuring) energies or by qualified a priori guesses.

So for $\ce{Hg^{+2}}$ the $\ce{6d}$  orbitals are accessible?Actually it has something to do with oxymercuration demercuration mechanism, so bear with me.

@yasir - There is a finite probability that the electrons are in any number of infinite electronic states - which includes those higher energy "orbitals" - but the probability is exceptionally low in the ground state. That probability can change - higher or lower - depending on the environment (applied electric field, for instance). I'd say all the "orbitals" are accessible but some of them (the ones we see for ground state electronic configurations) are way more likely to be populated than the ones we'd expect to see populated in an excited state.

Todd is correct in that I maybe used a bad word when saying accessable. But @yasir no; an orbital of the same subshell of a higher shell is generally not accessable; $\ce{Hg^2+}$ is a $\mathrm{5d}$ element thus it can easily access $\mathrm{6s}$ and maybe $\mathrm{6p}$ but never $\mathrm{6d}$.

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What is the evolutionary reason for the existence of rings in molecules?
Just to give you a bit of context, I don't have any background in Chemistry. I'm actually a Computer Scientist adventuring myself in the world of Chemoinformatics (more specifically, we're developing algorithms to improve drug discovery).
I noticed that many molecules (or at least drug-like molecules) have rings, but I was wondering: from an evolutionary perspective, what does it mean? Why did molecules evolve to have this "feature", do molecules with rings have better properties then similar ones without rings?

Molecules do not evolve. As for the rings, they are there for purely combinatorial reasons. Also, not even all classes of biomolecules have them; fats, for example, have none.

Welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site.

I noticed that most molecules (or at least drug-like molecules) have rings, but I was wondering: from an evolutionary perspective, what does it mean?

First note: Delete the ‘most molecules’ bit. The numbers are insane and nigh uncountable, but non-cyclic molecules are similarly common as cyclic ones are.
From an evolutionary persective this means: nothing.
Drugs are designed to inhibit certain enzymes as specifically as possible. The enzyme has a shape that happens to best fulfil it’s purpose according to evolutionary principles. (my draft had ‘design’ here for a second, but it would be the wrong word choice. Nature and evolution do not design.) This will render an active pocket having a certain orientation, size and hydrophobic pattern that drug designers attempt to mimick. If a (poly-) cyclic molecule fits well, it fits well. If a linear molecule fits better it fits better. Think of it as trial and error.

Come to think of it, your explanation actually provides a reason why the cyclic structures may be somewhat preferable. See, a (poly)cyclic molecule tends to have less conformational freedom than a similarly large non-cyclic one. If it fits, it fits.

@IvanNeretin I’m pretty much d’accord with you there. But if a (strained?) linear one fits, it’ll fit similarly and some macrocycles are rather flexible, so it’s not part of my argument ;)

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Competition between substitution and elimination (uni- or bimolecular)
Whats is/are the main product/s of the following reaction:

I think the main product would be the one obtained by $S_N2$, and then the one pair of enantiomers obtained by $E2$. Is $E1$ with subsequent rearrangement possible?

I believe that E1 mechanism is not possible, as there is a good base present and reaction is carried out in diluted base on room temperature.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Marko improved the typesetting of your formulae with MathJax, for more information on how to do so yourself, check out the [help], this meta post or this one.

"E2 gives enantiomers" - are you sure? Draw that molecule in the chair conformation.

Jan, thank for the welcome. @orthocresol If you take the beta-proton from the left side, or the beta proton from the right side, you get two products, which are enantiomers.

Oh, ok, my bad :D

For an $\mathrm{E1}$ mechanism, we would need to generate a carbocation. Our halide is attached to a secondary carbon so the generated carbocation would be secondary. A quick check of neighbouring stabilising groups is negative. And as a professor of mine once put it (possible slightly oversimplified but with some general truth in it):

Exactly one secondary carbocation exists and that is the prop-2-yl cation.

Thus, $\mathrm{E1}$ is not possible since the carbocation will not form in any measurable amounts. For the same reason, $\mathrm{S_N1}$ is not an option. For everything else, let’s consider the molecule’s 3D structure:

The tert-butyl group acts as a conformation ‘anchor’ meaning that the other chair form is not viable. Luckily, bromine is pointing upwards, which gives us access both to $\mathrm{S_N2}$ and $\mathrm{E2}$ mechanisms without having to invert or use a boat conformation.
Our base/nucleophile is methanolate which is both a good enough base and a good enough nucleophile, able to access both pathways. The centre for nucleophilic attack is also well accessable. If I had to choose, I would indeed go with your choice of saying the main products will be $\mathrm{S_N2}$ adduct and two enantiomers of elimination product will also be formed. (They stem from either abstracting the proton I drew or the enantiotopic proton on the back half of the six-membered ring.)

@UrosTodorovic Glad to be of assistance ^^


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Elimination with Amines
$\hspace{35 mm}$
ATTEMPT:
Since $\ce{Et3N}$ is a weak base and the carbon to which $\ce{Br}$ is attached is tertiary, we can rule out $\mathrm{E2}$ and $\mathrm{S_N2}.$ As we are providing heat, the reaction is likely to go via $\mathrm{E1}$ forming a stable tertiary carbocation.
Now as the base is bulky it will knock out the most easily available $\ce{H}$, giving us the less substituted alkene.
However my text suggests that this reaction proceed via $\mathrm{E2}$ with the base taking a $\ce{H+}$ and knocking off $\mathrm{Br}$ to give the same final product.
In general with amines out of $\mathrm{E1}$ and $\mathrm{E2}$ which reaction is preferred if there is a possibility of stable carbocation formation?

Didn't you post this a couple of days ago? Why did you delete the first question and repost it?

@bon , The question was not getting enough attention.So i repost it.

@yasir You should not repost your question if it is not getting enough attention. Instead, use a bounty.

@wythagoras, I tried using a bounty on this.HereBut nothing happened.


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How to determine the configuration of C atom bounded to two groups with same configuration?
What is the configuration of the central C atom in the following molecule and how to determine it?

What configuration? There is none. You have two identical substituents.

But there is no mirror plane passing through the center of the molecule?

See this earlier answer

No mirror plane, so what? Can you build another molecule that would have similar side arms (both R), but be different from this one in the central atom's stereo configuration alone?

I see. It is chirotopic.


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Gravimetric analysis of barium or sulfate
Which is a better way to obtain barium sulfate precipitate for gravimetric analysis: by adding $\ce{Ba^2+}$ ions to sulfate ion solution, or the other way round? I think it is better to precipitate sulfate ions by $\ce{BaCl2}$ solution, as this causes chloride ions to coprecipitate.

@Marko use \ce to format chemical expressions. Plain math mode gives italicised formulae. See the "basic chem" section

I think $\ce{BaCl2}$ should be added later. This is because $\ce{BaCl2}$ is soluble in water, but the metal sulphate you will be given may not be soluble. So if you add $\ce{M2SO4}$ later, you may get a wrong result as you have to make sure that the whole of $\ce{BaCl2}$ is used up. On the other hand all chloride salts are soluble. So while filtering the $\ce{MCl}$ will be removed easily along with the solvent and you will get $\ce{BaSO4}$.

I don't think so. Not all chloride salts are soluble and some sulfates are soluble enough. I want to hear an opinion from the precipitation point of view.


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Formation enthalpy vs. bond enthalpy
I have been working with thermochemistry at school for a while. I have been introduced to the two terms: formation enthalpy and bond enthalpy. However, I have not quite understood the difference between the two and would be very glad if someone could explain it!

Is there a connection between form. entalphy and bond entalphy? For example if we have: C + O2 => CO2. Can the formation entalphy be calculated if one knew the bond entalphy for each bond?

There is a connection of sorts, but not a direct one. As to your example: what bonds exactly do you include in "each bond"?

The double bond in the O2 and the 2 double bonds between the C and O in CO2

OK, but that's not enough: you have to consider the bonds in elementary carbon as well.

Ok, so let's say you take that in account as well. Then how would the form. entalphy be calculated?

OK, here is the difference: formation enthalpy refers to the formation of a compound from elementary compounds, while bond energy is applicable when considering the formation of a molecule from isolated atoms. The relation between the two is pretty much explained by the Hess's law: the enthalpy of a reaction does not depend on its pathway, but only on the initial and final states.
Now consider the following two pathways for your example.
Pathway 1: we burn graphite in oxygen, just like $\ce{C + O2 -> CO2}$. There is some change of enthalpy associated with this reaction.
Pathway 2: we vaporize graphite to get atomic carbon (see how many C-C bonds are broken, take that into account), atomize $\ce{O2}$ to get elementary oxygen (again, some bonds are broken), and finally produce $\ce{CO2}$ (this time some bonds are formed). Now if you'll look up the energies of all bonds involved and carefully add/subtract them all together, you'll have a decent approximation of the reaction enthalpy obtained in the pathway 1.
(Normally it works the other way around, that is, we use reaction enthalpies to calculate bond energies, but... whatever.)

The term "elementary compound" is erroneously used. A compound is made of two or more types of atoms. Thus, an element cannot be a compound.

Formation Enthalpy: energy it takes to form a compound from elements
Bond Enthalpy: an approximation of the energy it takes to form a compound from anything by calculating the bond energies(bond energies of the reactants - bond energies of the products)

authors: Дария Бектурсын


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Double alkylation of benzene to form tetralin
I have been asked to suggest a synthesis for the following compound (tetralin) from benzene:

I am unsure how to do this. It surely can't be Friedel–Crafts alkylation because of carbocation rearrangement and polyalkylation, right? Possibly a double acylation followed by a double reduction? Is there an issue with this; maybe a competing aldol type reaction in the double acyl chloride reagent that would be required.

Your idea with acylation is good imo

So many catalysts have been derived for selective hydrogenation of naphthalene to form that; why should one bother starting from benzene? ;)

You've got it right, double acylation and double reduction - but the first reduction occurs after the first acylation.  FWIW, the name of the reaction is the Haworth reaction and is commonly used to synthesize substituted tetralins, naphthalenes and higher aromatic polycyclics.
In your particular case, here is a diagram of the first 3 steps in the reaction

Friedel-Crafts acylation of benzene with succinic anhydride
reduction of the ketone using a Wolff-Kishner or Clemmensen reduction
a second Friedel-Crafts acylation using strong acid; it is an internal cyclization and forms the needed ring
and finally (not shown in the above diagram) a second Wolff-Kishner or Clemmensen reduction of the resultant tetralone to produce the desired tetralin

Why cant both Friedel-Craft additions be done before the reduction?

Friedel-Crafts alkylation and acylation reactions do not take place (at least not in good yield) with a deactivated aromatic ring. Once the first acylation occurs, the ring becomes deactivated by the attached ketonic carbonyl.  In fact, this is one of the advantages of the F-C acylation reaction over the alkylation reaction. In the acylation case, only one acyl group can attach before deactivation stops the reaction.  In the alkylation case, the ring becomes more activated and undesired polyalkylation often results.

authors: Артём Павлов

authors: محمد الصمادي


text:
Solubility of Ag3AsO4 in 0.02 M K3AsO4

Calculate the solubility of $\ce{Ag3AsO4}$ in $\pu{0.02M}~\ce{K3AsO4}$ neglecting the activity coefficients. Find the relative error. $K_\mathrm{sp}(\ce{Ag3AsO4}) = \pu{6.0e-23}$

I know how to calculate the relative error but I get a very complicated equation finding the concentration solubility product constant ($K'_\mathrm{sp}$). There should be a quicker way to solve this since it is a midterm question.
I tried this:
$$\ce{Ag3AsO4 -> 3Ag+ + AsO4^3-}$$
$$K_\mathrm{sp} = 27x^4$$
$$x = \pu{2.78e-6}$$
$$[\ce{Ag+}] = 3x = \pu{8.34e-6M}$$
$$[\ce{AsO4^3-}] = x = \pu{2.78e-6M}$$
$0.02~\mathrm{M}\ \ce{AsO4^3-}$ comes from $\ce{K2AsO4}$. So there should be an equation like:
$$K_\mathrm{sp} = (\pu{8.34e-6} - 3x)^3 \times (\pu{2.78e-6} + 0.02 - x) = \pu{6.0e-6}$$
And things get complicated. After finding $x$, I will also have found the final concentrations of silver and $\ce{AsO4^3-}$ ions. Then I will read the activity coefficients of them from the appendix table.
I have two questions:

Is my method true or false?

What is an easier way of solving this problem?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I improved the formatting of your post by adding MathJax markup; for more information on how to do so yourself, check out the [help], this meta-post or this one. As per our homework policy, this is a [tag:homework] question; but this is okay for you, since you already showed your work.

The correct solution:
$$K_\mathrm{sp} = a_\mathrm{Ag^+}^3 \cdot a_\mathrm{AsO_{4}^{3-}} = 6.0 \cdot 10^{-23}$$
$$K_\mathrm{sp} = K_\mathrm{sp}^{'} \cdot γ_\mathrm{Ag^+}^3 \cdot γ_\mathrm{AsO_{4}^{3-}}$$
$$K_\mathrm{sp}^{'} = \frac{K_\mathrm{sp}}{γ_\mathrm{Ag^+}^3 \cdot γ_\mathrm{AsO_{4}^{3-}}}$$
Ionic strength:
$$μ = \frac{1}{2} \cdot (0,06\cdot1^2 + 0,02\cdot3^2) = 0,12~\mathrm{M}$$
$$0,12~\mathrm{M} ≈ 0,10~\mathrm{M} $$
Activity coefficients at ionic strength $0,12~\mathrm{M}$ are:
$0.75$ for $Ag^+$
$x$ for $AsO_{4}^{3-}$
Then:
$$K_\mathrm{sp}^{'} = \frac{K_\mathrm{sp}}{0.75^3 \cdot x}$$
And the relative error is:
$$\%~error = \frac{|K_\mathrm{sp}-K_\mathrm{sp}^{'}|}{K_\mathrm{sp}} \cdot 100~\% = \frac{|6.0 \cdot 10^{-23}-K_\mathrm{sp}^{'}|}{6.0 \cdot 10^{-23}} \cdot 100~\%$$
I can't find the activity coefficient for arsenate. If someone can find it, please share it with us.

You start of by calculating solubility with one approximation: that all $\ce{AsO4^3-}$ ions are obtained by dissolving the potassium salt:
Let $S$ be the molar solubility:
$$K_\mathrm{sp} = 0.02 \times 27 S^3$$
$$S = \pu{0.0223 M}$$
Then, you calculate the solubility by taking in account the concentration of arsenate ion obtained by dissolution of silver arsenate:
$$K_\mathrm{sp} = 27 S^3 \times (S+0.02) = 27 S^4 + 0.54S^3$$
Solving this equation by applying the iteration method gives the value of $S = \pu{0.018M}$.
So the relative error is $23.9~\%$.

@Uros check this asnwer: http://chemistry.stackexchange.com/a/42231/23304

Your calculations are mathematically correct up to the second step and then I have no clue how you arrived at your value of $x$. Starting from there, $x$ would be:
$$K_\mathrm{sp} = 27 x^4 \\
\frac{K_\mathrm{sp}}{27} = x^4 \\
\sqrt[4]{\frac{K_\mathrm{sp}}{27}} = x \\
x \approx 0.021711\dots$$

However, to calculate the solubility you should be doing this:
$$K_\mathrm{sp} = \left[\ce{Ag+}\right]^3 \left[\ce{AsO4^3-}\right]$$
$$c\left (\ce{Ag3AsO4} \right) = x$$
$$K_\mathrm{sp} = \left(3x \right )^3 \left(x + 0.02~\mathrm{M}\right)\\
K_\mathrm{sp} = 27x^4 + 0.54~\mathrm{M} \cdot x^3$$
Note that this is the same equation as Uros proposed. I arrived there by saying:

$\left [\ce{Ag+}\right] = 3\ c\left(\ce{Ag3AsO4}\right)$ — silver ions all come from dissolved $\ce{Ag3AsO4}$
$\left [\ce{AsO4^3-}\right] = c\left(\ce{Ag3AsO4}\right) + 0.02~\mathrm{M}$ — $0.02~\mathrm{M}$ of $\ce{AsO4^3-}$ stem from $\ce{K3AsO4}$, the remaining from dissolved $\ce{Ag3AsO4}$.

Unfortunately, our equation is not (easily) solveable analytically to the best of my knowledge. You need some kind of estimation method. Uros’ estimate of $c\left(\ce{Ag3AsO4}\right) \approx 0.018~\mathrm{M}$ seems pretty accurate.

At the beginning you didn't account for the amount of arsenate produced by the dissociation of potassium arsenate. Why so?

@Marko At the beginning I was merely correcting OP’s calculations. But I’ll make that clearer.

Thank you. I really wonder how the lecturer will solve this problem in the class. If she shows an easier method, I will share it here.

The easier method is to neglect x in 0.02+x: Ksp=(3x)^3(x+0.02 M)=(3x)^3*0.02 M

@Jan check this asnwer: http://chemistry.stackexchange.com/a/42231/23304

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Why does a carbonyl reform in ester reactions
I have noticed that frequently whenever esters react with a nucleophile, the nucleophile will attack the carbonyl carbon and eventually in order for the carbonyl to reform the other oxygen will leave for example the first addition of the grignard reagent to an ester:
$\ce{PhMgBr + Ph-(C=O)-O-R}$
will result in
$\ce{(Ph)2-(C-O^{-})-O-R}$
as an intermediate, and then rather than the
$\ce{O^{-}}$ simply getting protonated with water for example, its favorable for the $\ce{O-R}$ to leave and the $\ce{C-O^{-}}$ to become $\ce{C=O}$. Why is the carbonyl reforming more favorable, isn't $\ce{O-R}$ a bad leaving group?

Alkoxide ion is only a "bad leaving group" in a relative sense (say compared to chloride). In this case, the tetrahedral intermediate resulting from the initial reaction between the Grignard reagent plus the ester is itself an alkoxide ion, and when the OR group leaves, that then becomes an alkoxide ion - so that step should be close to energetically neutral.
Saponification of esters by sodium hydroxide is another situation where alkoxide anion is a leaving group - again because alkoxide and hydroxide ions are close in energy.


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Deuterated solvents vs. regular solvents
Working on dissolving aspartame and i was wondering if it dissolves well in methanol will this mean that it dissolves well in deuterated methanol ?

And what do you think about it?

In general, it is safe to assume that solubilities in non-deuterated and deuterated solvents are much the same, and this would be the logical solvent of choice if looking for a NMR solvent.
However, there is a small difference in relative solubilities between H/D solvents, and you can gauge potential for differences based on the dielectric properties of the two. By memory, I think H2O/D2O have the largest difference (which is still pretty small).
Another factor that can cause differences in solubilities comes from the relative purity of the two solvents. For instance, many prepackaged deuterated solvents are often drier than bulk bench solvents, and this can sometimes cause significant solubility differences, especially in hygroscopic solvents like DMSO. I've seen countless examples of people claiming a compound was readily soluble in DMSO, only to have it sit like a brick when trying to dissolve in a freshly cracked vial of DMSO-d6.

I'm trying to find bp's for HMPA vs. HMPA-d18....

@Eric Brown - Reference dependent...check Sigma-Aldrich or CIL as a start. Here and here state 233 and 235 for D18 and H18 respectively.


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Chlorofluorocarbons rising
Chlorofluorocarbons are heavier than atmospheric gases. Then how are chlorofluorocarbons produced at the surface able to reach the ozone with such a heavy mass ?

Did you not read our responses (particularly mine) to your previous question http://chemistry.stackexchange.com/questions/41261/ozone-is-above-nitrogen-is-below, which is closely related to this one.  See also:  https://www.physicsforums.com/threads/if-air-is-a-mixture-why-dont-the-gases-separate.843424/

I suggest merging this question with your former question and then hoping for a comprehensive answer to both because they are so closely related …

Segregation in the lower atmospheric region is prevented by various mechanisms, mainly turbulent mixing of the atmosphere by rising and falling air streams. The main driving force of the large-scale atmospheric circulation is solar heating. The most important circular air flow pattern in the troposphere is the Hadley cell circulation. (Considerable information on the movement and mixing processes in the atmosphere was gained from monitoring following numerous tests of nuclear weapons in the atmosphere mainly during 1951–1962. The radioactive substances served as tracer material.)
Winds mix the atmosphere much faster than gaseous molecules can settle according to their weight. The natural composition of the dry atmosphere remains constant in the lower 80–100 km. Only above 100 km, segregation takes place because the concentration of light gases (e.g. helium) decreases more slowly with altitude than those of the heavy gases.
The same atmospheric mixing processes apply to chlorofluorocarbons. For example, trichlorofluoromethane $(\ce{CCl3F})$ is nearly uniformly distributed throughout the lower 15 km. At higher altititudes, the abundance of trichlorofluoromethane decreases – not because of its density, but because it is decomposed by UV radiation. By way of comparison, the less reactive tetrafluoromethane $(\ce{CF4})$ is uniformly distributed up to at least 50 km.
(Source: Frequently Asked Questions About Ozone (1): How Can Chlorofluorocarbons (CFCs) Get to the Stratosphere If They’re Heavier than Air? in Assessment of Ozone Depletion: 1998 – World Meteorological Organization Global Ozone Research and Monitoring Project – Report No. 44 [html] [pdf])


text:
When it is said 'at constant pressure, at constant pressure ', is it about pressure & temp. of the system or surroundings?
These are the following statements:

[...] the thermal decomposition of $1.0$ mole $\ce {CaCO3}(\text{s})$ at $1$ bar results in an increase of volume of nearly $90~\text{dm}^3$ at $800^\circ \text{C}$ on account of the carbon dioxide produced.

Is the system at $1$ bar or the surrounding exerts pressure of $1$ bar? Is $800^\circ \text{C}$ the temperature of the system or the surroundings?

The standard state of a pure substance is the pure substance exactly at $1$ bar & at $298.15~\text K$.

I think the pressure mentioned is of surroundings but the temperature? Is it of the system or he surroundings?

We see that at constant temperature & pressure, the change in Gibbs energy of a system is proportional to the overall change in entropy of the system plus its surroundings.

Now is the system at constant temperature & pressure or the surroundings?
Could anyone please explain me to clear this ambiguity?

When we say
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
we are referring to $\Delta G^\circ$, $\Delta H^\circ$ and $\Delta S^\circ$ of the system. As such, the variable $T$ in the equation refers to the temperature of the system. The condition for constant pressure $p$ is implicit; the equation above does not directly show why $p$ has to be constant. For more information, refer to this question and answer.
The sentences that you have given are not the best examples to explain this, because they do not relate to Gibbs free energy. I will give you a different example:

When hydrochloric acid is added to sodium carbonate at atmospheric pressure, the standard enthalpy change $\Delta H^\circ$ and the standard entropy change $\Delta S^\circ$ are measured to be $x~\mathrm{kJ~mol^{-1}}$ and $y~\mathrm{J~K^{-1}~mol^{-1}}$ respectively. Calculate the Gibbs free energy of the reaction and hence determine the equilibrium constant. (The temperature at which the reaction is carried out is $25~\mathrm{^\circ C}$.)

The wording of the question may seem to imply that it is talking about the temperature and pressure of the surroundings, and that is probably not a coincidence. For a simple chemical reaction like the one above, we would probably carry it out in a beaker. We weigh out a certain mass of $\ce{Na2CO3}$, and add a certain volume of $\ce{HCl}$, and pour the acid into the beaker and watch the bubbles.
Under such conditions, we can speak of the system being at mechanical and thermal equilibrium. Without going into too much detail about reversible and irreversible processes (I know there are loopholes in the explanation):

Mechanical equilibrium, in general, means that the net force acting on the system is zero. Correspondingly, this must mean that the pressure of the system and the pressure of the surroundings are equal at all points in time. If, at any point in time, this was not the case, e.g. if we had $p_\mathrm{syst} < p_\mathrm{surr}$, then the surroundings would "push" on the system and cause the system to contract until it reaches the point where $p_\mathrm{syst} = p_\mathrm{surr}$.
Thermal equilibrium means that there is no heat transfer into, or out of, the system. Analogously to above, this must mean that the temperature of the system and the temperature of the surroundings are equal at all points in time. We conventionally assume the surroundings to be a "reservoir" which maintains a constant temperature, and this is pretty true as far as our simple example is concerned. Therefore $T_\mathrm{syst} = T_\mathrm{surr}$.

Does that mean all processes need to be isothermal & isobaric? The standard enthalpy of ionization of hydrogen is +1312 kJ . Does that mean this energy is to be supplied to ionize hydrogen which is at 1 bar & 298.15 K? Does that mean during the process, the temperature must remain same? If so, then how could even there be exchange of heat, if the hydrogen remained at the same temperature as that of the surroundings all the time during the process? Gibbs Free energy is applicable when the process is isothermal-isobaric. Is it so? This is what I'm not understanding. [contd.]

If the system & surroundings are at same temperature 298.15 K throughout the process, how could there be supply of heat after-all this requires temperature-gradient, isn't it? Do every chemical reactions happen in the same way where throughout the process, the system & surroundings remain at same temperature? Can you please clarify that?

Your first two statements mean that, if T and p are the temperature and pressure of your system in its initial thermodynamic equilibrium state, the surroundings will be maintained at the temperature T and the pressure p throughout the change from the initial thermodynamic equilibrium state to the final thermodynamic equilibrium state (say by using a constant temperature reservoir and an external gas pressure).  Thus, in the final thermodynamic equilibrium state of the system, the temperature and pressure will also be T and p.  However, during the change between the initial and final thermodynamic equilibrium states, the temperatures and pressures within the system can vary spatially and with time.  Only at the boundary with the surroundings will the temperature and pressure match that of the surroundings throughout the process.  For more details, see Section 2.4 Properties of the Gibbs Function, in Denbigh, Principles of Chemical Equilibrium.

So, throughout the process, the substances needs to be at the same temperature as that of the surroundings, ?I s it so? There should be no increase or decrease in temperature of the substances throughout the process. Is it so? But how could there be exchange of heat if the system & surroundings all remain at the same temperature throughout the process? There needs to be a temperature-gradient. Isn't it? Also, if heat is being supplied or taken away from the system, wouldn't its temperature increase or decrease & thus remaining at the same temperature as that of the surrounding? Confused.

What do the words "However, during the change between the initial and final thermodynamic equilibrium states, the temperatures and pressures within the system can vary spatially and with time" mean to you?

So, during the process, the temperature & pressure within the system doesn't remain uniform? I thought Gibbs energy is applicable only for isothermal-isobaric process. Also, the equation $$G= H- TS$$ is a state function which implies $T$ is the temperature of the system, isn't it? So, $$\Delta G = \Delta H-T\Delta S $$ doesn't imply the change in entropy at constant temperature $T$ of the state?

The statements I was addressing didn't say anything about the thermodynamic functions.  But, irrespective of how the process is carried out, the change in the thermodynamic functions depends only on the initial and final thermodynamic equilibrium states. So the Gibbs energy doesn't only apply to isothermal-isobaric processes.   For a process in which the pressure in the system doesn't remain uniform, even though it is held constant at the boundary, q is still equal to $\Delta H$.

Yes, they are state-functions. But could you tell why the site I linked wrote isothermal-isobaric system?

I need more context than this.

Have you not understood my query, sir?

Let us continue this discussion in chat.

It is important think of the thermodynamic functions U, H, S, G, and A as physical properties of the material(s) comprising the system, not determined by the path of any particular process.  But note that, since the only way we have of evaluating the change in S between two thermodynamic equilibrium states of a system is to calculate the integral of dq/T for a reversible path.  This is independent of the actual path that took the system from its initial state to its final state.  Since G and A are also functions of S, the only way we have of evaluating the changes G and A are also to determine them for a reversible path.  This too is also independent of the actual path that took the system from its initial state to its final state.

authors: кирилл баканов


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Why do the melting points of Group 15 elements increase upto Arsenic but then decrease upto Bismuth?
The boiling points of group 15 elements increase on going down the group (or, as size increases) but the same is not true for the melting points. The melting points increase from $\ce{N}$ to $\ce{As}$ and then decrease from $\ce{As}$ to $\ce{Bi}$. Why is this the case?

A lower melting point is an indication of a lesser organised solid structure. So the question is actually:

What in the structures of arsenic and phosphorus make them less prone to melting?

Let’s start off by separating nitrogen from the entire group. It is a prime example of a diatomic gas such as hydrogen with a very inert $\ce{N#N}$ triple bond. As throughout the periodic table, multiple bonds with s and p orbitals between atoms of the same element are magnitudes more stable in the second period than anywhere else due to much better p-orbital overlap (smaller size, smaller interatomic distance). Therefore, we can consider phosphorus, arsenic, antimony and bismuth more or less as similar while nitrogen is an odd one out.
Phosphorus has four stable allotropes with vastly different structures. The white phosphorus allotrope, $\ce{P4}$ molecules, basically falls into the same class as dinitrogen and melts at around 40 degrees. Red, purple and black phosphorus build up network-like structures in two dimensions, either in tube-like chains (red) or as ‘wavy graphite’ (black). Only the black modification is clearly defined. Red phosphorus has melting points of around $600~\mathrm{^\circ C}$ while black phosphorus sublimes in vacuum at $400~\mathrm{^\circ C}$ — probably meaning that its melting point at standard pressure is much higher, maybe even higher than arsenic’s sublimation point. I found no references to liquid phosphorus so I have no idea whether the bonds of the network structure need be broken to liquefy or not.

Figure 1: crystal structure of black phosphorus, taken from Wikipedia.
Henceforth, we have established a structure element which we will be sticking to with only moderate modifications. The transition phosphorus — arsenic — antimony — bismuth is more or less the non-metal—metal transition. What explicitly changes is the distance between the sheets as shown in table 1.
$$\textbf{Table 1: }\text{Comparison of bond lengths and inter-sheet contact distances for pnictogens}$$
$$\begin{array}{lrrrr}
\hline
\text{ } & \ce{P_{black}} & \ce{As} & \ce{Sb} & \ce{Bi} \\
\hline
\ce{E-E}\text{-bond/pm} & 223.0 & 251.7 & 290.8 & 307.1 \\
\ce{E\bond{...}E}\text{-contact/pm} & - & 311.9 & 335.4 & 352.8 \\
\text{d}_\text{contact}/\text{d}_\text{bond} & \approx 1.5 & 1.239 & 1.153 & 1.149 \\
\hline
\end{array}$$
So while the phosphorus structure can be described well as consisting of semi-isolated sheets stacked above each other; when we have reached bismuth, these sheets are much closer to an overall three-dimensional network. Therefore, bismuth atoms are in a distorted octahedric environment as other metals would be. Arsenic, where the sheets are still well separated, exhibits almost metallic conductivity perpendicular to the sheets.
Putting this together and taking the extreme points, we can say:

The phosphorus structure is highly organised in that two directions are strictly discernible. Melting along one of these is no problem (perpendicular to the sheets) but melting within a sheet is a huge problem as $\ce{P-P}$ bonds would need to be broken.

The bismuth structure no longer has a true distinction of directions; all three axes are more or less equal. Therefore, it takes similar energies to break bonds in one direction as it does to break them in another direction meaning the overall process is easiest.

Furthermore, liquid phases are generally those that display a high local ordering but a low long-range ordering. The high local ordering can be preserved in bismuth even if the long-range ordering present in the solid state is lost. Conversely, the local ordering is phosphorus is a lot stricter and cannot easily be transferred into a state where long-range ordering is low.

Finally, note that non-metals whose solid phase does not consist of small, well defined molecules generally have higher melting points than metals. (Compare carbon’s sublimation point of c. $3600~\mathrm{^\circ C}$.)

The reason is that after arsenic, there is shielding. This causes the intermolecular forces to become weak. Thus, the melting point decreases. I hope you know about shielding of d and f electrons

But d orbital is present in arsenic too then why arsenic higher boiling point than phosphorus

As we move down the group no.of shells increases so the energy to release the electrons also increase (size is directly proportional to the Vander walls forces ) as we move down the group size increases so Vander walls force also increases therefore we have to boil it for a long time at a high temperature to break the forces and release electrons

Energy to release electrons decrease and not increase

Please be more diligent when writing answers. This is the famous scientist your are trying to name here: https://de.wikipedia.org/wiki/Johannes_Diderik_van_der_Waals


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Alkenes express different reactivity with acids depending on concentration and temperature
The below images show when 1-butene reacts with different reagents (in d it's also H2SO4 not H2SO). My question is why are the two products different?
I can get that in c), because it's more dilute there's more water to act as the nucleophile, and in d) there's more H2SO4 so that would act as the nucleophile but what's the role of temperature?

If hot concentrated sulfuric acid were used it may cause the butene molecules to react with each other to form an ether or another possible compound due to the dehydrating properties, high reactivity and protic properties of the sulfuric acid. The reaction to form butanol at a specific temperature must need water to stop this occuring. The second product must require the sulfuric acid to be stronger , so to prevent a reaction between the butene molecules and favour the ester formation over the alcohol formation the reaction is cooled.


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H2Te dipole and electronegativities of H and Te (high school level chemistry)
I saw somewhere that $\ce{H2Te}$ is a polar molecule, but as far as I know the only time dipole-dipole forces arise (giving molecules the possibility of becoming polar) are when there is a difference in electronegativity. I know $\ce{H2Te}$ is not a linear molecule but if there is almost no (less than 0.4) difference in electronegativity values the shape shouldn't matter as electrons wont be more drawn to either atom (or 'side' of the molecule). In other words, it seems to me the intermolecular forces should be induced-dipole induced-dipole (making the molecule non-polar), not dipole dipole.
I should clarify, I first saw this statement in the answers to a high school chemistry exam. I then googled it, and got the same answer (i.e that the molecule is polar).

I think you mean $\ce{H2Te}$. Electronegativity of  $\ce{H}$ is 2.2.  Electronegativity of  $\ce{Te}$ is 2.1. You can have Van-der- Waals forces between  non-polar molecules, such as $\ce{I2}$

sorry yes meant H2Te. Yes van der Waals forces do exist, but shouldnt they be London forces? not dipole-dipole, making the molecule polar

It would be educating for you to learn about ozone dipole moment. I guess, that case of azulene would be interesting too.

BTW, hydrogen-deiterium molecule has non-zero dipole (about 0.001 D) =)/

@Rani - what's the electronegativity of a lone pair?

Information in this question is all over the place and often out of context. Let me get a few things straight:

Van der Waals forces, also known as London interactions, occur between every pair of molecules regardless of polarity. They are the only intermolecular force observed for non-polar molecules such as $\ce{I2}$. They are explained by induced and spontaneous dipoles at a molecular level and femtosecond timescale.
Dipole interactions are present between macroscopic dipoles and account for the boiling point difference between e.g. tetrahydrofurane and cyclopentane ($\Delta \vartheta_\mathrm{b} \approx 15~\mathrm{^\circ C}$). For those, you need a notable difference in electronegativity resulting in an overall dipole (so $\ce{CO2}$ is not a dipole because the dipole moments of both $\ce{C=O}$ bonds cancel each other out). For strong dipoles, these interactions are orders of magnitude larger than the underlying van der Waals forces.
Electronegativities are, of course, atomic properties (and also depend on things like oxidation state if one is truly accurate). For no two different elements is the electronegativity difference ever zero. Every $\ce{X-Z}$ bond is always polarised ever so slightly in one of the two directions. Sometimes, as in the case of $\ce{NCl3}$, this is best determined experimentally. ($\ce{N-Cl}$ bonds in $\ce{NCl3}$ are polarised towards nitrogen.)
However, only bonds which are polarised to a certain extent (usually a cutoff value of $\approx 0.4$) are called polar on a high school level as per convention.
$\ce{H2Te}$ has non-polar bonds as per convention of the fourth bullet point. Therefore, you should consider it a non-dipole and hence only consider van der Waals forces/London dispersion forces. (DavePhD’s source is likely correct, but the dipole moment is neglegible at high school level.)
I would love to see a link to the pages that say $\ce{H2Te}$ be notably polar/display dipole dipole interactions. (Compare the $0.2~\mathrm{D}$ to the $1.78~\mathrm{D}$ dipole moment of ethyl acetate.) Calling van der Waals forces dipole interactions is usually considered wrong (although they build up upon the same principles: electrostatic interaction).

I tried to stick to a high-school level as well as possible. Hence a number of simplifications in this answer. At research levels, every molecule without improper rotation (of which inversion and symmetry planes are a special case) should be considered to have a non-zero dipole moment as DavePhD’s answer states. Thus, every molecule of that type will display dipole-dipole interactions however weak they may be.

"An ab initio Calculation of the Dipole Moment Surfaces and the Vibrational Transition Moments of the H2Te Molecule" http://www.sciencedirect.com/science/article/pii/S0022285296971453 is the latest I can find, they have 0.298 D calculated and experimentally "certainly larger than 0.19 Debye and very probably smaller than 0.26 Debye"

You can indeed have induced dipole-dipole between non-polar molecules, especially between molecules/particles with a large and "easily polarizable" electron cloud.
The electron cloud is not stationary in time, but dynamic. This means that the electron cloud is subject to fluctuations in its distribution. So, at some point in time, the electron cloud will be shifted in one direction. Hence, the molecule has become slightly polar. These fluctuations take place on a very, very, very short timescale, and basically the "non-polar" molecule is not really non-polar (although it may be close).
When this slightly polar molecule approaches a non-polar one, the small dipole moment induces a shift in the other molecule's electron distribution (due to electron-electron repulsion). The slightly negative part of molecule A will be attracted to the slightly positive part of molecule B, and a lower-energy state is achieved.
This is called induced dipole-dipole interactions, and is a simple explanation of how certain melting-point trends can be observed for groups in the periodic table. $\ce{He2}$ is another example of a molecule held together by induced dipole-dipole interactions.
Here is a (not very good) picture depicting the process in forming $\ce{He2}$.
See this youtube video for an introduction to various dipole-dipole interactions (skip to 4:35 to start on the induced dipole-dipole interactions)

Firstly, $\ce{H2Te}$ has a permanent dipole moment of 0.2 Debyes.
More generally, the symmetry of the molecule should be first considered to decide if the molecule has a dipole moment.
If the symmetry group is $C_1$ (only $E$ meaning identity symmetry operation), $C_\mathrm{s}$ (only a single plane of symmetry and $E$), $C_n$ or $C_{n\mathrm{v}}$, then think that the molecule has a dipole moment regardless of what particular atoms are involved. Someone could argue that these could still hypothetically have no dipole moment at some particular state, such as a specific rotational and vibrational state, but realistically these molecules will have a dipole moment. Then the specific elements and electronegativities could be considered to quantify the dipole moment.


text:
Room-conditions supercritical fluids?
Are there any reasonably obtainable supercritical fluids that I could, say, run my hand through? Wikipedia makes it sound like there are plenty of room-temperature ones, but no room-pressure fluids.

It looks that indeed there are no such fluds. Helium has probably lowest critical pressure - over 2  atm.

I mean, 2 atm is survivable, so if there aren't any others, you could post that as an answer.

See also http://physics.stackexchange.com/q/268470/26969

If exceeding the "liquid–liquid critical point" for a solution suffices, then perhaps you could survive contact with methane hydrate for a short while. Though He is supercritical at ~two atmospheres, running you hand through it would be a chilly experience at about 5 K.

The question asks for "supercritical" fluids.  So helium at 25 atm and 25 °C qualifies.  It would feel exactly like a gas at those conditions, because supercritical fluids are just gasses.  There isn't really anything too exotic about them.

As you can see on this data page on Wikipedia, helium-4 has critical pressure of 2.24 atm, but helium-3 has 1.13 atm - only slightly above atmospheric pressure. Hyperbaric chambers work with 1,3-1,5 atm, so you could try there.
There is, unfortunately, question if it makes sense, as you wouldn't notice any serious difference in properties between subcritical helium-3 and slightly above critical point in pressure, but so far above in temperature. It would still be lighter than air, for example.
I don't think that much higher pressure would be a huge problem - the highest pressure survived by human is supposedly 54 atm - so says Wikipedia, but I didn't find proof. If there is a compound with critical parameters sufficiently near standard, you could touch liquid-like, not gas-like supercritical fluid.
Ionic liquids have relatively low critical pressures (even only 14 bar), but too high crit. temperatures (source). On the other hand ethane's crit. temp. is below 30°C, but pressure over 48 atm. Maybe there's sweet spot somewhere, but it would be still high pressure and may be quite hot or cold.

Quick googling got me to $\ce{CF3Cl}$, that has cr.p. at $\pu{28.8^\circ C}$ / $\pu{38.6 bar}$. $\ce{C1}$-$\ce{C2}$ freons look to have critical point in $\pu{30 .. 60 bar}$ / $\pu{0 .. 100^\circ C}$ range, so some of them probably are survivable. However, some compounds of this class are used for general anesthesia, so $\ce{He/O2}$ mixture respirator would be required.

A number of long carbon chain molecules with large numbers of fluoride atom, have $P_c < \pu{10 atm}$ and $T_c < \pu{1000 K}$. That is within the reach of a good bicycle pump and a burner. In particular $\ce{C12F26}$, $\ce{C15H4F28O}$ and $\ce{C20F42}$:
\begin{array}{lll}
&\ce{C12F26} &P_c = \pu{912 kPa} &T_c = \pu{417 K} \\
&\ce{C15H4F28O} &P_c = \pu{784 kPa} &T_c = \pu{701 K} \\
&\ce{C20F42} &P_c = \pu{463 kPa} &T_c = \pu{700 K}
\end{array}
The critical point of a fluid is determined by its van der Waals constants.
For a fluid to be critical at STP it needs these constants to have the values
$a = 25$ and $b = 0.003$ in SI units. The value of $b$ corresponds to a molecule with around 300 atoms, but the value of $a$ is outside the range of published compounds.
However, something compounds like $\ce{C100F202}$ could be critical at around ordinary pressure and an easily achievable temperature, i.e. it might be possible to get a critical fluid by heating some goop in an open saucepan.
It also seems that fluorocarbons tend to be stable and inert, so they may be safe enough to handle.
@Diracology and @Floris provide an in depth exposition of the relationship between critical points and van der Waals constants in their answers to my Physics.SE question Fluids with critical point at ordinary temperature and pressure. Their answers with internet search of van der Waals constant data lead me to start looking at at large fluoroacarbons. The ones listed above are the largest I could find data for. It should be possible design more suitable molecules computationally. See Computing van der Waals constants from molecular structure.

C100F202 would be a Teflon oligimer


text:
What in Peanut Butter dissolves glue?
There is a common belief that Peanut Butter can dissolve certain types of glue. I've tested it and it does on a few common stickers. A forensic science crime procedural TV show (Bones) made a statement that

The oils in the peanut butter will break down the cohesion of the cyanoacrylates.

The most common cyanoacrylates based glue is Super Glue, which hardens quickly.
How does Peanut Butter interact chemically with the CA glue and why? Does it make a difference when the CA has hardened or not?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

The superglue is originally the cyanoacrylates suspended in a solvent such as acetone or MEK, as the solvent evaporates after application the glue dries. The natural oils in the peanut butter or other adulterants present would act to dissolve the cyanoacrylates or possibly convert them to a different compound. If it's crunchy peanut butter the small nut pieces would act as an abrasive and speed up the process. The cohesion is referring to how the materials are bound together.

I would add to what Joel wrote to say that it may be necessary only to weaken the interface between the glue and the substrate (surface it is attached to), rather than actually react with the cyanoacrylate resin. Besides the oils in peanut butter, which can lubricate between the glue and the substrate and allow slippage, the proteins in peanut butter can act as soaps to disrupt attractive forces between polar groups on the cyanoacrylate resin and the substrate.


text:
How is volume a State function?
My question as stated above, is how Volume a state function? As described by the  definition of a state function, that it is independent of the path adopted to bring about a change in it, how can volume be one, wouldn't volume change differently if we brought about a change in it by say increasing/decreasing temperature rather than say pressure?
Thanks
PS:Please explain in simple terms, as i am beginner in thermodynamics.
Thanks again.

You tell me the temperature and pressure, and I'll tell you the volume (per mole).  Does this have anything to do with what path you used to get to that state?

As described by the definition of a state function, that it is
  independent of the path adopted to bring about a change in it [...]

This is not the definition of a state function, since it is missing the crucial first part of it. Here is the definition: a state function is a property of a system that depends only on the current equilibrium state of the system, not on the way in which the system acquired that state.
The point is that something is a state function if you can tell its value by knowing just the equilibrium state of a system and nothing else. As Chester Miller said in his comment: tell me the temperature and pressure (i.e. specify the state of a system), and I'll tell you the volume. It is irrelevant how the system acquired that equilibrium state, the volume is determined by just the state.

Wouldn't volume change differently if we brought about a change in it
  by say increasing/decreasing temperature rather than say pressure?

In general, of course, when changing the system differently, its volume will also change differently. But, if by doing different changes you eventually will reach the same equilibrium state, all the state functions (and not just volume) will have the same values.


text:
Method for increasing concentration of extracts
I am using reagent grade propylene glycol (99.9% pure) and vegetable glycerin (99.7% pure) separately as solvents to extract certain active essential oils/molecules of various plant flowers.
The question is about after saturating the solvents with the essential oils. Can you further reduce the infused solvent solution volume?  I am starting with 50 fluid ounces and want to reduce the solvents to 30 fluid ounces.
The goal is to concentrate the the solvent solution and not harm the active ingredients suspended in the the two solvents.  The solvent boiling temperature is known and the essential oils/molecules are destroyed 80 degrees below the solvent's boiling points.


text:
What is the stereochemistry resulting from hydroboration of 1-methylcyclopentene?

What is the major product of the following reaction sequence?

I'm supposed to work out the structures of both 2 and 3, but I already know 2 to be 1-methylcyclopentene (as the most substituted alkene is formed during elimination).
For the hydroboration step, boron should add to the less substituted carbon (in an anti-Markovnikov fashion). However, I'm not sure which of the two choices B and C is correct. Why should one diastereomer be formed preferentially over the other?

The second step is a syn addition of the boron hydride. As depicted below, it will result in an anti-Markovnikov product with the boron on the less-substituted carbon. Both hydrogen and boron are on the same side of the former double bond and these two are the ones we refer to when we say syn addition.
The final step, the oxidation by hydrogen peroxide, is stereospecific with retention. So the newly-formed $\ce{C-O}$ bond is on the same side as the former $\ce{C-B}$ bond. Thus, an anti relationship is established between the methyl group and the hydroxy group.
Since 1-methylcyclopentene is planar, a racemic mixture of enantionmers is formed; neither re nor si side are favoured. The scheme below shows the attack from the si side of the $\mathrm{C2}$ atom.


text:
Can a stereospecific / stereoselective reaction produce a racemic mixture of enantiomers?
I think I have my terms mixed up but I'm confused about the concept of stereoselectivity / stereospecificity as they relate to reactions with alkenes.
For example, hydroboration is a stereospecific reaction. It leads to syn addition across an alkene. However, you can form a racemic mixture from it because you can add from either the top or bottom faces.

I think this is a good depiction of what I mean. It seems to me that hydroboration is called stereospecific because it does syn (rather than anti and syn). But it still forms two different stereoisomers, so how can we still call it stereospecific?

You haven't mixed up your terms: a stereoselective reaction is defined as one which preferentially forms one stereoisomer over another. Likewise, all stereospecific reactions are also stereoselective, also stereospecificity has stricter criteria then plain old stereoselectivity.
However, you have to remember that the term stereoisomer encapsulates both enantiomers and diastereomers. The hydroboration reaction is stereospecific because reacting one particular diastereomer of the alkene (either cis or trans) leads to one particular diastereomer of the product. It doesn't necessarily lead to a particular enantiomer. Basically, hydroboration leads to the formation of a racemic mixture of a particular diastereomer.
The alkene you have chosen, 2-methylbut-2-ene, is not a good example substrate that can actually show how the reaction is stereospecific. Since one of the alkene carbons is symmetrically substituted, there's no diastereomerism present in the products. If we use (2E)-3-methylpent-2-ene as an example substrate, then it becomes clearer:

Note how neither of the (2R,3R) or the (2S,3S) stereoisomers are formed. However, the two products that are formed are enantiomers of each other.

authors: Виктория Гребёнкина


text:
Choosing Stationary Phase Column Chromatography
When using column chromatography to purify an amine, alumina is the best choice of stationary phase (rather than silica gel.) Could this be due to alumina being slightly basic and thus the impurities are retained in the column longer?

I am not a specialist but it may be due to the lone pair of the nitrogen and the electronic gap of the aluminium. Don't forget that Al is on the third line of the PT so it can be hypervalent. He has no electron in his 3d-orbital and "a part" of an electron can be in.


text:
Will a strong acid still dissociate "completely" even in the presence of a stronger acid?
For example, nitric acid or sulfuric acid in an equal-molar solution of perchloric acid. Won't perchloric protonate a significant concentration of nitrate ions?

You’re in aquaeous solution? Yes, as long as the overall concentration isn’t too high.

Nitrate ions, not nitrite ions. Nitric acid -> nitrate; nitrous acid -> nitrite. But what Jan said. Unless you make the acids really concentrated, it's not likely to happen.

What if the solution was highly concentrated acid?

As far as solutions in highly concentrated acid, Nitric acid acts like a base in concentrated sulfuric acid (it forms the nitronium ion).

Suppose you have nitric acid (about Ka=20) and perchloric acid (Ka > 100,000).
$\frac{[A-][H+]}{[HA]} = K_a$
In 1M nitric acid alone, let x be [$\ce{NO3-}$], that is dissociated nitric acid
$\frac{x^2}{(1-x)} = 20$
x = 0.954M
the nitric acid would be about 95% dissociated.
If the solution is 1M nitric and 1M perchloric, with all the perchloric acid dissociating, the nitric acid equilibrium would be:
$\frac{x(1 + x)}{(1-x)} = 20$
x= 0.913
the nitric acid would be about 91% dissociated.
For 2M nitric acid alone:
$\frac{x^2}{(2-x)} = 20$
x= 1.83M
or 92% dissociated.
For 2M nitric and 2M perchloric:
$\frac{x(2 + x)}{(2-x)} = 20$
x = 1.69M
or 84% dissociated.
So yes it can make a difference and it is more apparent at higher concentration.

It might be worth mentioning the "leveling effect," e.g. something in concentrated sulfuric or acetic acid.


text:
Help solve this specific linear schrodinger equation, please
$$\left[-\frac{1}{2}\nabla^2 - \frac{2}{r} + C\right]\phi(r) = E\phi(r)$$
where $C$ is a known function. I am just looking for some help on the strategy to solve explicitly  for $\phi(r)$. One thing I am having trouble understanding is how we can solve this equation without expressing $\phi(r)$ in terms of $E$, but it seems like it is done all the time:
http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/05.5%3A_Particle_in_Boxes/Particle_in_a_1-dimensional_box
Thanks!

http://math.stackexchange.com/

@Shadock thats legit, but I feel like people here will probably know the specifics of SE better

This is not a problem, we need to use maths in chemistry and physics and in a lot of different subjects. If you want a good answer you may ask the question in ME. :-)

The particle in a box is a "simple" system in that the SE reduces to basically $\mathrm{d}^2\psi/\mathrm{d}x^2 = -k\psi$ and the general solution is quite obvious. When you have that $-2/r$ term, it gets a bit more irritating. Exactly what context are you trying to solve the SE in? The form of the SE that you are using makes me think the $\ce{He+}$ ion (where that function $C = 0$) - if I am not wrong you can use the power series method. There is a name for the solutions, I think spherical Bessel functions.

@orthocresol Hey! Good question...the form of that function is actually the hartree fock mean field approximation integral, so you are spot on, I am trying to do the HF method for helium.

Ok, I'd better let someone who knows this topic answer :p

I don't know if that someone exists :), all these quantum chemistry books say this is the go to method besides DFT yet it seems pretty complicated to the "go to" method

Don't worry, there are lots of smart quantum people around; I'm sure they'll notice.

@user2879934 As orthocresol mentioned this problem looks like the Schroedinger equation for a hydrogen-like atom, apparently the $\ce{He+}$ ion. For this kind of problem you can't simply use the simple approach that works for the particle in the box. Traditionally this sort of equation was solved via the Frobenius method (a.k.a. power series solution). Nowadays you might rather want to employ the newer
procedure using ladder operators. Both of these methods should be covered in great detail in Quantum mechanics textbooks.


text:
Reasons which explain why the efficiency of a reaction increases as the average rate decreases?
Reasons which explain why the efficiency of a reaction increases as the average rate decreases? in other words, why does a slow chemical reaction is more effective than quick ones?
I done an experiment and the reletionship I found was that. I added a catalyst to vary the average rate (speed) of my reaction and the slower it took to react, the most product was got. Cant find a possible reason for that. any help??

In most cases this is because temperature enhances all kinds of side reactions. For the second major reason, see the Nanoputian's answer.

Well it depends on the reaction. A notable example is the Haber process. The Haber process is the production of ammonia gas from nitrogen and hydrogen gas: $$\ce{2N2 + 3H2 -> 2NH3 + heat}$$ This reaction is extremely important in the industry sector as ammonia gas is used for a wide range of products such as fertiliser, bleach, cleaning agents and in WW2, it was used to make explosives.
Lets go off on a tangent for a moment:

In WW1 the Germans had very little resources due to British ships blocking Germany's allies from providing them materials. This meant that they had to produce great quantities of ammonia themselves as they desperately needed it to make explosives. Thankfully German chemistry, Fritz Haber, created the Haber process which allow the Germans to mass produce ammonia. It is estimated that WW1 was prolonged by 2 years due to the creation of the Haber process. Haber later received the Nobel Prize in 1918 for his production process of ammonia and 'contribution to agriculture'.

So the question you might ask is why was creating ammonia so hard? After all it is a pretty simple reaction. Well this has to do with your question.
You see, producing ammonia gas is actually quite easy, however the main issue is that it is very difficult to get a high yield at a fast rate. The reason is due to the extremely high activation energy of this reaction. This is because for the reaction to occur, the extremely strong triple bonds of nitrogen must first be broken, which requires quite a lot of energy. This high activation energy means that the reaction is going to be really slow. This shown by the Arrhenius equation:
$$k = A\mathrm e^{-\frac{E_\mathrm A}{RT}}$$ However we are able to increase the rate of the reaction. If we were to increase the temperature, according to Arrhenius  equation, the rate of the reaction will increase. However now there is another problem, this time about the yield of the reaction.
Note above in the chemical equation, the Haber process is an exothermic reaction, meaning that heat is released. Now if we were to increase the temperature, according to Le Chatelier's principle, the equilibrium will shift to the left. This means that less ammonia gas will be produced. This is shown by the graph below:

So the above conundrum describes your question. By increasing the temperature of the sytem, you can increase the rate of the reaction, however by doing so you are going to decrease the yield.
Note: this is only true if the reaction is exothermic. If it is endothermic, then the yield will increase with temperature.

This answer was extremely useful for my work and understanding of the topic. Thank you a thousand timess!!!!!! Regards, Canacana

@CANACANA Thanks, I am happy that  I was able to help you.


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Why does two nails of the same material not produce electricity when placed in saltwater?
If I put two the same nails in a saltwater why could it not produce electricity when I put the metal aluminum and copper it produces electricity?

First, you need to understand what is electricity. Electricity is simply the flow of electrons. Now, what I really suspect you are asking is why these electrons flow.
Electrons flow in an electric current due to a potential difference which is the difference in potential energy of the electron at two different places. Like everything else, electrons like to lower their potential energy (imagine water falling in a waterfall) hence in an electric current, electrons move because they are moving from a place of high potential energy to a place of lower potential energy.
Now when you place two of the same nails in a salt solution, there is no potential difference between the two nails as the potential energy of the electron moving from the two nails remain the same (since they are made of the same material).
However when you place two different metals into salt water, such as aluminium and copper, there is now a potential difference between the two metals. Hence this will cause the electrons to move creating electricity (an electric current).
For more detail about the voltage change and what the chemical reactions that are actually occurring and you google electrochemical cells and electrochemical series. Hyperphysics and Chemguide are both very good resources which provide in-depth explanation on what is actually going on.


text:
Why is the F-F bond in fluorine a p-p sigma bond?
Won't the lone pairs hybridize to $\mathrm{sp^3}$ making it an $\mathrm{sp^3}$-$\mathrm{sp^3}$ $\sigma$ bond? If it still is $\mathrm{p^\sigma}$-$\mathrm{p^\sigma}$  bond what is the spatial location of the lone pairs?

It certainly won't. What you can do is to use the hybridization model to describe the bonding in F-F. The bond is surely not made only by pure p-p overlap.

@Marko Yes it is …

There is contribution from 2sigma and 1pi(u) MOs to the F-F bond: http://image.slidesharecdn.com/videolectureforb-141228060539-conversion-gate01/95/video-lecture-for-btech-13-638.jpg?cb=1419768417

Hybridisation is a mathematical concept chemists use to better understand spatial distribution of bonded atoms. It should only be used if we cannot explain the structure will without resorting to hybridisation.
The energy of a 2s orbital is lower than that of a 2p orbital in atoms with a fully populated 1s orbital. Therefore, there is a desire to keep its energy low. A bond between two p orbitals (of different atoms) will also lower the energy of one of these p orbitals (and create an antibonding σ* unpopulated one, too).
Furthermore, p orbitals extend into space in a directed manner: Along their coordinate axis, while s-orbitals are spherical in shape. The p orbital is already in a nice direction for bonding; any mixing with s orbitals will make it extend into space slightly less — potentially unfavourable for orbital overlap.
Finally, the lone pairs on each fluorine atom will be a set of two p type orbitals, each carrying a lone pair perpendicular to the bond axis, and an s type orbital. Experimentally, one should be able to detect two different energy levels for the two different lone pairs. (However, I didn’t go looking for papers and I can’t remember any lecturer mentioning any off the top of my head.)
So the $\ce{F-F}$ bond is indeed a $\mathrm{p^\unicode[Times]{x3c3}\ce{-}p^\unicode[Times]{x3c3}}$ bond because it makes the overall system energetically favourable.

There are so many papers trying to explain that bond, there is no ultimate truth, and I don't think there is a general consensus. I would expect at least some contribution from the s orbitals, as the in-axis lone pair will definitely have to be polarised away from the bonding pair. (In pictorial terms of MO theory.) If I recall correctly, F2 is used as a point in case example for the charge shift bond.


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If 130.0 mL of 3.0×10−3 M Na2SO4(aq) is saturated with CaSO4, how many grams of would be present in the solution?
If 130.0 mL of 3.0×10−3 M Na2SO4(aq) is saturated with CaSO4, how many grams of would be present in the solution? [Hint: Does the usual simplifying assumption hold?]

My answer is still wrong. What have I done wrong?

Your answer and method used appears to be correct. I don't see why it would be wrong.

"how many grams of would be present in the solution?" - how many grams of what? Are you sure it's CaSO4 and not Ca or SO4?


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Validity of rationale for balancing chemical equations
The reason given for balancing chemical equations is the law of conservation of mass. They say that we balance equations with keeping in mind that atoms of every element included in the reaction must be equal on both sides of equations. And they just add some coefficients for this purpose.
But I apprehend why is the amount of substance not brought under consideration while doing this?
For example when we write,
$$\ce{H2 +O2->H2O}$$
That is; until we don't know how much of $\ce{H2}$ and $\ce{O2}$ was included in the reaction, how can start to balance it? How can we decide that atoms are not equal on both sides?  However, the molecular formulas given above are only representing the substances.

The concept of chemical equilibrium should answer a few of your questions.

When you write the equation
$$\ce{H2 + O2 -> H2O}$$
you are implicitly saying that the stoichiometric coefficient of every substance is $1$. It is like writing the equation $a + b = c$. By not having coefficients in front of $a$, $b$, or $c$, you are implying that $1a + 1b = 1c$ and not $2a + 3b = 4c$ or anything like that.
Obviously that cannot be true for the reaction above because if you react one $\ce{H2}$ molecule and one $\ce{O2}$ molecule, you get one $\ce{H2O}$ molecule and one leftover oxygen atom that doesn't know what to do with itself. The only balanced equation that does not lead to stray atoms is
$$\ce{2H2 + O2 -> 2H2O}$$
or any integer multiple of that, i.e.
$$\ce{$(2n)$H2 + $(n)$O2 -> $(2n)$H2O}$$
Humans prefer to work with small integers or half-integers, so $n$ is usually either chosen to be $1$ or $1/2$.

Suppose if I have a balanced equation, $\ce{C + O_2 -> CO_2}$ then It means that 1 atom of C reacts with 1 molecule of oxygen to give 1 molecule of $CO_2$. From where does come Carbon Mono-oxide as a product of side reaction? However we earlier said that equation is balanced.

That equation simply does not reflect the existence of the byproduct CO. You would have to write another equation for that.

And since you said, we could also write that equation in the form $$\ce{H_2 + O_2 ->  H_2O + O}$$ then it would be said balanced again. Why we add coefficient instead?

@Truth-seeker this is irrelevant. When Hydrogen molecule react with oxygen molecule, it produces Water vapors. Not Oxygen. When we add 2 there, we mean two molecules of $H_2O$ and therefore the compound is not changed by this edition. So, it is lawful to write 2 before $H_2O$ but it is not to write $O$ with $H_2O$.


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Heavy Metal Toxicity
Ok, so I am working on a project where I have to make a tax for manufacturers using non recyclable materials. So for this tax you multiply the weight (in kg) by how toxic the material is, this gives you an answer in pence. So I have a list of heavy metals that are dangerous to the environment but I don't know what one is the most toxic and the least toxic.
This is what I think the order from most dangerous to least dangerous is:

Selenium (most harmful to the environment)
Arsenic
Mercury
Copper
Lead
Cadmium
Nickel
Chromium
Antimony
Zinc (least harmful to the environment)

Could someone please share their opinion on this order and suggest a better order. Also could someone suggest a suitable number to multiply the weight by to get price in pence.
Thanks

and I would also like a suitable number to multiply the weight by for each metal. This sounds like an important part of your project that you should not be asking someone else to do for you. I think part of the goal of your project is to devise your own scale and justify it.

Can you at least put them in order because I really need that part

My answer below explains how figure out that order. It is not so straightforward that the average chemist has committed it to memory. I can group them into three categories (low, moderate, and high toxicity) by intuition, but I do not know the specific order. Given that effort has to be exerted to generate an order, it would be better for you to do it so that you understand your order when you need to explain it. In fact, I guess that the order you choose is less important than the logic of your rationale for this assignment.

This is probably a more complex problem than your instructor intended. In addition to the various metals having different toxicities, different compounds of these metals have different toxicities.
For example, nickel metal has low toxicity, but the compound $\ce{Ni(CO)4}$ is a poisonous, flammable, volatile liquid.
Likewise alkyl mercury compounds, e.g. $\ce{(CH3)2Hg}$ are vastly more toxic than the low solubility mercury salts like $\ce{Hg2Cl2}$.
If you just want to know about the pure metals, then this can be easier. Look up a Safety Data Sheet for each metal. It would be better if you pull all your SDS from the same source, that way they are all using the same data sources.
For example, here is a link to the SDS for mercury from Fisher Scientific. Section 12 contains ecological information, including LC50 (concentration that kills half of the population in the time period given) data for various aquatic organisms. Section 11 contains toxicological information, including LD50 (dose that kills 50% of the population) for various organisms.
The best way to do this is to try to find one set of data for all the metals with the same organism for the best comparison. Then, devise your scale.


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Solubility involving precipitation

Question
  How do I found out how much $\ce{AgNO3}$ can be dissolved in 1 liter of solvent without precipitation of $\ce{AgCl}$.
  Solvent has $$M_{\ce{\,Cl-}} = 0.1\,{\rm M}\\M_{\ce{\,NH3}} = {\rm 1.00\,M}$$
  Other given values are $$K_{\text{sp}_{\Large\ce{AgCl(s)}}} = 1.8\times 10^{-10}\\K_{\mathrm{f}_{\Large\ce{[Ag(NH3)2]+}}} = 1.6 \times 10^7$$
  Answer: $4.4\rm\,g$

My effort:
 Firstly I tried to find out $\ce{[Ag]}$ in saturated solution from eq.
$$K_{sp_{\Large\ce{AgCl}}} = \ce{[Ag][Cl]} 1.8 \times 10^{-10} = [\ce{Ag}] \times 0.1{\rm\,M}\times [\ce{Ag}] = 1.8 \times 10 -9$$
Then I placed that value in formation eq.
$$K_f = \frac{\ce{[Ag(NH3)2]+}}{\ce{[Ag] * [NH3]^2}}$$
$$1.6 \times 10^7 = \frac{x}{(1.8\times10^{-9} \times 1.0{\rm\,M}-2)}$$
$$x = 0.0288\,\rm M$$
and tried to find out $\ce{AgNO3}$ mass by using that by
$$m~\ce{AgNO3} = 0.0288\rm\,M \times 1.0 {\rm\,dm^3} \times 169.9 \frac{g}{mol} = 4.89\,\rm g$$
That value doesn't seem to be the answer. What did I do wrong?

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Edited. Hope this satisfies

OK, I retracted my close vote. Let me edit this post for formatting, and do not edit until I have done.

The formatting of some of the work is still weird, but I get the same answer - 4.89 g AgNO3.

Step 1: Find equilibrium concentration for Ag+
As you correctly showed in the first step, we need to find the maximum solubility equilibrium concentration of $\ce{[Ag+]_{eq}}$ that is still allowed. Your calculation is correct, the result is
$$ \ce{[Ag+]_{eq}} = \frac{K_\text{sp}}{\ce{[Cl- ]}} = 1.8\cdot 10^{-9}$$
Step 2: Find equilibrium concentration for [Ag(NH3)2]+
We now need to know how much of the $\ce{AgNO3}$ we put into the solution actually ends up in the complex $\ce{[Ag(NH3)2]+}$, which is not in danger of precipitating. The formation equilibrium is according to the reaction equation
$$ \ce{Ag+ + 2NH3 <=> [Ag(NH3)2]+} \,,$$
and reads
$$ K_\text{f} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+]_{eq}}\ce{[NH3]^2}} \,.$$
Substituting $x=\ce{[Ag(NH3)2+]}$ we quickly reach
$$ K_\text{f} = \frac{x}{\ce{[Ag+]_{eq}}(1-2x)^2} \,.$$
This is the step where you went wrong: You also need to consider that the concentration of $\ce{NH3}$ is changed by the formation of the complex. In principle we should also have taken this into account in the section above, but there the error is many more magnitudes smaller than here.
Solving the above gives $x = \ce{[Ag(NH3)2+]_{eq}} = 0.0258942$.
Step 3: Calculate the total amount of Ag+ in the solution
Armed with the knowledge from the two previous sections, this step is pretty trivial. As the contribution of free silver ions is many orders of magnitude smaller than the concentration of the complex, we can safely ignore it and approximate:
$$ \ce{[Ag+]_{tot}} = \ce{[Ag+]_{eq}} + \ce{[Ag(NH3)2+]_{eq}} \approx 0.02589 $$
Step 4: Calculate the amount of AgNO3 to add
If we assume that the activity coefficient is 1, we can directly infer the concentration
$$ c_{\ce{Ag^+_{tot}}} = \ce{[Ag+]_{tot}} ~ \mathrm{mol\, L^{-1}} \,.$$
From the concentration we can calculate the amount of $\ce{AgNO3}$ to add, as you did correctly,
$$ \begin{align}
m_{\ce{AgNO3}} &= c_{\ce{Ag^+_{tot}}} V M_{\ce{AgNO3}} \\
&= 0.02589 ~ \mathrm{mol\, L^{-1}} \cdot 1~\mathrm{L} \cdot 169.87 ~ \mathrm{g\, mol^{-1}} \\
&= 4.398~\mathrm{g}\,,
\end{align}$$
which, after rounding to two significant digits, leads directly to the result your solution says you should have gotten:
$$ m_{\ce{AgNO3}} = 4.4~\mathrm{g} $$

Tschoppi gave all the detaols, but it boils down to this:  when the silver forms a comes with the ammonia, some of the ammonia is used up so that the remaining ammonia concentration is less than 1 M.  The difference is enough to affect that second significant digit.


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What are some factors that influence the voltage of voltaic/galvanic cells and why?
Question: (2014 USNCO local exam Q#41):

What types of changes would increase the voltage of a voltaic cell.

I have learned a bit about voltaic cells, but I haven't been able to find information about this anywhere.
What I know about voltaic cells is that to calculate voltage, you would need to know the standard reduction potential for the half reaction. Then, by adding the E(reduction) and E(oxidation) together, you would get the E(cell). The only way I can think of to change the voltage is by changing the reaction itself, or to change the standard conditions (temperature?). Electrons also flow from anode to cathode, so if more electrons are flowing across does that mean there's a greater voltage? I need help figuring this out.


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Gas of independent molecules
$$\begin{align}
p &= kT\left(\frac{\partial \ln Q}{\partial V}\right)_T \\
&= \frac{kT}{Q}\left(\frac{\partial Q}{\partial V}\right)_T \\
&= \frac{NkT}{q}\left(\frac{\partial q}{\partial V}\right)_T \\
&= \frac{NkT\Lambda^3}{V}\cdot\frac{1}{\Lambda^3} \\
&= \frac{NkT}{V} \\
&= \frac{nRT}{V}
\end{align}$$
Looking at the first equation in this series I'm really confused how the $\partial \ln Q/\partial V$ has been simplified to $\partial Q / \partial V$.
My hunch is that it's something to do with the derivative of $\ln Q$ being $1/Q$ however I'm not sure how this ties in.

The expression $kT \left( \dfrac{\partial \ln Q}{\partial V} \right)_T$ is simply multiplied with $\dfrac{d Q}{d Q}$ (which equals 1, so it is okay). Then you rearrange the denominators. See orthocresol's answer.

This is more of a maths question, but I can give a quick answer.
As you said,
$$\frac{\mathrm{d}}{\mathrm{d}Q}\ln Q = \frac{1}{Q}$$
You could write this as
$$\frac{\mathrm{d}(\ln Q)}{\mathrm{d}Q} = \frac{1}{Q}$$
So, by the chain rule,
$$\begin{align}
\frac{\partial (\ln Q)}{\partial V} &= \frac{\partial Q}{\partial V}\cdot\frac{\mathrm{d}(\ln Q)}{\mathrm{d}Q} \\
&= \frac{1}{Q}\frac{\partial Q}{\partial V}
\end{align}$$
If you are studying physical chemistry, it is a good idea to remember this trick. It appears everywhere.

This is called the chain rule in case the OP wants to see an ultra formal Wikipedia page about it.


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Rates of solvolysis of p-dimethylamino benzyl halides
What will be the order of $\mathrm{S_N1}$ solvolysis rates for these?

I'm confused between the second and third options. $\ce{I-}$ is a better leaving group than $\ce{Br-}$, but the carbocation formed in 2 is more stable than 3. So how are we supposed to decide between them?

In the second compound, steric repulsions (denoted by the curly lines) between the N-methyl groups and the methyl groups on the ring occur:

This forces the $\ce{-NMe2}$ group to rotate such that the nitrogen lone pair is no longer parallel to the ring's $\pi$ system. As a result, the nitrogen lone pair can no longer contribute to the mesomeric stabilisation of the second carbocation.
The rate of solvolysis therefore increases in the order: $2 < 1 < 3$. The effect described above is called "steric inhibition of resonance".
A similar effect is observed in azo coupling:

In the second reaction, the nitrogen lone pair is not delocalised into the ring and the ring is therefore not nucleophilic enough to attack the diazonium ion.

authors: Анастасия Слепичева


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Titration of NaOH with acetic acid
During a titration lab where 0.1 N of NaOH is added to 20 mL of acetic acid with the same concentration, the equivalence point occurred after adding about 70ish milliliters of NaOH (different peers acquired different values, ranging from 65 mL and 75 mL of NaOH). This completely puzzled me, because after calculating a theoretical titration curve, my equivalence point occurred after adding about 20 mL of NaOH.
My teacher could not explain why this happened, except that acetic acid can act as a buffer – but how do you calculate this and incorporate it into a theoretical titration curve? I just don't understand how 20 mL of acetic acid would require 70 mL of NaOH to neutralise it (both have the same concentration).

There is nothing about acetic acid that would make your titre value over 3 times the expected value. I would be willing to bet that someone messed up somewhere with the chemicals you were given.

That is the only explanation I can come up with as well. It is just unfortunate that I will loose marks for not having a titration curve that does not reflect the curve acquired during the lab. I will try and convince my prof that the NaOH given to me must have been diluted incorrectly.

It can't happen.  Neutrality in aqueous solution means equal quantities H+ and OH-; since your solution is unlikely to have a net electric charge, so too must your anion and cation quantities be equal.  Buffering from acetic acid being a weak acid is meaningless if you don't have other anions around to make up that balance.
Solution is electrically neutral:
[H+] + [Na+] = [OH-] + [AcO-]
Solution is pH neutral:
[H+] = [OH-]
Substitute the latter in the former, subtract, you can only get:
[Na+] = [AcO-]
If the concentrations of both solutions were the same, and you're titrating by adding both to the same volume, you can only have equal concentrations if you add equal volumes.
If this was not the case (as you describe) we have to violate one of our assumptions:

Solution actually had a net electric charge (you're doing the titration in space and you're bombarded by beta radiation, i.e. unlikely)
Solution was not truly neutral.  Maybe someone calibrated your pH meters wrong, maybe everyone titrated with a faucet while texting and over-shot by a factor of 3.
Contamination with a strong acid added a variable we don't know about to the ion balance.
Most likely, the solutions were not truly the same concentration.  See if your teacher made her NaOH solution from solid NaOH that she let sit around in open air.  It can pick up a lot of moisture from the air, i.e. it's hygroscopic, and before you know it the weight you're measuring on a balance is half water.

Thank you! I agree with everything you said. I was looking at the data acquired from a titration using the same NaOH with HCl, and I am beginning to think that the acetic acid may have been diluted incorrectly...  but then again, I remember doing a previous lab with the NaOH and I acquired weird results... so the NaOH may very well have absorbed moisture from somewhere (we use the pellets)


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Tables/database/literature of specific or optical rotation
I'm looking for a comprehensive reference of experimental specific rotation or optical rotation data. It can be a published literature review, a free web database with API access or one that permits scraping, or even a table on a web page with references.
ChemSpider pages link to outside sources (I think only chemical suppliers) for optical rotation, but this is problematic:

it's unclear what molecules contain this information (not all stereoisomers do),
not all the product specification pages on the suppliers' websites actually contain the data,
the values aren't available through the ChemSpider API, and
it's forbidden to screen scrape ChemSpider pages.

The story seems to be similar for PubChem, though they allow scraping within reason.
For example, here is a review collecting experimental electron affinities for a large number of small molecules and elemental clusters. It's fine that this isn't web-based, because all the values are together and in bulk; writing a PDF scraper is worthwhile because there are so many entries.
Also acceptable would be a book, such as the comprehensive reference from Huber and Herzberg that contains rotational constants, bond lengths, and other constants for diatomic molecules.

Thanks for sharing that reference on electron affinities, it looks quite interesting

Have you already had a look at the Handbook of Chemistry and Physics? I am not too familiar with the work, so this is just a suggestion. They have many different values, if the ones you are looking for are among them, I don't know.

The sigma-aldrich website? They provide optical rotation for most of their chiral compounds

@pentavalentcarbon did you find an answer?

@Polydynamical at some point I discovered my dead tree copy of the Merck Index has many, but they are associated with each entry and are not in a separate table. I never did look at the HBCP or S-A website, nor did I find a "holy grail" type resource :(

@pentavalentcarbon Wow I didn't even think you would respond. I guess I'll just compile my own dataset. If I ever do compile it, I'll drop you a link :)

@Polydynamical It is not so easy to get rid of me. I had also found ~soon after asking this question https://brunl01.users.greyc.fr/CHEMISTRY/. Converting this into a pile of JSON would be a great starting point.

@pentavalentcarbon YOO this is exactly what I was looking for. The format, the data... perfect. The only thing is, that I need about 1000x the data . . . :)

@pentavalentcarbon UPDATE: I was just reading a paper and they extracted the optical activity data from the Reaxys database: https://arxiv.org/pdf/2110.04383.pdf (middle of page 8 & page 30)

There seems to be a way of searching for the compounds with optical rotation with Pubchem at this link

I've made some progress! Here's a book from 1932 that contains a whole bunch of optical rotation tables.


text:
Effectiveness of getters to keep gasses sulfur free
My engineering task poses a problem with sulfur damaging electronic components. There are enclosed gas volumes which contain sensitive electronic devices as well as NBR sealings and gaskets. The sealings can evaporate sulfur in different compounds, the most dangerous seemingly beeing hydrogen sulfide. I have two options protecting the electronics.

geometric separation by confining the electronics in separate housings. This might be possible but is very difficult due to the geometry of all devices.
removal of evaporating sulfur compounds from the air surrounding the electronics.

I want to evaluate option 2.
My first question is about the getter material in general. I know of only two usable materials: silver and zinc carbonate. Are there other usable (solid) materials?
Silver could be applied as a foil inside the common housing exhibiting a considerable surface to the gases. Zinc carbonate must be applied in a permeable containment like a paper or Tyvek bag, similar to desiccant bags.
I assume, the number of sulfur compounds which can eventually evaporate from some sealings, is rather limited. But the evaporation will probably never cease to happen. Hence the getter must be active for the whole product lifetime. Is there any mechanism that could prevent this? E.g. oxidation of silver or weird reactions of zinc carbonate?
I understand by using a getter, I cannot prevent the effect of the detrimental gasses on the sensible equipment in total, but I can limit concentration and time to extend the period until the destruction of the electronics. I assume, that by exposing a silver foil to the gas mixture with a surface n-times as big as that of the semiconductor, the resorbed sulfur will at least n-times as much as that of the semiconductor? So if I increase the surface of the silver foil the amount of sulfur likely absorbed by the semiconductor will decrease by the same factor, given a constant evaporation rate (ev=const [mol/s]).
I guess this is difficult to formulate for a bag of zinc carbonate, isn't it? Might the surface of the bag correlate the same way?
It seems, after installing sulfur sources and sinks, an equilibrium will establish. Given a constant gas concentration around getters and sensitive equipment, I want to make further considerations. Silver surfaces can bind sulfides directly while the housing of electronics has to be traversed by diffusion. Is it fair to assume that due to the presence of uncovered getters the mass flow by diffusion through the housings is reduced by much a greater degree than just by the proportion of surfaces between getter and electronics? Can I hope that this reduction is one or more orders of magnitude?


text:
Is there relation between “Plane wave basis” “Correlation-Consistent” basis and “Fourier basis”
This question is related to the following link I found while browsing the internet: http://www.gutenberg.us/articles/Basis_set_(chemistry) The question: Is there a relation between "Plane wave basis" "Correlation-Consistent" basis and "Fourier basis"?  Would anyone please add appropriate tags as well? Any comment would be appreciated.
One can think of Fourier basis as constructing the function with different frequencies. I was looking for constructing the function with vectors with same "correlation-consistent" . Is such concept possible? Is "correlation-consistent basis" is something similar?

To help others get you an answer or answers that are informative, can you give us an idea about your level of understanding/education in regards to quantum chemistry and computational methods? As it stands, the question is a little bit broad or likely to elicit answers that might not be pertinent, so it would help if you can amend your post to narrow the scope a bit.

Do not post the same question on few SE sites. 2) What is the actual meaning of "correlation" in your case.

@Wildcat correlation as in co-variance matrix. Fourier can be expressed by exponential, can we easily derive "correlation-consistent basis" similarly?

@Creator - In quantum chemistry (and related fields), the term "correlation" describes electronic correlation and has nothing to do with the statistical term you reference - see this Wikipedia page for electronic correlation for some clarification.

@Creator, I still don't understand what you're doing. "Correlation" in "correlation-consistent bases" has nothing to do with correlation of two random variables.

@Creator, I have no idea what do you mean by "the cc basis in relation to Fourier", but I'm more and more confident that whatever you're looking for you're doing it in the wrong place. This is a site about chemistry, not signal processing.

This was supposed to be a comment, but there is too much text, so it will be an "answer".

Is there a relation between "Plane wave basis"
  "Correlation-Consistent" basis and "Fourier basis"?

Well, mathematically, they are entities of the same type that serve for the exact same purpose:

they all are concrete examples of a well-known idea of linear algebra & functional analysis, the idea of a basis;
and they all are used to expand square-integrable functions.

Does this counts as a relation?

I was looking for 'a kind' of basis (instead of Fourier) for some work
  and I personally gave the name for my basis as "correlation basis".
  When I typed my thoughts in google to my surprise I got the link
  mentioned in my question and I tried to understand at what context it
  is using the basis. From their I tried to relate the three basis to check whether I can use it for my work.

Are you just afraid of a terminological ambiguity out there? If so, don't worry too much. First, only the phrase "correlation-consistent basis", not just "correlation basis" has a particular well established meaning in quantum chemistry community. This is reflected even in the names of correlation-consistent bases that all starts with "cc" prefix. Secondly, "correlation-consistent basis" has a well established meaning only in quantum chemistry community concerning electron correlation and complete basis set extrapolation, so that unless you doing quantum chemistry, you shouldn't be interested in correlation-consistent bases.

Electron correlation is interdependency of the states of individual electrons in a many-electron system. Consider the simplest many-electron system, a two-electron one, and the following two events:

$1$ - the event of finding electron-one at point $\vec{r}_{1}$;
$2$ - the event of finding electron-two at point $\vec{r}_{2}$.

Then mathematically the idea of electron correlation can be simply expressed as the inequality of the joint probability of $1$ and $2$ on the one side and the product of unconditional probabilities of $1$ and $2$ on the other side
$$
    \Pr(1 \cap 2) \neq \Pr(1) \Pr(2) \, .
$$
Similarly, in terms of electron densities, we could indeed write
$$
    \rho_2(\vec{r}_{1}, \vec{r}_{2}) \neq \rho(\vec{r}_{1}) \rho(\vec{r}_{2}) \, ,
$$
where

$\rho(\vec{r}_{1})$ is the probability density of finding electron-one with arbitrary spin at $\vec{r}_{1}$ while having electron-two with arbitrary spin at arbitrary position.
Similarly, $\rho(\vec{r}_{2})$ is the probability density of finding electron-two with arbitrary spin at $\vec{r}_{2}$ while having electron-one with arbitrary spin at arbitrary position.
And $\rho_2$ is the joint probability density of finding a pair of electrons one and two with arbitrary spins simultaneously at $\vec{r}_{1}$ and $\vec{r}_{2}$ respectively.

I think $\vec{r}_{1}$ and $\vec{r}_{2}$ can indeed be thought of as random variables and that essentially electron correlation is about these variables being in a statistical relationship with each other (they are unlikely to be close to each other and likely to be as different as possible). But if you think that a special basis set is needed to account for electron correlation, that is wrong: taking electron correlation in account has very little to do with a basis set choice. There is no need to use a correlation-consistent basis set to treat electron correlation: Pople basis sets can be used to treat it, as well as correlation-consistent one.