text:
Le Chatelier's principle: what are the effects of decreasing volume?

Which of the following is increased by decreasing the volume of the
  reaction system in the following reaction: $\ce{2H_2S(g) +3O_2(g)<=>
 2SO_2(g) +2H_2O(g)} + {\text{heat}}$
I. Rate of Reaction
II. Equilibrium concentration of reactants
III. Value of $\ce{K_{eq}}$

My attempt: By the gas laws, decreasing the volume of the container will increase the pressure, so equilibrium should shift to the right, as there are less gases on the right. Thus III is true. Additionally, heat is also generated by the shifting of the equilibrium, and the increased pressure on the system due to the decreasing of volume would also speed up the reaction, so I is also true. II is not true because the Equilibrium concentration of the products increases, not the reactants. So my final answer is I and III. Yet the answer key says I only. Why is that?

I think a good place to start is with the idea that that $K_{eq}$ will not change in value unless the temperature changes (see the second answer here for a good explanation). With this in mind we can say that statement III is false because the temperature is not being changed.
Statement II is shown to be false via Le Chatelier's principle, as the equilibrium concentration of reactants should decrease rather than increase due to the decrease in volume because the products have fewer gaseous moles and so would exert less pressure on the container.
Statement I can be shown to be true via Le Chatelier's principle by the same logic as Statement II: for the reaction to shift towards the product, the rate of formation of product must increase.
Therefore, only Statement I is true.
You might wonder how $K_{eq}$ can stay constant, but the rate of forward reaction can increase. Its important to remember that $K_{eq}$ is the ratio of the forward and reverse rate constants, not the forward and reverse rates. The rate constants have no dependence on volume (or pressure) so they won't change.

I am having some confusions. If the $K_{eq}$ remains constant, the rate of backward reaction should too increase in proportion with the forward. So why we observe more formation of the products?

Few links that are relevant :
chemguide
chemistry.purdue.edu
wikipedia
As given in the first link $K_{eq}$ will change only if you change temperature. What changes here is reaction quotient, $Q$.
$Q$ is calculated just like $K$ : $$\frac{(concentration\, of\, products)^a}{(concentration\, of\, reactants)^b}$$   where a and b are the respective stoichiometric coefficients.
It is clear that when you will decrease the volume of the container you will increase the concentration of both the reactants and products. But $Q$ will decrease as $\ce{O_2}$ in the reactants is cubed while both the products are squared.
If the system has to attain equilibrium $K$ has to be equal to $Q$ thus $Q$ has to increase which means that the concentration of the reactants has to decrease and that of the products has to increase.
Therefore the reaction shifts in the forward direction and the rate of the reaction increases.
Thus only I is right.

concept:
You should base it on the amount of moles on both sides

so if you wanted to figure it out u know that the side with the more moles is going to be affected i.e. if you decrease the pressure the side with the more moles of gas is going to decrease and shift the other way
to clarify:
1) You said that I is true but it isn't because we have to understand that only temperature changes the rate of the reaction so for rate of reaction we don't care about the decrease in volume.
2)For equilibrium concentration of reactants, we know that there is 5 moles of gases on the left side and there is four moles on the right side with heat, because you are not chaining the heat, when you decrease the volume that means that you are forcing the moles from the left to the moles on the right side.
3)For $K_{eq}$ we will go with your explanation of why it is true. But then we can rule out number 1. But it seems to me that your textbook either has a typo that tried to say "which of these are not true" or that you have not read the problem correctly.

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review

@Mithoron. fixed

I'm gonna click 'Looks OK' in review, but it would be nice if you elaborate a bit. Answers here are expected to be as elaborate as possible. If you edit, notify me and I'll consider upvoting your answer.

Chem SE as a site is trying to move away from giving hints as answers. If you want to just leave a hint, you can leave it as a comment. Answers need to be more fleshed out and give all the information to solve a problem (though that doesn't necessarily mean doing the whole problem step by step).

@M.A.R. It is fixed

The answer to the question given by the OP is correct.

Mind well, equilibrium constant $K_c$ as well as $K_p$ depends only on the temperature, they remain unaffected even though pressure is changed, or concentration of reactant or product is changed.


text:
How to determine the amount of barium in a sample of contaminated water?
Here is the assignment I have:

You are working for Matrix Pro - a company involved in analytical assessment and method development for the clean-up of contaminated waters.
You received water samples from a potential client. The samples may contain the following ions: $\ce{H+, Ba^2+, OH^─}$
The potential client requested a qualitative analysis of the sample as well as the development of a treatment technique that will bring the pH value to neutral and eliminate the barium ions. For comparison another water sample (considered to be non-contaminated) was also sent for similar qualitative analysis.
Assignment

Develop an experimental procedure to comply with the request of the potential   client.
Draw a flow chart.
Tabulate all your observations and data.

What I'm not sure about is finding the amount of barium present in the unknown sample so I can determine the amount of precipitating reagent to add.  I just want to know what type of analysis to do.  Chromatography? Spectrophotometry? Something else?

You could add some sodium sulfate solution to a sample of the water to see if barium sulfate precipitates. It wouldn't be difficult or expensive to do this quantitatively for that matter.

Because the client is ultimately interested in removing the barium by precipitation, gravimetric analysis is likely a suitable solution. This can be done in a few hours in a lab with basic equipment like a filter flask, a drying oven, an analytical balance and the appropriate reagents.
You can take advantage of the fact that barium sulfate is very insoluble in water by adding some saturated solution of sodium sulfate to a known volume of a sample of the water. Then just filter out the barium sulfate onto some pre-weighed filter paper, dry, weigh, and calculate the mass of barium (from the measured mass and the molecular weight of barium sulfate) per volume of water. You should repeat the analysis to be sure you added enough sodium sulfate solution to precipitate all of the barium.
If there is little or no precipitate found, you have a couple options. Firstly, you can optimize the gravimetric analysis by using large volumes of sample with a small filter and a good analytical balance. Also be sure to pre-filter the sample prior to adding the sodium sulfate in case there are suspended particulates, which would get measured just as if they were barium.
If a gravimetric analysis does not provide the detection limit required by the client, the next best options are atomic absorption spectroscopy (AAS) and inductively coupled plasma - atomic emission spectroscopy (ICP-AES). As these analyses can be done very cheaply (~\$10-$30 per sample) you may actually want to do this to begin with. The down side of outsourcing is that the turnaround time could be weeks, or the price could quadruple if you need results fast.

maybe... It depends on what the Ba level is in the contaminated water, and what the highest level acceptable in the "cleaned-up" waste.

Gravimetric analysis of $\ce{Ba^{2+}}$ was one of the practice exercises in my quantitative analytics lab. We had to use filter crucibles because $\ce{BaSO_4}$ is usually very fine-grained and to allow for heating at 900 K for several hours to remove water.

@TAR86 , My first response is: 900K ?! I guess that would get it dry in a hurry. The way I've described it is basically an ASTM standard method that my lab has probably done thousands of times over the years. So long as there is sufficient $\ce{Ba^2+}$ in solution, gravimetric analysis can actually give much better precision that AA or particulary ICP.

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text:
Has it been observed that the dihydrogen molecule dissociates into ions?
As we know from the improved methods valence bond theory (VB), as well as molecular orbital theory (MO-LCAO), the wave function of the $\ce{H2}$ molecule consist of two terms, which belong to covalent and ion structure, respectively:
$$\psi =\psi_\mathrm{cov} + \lambda\psi_\mathrm{ion}.$$
According to quantum mechanics, there is a nonzero probability $\propto\lambda^2$ for the ionic state wave-function term. Such reaction should lead to the $\ce{H2}$ molecule dissociation products $\ce{H+}$ (proton) and anion $\ce{H-}$:
$$\ce{H2 -> H+ + H-}.$$
Have such processes been observed in nature?

Surely1. At the end of the day, it is just the matter of supplying enough energy: the homolysis requires approximately 4.5 eV per hydrogen molecule, while for heterolysis more than 17 eV per molecule must be provided.

1) G. Busca, A. Vaccari, Heterolytic dissociation of hydrogen on high-temperature methanol synthesis catalysts, Journal of Catalysis, 108(2), 1987, 491-494, http://dx.doi.org/10.1016/0021-9517(87)90198-9.

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text:
Alternative definition of the mole?
1 amu (atomic mass unit) is defined as the one twelfth of the mass of a carbon-12 atom. The mole is defined as the number of carbon-12 atom in 12 g of carbon-12. In other words,
$$\require{cancel} 1 \textbf{ mole} = \dfrac{12 \textbf{ g}}{m(^{12}C)} = \dfrac{\cancel{12} \textbf{ g}}{\cancel{12} \textbf{ amu}} = \dfrac{\textbf{ g}}{\textbf{ amu}} $$
I believe that the 12 g were chosen so that the mole could exacly express the ratio between the g and amu.
Consider now this alternative definition : the amu is defined as the exact mass of a proton and the mole is exact ratio between the g and the amu. Wouldn't it be simpler this way?
What are the advantages of one definition over the other? Is there any reason we would want to keep the former? Why scientists decided to define the mole like that and not pick a simpler definition like the one I proposed?

I wasn't very clear. See my edited question :)

Ah, ok. That's slightly different.

One of the reasons to take Carbon-12 could be that it's hard to get your hands on one mole of pure hydrogen (and to weigh it). Carbon-12 is a (room temperature) solid which is relatively easy to produce, keep pure, store and weigh.

I believe that the mass of a proton has been derived experimentally with high precision. Why not define this as 1 amu = 1 g/mol and then express the mass of every atom accordingly? I don't understand where is the need to either weigh Carbon-12 of pure hydrogen.

@Corvinus - In addition to altering the molar masses of everything, the definition of the Avogadro constant is dependent on the definition of the unified atomic mass unit. Changing the definition of $\mathrm{u}$ would change the definition of $N_\mathrm{A}$, which would change the definition of the ideal gas constant $R$, since $R=N_\mathrm{A}k_\mathrm{B}$ (the latter term being the fundamental Boltzmann constant).

Actually, since $R$ is measurable (and $k_\mathrm{B}$ is not so easily measurable), changing $\mathrm{u}$ would change the Boltzmann constant, and all chemists and physicists would be upset.

I understand that it wouldn't make any sense to change the definition now. I see now that I wasn't very clear in my question : what I would like to understand is why scientists decided to define the mole like that and not pick a simpler definition like the one I proposed? (I'm editing my question to make that clearer)

I seriously doubt that any appreciable quantity of pure $\ce{^{12}C}$ exists. The mass of $\ce{^{13}C}$ is known to 13(?) significant figures and is about 1% of non-radioactive carbon. So you'd have to have $\ce{^{12}C}$ pure to about 12 significant figures to not perturb the mass. That just seems impossible for any reasonable mass of C.


text:
Calculating with Method of Equal-area differentiation
I have been trying to understand this method although I admit, don’t have a very good calculus background just yet.
For a given reaction where $x$ is time, $f(x)$ is cell concentration and assuming it to be first order (autocatalytic), we obtain the table:
\begin{array}{rrrrr}\hline
x  &
f(x)  &
\frac{\Delta f}{\Delta x}  &
\frac{\mathrm{d}f}{\mathrm{d}x}  &
\text{actual}\frac{\mathrm{d}f}{\mathrm{d}x}\\\hline
0    &    0    &    -    &    1010    &    1000    \\
0.2  &  182    &    910  &     805    &     818    \\
0.4  &  332    &    740  &     670    &     670    \\
0.6  &  451    &    605  &     550    &     548    \\\hline
\end{array}

For this data: Is it possible to calculate the $\mathrm{d}f/\mathrm{d}x$ column with the rate formula? Or maybe $\mathrm{d}f/\mathrm{d}x$ can only be calculated by using the equal areas given in the plot?

Could you clarify where you obtained this table? Also, a first order reaction is not the same as an autocatalytic reaction. Welcome to Chem SE and best of luck getting an answer!

Hello, and thank you for answering!
I got this method from “elements of chemical reaction engineering” cd-rom manual, available in link. I’ve been investigating more and this method consist on creating a histogram like graph and ‘split each in half’. However, I find this method to be very unreliable and I believe ‘automating it’ a bit with a formula might improve the precision.

In using this method, whether or not its worthwhile seems to depend a lot on how many points you have. For only a few points, this is a decent approximation scheme, though it requires a bit of judgement on where to draw lines to cut off equal areas. Typically, you will be working with more points and so it might be better to have a method thatch be applied more formulaically.  One method I would suggest with more points is finite differences. Alternatively, you can get a decent guess by creating a polynomial or exponential...

regression line in a spreadsheet or online calculator and then differentiating that equation. For this data, I was able to get slopes of 987, 830, 673, and 515 using the best R^2 quadratic curve.


text:
Oxidation of water — what is the half reaction for water?
During the electrolysis of a solution of copper sulfate, copper is reduced to form a solid on the inert electrode while water is oxidised at the anode. What is the half equation for water?
\begin{align}
\text{Is it}&&
\ce{4 OH−(aq) &-> O2(g) + 2 H2O(l) + 4 e−}\\
\text{or}&&
\ce{2H2O(l) &-> 4H+(aq) + O2(g) + 4e–}?
\end{align}
What is the difference? And what is the standard reduction potential in an electrolytic cell?

Both your reactions are equivalent. Here’s why. We start from $(1)$.
$$\ce{4 OH- -> O2 + 2 H2O + 4 e-}\tag{1}$$
(I have omitted the phase descriptors for clarity.) In any chemical reaction, we can add or subtract spectators as we see fit; this is akin to mathematical equations where we can add or substract expressions as we see fit as long as we do the same to both sides. For example, I want to add four protons to $(1)$ to give $(2)$.
$$\ce{4 OH- + 4 H+ -> O2 + 2 H2O + 4 e- + 4 H+}\tag{2}$$
If this were now a mathematical equation, I could combine (or divide) parts as I see fit. In a chemical reaction, we need to be a bit more careful. But as long as our focus is not on proton transfers, we can combine anything basic enough with acidic protons. Here, you can combine the four hydroxides and four protons on the reactant side of $(2)$ to give $(3)$.
$$\ce{4 H2O -> O2 + 2 H2O + 4 e- + 4 H+}\tag{3}$$
Now, we notice that we again have spectators. Two water molecules turn up both on the reactant and on the product side (an additional two are unique to the reactant side). We can remove them like we did the protons before, giving us $(4)$. I will also perform some rearranging.
$$\ce{2 H2O -> 4 H+ + O2 + 4 e-}\tag{4}$$
It so turns out that $(4)$ is your second suggested reaction. It is the same equation as $(1)$ as I have just shown; the difference is merely in the type of water molecules present.
Now which one should you use? Notice that one reaction consumes hydroxide while the other produces protons (formally! The consumption of hydroxide is equivalent to the production of protons). Thus, if your reaction medium is basic, you should use $(1)$ for your redox reaction. If, however, your reaction mixture is acidic, $(4)$ is more appropriate. In short, choose the one with the same charged particle as is used in the other half-reaction to make your life simpler.

You are looking for a reaction where water is oxidized, i.e. loses electrons. In the chemical equation, the electrons must appear on the opposite side as $\ce{H2O}$, since electrons are products of an oxidation half-reaction.
On the other hand, any time we write $\ce{X + e-}$ we are depicting a reduction of $\ce{X}$. Which of your proposed reactions shows water being oxidized (and not reduced)?

Strictly speaking, both of them. It is just a matter of pH.

You're arguing that although the first reaction produces two water molecules, it consumes (oxidizes) four water molecules, which are represented here by hydroxide intermediate derived from the autoprotolysis of water?

Well, yeah, you may put it like that.

What is hydroxide, if it is not a form of water minus a proton?


text:
Lassaigne's test for nitrogen
Why is it that benzene diazonium salts don't give the Lassaigne's test for nitrogen. After all they've got both carbon and nitrogen.

Benzene diazonium chloride is stable upto 5 degree Celcius, while in Lassaigne's test we need to heat the organic compound with sodium till red hot, before breaking it by plunging into cold water. All the nitrogen will be liberated before it finds a chance to combine with Na and form any NaCN, which is responsible for the test ultimately.

Is there any method other than the coupling test to test for the presence of benzene diazonium salts?

And in case we use benzene diazonium fluoroborate, will it gie positive result for Lassaigne's test? This is because I read that it is stabler than other diazonium salts at higher temperatures.

It would not be possible because fluoroborates again give of N2 while converting into fluorobenzene upon being HEATed.

@CheMistry This test also fails for the organic compounds having nitrogen in the ring because sodium cannot extract this nitrogen atom. But what about nitrobenzene?


text:
Isomerism in amide and nitrile aldehyde
I have the following question,

The given answer is c but I have trouble understanding why a would not be a correct answer. Because from my understanding in part a, one is amide functional group and the other is aldehyde with a -Nh2 nitrile group as a substituent. Then it should be a functional group isomerism too, right?

Nitrile is -CN. Other than that, I agree with your reasoning. These are compounds with the same molecular formula, but different functional groups, hence they must be f.gr. isomers.

What's wrong with choice b? I'm leaning towards All of the above.

Incidentally, the aminoaldehyde would probably react with itself almost immediately. Notwithstanding that I agree that "all of the above" is the correct answer.

@orthocresol I thought of that as well, but I figured classification of structure occured in the abstract, at an instantaneous moment not beholden to whatever reaction just occured to form form the compound nor by what route it may be about to decompose. But that aside, how likely is it to sacrifice the carbonyl for a three-membered ring? The aziridine wikipedia page mentions several syntheses, and I didn't see intramolecular attack of alpha- amino aldehyde (https://en.m.wikipedia.org/wiki/Aziridine?wprov=sfla1?

I think choice D, All of the above may be the correct answer. All three pairs are structural isomers, with different functional groups, thus functional group isomers (FGIs).
This brief overview uses a carboxylic acid and ester pair as an example of FGI, which validates choice B. It also cites an aldehyde $\ce{CH3CH2(C=O)H}$ and a ketone $\ce{HCH2(C=O)CH3}$. If the methyl group is swapped out for an amino group (not called a nitrile), we have your compounds $\ce{NH2CH2(C=O)H}$ and $\ce{HCH2(C=O)NH2}$. Not only should this pair be FGIs by analogy with the aldehyde-ketone, but they satisfy the definition of functional group isomers: Structural isomers (same atoms, different connectivity) that have different functional groups. This validates choice A.
Wikipedia cites your exact ether-alcohol pair as example of FGIs, validating choice C.
Thus, all of the above must be the correct answer. If you are in a class, I would ask your instructor about this problem.


text:
Wurtz reaction with two different alkyl halide
$$\ce{C5H11Cl + C4H9I + 2Na ->}$$
Is this possible in a Wurtz reaction? I want to add 2 different alkyl halides with two different halogens. I'm curious if two different types of halogen on the alkyl reactants is possible.

This is going to be pretty messy. Each reagent will react not only with the other one, but also with itself.

As a side note, I would also mention the Corey-House synthesis, a method to couple asymmetrical halides. We transform one of the alkyl halides into a lithium dialkyl cuprate (R2CuLi) and react it with the other alkyl halide (R'X). The result would be a alkane R-R'.

There are going to be issues concerning separation as the products will have nearly same boiling points making fractional distillation difficult.

The alkyl groups mentioned could be primary, secondary or tertiary
The last one's will not be going by Wurtz reaction (these will rather undergo elimination to give alkenes); only primary & secondary can.
Even here, we can have a mixture of products:


text:
Display of short contacts with Avogadro
For a small group of small organic molecules, I would like to use Avogadro to display short intermolecular contacts directing their packing patterns; separately from normal Hydrogen bonding pattern.  For example, to indicate intermolecular C(aryl)-H interactions.
It is known to me that this may be achieved with the freely accessible software Mercury by the CCSD (described here), too.  Yet, both for the sake of consistency with other drawings already prepared, as well some curiosity if Avogadro equally may provide this, I whish to address the question here.  The sole contribution in the software's mailing list partially relevant to this topic (here), dating back 2013, exludes even the indication of non-classical H-bonds.
To solve this problem, both the "classical version" (currently 1.2.0), as well as the more recent version Avogadro² (1.90.0) were accessed.

So, what's the question? How to represent an interaction between any two atoms of your choice?

It's a good feature suggestion, but it's not implemented in either version at the moment. I'd suggest, for example, suggesting the feature here for Avogadro v1.x or here for Avo2.

And as you seem to know, there's a mailing list and a discussion board specifically for Avogadro questions.

I'm voting to close this question as off-topic because it belongs on the Avogadro discussion forum.


text:
If a temperature of a reaction is raised from 300 to 320K, what factor will the reaction rate increase by

In general, if the temperature of a reaction is raised from 300k to 320k, the reaction rate will increase by how much?

So the first thing I do is try out the Arrhenius equation:
$k=Ae^{\frac{-E_a}{RT}}$
so I ignore the activation energy and A because they will cancel out to get the following:
$k=e^{\frac{-1}{300}}$
$k=e^{\frac{-1}{320}}$
and I divide the two k's to get 1.000208
but that can't be right.
It seems to me that I do not know how to answer this question and would like some assistance
I do not want to know a proof of anything but rather to know how to execute methods in solving this problem.

Your Arrhenius equation lost the negative sign on the exponent.

@Tyberius fixed, also not a duplicate, the link that you provided was about the correlation of the commonly accepted idea that if we increase temperature by 10 degrees then we double the K, but in this question rather than doubling, we are trying to calculate the ratio.

While that is true, I would argue that the other question effectively shows that this can't be answered in general. You can pick one or maybe a few different activation energies and see what you get, but there isn't a general answer without being given a specific activation energy. If you went by the general rule of thumb, your answer would be 4, since you increase the temperature by 20 degrees, so the rate at 320 would be 4 times the rate at 300 (again, just by that estimation scheme).

Agreed with out activation energy it vague and could vary by orders of magnitude. The exponents' use is the activation energy.

The amount of increase is dependent on the activation energy. For a very low activation energy, there will be almost no change (this is what you obtained by removing $E_a$ and $T$). For a high activation energy, the change will be quite large. For example, with an activation energy of $110\mathrm{\frac{kJ}{mol}}$, you the rate will increase by about $16$ times. $$\frac{\exp({\frac{-E_a}{R*320}})}{\exp({\frac{-E_a}{R*300}})}=\frac{\exp({\frac{-111000\text{J}}{8.314*320}})}{\exp({\frac{-111000\text{J}}{8.314*300}})}=\frac{7.6\times10^{-19}}{4.7\times10^{-20}}=16.14$$
So there really isn't an "in general" answer. However, many text books give a crude estimate that the rate will double when the temperature is increased by $10$. This isn't in general the case, as it depends on the temperature range (a change from $300$ to $320$ is more significant that a change from $600$ to $620$) and the the value of the activation energy.


text:
Why is the plateau on a heating curve in equilibrium?

From the interval of 100 to 140 minutes, why is the substance in equilibrium?
I believe it's because at this interval, freezing and melting begin to occur at the same rate and, when two opposing reactions occur at the same rate, equilibrium is present. However, since the potential energy is decreasing (since the substance is freezing), the equilibrium favors the exothermic reaction causing the substance to eventually become solid in the interval of 140 to 160 minutes.

Can you make it clear what the graph is ? I mean what is the substance you are measuring temperature of ?

@Raghav The substance was not given. Is that necessary? The graph is a heating curve.

You are right about the equilibrium existing because both freezing and melting occur at the same rate.
Imagine ice and water kept in a thermos flask under normal atmospheric pressure. You will observe that there is no change in the mass of both the phases.
But there is actually a lot of activity going on. Some molecules of ice enter the liquid phase and some molecules of water enter the ice phase. As both these processes occur at the same rate you don't see any change in the mass of both the phases.
In fact such equilibriums exist for all pairs of solids in their own liquids and gases above their own liquids at particular temperatures.
In your case I think there is a mechanism to extract heat from the system which shifts the equilibrium towards liquid changing to solid ie freezing.

The flat regions are where the solid is changing into liquid and liquid into gas at the higher temperature. These regions persist whenever there are still two phases in equilibrium. The heat necessary to cause the melting used to be called the Latent Heat of Fusion but nowadays is called the enthalpy of fusion (or melting) and similarly for vaporisation.
The solid is held together by a cohesive or lattice energy, generally called intermolecular energy (potential energy with attractive and repulsive parts) which has to be overcome for the solid to melt, and the heat absorbed by the solid is used to do this which means that the temperature does not rise until all the solid has melted. A liquid also has intermolecular forces but weaker than in the solid and thus has increased entropy. This increase in entropy has the effect of lowering the Gibbs free energy.
( The change of enthalpy on fusion can be calculated from the definition of enthalpy $H=U+PV$. Between the two phases, liquid to solid, the change is
$\Delta H = U_l-U_s +P(V_l-V_s)$
but as the change in volume between liquid and solid is small we can ignore the PV work term and so
$\Delta H = U_l-U_s $.
Using the heat capacity allow this to be evaluated since $U=U^{\mathrm{o}} +\int_0^T C_p(T)dT$
with the result  that
$ \Delta H = U_l^{\mathrm{o}}-U_s^{\mathrm{o}} +\int_0^T C_p(T)_l-C_p(T)_s dT$.
In vaporisation the PV term ignored in the solid-liquid change has to be added and becomes $PV_g$ for the gas and which for an ideal gas is RT  )


text:
How does adding acids (H+ ions) to limited soluble salts help it further dissolve?
For example, adding $\ce{HCl}$ to the very insoluble $\ce{CaF2}$ will help it further dissolve. How?
I believe the concept deals with equilibrium and Le Chatelier's Principle:
$$ \ce{ CaF2 <=> Ca^2+ + 2F^-} $$
Does adding $\ce{H+}$ react with the $\ce{F-}$, which shifts the equilibrium to the right-hand side and therefore further dissolve the salt?

In general, the solubility of a sparingly soluble salt having a basic anion will increase with the addition of a strong acid.  Given:
$$\ce{X+A- <--> X+ + A-}$$
where $K_{sp}$ is small and A- is a weak base, the addition of a strong acid will form HA, driving the reaction to the right and increasing $K_{sp}$.
In the specific case of
$$\ce{CaF2 <--> Ca^2+ + 2F−}$$
the addition of $\ce{HCl}$ gives
$$\ce{CaF2 + 2HCl <--> Ca^2+ + 2Cl- +2HF}$$
driving the reaction to the right, thus increasing the solubility of $\ce{CaF2}$.  Note that $\ce{HF}$ itself is a weak acid, having a pKa of $3.17$, and will be in equilibrium with $\ce{H+}$ and $\ce{F-}$.


text:
Effects of pressure on equilibrium

Which system at equilibrium will not be influenced by a change in pressure?
$$
\begin{align}
\ce{3 O2(g) &<=> 2 O3(g)}\tag{A}\\
\ce{N2(g) + 3 H2(g) &<=> 2 NH3(g)}\tag{B}\\
\ce{2 NO2(g) &<=> N2O4(g)}\tag{C}\\
\ce{H2(g) + I2(g) &<=> 2 HI(g)}\tag{D}\\
\ce{2 W(g) + X(g) &<=> 3 Y(g) + 2 Z(s)}\tag{E}
\end{align}
$$

My attempt: Equilibrium will be influenced if the number of moles on one side of the equation is different than the other side. Thus, I can eliminate option A,B, and C. But how to differentiate between options D and E? They both look right to me, because there are the same amount of moles of gases on both sides, and solids do not play a factor in determining equilibrium.
For reference, the answer given in the book is D.

The solid is slightly compressible, so there will be a change in volume. It's minimal compared to the gases, but it's not zero.

@Zhe But doesnt the equilibrium constant only depend on gases, as they are the only ones that occupy a significant amount of volume?

I  think that the answers should be D and E for as @Zhe writes the change in volume of a solid is minuscule and can be ignored.  There might be typo in E which is why the answer is D only.

As Zhe explains in the comments, the answer is D. Even though small changes in pressure are not expected to significantly affect the chemical potential of a solid, it is a function of pressure:
$$\mu(\mathrm{s}) = \mu^\circ(\mathrm{s}) + \int_{p^\circ}^{p}V_\mathrm{m}\,\mathrm dp \qquad T = \mathrm{const}$$
This means that a small change in the chemical potential of the solid also contributes to the total free energy change of the system when the pressure is altered. Usually this response is ignored as $V_\mathrm{m}$ is orders of magnitude greater for gases than for condensed phases.

Answer:
Well it's about the amount of moles of gases on left and the right sides of the equal/equilibrium sign
for A, it is 3 moles vs 2, so then if we apply pressure it shifts to the right, to make this more visual, imagine a tower with 3 blocks stacked and another tower stacked 2 blocks high, if we were place them side by side and push them down with a plate, obviously the tower with the 3 blocks is going to be compressed First
for B, it is 4 vs 2 moles
for C, it is 2 vs 1 moles
for D, it is 2 vs 2 moles
for E, it is 3 vs 2=3 moles (with solid)
From the list, we can see that in choice D, we have the same amount of moles on both sides. Adding pressure or decreasing them wouldn't affect in which direction it would go to.
You said that you wanted help between choice D and E, but again if we look at the problem, for choice E, we have a tower stacked 3 blocks high vs 3+solid blocks high, so that means that the one with the 3+solid blocks is higher and thus means unbalance
to convert that into chemistry terms, because you have an imbalance of moles of gases, that means that if you apply pressure, you are going to be disrupting the side with 3+solid moles as compared to 3 moles.
So
you want to have choice D because there is no imbalance of moles of gases
Also I understand that you are wondering about the solid being there, I believe it is not choice D because we have the same amount of moles but the definition of an equilibrium is that it goes both forward and back so then that implicates that the Z turns into products as well. So by adding pressure you actually favor one side more because either adding or decreasing pressure means that it becomes harder for Z to become a gas or to turn into a solid depending on the imbalance created. Z will eventually become a gas at one point and the added pressure or decreased pressure might make that harder. Z turns into gas molecules at one point in time, so your going to have more than 3 moles gas on left side, it does that naturally because definition of equibrium, but if you apply pressure it becomes harder to do that because RXN shifts right

But only gases apply to equilibrium, answer choice E has 2 mols of solid, making it 3 vs 3 moles.

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text:
How to solve titration problem?
A sample of 40.0 mL of a .100 molar HIO solution is titrated with a .150 molar NaOH solution. Ka
for hypoiodous acid = 2.3 x ^ (-11)
What is the pH of the solution?
I have the following work:

As you can see I was given half credit. Can anyone explain why?

Have you heard of the Henderson Hasselbalch?

Why would I use the Henderson Hasselbalch equation? According to my professor's handout the pH is determined by the excess of hydroxide--is this wrong?

It should depend on the volume of $\ce{NaOH}$ used, but that isn't given to you. As far as I know, you would need a volume of $\ce{NaOH}$ to know that you have an excess.

So the conjugate base determines the pH in this case? Why would it be so when there exists a concentration of OH- ions?

I'm not certain. It would if the moles of $\ce{NaOH}$ added was less than the moles of HIO, but, at least as presented, the problem doesn't give enough information to find moles $\ce{NaOH}$, since we don't have a volume.

Well, I still don't understand what exactly the pH is being determined by since I can't assume that NaOH is in excess without a given volume.

Can you ask your professor? The question could be missing information.

This is a terrible question! There are simply too many ways to interpret it. I would assume that the lab jargon "titrated" means that the equivalence point is reached, but then the concentration of the NaOH is almost completely irrelevant. It would be best that you asked for the master solution from your teacher.

OK. So I talked to my professor, and apparently, I was supposed to assume that by "finding the pH of the solution" is the initial pH of the solution (that is, the acid is only in solution). That's what I was thinking too--that this is a terrible question! There are many ways to interpret it. More things than just the acid can be present in a solution. I asked if there was missing information but he said no--that this was supposed to be a barrier that I was supposed to overcome. I wasn't expecting the latter conclusion, but this, personally I think was intentionally misleading.

He also stated that if I had done this, I should be using the Kb value of the acid instead of the hydroxide concentration--is what he said correct, and if it is, is it because the conjugate base is stronger than the hydroxide ion?

This is a case of salt hydrolysis of a salt of weak acid and strong base
$$\ce{HIO + NaOH -> NaOI + H2O}$$
$$\ce{NaOI -> IO^- + Na^+}$$
$$\ce{IO^- + H2O <=> HIO + OH^-}$$
You can calculate the final concentration of $NaOI$ to be $0.06M$ (denote this by $c$).
Let $h$ be the hydrolysis constant.
Thus in the equilibrium state, $[IO^-] = (c-ch)M$, $[HIO] = (ch)M$ and $[OH^-] = (ch)M$.
Now $$ k_h = \frac{[OH^-][HIO]}{[IO^-]} = \frac{(ch)(ch)}{(c-ch)} = \frac{ch^2}{(1-h)}$$
Here $k_h$ is the hydrolysis constant. As the hydrolysis constant is small hence we can approximate $(1-h)\approx 1$.
Therefore $$h = \sqrt{\frac{k_h}{c}}$$
So $$[OH^-] = ch = \sqrt{ck_h}$$
As $k_h = \frac{k_w}{k_a}$
Thus $$[OH^-] = \sqrt{c\frac{k_w}{k_a}}$$
As $[H^+] = \frac{k_w}{[OH^-]}$
So finally we can write $$[H^+] = \sqrt{\frac{k_{w}k_{a}}{c}}$$
Thus $$pH = -\frac{1}{2}log(\frac{k_{w}k_a}{c}) = \frac{1}{2}(-log(k_w)-log(k_a)+log(c)) = \frac{1}{2}(pk_w + pk_a + log(c))$$
Putting the values of $k_a$, $k_w$ and $c$, we have
$$ pH = \frac{1}{2}(14 + 10.64 - 1.22) = 11.71$$

In case you haven't studied salt hydrolysis, remember $k_h*k_a = k_w$ for salt of weak acid and strong base. You can derive it easily by writing the equations of $k_a$ and $k_h$ and multiplying them.

How are you obtaining the concentration c? Without knowing the volume of  NaOH added, we can't know the concentration of either HIO or NaOH in the final solution.

Well in titration we add NaOH drop wise and stop at the equivalence point. You can easily calculate the volume of NaOH needed for complete neutralization by equating the moles reacted. Then divide the number of moles of NaOI formed and divide it by the total volume (HIO + NaOH added) to get $c$.

I would argue the problem isn't clear as to where the titration goes to(whether it stops before or after equivalence), but equivalence would be a reasonable point.

In titration problems, it is already assumed that the student knows that titration stops when equivalence point is reached and hence I gave the solution.

But if you look at a typical titration curve, they are rarely if ever drawn only to equivalence. I think the question needs to be more specific to what point in the titration.


text:
Is Body Centred Cubic arrangement also a kind of packing in solids?
I am confused about the packing thing in solids . I read about Hexagonal Closed packing(HCP) and Cubic closed packing(CCP) then realised this stuff named as Body Centred Cubic(BCC) under the heading of "Arrangement in solids" . Now what is the real difference between packing and arrangement? After all they are arranged in 3d and finally form a crystal
IT would be very kind if someone can clear my doubt... :)

The real difference between packing and arrangement is that some authors prefer the former term and some the latter.

Perhaps better put, body centered cubic is a Bravais lattice, and many pure metals exist as bcc crystals.


text:
How does physical adsorption of mixed gas occur on solid? (eg I mean how the gas molecules of compete each other for a slot on solid surface)
I am working on sorption phenomenon of gases (including inert gas) on coal seam surface.
For a mixture of 2 gases at a certain pressure, carbon dioxide and methane, for instance, $\ce{CO2}$ is definitely adsorbed quicker and more comparing to methane. How come? I try to explain by Langmuir Isotherm and IAS model but it doesn't help much. For Langmuir isotherm, the affinity of gases to solid is determined by "b=1/PL" which comes from experiment. Langmuir isotherm assumes that all gas have equal access to the solid surface; however, this is untrue in reality.
Can you help me explain it in the thermodynamics point of view or molecular forces please? ($\ce{CH4}$, $\ce{CO2}$ are both non-polar gases)

CO2 is polar, in that its bonds are polar. It is just that its overall dipole moment happens to be 0.

Your supposition that gases do not have equal access to teh surface does not seem to me to be correct.
At the surface  physisorption and chemisorption can occur. The former is due to inter-molecular interaction with the gas molecule and the surface molecules so clearly will depend on the nature of both of them. In chemisorption a chemical bond is formed. The physisorption occurs at longer range and has a smaller potential well (as measured by potential energy vs distance from surface) than chemisorption. (See Atkins & DePaula 'Physical Chemistry' for a figure.)
The gases approach the surface at random and the number of collisions of each will be in proportion to their partial pressure. What happens on collision depends, as mentioned above, on the strength of the intermolecular interaction of gas and surface and so depends on what they are. At higher temperature the collision energy is greater than at lower ones and so one would expect less physisorption as the collision energy will be greater than that of most available physisorption sites.
It is also possible for multiple layers to be formed (as in BET isotherm) and for a physisorbed molecule to diffuse around on the surface until it leaves again (by random thermal activation) of finds a more stable site or becomes chemisorbed.
As to your particular case we can only surmise but $\ce{CO2}$ is far more polarisable than methane so should have the larger dispersion forces so larger induced dipole - induced dipole intermolecular interaction and so preferentially occupy the surface. Both gases have transient dipoles due to molecular vibrations of which the $\ce{CO2}$ should be the larger. This may also be important.

You are talking about a porous solid. This is very different to a flat surface where adsorption of all molecules can occur without any barrier. To be adsorbed in the pores, the molecules need to be transported from the gas bulk to inside the pores, this process can be very slow depending on the diffusion rate.
By measuring an adsorption isotherm, you just get the amount adsorbed at equilibrium, that is, after all the molecules entered inside the pores.
If your solid has very narrow pores, say, micropores, then the size of the pore is very near the size of your molecules, and adsorption will be slow. In your case, the size of the $\ce{CO2}$ and $\ce{CH4}$ is very different, $\pu{0.33 nm}$ for $\ce{CO2}$ and $\pu{0.38 nm}$ for $\ce{CH4}$. So the diffusion inside the pores is much more restricted for $\ce{CO2}$, then it is expected to be slower. Probably your sample have many narrow micropores that result in this behavior.


text:
What do we call ions without basic and acidic properties
According to Brønsted theory,

Acids are substances (molecules and ions) donating $\ce{H+}$
Bases are substances (molecules and ions) receiving $\ce{H+}$

I've been trying to find a complete Brønsted-Lowry's Base and Acid charts. But I cannot find the chart mention "substances without basic or acidic properties" such as $\ce{CO}$, $\ce{Na+}$, $\ce{K+}$, $\ce{Cl-}$, etc..
Could anyone help me with what I should call these substances? And a keyword for a Brønsted-Lowry chart that also includes the substances with no basic and acid properties? All my learning is in my non-native language English. Right now my current English lecture composed by the University doesn't mention the category for the matter I've been looking for.

CO is a neutral compound.

I believe that the word you are looking for is a pH-neutral compound, or simply a neutral compound.

I think your best bet would be neutral salts. This term describes compounds such as $\ce{NaCl, KNO3, CaBr2, CsClO4}$. This page from the University of Waterloo gives a brief table of ions that when paired produce a neutral salt.


text:
Why doesn't hydroxide concentration equal concentration of carbonic acid and bicarbonate in a sodium bicarbonate solution?
To find the pOH of  solution of  $\pu{0.42 M}$ of $\ce{NaHCO3}$, given a   $\ce{Kb}$ of $\pu{2.4E-8}$, my solution manual does the following steps:

Reaction is similar to : $\ce{HCO3- + H2O <=> H2CO3 + OH-}$
$\displaystyle \ce{Kb} = \pu{2.4E-8} = \ce{\frac{[OH-][H2CO3]}{[HCO3-]}}$
$\displaystyle \ce{Kb} = \pu{2.4E-8} = \ce{\frac{[OH]^2}{[0.42]}}$ , as we have $\pu{0.42 M}$ of $\ce{NaHCO3}$ so approximately $\pu{0.42 M}$ $\ce{HCO3-}$, and $\ce{[OH-] = [H2CO3]}$
Solve equation for $\ce{[OH-]}$, giving $\ce{[OH-]} = \pu{10^{-4}}$  and hence $\ce{pOH} = 4$

What I don't understand is: Given the equation $$\ce{HCO3- + H2O <=> H2CO3 + OH-},$$ doesn't this show $\ce{HCO3} = \ce{H2O} = \ce{H2CO3} =\ce{OH-}$, as they are all in a $\mathrm{1:1:1:1}$  molar ratio in the equation.
So wouldn't $\ce{OH-}$ also just equal $\pu{0.42 M}$?

No, it doesn't show that, much like it doesn't show that you have precisely one molecule (or one mole, or one anything) of H2O in your reaction based on the fact that there is one H2O in the equation.

Thank you. I must have really misunderstood something when learning this topic. I will have to do some more reading to make understand why your comment is true.

Could you please point out why we can take [HCO3-]=0.42, but not [OH-]=0.42? We are only told that we have 0.42 M of NaHCO3.

We can't. We only know that there were 0.42 M of NaHCO3 before the reaction. Then again, judging from experience and looking at the constant, we may say that the reaction occurs to a very little extent, so in the end not much on this NaHCO3 is spent, and its concentration is still almost 0.42M.

Thanks Ivan. I think I understand now: Recapping what you and other answer said: The reaction has a low Kb, so little of it dissociates, so HCO3- approx. = NaHCO3 = 0.42M (approx. equal to our starting point as Kb is so low - and so products <<<<< reactants).
[OH-] = [H2CO3], as they are both produced 1:1 as both have a coefficient of 1 and both appear as products. BUT we can't say [HCO3-]=[OH-] = [H2CO3] because Kb is not =1, and so 1 mole of HCO3- does not get fully used up and (in combination with H2O) become 1 mole of H2CO3+ 1 mole of [OH-]. Does that sound right? Thanks

Change "dissociates" to "hydrolyzes". Other than that, you are right.

Ah okay.  That also answers how I include the H2O. Thank you for your guidance!

@IvanNeretin   I have  a follow up to this if you don't mind.
For the reaction NH3 + H2O <=> NH4+  + OH-
Given 0.5 mole of amoonia dissolved in 250 mL of water, I am told [OH] is 2 mol/L
Why is this the case i.e. 2 mol/L [NH3] is taken to give 2 mol/L [OH-] but in my original question we were not allowed to do this, because the reactants don't necessarily all get used up to form products.

Note I have opened another Q which asks about this https://chemistry.stackexchange.com/questions/88663/why-does-2-mol-l-of-nh3-become-2-mol-l-of-oh

Your misunderstanding appears to be some confusion between stoichiometry and equilibrium concentrations. The chemical equation describes the stoichiometry or the ratio of the species in the reaction. In other words, each time the reaction happens, the equation describes how many of each reactant are consumed and how many are produced. In your example:
$$\ce{HCO3- + H2O <=> H2CO3 + OH-}$$
Each time this reaction occurs in the forward direction, one bicarbonate anion and one water molecule react to form one carbonic acid molecule and one hydroxide anion. In the reverse direction, one carbonic acid molecule and one hydroxide anion react to form one bicarbonate anion and one water molecule.
Since this reaction is reversible, the equilibrium constant tells you something about the extent of the reaction. In other words, what portion of the overall system is sitting on the reactant side and what portion sis sitting on the product side. Put another way, the system is at equilibrium when the rates of the forward and reverse reactions are equal. These rates are dependent on the concentrations of the species on the reactant and product sides of the equation and on rate constants.
$$\mathrm{rate_{forward}=rate_{reverse}}\\
k_\mathrm{f}[\ce{HCO3-}][\ce{H2O}]=k_\mathrm{r}[\ce{H2CO3}][\ce{OH-}]$$
The equilibrium constant is the ratio of the rate constants:
$$K=\frac{k_\mathrm{r}}{k_\mathrm{f}}=\frac{[\ce{H2CO3}][\ce{OH-}]}{[\ce{HCO3-}][\ce{H2O}]}$$
The equilibrium constant is thus a law of mass action expression for the situation where the forward and reverse rates are equal. At this state, both forward and reverse reactions are occurring following the stoichiometric ratios, but the concentrations of each species has reached a steady state.

Thanks Ben. I think I understand now: Recapping what you and other answer said: The reaction has a low Kb, so little of it dissociates, so HCO3- approx. = NaHCO3 = 0.42M (approx. equal to our starting point as Kb is so low - and so products <<<<< reactants).
[OH-] = [H2CO3], as they are both produced 1:1 as both have a coefficient of 1 and both appear as products. BUT we can't say [HCO3-]=[OH-] = [H2CO3] because Kb is not =1, and so 1 mole of HCO3- does not get fully used up and (in combination with H2O) become 1 mole of H2CO3+ 1 mole of [OH-]. Does that sound right? Thanks

Hi Ben, I have  a follow up to this.
For the reaction NH3 + H2O <=> NH4+  + OH-
Given 0.5 mole of amoonia dissolved in 250 mL of water, I am told [OH] is 2 mol/L
Why is this the case i.e. 2 mol/L [NH3] is taken to give 2 mol/L [OH-] but in my original question we were not allowed to do this, because the reactants don't necessarily all get used up to form products.

Note I have opened another Q which asks about this https://chemistry.stackexchange.com/questions/88663/why-does-2-mol-l-of-nh3-become-2-mol-l-of-oh


text:
Oxide and chloride of a metal
I found this problem:

The oxide of a metal contains 60.0% metal and the chloride of the same metal contains 25.26% metal (in mass percents). What are the possible formulas for the oxide and the chloride?

I looked up the solution and it states that the results are $\ce{MgO}$, $\ce{MgCl2}$, $\ce{TiO2}$ and $\ce{TiCl4}$, but I don't understand how they got it. How should I approach this kind of problem?

Because an oxygen ion is $\ce{O^2-}$ and a chlorine ion $\ce{Cl^-}$, you know that the salts have to be of the form $\ce{M_xO_y}$ and $\ce{M_xCl_{2y}}$.
Let $o$ be the atomic mass of oxygen, $c$ of chlorine and $m$ of the unknown metal M. The problem states that
$$0.6(xm+yo)=xm$$
which implies
$$0.6yo = 0.4xm \implies m = \frac{3y}{2x}o$$
If you recall some formulas for salts, you know that $x$ can only be 1 or 2 (e.g. $\ce{M_5O_3}$ makes no sense). This means that $m$ is a multiple of $\frac34o$ which is (approximately) 12. So we only need to check the elements with an atomic weight which is a multiple of 12:

12: Carbon (C) isn't a metal
24: Magnesium (Mg) is, and you can check the oxide/chloride satisfy the equations
36: n/a
48: Titanium (Ti) checks out
60: n/a
72: n/a
84: Krypton (Kr) isn't a metal

and so on. Even if the atomic weight corresponds to a metal, you can verify the value of $y$ doesn't match with the oxide that's formed.


text:
Does dissolving an ester shift the chemical equilibrium?
Esters are fairly insoluble in water , however one of the products of an esterification reaction is water , hence by washing the ester as a method of purification you are adding more water and increasing the concentration of water , would this not cause an equilibrium shift and hence result in a lower concentration of the ester ?

It would, if not for the fact that the reaction is pretty slow.


text:
Why doesn't liquid propane in a closed coke bottle burst?
I was watching this YouTube video in which the original experiment was to make a small rocket out of liquid propane and Coke. When that failed, the person doing the experiment decided to try pouring liquid propane into the Coke bottle and closing the cap on it to see what would happen.
I would have expected (and I believe he did too) that the bottle eventually would build up enough pressure to explode. However, that did not happen and along with that not happening, the liquid appeared to warm up to about air temperature and stopped boiling. When he opened the lid of the bottle, it quickly depressurized, the liquid began boiling and returned to its normal, cold state.
My guess is that once the propane vapors filled the bottle and expanded sufficiently, it put enough pressure on the liquid to keep it from boiling. So my question here is, is my guess correct and, along those lines, if it had been warmer outside, would that bottle have built up enough pressure to explode?
Note: Even if my guess is correct, I want to see the hows and whys of it in an answer.

What I learned is that the youtube poster Grant Thompson is an idiot. It is hard to imagine anything more stupid than heating a plastic coke bottle filled with propane with a blowtorch. // Not knowing the burst pressure of a coke bottle it is impossible to give L.B. an answer.

@MaxW Well, that's not the point here because my original question is more related to why the propane stopped boiling!

@MaxW Doing the same thing in a plastic water bottle (not designed to be pressurized at all and generally made of as little plastic as possible) would be more stupid.

The ideal gas law states that PV=T If volume is constant, then an increase in temperature will cause an increase in pressure. "Boiling occurs when the equilibrium vapor pressure of the substance is greater than or equal to the environmental pressure ... The boiling point varies with the pressure of the environment," or, boiling point increases as pressure increases. Vaporization It's a feedback loop. In other words, yes, you're correct.

Yes, based on what we can see in the video, your guess appears to be correct: as the propane-filled bottle warmed up, just enough propane evaporated to keep the pressure inside the bottle equal to the equilibrium vapor pressure of the liquid propane.
According to the video, the ambient temperature outside at the time it was recorded was "about 45 °F", or about 7 °C.  Using the formula given here, I calculate the vapor pressure of propane at that temperature to be about 4400 mmHg, or about 590 kPa or about 5.9 bar.
Meanwhile, according to this page, the pressure inside a warm can or bottle of Coke can reach at least 380 kPa, or about two thirds of the vapor pressure of propane on a cold day.  As the bottles are certainly designed with a considerable safety margin, to make sure that they won't burst even if handled carelessly or slightly damaged, it's not surprising that they can easily withstand the pressure of the propane in the video.
BTW, this is exactly how aerosol spray cans work: they contain a mixture of the liquid being sprayed and a propellant substance (quite often propane) that has a boiling point at 1 atm only slightly below room temperature (or, equivalently, that has an equilibrium vapor pressure only slightly above 1 atm at room temperature).  Thus, as the can is drained, the partial boiling of the propellant maintains the pressure inside the can at the propellant's vapor pressure, which is high enough to propel the spray out of the nozzle, but not so high that it would require an excessively sturdy and expensive can to contain it.

As for what would happen at higher temperatures, at room temperature (i.e. 25 °C), the vapor pressure of propane would be about 7100 mmHg or 950 kPa (according to the formula, or about 7600 mmHg or 1000 kPa according to the table, which seems to be taken from a different source).  According to this random forum post, the small ½ liter Coke bottles used in the video can apparently withstand at least 180 psi, or 1250 kPa, so the propane-filled bottle probably wouldn't burst even at room temperature (unless it happened to be damaged or otherwise particularly weak).  If the temperature was raised to, say, 45 °C (113 °F, a very hot day), the vapor pressure of the propane would rise further to about 1500 kPa, which just might be enough to make the bottle fail.  Also, elevated temperatures will soften and weaken the plastic somewhat, making failure more likely.
In any case, in my personal experience, the weakest point of such bottles seems to be the relatively thin and weak cap, which would likely fail at some point before the bottle itself did.  I wouldn't be surprised if that was by design, to make the typical failure mode relatively safe and predictable.

The weakest point of PET bottle is in the middle of the bottom.

The person in the video said it was about $\pu{45^oF}$. At this temperature propane has a vapor pressure of about 8 atm, so that is the pressure that built up inside the Coke bottle. Once this pressure had built up, it quit boiling.
The fact that he could still fairly easily squeeze the bottle suggests that it was probably not very close to bursting from the pressure, and I very much doubt that even a hot day would have caused enough pressure for it to burst. I once saw the aftermath of a similar experiment with a bottle like that in which the lucky-to-be-alive idiot had used liquid nitrogen to blow it up. The bottle had badly deformed and stretched out before bursting.
Bonus Answer: Why it didn't work
Since propane is less dense than, and immiscible with, water (or Coke), it just floated on top and was the first thing to pour out when the bottle was inverted. If he had held his glove-covered hand over the opening and inverted it so that the propane was on top inside the bottle, rapidly expanding in a confined space, it would have at least had a chance of working.

Thanks for including why his rocket didn't work... I kinda thought of that but hadn't actually thought it through yet

And, if you look at the video he got the idea from, the guys there use quite a distinctive technique for turning the bottle upside-down. They essentially allow the bottle to topple about its centre of gravity, which I guess allows some of the propane to float to the top. Thompson just turns it upside down, pouring the propane onto the floor.

Thinking about it further, as the propane floats through the Coke, it will be heated quite a bit, helping it to boil.


text:
Electrochemical concentration cells -- how anion levels change
It seems clear that in the anode, the anion concentration will increase as it enters from the salt bridge to balance the cations leaving from the oxidized electrode. But what is happening to the anion concentration in the cathode?  Cations are leaving the salt bridge to replace those being reduced on the electrode.  So is the anion concentration just constant?

Considering you have chosen a $\ce{Zn}$ concentration cell:
The cell representation is:
$$\ce{Zn(s)/ZnSO4(M1)||ZnSO4(M2)/Zn(s)}$$
Anodic reaction puts more $\ce{Zn^2+}$ ions in the solution:
$$\ce{Zn(s) -> Zn^2+(aq) + 2e-}$$
To balance these extra $\ce{Zn^2+}$ ions salt bridge pulls some anions into anodic compartment.
Cathodic reaction removes $\ce{Zn^2+}$ ions from cathodic compartment:
$$\ce{Zn^2+(aq) + 2e- -> Zn(s)}$$
To make up this equivalent amounts of cations are pulled off from the salt bridge.
So activity of $\ce{SO4^2-}$ is not changed in either of the solutions. It is only those of $\ce{Zn^2+}$ ions in both the compartments, which undergoes a change leading to redox reaction.

@Equinox - I have properly typeset the superscripts for the zinc cation and sulfate anion. Please see the body of my edit to see how you can format superscripts in the future.

Ahh, you added a carat. Thank you, I had assumed that the Chemical Equation environment would have taken care of that.


text:
The effect of temperature on equilibrium
If the forward reaction is exothermic, and the reverse is obviously endothermic. I getting rather confused. If the temperature is increased the equilibrium will want to decrease the temperature. However, (this is the bit im struggling with) if I want to decrease the temperature wouldn't you shift it to the left? As the reverse reaction is endothermic, and an endothermic reaction causes a decrease in the temperature or is this incorrect?

True, the reverse reaction would be favored, hence the equilibrium would shift somewhat to the left. Also, welcome to Chem.SE.

In order to understand an answer to this you need to be able to differentiate between system and surroundings
The system is the reactant mixture , now for an endothermic reaction the temperature of the reactant mixture will increase , however the temperature of the surroundings will decrease since heat is lost by the surroundings to the system , let's assume that an endothermic reaction took place in a test tube , if we put our hand around the test tube our palm will feel cooler since heat was transferred from our palm , however that heat was transferred to the reaction mixture hence the temperature of the mixture was raised . Hence this increase in temperature means we have altered the equilibrium conditions and thus excess heat needs to be lost therefore the eq will shift towards the exothermic reaction . However a decrease in temperature means the eq needs to gain heat which is in the form of an endothermic reaction


text:
RO (alkoxide) as a leaving group
Why is RO (alkoxide) a better leaving group than OH, despite RO being more unstable due to the electron donating effect of the alkyl group on RO?
I read that the suitability of leaving groups is dependent also on their nucleophilicity, as good nucleophiles will tend to re-attack the molecule it left from. To what extent is this true?

Both aren't viable leaving groups at all.

The alkoxide and the hydroxides aren't good leaving groups. Consider an alcohol, the $\ce{OH}$ group never leaves on its own. Oxygen donates a lone pair to the hydrogen of a hydronium ion (considering it to be in an aqueous solution). The water molecule now attached is a good leaving group (oxygen has a positive charge).
Nucleophilicity does not determine the suitability of a leaving group. A group is said to be a good leaving group when it can leave as a relatively stable ion. For example: iodine is a very good leaving group for an $\mathrm{S_N2}$ reaction.

Both alkoxides and hydroxides are not great leaving groups, but since we are just comparing their leaving group character...
First of all, leaving group tendency does not depend on nucleophilicity. Rather, it depends on basicity. Both of these are often confused to be the same thing. Yes, it's true that coincidentally both of these, in most cases, suggest similar things, but they are not the same. Let us compare the two for OH and OR.
Nucleophilicity:
OH is a better nucleophile than OR since it has more mobility(due to its small size)than OR. A good nucleophile should be small in order to attack sites easily.
Basicity:
This does not depend on mobility. It only depends on electron donating tendency (inversely proportional). Here, OR is a weaker base, hence it is a better leaving group

You are confusing hyperconjugation effect on a carbocation with basicity. Basicity mainly depends on the following factors in order of significance: Atom → Resonance → Inductance → Orbital. Atom is the same (oxygen), there is no resonance, so inductance is key. Carbon is more electronegative than H, so OR will better stabilize a negative charge, thus OR is a better leaving group stable weak base.

Remember this is true for an aprotic solvent like DMSO.In acuose solution you can get confused because solvation takes huge effect in protic solvents. So in water -OH is better stabilized by solvation thus making it weakbase or better LG. But guess what most likely you are exploring the Trans-Esterification here either hidroxy or alcoxy can be expelled (even if -OH is a better acuose LG)but notice that when you expell hidroxy you are left in the same previous ester state. So no change. In contrast if you expell -OR you will form a carboxilic acid COOOH. Cheers!

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text:
Why delta G = -nFE and not -F * integral(dn*E)?
$\Delta G$ equals the maximum work that a system can do and in redox reactions it's the work done by electrons. According to physics, $W=E\cdot q$; but that's true only if $E$ is constant, however during the reaction, $E$ decreases until it's zero, so mustn't it be like the integral $\int \mathrm dn\cdot E$? Why do we instead say that $\Delta G=-F \int \mathrm dn\cdot E$?

Because $\Delta G$, or more properly speaking $\Delta_\mathrm{r}G$, is not referring to the Gibbs free energy change of the system as $E$ slowly decreases to $0$. It's referring to the slope of $G$ when plotted against $\xi$, the so-called "extent of reaction". In this context, when you transfer $\mathrm{d}n$ moles of electrons, the Gibbs free energy change is $\mathrm{d}G = -FE,\mathrm{d}n$. The quantity $\Delta_\mathrm{r}G$ is not related to $\int \mathrm{d}G$, but is instead related to $\mathrm{d}G/\mathrm{d}n$ (that's also why it has units of kJ/mol and not kJ).

Since it is a gradient, $\Delta_\mathrm{r}G$ is defined only at a single state of a system, and therefore a single instantaneous value of $E$. The explanation of the meaning of $\Delta_\mathrm{r}G$ should be in most physical chemistry textbooks, and there is also a good (but fairly involved) article on it: J. Chem. Educ. 2014, 91, 386

The Anslyn / Dougherty book also covers this concept quite well.

n is not real number with meaning of molar amount but integer parameter with meaning of number of exchanged electrons per 1 ion or molecule. Therefore it is not differentiable.

We can find too many descriptions of how to use the formula ΔG = - zFΔE, but I have never seen how to derive it. However, I believe that this equation is probably not that difficult to derive. Perhaps the following logic can be used to derive it;
Let the W be  "the energy obtained as a result of the transformation of 1 mol of a certain reactant into a product in an isothermal and isobaric process".
And, we assume that all this energy was used to accelerate z mol of electrons. Suppose that this effect is represented by the potential difference ΔE,
W = - zFΔE
Conversely, if ΔE is given, and W is the "energy of z mol electrons moving to the location of potential difference ΔE," then
W = - zFΔE
Here,
ΔG = G(product) - G(reactant)
and theΔG means "maximum energy obtained in the process of changing 1 mol of a certain reactant to a product in an isothermal and isobaric process"
So, assuming that the maximum energy can be extracted in this isothermal isobaric process and that it is all used to accelerate z mol of electrons, ΔG causes the same effect as the following ΔE occurs between the anode and cathode
ΔG = -zFΔE
Consider the situation where a beanbulb is connected to a battery. An isothermal isobaric process is taking place in the dry cell, This isothermal isobaric process produces ΔG, but what is flowing through the beanbulb is electrons, not ions.
So, when considering the electromotive force of a battery, it would be natural to consider the flow of electrons. In the 18th and 19th centuries, ignorant mankind decided that current flows from the anode to the cathode. But actually, electrons flow from the cathode to the anode　（The charge of the particles mediating the current was positive, not negative）. Therefore, when thinking about electrons, the ± signs of potential always tends to cause such confusion.

The equation is simply the definition of electrical work: charge times voltage. In this case, written for a mole of electrons (Faraday constant).

@Matt Hanson You are probably right in explaining the simplest ΔG=-nFE. However, the subject of this question is not the derivation of this equation itself, but why the sign is "negative".
So the simplest answer to this question would be to explain the process of derivation with attention to the sign.


text:
How well do d-block electrons actually shield?
I've been using 'Chemical Structure and Reactivity: An integrated approach' by James Keeler and Peter Wothers to study periodicity, among other sources. However, there seem to be some commonly accepted contradictions used, and I don't understand how to reconcile these ideas:
1) D-block electrons shield poorly (s>p>d>f)
2) Using Slater's scale, d block electrons would, in fact, shield as well as most other (n-1) orbitals. Even though this is a rough scale, I take it to mean that d block electrons do have a notable shielding effect.
3) D-block contraction is the effect of increased Zeff due to the poor shielding of d-block electrons, and the addition of an equal number of protons.
Therefore, for example, down Group 13: the filling of the d-block orbital corresponds to an extra increase in Zeff with respect to the 4s/4p compared to the 3s/3p orbital. The 4s/4p orbitals will therefore be lower in energy than expected: shielding does not completely offset the extra protons.
4) Across the first row d-block elements, the energy of the 4s orbital does not fall very much. This is because the increase in nuclear charge is offset almost entirely by the addition of a shielding d-block electron, implying almost perfect shielding.
These four points seem, to me, to be essentially using d-block electrons at both ends of the spectrum when convenient. Is there a way to reconcile these ideas, or where have I gone wrong in my logic?

Related: If d-electrons are such poor shielders, why do trends increase more gradually across the d-block than the s or p-block? (I do not entirely approve of the answer there)

Let us read Keeler and Wothers's Chemical Structure and Reactivity carefully. (It's really a great book. I wish I had used it in the past.) We excerpt the relevant paragraphs that detail the points you mention, and highlight particularly important statements regarding screening. All page numbers are taken from the second edition.

Point 2, on Slater's rules.

[Energy ordering:] (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) ...

If the electron being considered is in an ns or np orbital, then electrons in the next lowest shell (i.e. that with (n-1)) each contribute 0.85 to $S$. Those electrons in lower shells (i.e. (n-2) and lower) contribute 1.00 to $S$.

If the electron being considered is in an nd or nf orbital, all electrons below it in energy contribute 1.00 to $S$ [the shielding constant]. (p. 262)

Point 3, on Z$_\text{eff}$ and d-block contraction.

...there is a downwards kink in the [s-block] orbital energy on moving from the 3s to the 4s [orbitals]. [...] This difference is due to the filling of the 3d orbitals between calcium and gallium. The effects of all the extra protons in the nucleus is to cause an additional lowering in the energy of the valence orbitals for the p-block elements in Period 4. However, the effect is not really drastic: the 4s and 4p orbitals in gallium do not seem to have experienced the full effect of the ten extra protons. We interpret this by saying that the effect of the extra protons on the orbital energies of the 4s and 4p has, to a large extent, been cancelled out by the electrons that have been added in the 3d orbitals.
We could rephrase all of this by looking at the problem from the perspective of effective nuclear charges rather than orbital energies. The increase in Z$_\text{eff}$ for gallium is slightly greater than might have been expected by comparison with boron and aluminium because of the filling of the d-block. (p. 266)

Point 4, on orbital energies.

...between scandium (Sc) and copper (Cu) electrons are being added to 3d orbitals, and such electrons form an effective screen for the 4s electrons. To put it another way, as far as the 4s electrons are concerned, on moving from one element to the next between scandium and copper, the effect of the extra proton is largely cancelled out by the addition of the electron to the lower 3d orbital. Consequently the energy of the 4s falls rather slowly. (p. 261)
The outer electron in potassium occupies the 4s AO, rather than the 3d. This is because the 4s orbital penetrates to the nucleus more effectively than does the 3d, resulting in the energy of the 4s being lower than that of the 3d. [...] ...both the 3s and the 3p have subsidiary maxima close in to the nucleus, but the first maximum for the 3s is very much closer in. The 3d has no such subsidiary maxima, and therefore is much less penetrating than 3s or 3p.
...the electrons in the 4s do not screen the 3d particularly well since much of the electron density from the 4s is further out from the nucleus than that from the 3d.
From scandium onwards the energies of both the 4s and 3d AOs drop steadily: this is simply the result of the increase in nuclear charge not being quite offset by the increase in electron-electron repulsion. Put another way, the electrons in the 3d sub-shell do not screen one another particularly well. Although both AOs fall in energy, the 4s falls less steeply, which can be explained by noting that this AO is quite well screened by the 3d electrons, not least as these are in a lower shell. (p. 627)

We see that Keeler and Wothers are at least self-consistent regarding points 2, 3, and 4: 3d electrons are effective at shielding 4s and 4p electrons and not effective at shielding 3d electrons. I believe, therefore, that the contradiction lies in point 1, which should be corrected as follows:

d-block electrons shield poorly s or p electrons in the same shell

I suspect the statement that d-block electrons shield poorly is taught in reference to the concept of penetration. If we look at plots of radial probability density for the 3s, 3p, and 3d orbitals, we might naively conclude, on the basis of mean distance from nucleus, that 3d electrons are closest to the nucleus, which means that they should shield 3s and 3p electrons well. In reality, because of penetration, this is not the case. Thus we conclude that 3d electrons shield (s or p electrons in the same shell) poorly, and the misconception arises if we try to extend this poor shielding to 4s electrons and the like.

If u see the radial probability distribution function of nd orbital and that of ns orbital u will notice small humps are more in case of s orbital than that of d orbital. Because of this reason s orbital are more 'penenetrating' towards nucleus than that of a d orbital. So the screening constant value for a d orbital is much less compared to that of s orbital.
 Again if we consider the angular probability distribution function then we can see that d orbital are very much distributed in space compared to that of a s orbital which is spherically symmetric. Due to this reason d orbital registers a lower value of screening constant.


text:
Precipitation reaction stoichiometry?
I have a question where I do not understand the final step to solve...here it is:
What will be the Strontium ion concentration remaining after 30.0 mL of 0.10 M Na2SO4 solution are added to 70.0 mL of 0.20 M Sr(NO3)2 solution?
First, I wrote the equation: Sr + SO4 -> SrSo4 (I know this is excluding ions, because I am primarily focused on the stoichiometry part)
I figured moles of Na2So4 and Sr(NO3)2, which correspond to moles of strontium. These are .003 moles Na2SO4 and 0.014 moles Sr(NO3)2. This is where I got stuck, looked to my book's answer explanation, and got more confused. The book states
"0.0030 mol of sulfate ion will combine with 0.0030 mol of strontium ion, leaving 0.011 mol of strontium in a total volume of 100.0 mL."
Why is the reaction leaving 0.011 moles of strontium? What was the calculation that obtained this number?
Thank you!

Sr(NO3)2 + Na2SO4 = SrSO4 + 2NaNO3
So, here we have, 1 mol Sr(NO3)2 reacts with 1 mol of Na2SO4 by unitary method, you can say, 0.003 mol Sr(NO3)2 reacts with 0.003 mol of Na2SO4.
But checking the reactants we have 0.003 moles Na2SO4 and 0.014 moles Sr(NO3)2
So only 0.003 moles of each reacts with the other to leave out any excess in either of the reactants, as in this case, Sr(NO3)2.
When 0.003 mol reacts of Sr(NO3)2, we have $$0.014 (initial) - 0.003 (used) = 0.011mol$$ of Sr(NO3)2 left.

If you liked the answer, you could also upvote it. :) @tyger2020


text:
How to find mole fraction of solvent and solute in a vapor pressure lowering problem?
At $25$ degrees celsius, the vapor pressure of pure benzene ($\ce{ C6H6}$) is $93.9$ torr. When a non-volatile solute is dissolved in benzene, the vapor pressure of benzene is lowered to $91.5$ torr. What is the concentration of the solute and the solvent, expressed in mole fraction?
Through Raoult's formula, I found that the mole fraction of the solvent, benzene, is:
$$P_{solution}=x_{\ce{C6H6}}\cdot P^{o}_{\ce{C6H6}}$$
$$91.5=x_{\ce{C6H6}}\cdot 93.9$$
$$0.974 = x_{\ce{C6H6}}$$
Now the only thing I am having trouble finding is the mole fraction of the solute:
$$x_{\ce{C6H6}}=\frac{n_{\ce{C6H6}}}{n_{\ce{C6H6}}+n_{\ce{solute}}}$$
I was wondering, is it even possible to calculate the mole fraction with just the given info? I think that it might not be possible, since I need the grams of solute dissolved or grams of solvent to do this.

Hint: What does the sum of all the mole fractions within the system have to equal?

$\displaystyle \ce{x_{solute}} = \ce{\frac{n_{solute}}{n_{solute} + n_{solvent}}}$
Thus, we have that $\displaystyle\ce{x_{solute}} + \ce{x_{solvent}} = \ce{\frac{n_{solute}}{n_{solute} + n_{solvent}}} + \ce{\frac{n_{solvent}}{n_{solute} + n_{solvent}}} = \ce{\frac{n_{solute} + n_{solvent}}{n_{solute} + n_{solvent}}} = 1,$ so $\ce{x_{solute} + 0.974 = 1\implies x_{solute} = 0.026}$


text:
How to determine which combination of substances will give a buffered solution given target pH and Kb of each substances

What combination of substances will give a buffered solution that has a pH of 5.05? (Assume each pair of substances is dissolved in 5.0 L of water.) (Kb for NH3 = $1.8 × 10^{–5}$; Kb for C5H5N = $1.7 × 10^{–9}$)
a)1.0 mole NH3 and 1.5 mole NH4Cl
b)1.5 mole NH3 and 1.0 mole NH4Cl
c)1.0 mole C5H5N and 1.5 mole C5H5NHCl
d)1.5 mole C5H5N and 1.0 mole C5H5NHCl

According to my text book it is choice C but I do not understand why and here was my thought process:

For starters I change the $K_b$ into $K_a$ to get : $5.6*10^{-10}$ for Ammonia and for the other I got: $5.9*10^{-6}$
It seems to me that this has to do with the Henderson Hasselbach Equation so I set up the following:
$5.05=5.9*10^{-6}+log{\frac{Base}{acid}}$
$5.05=5.9*10^{-10}+log{\frac{Base}{acid}}$
because I already have two substances.
But from here I feel like that wasn't the best route I should take because I got stuck right here and would like some assistance.

btw please don't down vote because I didn't "show enough thought" Im a beginner chemistry student and I really would like assistance

Using the Henderson-Hesselbach equation is a good idea. As you want the buffer's pH to be acidic, then using the $\ce{NH3/NH4+}$ buffer is not a good idea, as that would be basic, so we should use the $\ce{C5H5N/C5H5NHCl}$. Then, by Henderson-Hesselbach, we have that $\ce{5.05 = pH = pKa + \log \left(\frac{[C5H5N]}{[C5H5NH+]}\right)}\implies \ce{5.05 - pKa = 5.05 - 5.23 = \log \left(\frac{[C5H5N]}{[C5H5NH+]}\right)} \implies \ce{\frac{[C5H5N]}{[C5H5NH+]} = 10^{-0.18}\implies \frac{[C5H5N]}{[C5H5NH+]} \approx 0.66\approx \frac 23}.$ Thus, we want a buffer solution where the ratio of the base to the acid is approximately 2/3, which answer choice C satisfies.

I understand everything except why it has to be 2/3 to satisfy the fact that we want 5.05, if I understand ill check your answer!

@JohnRawls If you use the henderson-hesselbach equation and substitute in known values, then you can solve for the mole ratio of 2/3


text:
How does Fermi's golden rule yield a specific rate?
According to Fermi’s golden rule, the rate of an electronic transition is proportional to the magnitude squared of $\langle i|\hat{H}|f\rangle$.  Since energy is always relative to some reference point, I could add an arbitrary constant to the Hamiltonian, and it would do nothing but shift all the energy eigenvalues by that constant.  So I don’t understand how Fermi’s golden rule can yield a specific rate – changing the Hamiltonian by an arbitrary constant will (despite not changing the system) change the calculated rate.  Can anyone explain?

According to my quick read on wikipedia, the Hamilton operator to be put in is the perturbation. As such, it is not arbitrary and you may not add a constant. Furthermore, and someone may correct me on this, are constants not only applied to the diagonal?

Good point, oops.  thanks for pointing that out.  What about the electronic coupling in Marcus Theory?

I'm fairly sure the states $\require{\begingroup} \begingroup \newcommand{\ket}[1]{|#1\rangle} \newcommand{\braket}[1]{\langle #1 \rangle} \ket{i}$ and $\ket{f}$ are eigenstates of the unperturbed Hamiltonian $\hat{H}_0$. Therefore, by virtue of the Hermiticity of $\hat{H}_0$, they are necessarily orthonormal
$$\braket{i|f} = \delta_{ij}$$
therefore if we define $\hat{V'} = \hat{V} + k$ where $\hat{V}$ is the perturbation Hamiltonian and  $k$ is some arbitrary constant
$$\begin{align}
\braket{i|\hat{V'}|f} &= \braket{i|\hat{V}|f} + \braket{i|k|f} \\
&= \braket{i|\hat{V}|f} + k\braket{i|f} \\
&= \braket{i|\hat{V}|f}
\end{align}$$
if $\ket{i}$ and $\ket{f} \endgroup$ are different states. If they are the same state, then you are calculating the transition probability from one state to itself - not a very useful concept.

Thanks for the reply.  If  |i⟩ and |f⟩ are both eigenstates of the system, though, then $⟨i|H|f⟩=⟨i|E|f⟩=0$, which isn't a case I'm interested in.  So what about when |i⟩ and |f⟩ aren't orthonormal?

@Felipe see TAR86's comment on your question; I forgot that $\hat{H}$ in your question refers to the perturbation. I'll remove my last sentence and update the notation. They have to be orthonormal because $\hat{H}_0$ is hermitian.

In the case where the states are not eigenstates, you would need to expand them in the basis of eigenstates of the perturbation operator.


text:
Does addition of HCl favour, or disfavour, formation of tetraamminecopper(II)?
Which way does the equilibrium
$$\ce{[Cu(H2O)4]^2+(aq) + 4 NH3(aq) <=> [Cu(NH3)4]^2+ (aq) + 4 H2O (l)}$$
shift if HCl is added to the system?
When I did this experiment, I found that the original solution was rather cloudy, but adding HCl made it clear. I suspect that it was moving towards the more soluble side of the reaction, but I can't find the solubilities of the copper compounds.

What type of reagent is HCl? What type of reagent would HCl react with? Is there anything in this system with that property?

HCl reacts with ammonia and removes it from the equilibrium above:
$$\ce{HCl + NH3 <=> NH4+ + Cl-}$$
therefore the equilibrium will shift to the left.

Can't HCl also do this with water?

Yes it could, but then you form $\ce{H3O+}$ which would react with NH3 anyway... If you are adding aqueous HCl, it is already fully dissociated, so in fact you are basically adding $\ce{H3O+}$ to your solution.


text:
Dielectric constant and dissolution of ionic salts
We say that salts like $\ce{NaCl}$ get easily dissolved in water because the dielectric constant of water is very high. I know the dielectric constant of water is 81 and if a salt is put into water the force between $\ce{Na+}$ and $\ce{Cl-}$ ions gets reduced by 1/81 times, after that the salt can easily be dissolved by simply stirring.
But how can we say that the medium between two very small particles which are bonded together is water? How can water molecules be present between two strongly bonded ions?

Have you heard about solvation process if not then go check it

I guess I've never seen it said that salts dissolve because of the dielectric constant of water (which really isn't particularly high). There is some confusion somewhere in your line of reasoning.

Related: Are Na and Cl ions from NaCl in water separate charged particles?

Do not use a too mechanical model.  I see the point of your concern. This happens all the time at such a scale. Is like asking how a reaction "knows" how to proceed towards lower energy before actually doing it.

The dissolving of a substance into a solvent and the avoidance of contact ion pairing depends on several things. The dissolving relates to the change in Gibbs free energy between the crystaline solid and the solution of the cations and the anions from a salt. A Born-Haber cycle can be used to explain this.
You should be aware of the fact that the Debye-Huckel and extended Debye-Huckel equations for the activity coefficent for ions in solutions have terms in them for the relative dielectric constant of the liquid.
The higher the activity coefficent then the more the ions are able to move freely of each other. The extended Debye-Huckel equation can be modified into the Bromley equation by adding another term for ion-ion interactions.
When considering the effect of changing the relative dielectric constant.
I think that one of the best examples is sodium bromide, in water sodium bromide is a strong electroyte where the cations and anions are well separated. But in acetone where the relative dielectric constant is lower the same salt is a "weak electroyte" where the cations and anions tend to associate with each other.
I think that the formation of contact ion pairs is the first step when going from a perfect solution of ions to a crystaline ionic solid. But the water does more than act as a simple dielectric. What the water molecules will do is to bind to the ions. In particular the water molecules will coordinate as Lewis bases to the cations and then enlarge the size of the cations. This binding of waters can act as a bridge between the cation and the water.


text:
Forming hydrogen sulfide from electrolysis of sodium thiosulfate
So I was doing electrolysis on two silver electrodes and sodium thiosulfate.
Some observations

The voltage drops extremely fast from 2 V to less than 1 V in 15 minutes, possibly because of the thiosulfate reduction.
Gases are liberated out of solution.
The solution turns black quickly.

Alert: I am using a fume hood.
Since I'm afraid of this spoiled eggs smell I just want to know if I am in any way creating hydrogen sulfide? Just want to be sure that the gas is tetrathionate and not hydrogen sulfide.
My other concern is with the voltage. How can I keep it from dropping so fast?

Yup, you're forming sulfide.
Render the thiosulfate ion as $(\ce {S-}-\ce {SO3-})$.  Then at the cathode:
$(\ce {S-}-\ce {SO3-})+2e^- \rightarrow \ce {S^{2-}}+\ce {SO3^{2-}}$
Both products are basic, with sulfide ion being strongly so,  and they will react further with protons; so the sulfide ion gives some hydrogen sulfide.  Sulfide ions can also combine with silver ions, from the anode, giving a black precipitate.  The other product, sulfite ion, can be reduced further, possibly forming still more sulfide ion/hydrogen sulfide.
So the reaction is not safe.  But kudos for using a fume hood.


text:
Why change in enthalpy is negative?
I noticed that the enthalpy of the products of an exothermic reaction (bonds formed are stronger than bonds broken) is lower than the enthalpy of the reactants. Enthalpy equals internal energy plus $PV$. In a constant pressure environment, change in enthalpy equals to change in internal energy as $PV$ in constant. Internal energy consists of $E_k$ (kinetic energy) of particles and $E_p$ (potential energy) of bonds between particles. When stronger bonds are formed, $E_p$ is bigger. And the energy of the heat released to environment equals to the energy gained in form of potential energy, so why is the internal energy lower?
It would be helpful if someone shows me my mistake :)

Frequently the convention is $KE$ and $PE$ or if you have a physics bent, $T$ and $U$, instead of $E_{k}$ or $E_{p}$, respectively.

These are simply conventions. For more grasp, consider it like this: energy goes from high to low, that is energy flow is downwards, which means a negative enthalpy change

Bonds actually have negative potential energy. Think about it this way: in order to break a bond you have to expend energy so when a bond is created, energy must therefore naturally be released. Hence internal energy is lower.

When stronger bonds are formed, *Ep* is bigger.

I take issue with this statement.  When stronger bonds are formed, the internal energy becomes more negative, not just "bigger".  It is this decrease in internal energy that makes the products' bonds so strong, since now the activation energy will be very large for a reaction to occur.
This link might help to explain: http://www.rsc.org/Education/Teachers/Resources/cfb/energy.htm
My favorite passage from this link is, looking at a one-step exergonic reaction: "The height of the peak from the reactants side is a measure of the strength of the bonds in the reactants and the height of the peak from the products is a measure of the strength of the bonds in the product. Thus we can consider the bonds in the product to be stronger than the bonds in the reactants."
If PV constant, this means Ek is the same.  Ep gets more negative.  So naturally enthalpy decreases (negative sign).

When energy goes off from a molecule, we take the sign as negative - this implies that the sign of PE which we are considering in formation of stronger bonds is MORE, albeit WITH NEGATIVE SIGN, hence, it is going down in energy coordinate.
The magnitude of change in PE, with formation of stronger bonds is more, this will lower the Internal Energy.

The change in enthalpy is negative in an exothermic reaction because energy is "lost" through the reaction (because there is more energy on the products side than on the reactants side).
Another way to think about this is by calculating the enthalpy before and after a reaction, for example - and this is a synthesis [and exothermic] reaction:
$$\ce{2H2 + O2  <=> 2H2O}$$
The ΔH [change in enthalpy] is -484 kJ (you can calculate this using standard enthalpy of formation numbers).
You can also think back to the definition of enthalpy - disorder: When two molecules that were 'freely' moving around, come together, there is less disorder [so the disorder in the system has decreased, and consequently is negative].


text:
How do you draw nitrogen in a newman projection?
I have to draw a Newman projection for a molecule, sighting along the bond between a carbon and nitrogen. Given that the nitrogen is behind the carbon from this viewing angle, how would I depict the nitrogen? I know that I cannot draw it as a circle, as that represents carbon, so I am assuming that it is not possible to draw a Newman projection of this molecule from this particular angle. Is that a valid assumption?
And, if that is true, what would be considered an acceptable answer to that question? Should I just draw a standard 3D structure from the proposed viewing angle?

The forward atom is represented as a dot. The atom farther back is the circle.

If it were up to me, I would:

Draw the nitrogen atom as if it were carbon but label it as "N".
Assuming normal valence  (e.g. an amine but not a nitro conpound), include the formally nonbonded electron pair as one of the substituents on nitrogen.

You can just draw the nitrogen from the centre of the circle using a lesser "wedge" angle on the circle to indicate an eclipsed bond.


text:
Can sodium bicarbonate be considered an acid?
I was reading about acids and bases today and finally decided to question the statement that "baking soda is a base."
Let's start with the dissolving of baking soda, $\ce{NaHCO3}$.  The equation to represent its dissolution in water is $\ce{NaHCO3 + H2O <=> Na+ + H3O+ + CO3^{-2}}$.
In this "reaction," baking soda is a proton donor, creating hydronium ions and a conjugate base, $\ce{CO3^{-2}}$.  Doesn't this make baking soda an acid?
If what I have written here is correct, then baking soda will create a base when dissolved, but baking soda itself is not the base.  Is this accurate?

It's "amphoteric", meaning it can act as an acid or a base. It's usually called a base because it acts as a weak base in aqueous solutions giving a slightly basic pH. But, if you add a strong base to a sodium bicarbonate solution, the bicarbonate will give up a proton (meaning it's acting as an acid) to the base.

@airhuff But water is not considered a strong base, right?  Then is my above equation still accurate?  Or do we tend to ignore dissociations when we are interested in acid-base reactions?  Dissociation is sort of a reaction, no?

Really it's: $\ce{NaHCO3 + H2O  <-->  H2CO3 + OH− + Na+}$. Bicarbonate is a stronger base than water.

And $\ce{H2CO3}$ (carbonic acid) decomposes to $\ce{CO2}$ and water. Note that tap water can have a pH of 8.5 or more, so you may not see CO2 bubbling out on addition of Na bicarbonate. Tap water can also be acidic, it just depends on where you live and your water system. (I don't know if you're trying this or if you were already well aware of that, just FYI.) Hopefully someone here has time to lay this all out clearly for you in a good answer ;)

And yes, this dissociation absolutely is a chemical reaction, of the acid-base type.

The statement you read that "baking soda is a base" comes from the fact that a solution of sodium bicarbonate (baking soda) and water has a pH of around $8.3$.
However, sodium bicarbonate is amphoteric with respect to Brønsted–Lowry acid/base theory, which means that it can act as either an acid or a base. More specifically, the bicarbonate ion is amphiprotic, meaning that it can either donate a proton, acting as an acid, or accept a proton, making it act as a base. The reason that a sodium bicarbonate solution is slightly basic is that it's ability to take a proton from a water molecule is greater than its ability to donate a proton to a water molecule. In other words, the chemical reaction causing the solution to be basic is:
$$\ce{HCO3- + H2O <--> H2CO3 + OH-}$$
The other equilibrium reaction at play, the reaction "trying" to make the system more acidic, is:
$$\ce{HCO3- + H2O <--> CO3^2- + H3O+}$$
Again, in an otherwise pure aqueous system, it is the first reaction that dominates, and thus giving the observed slightly basic solution.
Note that the $\ce{H2CO3}$ produced in the first reaction is also part of yet another equilibrium process:
$$\ce{H2CO3 <--> CO2 ^ + H2O}$$
Where the $\ce{CO2}$ is liberated as a gas. This is the cause of the well-known "fizzing" when baking soda is added to vinegar (acetic acid).

Sodium bicarbonate can indeed act as an acid.  You just need a strong enough base with which it can react.
Take 0.1 mol sodium hydroxide and add it to a 100 mL distilled water sample, then add 0.1 mol sodium bicarbonate.  The pH goes down in the second step as the solutes are combined to make the almost-but-not-quite strong base sodium carbonate.  Compare this reaction:
$\ce{NaHCO3 + NaOH -> Na2CO3 + H2O}$
with this one where we see the same proton transfer process:
$\ce{HCl + NaOH -> NaCl + H2O}$