text:
Problem regarding failure of equivalence concept

Chile salt peter, a source of $\ce{NaNO3}$ also contains $\ce{NaIO3}$. The $\ce{NaIO3}$ can be used as a source of iodine, produced in the following reactions.
\begin{align}
\ce{IO3- + 3 HSO3- &-> I- + 3 H+ + 3 SO4^2-} \tag{i}\\
\ce{5 I- + IO3- + 6 H+ &-> 3 I2 (s) + 3 H2O} \tag{ii}
\end{align}
One litre of chile salt peter solution containing $\pu{5.80 g}$ $\ce{NaIO3}$ is treated with stoichiometric quantity of $\ce{NaHSO3}$. Now an additional amount of same solution is added to the reaction mixture to bring about the second reaction. How many grams of $\ce{NaHSO3}$ are required in step (i) and what additional volume of chile salt peter must be added in step (ii) to bring in complete conversion of $\ce{I-}$ to $\ce{I2}$?

My working out using Mole concept:

My working out using equivalence:

Why does the equivalence concept not work?
Please tell me what is wrong and what is the correct method to use equivalence.

Just a note: capitalization matters in chemistry. Io – Jupiter's moon, whereas IO – iodine monooxide.

Sorry I find typing really hard hence the pictures

I did an OCR on the photo from the textbook and it would be nice if you actually proceed with typing your attempt using MD and MathJax as well. In the last sentence, you ask whether eq. of $\ce{NaIO3}$ = eq. of $\ce{NaIO3}$, which seems to be a typo.

I edited it and re wrote it tell me if it's better


text:
What is the process for chemical engineering?
I will be upfront -- I am not a Chemist, nor do I have much experience beyond high school Chemistry.
I have always had a vested interest in chemistry, but my engineers brain has always left me wondering -- Where do I start if I want to design a new chemical with certain properties? I have attempted to research via Google, but to no avail.
Like I mentioned above, I am very mechanically minded, and to design say a transmission is simple enough - just select what type of gear(s) (..or design your own) put them all together, and BAM! You've got a transmission! Chemicals...could possibly kill you if you screw up...
I have a strong feeling that this question is a field of study in and of itself, so I do appreciate any answers.
Thank you!

This depends heavily on what field of chemistry you are in. I am in organic chemistry and in particular total synthesis. So what I have to do is select the compound and then plan a synthesis route that usually has a lot of already known synthetic steps to it. Then you try if the planned route works and on the way change a few things if it doesn't. In our lab total syntheses usually takes two to three people more than three years, because our molecules are quite large polyketides.

Thank you for your comment. I am from stack overflow, and while I don't know if it is normally like this, the Chemistry Stack exchange seems quite quiet.

I haven't been here long and I know what you mean. I don't believe, there are too many active people here. I can mainly compare it to tex.SX. There are a lot of bad questions here, too. Loads of new accounts just copy-pasting their homework without any own input and people not selecting answers. Quite tedious if you ask me. However, this is off topic, so... yeah. ;)

You specifically mean a process in chemical engineering? There's software like Aspen which is helpful in this.

One thing to note is that "traditional" chemical engineering is synonymous with process engineering. In these roles, it's all about efficiently making a certain product at industrial scales and does not involve designing new chemicals. Engineering new chemicals is really the job of research scientists (both those with chemistry and chemical engineering backgrounds). How one goes about this depends on the field. For instance, one can do high-throughput computational screening of experimentally synthesized materials to find promising candidates for a given application to later test.

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text:
Why do some elements with high electronegativity exhibit metallic properties?
Metals typically have low electronegativity, which makes them form ions easily and prefer making metallic bonds to covalent. However, some of them seem to disprove that. Take, for example, gold, lead and phosphorus (for comparison). Their electronegativity values are 2.54, 2.33 and 2.19 respectively. Phosphorus is a typical non-metal, while gold and lead should have higher electron affinity and be even stronger non-metals that phosphorus. Why isn't that true?

Chemistry is about emergent properties from simple components. While we have general trends, it is a mistake to assume that the trends always hold or that outliers are easy to understand.

There are many "electronegativities" - they aren't properties like density but invented parameters which, as you should already see, are rather lame.

The Van Arkel–Ketelaar triangle is relevant to this question.  Not only does the absolute electronegativity of an element influence metallicity, but also the difference in electronegativity between the elements in a material. See the axis labels on this diagram. That said, electronegativity is a concept which extends rather poorly to transition metals under most definitions.

Whether an element adopts a metallic or nonmetallic structure does not depend on electronegativity alone.  Aside from electronegativity at least two factors promote formation of a nonmetallic structure:

If the atoms are close to filling their valence shells they are more likely to do so through covalent bonds.  Iodine needs only one more valence electron per atom and can get it by forming covalently bound diatomic molecules; gold needs seven valence electron per atoms to complete the valence shell and goes the metallic bonding route instead.

Elements in earlier periods tend to form stronger covalent bonds favoring the nonmetallic structure.  Silicon forms the familiar tetrahedral semiconducting structure but lead, below it with similar electronegativity and valence electron configuration, is metallic.

Other factors may enter, too, and Group 14 is an excellent place to look for subtle effects.  Above I said that silicon has the tetrahedral structure we all know and love, but that is under ambient conditions. The tetrahedral structure of silicon is energetically more stable than the metallic one, but the latter has more entropy and thus silicon may be converted to a metal upon heating.  In the case of silicon the temperature for this switchover is high enough so that the metal is molten, so we have simultaneous metallization with melting -- and, therefore, contraction with melting.  Germanium does the same thing, whereas with lead the metallic structure is more stable all the time.  Tin is a transitional case where it's a semiconductor when cold but switches to a metal below the melting point.
So there is indeed more to what makes a metallic element or a nonmetallic one than just electronegativity.

Thanks! BTW, is there a strict formula/algorithm which lets one determine the type of a chemical element or we need a table for that?

Probably a table is better.  As you see we are dealing with a complex phenomenon to describe theoretically.

@ВасилийСвинко Elements has no fixed properties. Allotropes only , see eg diamond vs graphite vs carbon nanotubes.

I have a question related to the answer above. If gold needs seven valence electron per atoms to complete the valence shell, why is the electronegativity of gold high? If it is easier for gold to just lose one electron than to gain seven, isn't it right to have a lower electronegativity?

Losing just one electron does not give a highly stable configuration.  There really is much more to electronegativity than electron count.

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How to balance this equation?
Here is my problem
$$\ce{2X + 16H^{+} + 10Cl^{-} \rightarrow 2Mn^{2+} + 8H_{2}O + 5Cl2}$$
I want to find the $X$ in this reaction.
According to me
$$Mn_{2}O_{3}$$
Also I see that the values of $Cl_2$ and $H$ are equal. There's $8$ Oxygen in subtitles as well $(8H_{2}O)$.
Regards!

Mn2O3 does not balance Mn since there are 2 Mn2O3. Also that has 6 O on left and 8 on the right. That doesn't seem quite right....

The amount of oxygen does not add up with that solution!

I'm voting to close this question as off-topic because questions asking ‘am I right’ with respect to a specific exercise are a bad fit for the site.

If these numbers are correct and there is just one compound X missing, it is quite simple:
You have 2 manganese on the right, so you need 2 on the left. 8 oxygen on the right, so 8 left. Because stoichiometry is 2, divide these numbers by 2.
This gives you $\ce{1Mn}$ and $\ce{4O}$, forming $\ce{MnO4}$.
However, we are still missing the charge!
If you look carefully, you see that on the right you have $2~\cdot~+2=+4$. On the left, you have $+16 -10=+6$. That is $+2$ too much, so you have to add 2 negative charges. Divide by stoichiometry of your compound $\ce{X}$ and add to what you have already.
So your compound $\ce{X}$ must be $\ce{MnO4-}$. Your equation is:
$$\ce{2MnO4- +16H+ + 10Cl- -> 2Mn^{2+} + 8H2O + 5Cl2}$$
This, however, is usually not the way to solve this kind of question.


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Synthesis Golf VI: Muscarine
October's synthesis golf is rather late - apologies. I've chosen a slightly different-looking target this round, with the aim of inspiring a wider variety of routes.
Muscarine is a natural product found in some mushrooms. Most notably, it is a good agonist of one class of acetylcholine receptors - these receptors are therefore called muscarinic acetylcholine receptors.

The only conditions for this round are that:

the synthesis must be enantioselective; and
no more than one chiral centre may be purchased in the form of a building block. (Chiral catalysts, ligands, auxiliaries, etc. do not fall under this rule.)

Otherwise, any commercially available starting material (as usual, in the Sigma–Aldrich catalogue) is fair game! There is also no restriction on the counterion.

1. Retrosynthesis
This proposed route relies on being able to make the tetrahydrofuran via some type of ring closing etherification.
Through disconnection of the C-O bond, a linear diol is revealed which may be accessed in several ways.
Initially, I'd thought about doing a Sharpless dihydroxylation of  (Z)-hex-1,4-diene, however the starting material is surprisingly expensive and there are few literature reports of doing this dihydroxylation (and in those instances where it has been carried out, the e.e. and regioselectivity has been moderate to poor).
A different disconnection, using the chiral pool (less elegant, perhaps, but the starting material is cheap and ensures good selectivity) allowsthe diol can be disconnected across the C-C bond, revealing an aldehyde which is derived from a lactate ester.

2. Forward synthesis

Reagents and conditions: (a) TIPSOTF, 2,6-lutidine (b) DIBAL; allyl-MgBr (c) BnTCA, PPTS (d) Shi epoxidation (e) HF-Py, Py (f) TsCl, Py (g) NaN3 (h) H2, Pd/C (i) MeI
The synthesis starts with protection of methyl lactate with a TIPS group - a silyl group is chosen here as upon deprotection under fluoride conditions later in the synthesis, the THF should spontaneously close.
The TIPS protected methyl lactate may then be treated with DIBAL to afford the lactaldehyde. This must be used immediately to prevent epimerisation. In all likelihood, this could be achieved by adding a solution of allyl magnesium bromide to the same pot as the DIBAL reduction but failing this, there is good precedence for a reagent controlled boron mediated allylation (Tet. Lett. 2003, 44, 1737).
The terminal olefin installed by allylation is now setup to do a Shi epoxidation (Sharpless is terrible at terminal olefins) providing the final stereocentre. Treatment of the substrate with HF-Py should then deprotect the TIPS ether with concomitant formation of the THF ring system.
Finally, functional group transformations provide the desired natural product as shown in the scheme. Some issues may be observed when attempting to methylate the amine in the presence of the alcohol however from experience of similar substrates, the amine is likely to quaternise faster than the alcohol methylates and as such careful control of stoichiometry and conditions should allow the desired product to be isolated as the major component.
Summary
A synthesis of muscarine is proposed in 9 steps (linear) starting from methyl lactate, with the remaining stereocentres being installed using a substrate controlled allylation and a reagent controlled asymmetric epoxidation.
Addition steps could likely be cut out by (1) not protecting the alcohol that is currently protected with a benzyl group and (2) making use of an asymmetric aziridination in place of the Shi epoxidation.

Well done. The only problem I see is the possible epoxide opening with HF.

In all likelihood thats not an issue- buffered HF isn't really acidic. If it was a problem, a switch to TBAF would sort things out though.

Yeah, I would rather go with TBAF.

I would suggest protecting the alcohol as a functionalised benzyl, possibly p-MeO-benzyl as they are easier to remove in practice

@Waylander - I wanted the OBn so it could be cleaved as the azide was reduced under hydrogenation conditions (saves a step). Honestly I'm not even sure it needs protecting but thought I'd air on the side of caution. I also had concerns (though perhaps unfounded) that the ring wouldn't enjoy the oxidative conditions used to take a PMB off (aromatisation?).

The p-MeO-Bn will come off under hydrogenation - more easily than the OBn, which was my point. I've often struggled to get Benzyl ethers to completely deprotect under hydrogenation

My bad, I'd never thought of hydrogenating an O-PMB – always DDQ in the kind of molecules I work with. Useful tip!

Allylation of ethyl acetate should provide the unsaturated ester. Sharpless asymmetric dihydroxylation should proceed with the formation of the lactone. The free alcohol was then tosylated, substituted with azide and the azide was reduced by Straudinger procedure and the obtained amine was protected with Bn group. Davis hydroxylation should be reagent controlled to give the hydroxyl group with the correct stereochemistry. Tebbe olefination, was followed by OH-directed hydrogenation and Bn deprotection. Selective methylation of the amino group should provide the target molecule.

I'm wondering if some type of stereoselective azide thermolysis (of benzyl azide) could be used to conserve some steps instead of using Sharpless dihydroxylation.

Use malonic ester in first step. Better for scale up. Even better, unsat. ester from allyl alcohol and ethyl orthoacetate via Johnson-Claisen rearrangement. The first lactone also available from diazotization of L-glutamic acid (retention) followed acid-catalyzed lactonization and diborane on the carboxyl group. A lot less expensive.

ethylene glycol, pTsOH, reflux
mCPBA
(S,S)-salen Co (Jacobsen kinetic resolution)
$\ce{NaN3}$. Fortunately, this epoxide will open the right way since it's not a styrene...
toluene, pTsOH, water, Dean-Stark reflux. I'm trying to do this one-pot. The conditions for hydrolysis of the acetal are so close to the conditions for dehydation to form the dihydrofuran after intramolecular ring closure. We might need to play with solvent to get enough water in there for hydrolysis.
mCPBA. I'm hoping to get some diastereoselectivity here.
MeMgBr. We will likely need to optimize reaction conditions here. I think low temperature plus the possible addition of a Lewis acid will help selectivity.
$\ce{PPh3}$. Staudinger reaction
2.9 equivalents of methyl triflate. Slow addition at low temperature.

Nice one. I really liked the regio and stereoselectivity control of reaction 7.  I only think that the chemoselectivity could be a problem as azides could potentially react with Grignard on the terminal nitrogen.

@marko The note on reaction 4 is because I've been burned twice by that reaction on a styrene like substrate, where the azide preferentially adds at the benzylic position... :)


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Work done by adiabatic expansion of gas
The formula for the work done by adiabatic expansion of gas is
$$W = C_V\,\mathrm dT.$$
Why the volumetric heat capacity $C_V$ is constant and not the heat capacity at constant pressure $C_p?$ If the system is at constant volume, it is a isovolumetric process and hence no work can be done in that case.

Probably better to write $dW  = C_V dT$, since it is a differential amount of work.

Pretty simple, $\Delta U = nC_V\Delta T$ no matter what the process is. This is called a state function, i.e. it is independent of the path taken and only depends on the final and initial conditions here initial and final temperature.
However, work done is path dependent, so it depends on how you change pressure and volume. Let's say you take a gas from $p_1,$ $V_1,$ $T_1$ to $p_2,$ $V_2,$ $T_2.$ You can say $\Delta U = nC_V(T_2 - T_1),$ however we need to know the process (isothermal, adoabetic, polytropic) for finding work done.

One caveat... This only applies to an ideal gas.  In a real gas, the internal energy is also a function of volume.

Part 1. Let's consider first law of thermodynamics.
$\delta Q = dU + \delta W$ (classical sign convention)
or
$\delta Q = dU - \delta W$ (IUPAC sign convention)
In the case of adiabatic process, $\delta Q = 0$.
$0 = dU - \delta W$
$dU = \delta W$
$dU = C_VdT$ (by definition)
$C_VdT = \delta W$
After integration, assuming CV is constant.
$C_V\Delta T =  W$
Part 2. Why not Cp?
By definition, $\delta Q_p = dH = C_pdT$. Obviously, we don't have an isobaric process here.
P. S. Yes, those sign convention differences also make me feel bad.

The total differential of the internal energy (at constant composition and allowing only pV work) is usually written as
$$dU = -pdV + TdS$$
However the differential $dU$ can be written in terms of other variables, for instance p and T, in which case:
$$\begin{align} dU &= \left(\frac{\partial U}{\partial V} \right)_T dV + \left(\frac{\partial U}{\partial T} \right)_V dT \\ &= \left[-p + T \left( \frac{\partial S}{\partial V} \right)_T \right] dV + C_V dT \\ &= \left[-p + T \left( \frac{\partial P}{\partial T} \right)_V \right] dV + C_V dT \end{align}$$
The last equality follows from one of Maxwell's relations.
This expression btw is known as the thermodynamic equation of state.
For an ideal gas it follows that
$$dU  = C_V dT $$
Note that the change in energy is independent of changes in pressure and volume, it depends only on changes in T.
If the heat capacity is constant over the temperature change of interest then you may write
$$ \Delta U  = C_V \Delta T $$

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Why are group 1 elements so low in density?
I was studying the s-block elements and found that they extremely low in density. Lithium is said to be the least dense solid in the entire periodic table and their Cohesive Energies are also very low. Could anyone please explain in detail the reason for this?

Because it is energetically easier for an atom (other than hydrogen) with only one outer electron to lose that electron, than to gain seven more, in order to have an inert gas configuration, that electron has a high probability of wandering off. In bulk samples, that causes metallic conduction. Having that lone, loosely-attached electron increases the effective atomic radius.
Hydrogen is an interesting special case. At the ionization potential is more than twice that of lithium, for example, and $\ce{H2}$ has a covalent, rather than metallic, bond. The gas is diatomic, with very little attraction between molecules, so its density is low. However, at sufficient pressure, it is believed that hydrogen would behave as a metal.


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Find increase in OH- ion concentration after base is added

The pH of a solution is 7.00. To this solution sufficient base is added to increase the pH to 12.0. the increase $\ce{OH-}$ ion concentration is:
(A) $5$ times
  (B) $1000$ times
  (C) $10^5$ times
  (D) $4$ times

The answer is (C) $10^5$ times.
Now, the temperature hasn't been mentioned ergo we can't assume that it's a neutral soultion, right? So, I tried doing it this way:
\begin{array}{llll}
&\mathrm{pH} &= 7 &\qquad &\mathrm{pH} &= 12 \\
&[\ce{H+}] &= 10^{-7} &\qquad &[\ce{H+}] &= 10^{-12} \\
& & &[\ce{OH-}]~\text{added} \\
& & &= 10^{-7} - 10^{-12} \\
& & &= 10^{-7} (1 - 10^{-5}) \\
& & &\approx 10^{-7}
\end{array}
$[\ce{OH-}]$ added is the increase in its concentration, or so I think. I don't know what I'm doing wrong. Please, help.
Also, can you please point out the loopholes I might have in my concept if it's obvious from my doubt, so I can look up and take care of it?

The problem is that you are trying to subtract the values, whereas all you need to do is find an increase in ratio $\frac{[\ce{OH-}]_2}{[\ce{OH-}]_1}$ knowing that $\mathrm{pOH} = - \log{[\ce{OH-}]}$ and $\mathrm{pOH} = \mathrm{p}K_\mathrm{w} - \mathrm{pH}$:
$$\frac{[\ce{OH-}]_2}{[\ce{OH-}]_1} = \frac{10^{-\mathrm{pOH_2}}}{10^{-\mathrm{pOH_1}}} = 10^{\mathrm{pOH_1} - \mathrm{pOH_2}} = 10^{\mathrm{p}K_\mathrm{w} - \mathrm{pH_1} - \mathrm{p}K_\mathrm{w} + \mathrm{pH_2}} = 10^{\mathrm{pH_2} - \mathrm{pH_1}} = 10^{12 - 7} = 10^5$$

pOH=14−pH is true only for 25°C. So,unless it's specified otherwise, are we to assume that the dissociation is taking place at 25C?

@JamilAhmed Yep, you are right, but considering the problem is not mentioning temperature at all, I assumed it's safe not to overthink and go for the easiest solution:)

how did you edit my question so perfectly?(if you get what i mean)

@JamilAhmed It's just basic MarkDown and MathJax, if you feel like you want to know a bit more, I'd suggest to read MathJax basic tutorial and quick reference and How can I format math/chemistry expressions here?.

@JamilAhmed You only need to know that $\mathrm{pOH_1} + \mathrm{pH_1} = \mathrm{pOH_2} + \mathrm{pH_2}$, i.e. that the constant (the logarithm of the self-ionization constant) that is 14 at 25°C is the same before and after adding $\ce{OH-}$.

@wythagoras Good point, I changed $\mathrm{pOH} = 14 - \mathrm{pH}$ to $\mathrm{pOH} = \mathrm{p}K_\mathrm{w} - \mathrm{pH}$ in order to address the doubt about the temperature. Thank you!


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What is the difference in the effect of DTT (Dithiotreitol) and SDS (sodium dodecyl sulfate) for PAGE?
In class, we have studied that when running a polyacrylamide gel with our protein sample to separate the proteins by molecular size, we can use several techniques, among which we have denaturing PAGE and SDS-PAGE. Basically, we focused on the latter, in which SDS is a detergent added to negatively charge all proteins in the sample (at a constant mass:charge ratio), as well as denature them. Then, the professor mentioned that we also add reagents such as DTT or beta-mercaptoethanol to reduce the disulphide bonds in between proteins.
My question is: if SDS already denatures proteins (so we have primary structure only), why would we be interested in adding DTT?

Proteins are thought of as long polyamide chains. This long chain of amino acids is known as the primary structure. Due to hydrogen bonding of the backbone amide groups, a protein chain will organise most of itself locally either as an α-helix or a β-sheet; which one depends on the angles to the α carbon. These secondary structure elements then cluster to form a tertiary structure in which hydrophobic residues typically cluster together, aromatic residues may form π stacks, positively and negatively charged residues come together to build up ionic interactions, cysteins come together in disulphide bridges and hydrogen bonds are also formed where applicable. This last step is repeated if a protein consists of multiple subunits to form the multi-protein complex’s quarternary structure.
Almost all of the structure elements I mentioned that explain tertiary structures are intermolecular, meaning that they do not constitute a chemical bond between the connected fragments. Therefore, these structure elements are easily broken by the action of SDS whose aliphatic chain inserts into these tertiary structures, interacts with them and practically captures them on its inside.
The only exception to the former paragraph are disulphide bridges between cysteins. These are actually a chemical bond conneting the two fragments. In a simplified reaction scheme, the synthesis of a disulphide bridge looks like this:
$$\ce{\underset{\text{cystein}}{2 R-S-H} ->[][- 2e- \ -2 H+] \underset {\text{disulphide bridge}}{\ce{R-S-S-R}}}$$
Chemically, it is an oxidation of two cysteins.
Therefore, these disulphide bridges cannot be broken by the action of SDS alone. It takes a reductive agent such as DTT to resupply the electrons to reform the two cysteins. If no reducing agent is present (but there are disulphides in the protein structure), the disulphide bridge will remain intact and the protein will move not through the PA gel as it should according to its size since it is cyclic.

Thank you very much. I now understand the logic behind what was stated in class. From your explanation, if we had a protein with several subunits that were brought together through weak interactions such as van Der Waals forces or hydrogen bonds, would they show up as separate bands under the sole action of SDS (assuming no disulphide bridges were present)?

@Bee If the subunits are indeed only held together by intermolecular forces, yes.


text:
Adiabatic mixing of liquids

One mole of benzene at $\pu{20^\circ C}$ is adiabatically mixed with 2 moles of toluene at $\pu{40^\circ C}$. The heat capacity of benzene is $\pu{84 J K-1 mol-1}$ and the heat capacity of toluene is $\pu{96 J K-1 mol-1}$. If benzene and toluene form an ideal solution, compute the entropy change for the process.

I tried to use
$$\Delta S = n C \ln{(T_\mathrm f/T_\mathrm i)}$$
where $C$ is the heat capacity and $T_\mathrm f$ and $T_\mathrm i$ are the final and initial temperatures of the liquids, respectively. However, I am unsure of what the final equilibrium temperature of the mixture comes to.

In addition to the temperatures of the liquids equilibrating (which results in an entropy change), you also need to include the entropy change due to mixing.

So what do you get for the entropy of mixing, and how does that compare with the entropy change due to heat transfer?

Let's say the final temperature is $T$. Also we assume the system is isolated and the process is adiabatic, as you said. Also all energy that is released by toluene or gained by benzene is heat.
The absolute amount of heat then released by toluene equals to the absolute amount of heat gained by benzene:
$|Q_\mathrm t| = |Q_\mathrm b|$
Considering signs, it will look like this:
$Q_\mathrm t = -Q_\mathrm b$
By definition, $Q = nc\Delta T$. Thus,
$Q_\mathrm t = 2\cdot96\cdot(40-T)$, and
$Q_\mathrm b = 1\cdot84\cdot(20-T)$
Plug in the latter two equations into the second one, and here's what you get:
$2\cdot96\cdot(40-T) = -1\cdot84\cdot(20-T)$
Now just solve it for $T$. And plug that in your equation as $T_\mathrm f$.
Note that it's not really important to do the temperature in kelvin, since we are dealing with changes in temperature. And yeap, your $T$ should be between $20\ \mathrm{^\circ C}$ and $40\ \mathrm{^\circ C}$, because there are such things as thermodynamics laws. So if it's not, you know for sure you did something not really correct.

This does not provide the complete solution to the problem.  The entropy of mixing also needs to be considered.

@ChesterMiller It's homework so not a problem usually.

The entropy of mixing still needs to be included.

@ChesterMiller how might one calculate the entropy of mixing, then?

@JonathanSmith well that's easily googleable https://en.wikipedia.org/wiki/Entropy_of_mixing


text:
Can a chemical reaction cause a cigarette to taste bad?
I'm currently researching an idea where I'd like to find out if there is any chemical we can put in toothpaste that would react badly with tobacco smoke?
The idea is that smokers can just brush their teeth as anyone would normally do, but when they light up a cigarette it would taste awful. So the actual question is, of all the stuff that's in tobacco, is there anything we can make react that is

safe to put in toothpaste.
reacts solely by giving a bad taste.
doesn't let that bad taste last for longer than 10 minutes.

I just want to know if this is technically possible since I don't have a background in chemistry. Thanks for your help!

I'm afraid this is not going to work, even if the combination of tobacco smoke and toothpaste would produce an awful taste. Cigarettes themselves already taste nasty, but it is tolerated by the consumer's body because of nicotine dose is delivered and organism makes it feel like it was worth it. The same will happen with the toothpaste. The only possibility that still works in an absence of willpower is the nicotine patches which are gradually decreasing nicotine addiction. You need to tackle the root of the problem and not the symptoms.

Thanks for answering! The idea came to us since most stop-smoking products that are out there actually taste nasty themselves or require an extra effort. And although to a non-smoker a cigarette is nasty either way, to a smoker actually enjoys the taste (I myself am a smoker) The principle itself is actually based on classic conditioning, a method that has proven it's worth, since the entire western civilization is based on it. (e.c.:If you steal, you go to jail.) The question I am trying to find an answer to is whether it is or is not chemically possible.

You can find a list of the components of tobacco smoke on wikipedia (https://en.wikipedia.org/wiki/Tobacco_smoke). They are all several ng only. I would guess that tobacco smoke could lead to an oxidizing environment, so maybe that would be a starting point (e.g. silver spoon + egg -> bad taste). But I would fear that other food could contain similar ingredients,so that it wouldn't be really specific. You would probably want something to react on a specific ingredient of tobacco smoke, but in ng range it would be difficult.

I can sympathize with your lack of a "background in chemistry", but the question is really kinda broad. Good question, but that doesn't change the fact that it is too broad for this forum and hence should be closed as such. However, if you take user @deus 's cue and (try) zeroing down on a particular compound which can be "studied" by other users, it would greatly improve the question. Intriguing idea by the way; have an upvote! +1 ;)

The problem with conditioning is that it works both ways. The smoker will simply condition themselves away from the bad-tasting toothpaste. (Also, any toothpaste effect should wear of by lunchtime at the latest.)

There exist a commercial product sold in form of liquid. A drop of it on the tongue makes smoking almost impossible. You should be able to find it in the web as for it exist for sure. Saw a demonstration at Manchester airport. My friend tried. I, after him, did not because I wanted to smoke one cigarette before entering the non smoking zones....!


text:
How does water evaporate completely when at room temperature?
If water evaporates at room temperature because a small percentage of the molecules have enough energy to escape into the air, then why does a kitchen counter with a small amount of water eventually evaporate completely when at room temperature?

In the system (glass of water etc.) the water molecules have a distribution of different kinetic energies. Some of the molecules with a higher kinetic energy are therefore able to break free from the bulk and evaporate.

This is because of diffusion. It not "evaporates", it "diffuses" :-)

It doesn't.  There's still a microscopic amount of water present.

@Bdrs: your explanation doesn't describe how all the water can eventually evaporate, by your explanation all of the molecules have enough energy since the beginning.

While I understand the underlying question(Why does water evaporate at room temperature?), I'm really confused by the wording of this question, to me it essentially equates to "If water evaporates at room temperature, then why does more water evaporate at room temperature, too?"

As your small percentage of molecules with high enough kinetic energy evaporates, the remaining liquid water cools down. But in doing so, it drains heat from its surroundings and thus stays at room temperature (or close to it), so there is still some fraction of molecules that can evaporate, and they do so, and more heat is transferred from the surroundings, and so it continues until all water is gone.

And, in particular, a small amount of water evaporating won't lower the temperature or raise the humidity of the room in any measurable way.

Thanks for the answer ! Still got a question, because it seems that everything relie on The system had energy at start. How the layer of molecules get this starter energy from ? Or is it unrealistic to imagine not this case ?
Edit Might be the answer ?

Everything has thermal energy, unless it is cooled down to absolute zero. If you do that with your water (and prevent thermal exchange), then of course it won't evaporate.

This happens because the rate of evaporation is higher than the rate of condensation.
$$\ce{ H2O (l) <=>> H2O (g) }$$
This is also due to the fact that you have an open system:  matter and energy can be exchanged with its surroundings. The evaporated water can evaporate from the glass and condense somewhere else.

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

Then when is the rate of evaporation equal to the rate of condensation?

This depends on a number of factors (see https://van.physics.illinois.edu/qa/listing.php?id=1440&t=water-evaporation-rate), e.g. the temperature of the water surface, the temperature of the air, the humidity, and of course the area of the surface. Also, you have to take into account the airflow past the surface as well as the water currents convecting heat away from the surface.

Also a good read: https://www.e-education.psu.edu/meteo3/l4_p4.html

The water on the surface does not exist in isolation it is in contact with the air and with the surface. Random higher energy molecules in the surface and in the air will add energy by collision to the water molecules leading some of them to escape the liquid (evaporate).
This is why evaporating water leads to cooling of the air and surfaces around it.

This answer almost hits the point. It is not a "percentage of the total water" which escaped the liquid state, but a percentage of the water molecules close enough to the surface. As the surface retains a constant size (assuming a standard-issue cylindrical pot), we observe a more constant rate of evaporation (in the "percentage of total water" case we would observe some sort of half-life - an evaporation rate going down to zero asymptotically as the amount of water decreases).

Let's say, $q\in {]}0,100{]}$ is the minimal net percentage of the volume (or of the mass) that evaporates at any second $t$ (for every $t>0$). Saying "net", we assume that more water molecules leave the kitchen counter than return, and that the fraction of the molecules leaving the surface in relation to the number of molecules getting back has a constant, positive lower bound. (Other answers explain why it is likely to be so in kitchen conditions.)
Then, at most $100-q$ percent are left per time unit. So, after $t$ time units, the amount of water left will be at most $\mathrm{a_0}\Bigl(\frac{100-q}{100}\Bigr)^t$, where $a_0$ is the initial amount. Since $100-q<100$, we obtain  $$\lim_{t\to+\infty}\,\mathrm{a_0}\Bigl(\frac{100-q}{100}\Bigr)^t \ = \ 0\,.$$ In particular, after a certain point of time, the amount of water will be lower than the minimal possible amount (the volume of one $\mathrm{H}_2\mathrm{O}$ molecule or its mass, simplified, of course).
If the assumption made is not valid (say, due to great humidity somewhere in Asia), the result would be wrong: the water will NOT fully evaporate.
(An aside has to be made. Note that the above mathematical treatment is a gross simplification. To get a more realistic evaporation model, we should take into account that the evaporation happens only from the surface, and not from the whole volume, and that both the surface and the volume change with time. Also, bear in mind that even within one second, the evaporation rate changes.)

I think this ignores the physics aspect of the question. The question seems to me to be asking, "Wehn water evaporates, it's because the high-energy molecules leave the body of the liquid. OK, but once they've all gone, why doesn't evaporation stop?"

@DavidRicherby The question doesn't say it that way. What you wrote is your purely subjective interpretation of the question.

The question talks about the small percentage of molecules that have enough energy to escape and asks how water is able to evaporate completely. Having said that they understand how the energetic molecules are able to escape, it's certainly a natural interpretation that the question is asking how the unenergetic ones can escape, too.

@DavidRicherby It seems "natural" only subjectively, to you, but not necessarily to the others. If you wish to go with your own interpretation, post your own answer.

I don't need to post an answer because it's already been posted and received a lot of upvotes.

@DavidRicherby If it has been said, your comments are obsolete. Notice that I do refer to the other answers in my post.

Water always evaporates when above 0 degrees Celsius at normal atmospheric pressure.
Which means when above 0 deg there always are molecules with high enough energies to leave the liquid phase.
See this Quora answer (note the Quora answer points out evaporation at 0 degrees C is possible with the right pressure), this Physics.SE answer

Just a comment on temperature and evaporation: there is nothing magical about the temperature of 0C when we're talking about the evaporation of water. So long as the relative humidity is less than 100%, water at any temperature, above or below 0C, will evaporate.


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Aza-enolates: why no reaction at N
My understanding is that aza-enolates have an advantage over enamines in enolate-type alkylations because the reaction at N is less likely. As a result, unactivated electrophiles can be used (i.e. electrophiles that are not allylic / benzylic / $\alpha$-halo, source: Clayden et al., p. 593)
Question: why is the lone pair on N in aza-enolates less reactive?

From how I understand it, the problem with enamines + unactivated electrophiles is self-condensation. This naturally prevented in the N-enolate.

@deus ex machina So would you say that the raction at N isn't a problem at all? Why?

Because the n-enolate prevents an  electrophilic site

@deusexmachina No I mean, the enolate / enamine acting as a nucleophile at N. Because although the carbon will be the soft nucleophilic centre, the electron density on N will be greater, so why won't "harder" electrophiles react with it?

@deusexmachina From Clayden, Ed. 2, p. 592: "There is, however, a major problem with enamines: reaction at nitrogen. Less reactive alkylating agents -- simple alkyl halides such as methyl iodide, for example -- react to a significant degree at N rather at C. The product is a quaternary ammonium salt, which hydrolyses back to the starting material and leads to low yields".

Less electronegativity on N = more delocalization into C


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Are compounds containing active methylene groups positive for the haloform test?

Are compounds containing active methylene groups (e.g. 1,3-diketones) positive for the haloform test? Support your answer with a suitable explanation.

I was stuck on a certain question wherein a diketone was formed and I had to tell whether or not it shows the haloform test. It is clearly mentioned in my class notes that active methylene compounds don't show the haloform test; but in this problem the solution says that the diketone shows positive haloform just because it has an active methylene group. Please help!
Here's a possible mechanism that I came up with. Don't know if its correct. Sorry that it's handwritten; I don't know how to type chemistry.

At least the wikipedia article (https://en.wikipedia.org/wiki/Haloform_reaction) states that diketones are suitable substrates for the haloform reaction. Therefore I  see no reason why the haloform test shouldn't work. In fact I would suspect it to work even better due to the higher acidity in the alpha position.

@deusexmachina, as you say, the iodination at the acidic position is likely to proceed fine. The issue I see is that the positive result of the iodoform test (at least) is given by the formation of $\ce{CHI3}$, both the colour and the odour of which can be identified. With 1,3-dicarbonyls, the final step results in the expulsion of a diiodoenolate, as opposed to $\ce{CI3-}$, which is required for the positive result. I am busy tonight, so anybody is welcome to take this and write up an answer.

Please use the built-in image uploader; images included that way are not subject to deletion on imgur.

Hmm, seems I didn’t think far ahead enough in my last comment. Your mechanism has some points for nitpicking, but I agree that the dihaloenolate expelled should give a positive result.

@orthocresol please point them out. I think the addition of $Br_2$ is ambiguous, but it gives me the product I want!

For an update on the iodoform reaction on some 1,3-diones, see: https://chemistry.stackexchange.com/questions/135626/do-cyclohexane-1-3-diones-give-positive-haloform-reactions

As orthocresol already commented, apart from some nitpicks, the general direction is correct.
The only issue really is the addition of bromine, which does not occur via a [2+2]-cyclo-addition (2nd from below in scheme).
Ordinarily you would expect addition the addition of bromine to a π-bond to occur first via a polarisation complex and then via a bromonium ion (bottom of scheme).[1] In the case of the enolate, the π-bond is already readily polarised, and sufficiently asymmetric that a direct addition will occur.[2] The following depicts the HOMO -1 calculated at the DF-BP86/def2-SVP level of theory, the HOMO is an (delocalised) in-plane lone pair of the carbonyl-oxygen atoms in the C2v conformation (1st row). In the Cs conformation the π-bond is the HOMO, it is also the low-energy conformer (2nd row). (On mobile the rows might differ.)

Please note that the following scheme is just a proposed mechanism, and as such probably a reasonable simplification; the actual one might depend on more factors like solvent, temperature, pressure, etc., and might have more, or even less steps, other intermediates (ion pairs), hydrogen abstraction might be via water/base bridges, etc. For educational purposes I think this is sufficient though.

See for example Aromaticity-Bromonium Ion, Mechanism for bromoetherification of olefinic alcohol, Why does cyclopropane react with bromine?, What is trans-addition? and more...
Thanks to orthocresol pointing it out in the comments.

Enolate reaction with bromine (as opposed to ordinary alkene) is traditionally taught to not go via a bromonium ion (img), presumably because the electronic character of the two doubly bonded carbons is very different. But I have not consulted any literature, this is just based off undergrad teaching + textbooks, so...

Since I was unlucky to find any specific literature on the topic (I did not search very thoroughly), I have tried optimising a bromonium analogue at the DF-BP86/def2-SVP level of theory. I have been unable to obtain a stationary point (no local minimum and no transition state). Which is not unexpected given the HOMO/ HOMO -1 is strongly polarised.


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Experimental Difference b/w Graphite Oxide & Graphene Oxide
I'm looking forward to making a graphene supercapacitor using LightScribe method for 12th Grade project. I need Graphite Oxide, to start working although all I can find on the web is Graphene Oxide.
I've read What is the difference between graphite oxide and graphene oxide?, but I don't seem to understand whether I can use it in my application or not.
The paper I'm referring to regarding the project is: Making Graphene-Based Supercapacitors by LightScribe, Theint-Aung, Cornell College, Department of Physics, Mount Vernon, IA
Questions

Q1. Can I use graphene oxide instead of graphite oxide to accomplish the end goal of producing graphene supercapacitors (production method could be different)?
Q2. Can I convert graphene oxide into graphite oxide, using some reaction?
Note: I have access to an "alright" quality chemistry lab at school (not US), so processes involving ultrasonic baths are out of question, I have access to only cheap equipment.
P.S.: If possible, could you also provide a link where I could learn a bit about the chemistry behind the reactions you are talking about in your answer?

You should edit your title to increase your question's potential!

@Mockingbird Is that better? Titles are one of the things I’m not strong at...


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Do non-ergodic chemical reactions exist?
Do reactions whose final outcome is not even theoretically predictable by any means possible exist in Nature or does the outcome of all possible chemical reactions can (at least in theory) be predicted if we could get enough information about the conditions the reaction system is in?
My question isn't about a well-defined reaction system whose behavior is unknown to a certain degree but about the possibility in principle to discover a chemical reaction or a system of chemical reactions able to produce compounds which can't be predicted even in theory, even if we manage to acquire all the possible information about the system that can be acquired by all experimental means possible? What I want to ask is whether we can, at least in theory, know all possible products a reaction system can generate or do we "hit a wall" limiting our ability to predict certain chemical reactions over a certain limit?
I know very well the difference between a stochastic and a deterministic process, but what I'm asking isn't about stochasticity. It is about the possible predictability of all chemical reactions that could exist. As far as I know if we use the concept of stochasticity we may be able to derive possible, albeit only probable predictions about the state the system can be in. Therefore we can at least compute some probabilities different compounds can have to emerge under certain conditions. What I'm asking about is the validity of the concept of ergodicity, therefore I want to know is it possible to know the end state and all the reaction products of all possible chemical reactions, or there are some which could generate products that are impossible to predict in both theory and reality? Do reactions generating such unpredictable behavior exist in the real world or are they possible only in theory?

What does ergodicity have to do with predictability?  I'm not clear on the distinction between predictability or stochasticity you are drawing either.  Nor is the precise meaning of "predictability" as you use it.

@CurtF., it means even if we can only predict the probable outcomes of a reaction it still is a prediction. In the context of the question it means are there reactions whose outcomes we can't predict even to a some probability. Are there reactions we can say nothing about even if we get to know all the conditions under which the reaction is occurring? Can there be reactions we can't know the outcome of?

If the system gets chaotic enough, it should be possible to get a system that is practically unpredictable (https://en.wikipedia.org/wiki/Chaos_theory). But you really need to refine your question. The way you state it here, it could be interpreted in lots of different ways, so I think it is really unfit to SE.

Maybe it would help if you shorten your question to a single, clearly defined point. For example here, it is not clear to me if you mean with "all conditions known" that all information (e.g. atom positions, velocities, exact wavefunction etc.) is known, or only reaction conditions (e.g. temperature, concentration, reaction time) and so on. What is the point of the question? What do you want to achieve?

I want to know are there reactions able to produce compounds we can't know of advance, even only in theory? Can there be a reaction able to yield totally surprising results, so we couldn't build a model predicting its behavior no matter how precisely we can measure the conditions it is in?

Practically there is, if the system is chaotic. And theoretically (even for "Laplace's demon") there is for the case of radioactive decay provided below. Everything else is philosophy and not really fit for discussion here.

I don't think the philosophers on the Philosophy SE section are fit for so scientific questions. I don't think they can even understand the question, too. It contains too much chemistry to be moved to the philosophical community.

@YordanYordanov Your inability to translate the question to philosophers doesn't make the question a good or well defined Chemistry question

@YordanYordanov Surprising or unpredictable are not defined in the context.Literally any chemical reaction is a probabilistic event, therefore unpredictable and there is a probability that leads to surprising results. If you create complex reaction networks, chaotic systems etc it is more pronounced, but it still there always.

Life is an example of a series of chemical reactions with unpredictable results.
Forget, for a moment, the long term evolution of living creatures and consider the chemistry behind inheritance and selection (at least one sexual activity has started). The key chemical reaction that drives the DNA code for the next generation is random mixing of the genetic material of the parents to generate the DNA of the next generation. This process clearly demonstrates that the result of a chemical reaction is not entirely predictable given the components (the code is written in the same language, but the combination in children is a random combination of the parent's DNA).
Over a longer time period, we can also observe that entirely new types of life emerge via Darwinian random variation and selection. Future life is hard to predict given the existing life at a given point in time. And the underlying process is a chemical reaction involving DNA.
So maybe it isn't natural to think of this as chemistry. But it is built of chemical processes and demonstrates that, in a sufficiently complex system, chemistry can deliver results that are not easily forecastable. Perhaps simpler systems can demonstrate the same point, but this one is staring us in the face as an existence proof.

,I have never thought of crossingover in such a way. It really made an impression! Thank you for your thoughtful answer.

I only agree partially: Im principal you could calculate how the DNA evolves and recombines, or at least how probable a certain outcome is. In practice this just fails due to the huge amount of data points, but there are not infinitely many! You also could go the other way around: make the experiment and check whether you can calculate the outcome. Provided you have enough computer power and the apropiate level of theory, you can always calculate it. Because theory is supposed and designed to describe reality.

The time point of the radioactive decay of a single atom cannot be predicted, even given all experimentally available information, and is not even predictable theoretically by any possible means.
Just isolate a single atom and observe its radioactive decay, and you have the purely random system you strive for. Could also take more than one atom and observe fluctuations with Geiger counter.

I agree with your example @deusexmachina, but my question is can we do the same for a chemical reaction?

@YordanYordanov You asked if processes occur in nature and radioactive decay reactions happen naturally in our environment all the time.

Then, do radioactive decay fall into the category of chemical reaction?

Of course: if 17F decays to 17O it has chemical consequences. Decay chains are in fact very important chemically, if you want to synthesize certain radionuclides.

I'm a little confused where is the border between chemistry and physics here. That is why I'm asking.

I think the radioactive decay itself is a physical process which leads to acceleration of the remaining nucleus. This acceleration will thereby cause a number of subsequent chemical changes.

The final outcome of radioactive decay is known in principle, event though there might be some uncertainty about decay chains due to paths that might not have been discovered yet. The change of the composition of radioactive material over time can also be predicted by the decay chains an nuclide life times.

@aventurin I know research evolves fast, so maybe I am not at the latest state of research. But I thought that it is not known when a single atom decays, but only the nuclide-specific half-life of a decay which is of course just a statistical property of many nuclides. AFAIK it  is not known if the decay is truly non-deterministic or caused by a hitherto unknown inner property.

@deusexmachina I think a way to restate your comment is that for any infinitesimal amount of time, we don't know if an atom will decay; we only know about decay (very well) in the limit of long time intervals.

@deusexmachina We don't know when any individual atom will decay, but we do know bulk properties such as "after one half-life, almost exactly half the atoms in a decently sized sample will have decayed with overwhelmingly high probability." The bulk process is completely predictable, so it's not clear (to me, at least) whether this really answers the question. One isn't usually looking at just a handfull of atoms.

@DavidRicherby Well you have two ways to change that: a) lower the amounts of atoms, e.g. by diluting with inactive nuclide or b) lowering the time frame (random fluctuations in time). Essentially, radioactive decay shpuld be a perfect random number generator as the decay processes are indeterministic and perfectly decoupled.

The ultimate fate of any isolated system can be predicted since it tends to thermodynamic equilibrium. However, it might be that the time to reach the end state is such long that no one ever will be able to observe it.
For closed and open systems things are more complex. But in principle it is possible to predict the "state" when the flow of matter and energy over time is known. Some of these systems may exhibit chaotic behavior, such that said "state" is actually a probability distribution over the possible states. That may be the reason why you have cited the concept of ergodicity.

How about kinetically controlled reactions then?


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Dipole moment of ferrocene conformations

Which conformation of ferrocene has a zero dipole moment?

I thought that the answer should be the staggered conformation because it would have the cyclopentadienyl groups oriented away from each other. However, the answer given is the eclipsed form. I can’t understand why that happens.

Now that's a cryptic question indeed. You are right in that the staggered conformation has zero dipole moment because of symmetry. So does the eclipsed conformation. So does every other conformation as well.

I would also argue both the eclipsed and staggered form have zero dipole-moment due to symmetry, by the argument that, if ferrocene is aligned along the z axis, then for any dipole $v = (v_x, v_y, v_z)$ there is a dipole $v' = (v_x, v_y, -v_z)$ so that the net dipole moment cancels. Maybe it is about polarizability?

To find out whether a compound can have a nonzero dipole moment, consider its symmetry. A dipole moment is a vector which must be transformed onto itself by every symmetry operation present in the molecule’s point group.
It should be easy to see that te eclipsed conformation has five planes of symmetry perpendicular to the cyclopentadienyl rings ($5\sigma_\mathrm v$). Another sixth plane of symmetry is sandwiched between the two rings and passes through the iron atom ($\sigma_\mathrm h$). The only vector that will be transformed onto itself by all these planes of symmetry is the null vector $\vec{0}$. Therefore, this conformation has zero dipole moment.
The staggered conformation also has five planes of symmetry perpendicular to the ring planes as the eclipsed conformation ($5\sigma_\mathrm v$). It also features a centre of symmetry in the iron atom ($i$) as well as improper rotation ($S_{10}$). The inversion centre alone is enough to determine that only the null vector $\vec 0$ will be transformed upon itself. Therefore, this form also has no dipole moment.
Both you and the book are correct.


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Solving displacement of volume in water after a object is dropped inside the water container?

A $\pu{53.7 g}$ object having a density of $\pu{5.39 g mL-1}$ is added to a graduated cylinder containing water and sinks to the bottom. The water level is now at the $\pu{82.2 mL}$ mark. What is the volume of the graduated cylinder?

My work so far:
$$V = \frac{m}{\rho} = \frac{\pu{53.7 g}}{\pu{5.39 g mL-1}} = \pu{9.96 mL}$$
Should I just take the displacement of the water with the object sunk to the bottom to arrive at the original volume of water like so:
$$\Delta V = \pu{82.2 mL} - \pu{9.96 mL} = \pu{76.2 mL}$$
If not, what do I need to do to finish this problem?

I am a little confused by the assignments question What is the volume of the graduated cylinder. Are you sure, these are the words? Because to me, the volume of the cylinder does not have anything to do with its actual contents. For example: I may fill my 250 mL cylinder with only 100 mL of liquid.

This is not chemistry because it speaks about water. At least we can consider it as physics because it speaks about volume. By the way this is only math because everybody should know a liquid takes the shape of its container.

I'm voting to close this question as off-topic because it is not chemistry in the slightest.

Short answer: yes, you are correct.
You calculated the Volume of the object, which was added to the water in the cylinder to give the total volume of 82.2 mL. So deduct this and get your solution... which you did.


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Why doesn't the sulfate anion in copper(II) sulfate have a subscript of 2?
So I'm being asked to find the chemical formula for copper(II) sulfate.
I thought it would be written as:
$$\ce{Cu(SO4)2}$$
Apparently it's written as:
$$\ce{Cu(SO4)}$$
Can anyone please explain to me why the latter answer is correct?

The sulfate anion, $\ce{SO4^{2-}}$ has a charge of -2. Copper can have cations with either +1 or +2. So the two possible copper sulfates are cuprous sulfate (copper(I) sulfate), $\ce{Cu2SO4}$, and cupric sulfate (copper(II) sulfate), $\ce{CuSO4}$.


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What can be concluded from two different values for the interplanar distance between planes of atoms in an X-Ray diffraction experiment?
I analyzed an unknown Alkali Halide crystal using x-rays from a copper source. I need to find the d-spacing for this crystal, i.e. the interplanar distance between planes of atoms.
I found my $K_\alpha$ and $K_\beta$ peaks and used Bragg's Law to calculate the d-spacing. However, these values do not match. Is this an experimental error or could the non-matching values be correct somehow?
For $K_\alpha$:
$$d_{hkl}=\frac{λ}{(2 \sin(θ))}=\frac{(1.54184 Å )}{(2 \sin((28.3°)/2) )}=0.771 Å$$
For $K_\beta$:
$$d_{hkl}=\frac{λ}{(2 \sin(θ))}=\frac{(1.39222 Å )}{(2 \sin((25.7°)/2) )}=2.488 Å$$
My attempt: in both cases, I assumed that these were first order peaks. Could it be that the $K_\beta$ peak is actually a third order peak? If so, then $n=3$ and $d_{hkl}=2.313 Å$. This is reasonably close to the other value. But then what would that tell me about the crystal?

You analyzed it with... what? XRD? I have never seen Bragg's law applied to $K_\alpha$ and $K_\beta$ directly like that. What peaks? can you show the entire diffractogram, if indeed it was an XRD that was done?

@StianYttervik I used a TEL-X-OMETER to collect the intensity measurements. So yes, x-ray diffraction.

However, I only collected 2 peaks, not the entire range of angles. I was told that these two peaks corresponded to the $K_\alpha$ and $K_\beta$ peaks.

The item you linked to is both an XRF and an XRD. Are you sure you have recieved a diffractogram problem? I ask, because $K_\alpha$ and $K_\beta$ is notation used for emission of x-rays (for instance, in fluorescense), NOT their reflection. Which leads me to believe you have been handed an XRF problem, but I won't speculate without information... In either case, the smallest lattice spacing (a) for alkali halides is $\ce{LiF}$ at a > 4.0 Å so there is that...


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What is the crystal structure for a crystal where the ratio of the squares of the angle peaks is 0.825?
I followed this and it gives an example of how to identify a crystal structure using XRD data.
I have an unknown alkali halide where two peaks were found at $2\theta =25.66$ and $2\theta = 28.3$. I used a copper source for the x-rays. I need to figure out the crystal structure from this. Using the example in the link above, I got
$$\sin ^2 (\theta _1)=0.04931022$$
$$\sin ^2 (\theta _2)=0.05976132$$
Therefore,
$$\frac{\sin ^2 (\theta _1)}{\sin ^2 (\theta _2)} = 0.82511934$$
The PDF only mentions what the result will be for BCC (0.5) and FCC (0.75). Can I reliably conclude that this must be a FCC crystal?

No, you can't. Two peaks are way below enough to make any sense.


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Oxidation of Magnesium with Nitric Acid

Write an unbalanced equation for the following: Magnesium metal is added to dilute nitric acid.

My attempt: Nitric acid commonly oxidizes metals and gets itself reduced in the process - to $\ce{NO}$ if the solution is dilute, and to $\ce{NO2}$ if it is concentrated. Therefore, I thought that the reaction would proceed as follows: $\ce{Mg + HNO3 -> Mg^2+ + NO + H2O}.$
However, my book says that instead of being reduced to $\ce{NO},$ the nitric acid is actually reduced to ammonium in the following fashion: $\ce{Mg + HNO3 -> Mg^2+ + NH4^+ + H2O}$
Why is the reaction with dilute nitric acid different specifically for Magnesium (as compared to nitric acid reactions with other metals, such as copper) ?

There is nothing specific about Mg. Other active metals (Al, for instance) would behave the same way. Then again, in fact you'll get a bunch of products containing nitrogen in all oxidation states from -3 to +4; which of them would be the major one depends on many things.

Reading ‘dilute’, I would have assumed that the proton reduces magnesium liberating hydrogen (and leaving the nitrate untouched). $\ce{NO}$ is produced by half-concentrated nitric acid as far as I can remember.

https://chemistry.stackexchange.com/questions/6531/attempting-to-balance-a-redox-reaction-using-half-reactions-leaves-an-unbalanced


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How to identify the asymmetric carbon in [4-(1-bromoethylidene)cyclohexyl]acetaldehyde?
How to identify an asymmetric carbon in the molecule below? Is the carbon identified by the asterisk asymmetric or not? I would say no as it has no 4 different groups attached.

Is this correct?
I will greatly appreciate if anyone can confirm or otherwise explain why it can be asymmetric.

A simple test. Create the mirror image, using wedge and hash bonds to indicate configuration if required.. See if it superimposes on the original. If it doesn't, it's a stereoisomer (the enantiomer if there only one chiral centre).  The "four different group test" can fall down for compounds with internal mirror planes e.g. meso-tartaric acid

@Beerhunter How does meso-tartaric acid fail the four different substituents rule? It has two stereo-centres, i.e. asymmetric carbons, it just isn't chiral due to the mirror plane.

related: https://chemistry.stackexchange.com/q/59124/4945

How do CIP rules deal with (E)- / (Z)- relative to a group? i.e., is this stereocenter (R) or (S)?

Exactly, the mirror plane criteria is better

There is probably little reason in discussing the hydrogen and the $\ce{CH2CHO}$ group which are definitely different residues. The question remains whether the sides of the ring are identical or not.
If you go along the top side of the ring, you will encounter:

$\ce{CH2}$
$\ce{CH2}$
$\ce{CR=CMeBr}$

Looking into it a little deeper, we see that we are coming from the (Z) side of the double bond.
Going along the bottom side you will encounter:

$\ce{CH2}$
$\ce{CH2}$
$\ce{CR=CBrMe}$

Here, however, we are arriving from the (E) side of the double bond.
Since (E) is not equal to (Z) the two sides of the ring are not equal, therefore the carbon has four different residues, therefore it is chiral.

This answer finally clears up my confusion....thank you!!

Yes it does have 4 different groups attached. The sides of the ring are not equivalent. One of them is in trans position to bromine, another is in cis, and that can't be changed easily.


text:
Does chemistry change under high voltage?
Do chemical reactions change when you charge the entire reaction vessel plus or minus $\pu{1 MV}$ or more?  Is there a name for such chemistry?
I was looking at "electrochemistry" expecting to see a host of information on charged reactions (either statically or with a constant charge applied) but it all just appears to be focused on batteries, not so much on synthesizing novel materials.

So what you want more precisely is high electric fields (V/m). This has some interesting facets. It is possible to generate massive fields (> 1 MV/m) by submitting a thin sample sandwiched between two electrodes to a very mundane single-digit voltage difference. Also, the kinds of electric fields which build up on the interface between electrodes and solvents (electric double layer) may be relevant. I don't know much about these things, though.

It is also worth noting that a static electric charge on the outside of a vessel will make no difference to the electric field gradient inside the vessel. If you wanted to explore an effect, you need to create a field gradient and that means two electrodes at different potentials.

Long story short, no, they don't.
Charging a vessel to 1 MV is not a big deal, if you look at it from the inside. To put things into perspective, imagine a vessel of about 10 cm across, which is kinda OK for a flask. Imagine it spherical to make the calculations easier.

Now, the capacitance of a sphere is $C=4\pi\varepsilon_0R$, which evaluates to 5.56 pF (yes, "p" stands for "pico").
So charging your vessel to 1 MV implies putting $5.56\;\mu\rm C$ of electric charge on it.
That's $3.48\cdot10^{13}$ extra electrons (or some negative or positive ions, that doesn't matter) ...
... which may seem big until you compare it to the amount of water that would fit into the vessel: $1.75\cdot10^{25}$ molecules.

So it is like 1 electron per quite a few billions of molecules. That would not make much of an influence. The molecules will just sit there as usual, feeling nothing; most of them have never seen an electron in their entire lives, and have no idea that their vessel is being charged.
Things will change a great deal if you touch the vessel from the outside, but that's just ordinary electrochemistry which you seem to put out of scope of your question.

Plus the charge will tend to accumulate on the outside of the vessel, not internally, if I remember my E&M correctly.

Yes, that too. Moreover, for the same reason of repulsion, they will tend to leave the vessel and fly away.

What about with a much smaller reaction vessel - something on the order of a capillary tube.

@CoryG if the container is open, you could do things like electrospraying. The drastically increased surface area it provides could change rates and such, but electrospraying megadalton-sized molecules (proteins) is surprisingly gentile to them.

Cannot one in principle think of the reaction between two opposite charged walls? Like a dielectric layer. Whatever has electrical dipole  align with the field and possibly this hinder the reaction (or affect the reaction). Plus effect like inhomogeneous concentration. Never thought of this.

@Alchimista You've just described an electrolytic cell. Such a thing surely does exist, and gives rise to an entire field of chemistry.

@Ivan Neretin. I mean whatever you do in apolar solvent without ionic content. I thought it is implicit.

I said walls else I would have called them electrodes

@Alchimista When you apply voltage to the walls, they become electrodes, or at least try to. If that can't be done, they hardly influence anything. It is not that easy to align the dipoles. Related: https://chemistry.stackexchange.com/questions/60289/in-practice-do-polar-molecules-actually-align-in-an-electric-field/60297#60297

@Ivan Neretin I know this. I said in principle. You can think of a very strong field and I wounder if it could affect the reaction. For instance the kinetics. Statistically colliding molecules will have an orientation different that that without field. Assuming most of mechanisms, the reaction should slow down. It is now clearer? I am not thinking of getting totally different chemistry but some kinetical effects, and perhaps different "selectivity" if more than one product is expected. And again in principle. Surely we will never get perfectly aligned molecules, and more so in a liquid env...

@Alchimista OK, then I'll repeat the obvious for the benefit of others: yes, the alignment of molecules could theoretically result in some unusual chemistry, but it is impossible, because the required fields are too great and would cause a breakdown.

This is far from repeating the obvious because 1) you did not even mentioned it in the previous comment and 2) there is nothing too great here.*  The breakdown will be chemistry, and indeed a radical change :) No, I wasn't considering field able to strip apart a molecule. But you suggest an even drastic chemistry. Cheers. *tilting a dipole will hardly exceed splitting two charges

Well, I explained this in an earlier answer elsewhere, see the link in one of my comments.

Also worth noting that a vessel with a large charge will still have a zero electric field gradient on the inside. So there will be no field for the molecules to feel.

If the static charge on your "reaction vessel" is as high as 1MV (megavolt), you may find that parts of your lab that you wouldn't otherwise consider to be your reaction vessel start to take part in the reaction.
If we suppose that your reaction vessel is made of "window glass", the dielectric strength is only about 10MV/m, and assuming that your vessel is the thickness of conventional glassware, a static discharge through your vessel is not out of the question. Such an event will definitely change the "chemistry" of your experiment significantly, as lightning bolts often do.
As far as terminology goes, I'm not sure if there's a name for the sub-discipline of chemistry dealing with chemical changes under high electric field strengths, but electrochemistry might still be pretty close to what you are looking for. Note that the field strengths typical in electrochemistry are usually on a much smaller scale than what you describe.

I really like Ivan's answer above, but wanted to add one other perspective.
Static discharge is a significant possibility, and that might change the chemistry (e.g., redox events). If that doesn't happen, I'd argue that very little will happen because these fields are very common on the nanoscale.
Let's imagine we have ~1MV/m applied to the vessel. That works out to $1\times 10^6\; \mathrm{V/m}$. So about 1V per µm. Even if you have a much smaller reaction flask (10 cm maybe), that's 1V per 100 nm.
Consider the molecular scale. If I have a monocation separated 1.0 nm from a molecule, the molecule experiences ~1V/nm in vacuum. That would be $1\times 10^9\; \mathrm{V/m}$. Granted, in any reasonable environment, the dielectric constant of the solvent will decrease the field substantially, but even in water, the field would be ~$1\times 10^7\; \mathrm{V/m}$.
There's an atomic unit of electric field: $5.142 \times 10^{11}\; \mathrm{V/m}$. That corresponds to the interaction energy between a proton and an electron in a hydrogen atom.
Your fields are much lower than common ion-molecule interaction fields or proton-electron fields.

I'll point out that you can get interesting interactions (polarization, induced dipole, molecular rotation/alignment, even electromechanical effects) at much lower voltages.

I'm wondering if the static discharge can effectively "steal" energy from a reaction and halting it or lowering yield. I'm thinking of a situation that would parallel this -- maybe sort of like discharging a capacitor or something if that makes sense.

I heard that strength of molecular field was even used to test Standard Model, IIRC.

If you had 1MV on a capillary tube wouldn't that cause all the polar molecules to align radially though?  That seems to influence proteins-driven pretty radically when you have hydrophobic portions of the molecule on the inside and hydrophobic portions on the outside.  I can't claim to have experimented with it or read of any experiments with it, but wouldn't that alignment via electronegativity have a pretty radical effect on any polymers?

@CoryG - I'd have to do some calculations, but electrochemistry demonstrates that screening occurs pretty rapidly. See, e.g. https://en.wikipedia.org/wiki/Double_layer_(surface_science)

High electric fields can indeed influence chemistry.
E.g. the First Wien Effect describes how high electric fields ($10^7~ \pu{V~m^{-1}}$) increase the conductivity of electrolytes.
The Second Wien Effect can lead to an increased dissociation of weak electrolytes. There's some speculation that this non-linear effect may play a role in the conduction of electrical impulses in neurons where strong electric field exist at the membrane.

I thought about this too but knew it will work only in space so that the different charges would not be involved into a developing field. If you study idle or colder gasses, they show that negative ions become more stable because of lesser collisions. And the same for positive ions but that is kind of obvious even with collisions. I find this challenging opportunities when studying elektrochemistry (and it's influence of temperature differences which are also more easy to install in space). Think of all molecules sharing electron; one could think of a shortage that could become quite different in space if we could add or diminish charge in processes. My interest has come from discovering the non-existence of electrical repelling. When millions of volts can be stable just being enough isolated; then how could one say that electrons are repelled when charging such spheres??? Eric Dollard proposes the same in his youtube story: the history of energy synthesis (47th minute). Repelling is a postulate and never investigated. The definition of an electrical field is OK but repelling is being "explained" outside this definition. Make two big plates (1 m2) charged equal and have a probe charge in between them (equal charged) and nothing will happen (untill you approach the borders of the plates when charges outside of the plates get their influence. The electrometer is a fraud since the quality of our earth to attract any kind of different charge is being perfected by the earthed shielding and so explains the attracting forces on the blades. Fieldlines don't go to likewise charged items because there is no voltage difference and so no attraction and this last argument is the one from Eric Dollard and his website. If people only would understand this extremely dumm and simple mistake it could revolutionize electrochemistry perspectives for our future. Also get aware of the fact that all elements are elektro negative, meaning they attract electrons (there is no such quality as "electro positiveness!!). This together with the fact that it costs at least around 13 ev to loosen an electron out of it's bonded state and the notion that our earth would be on zero is a total misrepresentation of the truth !!!!

This is very convoluted and extremely confusing.  Can you break it up into paragraphs and substantiate your claims via mainstream references, please?

The problem is that it is very contrarian against main knowledge, so one should check this out. My "problem" is that I never did the official courses in chemistry or physics because of this huge misconception of repelling forces which is the beginning of both these mentioned sciences. Boy have I been right on this! Have a look on the site of Miles Mathis where he corrects numerous other mistakes. Mistakes I guessed to exist from off the beginning. I apologize for my style of cramming important facts and considerations; please ask what is wrong about them.

Mainstream references are important for answers on the site.  Even if there were to be some truth to your claims, it's important for students to learn the fundamentals first.

Jonsca, I fully understand and appreciate your attitude and I did mention Eric Dollard who had a lot of views on his enormous body of work on the history of electricity, but I cannot help that no one started thinking about his sole remark on that 47th minuit and I'm trying since almost 50 years. Everyone is stuck on mainstream knowledge and math can "correct" evils in understanding such that people lose track. As long as formula's make sence no one doubts the axioms. Have like 20 barmagnets organise themselves and repelling disappears. Counterintuitive but not to be found in scholarly books


text:
How to prepare phosphate buffer solution from mono- and disodium phosphate with a specific pH?
I want to prepare an aqueous solution of a mobile phase of $\pu{0.04M}$ Brij-35. The mobile phase has to be adjusted to pH $7.4$ with $\pu{0.5M}$ phosphate buffer which will be prepared with $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$ solutions.
Can anyone help me to prepare a phosphate buffer solution of $\pu{0.05M}$ using $\ce{NaH2PO4.2H2O}$ and $\ce{Na2HPO4.7H2O}$ with pH $7.4$?

@fl.pf. The shorter $\ce{NaH2PO4.2H2O}$ works perfectly fine, mhchem handles quite a lot of shorthand. (I am not saying that that would have been my reason to further edit, but it might help you save some time for future edits. I simply wanted to fix the title in this case.)

One possibility is to use the Henderson-Hasselbalch equation $(1)$.
$$\mathrm{pH} = \mathrm pK_\mathrm a + \lg\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]}\tag{1}$$
Plugging the $\mathrm pK_\mathrm{a2}$ value of phosphoric acid ($7.198$) and the desired $\mathrm{pH}$ value will give you a ratio of hydrogenphosphate and dihydrogenphosphate concentrations which you can solve to the desired total concentration $c_\text{tot}$ by simple algebra.
However, SigmaAldrich also maintains a nice buffer reference centre that includes ready-calculated masses for different $\mathrm{pH}$ values. All that remains is to make sure your final concentration is the desired one.

as you can see using SigmaAldrich they provided calculations using concentration of 0.2M but mine is o.o5M according to my question above, how can I convert concentration from o.2M to o.o5M ? will I still use the same amount of ml provided ?

@UHAGAZED.Serge By dividing by four, obviously.

can you divide concentration (in terms of g/l)by four or you divide volume by four ?


text:
Does a covalent bond constitute of 1 or 2 electrons?
A covalent bond is pictured as two electrons shared by two different atomic nuclei, which means that each of the two nuclei has two electrons. The problem I see with it is that it implies that if a hydrogen atom is neutral then a hydrogen nucleus in a covalent bond must be negatively charged since it has one more electron than a hydrogen atom. Which implies that a hydrogen molecule is negatively charged which is a contradiction (molecules are neutral).
Alternatively, one can say that the covalent bond is composed out of a single electron, which acts as a negative potential for both hydrogen nuclei hence it exerts negative potential of two in the molecule. But then the question is, where does the second electron go when two hydrogen atoms form a covalent bond?
My question does not relate only to this specific example, it's about covalent bonds in general.

You are overthinking it. A covalent bond kinda "consists of" two electrons, but it belongs jointly to two atoms.

This would be an extremely conservative view saying that a covalent bond consists of two electrons (Lewis mad that famous). However, bonding interactions are not localised and it is impossible to tell how many electrons form a bond; or what is a bond in the first place. In this specific example though, there are two protons, two electrons, everything is neutral. By necessity the bond consists of two electrons.

@Martin-マーチン What you say is reasonable. I was interpreting the pictures literally since both atoms are removed an equivalent distance apart from the electrons. So, I assume, I must see the electrons completely delocalized? There is no geometry involved then?

As you may know, electrons are not only particles. They are not at all like in the picture

@Martin-マーチン Is the covalent bond formed because it is energetically optimal for both atoms?

You may also further learn about oxidation numbers and partial electric charges. They tell you that the $H_2$ molecule is perfectly neutral.

The bond in your picture is made up of two electrons, and in the Lewis formalism so is every single bond. Yet there is no negative charge. We can think of this in two ways:

Remember that the bond electrons are shared. So each atom contributed one electron and, upon dissociation, would usually take back that one electron. Counting only that one electron will give a neutral hydrogen atom, not a hydride ion.

Charges on an atomic or molecular level are due to a mismatch of the numbers of protons and electrons. In the hydrogen molecule, you have two nuclei consisting of one proton each, meaning an overall $2+$ charge. Furthermore, you have two electrons that get shared in the covalent bond giving an overall $2-$ charge. $+2 + -2 = 0$, so the overall molecule is neutral.
If you wanted to place both electrons on one atom to form a hydride anion, that would mean the other atom is left with zero. Instead of a hydrogen molecule $\ce{H-H}$ you would have a separated ion pair $\ce{\overset{-}{H}\bond{...}\overset{+}{H}}$.

Once you leave the very simplified Lewis formalism and turn to theories such as molecular orbital theory which explain molecules much more accurately, this way of thinking will be lost altogether since there is no longer an obvious localisation of two electrons to a single bond.


text:
Na+ or Ca2+ . Which ion has the larger radius?
Which ion has the larger radius? Some sites say $r(\ce{Na+}) > r(\ce{Ca^2+})$. Any explanation would be appreciated.

The thing is that you need to know the coordination environment in the first place as ionic radii are C.N.-dependent. From Shannon's canonical paper "Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides"[1]:
\begin{array}{cc}
\hline
\text{Ion} & \text{C.N.} & \text{C.R., Å} & \text{I.R., Å} \\
\hline
\ce{Na+} & 4 & 1.13 & 0.99 \\
& 5 & 1.14 & 1.00 \\
& 6 & 1.16 & 1.02 \\
& 7 & 1.26 & 1.12 \\
& 8 & 1.32 & 1.18 \\
& 9 & 1.38 & 1.24 \\
& 12 & 1.53 & 1.39 \\
\hline
\ce{Ca^2+} & 6 & 1.14 & 1.00 \\
& 7 & 1.20 & 1.06 \\
& 8 & 1.26 & 1.12 \\
& 9 & 1.32 & 1.18 \\
& 10 & 1.37 & 1.23 \\
& 12 & 1.48 & 1.34 \\
\hline
\end{array}
where C.N. - coordination number; C.R. - crystal radius; I.R. - ionic radius. The data is also available online for free. Comparing similar coordination environments, it's easier to notice that overall $\ce{Na+}$ is indeed larger than $\ce{Ca^2+}$:

Reference

Shannon, R. D. Acta Cryst A, 1976, 32 (5), 751–767 DOI: 10.1107/S0567739476001551.

If there is no such thing as a non-integer CN then drawing any lines is strictly speaking wrong... apart from that this is a really useful answer!

According to the diagonal relation, $\ce{Na}$ and $\ce{Ca}$ have similar radii.
$\ce{Na+}$ and $\ce{Ca+}$ have thus similar radii as well but, when one more electron is removed from $\ce{Ca+}$, making it $\ce{Ca^{2+}}$, its effective nuclear charge increases, which will make its radius decrease.


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Relation of conductivity with electrode distance
Conductivity is defined as the conduction of ions when the distance between the electrodes is 1 cm and length is 1 cm.
Say for instance we have a solution where the distance between the electrodes is 2 cm and its height is also 2 cm, what will be the resulting conductivity of the solution?

sigh... lifted out of Definitions section of wikipedia article on Conductivity (electrolytic)
Resistance, $R$, is proportional to the distance, $l$, between the electrodes and is inversely proportional to the cross-sectional area of the sample, $A$. Writing $\rho$ (rho) for the specific resistance (or resistivity),
$R={\frac {l}{A}}\rho$
In practice the conductivity cell is calibrated by using solutions of known specific resistance, $\rho^*$, so the quantities $l$ and $A$ need not be known precisely. If the resistance of the calibration solution is $R^*$, a cell-constant, $C$, is derived.
$ R^{*}=C\times \rho ^{*}$
The specific conductance (conductivity), $\kappa$ (kappa) is the reciprocal of the specific resistance.
$\kappa ={\frac {1}{\rho }}={\frac {C}{R}} $
So if $l_1 = 1 \text{ cm}$, $A_1 = 1 \text{ cm}^2$ but $l_2 = 2\text{ cm}$, $A_2 = 4 \text{ cm}^2$
$\kappa_1 ={\frac {C}{R}} $
$\kappa_2 ={\frac {2C}{R}} = 2\kappa_1 $


text:
Can you heat Bismuth crystals to reset the oxide layer for new colors?
I'd like to get a bismuth crystal with small structures and deep blue colors for a ring I want to make. I can find these properties in different crystals online but never together. Can I just take a crystal with the right shape and blowtorch it to create a new oxide layer? Would I first have to chemically remove the thin oxide layer?

I have made Bismuth crystals where I etched the oxide layer off and tried heat treatment to reproduce the colored oxide layer. It did not work in a $200^\circ C$ furnace, but that doesn't mean a torch couldn't, I would suspect that it is non-trival though.

Just a thought, but perhaps a thin oxide could be deposited by making bismuth the anode in a dilute sodium or potassium hydroxide bath.

Would that bath "redo" the oxide layer though? And wouldn't that create a thicker, more even layer than one resulting from the uneven cooling of a crystal? Also, do you know if nail polish would work to preserve the layer without dissolving it?

thickness is a function of time and current Nail polish remover will not dissolve the oxide.


text:
Looking for a theory predicting conductivity values of electrolytes at higher concentrations
I'm looking for a theory to calculate the electrical conductivity of electrolytes, namely $\ce{NaCl}$ and tartaric acid ($\ce{C4H6O6}$) dissolved in water, depending on their concentration.
The only equation I could find was the one derived from the Debye-Hückel theory, which I read works great for 1:1 electrolytes like $\ce{NaCl}$ (not sure about tartaric acid 1:2 electrolyte), but only until about 0.1 M.
The concentrations in my case vary from 0.2 M up to 4.0 M.
Is there a theory I can use to calculate conductivities in that range, or is the Debye-Hückel theory the only one?

Related: Conductivity as a function of acid concentration.


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How to Predict the Formation of Racemic Mixtures or Single Products from a Diels Alder Reaction?
So far, I understand the mechanism of the Diels Alder reaction and am able to correctly predict the product of some slightly complex examples.
However, it seems that in some examples, a racemic mixture is formed, whereas in others that's not the case.
I have basic understanding of the Endo Rule (endo product favored over exo product), but have trouble with the stereo chemistry involved.
I also know that the symmetry of the diene and dienophile has to do with it, but can't seem to find a consistent pattern.
Any help would be appreciated.

From the phrasing in the question, it's not clear whether you mean to be asking about enantiomers or diastereomers. For example, the endo and exo products are diastereomers, while racemic mixtures are made up of enantiomers. Perhaps it is both. Please clarify.

It is very simple: if you do not have any enantiopure compound present at the beginning of the reaction, the end product will either be inherently symmetric (contain a plane of centre of symmetry) or racemic. This will be because there are two symmetry-equivalent paths of attacks that will lead to the two enantiomeric products (if enantiomers are even generated) and neither of the two paths is inherently more probable. It is a great exercise to try getting both enantiomers for a stereouncontrolled reaction!
If you do have any enantiomeric excess in your reactants, your products will also have some enantiomeric excess. If the chiral centre is far from the reactive site, the ee will be low; if it is very close or part of the reactive site the ee will typically be much higher. This is because the transition states will now be diastereomeric rather than enantiomeric, meaning they differ in energy and one will be more likely.
While it is still a great exercise to draw the two different transition states leading to different products it can be much harder to distinguish the more likely major product; especially if it is a chiral catalyst giving enantioselectivity.


text:
How to calculate the pressure required to converted graphite to diamond at room temperature (298 K )?
I want to calculate the pressure required to convert graphite to diamond. The information that I have is the density of diamond and graphite, enthalpy change of diamond to graphite at room temperature and entropy of diamond and graphite at room temperature. Which thermodynamic relation can we use to calculate the pressure?

Hello and welcome to Chemistry.SE. Homework type of questions are welcome here, but must be presented according to the guidelines described here. The main thing is that you need to show some attempt to solve the problem yourself. That ensures that we aren't doing your homework for you, and shows us where you are getting stuck. As it stands, your question will be closed unless you [edit] it accordingly. Anyway, welcome again and best of luck!

At constant temperature, the change in free energy is vdP.