Organic molecules with 21.62% oxygen I have been told to answer the following question:

  Seven different compounds, all either alcohols or ethers,
  contained 21.62% oxygen. Find the structural formula for all of these compounds and name the alcohols.

So far I've tried using empirical formula calculations to work out a simple formula but this has yielded no results because I do not know the exact ratio of hydrogen to carbon in the formula nor do I know any masses which I could use to scale up the empirical formula into a molecular formula. It seems to me that there just isn't enough information to answer the question - after hours of searching I can't even find one, let alone seven. I'm sure the answer is stupidly easy but it's evaded me. Does anyone have any suggestions of things I might be able to try? This question is UK A-Level standard.

Disclaimer:

Here I take the molar masses of carbon, hydrogen and oxygen to be $\pu{12g, 1g, 16g}$ respectively, which is what would be expeccted of you at school.

For the sake of simplicity, let's assume you have $\pu{100g}$ of the said compound. Which would mean it contains $\pu{21.62g}$ of oxygen.

Employing a bit of basic math in the analysis of that fact, leads us to similar (but more useful) re-statement:

  $\pu{74g}$ of the said compound will contain $\pu{16g}$ of oxygen (the number also being the same as the molar mass of oxygen).

Now our substance possess a general formula of $\ce{C_{x}H_{y}O_{z}}$.

So in $\pu{74g}$ of it (of which $\pu{16g}$ is oxygen), the molar masses of $\pu{C}$ ( times the number of moles of it, which is $x$ ) and $\pu{H}$ ( times the number of moles of it, which is $y$ ) must add up to $\pu{74-16=58}$. Which gives us the equation:

$$\ce{x(12) + y(1) = 58}$$

Now neither $x$ nor $y$ can carry the value $\pu{0}$ (else we won't be dealing with a hydrocarbon anymore). This gives us three mathematically (not necessarily physically) plausible options:

$\pu{x= 2; y= 34}$
$\pu{x= 3; y= 22}$
$\pu{x= 4; y= 10}$

Even if you've only just started studying organic chem. you should be able to tell that the first two options are absolutely ridiculous (that's way too many hydrogen atoms that can be accommodated around 2 or 3 carbon atoms). So we go with the third option: $\pu{x= 4; y= 10}$, which sounds promising.

Which gives us the formula  $\ce{C_{4}H_{10}O}$

You may now work out what alcohols or ethers that could give you ;-)


What is the net ionic equation of the following? Aqueous chlorine is added to a solution of aluminum bromide in a single replacement reaction which forms no precipitate and bromine liquid. Determine the net ionic equation.

Since this is a single replacement reaction, it will take the form of  $$\ce{A + B-C -> A-C + B}$$

Since this reaction forms no precipitate, the new compound will not be a solid.

$$\begin{multline}
\ce{aqueous chlorine + solution of aluminum bromide ->}\\
\ce{ non-solid compound + liquid bromine}
\end{multline}$$
\begin{align}
\ce{Cl2(aq) + Al^{3+}Br3^-(aq) &-> \text{non-solid compound} + Br2(l)}\\
\ce{Cl2(aq) + AlBr3(aq) &-> \text{non-solid compound} + Br2(l)}\\
\ce{Cl2(aq) + AlBr3(aq) &-> Al^{3+}Cl3- + Br2(l)}\\
\ce{Cl2(aq) + AlBr3(aq) &-> AlCl3 + Br2(l)}
\end{align}

Since most chloride salts are soluble (except for silver halides), $\ce{AlCl3}$ is predicted to be aqueous.

\begin{align}
\ce{Cl2(aq) + AlBr3(aq) &-> AlCl3(aq) + Br2(l)}\\
\ce{3Cl2(aq) + 2AlBr3(aq) &-> 2AlCl3(aq) + 3Br2(l)}
\end{align}

Here is the complete ionic equation:

$$\ce{6Cl⁻(aq) + 2Al^{3+}(aq) + 6Br-(aq) -> 2Al^{3+}(aq) + 6Cl-(aq) + 3Br2(l)}$$

Cancelling out the spectator ions, we get

$$\ce{6Br-(aq) -> 3Br2(l)}$$

In lowest terms:

$$\ce{2Br-(aq) -> Br2(l)}$$

You are making a mistake splitting up your reaction equation from \eqref{full} to \eqref{wrong} in order to filter out the spectator ions.

\begin{align}
\ce{3Cl2(aq) + 2AlBr3(aq) &-> 2AlCl3(aq) + 3Br2(l)}\tag1\label{full}\\
\ce{6Cl⁻(aq) + 2Al^{3+}(aq) + 6Br-(aq) &-> 2Al^{3+}(aq) + 6Cl-(aq) + 3Br2(l)}\tag{wrong}\label{wrong}
\end{align}

What you are essentially doing is not conserving the charges. Dichlorine $\ce{Cl2}$ is a neutral molecule, and the following is what your transition implies, but is also wrong:
$$\ce{Cl2 -> 2Cl-}\tag{also wrong}$$

The correct way to transform \eqref{full} is to split (dissociate) only the ionic compounds $\ce{AlBr3}$ and \ce{AlCl3}, and then cancel the spectator ions:

\begin{align}
\ce{3 Cl2 (aq) + 2 AlBr3 (aq) &-> 2 AlCl3 (aq) + 3 Br2 (l)}\tag1\\
\ce{3 Cl2 (aq) + 2 Al^3+ (aq) + 6 Br- (aq) &->
    2 Al^3+ (aq) + 6 Cl- (aq) + 3 Br2 (l)}\tag2\label{split}\\
\end{align}

Next step is to cancel the aluminium ions on each side, and reduce the equation to \eqref{net-ionic}:

\begin{align}
\ce{3 Cl2 (aq) + 6 Br- (aq) &-> 6 Cl- (aq) + 3 Br2 (l)}\tag3\\
\ce{Cl2 (aq) + 2 Br- (aq) &-> 2 Cl- (aq) + Br2 (l)}\tag4\label{net-ionic}
\end{align}

I don't think the reaction will actually produce liquid bromine, but aqueous bromine instead; that doesn't matter to the solution though.


Can I use pure zirconium as an anode in electrolysis? I read that zirconium is not very reactive even when supplied with a current in an electrolyte. If I had a sizable piece of it, can I use it to perform electrolysis for rust removal of items? Would it have the same or similar results to graphite which results in a less messy process, namely the black film left on parts? Is it toxic?

No, you cannot use zirconium as an anode for electrolytic rust removal.
Zirconium is a member of the family of 'valve metals' (an archaic term), which form a strongly passivating oxide film when exposed to air and/or water.  From here:

Owing to their low electrochemical potential the group IVB and VB valve metals Ti, Zr, Hf, V, Nb and Ta readily react with water or oxygen to form a dense, protecting passive layer.

You're right that these metals, including Zr, tend to be exceptionally chemically resistant (same source):

Because of these protecting oxide films, the valve metals show an exceptional resistance towards corrosion in many aggressive environments. This explains why valve metals are widely used in the construction of chemical apparatus.

However, a feature of these metals is that the resistance of the passivating oxide film increases significantly when an anodic bias is applied to them (same source):

Valve metal oxides are formed conventionally electrochemically ... .

The Wikipedia article on anodizing concurs (emphasis added):

Anodic films are most commonly applied to protect aluminium alloys, although processes also exist for titanium, zinc, magnesium, niobium, zirconium, hafnium, and tantalum.

Thus, were you to construct an electrochemical cell with a zirconium anode and a rusty steel cathode, I expect you would get a brief current transient as the Zr-oxide passive film grew a bit, but then it would tail off to essentially zero.  While I haven't worked specifically with zirconium, the above is the observed behavior of all of the valve metals I have experience with: tantalum, niobium, titanium, and aluminum. I would be quite surprised if zirconium behaved any differently.


Predicting order of nucleophilic substitution reactivity What order of reactivity do you predict will be observed when each alkyl halide is mixed with sodium iodide in acetone?

  1-Chlorobutane
  1-Bromobutane
  2-Chloro-2-methylpropane
  Bromobenzene
  2-Chlorobutane
  1-Chloro-2-butene

I know that $\ce{I-}$ is the best leaving group, followed by $\ce{Br-}$ and $\ce{Cl-}$.
For the $\ce{NaI}$ reaction, tertiary halides should react fastest and primary halides should react slowest.

  What order of reactivity do you predict will b observed when each alkyl halide is mixed with silver nitrate in ethanol?

  1-Chlorobutane
  1-Bromobutane
  2-Chloro-2-methylpropane
  Bromobenzene
  2-Chlorobutane
  1-Chloro-2-butene

For the $\ce{AgNO3}$ reaction, primary halides should react fastest and tertiary halides should react slowest.

Which takes preference, leaving group or rank? Also, how do I rank 1-chlorobutane vs. 1-chloro-2-butene? They are both primary and the halide is $\ce{Cl-}$ for both.

Acetone a is aprotic polar solvent which is often used as solvent for $\ce{S_{N}2}$ reactions (notably in conjunction with $\ce{NaI}$ as in the Finkelstein reaction). For $\ce{S_{N}2}$ reactions steric hindrance and leaving group capability is of great importance, so primary halides will be usually easier replaced by iodine then tertiary.
$\ce{AgNO3}$ will probably react as a base forming nitric acid. Even though nitric acid is a rather strong acid this is more likely than $\ce{NO_3-}$ acting as nucleophile. Besides that, $\ce{Ag+}$ is known to have a unique selectivity for halogens and will attack them readily, forming $\ce{AgX}$ and a carbocation, hence promoting $\ce{E 1}$/$\ce{S_{N}1}$.


Standard Enthalpy of Formation of H2(g) vs H(g) My question is somewhat related to this quesion. When I look at a table of standard values there are two entries for Hydrogen:

$\ce {H2(g)} $  - 0 kJ/mol
$\ce {H(g)} $  -  218 kJ/mol

I understand the $\ce{H2(g)} $ is 0 because it is used as a reference, why is monatomic $\ce{ H(g)}$ given in the table and what does it mean? When would one use it in a calculation of enthalpy change?

From what I understand Hydrogen only exists as a monatomic gas at very high temperatures, the Standard enthalpies of formation are given at 1 bar, 298.15 K.

The value may be used whenever atomic hydrogen is present as a (probably intermediate) species. E.g. take the reaction

$$\ce{2H <=> H2}$$

which is the basic reaction in atomic hydrogen welding.


Where to find common inorganic/organometallic molecules? I'm working on a project to create a large open repository of quantum calculations, largely for teaching purposes.

I can get thousands of common organic compounds easily from sources like PubChem or drug databases.

I'm looking for sources for teaching or other uses with VSEPR examples or other types of inorganic and organometallic molecules. Ideally I'd want 100s or 1000s with good coverage of the periodic table.

Any ideas?

To be clear, there are lots of databases of solids. I know most of those, although feel free to suggest some for archival. I really want isolated complex ions, molecules, etc.

I don't think there are many, since I haven't turned anything up besides the Cambridge database, which is restrictive about reuse.

Thanks!

I appreciate the suggestions but some digging found a few possibilities:

As I mentioned in the question, there's the Cambridge Structural Database, which includes over 700,000 compounds (both organic and inorganic/organometallic). It's decidedly not free, but available at many universities.
There is the "Teaching Subset" of 733 compounds from CCDC which offers a Java web view of interesting compounds, including plenty of VSEPR examples and inorganic and organometallic species. This is free to access, but as far as I can tell, not free to redistribute.
Cool Molecules from St. Olaf college, including 900+ structures from experimental data, and lots of inorganic and organometallic species (most elements are reflected), including lots of "cool" or unusual shapes (e.g., hexagonal bipyramidal).
Crystallography Open Database as mentioned by @permeakra. I was aware of this for solid-state materials, but it seems as if they've merged in data from CrystalEye, a resource which grabbed crystallography data from journal articles, including many molecular species now, and can be browsed by journal (e.g., Organometallics).

The last two resources seem the best, since the data can be freely distributed. They also don't require using Java to view the structures.


Does halogenation by phosphorus tribromide lead to an inverted product? I think it follows the SNi mechanism and would not lead to Walden inversion.
Hence, the product should be (S)‐2‐bromobutane:

But the answer says otherwise (the problem didn't indicate the presence of pyridine in the reaction mixture though).

Yes, the haloalkane product will have the inverse stereochemistry of the alcohol that it was produced form. $\ce{PBr3}$ does this by converting the $\ce{-OH}$ into a good leaving group and providing a nucleophile all in one. This is can be seen in the two step mechanism:

$\hspace{2.4cm}$

When $\ce{Br-}$ ejects $\ce{PBr2OH}$ in an $\mathrm{S_N2}$ reaction, the stereochemistry is inverted:

$\hspace{3.1cm}$


Rationalising the reactivity of phosphoryl chloride $\ce {POCl3}$ is a reagent well-known for enacting various functions in organic synthesis. For example, together with pyridine, it is able to dehydrate alcohols to give alkenes. Also, reacting it with a substituted amide gives chloromethyleneiminium ions. Salts containing these ions are also known as Vilsmeier reagents and they have great practical use in synthesis. The above two reactivity patterns seem to suggest that $\ce {POCl3}$ reacts with oxygen atoms containing rather high electron density, such that such reactions with other carboxylic acid derivatives at the carbonyl oxygen is not really observed. Firstly, I would like to ask if my rationalisation of its reactivity here is correct.

Also, as suggested by the same site, $\ce {POCl3}$ is also able to perform the halogenation of alcohol, in a manner similar to how $\ce {PCl5}$ performs the reaction, but only in the presence of dimethylacetamide (DMA) or dimethylformamide (DMF). Why is that the case? What particular properties of the solvent allow the transformation to be effected?

AFAIK it's not a particularly common choice, but Vilsmeier reagents are sometimes used to convert alcohols to chlorides (see e.g. Tetrahedron Lett. 1992, 33 (34), 4901–4904). This explains the requirement for DMF/DMAc as solvent.

A reasonable mechanism is as follows:

(see also Org. Lett. 2002, 4 (4), 553–555 for a similar mechanism)


Why does dissolving a salt in water get a high pH? I dissolved some sodium carbonate in water and I measured the $\mathrm{pH}$. It turned out to be 11. I don't really understand why dissolving it in water increased the $\mathrm{pH}$. I mean, I know that $\mathrm{pH}$ of water was 7, so it must have something to do with the $\ce{H+}$ ions. We just covered acid and bases today so please bear with me.

Here are my thoughts. My salt had some $\ce{OH-}$ salt and this combined with the $\ce{H+}$ to produce $\ce{H2O}$ thereby decreasing the amount of $\ce{H+}$ ion. Or maybe one of my salt ions combine with the dissolved $\ce{H+}$, again thereby decreasing the $\ce{H+}$ concentration and increasing the $\mathrm{pH}$, so I'd have something like this could have something like:

$$\ce{Na+ + H+ -> NaH2+}$$

or maybe

$$\ce{CO3^2- + H+ -> HCO3-}$$

Which one do you think happened?

It's a little more complicated than that.. When you dissolved $\ce{Na2CO3}$ in water you had a solution of ions: $\ce{Na+}$, $\ce{CO3^2-}$, $\ce{H3O+}$ and $\ce{OH-}$ (the later are present in water even when there is no salt, since it dissociates too, however they are in equal quantity in pure water so the pH is 7).

Now when you mix all those ions, the equilibrium isn't only the equilibrium of water dissociation ($\ce{H2O + H2O <=> H3O+ + OH-}$). Now you have a whole series of equilibria occurring in the solution simultaneously:

1) Water dissociation:
$\ce{H2O + H2O <=> H3O+ + OH-}$

2) Salt dissociation:
$\ce{Na2CO3 <=> 2Na+ + CO3^2-}$

3) Base equilibrium*:
$\ce{NaOH <=> Na+ + OH-}$

4) Acid equilibrium*:
$\ce{CO3^2- + 2 H3O+ <=> H2CO3 + 2 H2O}$

The salt dissociation doesn't affect the pH directly. We have two equilibria though that do unbalance the proportion of $\ce{H3O+}$ and $\ce{OH-}$ ions, which are the "base equilibrium" and "acid equilibrium". We don't think about them at first sight because we didn't add any $\ce{NaOH}$ or $\ce{H2CO3}$ to the solution, but everything needed to form them is right there.

$\ce{NaOH}$ is a strong base, however $\ce{H2CO3}$ is a weak acid. So it's likely (and you witnessed it in experience) that more $\ce{H3O+}$ will be consumed in comparison with the $\ce{OH-}$, making the solution more alkaline.


How to clean glassware containing pesticide solutions I have pesticide (dimethoate) dissolved in methanol in volumetric flasks and graduated cylinders. How do I clean this glassware properly before moving on to the next dilution of a different pesticide.

I like this question because the cleaning of glassware is an often overlooked science in itself. You want to remove "contaminated glassware" as a possible factor in the interpretation of unexpected results in particular. A lot of brain-power and time can be wasted if you cannot assume that all the glassware (and any other surface coming in contact with your reaction) is clean with respect to the compounds of interest, and of any compounds that could interfere with their reaction.

For the given example, since dimetholate is soluble in methanol, methanol would seem to make a good solvent for rinsing the glassware before it dries.  So long as the solute is still dissolved in the solvent, a thorough solvent rinse should be effective with respect to removing dimetholate.  If the solute has dried, even though equilibrium thermodynamics say methanol should re-solublize  anything from the original methanol solution, the kinetics of this can be limiting/prohibitive. If this is the case (i.e. the solution has dried before rinsing), an aqueous detergent scrub followed by a methanol rinse is probably necessary. As water and methanol are miscible, this should not result in any mixed-solvent induced re-precipitation of anything originally present in the methanol-dimetholate solution.

Keeping to your example, as water an methanol are miscible, even in the case where the solution has not dried you should first rinse with methanol then with DI water. This will ensure any methanol that might precipitate the next standard material will be removed, and any remaining solvent (water in this case) will also not precipitate the water-soluble malathion.

The process is more complicated when you are using the glassware for a solvent in experiment 1 that is not miscible in experiment 2.  Say you a toluene  solution followed by an aqueous solution. As water and toluene are not miscible, any toluene left over from your toluene rinse would form a 2-phase solution in the aqueous solution.  While you always have the option of thoroughly rinsing then thoroughly drying (with the disadvantage of this discussed above), it is generally more desirable to do the following: thoroughly rinse the first solution in toluene, then rinse that with a solvent that is miscible with both toluene and water, like acetone. After the acetone rinse, then you can rinse the acetone with Di water.

If you have a reasonably limited number of solvents, it is good to keep a set of glassware dedicated to each solvent matrix. In my experience, the best procedure, and in fact it was part of an SOP for glassware cleaning, was to keep keep the glassware soaking in the solvent for which it will be used, rather than letting it dry.  Obviously this depends on the use case and there are certainly times that it is necessary to thoroughly dry your glassware. But where that is not necessary, to the degree reasonable, I recommend keeping the glassware separated based on what you use it for (i.e. solvent type), and keeping it soaking in that particular solvent.


How can Nernst equation apply in this redox reaction? $$\ce{CuSO4 + Zn -> ZnSO4 + Cu}$$
What is the cell potential of the voltaic cell with copper and zinc electrodes if the system is at 50 degrees and the solutions are both at 1 mol? How can it be equal to the standard potential from the nernst equation? $$E_\text{cell} = E^\circ – \frac{RT}{nF}\ln Q$$

If the solutions are at 1 mol concentrations each then $Q = 1$ and $\ln Q = 0$ resulting in $E_\text{cell} = E^\circ$.
But the above basically shows that concentration affects cell voltage but temperature doesn't have an effect on cell potential – which isn't the case. How can I calculate how different temperatures affect the cell voltage?

The flaw in your reasoning is that you are believing that $E^\circ$ is a constant, when really it is a function of temperature.

See Standard Electrode Potentials and Temperature Coefficients in Water at 298.15K for a linear approximation of the temperature dependence of $E^\circ$.


Why don't beryllium and magnesium dissolve in ammonia? From my textbook:

The [alkali and] alkaline earth metals dissolve in liquid ammonia to give deep blue solutions forming ammoniated ions. $$\text{M}+(x+y)\text{NH}_3 \rightarrow [\text{M}(\text{NH}_3)_x]^{2+} + 2[e(\text{NH}_3)_y]^-$$

However, my teacher has told me beryllium and magnesium are anomalous in that they don't dissolve in ammonia solution. I couldn't find a reference for this claim in JD Lee's Concise Inorganic Chemistry, Fifth Edition.
I would like to know two things:

What determines a metal's solubility in liquid ammonia?
Why don't beryllium and magnesium dissolve in ammonia?

Magnesium and beryllium are not really anomalous.  Their poor solubility or nonsolubility in ammonia can be correlated with the fact that their ions are more densely charged than those of the alkali and heavier alkaline-earth metals that do readily give solvated electrons in liquid ammonia.  However, at least for magnesium solvated-electron solutions can be formed in some other solvents or, in ammonia, with an electrolytic process.
Let's look more closely at magnesium, for which much more is known experimentally about solubility and solvated-electron generation than for beryllium.  Compared with lithium, calcium and the elements below these, the greater charge density of the magnesium ion (and by extension, the beryllium ion) inhibits solvated electron formation in at least two ways.

The metal gets passivated.  Solvated-electron solutions of metals in ammonia are not completely pure metal-ammonia solutions.  The highly electropositive metals will slowly react to form the amide and hydrogen, but the amide ion in solution stabilizes the solvated electrons and thus also slows the reduction reaction.  But with the doubly charged and compact magnesium ion, the hard-base amide ion form a salt with little solubility.  This inhibits metal dissolution rather than stabilizing solvated electrons.

What does dissolve, reacts faster.  With its compact size and double charge, magnesium ion polarizes the ammonia molecules and tends to displace protons.  These then react (as solvated ammonium ions) with the solvated electrons.

IUPAC has published a review by Lagowski [1] which includes a discussion of metal solubilities in ammonia solvent.  Magnesium is considered "soluble" in their table, but the text indicates that this is a borderline case and other solutes, such as ammono-bases, are needed to get a stable solvated-electron solution.  Magnesium is shown with a higher ion charge density than any of the clearly soluble metals, including the lanthanides europium and ytterbium.
Thus compared with metals that form those blue solutions, magnesium dissolves more slowly and reacts faster, and so few solvated electrons can accumulate in a proposed magnesium/ammonia solution.
Breaking through
Magnesium is certainly electropositive enough to generate solvated electrons.  Xu and Lerner [2] report that magnesium along with its heavier congeners calcium, strontium and barium generate solvated electrons in ethylenediamine, these electrons being captured by graphite to form blue or green intercalation compounds.  (Curiously, magnesium and barium appear to adopt a +1 oxidation state while the intervening metals go to +2.)  Significantly, magnesium requires more aggressive conditions of higher temperature and longer time to react, a difference the authors attribute to the effects of passivation (correlated with compact ion size) and compare with lithium versus other alkali metals.
More relevant to dissolution in ammonia is the electrolytic process described by Combellas et al. [3].  In this process, the magnesium is dissolved electrolytically into a magnesium salt solution, generating reasonably stable solvated-electron solutions.  Such solutions find use in reductive activating PTFE surfaces without undesirable carbonization that alkali metals tend to produce; and since the process is electrolytic it can be controlled by way of the electric current.  The electrolytic process breaks both barriers cited above.  With the polarization the magnesium dissolves much faster than it would as an isolated metal, and observations of solvated electron lifetimes made by the authors indicate that the dissolved salt is stabilizing the blue solution against hydrogen evolution.  (The authors do not give a solution-chemistry explanation for this, but the observations are consistent with the magnesium being incorporated into singly-charged ion pairs rather than remaining as independent +2 ions.)
References
1.
J.J.Lagowski, "Solution Phenomena in Liquid Ammonia", http://publications.iupac.org/pac/25/2/0429/pdf/index.html.
2.
W. Xu and M. M. Lerner, "A New and Facile Route Using Electride Solutions To Intercalate Alkaline Earth Ions into Graphite", Chemistry of Materials 2018 30 (19), 6930-6935
DOI: 10.1021/acs.chemmater.8b03421
3.
C. Combellas, F. Kanoufi, A. Thiébault, "Solutions of solvated electrons in liquid ammonia: Part 1. Chemical properties of magnesium solutions", Journal of Electroanalytical Chemistry ,
2001 499, 1, Pages 144-151.   ISSN 1572-6657,
DOI:  10.1016/S0022-0728(00)00504-0


Does this equation - involving calcium carbonate and a lead (II) cation - balance? $$\ce{CaCO3 + 2H2O + Pb^{2+} -> Pb(CO3) + Ca(OH)2 + H2}$$

Does this equation balance?
On the right you have lead with 2+ charge, whereas on the right its completely neutral. So my question is do you need to balance the charges in order to make the equation correct?

If that equation is wrong, what is the right equation?

I need to show that hydrogen gas is formed (or some other gas but I don't know what else it can be), that the solution will become more basic (proven with $\ce{Ca(OH)2}$), and that the lead has to replace the calcium.

Unfortunately your equation does not add up, let me briefly explain why.

Calcium carbonate is a not very soluble salt. There is ionic binding between the calcium cation and the molecular carbonate anion. If you dissolve it in water you, the following equilibrium will establish: $$\ce{CaCO3~(s) <=>[\ce{H2O}] Ca^{2+}~(aq) + CO3^{2-}~(aq)}$$

Lead carbonate is also a salt, it is even less soluble. Only a little bit will dissolve according to the following equilibrium:$$\ce{PbCO3~(s) <=>[\ce{H2O}] Pb^{2+}~(aq) + CO3^{2-}~(aq)}$$

Calcium hydroxide is also a salt. It is a little bit more soluble than calcium carbonate. The following equilibrium is established when put in water:$$\ce{Ca(OH)2~(s) <=>[\ce{H2O}] Ca^{2+}~(aq) + 2{}^{-}OH~(aq)}$$

You write the following:

  $$\ce{CaCO3 + 2H2O + Pb^{2+} -> Pb(CO3) + Ca(OH)2 + H2}$$

So in principle you have $\ce{Ca^{2+}}$, $\ce{Pb^{2+}}$ and $\ce{CO3^{2-}}$ on both sides of the equation. What is left is not a balanced equation, since there are two electrons missing:
$$\ce{2H2O ~\nrightarrow 2{}^{-}OH + H2}$$

You need a source for the electrons, one way of balancing the equation is
$$\ce{CaCO3 + 2H2O + Pb -> Pb(CO3) + Ca(OH)2 + H2}$$

Given the insolubilities of all the species involved, it might be hard to see any reaction at all. And I have not checked if the redox reaction is favourable, with potentials.


Do reactions carried out in light always follow free radical mechanisms? If light is used in a reaction, is it necessarily a free radical reaction? Or are there some other reactions as well which involve light but don't follow a free radical mechanism?

Often, yes; but sometimes not, and generalisations like that often don't work in organic chemistry.
It's a better idea to understand why there's a correlation between "light" and "radical reactions". One way of generating radicals is by breaking bonds, and one way of breaking bonds is to use light. Of course, this only really works for relatively weak bonds, such as the Br–Br bond, or basically any kind of bond between Group 16 and/or Group 17 elements (O–O, N–Cl, ...). Many of the classic radical reactions involve this kind of pattern, and this is probably why you're drawing this link.* Examples include: free-radical halogenation, the Hofmann–Löffler–Freytag reaction, and the Barton reaction.
Alternatively, if there aren't any weak bonds to break, it could be that you're trying to use light to promote an electron to an excited state. Now, because the vast majority of organic chemicals are closed-shell molecules (all electrons are paired), when you excite something you're going to get a radical species. A simple example is the Norrish type II reaction. This radical generation can be indirect as well, occurring via excitation of a photocatalyst, which then oxidises / reduces an organic substrate to form an organic radical. This is the idea behind photoredox catalysis, which has been a very, very hot topic in recent years.
It is not wise to make a sweeping generalisation, though. You could excite a molecule and have it react immediately, without generating an intermediate radical species. This is the case in photochemical electrocyclic reactions, for example. And there are many photoisomerisations which (as far as I am aware) occur in a single step, without going through a true radical pathway. A very important one is the cis-trans photoisomerisation of retinal. The mechanisms of these reactions are not well-described by the curly arrows of organic chemistry: you'd need to learn some photochemistry to get a good idea of how these work.
And not every chemical relevant to organic chemistry is closed-shell. Dioxygen, O2, is a diradical species; and when you excite it, you get an excited state where the electrons are paired up, usually referred to as singlet oxygen.† Singlet oxygen reacts with other organic molecules in a decidedly non-radical fashion: for example, it does classic Diels–Alder reactions and ene reactions.

* Note that many of these can be also accomplished by simple heating, which can break weak bonds too.
† Usually dioxygen is not excited directly. A photosensitiser like rose bengal is used.


How do I analyze 17 different gases from a sample? I am designing a test setup in which I heat a battery until it releases its gases, and I need to analyze the composition of the fumes. Below is the list of gases that are believed to be present in the battery's fumes. I was planning on just using a bunch of gas detection tubes, but they don't seem to be available for all these gases. I was thinking about moving on to buying a gas chromatograph instead since, so far, it seems cheaper than getting a bunch of tests, but I have never used a GC, so I don't know if this would even work for so many gases.
If I were to go the GC route, what would I need to learn and buy in order to test for these gases? Is there a handbook that demonstrates the proper setups for different gases since I think you need to change some physical components for proper testing of different gases (I didn't find any)? Or is there a better method to test for this?

You could use a permanent gas analyzer that has an FID channel for hydrocarbons and CO2, with a TCD detector for H2 and HF and a third MS channel if needed. Basically, this would mean you split your sample into three different columns in the oven to analyze them for different things. I agree with Karl and don't reinvent the wheel. I've done the hydrocarbons (C1-12) and CO2 and permanent gases using a Thermo Trace 1310 with an FID and TCD detector and would recommend that for ease of use. Then I would probably add an MS detector to it in case you need it.
How are you hoping to get the battery gases into the GC? I think you could fit a battery into a headspace vial and heat it up to release the gases then directly inject the headspace into the GC.


Oxidation numbers of phosphorus pentachloride I read detailed explanations of what oxidation numbers/ states represent and it boils down simply to this: If it's positive, it shows how many electrons the element lost, and in the case of a negative oxidation number, it shows how many electrons the element gained. That I understand.

What I don't get is, in Phosphorus pentachloride $\ce{PCl5}$, phosphorus forms 5 single bonds with chlorine atoms and 'gains' $5$ more electrons to accommodate a total of $10$ electrons. Shouldn't its oxidation number be $-5$ instead of $+5$?

When considering the oxidation state of covalent compounds, we need to pretend they are ionic. When we do so, we assign all of the electrons to the more electronegative element. The electronegativity of phosphorous using the Pauling Scale is 2.19 and the electronegativity of chlorine is 3.16. Since chlorine is the more electronegative element, we assign all electrons in the bonds to the chlorine atoms. Thus $\ce{PCl5}$ is treated as the fictitious ionic compound comprised of $\ce{P^{5+}}$ ions and $\ce{Cl-}$ ions.


Will the addition of salt further cool down ice that is already at -30 °C? I know that ice gets colder when salt is added. But how big is the effect if it is already −30 °C?

Can it cool it down more or does it just work with −1 °C cold ice?

Another question: If I have a water cooling circuit, does adding salt help it cool down?

If you add salt to icy water, it's the water that gets colder. Pure water under atmospheric pressures will be $0~\mathrm{^\circ C}$ - even when the ice melting in it is colder (for example if it's just out of the fridge). When you then add salt to that water, it lowers the freezing point of your water-salt solution, all the way down to $-21~\mathrm{^\circ C}$ once it contains $5.2\ \mathrm{mol}$ of NaCl per kg of water. That's the maximum amount of salt this solution can contain at that temperature. (Find the thermophysical data at NIST (page 30).

Now, if you add salt to liquid water (containing no ice), you will actually decrease its temperature slightly because of the energy required to solvate the salt in the water. This won't help you in a 'water cooling circuit' because you also lower the heat capacity.


Role of methanol in NaBH4 reduction In reduction of carbonyls (aldehyde and ketones) to alcohols using NaBH4/Methanol, alkoxides are formed with the Boron which usually gets protonated on addition of water/acid.

My question is what is the function of methanol in this reaction as methanol reacts with NaBH4 to form sodiumtetramethoxyborate and hydrogen gas.

Some sources say methanol protonates the alkoxide ion and after doing series of reduction experiments I find that the alkoxide formed only gets protonated to an alcohol on addition of water.

So what does methanol do in this reaction/what is the effect of methanol?

I believe methanol does protonate the alkoxide. The resulting methoxide coordinates to $\ce{BH3}$ forming $\ce{(MeO)BH3-}.$ The electron donating character of the methoxy group makes the remaining protons more hydride which results in a more reactive hydride source.


Why do cyclic ethers have higher boiling points than their acyclic isomers? TL;DR version is the question title. Some context and data follow.

I was creating an assignment for my organic chemistry students in which they would need to draw as many isomers as they could from a simple formula, for example $\ce{C3H6O}$ (which fits 7 structures - 9 if you count minor enol tautomers and 11 if your count stereo). In this set (ignoring stereo and enols) there are:

  Two alcohols

Cyclopropanol - boiling point $=101-102\ ^\circ\text{C}$
Allyl alcohol - boiling point $=97\ ^\circ\text{C}$

  Two carbonyl compounds

Propanal - boiling point $=46\ ^\circ\text{C}$
Acetone - boiling point $=56\ ^\circ\text{C}$

  Three ethers

Propylene oxide - boiling point $=34\ ^\circ\text{C}$
Oxetane - boiling point
$=49-50\ ^\circ\text{C}$
Methyl vinyl ether - boiling point $=6\ ^\circ\text{C}$

The boiling point of methyl vinyl ether immediately stood out as anomalous. The fully saturated ethyl methyl ether has a boiling point of $7.4\ ^\circ\text{C}$.

Here are some more data comparing cyclic ethers to their acyclic isomers (and some analogs). The gap appears to be closing as the number of carbons increases.

  Four carbons

Tetrahydrofuran - boiling point $=66\ ^\circ\text{C}$
1-Butene oxide - boiling point $=61-65\ ^\circ\text{C}$
Ethyl vinyl ether - boiling point $=33\ ^\circ\text{C}$
Allyl methyl ether - boiling point $=42-43\ ^\circ\text{C}$
Diethyl ether - boiling point $=34.6\ ^\circ\text{C}$
Methyl propyl ether - boiling point $=38.8\ ^\circ\text{C}$

  Five carbons

Tetrahydropyran - boiling point $=88\ ^\circ\text{C}$
2-Methyltetrahydrofuran - boiling point $=80.3\ ^\circ\text{C}$
Allyl ethyl ether - boiling point $=65-66\ ^\circ\text{C}$
Butyl methyl ether - boiling point $=70-71\ ^\circ\text{C}$

The data I could find suggests that cyclic ethers have higher dipole moments than their acyclic counterparts.

$$\mathbf{Four~carbons}$$
\begin{array}{c @{} c} \hline
\text{THF} & \mathrm{1.63~ D ^1, 1.75~ D ^2} \\
\text{1-butene oxide} & \mathrm{1.89~ D^2} \\
\text{ethyl vinyl ether} & \mathrm{1.26~ D ^2} \\
\text{diethyl ether} & \mathrm{1.15 ~D ^3, 1.15 D ^2} \\
\text{methyl propyl ether} & \mathrm{1.11 ~D ^2} \\\hline
\end{array}

$$\mathbf{Five~carbons}$$
\begin{array}{c @{} c} \hline
\text{Tetrahydropyran} & \mathrm{1.58~D ^2, 1.87~D ^4} \\
\text{2-Methyltetrahydrofuran} & \mathrm{1.38~ D ^5} \\
\text{dipropyl ether} & \mathrm{1.21~D^2, 1.00~D^6} \\\hline
\end{array}

This seems consistent with intuitive expectations based on conformational models where, in the cyclic ethers where the molecular geometry is constrained by the ring, the lone pairs are pointing in the opposite direction from the carbon skeleton. Whereas the lone pairs and carbon chains in the conformationally more mobile acyclic isomers are less geometrically fixed and therefore less "directed" in space.

It seems likely that the higher dipole moments in the cyclic compounds would lead to greater alignment / ordering in the liquid phase, which in turn would lead to higher boiling points.

References:

  1: http://en.wikipedia.org/wiki/Tetrahydrofuran
  2: https://physicalchemistryrosamonte.wordpress.com/material-balances/material-balances-on-a-crystallizer/physical-properties-of-pure-methanol/dipole-moment/
  3: http://en.wikipedia.org/wiki/Diethyl_ether
  4: http://www.drugfuture.com/chemdata/tetrahydropyran.html
  5: http://www.stenutz.eu/chem/solv28.php?s=1&p=20
  6: https://books.google.com/books?id=G6jaBwAAQBAJ&pg=PA50&lpg=PA50&dq=propyl%20ethyl%20ether%20dipole%20moment&source=bl&ots=VA52gqJ0kn&sig=GM8tua_QpXccFW3VXpy9qiNf7rk&hl=en&sa=X&ei=rzRvVZ__Cc32yQT9gIPACQ&ved=0CDsQ6AEwBQ#v=onepage&q=propyl%20ethyl%20ether%20dipole%20moment&f=false


Making artificial sweat I'm currently working on a chemistry investigation wherein I need to dilute a compound with artificial sweat. I have looked at various forums, including this, which gave answers, but is very outdated.

I have also checked on ISOs standards, but it seems that I cannot get the information I want - ISO 105-E04 - without paying a huge sum.

Do any of you know of a recipe for artificial sweat that I can easily conjure in a lab, that's also backed up by various sources?

Thanks

P.S. I cannot just use water + salt, as that would be too simple and cause too many uncertainties for my experiment.

The international standard ISO 105-E01:2013 Textiles – Tests for colour fastness – Part E01: Colour fastness to water includes specifications for two different solutions:

Alkaline solution, freshly prepared, using grade 3 water complying with ISO 3696, containing, per litre:

0.5 g of ʟ-histidine monohydrochloride monohydrate ($\ce{C6H9O2N3.HCl.H2O}$);
5 g of sodium chloride ($\ce{NaCl}$);
and either
5 g of disodium hydrogen orthophosphate dodecahydrate ($\ce{Na2HPO4.12H2O}$)
or
2.5 g of disodium hydrogen orthophosphate dihydrate ($\ce{Na2HPO4.2H2O}$).

The solution is brought to pH 8 (± 0.2) with 0.1 mol/l sodium hydroxide solution.

Acid solution, freshly prepared, using grade 3 water complying with ISO 3696, containing, per litre:

0.5 g of ʟ-histidine monohydrochloride monohydrate ($\ce{C6H9O2N3.HCl.H2O}$);
5 g of sodium chloride ($\ce{NaCl}$);
2.2 g of sodium dihydrogen orthophosphate dihydrate ($\ce{NaH2PO4.2H2O}$).

The solution is brought to pH 5.5 (± 0.2) with 0.1 mol/l sodium hydroxide solution.


How to make by dilution a certain ion concentration The question I'm trying to answer is this:

  What volume of $0.25\ \mathrm{mol/l}$ calcium nitrate is required to make by dilution with water $500\ \mathrm{cm^3}$ of a solution with a nitrate ion concentration of $0.1\ \mathrm{mol/l}$?

My attempt:
Using $n=cv$, we find the the number of moles of nitrate ions to be $0.05$ moles. Now from the chemical formula of calcium nitrate, we know that $1$ mole of calcium ions gives two moles of nitrate ions. Therefore $0.05$ moles of nitrate ions gives us $0.025$ moles of calcium ions. Now, using $v=\frac{n}{c}$, where $n$ is the no. of moles of calcium ions, we obtain the volume of $0.1$ litres.

My attempt actually managed to give me the correct but I don't understand a part of it; I can't understand why when I'm trying to find the volume of calcium nitrate I have to use the number of moles of calcium ions and not the number of moles calcium ions added to the number of moles of the nitrate ions.

(Note: I asked this question before, deleted it, changed it and re-published it as of now.)

When you use the number of moles of calcium ions, you are actually using the number of moles of calcium nitrate. You have done the 1:2 conversion already in your question. You found the amount of nitrate ions needed to be 0.05. But then instead of converting that amount to moles of calcium ions, you were actually converting it to a number of moles of calcium nitrate, since the ratio of moles calcium ($\ce{Ca}$) to moles calcium nitrate ($\ce{Ca(NO3)2}$) is 1:1.


Haloform Reaction of Furfural My instructor pointed out that furfural undergoes haloform reaction - I don't understand how. I couldn't figure out the mechanism on my own, and couldn't find much online either. (regarding the products, or the mechanism) Could someone please help me out?

As far as I know, only methyl ketones and alcohols  oxidisable to the same undergo haloform reaction - so something unexpected must be happening here.

It'd be great if you could provide a detailed mechanism and explanation, and if possible provide sources.

Aldehydes can be oxidized by bromine under acidic conditions per the mechanism is given below.  The haloform reaction, however, occurs under basic conditions and requires a methyl ketone.  Your professor probably forgot that the haloform is for methyl ketones in basic conditions and called the reaction below a haloform even though it isn't. Given the similarity in the oxidation of the carbonyl group to a carboxylate with a halogen, (and the thousands of organic reactions to remember,) it seems like a simple easy mistake.


Formulae for energy of photon (E=hf and E=hv) I have seen the energy of a photon given by the formulas:

$$E = h \cdot f \tag{1}$$
Where $E$ = energy of the photon, $h$ = Planck's constant, $f$ = frequency of radiation (Source: BBC article)

I've also seen it given as

$$E= h \cdot \nu \tag{2}$$
Where $\nu$ stands for frequency (Source: Wikipedia article)

But in this topic I've also seen

$$f=\nu/\lambda \tag{3}$$
Where $\nu$ = speed, $\lambda$ = wavelength (Source: Simple English Wikipedia article)

Is there any difference between equation 1 and equation 2? And is equation 3 unrelated?

I understand your confusion. They all are actually the same formulas. It is because lets start with the first equation,

$$E=h \cdot f$$

But we know that frequency can be written in terms of speed formula,
$$ v = f \cdot \lambda$$

Which can be rearranged into the following,
$$ \frac{v}{\lambda} = f $$

If you put that expression into the above equation, then that becomes

$$ E = f \cdot \frac{v}{\lambda}$$

Hence the same formula can be written in many ways.

Your first two equations are just the same. $f$ usually stands for frequency and in the second equation you have just used $\nu$ for it. That is the frequency of the photon that is coming in or being released.

Whereas your third equation actually comes from the speed formula as I have written above,
$$ v = f \cdot \lambda$$

Rearranging for frequency, you get:
$$ \frac{v}{\lambda} = f $$


Why should the salt being used in salting out process be neutral? When doing salting out it's been said that the salt being used should be neutral, meaning both acid and base which make the salt should be categorized as strong acids or bases. Why is that?

For example, $\ce{NaCl}$ could be used as a salting out salt in principle, because $\ce{NaOH}$ is a strong base and $\ce{HCl}$ is a strong acid but using $\ce{NaCH3COO}$ is not right because $\ce{CH3COOH}$ is not a strong acid.

"Its [sic] been said" is inadequate contextualization. Many organic solutes are potentially capable of being ionized, especially in aqueous solution. Maintaining aqueous media at a neutral pH will for most solutes minimize this (undesired) effect. For instance, in the following reference (note date and distinguish between what is observational (and thus probably "true") and theoretical (based on what is likely to be incorrect chemistry)), read the last sentence on page 531 over to 532. http://www.jbc.org/content/33/3/531.full.pdf


What are the oxidizing and reducing agents in the reaction? What are the oxidizing and reducing agents in the reaction?
  $$\ce{Pb(s) + PbO2(s) + 2H^+ + 2HSO_4^- -> 2PbSO4(s) + 2H2O(l)}$$

I thought that, since the solid lead is going from $0$ to $2+$ it is being oxidized, and in the half reaction it works with the $\ce{HSO4^-}$, so the $\ce{HSO4^-}$ would be the oxidizing agent.

You are correct, the Pb(s) is oxidized, which makes it the reducing agent. However, there is no change in the oxidation state of the H or the SO4−, so it is not the oxidizing agent. Rather the PbO2(s) is the oxidizing agent, as the Pb goes from +4 in the PbO2(s) to +2 in the PbSO4(s).


Why do batteries drain so quickly if they are forced to draw more current than they normally do? When a battery is pushed to use twice the current it normally does, it lasts for less than half as long before dying...
In fact, batteries often come with a 'C' rating that gives you an idea of how long they might last when run at a high current level...
What is going on inside the battery when this occurs?  What is the chemistry, so to speak?

The capacity of a cell is measured toward the terminal cut off voltage. If e.g.a double current is drained from a cell, the external cell voltage is decrease by the doubled voltage drop on the cell internal resistance: $$U_\mathrm{e}=U_0 - R \cdot I$$
That caused reaching the terminal voltage sooner then in the half time.
It is additionally leading to the ohmic internal energy wasting $$P=R \cdot I^2$$
But this energy loss affects the useful energy output, not the capacity ( directly).
Additionally, higher than optimal current causes lowering of the immediate cell open voltage. Limited rate of diffusion of electroactive components and/or of electrode reactions are not able to maintain electrode equilibrium potentials.
In result, even the $U_0$ (non linearly) decreases with the higher current and the terminal $U_e$ is reached after providing smaller total charge.


Is aluminium amphoteric? In my book it says that aluminium can react with water as a base when it's in the form $\ce{[Al(OH)2(OH2)4]^+}$.

And this way as an acid: $\ce{[Al(OH2)_6]^3+}$.

So ... does it depend on with which ion it's accompanied by? Let's say, $\ce{AlCl_3}$ will react as an acid?

Yes, you are right. The reaction depends on with which ion it is accompanied with. Note that these species are in equilibrium and present in the solution  and that particular reaction proceeds according to which ion is present in the solution in concentrated form.

Aluminum hydroxide is formed when soluble salt of $\ce{Al^3+}$ i.e
$\ce{[Al(H2O)6]^3+}$ when it is reacted with base.

$$\ce{[Al(H2O)6]^3+(aq) + 3 OH-(aq) → Al(H2O)3(OH)3(s) + 3 H2O(l)}$$

This insoluble metal hydroxide can act as a base, since it can be
redissolved by reacting it with an acid.
As Brønsted base

$$\ce{Al(H2O)3(OH)3(s) + 3 H3O+(aq) → [Al(H2O)6]^3+(aq) + 3 H2O(l)}$$

Alternatively, the metal hydroxide can act as an acid, since it can
react with a base.
Lewis acid

$$\ce{Al(H2O)3(OH)3(s) + OH-(aq) → [Al(H2O)2(OH)4]-(aq) + H2O(l)}$$

Thus, all aluminium compounds are said to be amphoteric.


PIB wave functions and the effects of e- delocalization on UV absorption in conjugated systems I'll try to keep this short and concise. What made me start thinking about this question was the material we have been covering regarding PIB wave functions in my quantum chemistry class, and material we have been covering regarding ozone absorption in my climate class. I would like to try to understand the relationship between the increased wave function domain that delocalization yields for conjugated pi electrons, and the effect that this has on the wavelengths of light that they can absorb. Like ozone, I understand that most other conjugated pi systems have strong UV absorption, so I would assume there is some correlation there.

Specifically though, I would like to know if this can be explained (at least to some extent) by a simple QM model like the 1D PIB, and if how electrons' wave functions are altered when their spacial domain is extended across the whole molecule by delocalization can explain UV absorption in conjugated pi systems.

You can use the PIB model as a very approximate way of thinking about energy levels in conjugated ($\pi$ electron) molecules. Absorption in triatomics is much more tricky, however, and usually a Walsh diagram is necessary.

The general idea is that as the conjugation gets larger the absorption wavelength for electronic transitions gets smaller and so shifted to the red part of the visible spectrum. (A similar argument applies to particles on a ring for aromatics)

The energy levels of a PIB vary as $E_n \approx (n/L)^2$ where n is the quantum number and L the length of the box, or conjugation. Take this to be sum of all conjugated bond lengths. Thus as L gets larger the energy levels become closer together. Secondly as the number of conjugated bonds increases the number of quantum levels filled also increases, as there are more electrons. The optical transition is from the highest filled level to the next one up (HOMO-LUMO), the energy difference is $E_{n+1}-E_n$ at each value of L. If you take the length of the box to be the number of bonds m of length a, then $L=ma$ it is easy to show that the absorption spectrum shifts to the red as the conjugation increases, although you should definitely not expect accurate values in terms of energies.


Gelatination of solution in saponification When I performed a saponification experiment in school, I added some ethanol to the solution (glycerol ester fatty acid + sodium hydroxide) to change the equilibrium of the reaction. (Molecules of salt of fatty acid made some chunks so they don't react efficiently.) But accidentally I spilled a lot of ethanol and the solution became hard, like a hair gel. And after I added some water to it, it became liquid again.

What I want to know is why the solution became a gel. I know that ethanol is used for an efficient reaction because ethanol can be a common solvent of grease and water and it can help sodium hydroxide and glycerol ester meet, but I don't know why they became harder when ethanol was overused.

Of course, the solution was in the water bath.

Sodium stearate is not very soluble in water (above about 2%, it becomes viscous but not a firm gel). Ethanol can reduce the viscosity at first, but sodium stearate is not soluble in ethanol either, so an excess of ethanol causes a precipitation of solid that firms up the aqueous gel. More water  dissolves that precipitate and restores the fluidity.
The solubility and rheology of fatty acid soap solutions depends on temperature, concentration, solvents and the structure and purity of the fatty acid group. Stearate is very different from oleate, and both are different from laurate.


What is "physicist's water"? There is a statement on page 60 of this dissertation:

  the "physicist's water molecule" ($\ce{O-H = 1.1 \mathrm{\mathring{A}}, \angle HOH = 104^\circ}$)

I understand it to be a model similar to the "spherical cow in a vacuum,", in that it's meant to be an easy-to-remember approximation, but

What is the etymology of this phrase or definition?
Why this specific choice of parameters? It could just as easily be $\pu{1.0 \mathrm{\mathring{A}}}$, according to Wikipedia, and the angle could vary as well.

I suspect the author might refer to "physicist's water molecule" term in order to underline the fact the geometry and energy of this molecule has been introduced as is, and hasn't been affected by intra- or intermolecular interactions.
I believe there is an allusion to Ancient Greece's philosophical schools which treated natural phenomena such as air, fire, water etc. as the purest and simplest entities. The term "physicist" itself refers to pre-socratic philosophers (Ionians, mostly) which abandoned religious attempts to formalize the reality [1, p. 20]:

... Pre-Socratic philosophers [...] have to be called Physicists because they abandoned the civil religion, whose insistency deified the city’s eponym, the royal function or the legislator’s role, the valiant warrior or the fertile woman, because they disdained what the social projects into the religious, whereas they kept or invented a global religious, that divine that’s immanent to the universe.

For instance, based on what's been considered the most fundamental element, physicist’s water has originally been introduced by Thales of Miletus, just as physicist’s fire "belongs" to Heraclitus.
Reference

Serres, M.; Burks, R. Geometry: the third book of foundations; Bloomsbury Academic: London; New York, 2017,. ISBN 978-1-4742-8139-3. (Google Book)


Why is the standard enthalpy of formation of elements in their native forms zero? Why is the standard enthalpy of formation of $\ce{O2}$ and other reference forms of elements equal to zero?

Well, from what I know, bonding releases energy.

Also, does anybody know a source for finding the reference form of an element at standard conditions?

It is purely a matter of definition. A standard enthalpy of formation describes the change in enthalpy during the formation of 1 mol of a target compound by reacting the (pure) elements it consists of, whereby each element is expected to be in its most stable modification for the given temperature. Now take the case of oxygen as an example: At standard conditions $\ce{O2(g)}$ is the most stable modification of oxygen. That means at standard conditions the reaction

$$\ce{2O <=>> O2(g)}$$

is defined in such a way that each of the elemental $\ce{O}$s are defined to be exactly the same state as an oxygen atom that is part of an $\ce{O2(g)}$, i.e. their enthalpy $2H^{\circ}(\ce{O})$ matches exactly that of $\ce{O2(g)}$: $H^{\circ}(\ce{O}) = \frac{1}{2}H^{\circ}(\ce{O2(g)})$. This gives you a reference point for $H^{\circ}(\ce{O})$ which is otherwise nearly impossible to measure much as the Standard Hydrogen Electrode is used to set up a reference system for the electrode potential.

A table showing some standard enthalpies of formation can be found here. There it is also indicated which phases are used as reference form.


Flame test: Is the metal atom or the metal ion responsible for the flame colour? A sample of $\ce{CaCl2}$ is placed in the flame of a Bunsen burner. The component of the substance which is responsible for the production of the flame colour is:

$\ce{Ca^2+}$
$\ce{Ca}$
$\ce{Cl-}$
$\ce{Cl}$

Why is the textbook answer $\ce{Ca}$ and not $\ce{Ca^2+}$ ?

The phenomenon that is witnessed during a flame test is an "atomic emission".
This statement may seem inappropriate, since it is a solution of metal ions (and not atoms) that is tested. The reason for calling it atomic emission lies in the process occurring in the flame. One of the steps of the process is an atomization step. That is, the flame converts the metal ions into atoms. When a solution of sodium chloride is placed in a flame, for example, the solvent evaporates, leaving behind solid crystalline sodium chloride. This evaporation is then followed by the dissociation of the sodium chloride crystals into individual ground state atoms - a process that is termed atomization. Thus sodium atoms are actually present in the flame at this point rather than sodium ions, and the process of light emission actually involves these atoms rather than the ions.


Is there a difference between equilibrium and steady state? The term equilibrium is used in the context of reversible reactions that reach a point where concentrations no longer change. The term steady-state is used in enzyme kinetics when the concentration of the enzyme-substrate complex no longer changes (or hardly changes, in case of a quasi steady state). It is also used to describe multi-step biochemical pathways. Is there a difference between the two, given that both concern a situation where concentrations don't change over time?

Yes, equilibrium and steady-state are distinct concepts.

A reaction is at equilibrium if reactants and products are both present, the forward and reverse rates are equal and the concentrations don't change over time. If this is the only reaction in a closed, isolated system, the entropy in the system is constant.

Steady-state implies a system that is not at equilibrium (entropy increases). A species is said to be at steady state when the rate of reactions (or more general, processes) that form the species is equal to the rate of reactions (or processes) that remove the species.

In both cases, there are rates ($\mathrm{rate}_1$ and $\mathrm{rate}_2$) that are equal. For an equilibrium, the forward and reverse rate of the same reaction are equal to each other. For a steady state, the rates of processes leading to increase of the concentration of a species are equal to the rates of processes leading to decrease of the concentration of the same species.

$$\ce{A <=>[rate_1][rate_2] B}\ \ \ \ \    vs \ \ \ \ \  \ce{source->[rate_1]C->[rate_2]sink}   $$

For an equilibrium, all concentrations are constant over time. For a steady-state, there is a net reaction, so some amounts change (the amount of source and sink), while at least one species - the one at steady state - has a constant concentration as long as the conditions of steady state prevail.


Recrystallization and purification techniques for getting a pure sample Could someone explain in detail as to why exactly recrystallization gives a more pure solid sample? As well as why is would be better 'technique' to allow for the crystal to form slowly?

I was thinking that it would be because being a purification technique it essentially breaks the lattice and allows the structure to reform again. As for why it needs to be done slowly, well, because rapid changes in heat would cause the structure to reform into the impure solid it was.

I'm looking for a more detailed, better expressed, answer to this.

Recrystallization is a purification method because it is a slow selective generation of a crystal framework, (mostly) free of trapped impurities. It is only effective as a purification method if done properly. Poor recrystallization techniques; rapid cooling, dramatic changes in solubility with temperature, bad choice of solvent, over evaporation of solvent, can all produce crystals of low purity which are not much better than the starting crop.

Impurities in crystals can occur via three mechanisms;

inclusion: impurities occupy a lattice site within the crystal framework; that is, they become part of a defective crystal unit themselves
occlusion: impurities become physically trapped within the crystal framework. This can be a common occurence for some solvent molecules
adsorption: impurities form a weakly bound interaction to the crystal surface.

The faster the crystallization takes place, the greater the chances these mechanisms have of occurring.  Crystals are specific odd-shaped units that need to pack in nicely to their neighbours to form a good crystal. This takes time. A new molecule may attach to the growing lattice, but if it is not the right fit (such as because it is an impurity), it will leave in favour of a better suited fit attaching to the lattice. I may be showing my age here, but we used to liken crystal growth to a game of Tetris. Heck you can still play it! Done slowly, there is time to rotate all the pieces and get the perfect fit. As you speed things up, it becomes harder to get the right fit, holes are left, and before long the whole crystal goes to the dogs. That is essentially what is happening with rapid crystal growth. It's why you get smaller crystals and greater levels of impurities.

Of course, there other methods for purifying solids which are effective also, if used correctly. Precipitation, sublimation and fractional crystallization are all methods for purifying solids, which can give high levels of purity, matching purities obtained be recrystallization.


How to deal with the unit "Coulomb per equivalent"? In a question that I came across they defined $1\,\pu{F}$ as 96500 coulomb per equivalent. Now as far as my knowledge is concerned $1\,\pu{F} = 96500\,\pu{C/mol}$. The question involved asks to calculate the current that needs to be passed (or may be a function of that) wherein I mistook the $n$ factor and the answer was just half of what it was supposed to be. However that taking $1\,\pu{F}$ as "coulomb per equivalent" would lead me to a antoher unit (equivalent per mole) in order to get the desired unit in the question.

In this context I would like to:

a)  Get an intuitive explanation into the number of equivalents (just like moles is Avogadro number of substances).

b)  Know what the unit "equivalents/mole" means (is it the same thing as $n$ factor?) How does the $n$ factor treat number of moles and equivalents so differently?

For an electrochemical reaction, you count the atoms / ions by mol, and use the coulomb as a counting unit of charge. For a more intuitive explaination of the $n$ factor in the Faraday equation, try this analogy:

The summer olympics include swimming in a pool with lanes $50\,\pu{m}$ long.  Among the typical competitions are runs about $50$, and $100\,\pu{m}$.  Now, for the sake of the analogy, think the swimmers were representatives of ions.  Each of the swimmers could represent

a $\ce{Cl-}$ to yield $\ce{Cl}$, a change by one unit of charge.  So they swimm only one lane in one competition, $n = 1$; the single discharge happens when the swimmers clap the wall with their hand.
a $\ce{Cu^{2+}}$ wanting to yield $\ce{Cu}$. As they start with two charges now, two discharges in this single race are required ($n = 2$). Their first discharge is when they contact and turn at the opposite wall ($50\,\pu{m}$), their second and final discharge when they clap the other wall with their hand ($100\,\pu{m}$).


What are actually the differences between Sidgwick & Powell theory and Nyholm & Gillespie Theory? The famous VSEPR theory was proposed by first Sidgwick & Powell. It was modified then by Nyholm & Gillespie.

I was reading Inorganic Chemistry by J.D.Lee where he discusses very briefly the theory of Sidgwick & Powell; I couldn't somewhat conceive the language. So, I googled it but to my misfortune I could nowhere get more info other than this:

  The idea of a correlation between molecular geometry and number of valence electrons (both shared and unshared) was originally proposed in 1939 by Ryutaro Tsuchida in Japan, and was _independently presented in a Bakerian Lecture in 1940 by Nevil Sidgwick and Herbert Powell of the University of Oxford. In 1957, Ronald Gillespie and Ronald Sydney Nyholm of University College London refined this concept into a more detailed theory, capable of choosing between various alternative geometries [1]

Everyone would give the historical info that Sidgwick & Powell was the first to introduce the theory which was later modified to the present VSEPR theory. I don't want that!

Can anyone please help me to bring me forward the original Sidgwick-Powell theory? Also the million-euro question: What are the differences between the two theories? In another way to say: What were the modifications that Mr. Nyholm & Gillespie brought forward that "improved the original" theory?

[1]https://en.wikipedia.org/wiki/VSEPR_theory#History

Perhaps you should read the original papers: N.V. Sidgwick and H.M. Powell, "Stereochemical types and valency groups", Proceedings of the Royal Society of London Vol 176, No. 965, pp 153-180 (1940), and R.J. Gillespie and R.S. Nyholm, "Inorganic Stereochemistry", Quarterly Reviews, Chemical Society 11 (4) 339-380 (1957).

To quote out of Gillespie's paper:

  Sidgwick and Powell assumed that lone pairs of electrons and bonding pairs were of equal importance and that they arranged themselves symmetrically so as to minimise the repulsions between them. The ligand(or crystal)- field theory is concerned with the influence of the non-bonding electrons upon the arrangement of the ligands which are assumed to be negative ions or dipoles surrounding a central metal atom. Recently the influence of covalent binding has also been considered. These various theories are in fact closely related and an attempt has been made in this Review to integrate them. It appears that the simple theory of electron-pair repulsions receives considerable justification from quantum-mechanics and when refined to take into account differences between the repulsions between lone pairs and bonding pairs of electrons provides a very satisfactory theory of the stereochemistry of inorganic molecules.

That should get you going...


do candles burn wood varnish fumes? A friend of mine bought a piece of furniture that had a super smelly wood varnish, so I read around online and someone said to light candles to burn the chemical fumes.  Does anybody know if there is a chemical reaction between the wood varnish fumes and the candles? they are unscented candles and I assume they can break down the hydrocarbons in the wood varnish fumes.  Are there any other harmful chemical in the wood varnish fumes that are not oxidized by the candles that cannot be smelled by humans?

Smelly compounds can be destroyed by free radicals wich are formed by a burning candle. But you need a lot of candles for your apartment.
Candles were also used in bathrooms (after toilet visits) to destroy smelly compounds like $\ce{H2S, CH3-SH, CH3-S-CH3}$ ...

A better and safer way is it too use cyclodextrin. This compound absorbs molecules and destroys them as well. Cyclodextrin is the working stuff inside the air-improvement sprays.


How are these compounds strong bases, when they are sparingly soluble in water? How is it possible that some insoluble compounds in water, e.g. $\ce{Mg(OH)2}$, $\ce{Ca(OH)2}$, $\ce{Sr(OH)2}$ are considered strong bases?

Are they even strong bases? How can I tell if a compound is a strong base or not?

Is it possible to explain with equations and/or formulas what happens when these compounds are dissolved into water?

The typical definition of strong base that you will see at an introductory level is that there is complete dissociation to hydroxide ions, for example that NaOH completely dissociates into Na+ and OH-.

This is a terrible definition because reactions never go to completion.  The definition contradicts the concept of chemical equilibrium.

In reality:

$\frac{[Na+(aq)][OH-(aq)]}{[NaOH(aq)]} = K_B = ~4M$ (from table discussed below)

So even dissolved aqueous NaOH is not all dissociated.  It can exist as ion pairs or molecules in solution without being dissociated.

$K_B$ is quantification of how strong or weak the base is.  Strong base could be defined as $K_B$ exceeding a certain value. Let us say $K_B>1$ and therefore $pK_B<0$

So the strength of the base only refers to the equilibrium between the undissoicated dissolved species and the dissociated ions.  It is independent of how much or how little solid dissolves. Some hydroxides, like NaOH and KOH have high solubilities and some have low solubilities, but $K_B$ is a different equilibrium.

For quantitative data on the $K_B$ values of hydroxides, see:
DISSOCIATION CONSTANTS OF INORGANIC ACIDS AND BASES IN AQUEOUS SOLUTION
Look up the corresponding metal ion alphabetically, and a $pK_B$ value is given.

However, keep in mind that for each of $\ce{Mg(OH)2}$, $\ce{Ca(OH)2}$, $\ce{Sr(OH)2}$ there will be two $K_B$ values, since there are two hydroxides that can dissociate.

The $pK_B$ values for the second hydroxide dissociating are Mg 2.6, Ca 1.4, Sr 0.8, Ba 0.6 becoming stronger bases as you go down the column in the periodic table, increasing radius stablizing the charge of the ion.  The first $pK_B$ would be even lower, but I don't see any quantitative values.  It would be so low that it would be difficult to measure.  So, yes these are strong bases.

As far as how to tell, other than experimentally, how strong a base of a hydroxide is, there is a good lecture The Acid-Base Character of Oxides and Hydroxides in Aqueous Solution by Pilkington.  As you go from left to right across the periodic table (excluding the noble gases), the hydroxides progress from being strong bases to being acids.


LogP of halogen derivatives of aromatics and aliphatics explanation Benzene logP = 2.13 (logP=log([benzene in octan-1-ol]/[benzene in water]))
Fluorobenzene logP = 2.27 (slight increase, slightly more hydrophobic)
Chlorobenzene logP = 2.84
Bromobenzene logP = 2.99

So as we go down group 7, logP or hydrophobicity increases. What can explain these trends? Furthermore on aliphatic compounds, we observe this trend which confuses matters more.

-Fluorine DECREASES logP slightly (by -0.17) on an aliphatic compound. The book doesn't give a reference as to which alpiphatic compound this is and which hydrogen is substituted, but oh well.
-Chlorine increases logP (by 0.39) on the same aliphatic compound.
-Bromine increases logP even more (by 0.60) on the same aliphatic compound.

So a few questions:
1. Why does a fluorine substitution decrease logP slightly on the aliphatic but increase slightly on the aromatic?
2. Why do chlorine and bromine substitutions increase logP in aliphatics and slightly more in aromatics.

Answers to these trends would be greatly appreciated, thanks.

Here is a chart of the solubility of the species in water and the log of the partition coefficient you noted.

                           solubility       solubility
                  MW       g/100ml water    moles/liter   log(P)
  Benzene         78.11    0.179 (1)        0.0229         2.13
  Flororbenzene   96.103   0.155 (2)        0.0161         2.27
  Cholorbenzene  112.56    0.050 (2)        0.0044         2.84
  Bromobenzene   157.01    0.041 (2)        0.0026         2.99

(1) Wikipedia

(2) Cation Complexes of Compounds Containing Carbon-Carbon Double Bonds. VI. The Argentation of Substituted Benzenes
L. J. Andrews, R. M. Keefer
J. Am. Chem. Soc., 1950, 72 (7), pp 3113–3116
DOI: 10.1021/ja01163a085
weblink http://pubs.acs.org/doi/abs/10.1021/ja01163a085

There is a Wikipedia chart in article on Ionic radii.

 Ion     radii (picometers)
  F-      119
  Cl-     167
  Br-     182

The halide on a benzene ring is going to be relatively covalently bound. It definitely won't dissociate to form ions and only very moderately participate in hydrogen bonding.

So as the size of the halide increases the "volume" of the molecule increases. Thus the molecule would preferentially like to be in the non-polar phase.

Here is a scaled image of fluorobenzene molecule:

and a scaled image of the bromobenzene molecule.

Haloalkanes

The "anomalous" fluoroalkane is explained in a different manner. To get an n-alkane that has a boiling point above room temperature we need to go to n-pentane.

                   BP                 solubility      solubility
                   (deg C)   MW       g/100ml water   millimoles/liter
 n-Pentane         36        72.15    0.040 (1)         0.55
 1-Fluoropentane   63        90.14    1.270 (2)         14.1
 1-Chloropentane  108       106.6     0.197 (3)         1.85
 1-Bromopentane   130       151.05    0.127 (4)         0.838

(1) Wikipedia
(2) NIH
(3) Wikipedia
(4) Handbook of Aqueous Solubility Data, Second Edition
 By Samuel H. Yalkowsky, Yan He, Parijat Jain

There are two competing effects.

(1) The size of atom bound to the alkane.

    Br > Cl > F > H

    Scaled model of Fluoroethane

(2) The second effect is the electronegativity of the X group in the $\text{Alkane}\text{X}$ bond.

 F    >   Cl   >   Br   >   H
 4.0      3.2      3.0      2.2

The fluorine atom has a slightly larger radius in a fluoroalkane than the corresponding C-H group. But the halide groups allow more dipole forces. So by "volume" the overall effect is:

 fluoroalkane
 alkane
 choloroalkane
 bromoalkane


Why can sulfur act like a catalyst? In the reaction of combustion of gunpowder,

  Sulfur is used to speed up the reaction and acts like a catalyst in that it increases the rate of reaction. However, unlike a catalyst, it is used up in the reaction.

I understand that transition metals are often used as catalysts but why can sulfur act like a catalyst?

It's not a catalyst here. It speeds up the reaction, which is why it acts like a catalyst.

Here is the reaction without sulfur:

$\ce{6 KNO3 + C7H8O -> 3 K2CO3 + 4 CO2 + 2 H2O + 3 N2}$

And here it is with sulfur:

$\ce{4 KNO3 + C7H4O + 2 S —> 2 K2S + 4 CO2 + 3 CO + 2 H2O + 2 N2}$

(Source)

Sulfur decreases the ignition temperature, as well as probably contributing a bit to the rate. Here, it is effectively turning $\ce{CO2}$ and $\ce{K2CO3}$ to $\ce{CO}$ and $\ce{K2S}$. If you take into account Le Chatelier's principle, the conversion of the products of the basic reactants speeds it up. (The decrease in concentrations/activities of the products due to their conversion means that the equilibrium shifts in the forward direction to compensate for it.)

So it is speeding up the reaction and making it easier to carry out. Not a catalyst.

A note: There's no reason why any element/compound can't act as a catalyst. Transition metals are generally catalysts in a certain type of reaction, and they generally catalyze the reaction by adsorbing the gases. Catalysts can work in many other ways—generally, they form intermediates in the reaction and present an alternative path via the intermediate. ($\ce{V2O5}$, for example, is a catalyst in many reactions involving organic compounds since it forms cyclic intermediates with various groups).


Why do positive ions go to the cathode? In electrolysis why do positive ions go to the cathode? Since the cathode is positive, won't the positive ions face repulsion at the cathode and would require more energy?

I understand that reduction happens at the cathode and positive ions reduce but somehow it does not make sense as positive - positive charges repel.

TL;DR The cathode in an electrolytic process is considered to be
negative, so there is actually no contradiction. The cathode is a positive electrode in a galvanic cell.

There are different notations for the sign (±) of the cathode used in the literature, which are determined, in particular, by the nature of the process. A very broad definition of a cathode is that it is the electrode of some device connected to the negative pole of the current source.
For electrolysis it is commonly believed that the “−” cathode is the electrode on which the reduction process takes place, and the “+” anode is the one where the oxidation process takes place. When the cell works (for example, during copper refining), an external current source provides an excess of electrons (negative charge) at one of the electrodes — the cathode, where metal is reduced. On the other electrode, there is a lack of electrons and oxidation of metal takes place — this is the anode.
At the same time, during the operation of a galvanic cell (for example, copper–zinc), an excess of electrons (and a negative charge) on one of the electrodes is provided not by an external current source, but by the actual oxidation reaction of the metal (zinc dissolution). Following the definition given, this electrode is an anode. Electrons passing through the external circuit participate in the reduction of copper, so the cathode will be a positive electrode in this case.


Why do some crystal structure files not include bonds? Why do some crystal structure files not include bonds (the atoms are just placed in the respective position in the cell without bonds between any atoms)?
Shouldn't the bonds always be there?
Also, where can I look up these bonds?

If you solve an x-ray crystal structure, you simply get a set of atomic positions. No more, no less.

Often, the atomic positions don't include hydrogen atoms, only the "heavy" atoms.

Thus, most CIF (crystallographic interchange format) files, there are no defined bonds. Somewhere along the way, software perceives bonding information because most chemists prefer to have a valence-bond graph of atoms and bonds (and bond orders).

There are many algorithms for determining a bond graph from a set of atomic positions with varying "accuracy." I put the word in quotes because it's sometimes unclear what the goal should be when determining bond orders. For example, is a molecule in a keto or enol tautomer form? In some cases, the $\ce{C-O}$ bond is not clearly a single bond distance or a $\ce{C=O}$ distance. Maybe there's an ensemble average of both forms.

The standard seems to be if the CIF file looks more "ionic" (e.g., a salt like $\ce{NaCl}$) bonds probably will be ignored, but otherwise, bonds are interpreted from interatomic distances.


Assigning R/S configurations to Fischer projections of 3‐chloro‐2‐hydroxybutanoic acid A question requires you to assign the R/S configuration for 3‐chloro‐2‐hydroxybutanoic acid:

In the answer, while assigning the configuration to carbon #1, apparently the group below is given the third priority, lower than the COOH group above.
Using the CIP rules, doesn't the Cl atom in the bulky group give it a higher priority than the COOH group?

The compound mentioned is (2S,3R)-3-chloro-2-hydroxybutanoic acid. For Carbon atom #2, the CIP priority order is

$\ce{Cl}$
Carbon #1
$\ce{CH3}$
$\ce{H}$

The methyl group is lower priority because Carbon #1 has a hydroxyl group attached to it, which is higher priority than Hydrogen. This implies that the orientation at carbon atom #2 is R.
For more in-depth information, you might want to check out Wikipedia


Can we break the covalent bonds of water molecules using resonance frequencies? Can we break the covalent bonds of water molecules $\ce{H2O}$ using resonance frequencies? What is the wavelegth of the water's molecule? Can we create a vibration like this and use it to separate hydrogen from oxygen?

Yes this is possible. Your question does not state the source of the frequency, but in this internet article they have used laser light.

I am not sure at what frequency you need to irradiate water to break the O-H bond, but I would speculate that a good frequency would be close to the IR stretching frequencies of the O-H bond in water.


Intermolecular forces of attraction between positional isomers of alcohols This seemingly trivial question is as follows: Why is the boiling point of 1-butanol ($\ce {117.7 ^\circ C}$) higher than that of 2-butanol ($\ce {99 ^\circ C}$)?
The only reason I can think of is that the former molecule has a "straighter" chain than the other and a "straighter" chain has a greater surface area allowing for stronger van der Waals forces to exist between the molecules. As molecules in the liquid state do not pack themselves close to each other, ideas of "packing efficiency" would not be useful in explaining these boiling point differences. Perhaps, it is instructive for us to first observe if the trend is persistent for longer chains?
\begin{array}{|c|c|c|c|} \hline
\text {No. of C} & \text{1-alcohol} & \text{2-alcohol} & \text{3-alcohol}\\ \hline
\ce{5} &  138 & 119.3 & 116\\ \hline
\ce{6} & 157 & 140 & 135\\ \hline
\ce{7} & 175 & 159 & 156\\ \hline
\ce{8} & 195 & 178.5 & 173\\ \hline
\ce{9} & 214 & 194 & -\\ \hline
\end{array}
As can be seen from the above table, where the boiling points are all sourced from Wikipedia, with units of $\ce { _^\circ C}$, the trend is indeed persistent. The alcohol isomer with the $\ce {-OH}$ group attached to the 2-position consistently exhibits lower boiling points of around $\ce {16}$ to $\ce {20}$ $\ce {^\circ C}$. Shifting the hydroxyl group further along the chain by one more position also lowers the boiling point of the molecule but to a lesser extent.
The explanation I can come up with for this is the same as aforementioned: By placing the hydroxyl group at the end, there is the maximum surface area for interaction of the alcohol molecules with each other. Furthermore, the hydroxyl groups, being placed at the ends, would cause the least amount of disruption to the hydrophobic interactions between the alkyl chains. Does anyone have better explanations for the above trends?

It does seem that placing the hydroxyl group at the terminal carbon maximises the hydrogen bonding interactions between alcohol molecules and also the van der Waal forces between them. This is best illustrated with a crude diagram I have made:

Finer differences due to the position of the hydroxyl group along the chain, such as the differences between the 2- isomer and the 3- isomer, are not explained here. My explanation does explain the markedly higher boiling point of the 1- isomer.


solve steady state of simple product reaction consider these biochemical reactions (simpler than usual michaelis-menten setup):

$E + S \rightarrow^{k_{f}} ES$

$ES \rightarrow^{k_{r}} E + S$

$E$ and $S$ reversibly make $ES$. the forward/reverse rate constants are equal: $k_{f} = k_{r} = k$. how can you solve for steady state concentration of $[ES]$ given initial amounts $[E]_{0}, [S]_{0}$ and assuming $[ES]_0 = 0$?

$d[ES]/dt = k[E][S] - k[ES] = 0$

rate constant cancels:

$[ES] = [E][S]$

then $[E] = E_{t} - [ES]$ where $E_{t}$ is total enzyme concentration. by substitution:

$[ES] = [S]([E]_{t} - [ES])$

$[ES] = [S][E]_{t} - [ES][S]$

$[ES] + [ES][S] = [S][E]_{t}$

$[ES] = \frac{[S][E]_{t}}{1 + [S]}$

but if $E_{0} = 10, S_{0} = 5$ then $[ES] = (5 * 10) / (5 + 1) \approx 8.3$ but the right answer is closer to 4. what's wrong with this?

There are $3$ equations for $3$ unknowns $[E],[S],[ES]$ given by

$$\left\{\begin{array}{l}
k_f[E][S]=k_r[ES],\\
[E]+[ES]=E_0,\\
[S]+[ES]=S_0.
\end{array}\right.$$

We can use the last two equations to cancel $[E]=E_0-[ES]$ and $[S]=S_0-[ES]$ and plug them into the first equation to get a quadratic equation for $[ES]$ given by

$$k_f(E_0-[ES])(S_0-[ES])=k_r[ES].$$

Then we plug in numbers $k_r/k_f=1$, $E_0=10$, $S_0=5$ to obtain $[ES]=8\pm\sqrt{14}$. Since we have $[ES]\leq\min\{E_0,S_0\}=5$, the solution is $[ES]=8-\sqrt{14}\approx 4.26$.


Why does the viscosity of honey decrease when mixed with other viscous fluids? In order to make honey mustard, I mix honey and mustard.  But the resulting mixture is less viscous then both honey and mustard.  How does mixing two viscous substances give a less viscous one?

Honey is a homogeneous substance, a solution of around 98% sugar and water.

Mustard is a heterogeneous mixture, containing about 1:1:1 oils, proteins and water, all in very small particles. The proteins act as a stabiliser to the small oil droplets. They cant slide past each other easily, and act more like a dense network of small solid particles. Like a heap of wet sand.

Now if the aqueous phase gets much larger due to the honey, the oil+protein droplets loose direct contact, and you no longer have the solid-like behaviour of wet sand.

Your aqueous honey-phase however now contains perhaps 25 or 30 percent of water instead of 15 or so in pure honey, and that lowers its viscosity a lot.

You have a major phase with reduced viscosity, and a disperse phase which is no longer continous (not "percolated"), and so the whole thing flows rather easily.

(Btw. the viscosity of mustard is not that high, however it usually shows a yield stress, that is it does not flow at all e.g. under its own weight. You open a mustard glass and the inside looks just the same as it did when you last closed it. Honey however is just a boring Newtonian fluid, no shear thinning, no thixotropy, nothing.)


How is sulfuric acid with a concentration of 12 M also 72% w/w? According to LabChem, $12~\mathrm{M}$ of $\ce{H2SO4}$ is $72~\mathrm{w/w\%}$.

When I try to convert $12~\mathrm{M}$ of $\ce{H2SO4}$ to $\mathrm{w/w\%}$ I get $117\%$ which is obviously wrong.

$$\begin{multline}
12~\mathrm{M}\cdot\left(\frac{1~\mathrm{L}}{1000~\mathrm{ml}}\right)
\left(1~\mathrm{\frac{ml}{g}}\ \ce{H2O}\right)
\left(98.09~\mathrm{\frac{g}{mol}}\ \ce{H2SO4}\right)\\
= \frac{1177~\mathrm{g}}{1000~\mathrm{g}} = 1.177=1.177\cdot100\%=117.7\%
\end{multline}$$

I must be doing something wrong. How do you convert it correctly?

Your problem is that you've assumed the density is 1 gram/mL.

Remember that a molar is defined as a mole of solvent per liter of solution, not solvent. Usually, in introductory chemistry classes, we skip over the fact that adding solute to a solution increases its density, because it makes life more complicated.

As you just found out though, sometimes you really need to account for the change in density that occurs as you add solute. This webpage claims that the density of $12~\mathrm{M}$ sulfuric acid is about $1.634~\mathrm{g/mL}$.

Using this updated calculation, we still have the original $1177~\mathrm{g}$ of sulfuric acid. However, the solution mass is now $1000~\mathrm{mL}\cdot 1.634~\mathrm{g/mL} = 1634~\mathrm{g}$. Finally, we obtain that

$$
\frac{ 1177~\mathrm{g} \, \ce{H2SO4}}{1634~\mathrm{g} \, \mathrm{solution}} = 72.03\% \; \mathrm{w/w} $$

A note about this calculation: it's possible we're begging the question, since I don't know which values on that linked page are measured and which are calculated. It's possible that the density was calculated from measured data. The important thing here isn't how the data was obtained though, it's that the density of solution changes as you add more solute, and so you have to change how you calculate things from the molarity.


Does dissociation require energy to occur? I assume spontaneous separation to ions of salt solution (or acid) requires energy to occur, right?

Where does that energy come from? Is it from the heat of the solution?

Now, if someone devised a way to draw potential energy from those separated ions, could he make this process continuous (assuming solution would draw heat from its surroundings), or would it eventually stop to work?

When the ions of an ionic compound disolve, do not stay isolated. They form new bonds with the solvents. This makes the overall process spontaneus when a salt is soluble.

Dissociation of course requires energy. But bond formation between ions and solvent molecules release energy.
This can result on a possitive (endothermic disolution, heat is absorbed in the process, like for example, $\ce{KI}$ in water. Bonds formed release less energy than bonds broken) or negative disolution enthalpy (exothermic disolution, like the one for $\ce{H_2SO_4}$ and water. Bonds formed -hydrogen bonds, stronger that ion-dipole interactions-, release more energy that is consumed by bonds broken).

Remember that if you want to know if a process is or not spontaneous, you do not have to take into acount just enthalpy, but Gibbs free energy, which is $\Delta G= \Delta H + T\Delta S$, which also takes into account entropic factors and the influence of temperature.


Shift in the position of equilibrium by a catalyst to favor the forward reaction We were looking at the factors that affect dynamic equilibrium today. The teacher explained how catalysts only affect the rate at which dynamic equilibrium is achieved, and not its position.

My question is: will a catalyst specific to the forward reaction shift the position of equilibrium to favour the forward reaction? Or will it, still, only affect the rate?

Wikipedia says

  Catalyzed reactions have a lower activation energy (rate-limiting free energy of activation) than the corresponding uncatalyzed reaction, resulting in a higher reaction rate at the same temperature and for the same reactant concentrations.

This means that a catalyst will reduce activation energy as seen here

We clearly see that the activation energy of both forward and backward reaction equally. Hence by Arrhenius equation both the rate constants change by same factor thereby there is no change in equilibrium constant.


Possible carbocation rearrangement in substitution reaction to give two different products In the following substitution reaction, is compound I formed by $\mathrm{S_N}$2 and compound II formed by $\mathrm{S_N}$1 and carbocation rearrangement?

Or is there some rearrangement I'm not aware of?

Is compound I formed by $\mathrm{S_N}$2 and compound II formed via $\mathrm{S_N}$1 after a carbocation rearrangement? Or is there some rearrangement I'm not aware of?

For the first question, the answer is probably not because the mechanism of a given reaction is mainly based on the condition used in the reaction (at least at the undergraduate level of teaching). The the answer for the second question is yes, you are unaware of the mechanisms called $\mathrm{S_N}$1' and $\mathrm{S_N}$2' (as Shoubhik R Maiti mentioned in the comments section). This type of reactions are common when the allylic compounds are the substrate, which is appropriately called allylic rearrangement or allylic shift. Allylic shift is possible under either $\mathrm{S_N}$1 or $\mathrm{S_N}$2 conditions, and commonly known with $\mathrm{S_N}$1' and $\mathrm{S_N}$2' notations, respectively.

Strictly speaking, if you have used condition favoring $\mathrm{S_N}$1 reaction, the intermediate is a resonance stabilized allylic carbocation. Thus nucleophile would attack at two positions (either carbon 1 or 3) to give the resulting products, I and II. For instance, see your reaction in a protic solvent ($\mathrm{S_N}$1), which would given the same type of product mixture (for convenience, I have used the labelling to show the difference):

Molecular modeling calculations would shows you the allylic secondary carbon position is more stable than the allylic primary carbon (recall your ab initio calculations for kinetic versus thermodynamic control in Org.Chem.II lab course; a simple and effective lecture demonstration of thermodynamic versus kinetic control, see: Ref.1) in intermediate carbocation. Accordingly, compound II would be the major product.

Similarly, if you are using the condition favoring $\mathrm{S_N}$2 reaction (strong nucleophile in a polar aprotic solvent) in this type of reaction, it is also possible to get two product due to $\mathrm{S_N}$2' mechanism. Usually, a bulky leaving group in $\mathrm{S_N}$2 conditions or bulky non-leaving substituent on the framework which give rise to significant steric hindrance would thereby increase the conjugate substitution ($\mathrm{S_N}$2'). Following scheme show the mechanism for two products:

Late Edit:
Since it seems OP is still cannot see how compound II is formed, I included the arrow-pushing mechanism for $\mathrm{S_N}$2 versus $\mathrm{S_N}$2' products:

References:

I. J. McNaught, "Thermodynamic versus kinetic control: A lecture demonstration," J. Chem. Educ. 1978, 55(11), 722 (https://doi.org/10.1021/ed055p722.1).