Why can we smell copper? If I can smell an object, it means that molecules of it are getting separated from it, so they can reach my nose. As far as I know, metals don't sublimate, especially not in room temperature. However, copper has a very strong and characteristic smell. Does it mean that copper will degrade pretty quickly, or are just we so sensitive to it that a few molecules are sufficient? I assume it has to do something with oxidization, but it doesn't oxidize that much naturally as other metals, for example, iron.

This is a nice question, as it confronts a very replicable and common experience with a well established yet seemingly contradictory fact. As you expected, the smell of metal has nothing to do with the metal actually getting into your nose, as most metals have far too low of a vapor pressure at ordinary temperatures to allow direct detection. The characteristic smell of metal, in fact, is caused by organic substances!

There has been the focus on the specific case of the smell of iron (free-access article!). There are at least two ways in which iron produces a metallic smell. Firstly, acidic substances are capable of corroding iron and steel, releasing phosphorus and carbon atoms present in the metal or alloy. These can react to form volatile organophosphorus compounds such as methylphosphine ($\ce{H3CPH2}$ which have a garlic/metallic odor at small concentrations. From the article:

  The “garlic” metallic odor (see Supporting Information) of the gas product from the acidic dissolution of cast iron is dominated by these organophosphines. We measured an extremely low odor threshold for two key odorants, methylphosphine and dimethylphosphine (6 and 3 ng P/m³, respectively, garlic-metallic odor), which belong therefore to the most potent odorants known. Phosphine ($\ce{PH3}$) is not important for this odor because we found it has a much higher odor detection threshold (>10⁶ ng/m³). A “calcium carbide” (or “burned lime”/“cement”) attribute of the general “garlic” odor is probably caused by unsaturated hydrocarbons (alkynes, alkadienes) that are linked to a high carbon content of iron (Table 1, see Supporting Information).

Also, it turns out that $\ce{Fe^{2+}}$ ions (but not $\ce{Fe^{3+}}$) are capable of oxidizing substances present in oils produced by the skin, namely lipid peroxides. A small amount of $\ce{Fe^{2+}}$ ions are produced when iron comes into contact with acids in sweat. These then decompose the oils releasing a mixture of ketones and aldehydes with carbon chains between 6 and 10 atoms long. In particular, most of the smell of metal comes from the unsaturated ketone 1-octen-3-one, which has a fungal/metallic odour even in concentrations as low as $1\ \mu g\ m^{-3}$ . In short:

  Sweaty skin corrodes iron metal to form reactive $\ce{Fe^{2+}}$ ions that are oxidized within seconds to $\ce{Fe^{3+}}$ ions while simultaneously reducing and decomposing existing skin lipid peroxides to odorous carbonyl hydrocarbons that are perceived as a metallic odor.

In the supporting information for the article (also free-access), the authors describe experiments performed with other metals, including copper:

  Comparison of iron metal with other metals (copper, brass, zinc, etc.): When solid copper metal or brass (copper-zinc alloy) was contacted with the skin instead of iron, a similar metallic odor and GC-peak pattern of carbonyl hydrocarbons was produced and up to one μmole/dm² of monovalent cuprous ion [$\ce{Cu+}$] was detected as a corrosion product (Supporting Figs. S3 to S6). Zinc, a metal that forms $\ce{Zn^{2+}}$ but no stable $\ce{Zn+}$, was hesitant to form metallic odor, except on very strong rubbing of the metal versus skin (that could produce metastable monovalent $\ce{Zn+}$). The use of common color-tests to demonstrate directly on human palm skin the presence of low-valence ions (ferrous and cuprous) from the corrosion of iron, copper and brass alloys is shown in Supporting Figure S6. Alumina powder rubbed on skin did not produce significant odorants. These results provide additional evidence that it is not metal evaporation, but skin lipid peroxide reduction and decomposition by low valence metal ions that produces the odorants.

The last paragraphs of the article summarize the findings:

  In conclusion: 1) The typical “musty” metallic odor of iron metal touching skin (epidermis) is caused by volatile carbonyl compounds (aldehydes, ketones) produced through
  the reaction of skin peroxides with ferrous ions ($\ce{Fe^{2+}}$) that are formed in the sweat-mediated corrosion of iron. $\ce{Fe^{2+}}$ ion containing metal surfaces, rust, drinking water, blood etc., but also copper and brass, give rise to a similar odor on contact with the skin. The human ability to detect this odor is probably a result of the evolutionarily developed but largely dormant ability to smell blood (“blood scent”).

  2) The “garlic-carbide” metallic odor of phosphorus- and carbon-rich cast iron and steel under attack by acid, is dominated by volatile organophosphines. Corroding cast iron
  is an environmental source of C–P compounds that may lead to confusion in the verification and monitoring of the Chemical Weapons Convention (see also ref. [15])

As an aside, this may be why sometimes people recommend getting strong smells off your hands by rubbing them against a metal object. While it probably doesn't work for some metals and for some smelly compounds, it's possible that the metal catalyzes the decomposition of the malodorous substances into less strongly smelling ones.

You can read a little more in this press article on the study.


Mole Fraction and Atom Fraction I'm probably overthinking this but would appreciate a peer check.
Mole fraction of element $i$ in compound: $\frac{\mathcal{M}_i}{\mathcal{M}}$
I used script M so as not to confuse with molar mass (often denoted as capital M).
Atom fraction of element $i$ in compound: $\frac{N_i}{N}$
Now, if Avogadro's number is atoms/mol then isn't the mole fraction equal to the atom fraction?
$$
\frac{N_i}{N}\frac{[\text{atom}]}{[\text{atom}]} \cdot \frac{N_a}{N_a}\frac{[\text{atom}/\text{mol}]}{[\text{atom}/\text{mol}]} = \frac{\mathcal{M}_i}{\mathcal{M}} \frac{[\text{mol}]}{[\text{mol}]}\\
\frac{N_a}{N_a}=1\\
\frac{N_i}{N}=\frac{\mathcal{M}_i}{\mathcal{M}}
$$

The fallacy lies in the assumption that mole fraction and the number fraction are always the same.
For example, take $\ce{Na2SO4}$
what is the atom fraction of Na: 2/(2+1+4) = 2/7
What about the mole fraction of Na? It 2$N_A$/$N_A$(2+1+4)= 2/7
However, this "simple" formula of yours breaks down in the case of mixtures even if we mix them in 1:1 mol ratio. Calculate the mol fraction of Na ions in a mixture using your formula for $\ce{NaCl}$ and $\ce{Na2SO4}$ and see that you will not get the right answer.


Ionization of amino acids In my book it talks about amino acids. It says the general structure of an amino acid is as follows:

However, on the next page it lists specific amino acids such as glycine shown here:

My question is when comparing the first image to glycine, why does glycine or any of the other specific amino acids have an extra H atom as in "H3N+"?

Where does the extra H come from and why is it positively charged?

Also, on the general structure it shows the an amino should have a COOH but on glycine it is just C double bonded to an O and then another O- negative charge.

Does this have to do with the pH and ionization?

The two forms of amino acids that your question asks about are called zwitterions (technically IUPAC says they should be called dipolar ions but the term zwitteron is pervasive).

Since amino acids contain both acidic (the carboxylic acid) and basic (the amine) moieties the molecule is able to undergo what almost appears to be an intramolecular acid-base reaction in which the acidic proton of the carboxylic acid protonates the amine, this is what gives rise to the species on the right of the image below.

Its worth pointing out the this property of the amino acids is valid at physiological pH for the naturally occurring amino acids, generally in aqueous medium (which living organisms have. Also as Jan pointed out in the comments, other polar solvents would also allow the amino acid to show this characteristic).

At other pH values, the molecules become singly ionised, favouring either the cationic or anionic forms. The range of pH values at which the amino acid is zwitterionic will also vary depending on which amino acid you're looking at (the pKa values are different for every amino acid, even if only subtly)


Explanation of certain chemical bond diagrams Here's a "simple" bond diagram of 1,1,1,2-Tetrafluoroethane which I believe is a Lewis structure:

OK, that's easy enough - four bonds from each carbon atom, and each fluorine atom having one bond.

Here are some more "complicated" bonds which I have yet to do in my high school chemistry class, so I don't understand how to interpret them.

Here's caffeine:

And chloroform:

How would I go on interpreting these diagrams? Or is there missing information?

Image sources: Wikimedia

The second structure you posted is a line angle diagram. In this diagram, methyl groups (-CH3) are implied using sticks with nothing at the end.

Carbon atoms are also implied; carbons exist at every "corner" of the polygon which doesn't have an element explicitly identified. Hydrogens attached to carbons are also implied. So at the intersection of two lines, if there are no other attachments and no charge, you can expect there to be two hydrogens (carbon is generally tetravalent).

The third structure you posted is a wedge-dash diagram. Wedges depict substituents angled toward you; dashes are substituents pointing away from you; the lines are coplanar. The advantage of this diagram is that it shows the 3D nature of the molecule. Your first picture of the fluoro-carbon compound implies 90 degree bond angles. However, you should realize that the carbon atoms are sp3 hybridized and thus should not have 90 degree bond angles.

Further reading and exploration of other depictions of molecular structure:

http://www.masterorganicchemistry.com/2010/10/15/the-many-many-ways-to-draw-butane/


Regioselectivity in the formation of an imidazo[1,2-a]pyrimidine In the heterocycle synthesis below (taken from Org. Process Res. Dev. 2006, 10 (3), 398–402), a 5,6-fused heterocycle (an imidazo[1,2-a]pyrimidine) is formed by treating 2-amino-4-(trifluoromethyl)pyrimidine with the diethyl acetal of bromoacetaldehyde, with the cited paper claiming to get a 25:1 selectivity for 5 (minimal amounts of 5a also produced were able to be easily removed).

  Ref: Org. Process Res. Dev. 2006, 10 (3), 398–402.

There are (as ever with heterocycle syntheses) a few different mechanisms one could draw, however, no matter which way round one draws the mechanism, the major product is always the one that arises from what appears to be the less favoured intermediates.

By less favoured intermediates, what I mean to say is that to arrive at the major product, necessarily a cation ends up closer to the electron withdrawing trifluoromethyl group than it would do if the minor product was formed.

Given that all the steps are essentially reversible, is the product distribution arising from a purely ground state thermodynamic preference for 5 over 5a, or is there a mechanism that I'm completely neglecting (sorry for the lack of pictures, I'm not able to ChemDraw right now).

I propose the route in the following scheme. I don’t see any unfavourable intermediates. In short, the first step would be formation of a Schiff base with the extracyclic amino group and the aldehyde. This — in one of many orientations as shown below — places the bromide nicely so that one of the pyrimidine nitrogens can attack it nucleophilicly. Upon displacement, an intermediate nitrogen-centred cation is generated; loss of a proton and switching of electrons regenerates aromaticity and gives the desired product.

Scheme 1: Proposed reaction scheme. The attacking nitrogen $\mathrm{sp^2}$ orbital in step 2 is drawn explicitly.

We can draw structures of the Schiff base intermediate that would allow either of the pyrimidine nitrogens to attack. However, the top one experiences a stronger $-I$ effect by the neighbouring $\ce{CF3}$ group hence its electron density and nucleophilicity should be reduced. The para-nitrogen does not experience this inductive effect since it greatly weakens with greater distance.

The ortho-nitrogen’s lone pair is indeed affected since inductive effects affect both σ and π bonds while mesomeric effects mainly affect π electrons only.


Why group 13 elements don't show +2 as their major oxidation state? These are the things I understand:

  1) Group-13 elements show +1 and +3 as their major oxidation
  states(OS).

  2) The general configuration of G-13 elements is $ns^2$ $np^1$.

  3) In the excited state, their electronic configuration is $ns^1$
  $np_x^1$ $np_y^1$ .

On losing two electrons from their excited state,
their configuration will become $ns^1$ $np_x^0$ $np_y^0$

I don't understand why will this configuration be unstable?

The one p electron is energetically much higher than the two s electrons (and any d electrons, if present). So it is rather easily removed to give a monocation or the $\mathrm{+I}$ oxidation state.

Thereafter, to remove a second electron you need to rip out one of the paired s electrons. This is not going to be easy and requires a lot of energy. But once you have arrived there, at a hypothetical $\mathrm{+II}$ cation, you again have one single lone electron in a relatively (compared to the core orbitals) orbital that should, again, be easily removed. Obviously that can happen leading you quickly to the $\mathrm{+III}$ state.

Indeed, it would be rather easy to see that two $\mathrm{+II}$ atoms might come together and transfer a single electron from one to the other—a disproportionation, giving $\mathrm{+I}$ and $\mathrm{+III}$ from two $\mathrm{+II}$. This would be predicted to be an exothermic process.

This only applies to isolated atoms; where bonds are formed other oxidation states are accessible. For example, boron exhibits the $\mathrm{+II}$ oxidation state in bis(pinacolato)diboron as can be deduced easily.


What is the correct unit for the equilibrium constant? I have the following equation

$$K_a = \dfrac{c\alpha^2}{1-\alpha},$$

with units $c= \pu{mol/cm^3}$ and $\alpha$ is the same. I'm not sure how to get the correct units for $K_a$. What I have come up with so far is

$$K_a= \pu{mol^2/cm^6}.$$

Andselisk correctly identified the law of dilution and the name Ostwald is often connected with it.
$$K_\text{dissociation} = \frac{\alpha^2}{1-\alpha}\cdot c$$
However, the degree of dissociation is $\alpha$ and has "no" unit, i.e. dimensionless quantity. Therefore the unit for the equilibrium constant is that of a concentration, in SI that would be $\pu{mol m-3}$.

It is derived from the Law of mass action; the derivation can be found on Wikipedia and various other sources.

In principle there is no fixed unit for the (generalised) equilibrium constant, as it is simply defined as a product of the involved quantities according to the IUPAC gold book.

  Equilibrium Constant
  Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always dimensionless, as it is defined differently (IUPAC gold book).

  Standard Equilibrium Constant $K$, $K^\circ$
  (Synonym: thermodynamic equilibrium constant)
  Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{\mathcal{R}T}\right\}$$  where $\Delta_rG^\circ$  is the standard reaction Gibbs energy, $\mathcal{R}$  the gas constant and $T$  the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.


Why are 7-membered rings less likely to form than 5- and 6- membered rings? I see 5- and 6-membered ring products as the major products even when 7-membered rings are possible. Why is that 5- and 6-membered rings are more likely to form than 7-membered rings? Is it even true?

My textbook says a 5-membered ring is the major product and the 7-membered ring is the minor product in intra-molecular aldol condensation reaction of 6-oxoheptanal.

Major Product: 1-(cyclopent-1-en-1-yl)propan-2-one
Minor Product: cyclohept-2-ene-1-carbaldehyde and 2-methylcyclopent-1-ene-1-carbaldehyde

This is not the only reaction where I found a 5-membered ring as a major product when a 7-membered ring was possible.

There are two factors that need to be considered to answer your question - entropy and enthalpy.

Entropy favors the formation of smaller rings. If we consider a hydrocarbon chain, the number of possible conformations increases dramatically as we go from 3 carbons to, for example, 10 carbons.  Only a few of these conformations will be correctly arranged so that the two ends of the chain are close to each other, which is necessary if we are to form a bond. So, entropically speaking, it will be much more probable to form a 3-membered ring than a 10-membered ring. This is shown in the following graph where 3-, 4- and 5-membered ring formation is entropically most favored.

Note that a less negative $\Delta S$ favors the reaction since $\Delta G=\Delta H - T \Delta S$.

Enthalpy comes into play when we consider the strain in the transition state leading to the possible products. Here is a chart showing the relative ring strain as a function of ring size.

As expected, strain is the highest in 3- and 4-membered rings.

If we overlay (add) these entropic and enthalpic effects we would get a graph similar to the one pictured below.

In general, we find that 5-membered ring formation is usually the favored pathway.  This web article (J. Am. Chem. Soc. 1984, 106, 1051-1056.) is a nice review of this topic that may be useful for a more in-depth reading on the subject. The above web article is also the source of the above three images.


Products of Electrolysis of NaCl I have gone online to see many times that the electrolysis of $\ce{NaCl}$ dissolved in water produces Lye: $$\ce{2NaCl + 2H2O -> 2NaOH + H2 + Cl2}$$ Other times $\ce{NaClO3}$ or some other compound. I just want to know with one solid answer (anyone), what does this reaction produce?

Thank you.

The one solid answer is that everything you've heard is right, given the appropriate conditions.

If the cathode and anode spaces are properly separated with a porous membrane or a salt bridge, then it is just like that:
$$\ce{2NaCl + 2H2O -> \underbrace{2NaOH + H2}_{cathode} + \underbrace{Cl2}_{anode}}$$
If they are not, then the products inevitably mix with each other and react further:
$$\ce{2NaOH + Cl2 -> NaCl + NaClO + H2O}$$
or, at elevated temperatures,
$$\ce{6NaOH + 3Cl2 -> 5NaCl + NaClO3 + 3H2O}$$
See, chemistry is diverse. Oh wait, and there are more options. Think of the mercury cell.


Cold process to make glass Since glass is a mix of silicates of sodium and calcium produced by heating or fusing together $\ce{Na2O}$ (Lewis base) with sand or $\ce{SiO2}$ (Lewis acid). Since $\ce{Na2O/NaOH(aq)}$ is a strong base and is used to create water glass(es)$(\ce{Na2SiO3},$ $\ce{Na_xSiO_y}$ etc.).

Is there some procedure in place to create glass by mixing (a base rich) water glass at room temperature with calculated amounts of $\ce{CaO}$ in mould to prepare float glass of definite shape without heating or mild heating?

Not glass per se, but silica is often produced chemically by silicon alkoxide gelation to create aerogels.

The problem with creating glass by some cold process would be that the reaction side-products, such as water, would have to be removed, leaving holes in a froth of glass. This is fine for creating an aerogel, but is not helpful in making solid objects, such as test tubes or lenses. Of course, you could sinter the glass frit or froth, but that still requires heating.


Why do higher-mass isotopes have higher melting and boiling points than lower-mass isotopes? Different isotopes of the same element have the same number of electrons, so the nature of bonding between different isotopes of the same element should be the same, yet their melting points and boiling points vary. This would suggest that the interparticle forces and distances are different for different isotopes of the same element. Why is this exactly?

My best guess is that is has something to do with the kinetic energy (KE) formula

$$E_k = mv^2/2$$

If two samples of different isotopes of the same element were kept under the same pressure and temperature (so the KE of the particles in the different samples would the same), the particles of the heavier isotope would have a lower velocity than the particles of the lighter isotope, as a greater mass means a lower velocity (where KE is equal). Does it have something to do with the fact particles of a heavier isotope move more slowly when KE is equal?

As you already mention in your question, melting and boiling of a substance is related to the inter-particle forces. You also correctly assumed that the same number of electrons results in identical (electronic) bonding properties, which basically is a different formulation of the Born-Oppenheimer Approximation (as in the BOA the nuclear mass is assumed to be infinite, different isotopes should have the same electronic structure).

So what is the difference then?

Well, it's in the vibration between the different particles. Consider a number of (neutral) particles in a certain phase (it can be solid, liquid or gas) at a certain distance with respect to each other. Because of dispersion forces between the particles, they interact and to a first approximation, every particle experiences a harmonic force $F(x)=-k(x-x_e)$. The magnitude of this force depends on the distance between the particles and on the electronic structure of the particles and in principle not on the mass. However, as you might know, a harmonic force in quantum mechanics results in quantized energy levels given by

$$
E_v=h\nu_e(v+\frac{1}{2}),
$$
where $E_v$ is the vibrational energy of the $v$th level and $h$ is Planck's constant. What is important to realize is that the energy of the lowest state is not equal to zero but has a value of $h\nu_e/2$, which is referred to as the zero point energy. Of course the potential is not really harmonic but rather anharmonic so that at a certain distance between the particles the force becomes negligible and the particles dissociate (or melt or evaporate). The lower the zero point energy, the larger the binding energy. The constant $\nu_e$ is given by $\nu_e=\frac{1}{2\pi}\sqrt\frac{k}{\mu}$, where $k$ is the harmonic force constant and $\mu$ is the reduced mass. The constant $k$ is the same for the different isotopes as it depends on the electronic potential between the particles but the reduced mass is of course different for the different isotopes. As you can see from the definition of $\nu_e$, the heavier species have a smaller zero point energy and thus a larger binding energy and will require higher temperatures to melt and boil.


Should an entire compound or just the redox-active element be written in redox pairs? For oxidant/reductants and conjugate pairs, for example $\ce{MnO2 -> Mn^2+}$, do we write $\ce{Mn^4+}$ or $\ce{MnO2}$ as the oxidant? Websites conflicts what my text book says, because it appears that my text book would write the conjugate pair as $\ce{Mn^4+/Mn^2+}$ which implies that it considers $\ce{Mn^4+}$ is the oxidant not $\ce{MnO2}$.

I just realised my example wouldn't work for what I was trying to say since $\ce{MnO2}$ does not exist in ionic form. But using a compound that did exist in the ionic form, would I be saying that the oxidant is the entire compound or the ion that has a change in oxidation number?

You generally use whichever compound is actually present in solution.

For manganese(IV), it will practically always be $\ce{MnO2}$ so use that. Manganese(VII) will always be $\ce{MnO4-}$ so use that. Manganese(II) salts dissolved in water dissociate and coordination complexes of manganese(II) are very labile. Thus, it doesn’t make sense to assume any $\ce{[MnL_n]^2+}$ complex to predominate and you should simply use $\ce{Mn^2+}$.

It does, for example, make a difference if your iron redox pair is $\ce{Fe^2+/Fe^3+}$ or $\ce{[Fe(CN)6]^3-/[Fe(CN)6]^4-}$. Likewise, a reduction using tin(II) should better be written $\ce{Sn^2+/[SnCl6]^2-}$.

If you are in doubt whether something dissociates or whether it should be considered a single entity, the cleaner solution is probably to write the fully dissociated salt, so you can get away with using $\ce{Sn^2+/Sn^4+}$. However, as soon as molecules or oxyanions are involved you cannot resort to this; oxalate reductions have to be $\ce{C2O4^2-/CO2}$ and permanganate has to be $\ce{MnO4-/product}$.


Calculate Michaelis-Menten constant of enzyme catalyzed reaction I have:
$$\begin{array}{c|c}
\dfrac{v }{ [S]}/\pu{s^{-1}} & v \cdot 10^2/\pu{mol dm^-3 s^-1} \\ \hline
0.257\ & 5.15\\
0.895 & 4.48\\
2.00\ & 3.35\\
3.59\ & 1.8\\
4.82\ & 0.48\\
\hline
\end{array}$$
And I want to calculate $K_M$ graphically.
I know that I should use the equation:
$$\frac{1}{v}=\frac{1}{v_\mathrm{max}}+\frac{K_M}{v_\mathrm{max}[S]},$$
and that I should plot $1/v$ against $1/[S]$, and then the slope will equal $K_M$/$v_\mathrm{max}$.
I began with converting the data given from $v/[S]$ to $1/[S]$ by dividing each data point with $v$. This gave:
$$\begin{array}{c|c}
\dfrac{1}{[S]}/\pu{dm^3 mol^{-1}} & v \cdot 10^2/\pu{mol dm^-3 s^-1} \\ \hline
4.99*10^{-4}\ & 5.15\\
0.0019 & 4.48\\
0.0059\ & 3.35\\
0.0199\ & 1.8\\
0.1004\ & 0.48\\
\hline
\end{array}$$
Plotting $1/v$ against $1/[S]$ gives the graph:

But when I want to calculate $K_M$ through:
$$\text{slope} \times V_\mathrm{max} = \text{slope} \times \frac{1}{\text{intercept}}$$
I don't get the answer which is in our answer key. Unfortunately, only the answer for $K_M$ is given, and not how to get the answer. The answer is $K_M = \pu{0.0102 mol dm-3}$. Could someone explain where I am going wrong?

What you have given is:
$$\begin{array}{c|c}
\hline
\dfrac{v}{[S]}, \ \pu{s^{-1}} & v, \ \pu{mol dm^-3 s^-1} \\ \hline
0.257  & 5.15 \times 10^{-2}\\
0.895 & 4.48 \times 10^{-2}\\
2.00  & 3.35 \times 10^{-2}\\
3.59  & 1.8 \times 10^{-2}\\
4.82  & 0.48 \times 10^{-2}\\
\hline
\end{array}$$
Michaelis-Menten equation for enzyme kinetics is:
$$v = \frac{V_\mathrm{max} \cdot [S]}{K_M + [S]} \tag1$$
When rearrange this equation (reciprocal) to get Lineweaver-Burk relationship:
$$\frac{1}{v} = \frac{1}{V_\mathrm{max}} + \frac{K_M}{V_\mathrm{max}[S]} \tag2$$
When multiply the equation $(2)$ by $v \times V_\mathrm{max}$, you get:
$$V_\mathrm{max} = v + \frac{vK_M}{[S]} \tag3$$
When rearrange the equation $(3)$, you get $y = mx + c$ type equation (straight-line equation):
$$v  = - K_M \cdot \frac{v}{[S]} + V_\mathrm{max} \tag4$$
The data you have is $\frac{v}{[S]}$ and $v$, which fix the relationship given in the equation $(4)$. Thus, if you can plot them against each other ($\frac{v}{[S]}$ versus $v$) you'd get a straight line with negative lope. The numerical value of the slope is $K_M$ so you can calculate K$_M$ graphically. Also, the $y$-intercept of the graph is equal to $V_\mathrm{max}$:

From the equation of the graph: $\text{|The slope|} = 0.0102$ and $y\text{-Intercept} = 0.054$
Thus, $K_M = \pu{0.0102 mol L-1}$ and $V_\mathrm{max} = \pu{0.054 mol L-1 s-1}$
Once I found OP's mistake of plotting, I was curious to see how this set of data would show the Lineweaver-Burk relationship. Thus, I make the table for $\frac{1}{[S]}$ and $\frac{1}{v}$:
$$\begin{array}{c|c}
\hline
\dfrac{1}{[S]}, \ \pu{L mol-1} & \dfrac{1}{v}, \ \pu{L s mol-1} \\ \hline
4.99  & 19.42 \times 10^{-2}\\
19.98 & 22.32 \times 10^{-2}\\
59.70  & 29.85 \times 10^{-2}\\
199.44  & 55.56 \times 10^{-2}\\
1004.17  & 208.33 \times 10^{-2}\\
\hline
\end{array}$$
As predicted by  Lineweaver-Burk, the plot of $\frac{1}{[S]}$ versus $\frac{1}{v}$ is a straight-line with a positive slope:

From the equation $(2)$, which is the equation of the graph: $\text{The slope} = 0.1891 = \frac{K_M}{V_\mathrm{max}}$ and $y\text{-Intercept} = 18.379 = \frac{1}{V_\mathrm{max}}$
Thus, $V_\mathrm{max} = \frac{1}{18.379} = \pu{0.0544 mol L-1 s-1}$ and $K_M = 0.1891 \times V_\mathrm{max} = 0.1891 \times 0.0544 = \pu{0.0103 mol L-1}$
Both graph give the same values for $K_M$ and $V_\mathrm{max}$.


Simple procedure for showing presence of milk proteins in clarified butter (ghee) By clarifying butter, e.g. by melting butter and filtering it, we rid ourselves of most of the protein in the butter, and are left with mostly fat. However, the separation of protein and fat is not complete. Some people cannot eat milk proteins due to allergic reactions, and cannot therefore use unrefined butter in their cooking.

Is there a simple "home-kitchen" procedure to test whether the filtration sufficiently removed the milk proteins? Any proteins in the ghee can be precipitated by adding acetic acid, but this is not specific to milk proteins.

Does anyone have any ideas?

A classic test for protein is Biuret solution, using readily available reagents:  sodium hydroxide, copper sulfate and potassium sodium tartrate (Rochelle salt). The test is not specific for milk protein, but what other protein would be in pure butter?

Use caution with the chemicals -- the first listed is quite caustic, and the second is poisonous. BTW, save some leftover Rochelle salt if you'd like to make a piezoelectric crystal.


Chlorination of propene at 773 K The first step of preparation of glycerol
with propene involves the chlorination of of propene at 773 K. It results to form allyl chloride. However the halogenation of alkenes  generally give vicinal dihalides

$$\ce{CH3-CH=CH2 + Cl2 -> CH2Cl-CH=CH2}$$

The reaction proceeds via a SFR (Substitution Free Radical) mechanism and yields an allylic or benzylic halogenated alkene. This result is specific for low concentration of halogen and high temperature.

The results of the same reaction of halogens with alkenes but at different reaction conditions, i.e. high halogen concentration and room temperature result into anti addition of the halogen atoms on the double bond. Thus the reaction takes place faster in presence of $\ce{CCl4}$, i.e. non polar solvents. The reaction proceeds via formation of cyclic intermediate. The result of the same reaction in water instead of $\ce{CCl4}$ will result in to anti addition of halide ion and $\ce{OH-}$ ion on the double bond.


Why does SiH4 have unusually large bond angles? The bond angles of $n > 2$ hydrides are usually around $90^\circ$, such as $\ce{PH3, H2S},$ $\ce{H2Se},$ $\ce{H2Te}.$ I understand that this is from increased $\mathrm{p}$ character in the bonding with these compounds.

However, for $\ce{SiH4}$ and $\ce{GeH4}$ the bond angles are actually quite close to $109.5^\circ$. Why do $\ce{Si}$ and $\ce{Ge}$ not typically use unhybridized $\mathrm{p}$ orbitals, and instead use closer to $\mathrm{sp^3}$ hybrid orbitals for bonding?

In hydrides that contain two or three hydrogens connected to a central atom such as $\ce{PH3}$ or $\ce{H2Se}$, the number of bonding partners does not exceed the number of available p orbitals on the central atom. In an extreme application of Bent’s rule, the most ideal bonding situation would be to have pure p-type orbitals form the required bonds and at least one lone pair reside entirely in an s-type orbital. This allows stabilisation of the s lone pair (as an s orbital’s energy is lower than a p orbital’s one) and better bonding (as p orbitals are directional and thus offer greater overlap with hydrogen; furthermore, they are higher in energy which better matches hydrogen’s orbital energy and again allows for greater bonding energy).

This ideal situation already falls apart in the second period where the bond angles of ammonia and water are unusually large; this is because the central atoms are too small to accomodate two or three bonding partners at $90^\circ$ angles so the bond angles are increased in a delicate balance until the lowest overall energy is obtained. The results are hybrid orbitals with a nonzero s component.

In $\ce{AH4}$ molecules, the situation is different and worse yet again. There are not enough p orbitals to accomodate four bonds so the s orbital has to help. Symmetrically, the optimal arrangement is a tetrahedron and this just happens to almost perfectly match an $\mathrm{sp^3}$ arrangement. As silane (and methane, germane, etc.) has four identical substituents, there is no energy to be gained by reducing symmetry so the ultimate geometry is that of an ideal tetrahedron whose bond angles are $109.5^\circ$.


Does the enthalpy of solution formula (LE + Hyd) change depending on the question? Hey guys. I have been working on enthalpies and literally every video and website I’ve seen,
the formula for calculating enthalpy of solution = lattice enthalpy of dissociation + enthalpy of hydration, using which, I got a few questions correct. However, this particular question (picture added), I have used the same formula and gotten the answer -5273 $\pu{kJ mol-1}$.
The mark scene for this question states that the answer is -51 $\pu{kJ mol-1}$.
I know what they have done, I know that they have used        -(-2662) because the answer sheet has the working. I don’t understand why they gave done this Why have they done so?????
Thank you!!

Apparently your problem is related to the improper use of negative signs. Let's solve it differently.
Suppose you start from the separated ions $\ce{Ca^{2+}}$ and $\ce{2 F^-}$ in the vacuum. This state has the energy zero by definition.
Now suppose you join the ions to produce $1$ mole of solid $\ce{CaF2}$.  The process is exothermic, and the final state $\ce{CaF2(s)}$ is at the energy level $-2611 kJ/mol.$
Now suppose you are doing another transformation with the same ions. You put them directly into water, so that the ions become hydrated in solution. You obtain a solution of $\ce{CaF2}$ in water, or a solution of $\ce{Ca^{2+}(aq)}$ and $\ce{2 F^-(aq)}$. This transformation is also exothermic, and at the end $\ce{Ca^{2+}(aq)}$ + $\ce{2 F^-(aq)}$ are at the energy level $- 2662 kJ/mol$.
The final levels after these two transformations are $-2611 kJ/mol$ for the final solid, and $-2662 kJ/mol$ for the solution. These energy levels are not very different from one another.
So the passage from the solid state $\ce{CaF2}$ to the solution $\ce{Ca^{2+}(aq)}$ + $\ce{2 F^-(aq)}$ is slightly exothermic, with $\Delta H = -51 kJ/mol$


What kind of metal drums are used to store Hydrochloric Acid? I would expect Hydrochloric Acid to be stored exclusively in glass or plastic containers, but I was surprised to find Honeywell offers it for sale in 30Kg metal drums (Link).

I think Hydrochloric Acid would dissolve Nickel, Chromium and Iron in stainless steel, which leaves me wondering what kind of metal that could be, and whether this form of storage wouldn't (at least slightly) contaminate the acid?

We might as well consult the experts, namely storage tank manufacturers.  This site describes several types of storage tanks, none of them metal except for rubber-lined steel, in which (we hope) the rubber keeps the acid out of contact with the steel.  Rubber lining of steel is often used in processes such as acid rinsing tanks where the steel is needed for structural purposes, but for a storage tank as such the referenced site renders cross-linked polyethylene better.


Whether copper can combine with oxygen in equal proportion? Copper sulphide contains 66.6% copper and sulphur trioxide contains 40% sulphur. According to law of reciprocal proportions copper is 80% in copper oxide. The calculation: In the first, for 33.4% sulphur, there is 66.6% copper. In the second, there is 40% sulphur. Therefore, in the resultant there should be 80% copper. My question is why cannot we calculate the other way round:In the second, for 40% sulphur there is 60% oxygen; therefore, for 33.4% there should be 50% oxygen. Hence, the resultant combination should have equal proportion of copper and oxygen. Is there any such substance?

The Law of Reciprocal Proportions states: "If two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine."

Wikipedia has a different phrasing: "If element A combines with element B and also with C, then, if B and C combine together, the proportion by weight in which they do so will be simply related to the weights of B and C which separately combine with a constant weight of A."

Note this is a statement about the weights of substances, not about the molar fractions or volumes.

For example,

Copper sulfide ($\ce{Cu2S}$) contains 66% copper by moles. It does not, however, contain 66% copper by weight. Instead it contains $(2\cdot63.55)/( 2\cdot63.55+32.06) = 79.9\%$ copper by weight, or 20.1% sulfur by weight.

Likewise, while sulfur trioxide ($\ce{SO3}$) contains 25% sulfur on a molar basis, it contains $(32.06)/( 32.06+3\cdot16.00) = 40.0\%$ sulfur on a mass basis, or 60.0% oxygen by weight.

So when we apply the law of reciprocal proportions to copper oxide, we need to make use of the 20.1% and 40.0% numbers for sulfur. For example, if we had two masses of $\ce{Cu2S}$ and $\ce{SO3}$ each containing 1 g of sulfur, then we would have $1\rm{g}\cdot(79.9/20.1) = 3.98$ grams of copper and $1\rm{g}\cdot(60.0/40.0) = 1.50$ grams of oxygen. The ratio here is $3.98/1.50 = 2.65$.

Now, when we examine copper (II) oxide ($\ce{CuO2}$), for example we see that the ratio of masses for copper and oxygen are $63.55/(2\cdot16.00) = 1.99$ ... and 2.65 is very close to a simple multiple of 1.99 (specifically, 4/3 - the difference is due to rounding.)

The fact that this is always true is trivial, once you know atomic theory. So while the Law of Reciprocal Proportions is important in a historic context - as it helped lead to the development of modern atomic theory - basically no one uses it as such these days.

Say we have the compounds $A_hB_j$, $A_kC_l$ and $B_mC_n$.

In $A_hB_j$ the masses occur in the ratio $(j\cdot M_b) / (h\cdot M_a)$. In $A_kC_l$ the masses are in the ratio $(l\cdot M_c)/(k\cdot M_a)$, so for a fixed mass (say $M_a$g) of A, the ratio of the masses is $(k\cdot j\cdot M_b)/(h\cdot l\cdot M_c)$

Now for the compound $B_mC_n$, the masses occur in the ratio $(m\cdot M_b) / (n\cdot M_c)$. Taking the proportion of this with the previous proportion, we see that the masses cancel, leading to $(k\cdot j\cdot n)/(h\cdot l\cdot m)$. Now because molecules form with small, whole number coefficients in their molecular formulas, this final quantity will typically be a simple, small fraction. (a.k.a. a "simple multiple" of each other.)

So all the Law of Reciprocal Proportions is doing is saying that 1) all atoms of a given element have about the same average weight 2) molecules form from (typically) small, integer numbers of atoms. -- That's it.

This is why you don't see many people discussing it. It's not all that interesting to people who are accustom to thinking atomically.


Solubility product of HgS The reported solubility product of $\ce{HgS}$ is $4 \times 10^{-54}$. How is this possible? The minimum concentration that is possible is 1/avogadro's constant which shall be about $10^{-23}\ \mathrm{M}$. Hence the minimum value of solubility product for a diatomic specie like $\ce{HgS}$ would be about $10^{-46}$. What am I missing?

Imagine you have one googolplex ($10^{10^{100}}$) liters of water. Now let's dissolve one mole of sugar in your water. Setting aside where would you find so much water, the concentration would be way lower than the inverse of Avogadro's constant.

Long story short, there's no lower limit to the concentration of a substance. You can always add more solvent to the mixture thus making the concentration smaller.


Should the tip of the burette of an automatic titrator be immersed in the analyte? In all the tutorials I found online, the tip of the burette (containing the titrant - NaOH) was immersed into the analyte solution. My question is: shouldn't the tip be out of the solution? Isn't there a possibility for the analyte to crawl up the tip and react with the NaOH inside the burette that hasn't been dispensed yet?.

Everyone does that... Even in the official tutorials of the company

In classical titrimetry you would avoid dipping the burette into the sample. However, for autotitrators, the electronic dispenser tip (electronic burette) should be dipped in the sample. This is to avoid any error due to the a drop clinging to the tip. In manual titration one would wash it. Most autotitrators are like this design including Karl Fischer system. Your concern can be alleviated by the fact that you need pressure to cause a back flow of the sample into the tip of the dispensor.


Why does benzene have a D6h point group? I understand the fact that the molecule has a principal axis of rotation of C6 however when following the point gorup flow chart why is the answer to the second question NO when it contains two C3 axis? Hence the answer should be a YES right? (I understand the molecule has C2 axis as well)

Does it have two or more $C_2$ axes with $n>2$?

This question is asking if there is more than one $C_{n>2}$ axis not if there is more than one $C_{n>2}$ symmetry operation.

Yes, benzene has some $C_3$ operations. However, as pointed out in the comments, the $C_3$ axis lies along the $C_6$ axis - both run through the center of the ring perpendicular to the plane of the molecule. This axis is also a $C_2$ axis. There is only one $C_{n>2}$ axis - it just happens to be both a $C_6$ and a $C_2$ axis. All other rotation axes in benzene are $C_2$.


Acid-base reaction: Al(OH)₃ + NaOH Why is the product of $\ce{Al(OH)3 + NaOH -> Na[Al(OH)4]}$ and not $\ce{NaAlO3H2 + H2O}$? Is this an acid-base reaction?

All reaction sources: Chemiday

As I said in the comment section that aluminium hydroxide is amphoteric i.e. It dissolves in acid, forming $\ce{[Al(H2O)6]^3+}$ (hexaaquaaluminium) or its hydrolysis products and dissolves in strong alkaline solutions, forming $\ce{[Al(OH)4]−}$ (tetrahydroxidoaluminate). So, when it is dissolved in base, it forms a sodium aluminohydroxo complex salt. Why? Because as @Pritt mentioned, aluminium has vacant orbital for which free electron can reside there forming a complex salt.

$$\ce{Al(OH)3 + NaOH(conc.) → Na[Al(OH)4]}$$

$$\ce{Al(OH)3 + 3NaOH → Na3[Al(OH)6]}$$

No worries. There is also an acid-base reaction where the two combines to form a salt and water but this occurs at very high temperature and you won't find this type of reaction occurring in laboratory. This reaction can only be find in industries.

$$\ce{Al(OH)3 + NaOH ->[1000 C] NaAlO2 + 2H2O}$$

If only aluminium was used, there would be evolution of hydrogen gas from a very similar exothermic reaction. Note for simplicity, the species is written  $\ce{NaAlO2}$ but it actually contains $\ce{[Al(OH)4]−}$ ion or perhaps the $\ce{[Al(H2O)2(OH)4]−}$ all due to presence of vacant orbital(see sodium aluminate).


Basis sets - negative contraction coefficients? I was looking through Basis Set Exchange when I noticed quite a few basis sets (STO-3G to STO-6G, for example) have negative contraction coefficients for some primitive Gaussians. As (based on what I know) the overlap between two STO-nG basis functions is equal to the sum of the overlap integrals between the pGTOs used to approximate the STO, this means that the diagonalised overlap matrix may have some negative elements.

If this happens, what would the program do?

Diagonalization of a matrix means to find the eigenvalues of the matrix and put them into a diagonal matrix:

$$\mathbf{S}^\text{diag}=\mathrm{diag}\left( \lambda_1,...,\lambda_n \right)\tag{1}\label{diagmat}$$

for $\lambda_i$ being the eigenvalues of the matrix $\mathbf{S}$.
When we want to find the eigenvalues of the matrix $\mathbf{S}$ we solve an eigenvalue problem:

$$\mathbf{S}\mathbf{c}=\bar{\lambda}\mathbf{c}\tag{2}$$

for $\mathbf{c}$ being a matrix of the eigenvector, and $\bar{\lambda}$ being a vector of the eigenvalues.
This equation can also be written out in sum-form for the $i$the eigenvalue:

$$\sum_kS_{jk}c_{ik}=\lambda_ic_{ij}\tag{3}\label{eigproblem}$$

Now as you have stated we can define our atomic orbitals as a linear combination of primitive basis functions:

$$\chi_i=\sum_k c_{ik}\phi_k\tag{4}\label{AO}$$

This can be used to evaluate the self-overlap of the atomic orbitals:

$$S_{ii}=\int_\Omega\chi_i^*(\bar{r})\chi_i(\bar{r})\mathrm{d}\bar{r}=\int_\Omega\left|\chi_i(\bar{r})\right|^2\mathrm{d}\bar{r}=\left< \chi_i \left| \chi_i\right.\right>\tag{5}\label{selfoverlap}$$

This can now be expanded out using Eq. (\ref{AO}):

$$S_{ii} = \left< \sum_k c_{ik}\phi_k \left| \sum_j c_{ij}\phi_j\right.\right>\tag{5}$$

Now using the definition of the overlap-integral $S_{jk}=\left< \chi_j \left| \chi_k\right.\right>$ the above equation can be formulated as:

$$S_{ii}=\sum_{j,k}c_{ij}^*S_{jk}c_{ik}\tag{6}$$

Now using Eq. (\ref{eigproblem}) this can be expressed as:

$$S_{ii}=\sum_{j,k}c_{ij}^*\lambda_ic_{ij}=\lambda_i\sum_j\left|c_{ij}\right|^2\tag{7}$$

Now since $S_{ii}$ can be seen to be larger than or equal to zero from Eq. (\ref{selfoverlap}) and $\left|c_{ij}\right|^2$ must also be larger than or equal to zero, it follows that the eigenvalues $\lambda_i$ must all also be larger then or equal to zero.
I.e. from Eq. (\ref{diagmat}) the diagonalized overlap matrix cannot have any negative values.


What is the reason behind choosing the specific elements used for the synthesis of heavier elements? The following has been derived from the wiki page "Extended periodic table" and is one of the many examples d or f block elements being used in synthesizing new heavier elements (or at least this is the pattern I figured out by searching many other examples of the nuclear reactants in the way to synthesize heavy elements):

  Attempts to synthesize still undiscovered elements:

  \begin{align}
\ce{^254_99Es + ^48_20Ca &-> ^302_119Uue^*} \\
\ce{^249_97Bk + ^50_22Ti &-> ^295_119Uue + 4 ^1_0n} \\
\ce{^{nat}_68Er + ^136_54Xe &-> ^{298,300,302,303,304,306}Ubb^* -> no atoms} \\
\ce{^238_92U + ^66_30Zn &-> ^304_122Ubb^* -> no atoms}
\end{align}

The question is, why is it like this? How did the scientists reached an agreement on which elements to collide? Was it anything especial about those atoms?

P. S. If you are sure that there is an exception (exceptions omit rules; at least in this case), your answer will be accepted.

There are different strategies for synthesizing superheavy elements.

Two of the main strategies are referred to as "hot fusion" and "cold fusion".

The first two reactions are good examples of the "hot fusion" strategy. I will focus on these two reactions, as the other two are much older and I don't think they are good examples of current thinking.

Experimentally, a very heavy element target (the Es or Bk in the first two reactions) is bombarded by an ion beam of the second element.  The target element is choosen on the basis of being very heavy, availible and not rapidly decaying.

$^{48}_{20}\ce{Ca}$ is choosen for a very special reason.  This is not ordinary calcium.  The most common isotope of calcium has only 20 neutrons, but this isotope has 28!.  It is extremely neutron rich, espeically compare to other relatively stable isotopes in this mass-range of the periodic table.  $^{50}_{22}\ce{Ti}$ is also very neutron rich.

Why are neutron rich nuclei chosen? There are two reasons for this:

First, confining many protons in a small space causes electrostatic repulsion.  As atomic number increases, the stable nuclei deviate further and further from a 1:1 neutron:proton ratio, because having more neutrons keeps the protons further apart and decreases electrostatic repulsion energy.

Secondly, you will notice the asterisk (*) in the 1st, 3rd and 4th equations.  This denotes the compound nucleus which initially forms in the fusion reaction.  It is an extremely ($<10^{-21}$s) short lived intermediate state, not the ground state of a nucleus, but more like a highly excited state.  It's formation is not considered creating a new element.  It is critical to the strategy of forming an element that the intermediate state loss energy through neutron "evaporation".   Notice that in the second equation, the step of the neutrons evaporating is shown, while the intermediate state is not shown.  Generally speaking, in the hot fusion strategy, 3-5 neutrons will evaporate to cool the intermediate state.

For further reading see:

Future of superheavy element research: Which nuclei could be synthesized within the next few years? and

The discovery of the heaviest elements Rev. Mod. Phys. 72, 733


Can the regions of 1s and 2s subshells overlap? I was studying atomic orbitals and always had this question lingering.

As we can see in the image, the 1s and 2s subshells both are known to closely surround nucleus. The overlapping peaks in the graph (of probability vs radius) lead me to believe that the part region of 2s before the node should overlap with the region under 1s.

And by 1s and 2s I mean their contours ( regions which contain 90% probability of electron occurring).

Am I committed a blunder here or do subshells overlap?

Additionally, does the same logic apply when comparing the spatial distribution of say 2p and 2s aswell? Do they have overlapping regions of high electron probability density as well?

Subshells are not a real physical objects in the sense that we can't measure them, can't designate them. There are just a mathematical abstraction. So, there are meaningless in terms of physics to ask about their overlap. It can lead you to misunderstanding.

Let's look at the region where 1s-electron can be found with 90% probability. There are non-zero probability of 2s-electron being found there. Even if you'd consider the region where 1s-electron can be found with 10% probability, there are still non-zero probability of 2s-electron being found there. So, at first you should understand that 2s (2p, 3s, 3p and so on) electron have non-zero probability to be found anywhere except where its probability density function equals to zero.

Now let's say we have performed infinite number of measurements of 1s and 2s electrons coordinates. Let's discard 10% outliers in both cases. Now we have clusters which contain 90% probability of electron being found there. Will these regions overlap? Maybe. The answer is one can calculate how much outliers one should "remove" to obtain "non-overlapping regions" (I can't calculate this, unfortunately). For 1s and 2s it may be 95%-regions, or may be 89%-regions that not overlapped. And theoretically, one can perform such calculation for any pair of orbitals he wants.


Can toluene be oxidized to benzoic acid using Pyridinium Chlorochromate(PCC)? My question stands:

Is it possible to oxidize toluene to benzoic acid using PCC?

PCC is generally regarded as a milder oxidizing agent, used to effect selective oxidation of alcohols to aldehydes (from $1^{\circ}$ alcohols) or ketones (from $2^{\circ}$ alcohols). While over-oxidation to a carboxylic acid from a primary alcohol is definitely possible, PCC should not react with a simple unfunctionalized benzylic carbon.

However, oxidation at the benzylic position is easily achieved with more aggressive oxidizing agents. In particular, chromic acid (generated in situ from chromium salts in aqueous sulfuric acid, as in the Jones oxidation) as well as aqueous potassium permanganate will fully oxidize virtually any benzylic carbon all the way to the carboxylic acid, providing the benzylic carbon has at least one hydrogen. Of course, the rest of the chain is completely cleaved in the process.

Edit: Out of curiosity, I consulted my copy of March's Advanced Organic Chemistry to determine if any other common reagents were able to effect the same benzylic oxidation. Nitric acid is listed along with the two methods I described above. Naturally, a survey of various more exotic methods is also given. Intriguing was the use of sodium hypochlorite in acetonitrile, or NBS in aqueous sodium hydroxide under photochemical conditions, for the specific complete oxidation of aryl methyl groups.


Moles of Diatomic Molecules Does "Mole of Iodine" or any other element that normally appears as diatomic molecules refer to a mole of single atoms ($6.02\times 10^{23}$ iodine atoms) or a mole of the molecules ($6.02\times 10^{23}$ molecules of $I_2$)?
EDIT: If there is no specification of moles of atoms or moles of molecules, what do I do? Do I default to one of them?
e.g. in this question

For the combustion of sucrose:
$\ce{C12H22O11 + 12O2 -> 12CO2 + 11H2O}$
there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?

When we ask, for example, "how many moles of Iodine (or Oxygen or Hydrogen) is contained in 100g?" It ALWAYS refers to the molecule (e.g. I2, O2, H2 etc). Therefore, the number of 6.02 x 10^23 refers to the number of molecules, not atomes of Iodine and therefore you actually have 12.04 x 10^23 atomes of Iodine contained in one mole of molecular Iodine. However, when you look at the periodic system, they always refer to atomic mass not molecular. In case of Oxygen, the atomic mass of 16 g/mol refers to 1 mol of O, not O2 as molecular form. Further, when you want to know mass of the Oxygen (gas form, O2) then: 2 (number of atoms) x 16 g/mol = 32 g/mol - this is the mass of 1 mol of O2 (Oxygen-diatomic form - gas present in the atmosphere).


Why does strong Lewis acid-strong Lewis base interactions prevail over hard-soft acid-base interactions? In Pearson's paper, he mentions that strong acid-strong base interactions are more favourable compared to hard-soft acid-base interactions. He cites the example of  $\ce{H+ + H- -> H2}$. In this example, the hydrogen cation is a strong and hard Lewis acid while the hydride ion is a strong yet soft Lewis base. By just considering hardness and softness of the acid and base, we would not expect a particularly favourable interaction between the two species. However, Pearson says it is because of a more favourable interaction which is the strong acid-strong base interaction which is at work in this case.

Some background information:

When the HOMO of the Lewis base and the LUMO of the Lewis acid are similar in energy, the bonding molecular orbitals produced are significantly lower in energy and thus, electron occupation of these bonding MOs give rise to significant stabilisation. This occurs when hard Lewis acids and hard Lewis bases react and also when soft Lewis acids and soft Lewis bases react. This is the molecular orbital basis of Pearson's Hard Soft Acid Base Principle.

Can we explain the favourable interaction of strong Lewis acids and strong Lewis bases using molecular orbitals as well?

Reference: Pearson, R. G., & Songstad, J. (1967). Application of the Principle of Hard and Soft Acids and Bases to Organic Chemistry. Journal of the American Chemical Society, 89(8). Retrieved June 5, 2017 (link to paper)

The Klopman-Salem equation accounts for the energy gain (or loss) when the orbitals of one reactant overlap with those of another. The equation is

$$\Delta E=-\sum_{ab}(p_a+p_b)\beta_{ab}S_{ab}+\sum_{k<l}\frac{Q_kQ_l}{\varepsilon R_{kl}}+\sum_{r}^{\text{occ}}\sum_{s}^{\text{unocc}}-\sum_{s}^{\text{occ}}\sum_{r}^{\text{unocc}}\frac{2(\sum_{ab}c_{ra}c_{sb}\beta_{ab})^2}{E_r-E_s}$$
where $p_a$ and $p_b$ are the electron population in the atomic orbitals $a$ and $b$, $\beta$ and $S$ are the resonance and overlap integrals, $Q_k$ and $Q_l$ are the local charges on atoms $k$ and $l$, $c_{ra}$ is the coefficient of atomic orbital $a$ in molecular orbital $r$, where $r$ refers to the molecular orbitals on one molecule and $s$ refers to those on the other, $E_r$ is the energy of molecular orbital $r$ and $E_s$ is the energy of molecular orbital $s$.

The first term is a closed-shell repulsion term, the second term is the Coulombic repulsion or attraction, and the third term is the interaction of all the filled orbitals with all the unfilled of correct symmetry. This equation was derived using perturbation theory being the first term is a first-order perturbation term and the third the second-order perturbation term.

Now, the HSAB theory is affected by charge and orbital interactions. The h-h interactions is mainly electrostatic in nature, while in a s-s interaction the orbitals involved are close in energy. With the Klopman-Salem equation we can see that the second term is the hard contribution and the third is the soft contribution. There is a simplified form, which accounts only for frontier orbitals, that is
$$\Delta E=\frac{Q_{\text{Nu}}Q_{\text{El}}}{\varepsilon R}+\frac{2(c_{\text{Nu}}c_{\text{El}}\beta)^2}{E_{\text{HOMO}(\text{Nu})}\pm E_{\text{LUMO}(\text{El})}}$$
So, back to the question, a hard molecule or ion is characterized by a large HOMO-LUMO gap (definition of hardness) and the second term, in the simplified version, is very small, so the contribution is mainly electrostatic, i.e., the more charged the sites in the molecule/ion the greater will be the stabilization (or repulsion, in case of like charges). The electric field of the two parts in the molecules/ions would diminish $R$ and the term would be very large. In the case of soft-soft interactions, the orbital term is the dominant because the HOMO-LUMO gap is smaller.

But sure, the interaction has to be between to molecules or ions, not a single specie. You can apply this to nucleophiles and electrophiles as well. A hard nucleophile (negative charge localized and low-energy HOMO) would tend to react with a hard electrophile (positive charge localized and high-energy LUMO), leading to a great stabilization by the electrostatic term. Meanwhile a soft nucleophile (high-energy HOMO) would react with a soft electrophile (low-energy LUMO), stabilized by the orbital term, because $E_{\text{HOMO}(\text{Nu})}-E_{\text{LUMO}(\text{El})}$ is small.

If it is necessary, I can edit with a MO diagram to see this more visually.


How can a compound having a plane of symmetry be Chiral? The compound( Given in the image below) is treated with $\ce{excess H2 in Pt}$.
Here, (a) - The given compound. (b) - What the answer should be according to me.
(c) - What the actual answer is:

Q1 - How is the chirality center at the top right even a chiral center, as the carbon has 2 identical groups attached to it?

Q2 - How is the compound as a whole a chiral compound? As far as I'm aware, compounds having a plane of symmetry (passing through the two 'chiral' centers in this compound) are supposed to be achiral?

I think you are getting confused between chirality and E-Z isomerism in cyclic structures. So let's clear that up first.
According to IUPAC:

chirality
The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane,  σ = S1, a centre of inversion,  i = S2, a rotation-reflection axis, S2n). If the object is superposable on its mirror image the object is described as being achiral.

Chiral molecules are always dissymmetric (lacking $S_n$) but not always asymmetric (lacking all symmetry elements except the trivial identity). However, asymmetric molecules are necessarily always chiral.
For a basic E-Z isomerism definition:

A cycloalkane has two distinct faces, and any substituent on a ring lies toward one of two faces. When two substituents on a ring point to the same face, they are cis. When the two substituents point to opposite faces, they are trans.

Now, I'll address your second question first:

[OP]How is the compound as a whole a Chiral compound because afaik Compounds having a plane of symmetry(Which passes through the two chiral centers in this compound) are supposed to be achiral.

You're right. This compound does possess a plane of symmetry. However, that only means it can't form enantiomers. It can very well form diastereomers (in this case a pair of E-Z isomerism).
Some further examples to illustrate the point:

The trans isomer is achiral, meaning this structure doesn't have enantiomers. However, it still has a cis diastereomer which can show chirality.
An example closer home:
Image courtesy
I'm sure you must've figured out by now, that the carbon on top isn't a capable of showing optical isomerism. However, due to the dissimilar substituents present ($\ce{H\text{ and } CH3}$), the compound can show cis-trans isomerism.
Further reading:


How is tea affecting water's color? Black tea is an amazing product of the green tea leaves. A recent study has proven that;

  The anion profile for the tea solution indicated that oxalate and citrate were the major anions detected.

This resulted in the fact that a $\pu{1\!\% w\!/\!v}$ solution of normal (World Blend – Tetley USA Inc; you decide it for yourself!) tea has a pH of $\rm 4.9$. Non of these pieces of detailed info were an answer to my question.

My question is, definitely somehow oxalate and citrate managed to dissolve into water as we put tea leaves in it;

Where are they stored? (How does tea plant make them? I'm looking for the reactions that occur in the leaf)
How and why do they get released from tea leaf? Since you don't usually see any "big" differences between tea leaf and a soaked tea leaf? (I'm not looking for the how of solution)
What makes the color of water change? Why does the color vary with pH alteration? (There must be a reaction, what is it?)
Pt: These questions are additional questions to the title.

Where are they stored? (How does tea plant make them? I'm looking for the reactions that occur in the leaf)

Both citrate and oxalate are associated with ATP production via the citric acid cycle. As is obvious from the name, citrate is a critical component of the process. Oxalate is a hydrolysis product of oxaloacetic acid, which is also part of the cycle. Both of these ions are present within the plant cells.

  How and why do they get released from tea leaf? Since you don't usually see any "big" differences between tea leaf and a soaked tea leaf? (I'm not looking for the how of solution)

If your tea leaves are crushed or broken in any way, this really is just simple dissolution. If you have whole tea leaves, the oxalate and citrate are migrating into and out of the plant cells. The cell wall is permeable, so the ions (or corresponding acids) can simply flow through it. The cell membrane is semipermeable, and the diffusion of the ions across it is a biologically controlled process that I honestly don't have the background to explain.

  What makes the color of water change? Why does the color vary with pH alteration?

The pH and color are not related properties. The pH is affected by the citric acid and oxalic acid associated with the anions previously discussed, as well as other minor components that are likely present. The color comes from some of the many other compounds that are extracted into the water, which number in the hundreds, if not thousands. Some of these compounds are inherently colored - no reaction is required to produce color in the solution.


How to make Poinsettia pigment into ink? Like many families, this winter we had bought a red poinsettia plant to decorate our kitchen, and now that it's losing its leaves, I had the genius idea to extract the red pigment. The activity proved a success, and let my 5 yr baby girl amazed (and us too). The pigment was extracted by letting leaves soak in 96% alcohol overnight in a closed container in a hot spot, then filtering the solids and heating the solution until alcohol evaporated for the most part, also adding a few drops of lemon juice to increase acidity and transform the brownish liquid into a blood red tint the that now has the aspect of an ink. Unfortunately, we were unable to use this ink to basically anything, as most mediums like regular paper or white paint are neutral or high pH and turn the poinsettia ink from red to any other color, like blue or green.

As you might have gathered by now, I am not a chemist, and I have no knowledge in the matter beyond high school classes from the old days. I did understand thou after reading this article that changes composition and color with pH.

So what I ask for, if at all possible, is a method described in plain terms and using household accessible chemicals to stabilize this red ink in order to use it, for instance, to paint on paper without losing its intense red color. (at least for a few hours or days)

I think that would be hard, you could try to use a mordant to convert a dye to a pigment, you can use Alumina $\ce{FeSO4}$  or $\ce{CuSO4}$. There is an article that explains it in detail how to use Poinsettia for colouring silk fabric.

  Rawat, Bhawana, et al. "Color fastness properties of silk fabric dyed
  with poinsettia leaves." MAN MADE TEXTILES IN INDIA 48.11 (2005): 418.

You can try to soak a thick paper in a solution of 1-5% of $\ce{FeSO4}$  with your dye heating gently.

Another solution could involve a different medium, like arabic gum mixed with citric acid.


How many stereoisomers are possible for the compound in picture? How many stereoisomers are possible for the compound in picture? I'm getting 8. Is that correct?

My approach was as follows: The left hand methyl groups may be both up or one up and one down. For each of these cases the groups on the right may be in $(E{,}Z'),(E'{,}Z),(E{,}E'),(Z{,}Z')$ configurations. So $4\cdot2=8$.

I think it is quite evident that carbon having 2 π bonds must be linear. I apologize for the non-linearity in the picture. I hope you will be able to understand the question otherwise.

Follow these steps:

Identify all possible sources of stereochemistry. You got pretty far.
Determine how many isomers can derive from each. I think you went wrong here.
Do you have symmetry? Reduce accordingly.
Multiply all the numbers together.

In your molecule:

Double bonds:

the internal double bond in the residue on the top right
the cumulene, which also shows isomerism (E/Z for uneven double bond counts, aR/aS for even ones).

Asymmetric carbons:

the top methyl group
the bottom methyl group
the cumulene-containing side-chain
the other unsaturated side-chain.

Every one of these can be either R/S or E/Z.
The molecule does not have any elements of symmetry, so we cannot reduce.
All we have are factors of 2; two from the double bonds, another four from the asymmetric carbons. This leads us to:
$$n(\text{isomers}) = 2^{(2 + 4)} = 2^6 = 64$$


Why are entries missing on a solubility data chart for ionic compounds? There is a solubility chart in my college chemistry class text for "Solubility of Ionic compounds in water". I have copied and pasted image of chart below.  I was surprised to find a dozen ionic compounds marked "--" meaning "no data".

Among the entries marked "no data" are

iron(III) iodide
iron(III) sulfide
iron(III) carbonate
iron(II) chromate
copper(II) iodide
magnesium chromate
silver hydroxide
and carbonate, chromate, and nitrate for Sn2+

It's hard to believe there is no solubility data on said ionic compounds. They should be either soluble, insoluble, slightly soluble, or an impossible compound.  I asked my professor, and he said most of said compounds are too unstable to get reliable solubility data.  In other words, the said cations and anions would not form a stable ionic compound long enough to test solubility.  He also said it might not be the case for all the "--" no data compounds. Which of the said "no data" compounds are impossible compounds as distinguished from possible stable ionic compounds for which there is really no solubility data?

In the case of magnesium chromate, good solubility in water has been known since 1931.  From Hill et al. [1], citing [2]:

The only figure on the solubility of the salt [as of 1940] appears to be that of Kohlraush[2], who reported that at 18°[C] 100 ml of saturated solution contained 60 g of magnesium chromate.

Most of the other compounds listed, as described in comments to the question, do not exist or are highly unstable because the ions are incompatible in the proposed solid. Iodide and sulfide ions are oxidized by iron(III), iodide is also oxidized by copper(II), chromate ions are reduced by iron(II) and tin(II), and tin(II) also reduces nitrate.  Carbonate ion is not redox-active, but fails to form a stable salt with some weakly-basic or amphotetic oxide forming metal ions like tin(II) or iron(III); a similar failure occurs with proposed silver hydroxide.
References
1.
Arthur E. Hill, Glenn C. Soth, and John E. Ricci, "The Systems Magnesium Chromate—Water and Ammonium Chromate—Water from 0 to 75°", J. Am. Chem. Soc. 1940, 62, 8, 2131–2134. https://doi.org/10.1021/ja01865a059
2.
Joseph Mellor, Treatise on Inorganic and Theoretical Chemistry, Vol. II (London: Longmanns, 1931), p. 275.

Besides missing the reference on magnesium chromate, the chart has some errors associated with oxides and sulfides.
Magnesium oxide is listed as insoluble in water, which it is if it is calcined at a high temperature, but otherwise it becomes (kinetically) reactive forming the hydroxide; see this answer. Barium oxide is not "soluble", but reacts in the same way as calcium, strontium, and low-temperature calcined magnesium oxides.
Among sulfides, the $\ce{S^{2-}}$ ion is a relatively strong base, so the alkali and alkaline earth metal sulfides that "dissolve" in water are in fact all prone to hydrolysis.  Calcium and magnesium sulfides do undergo more complete hydrolysis, as the corresponding hydroxides have limited solubility and their precipitation limits the basicity of the solution. Sulfide ion is also too strongly basic to coexist with ammonium ion, so ammonium sulfide as $\ce{(NH4)_2S}$ properly does not exist.


I'm looking for something as strong as Acetone We use Acetone to remove nail polish and gel polish at our salon.  Is there another  alternative substance out there with less irritated smell like Acetone?

Mineral oil is often used to remove nail polish. Give it a try (J&J has a great one for babies with a soft smell).


Confused about energy change diagrams I have been reading about endothermic and exothermic reactions, as well as change in enthalpy.

From what I have read, an exothermic reaction is where more energy is given out as bonds are made when products form than energy taken in as the bonds of reactants are broken.

Wouldn't this mean that the products have more enthalpy? However, a diagram online shows this (https://www.bbc.com/bitesize/guides/zg84y4j/revision/2):

Why does this diagram show that more energy is taken in as the bonds of reactants are broken? Isn't it incorrect?

I am only just getting started with this topic so I'm just learning concepts. I am not yet familiar with the formulas.

Wouldn't this mean that the products have more enthalpy?

The products have less enthalpy, maybe the following diagram remediates misconception :

$ΔH=Σproducts-Σreactants = {-965.5 -(-74.87)}=-890.63$


Is it accurate to describe these radiation measurement units in terms of where-it's-measured and relationship-to-time? I'm reading about measuring radiation and trying to come up with a more intuitive retelling for an audience familiar with math but not necessarily chemistry or any physical science. Is any or all of the following inaccurate or oversimplified?

Radioactivity (Ci, Bq) - Radiation from the material's point of view. Measured as a rate of disintegration.

Exposure (R, C/kg) - Radiation from the air's point of view. Measured as a sum of radiation that's passed through.

Absorbed dose (rad, Gy) - Radiation from the absorbing target's point of view. Measured as a sum of energy deposited.

Dose equivalent (rem, Sv) - Radiation also in terms of sum energy-absorption, but adjusted for a medical point of view.

As you can tell, the 2 pieces I want to highlight are (1) where-it's-measured and (2) time dimensionality (rate versus summable quantity).
My objective question is, where have I got it wrong? What has potential to cause confusion?
But of course, interested in complete reformulations.

Maybe some numerical values may help.

Radioactivity. The activity is the number of disintegrations occurring per second in a given source. The unit is the Becquerel : $1 Bq = 1 s^{-1}$. For example, the experience shows that $3.7·10^7$ nucleus are decomposed per second in $1$ g $\ce{Ra-226}$. The activity of $1$ g $\ce{Ra-226}$ is $3.7·10^7 Bq$ which was previously called $1$ Curie ($1$ $Ci$). One Curie is a huge activity for usual measurements in the laboratory.

Exposure dose. If a gamma-ray radioactive source is in air, the gamma rays ionize the air around. As a consequence, they deposited some energy in a small volume of air situated at a distance from the source. If the small air sample is far away from the source, the amount of energy is smaller than if it is near the source. The amount of energy so absorbed by the air sample is called the dose (or exposure dose) and is determined with dosimeters, and it is measured in $J/kg$ air. The unit of dose is röntgen ($1~r$), equal to $0.01 J/kg$.

The experience shows that $1 Ci$ of a gamma-ray emitter delivers a dose of $1$ röntgen per hour ($1 r/h$) in air at $1 m$ distance from the source. This value is  somewhat dependent on the nature of the radioactive source. Let's forget about this difficulty for the present purpose. Most important, the dose decreases with the square of the distance. So, to be consistent with the units, the dose delivered by a gamma-ray source is $1 r·m^2 Ci^{-1} h^{-1}$. A small volume of air (some milliliters) situated at $1 m$ distance from a $1 Ci$ gamma source, absorbs in one day ($24 h$) a dose equal to $(1~r·m^2·Ci^{-1} h^{-1})·1 Ci·24 h/(1 m)^2 = 24~ r = 0.24 J/kg $. This energy is finally transformed into heat.

If the irradiated matter is not air, but another substance, like water, the amount of energy is called absorbed dose and is expressed in Gray, abbr. $Gy$ ($1 Gy = 1 J/kg$). The experience shows that an exposure dose of $1$ röntgen delivers an absorbed dose of $0.01 Gy$ in water, like in air.

Living tissues are mostly made of water. But they don't react the same way to a given absorbed dose. Some molecules like ADN are much more sensitive to a given adsorbed dose than water. To compare the biological effects of a given dose expressed in Gray, it is necessary to define a new parameter, the effective dose, which is expressed in Sievert, abbr. $Sv$. $1 Sv = 1 Gy$ in water. But $1 Sv$ = up to $20~Gy$ in sensitive tissues.

The present explanation is just a first approach of the domain of dosimetry and radioprotection. As quickly mentioned in the text, the energy deposited in a target depends also on the energy of the gamma ray, which was forgot here. And we have not spoken about alpha and beta rays, whose dosimetry is not similar to gamma rays. Dealing with these problems should take hours, and cannot be done in the present message.


Nomenclature of a cyclic ketone Why isn't this molecule named 2-fluorocyclopent-1-one?

The 1-specification for the ketone function is superfluous, there is no information obscured by omitting the '1'.  It is the most important functional group in molecule, and therefore establishes the 1-position.

Also, cyclopentone is not a proper name.  You are missing the 'ane' suffix to indicate saturation.  It would be a cyclopentanone.

Final name:

2-fluorocyclopentanone.


Can the change in the entropy of the surroundings always be obtained by dividing heat transferred by the temperature at which the transfer occurs? Consider $\pu{1 mol}$ of an ideal monoatomic gas going through reversible isochoric heating from $\pu{100 K}$ to $\pu{1000 K}$.
Calculate $\Delta S_\pu{sys}, \Delta S_\pu{surr}.$

$$\Delta S_\pu{sys} = nC_v \int_{\pu{100 K}}^{\pu{1000 K}} \frac{dT}{T} = \frac{3}{2}R\ln10$$
Now, as I have
$q_\pu{sys} = 1350R$.
I can put $\displaystyle \Delta S_\pu{surr} = \frac{-q_\pu{sys}}{T_\pu{surr}}$ to get
$$\Delta S_\pu{surr} = -\frac{27}{20}R$$ but in a reversible process $\Delta S_\pu{total} = 0 $ but that gives me $$\Delta S_\pu{surr} = -\frac{3}{2}R\ln10$$
Where am I wrong?

Image source: Physical Chemistry, 10th ed. by Atkins and de Paula, p. 116.
Here, $\displaystyle q_\pu{sur} =$ heat supplied to surroundings.

I disagree with the explanation given in the text.  The entropy change of the surroundings is given by the equation presented in your title only for an ideal constant temperature reservoir.  Such an ideal reservoir is characterized by (a) an infinite capacity to absorb heat without its temperature changing and (b) infinite thermal conductivity, so that temperature is uniform throughout the reservoir, including at the interface with the system (where all the heat transfer occurs). So all the heat transfer between the system and the reservoir occurs at the reservoir temperature $T_{Res}$. In this way, all the irreversibility of the process occurs within the system, irrespective of whether the process is reversible, and the entropy change of the system always satisfies $ΔS≥Q/T_{Res}$.

None of the entropy generation for the process occurs within the ideal reservoir, and the entropy change of the ideal reservoir is thus always $-Q_{sys}/T_{Res}$, irrespective of whether the process is reversible. I should also mention that, although often not stated, the Clausius inequality is supposed to be expressed in terms of the temperature at the interface with the surroundings $T_{Res}$ where all the heat transfer is occurring.

Typically, all the surroundings reservoirs you will encounter in most thermodynamics texts will be ideal constant temperature reservoirs.

The description in your post is inappropriate because, if the reservoir has a constant volume (i.e., constant mass), its temperature can change between the beginning and end of the process.  Its entropy change will not be given then by the equation presented.  In addition, the description omits the requirement that the reservoir most have infinite thermal conductivity (so that the temperature in the reservoir is spatially uniform).


How to calculate the equilibrium constant from the density of a nitrogen dioxide and dinitrogen tetroxide mixture? The density of an equilibrium mixture of $\ce{N2O4}$ and $\ce{NO2}$ at $\pu{1 atm}$ and $\pu{373.5 K}$ is $\pu{2.0 g/L}$. Calculate $K_c$ for the dissociation reaction

I did this:

Let initial amount of substance be $a$.
Initial vapor density $=(\text{mass of }\ce{N2O4})/2 =46$
For final vapour density I find effective molecular mass of the mixture $\ce{N2O4}$ + $\ce{NO2}$ at equilibrium using ideal gas equation which comes out to be $61.33$ and thus final vapor density is $61.33/2=30.665$.
Now supposing degree of dissociation to be $x$ , total amount of substance at equilibrium is $a\times(1+x)$.
As $\text{Vapor density} \times \text{total amount of substance} = \text{constant}$ at same temperature, $46\times a=30.665\times a\times(1+x)$
$x=0.5$ Using this in the equation
$$K_p=\left[\frac{4x^2}{1-x^2}\right]\times P$$ for this reaction I get $K_p=\frac43\pu{ atm}$ and then $K_c=0.043$.
The answer is 2 but how?

I confirm your answer.

The molar density of the mixture is

$$\frac{n}{V}=\frac{p}{RT} = \pu{0.0326 mol/L}$$

So, the average molecular weight of the mixture is

$$\frac{2}{0.0326} = \pu{61.35 g/mol}$$

If $x$ is the mole fraction of $\ce{NO2}$ and $(1-x)$ is the mole fraction of $\ce{N2O4}$, then the average molecular weight of the mixture is also

$$46x + 92 (1 - x) = 61.35$$

Solving for $x$ gives $x = 1/3$. So, $(1-x)=2/3$. These are also the partial pressures of $\ce{NO2}$ and $\ce{N2O4}$, respectively (in atm). This leads to the values of $K_p$ and $K_c$ that you calculated.


Deriving an expression for the constant pressure heat capacity Given the relation $G = H − T S$, show that:
  $$\left(\frac{\partial S}{\partial T}\right)_p=\frac{C_p}T$$

Now I realise that $T\mathrm dS=\mathrm dQ$ and that would I suppose get the result. However, I have been asked to use the expression $G = H - TS$ to find the result. If I just differentiate implicitly I end up getting:

$$\left(\frac{\partial G}{\partial T}\right)_p+\left(\frac{\partial S}{\partial T}\right)_p=\frac{C_p}T$$

What am I doing wrong?

You could start by dividing your whole equation by $T$:

\begin{align}
 \frac{G}{T} = \frac{H}{T} - S
\end{align}

and then take the derivative with respect to $T$ at constant $p$:

\begin{align}
 \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} = \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} - \left(\frac{\partial S}{\partial T} \right)_{p} \ .
\end{align}

That way you've already introduced the desired derivative of the entropy with respect to $T$ at constant $p$.
Then you can use that

\begin{align}
 \frac{G}{T} = \frac{H}{T} - S  \qquad &\Rightarrow& \qquad \mathrm{d} \!\left(\frac{G}{T}\right) &=  - \frac{H}{T^2} \mathrm{d} T + \frac{1}{T} \underbrace{\mathrm{d} H}_{= T \mathrm{d}S + V\mathrm{d}p} - \mathrm{d} S \\
&&&= - \frac{H}{T^2} \mathrm{d} T + \frac{V}{T} \mathrm{d}p \\
&\Rightarrow& \qquad \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} &= - \frac{H}{T^2}
\end{align}

and simply using the product rule

\begin{align}
 \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} = \frac{1}{T} \underbrace{\left(\frac{H}{\partial T}\right)_{p}}_{= C_{p}} + H \underbrace{\left(\frac{\partial \frac{1}{T}}{\partial T} \right)_{p}}_{=-\frac{1}{T^2}} = \frac{C_{p}}{T} - \frac{H}{T^2} \ .
\end{align}

Putting it all together gives you the desired result:

\begin{align}
 \left(\frac{\partial\frac{G}{T}}{\partial T}\right)_{p} &= \left(\frac{\partial\frac{H}{T}}{\partial T}\right)_{p} - \left(\frac{\partial S}{\partial T} \right)_{p} \\
- \frac{H}{T^2} &= \frac{C_{p}}{T} - \frac{H}{T^2} - \left(\frac{\partial S}{\partial T} \right)_{p} \\
\Rightarrow \left(\frac{\partial S}{\partial T} \right)_{p} &= \frac{C_{p}}{T} \ .
\end{align}


Why would two isomers have different solubilities in water? I was wondering as to why, e.g. $\ce{CH3CH2CH2CH2OH}$ and $\ce{(CH3)3COH}$, do not have the same solubility in water, considering they both have the same molecular formula.

In the most general terms: Molecules consist of nuclei and a surrounding electron cloud. That cloud is dependent on molecular charge and geometry - so it will be different for two different isomers. It follows that its interaction with the nuclei and electron cloud of a third molecule, say, water, will be different.

More specifically, one will find that properties such as

polarizability (related to van-der-Waals forces)
bond dipoles (related to H-bonds and static dipole interactions)
effective electronegativity (related to H-bonds and static dipole interactions)

On a very basic level, any solvation involves the formation of solvent shells. When varying the shape and surface of the solute, those shells will change as well. One will then find shells that work well and others that work less well, influencing the solubility.


NaOH to stabilize aqueous NaCN This patent notes in passing:

  It may be necessary to add $\ce{NaOH}$ to stabilize aqueous $\ce{NaCN}$. Anyone
  skilled in the art would know the amount of $\ce{NaOH}$ to add.

Obviously I'm not skilled in the art.  I assume the stabilization is against hydrolysis.  What is the mechanism by which $\ce{NaOH}$ would stabilize aqueous $\ce{NaCN}$, and how would one determine the correct amount of $\ce{NaOH}$ to add?

(I'm expecting it's some target pH level, but I'm too far along my "assumption" chain to know what the issues are.)

Hydrogen cyanide is an acid with a pKa of 9.2, meaning that at pH 9.2 in water you have about a 50-50 mix of cyanide anion and hydrogen cyanide. If you used sodium hydroxide to adjust the pH of a NaCN solution to, say, 12.2, you would inhibit the decomposition of NaCN by shifting the equilibrium to 99.9 percent cyanide anion and 0.1% hydrogen cyanide. Each pH unit higher gives you an additional factor of 10 lower concentration of HCN.

Another factor is temperature: since HCN boils at 28 C, it tends to evaporate from a room-temperature solution of NaCN, driving the reaction to the right. The solution would be more stable if kept cold.

Finally, it would be better to keep the solution in a tightly closed bottle, since carbon dioxide could be absorbed from the air, acidifying the solution and releasing hydrogen cyanide.


What is a "lower organic acid"? From the "Technical guide for the elaboration of monographs" by EDQM, edition 7, 2015, page 25:

  Where the counter-ion of an active substance is formed from a lower organic acid, a test for related substances of the organic moiety is usually not considered necessary (for example, magnesium lactate used as a source of magnesium).

What is the meaning of "lower organic acid"? An organic acid of simple composition, with a short alkyl group?

The term doesn't seem to have an official IUPAC definition however several sources agree that the lower organic acids are those that are soluble in water.

In the simplest case this could just refer to the first 5 carboxylic acids (C1-C5): methanoic (formic) acid, ethanoic (acetic) acid, propanoic acid, butanoic acid and pentanoic acid.

What is less clear is whether small water soluble acids such as lactic acid  and tartaric acid are counted. (The context of the quote in your question suggests it may well be, though it's wording isn't overly clear)

Since the phrase has no official definition I think it's probably safe to use it as you see fit, ensuring you define what you plan on calling a lower acid and use it consistently. You will equally see people refer to lower amines or halides, which in the same way isn't a well defined phrase but just lets people vaguely define a group of compounds.


Biologically inert water mixture (or other fluid) for heat-exchange purposes I have a steel chamber submerged in a 14 deg water bath to keep it at constant temperature. The setup needs to operate without intervention for a very long time (weeks or months).

The bath is exposed to open air in the lab and we have been having issues with mold development in the bath despite adding bleach to the water.

Can you suggest a different additive (or a different liquid altogether) to use to make sure the bath stays reasonably sterile?

Apparently an acceptable choice of additive is silverised hydrogen peroxide. Only a few 100ppm should be necessary to keep the water sterile and the half life is about 5 weeks.

I'll go with this and update this answer if there are any issues.

PS

This article on overclocking (http://www.overclockers.com/pc-water-coolant-chemistry-part-ii/) has a number of suggestion on water additives to improve the thermal conductivity and  keep the system clean.


Basic behavior of ammonia in aqueous medium Thanks to the reversible reaction
$$\ce{NH3 + H2O <=> NH4+ + OH-},\tag{R1}$$
we know that ammonia acts as a weak base. However, I was wondering why it ends up forming ammonium $\ce{NH4+}$ and not azanide $\ce{NH2-}$ according to
$$\ce{NH3 + H2O <=> NH2- + H3O+}.\tag{R2}$$

All what Karl and Poutnik are saying is right. But the real question is : Why ? Why is $\ce{NH3}$ more reluctant to loose one proton than $\ce{H2O}$, at the atomic level ? Why is $\ce{NH2^-}$ such a strong base ?
Let's compare this behaviour in the series $\ce{N-O-F}$. Starting with $\ce{F}$, we can observe that $\ce{F}$ holds one $\ce{H}$, only one $\ce{H}$, and it does not hesitate to give its unique $\ce{H}$ atom (as a proton) to a great variety of bases. The next in line, the oxygen atom $\ce{O}$ can hold two $\ce{H}$ atoms. This is one more than what does $\ce{F}$. Well, $\ce{O}$ in $\ce{H2O}$ is also more reluctant to loose one $\ce{H}$, and accepts to give it (as a proton) only to strong bases. The third in line is the nitrogen atom $\ce{N}$ which can hold three $\ce{H}$ atoms. It's still more than what does Oxygen. But $\ce{N}$ in $\ce{NH3}$ is still more reluctant to loose one of its three $\ce{H}$ atoms as a proton $\ce{H+}$. It even prefers stealing one $\ce{H+}$ ion from the neighborhood rather than giving it. Apparently the more $\ce{H}$ atoms the central atom has, the more it wants to increase this number. It looks like the human society : the more money you've got, the more you want to increase your fortune.
Edit :
Of course it would be interesting to see if this tendency $\ce{F-O-N}$ can be extrapolated to the carbon atom. If my reasoning is correct, the Carbon atom in $\ce{CH4}$ should be extremely reluctant to loose one $\ce{H+}$ ion for producing $\ce{CH3^-}$. Indeed I am not sure that the ion $\ce{CH3^-}$ does exist, and has a measured $p\mathrm{K_a}$ value. It should be extremely high. On the other hand, $\ce{CH4}$ should be eager to get one more Hydrogen. However the electronic situation is not similar to $\ce{F, O}$ and $\ce{N}$. In $\ce{HF, H2O, NH3, F^-, OH^-, NH2^-, H3O+, NH4^+}$, the central atom has always the electronic state of Neon. It is known that $\ce{CH4}$ does produce $\ce{CH5^+}$ ion in high pressure mass spectrometry. But the electronic structure of the carbon atom in $\ce{CH5^+}$ is unknown, to my knowledge. So I am not sure whether the tendency, valid for $\ce{F-O-N}$, can be extrapolated to the Carbon atom.


Chaos in chemical reactions I recently read about Chaos Theory and was wondering if a chemical reaction results in or shows characteristics of chaos (I found a few examples of such reactions here)

  Chaos theory concerns deterministic systems whose behavior can in principle be predicted. Chaotic systems are predictable for a while and then 'appear' to become random.

If the reaction becomes unpredictable towards the time of attainment of equilibrium we can't possibly determine how the reaction will proceed during the time frame so will the reaction attain a state of equilibrium and will it be at the time it should've without chaos being in the picture?

If the equilibrium shifts can we quantify by how much and how much more or less time will it take to attain it in general or we need to take specific reactions and analyze them?

Also after attaining equilibrium can the system again show Chaotic behaviour and if it does what can se say about the state of equilibrium?

A homogeneous reaction mixture can hardly show macroscopical chaotic behaviour. Well known cyclic counterexamples exist, and practically any reaction which proceeds faster than diffusion or mechanical mixing can homogenise it again (e.g. any reaction that generates a lot of heat!) is liable to show some intermediate chaotic concentration gradients. Its usually hard to notice, and most reactions are perfectly reproducible, if you set them up in the same way (educt impurities!).

If your reaction vessel contains inflows of different chemical composition, then you easily get chaotic behaviour. Think about a tube reactor. I believe chemical engineers are spending a lot of their time trying to control and avoid chaos.


What makes DNA/Protein in a gel run in straight lines? In a gel electrophoresis, let's assume an SDS-PAGE, I understand that the sample is loaded on a cuboidal well and the electric field is straight downward. It is easy to assume that the sample is going to run straight in the gel. Like it does here.

But when I observed the apparatus in my lab, the anode was all throughout the base - looking something like this (cathode not shown here and the anode is insulated on the side - so effectively present at the base of the gel).

What makes the sample run in the gel only in a straight line? You may say that the forces cancel out on the either sides but what about the well that's at one end? It's being pulled downwards and also the other end.

So how does this work?

What you are trying to achieve in the electrophoresis experiment is a homogenous electric field. If the electric field points downward homogenously, then the greatest force will be exercised in a downwards direction which will lead to a downwards movement of your charged sample.

While not being exactly equivalent, you can show a similar feature (homogenous field) with a horseshoe magnet. If you place a piece of paper or cardboard over the top of it and then sprinkle iron filings onto that, they should arrange themselves in a linear fashion between the magnet’s poles.


How does iron rust in basic conditions? I performed an experiment about the corrosion of iron. I used acid and base in the experiment.

I understand that the acid speeds up the corrosion of iron. But I don't understand why the base also corrodes iron. I can't find out the answer even after I've googled it.

In my experiment I mix an ulcer drug into a glass of water, then I put in an iron nail. After I let it sit for a few days, the iron nail becomes rusty. I don't know of any reaction between a base and iron that gives rust.

Iron can oxidize to "rust" (which is actually a mixture of hydrated iron(iii) oxide $\ce{Fe2O3*nH2O}$ and iron(iii) oxide hydroxide $\ce{FeO(OH)}/\ce{Fe(OH)3}$). The iron hydroxide species look like they could come from basic solutions.
Even if consider only the formation of $\ce{Fe2O3}$, this can occur by several pathways.
Using Wikipedia's Standard electrode potential data page as a source of known redox half reactions, the process may be something like this:
In acidic medium
The net reaction in acid is between the iron and the protons in the acid to produce $\ce{Fe^{3+}}$, which reacts with water in a redox neutral (but generates more acid) process to form the insoluble oxides/hydroxides. The initial redox reaction is spontaneous. The water is the net source of oxygen.
$$\ce{2Fe(s) + 6H+(aq) - 2Fe^{3+}(aq) + 3H2(g)} \ \ E^\circ=+0.04\ \text{V}$$
$$\ce{Fe^{3+}(aq) +3H2O -> Fe(OH)3(s) + 3H+(aq)}$$
$$\ce{Fe(OH)3(s) -> FeO(OH)(s) + H2O}$$
$$\ce{2FeO(OH) -> Fe2O3 + H2O}$$
In basic medium
The net reaction in base is between oxygen (in the air, which is not in contact with your submerged nail) and the iron in the nail. The iron initially reacts with the hydroxide ions, which are regenerated by the reduction of oxygen at the surface of the water.
$$\begin{align}
\ce{4Fe(s) + 8OH- &-> 4Fe(OH)2 + 8e-} \ \ \ &&E^\circ=\pu{+0.89 V}\\
\ce{4Fe(OH)2 + 4OH- &-> 2Fe2O3 +4e-}\ \ \ &&E^\circ=\pu{+0.86 V}\\
\ce{3O2 + 6H2O + 12e- & -> 12OH-}\ \ \ &&E^\circ=\pu{+0.40 V}\\ \hline
\ce{4Fe(s) +3O2(g)& -> 2Fe2O3(s)}\ \ \ &&E^\circ=\pu{+1.95 V}\\
\end{align}$$


What is the difference between cyanato, cyanido and cyano? When should i use which? I understand that cyanido is the newly adopted naming in place of cyano but what is cyanato? Can anyone please give examples to illustrate the difference?

Cyano- is related to the neutral CN group. Cyanido- is related to the CN- anion. Cyanato- is related to the CNO- anion.


How do I determine whether it is the *ferrous* or *ferric* ion that is implied in this displacement reaction? I'm supposed to write a balanced equation for:

$$\ce{Iron(s) + copper(II) sulfate(aq) -> ?}$$

If I use Iron(II)/ferrous I get

$$\ce{Fe + CuSO4 -> FeSO4 + Cu}$$

and if I use Iron(III)/ferric,

$$\ce{Fe + CuSO4 -> Fe2(SO4)3 + Cu}$$

(Before balancing).

Is there a way to determine which one I am supposed to use?

You can determine which reaction takes place by the standard electrode potentials of the half reactions. The relevant standard reduction potentials for the possible reactions (source of the values) are:

$$\ce{Cu^2+ +2e- ->~ Cu} \quad E_0=+0.342\mathrm{V}$$
$$\ce{Fe^2+ + 2e- ->~ Fe} \quad E_0=-0.447\mathrm{V}$$
$$\ce{Fe^3+ + e- ->~ Fe^{2+}} \quad E_0=+0.771\mathrm{V}$$

The overall potential of the electrochemical cell is defined as the difference between the standard reduction potentials of reduction and oxidation reaction.

$$E_{cell}=E_{red}-E_{ox}$$

For a spontaneous reaction to occur, $E_{cell}$ must be positive. With the potentials of copper reduction and oxidation of iron to $\ce{Fe^2+}$, you get

$$E_{cell}=E_{\ce{Cu^2+/Cu}}-E_{\ce{Fe^2+/Fe}}=+0.342\mathrm{V}-(-0.447\mathrm{V})=+0.789\mathrm{V}$$

It is obvious from the values that $\ce{Cu^2+}$ cannot further oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$ because $E_{cell}$ is negative for this reaction.

$$E_{cell}=E_{\ce{Cu^2+/Cu}}-E_{\ce{Fe^3+/Fe^2+}}=+0.342\mathrm{V}-(+0.771\mathrm{V})=-0.429\mathrm{V}$$

The balanced equation for the redox reaction between iron and copper(II) sulfate is therefore

$$\ce{Fe + CuSO4 ->~ FeSO4 + Cu}$$